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Lecture Notes in Mathematics A collection of informal reports and seminars Edited by A. Dold, Heidelberg and B Eckmann, Z0rich
311 Conference on Commutative Algebra Lawrence, Kansas 1972
Edited by James W. Brewer and Edgar A. Rutter The University of Kansas, Lawrence, KS/USA
Springer-Verlag Berlin.Heidelberg • New York 1 973
A M S Subject Classifications (1970): 13-02, 1 3 A 0 5 , 13B20, 13Cxx, 1 3 D 0 5 , 13E05, 13E99, 13F05, 13F20, 1 3 G 0 5 , 1 3 H 1 0 , 1 3 H 1 5 , 1 3 H 9 9 , 13J05
I S B N 3-540-06140-1 Springer-Verlag Berlin - H e i d e l b e r g - N e w Y o r k I S B N 0-387-06140-1 Springer-Verlag N e w Y o r k . H e i d e l b e r g • B e r l i n This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to the pubiisher, the amount of the fee to bc determined by agreement with the publisher. © by Springer-Verlag Berlin • Heidelberg 1973. Library of Congress Catalog Card Number 72-96859. Printed in Germany. Offsetdruck: Julius Beltz, Hemsbach]Bergstr.
PREFACE
This volume tributed
contains
to a Conference
The University 1972.
The Conference Foundation
like
on Commutative
of Kansas
Science
to express
a collection
during
Algebra
the week of May
received
under Grant
of articles
funding
held at 8 - May 12
from the National
No. G.P.
our appreciation
con-
33192
and we would
to the Foundation
for its
support. To make possible Proceedings, volume
the early
and thereby
to insure
has been prepared
manuscripts
supplied
the correction
delay
of minor
in the appearance
to include
the paper
appearance
by returning
therefore
their
timeliness,
In instances
would
have
of the Proceedings, rather
led to a
the editors
than delay
it to the author
take the responsibility
the
from the typewritten
errors
as received
of these
chose
its
for correction.
for all misprints
We
which may
the reader. We should
participants thanks
directly
by the contributors.
where
bedevil
appearance
must
Buchsbaum, who were
like to thank
for their
go to Professors Robert
involved
without
whose
Gilmer,
and cooperation.
Shreeram
Irving
Special
thanks
Phil Montgomery
also must
help.
Finally
it would
Abhyankar,
Kaplansky
in the Conference
assistance
Conrad,
efficiency
each of the Conference
enthusiasm
from
David
and Pierre
its outset
go to our Kansas and Paul McCarthy to the countless
colleagues,
and
Phelps details
involved
Edgar A. Rutter August,
and
for her
W. Brewer
Lawrence,
Paul
for their advice
such an undertaking. James
Samuel
not have been possible.
we wish to thank Deborah
in attending
Special
1972
in
TABLE SHREERAM
ABHYANKAR
OF CONTENTS
~Notes by A. SATHAYE)
On Macaulay's
Examples
1
JIMMY T. ARNOLD Prime
Ideals
J. W. BREWER, W. J. HEINZER Krull
in Power
P. R. MONTGOMERY,
Dimension
J. W. BREWER
Series
On a Problem D. DAVIS
PAUL EAKIN
Rings
26
46
in Linear Algebra
and ANTHONY
Ideals
Rings
57
HEINZER
Problem
for Rings
and E. GRAHAM EVANS,
Three Conjectures nomial Rings
50
V. GERAMITA
in Polynomial
and WILLIAM
DAVID EISENBUD
of the Functor
and DAVID EISENBUD
A Cancellation
ROBERT
and
and E. A. RUTTER
DAVID A. BUCHSBAUM
Maximal
17
E. A. RUTTER
of Polynomial
A Note on the Faithfulness S 8 R (-)
EDWARD
Rings
About Modules
61
JR. Over Poly78
GILMER
Prefer-like Conditions on the Set of Overrings of an Integral Domain ~DWARD
90
D. DAVIS
Integrally
Closed
Pairs
103
WILLIAM HEINZER Noetherian
Intersections
of Integral
Domains
II 107
MELVIN HOCHSTER Cohen~Macaulay
Modules
120
IRVING KAPLANSKY Commutative
Rings
153
Vl R. E. MACRAE On Euclidean
Rings
R. E. MACRA E and PIERRE Subfields Fields
of Index
of Algebraic
Functions
167
SAMUEL 2 of Elliptic
Function 171
JOE L. MOTT The Group tions
of Divisibility
and Its Applica194
JACK OHM Homological LOUIS
Dimension
J. RATLIFF,
JR.
Chain Conjectures JUDITH
209
and Euler Maps
222
and H-Domains
SALLY
On the Number Dimension O
of Generators
of Ideals
of 239
WOLMER V. VASCONCELOS Rings
of Global
Dimension
Two
243
PARTICIPANTS
Gover, E.
Orbach, K.
Albis, V.
Grams, A.
Orzech, G.
Arnold, J. T.
Greenberg,
Barger, S. F.
Hays, J.
Paeholke, K.
Bennett, B.
Hedstrom, J.
Parker, T.
Bertholf, D.
Heinzer, W.
Prekowitz,
Bittman, R.
Heitman, R.
Ratliff, L.
Blass, J.
Hinkle, G.
Razar, M.
Boisen, M.
Hoehster, M.
Robson, J. C.
Boorman, E.
Huckaba, J.
Sally, J.
Brown, D.
Kaplansky,
Buchsbaum, D.
Kelly, P.
Sathaye, A.
Butts, H. S.
Kohn, P.
Shannon, D.
Davis, E. D.
Kuan, Wei-Eihn
Sheets, R.
Eagon, J.
Larsen, M.
Sheldon, P.
Eakin, P.
Larson, L.
Shores, T.
Eggert, N.
Levin, G.
Smith, W.
Eisenbud, D.
Lewis, J.
Snider, R.
Enoehs, E.
Lu, Chin-Pi
Stromquist, W.
Entyne, S.
MacAdam,
Thaler, A.
Epp, S.
McDonald,
Evans, G.
MaeRae, R.
Fields, D.
Magarian,
Fischer, K.
Mott, J.
Wiegand,
Gabel, M.
Murthy, M.
Woodruff, D.
Geramita, A.
Nichols, J.
Van der Put, M.
Gilbert, J.
Ohm, J.
Vasconeelos, W.
Gilmer, R.
O'Malley, M.
Vekovius, A.
Abhyankar,
S.
B.
I.
S. B.
Orzeeh, M.
B.
Samuel, P.
Wadsworth,
A.
Wiehman, M. E.
Wiegand, R. S.
ON M A C A U L A Y ' S
EXAMPLES
Lectures by Professor Abhyankar N o t e s b y A- S a t h a y e
§l
Introduction. ideals m
Pm
INTRODUCTION
in the p o l y n o m i a l
ible c u r v e s
in a f f i n e extended
space will
3-space with to the
ous
time, b u t
After
individuals
The n o t e was
service
1 -- a r e s i d u a t i o n
In §2 w e c o n s t r u c t
(m - 2)
not h a v e any
f r o m the
f) so t h a t it does n o t
and the c u r v e s
of d e g r e e
fixed p o i n t o u t s i d e .
the d e t a i l e d
Let
Y, Z3,
k
each
m a 1
needs
at l e a s t
Note:
We note
irreducible
be
there exists m
a prime
of the set, raises
in
following
proof.
closed 3
ideal
theorem along
field and
variables p
m
c k[X,
over
A = k.
Y, Z3
Then so t h a t
for P m
generators.
that for
surface
original
ring
it do
in §4.
an a l g e b r a i c a l l y
the p o l y n o m i a l
through
of the e x a m p l e s ,
p r o o f of the
lines as in M a c a u l a y ' s
lie on any curve
(m - i)
This c o n s t r u c t i o n
are c o l l e c t e d
Theorem:
and p r o v e
t h e o r e m to be u s e d later on. 1 ~ m ( m - i) p o i n t s in the p l a n e
some q u e s t i o n s w h i c h
k[X,
Purdue S e m i n a r b y
P. Russell o f f e r e d
fix u p t e r m i n o l o g y
for the c o n s t r u c t i o n
the same
Also
it w o u l d
note.
of the note.
besides being useful
§3 p r o v i d e s
thought
a set of
(in fact on a c e r t a i n c u r v e of degree
in
t h e m to v a r i -
an e x p l a n a t o r y
Professor Abhyankar.
in the w r i t i n g
is w r i t t e n
'well known'
t o d a y and s e e m in n e e d of
Professor Abhyankar to p u b l i s h
In the r e m a i n i n g p a r t of §i w e Proposition
thus p r o v i d i n g e x a m -
and r e e x p l a i n i n g
p r e p a r e d b y A. S a t h a y e
and d i s c u s s i o n s w i t h
The 3-
3.
are r a t h e r m y s t e r i o u s
public
in a f f i n e
a l t h o u g h precise,
the p r o o f s
irreduc-
at the o r i g i n .
p r o b a b l y p a r t of the
o v e r the years,
many suggestions
generators,
that the proof,
rediscovering
be a w o r t h w h i l e
singularities
ring of the o r i g i n
m
assertions which were
Macaulay's proof.
local
n e e d at least
are ideals d e f i n i n g
l o c a l r i n g of d i m e n s i o n
It seems h o w e v e r
of p r i m e
ring in 3 v a r i a b l e s w h i c h
s t i l l n e e d at least
in a r e g u l a r
with many
given examples
In fact all the e x a m p l e s
same
ples
PRELIMINARIES
In [i] p. 36 M a c a u l a y has
generators.
ideals
AND
m = i, 2
one can take
and an i r r e d u c i b l e
curve
the ideal of an
respectively
and h e n c e w e
assume
i.I
m ~ 3.
After
fixing
m
we shall denote
General
Terminoloq Z
Let
be an a l g e b r a i c a l l y
k
k[X 0 . . . . . over
Xn]
k.
degree
Then i}
Let ible.
closed
field.
be the g ~ a d e d p o l y n o m i a l R (n) =
where
0
ring
(--). H
R (n)
{~I u, v E R (n), function Let r i n g of
i u, v
field of
coordinate
in p r o j e c t i v e homogeneous
c R (n)
H
at
homogeneous
is d e f i n e d b y
i.e.
those
Q(H)
r e s p e c t to
is i r r e d u c -
Q(H)
~(H)
H.
interchange
R (n) .
A divisor so that
divisors
clearly
We d e n o t e if
so t h a t
Let
minimal
prime
for w h i c h
H
for all
Thus
there
D(m)
denote v ~ 0],
the
D
m
on
D(m)
prime
ideals
in
m
The
local
are n o n s i n g u l a r
as a b o v e
R (n)
r e s p e c t to
and
Om
Om
in
is a
H
H = 0
is a m a p
D
for all b ~ t
for all
is d e n o t e d b y
m.
and p r i m e d i v i s o r s
given by
m <--> O m
d e p e n d i n g on the c o n t e x t .
: ~(H)
prime -->
ideals
Z,
in
m £ ~(H).
(D + E) (m) = D(m)
~(H).
Define
The
+ E(m).
D ~ E
if and
T h e n the s e m i g r o u p of d i v i s o r s
~(H)
and d i v i s o r s
in
.
Let
the g r o u p of
finitely many
form a group by defining
a E(m)
ideal.
is a one to one c o r r e s p o n d e n c e
the g r o u p of all d i v i s o r s by
D ~ 0
~(H)
which
the set of all m i n i m a l h o m o g e n e o u s
integers
by
O m = [~ E Q(H) I v ~ m } .
of the f i r s t k i n d w i t h
denote
H
and then in fact a p r i m e d i v i s o r of the f i r s t
between minimal homogeneous
Thus w e m a y
of
modulo
of the s a m e degree,
be a h o m o g e n e o u s
m
p r i m e d i v i s o r of kind with
variables
r i n g of the h y p e r s u r -
n-space.
We s h a l l c o n s i d e r only h y p e r s u r f a c e s i,
P.
H.
m
codimension
by
m
R (n) =
~ R! n) where R! n) = [u I u i=0 i 1 is h o m o g e n e o u s of all d e g r e e s .
is the h o m o g e n e o u s
of d e g r e e
P
(n + i)
H E R! n) for some i so t h a t H ~ 0 and H 1 the images of v a r i o u s q u a n t i t i e s in R (n)
face
only
Let in
Denote
a bar
of
the ideal
~(H)
D
are c a l l e d
effective. For with
O m.
m £ ~(H)
we
shall d e n o t e b y
F o r a set of h o m o g e n e o u s ~v E Om]
and
vm(S)
0 ~ u E R (n),
homogeneous,
define
If
u
< U > = . < u > - .
For
ideal g e n e r a t e d b y
define
{U E R(n) I U
S c R (n)
associated
define
= rain [vm(a) [ a E SOm].
U E R (n), U homogeneous u For 0 # ~ E Q(H) define D E ~(H)
the v a l u a t i o n
elements
S O n = {~I u E S,
vm({u]).
vm
< u > E 9(H) and
by
(m)
~ = u ~ 0,
(~) £ ~ ( H )
I(D) C ~ R (n) homogeneous,
~ = 0
=
define
so that
to b e
For
(~) =
the h o m o g e n e o u s or
z D].
For
m E ~(H)
we have
the d i v i s o r
Dm(m' )
Clearly
divisors
elements
of
a divisor We = C.
~(H)
: A --> degree We
defined
fix,
for
f = O(F)
It f o l l o w s
defines
an
~(U(F))
1.2
Pointsets
correspondence
a prime
Z] = B Y,
and
Y,
B
that
Y,
without
identity
shall
be denoted Y Z = W r h ( ~ , ~, ~),
Z))
Note
the
divisors.
R (3) = k[X,
of the
with
divisor.
of p r i m e
Z]
We have which
u(h(X,
are defined
on
Z, W]
the g r a d e d
map
from
by
~.
A Let
where
r =
= the
defines
F
an i r r e d u c i b l e
is h o m o g e n e o u s
and w e
which
shall
assume
is n o n s i n g u l a r
an i r r e d u c i b l e ,
F E A
of degree that
~(F)
in c o d i m e n s i o n nonsingular
i.
curve.
= K.
f on
on
f
being f
shall be
be
in the p r o j e c t i v e
a vector ~ = 0
define
Fi(L,
called
one d i m e n s i o n a l
s e t of p o i n t s
[L, ~] = [h I h E L, space again.
also
= F E C
divisors
by points
L c Bn
the notes,
f
since
~(f)
f
of
(i.e.
surface
~(f)
The e f f e c t i v e
We
A. B,
u(F)
then
= Q,
multiplicities)
Let
by
Dm
k[X,
, m ~ 3
m
that
that
Put
by
m = m,.
Y,
the r e m a i n d e r
E B
irreducible
It is c l e a r
~(f)
by
if
combination
ring
of)
i
by
h.
now
so t h a t m).
C
of
ring
m # m'
call
R (2) = k[X,
shall be denoted
(the u n d e r l y i n g
if
in o n e - o n e
linear
the u n d e r l y i n g
0
defined
[
are
also
a formal
shall write
Also,
type
and w e
is j u s t
structure to
of this
=
D m 6 ~(H)
on
For
< h > ~ 29]
29) = the
(with
divisors
pl~ne.
We
on
shall
f
denote
f.
space.
or
pointsets
prime
fixed
a pointset which
part
29
we
is c l e a r l y
of
L
on
define a vector
~
as
follows. If
h E [L, 29]
Fi (L, 29)(M)
= min
A pointset
Remark
I:
implies
~
then
[ < h > ( M ) I h E {L, 29], 29
is c a l l e d
It is e a s y
h E [B n, 29},
~ = O,
~ 0
to s e e then
29
n-free
that is
if
~ if
29
Fi (L, ~}) = 29. ~ 0]
for
Otherwise
M E @(f).
F i [ B n, 29] = ~).
is
n-free
(n + l ) - f r e e .
and
there
exists
With each pointset where
~
w e d e f i n e d e g r e e of
the sum is e x t e n d e d
pointset
if
~(M)
A point defines
~ 1
we have
into ~ ( ~ ( F ) ) D = @*
for
consists
for all
M c ~,
the m i n i m a l
E ~(f)
o v e r all
i.e.
M;
otherwise
~
homogeneous
prime
ideal This
M ~ c ~. induces
a * . Any
(0, 0, 0,
G E A
with
l)
~(G)
a simple
is s i n g u l a r . prime
ideal Thus
D E ~(~(F))
B,
from ~(f) so that
divisor,
in the p r o j e c t i v e = g E B
in
for e a c h
a map
shall be c a l l e d a h o m o g e n e o u s
through
It is c l e a r that if
is c a l l e d
a minimal homogeneous
w h i c h we d e n o t e b y
of lines
~
~* = M ~ E 9(u (F)).
@ E ~(f)
~ = deg ~ = ~ ~(M)
M E ~(f).
since
it
3-space.
for some
n
then
n
< ~ G > : *. A pointset d i m k [B n, ~] n-gg
for
1.3
~
is c a l l e d
= max
n = 0,
[0,
!:
+ @ = , g'
6 Bs_ t
Proof:
there
the f o l l o w i n g
Let
~,
that
exists
u
~
If
n
min
g,
in
vm(h)
to p r o v e
nB(m ~
Note
~ : , and there
0 # ~ E Bt' exists
of
and h e n c e
gB(m ) This
u E [; u / 0
all the a s s o c i a t e d
the u n m i x e d n e s s
(We a p p l y
t h e o r e m all of
the t h e o r e m to the
of
g
if
h B(~)
for all the a s s o c i a t e d being
if
Now
a valuation
= n N
~'
are c o n t a i n e d
in
and we
that w e h a v e
c
"B(m)"
prime
ideals
s i m p l y get from proved
where m
~
above
Thus m
of
a
none of the a s s o c i -
so that E~
a
prime
ring,
m i n [Vm(U) I u E ~ B ( m ) ] gB
ideal
f.)
for an a s s o c i a t e d
if and o n l y
B(m)
follows
for then so that
< g ' > : - = @.
and t h e n take r e s i d u e s m o d u l o
that
n'
s - t,
being homogeneous
~ E n
But
for some
E g B . g
if and o n l y
ideals
~ vm(g).
R e m a r k 2:
~
B
[v~(u) I u E ~ B ( m ) ] .
Hence_
~ = u~ of d e g r e e
is the p r i m a r y c o m p o n e n t
c nB(m )
ated p r i m e
that
[3] p. 203. f
h B(m ) c n B ( m ) .
~B(m)
s ~ t
that
m , then c l e a r l y
it is e n o u g h ~,
so that
g' : u
to s h o w
t h e m are m i n i m a l .
ideal
be p o i n t s e t s
are homogeneous, and by
generated by
theorem.
Then clearly
to p r o v e
It is w e l l - k n o w n of
is c a l l e d gg if ~ is 1 d i m k B n = ~ (n + i) (n + 2).
Note:
residuation
is h o m o g e n e o u s
g' E Bs_t,
Thus we w a n t
~
if
< g ' > : @.
It is e n o u g h
primes
@
0 @ h E V s.
so that
it follows
m - i.
generic)
f
We s h a l l p r o v e
proposition
(n-geometrically
dim k B n - deg ~].
i, 2 . . . . .
R e s i d u a t i o n on
n-gg
n'B (m) = B ( m ) " if and o n l y if
and h e n c e
I(-~) = g ~
~ E gB
.
whenever
the
only associated prime hold
in the g e n e r a l
1.4
The t a n q e n t s Let
G E A,
for all b u t Then
d
of
case w h e n
g B B
and this
is r e p l a c e d b y
G ~ 0.
Then
o(G)
finitely many values
~ (G) = the l e a d i n g S c A
=
of
the o r d e r of
For any set I c A, ~(I)B
are m i n i m a l
R (n)
result will
as in i.i.
at the o r i g i n
is c a l l e d
define
ideals
G
is a h o m o g e n e o u s
n.
Let
G = gd"
where
Ord G.
We d e f i n e
F E I.
Then
Z(I)B
I,
~ say,
t a n g e n t set of
: [~(H) I H 6 S]. -i ideal in B. ~ (£(I)B)
u
homogeneous].
1.5
Divisors
also defines
(Assume
on
~(F)
so t h a t
if
and the B e r t i n i
£ - E =
of
For any
l i n e a r s y s t e m if, w i t h
over
k.
If
or
defines
f
the
I.
Sup-
called
the
= m i n [vm(~) I u E ~(I)B,
fixed
on
40 E L,
~ @ 0,
~(F)
for some effective
(h) = £ - £0
~ c Cn,
ideal
Theorem
£, E
(g)
l e c t i o n of l i n e a r l y e q u i v a l e n t
[h E ~I h = 0
on
We
Z ( I ) B ~ 0.)
R e c a l l that two d i v i s o r s arly e q u i v a l e n t
a pointset
~(m)
0
Z (0) = O.
~(S)
t a n g e n t c o n e at the o r i g i n of the a f f i n e v a r i e t y d e f i n e d b y pose
gn
d = min In I gn ~ 0].
and is d e n o t e d by
form of
define
gn " gn E B n
are said to be
g E ~.
Let
divisors.
the set of
for some
L
space
be a c o l -
is c a l l e d
functions
£ EL]
is a v e c t o r
L
k,
a
!'
is a v e c t o r over
line-
= space
we obtain
a linear system putting
(If w e
fix
£0 = '
In general, L'
by
h0~'
denoted by
replacing
for s o m e h1 [~21
generated by
go E ~, 40
then by
£~
0 ~ h 0 E ~.
0 ~ h i E i']
i'
: [~
has
I g E i}.) 0 the e f f e c t of r e p l a c i n g
In p a r t i c u l a r ,
depends
on
L
the s u b f i e l d
of
o n l y and w i l l be
k(L).
Given a linear system the fixed c o m p o n e n t of
L,
L,
we define
the d i v i s o r
Fi(L),
called
by
Fi(L) (m) = rain [£(m) I A E L]
for any
mE
ar system,
~(~(F)). Fi(L')
Bertini's
= 0
Then and
L' = [4 - Fi(L) k(L)
t h e o r e m now states
A E L]
is a g a i n
a line-
= k(L'). the
following:
If
L'
is a l i n e a r
system without transcendence exists for
fixed c o m p o n e n t d e g r e e of
E 6 L'
In fact,
k(L')
such that
"most" m e m b e r s
of
of
the t r a n s c e n d e n c e
Prolx~stition 2: 1 deg ~ = 7 m ( m -
L'
Proof:
i:
set.
Also
of
k(L')
§2
THE
is at least
2,
then there
(This is true e v e n
4.2
[2] p. 61 w e can d e d u c e
i.e. not a p r i m e d i v i s o r ,
over
k
is at m o s t
that then
i.
POINTSET
shall prove
There exists ~
is
so t h a t
g E [Bm_l,
a simple pointset
(m - 2) gg ~ ~ 0,
p},
and
> ~
~ # 0
~
(m-
on
f
and for any
with
so that
i) free,
(M)
i.e.
there
M E @(f)
= ~(M).
In a d d i t i o n
m-free.
First we
Lemma
k
and s u c h that the
is a p r i m e d i v i s o r .
is r e d u c i b l e ,
degree
i),
g E Bm_ 2
is t h e n
over
= 0)
L'.)
In this s e c t i o n w e
is no
E
Fi(L')
f r o m the p r o o f of Thm.
if e v e r y m e m b e r
there exists
(i.e.
Let
shall p r o v e
~ c Bn,
let
~
the f o l l o w i n g
n < m - 1
be any point.
lemmas.
be a v e c t o r Then
space
d i m [~, ~]
and
~
a point-
- 1
d i m [~, ~ + 9~] < d i m [~, ~9].
Corollary
i:
d i m [i, ~}
Corollary
2:
If
d i m [~5, ~ + ~}
Lemma
2:
Lemma
3:
exists
~
~2
g EIBm_I,
and
a simple pointset
is
d i m [~, 29} # 0,
then
- i.
~
so that
deg ~ =
gg.
be a p o i n t s e t ~ ~ 0
~ m ( m - I).
with Also
of d e g r e e
1 ~ m ( m - i).
= ~2 + ~' ~
is
where
(m - 2) gg
~'
T h e n there is a p o i n t s e t
if and o n l y
if
F'
(m - 2) gg.
Lemma is
and
Let
of d e g r e e is
Fi [~, ~} (M) = ~(M)
= d i m [~, ~}
There exists
1 m ( m - i) 2
> d i m ~ - deg ~.
4:
(m-
Let 2) gg.
~
be Then
a s i m p l e p o i n t s e t of d e g r e e ~
is
(m-
i) free.
1 ~ m ( m - i)
which
Corollary
3:
£
is
m-free.
The p r o o f of P r o p o s i t i o n and C o r o l l a r y
We w i l l
d < m
then
there and
use w i t h o u t
~ ~ 0
i:
further
since d e g r e e
If
is n o t h i n g
exists
follow
from L e m m a s
3, 4
d i m {~, ~}
to prove.
h E [~, ~},
comment
of
then
0 ~ h ~ Bd
and
dim {~, ~ + ~]
d i m [i, ~]
h / 0
if
f = m.
= 0
If
that
Since
> 0
~ c B
then c l e a r l y with
n
= 0
n < m,
and there
~ ~ 0
> Jg. Clearly
To prove
[~, ~ + ~] c [~, ~}
the s e c o n d
inequality
d i m {~, ~ + ~] < d i m [~, ~]
clearly
3.
Note:
Proof L e m m a
2 will
~ ~,
i.e.
~ ~ + 9~.
(~) Let
then
g ~ [~, ~},
= ~(~)
(~) Thus
so that there
Then
~(~)
g / 0.
< d i m {~, ~].
to c o n s i d e r is
< (~)
Then
Otherwise
= (~).
> ~(~)
{~, ~}
This p r o v e s
d i m {~, ~ + ~]
the case w h e n
h E {~, ~9},
<
~ ~ 0,
(~ + ~) (9~) = ~(~)
+ I,
= ~(~).
g E [~, ~ + ~].
(~)
and
it is e n o u g h
Hence
= (~)
is g e n e r a t e d
since
there
~ ~9. (~)
exists
= (~)
(~)
~ ~(~),
~ 6 k
so that
as a v e c t o r
If
> ~(~)
+ 1
we h a v e
so that
g + ih E {£, ~ + ~}.
space by
h
and
[~, ~ + ~].
the result.
Proof C o r o l l a r y
i:
Follows
by
induction
on
deg ~
using
the i n e q u a l -
ity
d i m {~, ~}
Proof C o r o l l a r y there
exists
2:
Since
h 6 [!, ~],
[~, ~ + 9~] ~ {~, ~].
Thus
- 1 ~ dim {~, ~ + ~ .
Fi {i, ~] (~) = ~9(~) ~-~ 0
from L e m m a
d i m {~, ~ + ~] < d i m {~, ~},
Proof 1 Lemma
There so that
Put
2:
exists
~t
~0
is
= 0
so that
hence
We shall
prove
the
a simple
pointset
and
(~)
d i m [~, ~}
= ~(~).
1 it follows
that
~ 0
Then d i m {~, ~]
the result.
following
~t
more
of d e g r e e
general
t,
result.
t = 0,
1 ....
gg.
which
clearly
satisfies
the c o n d i t i o n s .
Assume
- 1
~O'
~i .....
~t-i
are
ns, d = d i m Vs, d /~t = ~ t - i such
that
that
gd
+ ~
already
O ~ s ~ t - i. is s i m p l e
and
V ( t _ l ) , d ~ [0}
Put
We shall
gg.
Vs, d = [Bd,
find
For each
choose
gd
a point
V(t_l), d
E V(t_l), d
/gs] ~
and
so t h a t
, 1 < d < m
so t h a t
so
gd # 0
~ 0.
Let
Nt
g £ Nt
be
the
implies
~t = ~t-I then
+ 9~"
s e t of all
(7~) We claim
= 0
If
such
and
that
nt, d = m a x
implies
defined.
induction
such
that
~ ( t - l ) (~) = 0.
~
Put
Vt, d : [Bd, ~ t ] ,
[0,
dim B d - deg ~t].
nt, d = 0.
nt, d = d i m Vt, d
Also
n(t_l), d = 0
hypothesis
O = max
so t h a t
Choose
that
if
n ( t _ l ) , d = 0, t h e n c l e a r l y by
gd"
such
[O,
dim B d - deg ~(t-l)
dim Bd - deg ~(t_l)]
~ O.
dim
Hence
Bd - deg ~t
~ O
and nt, d = m a x If
n ( t _ l ) , d a i, t h e n b y
n(t-l),d Also by Corollary there
[0,
exists
2,
induction
nt, d = n ( t _ l ) , d - 1
gd E V ( t _ l ) , d , gd @ 0
nt, d : d i m
2 Lemma
2:
curve
f
choose
the p i e c e
denoted k[~',
~' (f),
~']
An element defined
It is e n o u g h
contains
for
a pointset
outside has
where h E Bd
X'
hypothesis
= dim B d - deg ~(t-l)
F i ( V ( t _ l ),d ) (~) : 0 = ~ ( t - l ) (~)"
Proof
dim B d - deg ~t].
the
affine
induces by
the c h o i c e
of
= 0
so t h a t
affine
piece
B d - d e g 29t a O.
to p r o v e having
line
a
since by (~)
Hence
- ~Y
that
some
the d e s i r e d
Z = 0.
coordinate
= ~X , ¥'
~ E ~' (f)
and
> i.
and
k-valued
ring
properties.
(Note
Z ~ f.)
k[X',
f' (X' , Y') function
h' (~) = ~ - r e s i d u e
This
Y']/f'(X',
piece,
Y')
: f(X' , Y' , i). h'
of
o f the We
on h(X',
£' (f) Y',
i).
=
9 Now
deg
f
= deg
{h' I h E Bd~,
f = m,
a vector
and hence space
dimension
s(d)
= 71
(d + i ) ( d
of degree
< d
in
X'
h 6 Bd h'(~)
and = 0.
h 6 Bd ,
and
~ 6 @(f'). Thus
h' (~i)
It
follows
Y'.
that
{Bd,
They
~]
Bd
Bd
functions
Let
~ = ~i
if a n d o n l y
d < m,
k-valued
+ 2).
w I .....
on
a ~
+
@' (f)
for
B~.
if a n d o n l y
"'" + ~ t "
~i
of
be monomials
Ws(d)
form a basis
Then clearly
for s i m p l e
a ~
of
for
Let if
~ @'(f)"
and
if
= 0,
i = i,
2 .....
is i s o m o r p h i c
t.
to t h e s p a c e
of solutions
of
s(d) i~=~lliwi(7~j)
To
find
~
~i .....
with
the r e q u i r e d 1 (t = ~ m ( m - I),
~t
(wi(~j))
has maximal
d = 0,
...,
i,
= 0,
properties all
possible
m - i.
For
j = 1 .....
~
is
gg.
now
appeal
set and If
V
a vector
d i m V ~ t,
E V
such
that
Applying so t h a t ible.
to the
- rank
= s(d)
- min
then
this
s(d),
so that
s(d)]
the m a t r i x
for
there
exist
(w i(~j)) {t,
s(d)
s(d)]
- t]
elementary
k-valued
lemma:
functions
~i ....
' ~t
Let
on
6 S
S,
and
S k
be
a
a field.
~i'
....
~t
= 6ij. S = @' (f)
(wi(~j)),
Then clearly
i = 1 .....
of
to
the m a t r i x
{0,
following
space
Ui(~j)
{t,
= s(d)
= max
We
distinct)
min
to d e t e r m i n e
then
d i m {B d, ~]
and
it is e n o u g h
~i
rank
t.
f o r any
and
i,
V = B m _ 2,
j = 1 .....
d ~ m - i,
j = 1 .....
t
has
we
find
the m a t r i x
maximal
~i .....
t = s ( m - 2)
(wi(~j)),
possible
rank,
as
required.
Proof Lemma exists theorem d e g J'
3:
By C o r o l l a r y
0 ~ g 6 { B m _ I, J} deg = m(m 1 = ~ m ( m - I).
It is e n o u g h
-
1
with I)
to p r o v e
d i m {Bm_l, a
we have
that
if
j
/~
~ m
and since by
< g > = J + J'
J
so t h a t
is n o t
there
Bezout's
with
(m - 2) g g
~t
is i n v e r t -
then
~'
10 is n o t
(m - 2) gg,
Thus 3.
let
Then by
= dim
Bezout's
(Bm_3)
there
- 1
exists Then
1 there
exists dim
Proof
Lemma
g'
4:
6 Bm_ 2
We have
g 6 { B m _ I, ~]
~ =
Hence
~ {Bm_2, <~>(~) h~h~
£]
= 0.
NOW
~
R' (~) = 0
some and
h~ 7,
then
(~)
<~>
for any
R(~)
~ =
= J'
~ E @(f)
# 0
6 B1
R(~)
R(~)
E 9(f).
= i. Then
= 0
by
so t h a t
for s o m e
<~>(~)
= 0.
of
Fix
g
= 0 = R(~)
R' (~) m i.
~
= i,
with
we have
as a b o v e
0 ~ g E I B m _ I, ~]
follows
and we have
For each so t h a t
we may choose h ~
exists
= 1 : R(~).
(Existence
Then
+ J
so t h a t
a point with
= 0,
+ J'.
there
and h e n c e
E { B m _ 2, ~ - ~} (£ - ~)
(~)
(~)
= £ + £'.
Proposition
= R(~).
exists
~
Hence
(m - 2) gg.
(~ - ~)
that
= I.
and
I.
and by
d i m I B m _ 2, ~}
E { B m _ 2, F - ~] (~)
and
+ J'
1 and
with
assume
@ 0
where
there
be
Corollary
+ ~
< g > = £ + £'.
Thus we may choose
(~ - ~)
~ 0
pointset m ( m - 3) - i) = 2
by Corollary
Choosing
let
is n o t
(~)
and hence
E I B m _ I, @}
and write If
m 1
['
g'
that
by
+ J'
and
the c a s e w h e n
Then we may write d i m I B m _ 2, ~ - ~]
=
J'
J, J'. for some
1 - ~ m(m
- 2) m 1
so t h a t
to s h o w
so t h a t
consider
hypothesis.
~ 0
£}
so t h a t
+ £'
in
= J + £
deg J = m(m
£ { B m _ 3, J]
J']
symmetric
and
d i m {Bm_3,
= J + J + ~'
{Bm_2,
First
theorem
0 ~ h'
being
h @ 0
so t h a t
Hence
the r e s u l t
h E B m _ 2,
h~
~
nothing
with
(~)
6 B1
E [Br~_l, £]
as in L e m m a
and
to p r o v e .
R(~) = 0.
with
= 1 If f o r
<~>(~)
(~)
3.)
= 0
= 0 = R(~)
as r e q u i r e d . Thus we have all
~
with
thus
proving
only
~(~) the
to c o n s i d e r
= i.
We
shall
the
case when
deduce
(~)
a contradiction
~ 1 from
lemma.
We have
= (£ - 7) + ~ + Hence
h
deg ~r
= dim
{Bm_ 3,
Also
(Bm_3)
- 1
and as
£~.
in L e m m a
3 there
so that
=
(~ - 7)
+ ~ +
(~' - ~)
: + ~ '
10
-~
+ ~
so t h a t
+ ~ +~.
exists
for this,
11 By
Proposition
1 there
exists
g~ E [Bm_ 2, @'
:~' _ ~ + ~ By L e m m a d i m {Bm_2,
F'
2,
d i m [Bm_2,
- ~]
= 1
(up to a c o n s t a n t But then a @
k).
(~)
Proof C o r o l l a r y that by R e m a r k
3: 1
that
$(G)
exists
3:
d i m {Bm_2,
is
so that
ideal
and
have
Proof:
Consider
other hand
E B m) .
<~(H)>(m)
= 0.
and is
GW,
the v e c t o r
space
+ ~I
Fi(l~l)
GX,
~ 0,
determined
for all
= i,
that
is
a contradiction.
0 # g E {Bm_ I, ~],
F(~)
o(H)
Obviously and
if
~
> 0
D P
where G
implies
E IBm,
Fi(l~l)
so
~]
so that
] c C--re.
> ~*.
On the
<~ (H)> = * is not h o m o g e n e o u s
is h o m o g e n e o u s
and since
G Z
E ~
F
is
so that
degree
of
k ( l ~ I)
Theorem
there
exists
a prime divisor
+ H E
+ H--> = ~ *
II
+ m.
m
P
F' (~) = 0.
/~* ~ Fi(ILI)-
so that t r a n s c e n d e n c e
XGW
so
Then there
(F, G + H ) A = ~
with
0 ~ ~ (H) E ~
G Y ,
0 ~ G E A
F u r t h e r we can c h o o s e
= ~*.
3, c l e a r l y
Let
= ~ + ~'.
i.e.
H 6 A
Fi(l!l)
2.
over
k
2.
for some
by
~.
THEOREM
and
Put d i f f e r e n t l y ,
By B e r t i n i ' s i.e.
E B
points,
Hence
k ( l ~ I) : ~
and w r i t e
~(~)
m-free by Corollary Now
~]
there exists
(h = ~(H)
F(~)
3 there e x i s t s
no c o m m o n
that
with
PROOF OF THE
G(H)
= {X~W
We c l a i m
g~ = g'
- ~}
m free.
m = I(~).
and
~
F'
is u n i q u e l y
be as in P r o p o s i t i o n
= g E [Bm_l,
is a prime F'
~
d i m {Bm_2,
take
~] ~ i,
As in L e m m a ~
Let
H 6 A
i, and g~
for all
so that
+¢~. Since
So we may
z 1
§3
Proposition
= 0.
by L e m m a
in
and we get
F']
- ~]
E I£I ',
12 S
With each divisor
~ =
~ n i m i on ~ (F) w e can a s s o c i a t e i=l d e g m i = d e g r e e of the i r r e d u c i b l e c u r v e
deg ~ = ~
n i deg m i
where
on
defined by
the ideal
~(F)
faces y i e l d s
for
U 6 Cn,
m i.
Then Bezout's
theorem
for s u r -
U ~ 0
deg = nm.
Thus
deg <XGW+
H> = m
2
.
Also
for any
~ £ @(f),
deg
(~*) = i,
i.e.
the d e g r e e of a h o m o g e n e o u s p r i m e d i v i s o r is i. Hence clearly 1 1 d e g ~* = d e g ~ = : m ( m - i) so t h a t d e g (m) = : m ( m + i) > i. In particular
m
clude
k ~ 0
that
is not a h o m o g e n e o u s
since otherwise Now
k~W
to the p r i m e As
m
since otherwise = ~'* + < W >
+ H ~ W~
m
would
since
Hence we can con-
is h o m o g e n e o u s .
Also
H ~ 0
n o t be prime.
H 6 A
and h e n c e
W
does not b e l o n g
ideal of m .
in R e m a r k 2 we can c o n c l u d e
and s i n c e
prime divisor.
W
does n o t d i v i d e
XG W
that
(IGW
+ :
+ H)C = I(~*)
we have
n I(m)
in a f f i n e
3-space
(kG + H,
where
~(P)
that
~(~) Since
= I(m),
~(~)
F)A = ~ n P
= I(~*).
It is s t r a i g h t f o r w a r d
to c h e c k
= I(~) . k ~ 0
w e may
take
k = 1
and w e h a v e
the r e q u i r e d
G + H. F o r the last r e m a r k we with
R(Y~) = 1
there
is
simply have
0
to o b s e r v e
~ g~ 6 { B m _ I, ~]
t h a t for e a c h
so that
(~)
= i.
We m a y t h e n take a l i n e a r c o m b i n a t i o n and
(~)
= (Y~) = i.
~ C~g~ = g so t h a t C~ £ k R(~) =l It is c l e a r t h a t G = -l(g) is the
required element.
P r o p o s i t i o n 4: satisfies
Proof:
Let
<%(F')> ~(F")
~ ~
It is e n o u g h F" 6 A
<~(F")>(~)
does not pass
since,
as
so t h a t
as in P r o p o s i t i o n
Choose
and
~(F")
F~
P
F' 6 P.
> ~'.
~ 0
For
3 the t a n g e n t set
~
of
~'.
to p r o v e
so t h a t
= 0
~(F")
for all
7~
t h r o u g h any p o i n t of
F i { B m _ I, ~} (~)
= ~, = 0
for e a c h and
o(F~)
12
t h a t if
6 {Bin_ I, ~
with ~'.
Y~ w i t h
~(F')
~ 0 with
~' (Y~) > 0, This
i.e.
is p o s s i b l e
~' (Y~) > 0
6 {Bm_ I, ~}.
then
we
find
A suitable
P
15 (general)
linear combination
We c l a i m that = Now of
F" ~ P
is h o m o g e n e o u s
F'F"
6 P n ~ =
(F, G + H ) A
is
> ~' (~).
Since =
~ + ~'
(f, g)B
if
where
~ 0.
=
and n e i t h e r
E
's
m
serve
(m)
this w e
<~(F")>(7~)
= 0
<£(F')>
nor
Clearly
~
~',
~' (~) > 0
and h e n c e
to
~((F,
G + H)A)B
G + H)A)B.
as w e l l
assume
since otherwise
is zero.
set
as r e q u i r e d .
f, g E ~(F, w e may
~ ( U I F + U 2 ( G + H))
F". but
find t h a t
if
o u r c l a i m is e q u i v a l e n t
(F, G + H)A,
as
> 0;
is not.
~(Ul)f + Z(U2)g ~ 0
Z (U 2)
will
We c l a i m that the t a n g e n t
Assuming
But
g = q(G).
Then
and
This m e a n s
UIF + U 2 ( G + H)
~(U2)
F~
(F, G + H)A.
~ + ~'.
< £ ( F ' ) ~ ( F " ) > > ~ + ~'. <~(F')>(~)
of the
since otherwise
Also
that
Z(U2) g
= 0
It is t h e n c l e a r t h a t
= ~(UI) f + Z ( U 2 ) g
so t h a t ~((F,
Corollary
Proof:
4:
Let
and h e n c e being
p c
(f, g)B.
Ord(E')
F' ~ 0.
> m
2) gg,
or
£(F')
Then
~(F')
E Bd
~ 0
> ~'.
with
and e i t h e r
In the
d > m - 1
~(F')
l a t t e r case, so t h a t
= 0
~'
Ord(F')
>
i).
(m-
Proposition fact if
5:
U I,
For
~, ~',
.... U s
P
as b e f o r e ,
generate
P,
[ Bm_l,
~'~
are in
Bm_ 1
Proof:
F o r the f i r s t s t a t e m e n t
generate
gl E IBm_ I, ~'}, -i
=
((X, Y, Z)A) m-l.
F' E P,
(m-
G + H)A)B
(gl + hl) Now
Also
if
pointset exists
gl ~ 0
E P,
for t h e n
0 ~ gl E Bm_ 1 h = ~(H), 3'.
Now
0 ~ h I E Bm
there
IBm_l,
it is e n o u g h to p r o v e t h a t exists
h I E Bm
which
= ~' + 3.
< h g l > = < g > + J + 3' < h l > = I + 3'. m
h g I = gh 1 so that HG 1 = GH 1 + UF
13
E ~(P).
Also
= ~ + ~'.
= ~ + 3'
and b y Then
for
so that
+ hl))
we have by construction
with
and in
Z(Ui)
.
gl = ~ ( ~ - l ( g l
and
~'} c ~(p)
then t h o s e o f the
for some
Proposition
1 there
14 where -i G1 = o
-i (gl),
H1 = c
(G).
Hence (G + H ) G 1 = G(G 1 + H I) + UF. Now
G + H, F E P
the same r e a s o n as G1 + H1 E P
so that
F" ~ P
G(G 1 + HI)
E P.
Also
G ~ P
in the p r o o f of P r o p o s i t i o n 4).
(for Hence
as r e q u i r e d .
For the s e c o n d s t a t e m e n t U 1 .....
Us
of
P,
any
£(alU 1 + Now
"'" + asU s), £ (U i) E B r,
observe
that g i v e n a set of g e n e r a t o r s
0 ~ gl E {Bin_I, ~'}
a I, r >
is of the f o r m
.... a s E A. (m - i) b y C o r o l l a r y
4.
Since
0 ~ gl E
Bin_ 1 ,
~ (a i) Z (U i) • (U i ) EBm_ 1
gl =
(ai) Ek This p r o v e s
the s e c o n d s t a t e m e n t .
P r o o f of the Theorem: t h a t the p r i m e
By C o r o l l a r y
ideal
P
needs
§4
R e m a r k 3:
If the p o i n t s e t
1
at l e a s t
R e m a r k 4: ~'
One w a y
P
~'
~' (~) ~ 1 i n s t e a d of
(m - 2) gg.) Bertini's pointsets question
is
to p r o v e
can be a r r a n g e d
so that to
set of
m
It follows
generators.
~'
in §3 c o u l d be p r o v e d
to be
4 c o u l d be s h a r p e n e d
(m - i)
to p r o v e
that
~'.
that
~'
is
(m - i) free is to s h o w t h a t
to b e a s i m p l e p o i n t s e t d i s j o i n t implies ~.
~ m.
SOME Q U E S T I O N S
free t h e n the p r o o f of P r o p o s i t i o n the t a n g e n t
d i m {Bm_ I, ~'}
~(~)
(By L e m m a
This c o u l d be done
t h e o r e m on v a r i a b l e
= 0). 3,
~'
Then one m a y is
to the
g ~ 0].
0
~
(i.e.
apply L e m m a 4
(m - 2) gg
in c h a r a c t e r i s t i c
singularities
{ < g > - ~I g E { B m _ I, ~},
from
since
~
by a p p l y i n g
l i n e a r s y s t e m of
W e ask the f o l l o w i n g
in g e n e r a l :
Q i:
Does t h e r e e x i s t g 6 Bm_ 1 so that 1 d e g ~' = ~ m ( m - I) and b o t h ~, ~' are disjoint?
14
= ~ + ~', (m - 2) gg,
deg ~ = simple
and
is
15 Q 2:
g E Bm[ 1
Let
d e g @ = d e g @'
Another Lemma
2.
be
= ~ m(m
member
Does
it
i)
~
is
(m - 2) gg
~
is
(m-
way would
i)
be
and w r i t e
follow
ii)
possible
We
a generic - i).
to h a v e
Let
g E Bm_ 1
with
= ~ + ~'.
does
it f o l l o w
that
~'
is
Clearly we may
Remark
above
5:
a lemma
of the
type
of
ask
free,
in the
= F + F',
free?
Q 3:
gg
that
As
assume,
Given
(m - i)
if n e c e s s a r y ,
that
~
is
(m - i)
free?
that
~
is
(m - 2)
gg
question.
in R e m a r k
3,
if
~'
is
(m - i)
free,
then
~'
is the
t a n g e n t set of P. Thus if w e c o u l d a r r a n g e ~' to be of the 1 ~' = ~ m ( m - i)~ the r e s u l t i n g c u r v e t h e n w o u l d h a v e o n l y one (set t h e o r e t i c a l l y )
at the o r i g i n .
Q 4:
can we arrange 1 ~' = ~ m ( m - i ) ~ ?
Remark
6:
Just
pointsets distinct such
in the
have
fixed
no
through
set
~
on
simple
~
we
so that
~
and
point
outside
~.
then
the
f.
Thus
in g e n e r a l
image
For
a set
~
as above,
of degree
n
through
In c h a r a c t e r i s t i c one
can
a~o
and
of the
on a c u r v e as
f
prove
not
equal
a more
start with
there
curve
One has
may
f
form
we may
formal
clearly
0, that
of d e g r e e
If w e of
~
we
define
sums
of
corresponds
to
would
m - 2 through
a nonsingular be
the
curve
required
is
min
{n I t h e r e
is a n o n s i n g u l a r
?
using
Lemma
indeed
to z e r o
pass
f
1 ~ m ( m - i)
of
of d e g r e e (m - i)
could on
~
ask:
what ~]
a set
is no c u r v e
the c u r v e s
Q 5:
induction
free
on
tangent
ask
multiplicities) pointset
~,
characteristic
we
type
in the plane.
plane
through
(m - i)
pointsets
(without
Every
Hence
to b e
defined
to c o n s t r u c t
passing
f
as w e
points.
Thus points
~'
in the p l a n e
a pointset
or
4 and
a type
the m i n i m u m
the q u e s t i o n
generalized
version
15
is
of d e s c e n d i n g (m - i).
remains
open.
of Q 5, viz.
In
16 Q 6: For a pointset nonsingular
~
in the plane h o w to find
curve of degree
n
through
~}
min {n I there
is a
?
BIBLIOGRAPHY
[1]
Macaulay, Hafner
[2]
"The Algebraic
Service Agency
Matsusaka,
Memoirs
Ser A, Vol. XXVI,
[3]
Zariski,
Samuel,
Theory of Modular
Systems",
Stechert
(Reprint) ~, 1964.
of the College 1950,
of Science,
Univ.
of Kyoto,
p. 51-52.
"Commutative
Algebra"
Company.
16
Vol.
II, Van Nostrand
PRIME IDEALS IN POWER SERIES RINGS
Jimmy T. Arnold
Virginia Polytechnic Institute and State University
ABSTRACT. In this paper we wish to briefly review some known results concerning the ideal structure of the formal power series ring m[[x]]. As the title indicates, primary consideration will be given to prime ideals in R[[X]]. We begin by discussing some of the basic difficulties which arise in relating the ideal structure of R[[X]] with that of R. We then consider the Krull dimension of R[[X]] and, finally, we review some results on valuation overrings of D[[X]], where D is an integral domain.
i.
Introduction.
Our notation and terminology are essentially that of [8].
Throughout, R denotes a conm~utative ring with identity and D is an integral domain with quotient field K.
If R has total quotient ring T, then by an overring S of R
we mean a ring S such that R c S c T. w and w
The set of natural numbers will be denoted by
is the set of nonnegative integers.
If A is an ideal of R, then we let
O
A[[X]] = If(X) = ~'~l=o a'xil I a.1 E A for each i E Wo} and we denote by AR[[X]] ideal of R[[X]] which is generated by A. q.f.
the
We denote the quotient field of D[[X]] by
(D[[X]]).
2.
Quotient overrinss and extended ideals.
Let S denote a multiplicative
system in
R.
In studying the ideal structure of the polynomial ring R[X], one important and
widely used technique is to pass to a quotient ring R S of R and utilize the fact that Rs[X ] is a quotient ring of R[X] - namely, Rs[X ] = (R[X])s. 35.11 of [8] and Theorems 36 and 171 of [I0].)
(cf. Theorem
One particularly important aspect of
this technique is that in studying the integral domain D[X], one is able to make considerable use of the ideal structure of the Euclidean domain K[X].
For example,
one can easily describe the essential valuation overrimgs of D[X] in terms of the essential valuation overrings of D and K[X] [4, Lemma I].
17
Unfortunately,
one canno~
in general, employ such techniques in studying power series rings.
In fact, it is
clear from the following result, proved by Gilmer in [7], that D[[X]] and K[[X]] seldom share the same quotient field.
THEOREM I.
The following conditions are equivalent.
(I)
~.!.(D[[X]]) = ~.!.(K[[X]]).
(2)
K[[X]] = (D[[X]]) D _ (0)"
(3)
If [ai}i= 1 is a collection of nonzero elements of D, then Ni~ 1 aid # (0).
In [13, Theorem 3.8], Sheldon restates Theorem 1 in the following more general form.
THEOREM 2.
If S is a multiplicative system of D, then the following statements
are equivalent. (I)
~.i.(Ds[[X]] ) = ~.~ (D[[X]]).
(2) Ds[[X]] ~ (D[[X]])D- (0)" (3)
Fo___rrevery sequence [si}i= 1 of elements of S, Ni= 1 aiD # (0).
Even when the conditions of Theorem 2 hold, it need not be the case that Ds[[X]] = (D[[X]]) S.
In fact, Ds[[X]] may not be a quotient overring of D[[X]].
For example, if we let V denote a rank two discrete valuation ring with prime ideals (0) c p c M, then Vp[[X]] ~ (V[[X]]) V - (0)' however, Vp[[X]] is not a quotient overring of V[[X]].
In [13], Sheldon further investigates the relationship between
q.f.(D[[X]]) and q.f. (~[[X]]), where D 1 is an overring of D.
He shows, for example,
that D is completely integrally closed if and only if q.f.(D[[X]]) # q.f.(Dl[[X]] ) for each overring D 1 properly containing D [13, Theorem 3.4].
He also shows that if
= a iD = (0), then q.f (D[[X /a]]) has infinite transcendence a E D - (0) and n i=l degree over q.f (D[[X]]). Another useful technique in considering ideals in R[X] is the following: (*)
Let B be an ideal of R[X] and set A = B N R.
ideal in the polynomial ring (R/A)[X] ~ R[X]/AR[X].
Then B = B/AR[X] is an
Moreover, B A (R/A) = (0).
(cf. Theorems 28, 31 and 39 of [I0].) Even though we do have the isomorphism (R/A)[[X]] ~ R[[X]]/A[[X]] for power 18
series rings, it is not generally true that A[[X]] = AR[[X]]. of R[[X]] with A = B n R, it may happen that B ~ A[[X]].
Thus, if B is an ideal
The following result is
given by Gilmer and Heinzer in [9, Proposition I].
THEOREM 3.
Let A be an ideal of R.
Then A[[X]] = AR[[X]]
if and only if for
each countably generated ideal C c A, there exists a finitely generated ideal B such that C c B c A.
Gilmer and Heinzer also note that the equality A[[X]] = AR[[X]] need not imply that A is finitely generated.
For example, if V is a valuation ring with maximal
ideal M and whose set of prime ideals is of ordinal type ~, the first uncountable ordinal,
then MV[[X]]
= M[[X]], but M is not finitely generated
[9, p. 386].
Howeve~
in order that we be able to employ globally the method described in (*), we need that A[[X]] = AR[[X]]
for each ideal A of R.
The following result is an easy consequence
of Theorem 3.
THEOREM 4.
The equality A[[X]] = AR[[X]] holds for each ideal A of R if and
only if R is Noetherian.
If Q is a prime ideal of R[[X]] and P = Q n R, then Q ~ P 4 ~ X ] ] .
Thus, if we
wish to apply the reduction technique of (*) only to prime ideals of R[[X]], is sufficient to have P[[X]] = ~PR[[X]]
for each prime ideal P of R.
then it
In considering
radical ideals in R[[X]], some results concerning nilpotent elements in R[[X]] will be useful. Let f(X) = E. ~
i=0
a. X i E R[[X]]. 1
If R has characteristic n > O, then Fields
has shown in Theorem I of [5] that f(X) is nilpotent if and only if there is a k positive integer k such that a. = 0 for each i E w . l
Further, Fields gives an
O
example to show that one cannot drop the assumption that R has finite characteristic, k that is, in an arbitrary ring R it may happen that a i = 0 for each i E ~o' yet, f(X) is not nilpotent. f(X).
Let Af denote the ideal of R generated by the coefficients of
If we impose the condition that ak = 0 for each a E Af, then there exists
m E w such that mA~ = (0). characteristic"
This satisfactorily imitates the assumption of "finite
and, in fact, yields that g(X) is nilpotent for each g(X) E Af[[X]].
19
More generally, we have the following result [i, Lemma 4].
LEMMA.
Let A be an ideal of R and suppose there is a positive inteser k such
that ak = 0 for each a E A.
Then f(X) is nilpotent
for each f(X) E A[[X]].
The author does not know if f(X) nilpotent implies the existence of k E m such that a k = 0 for each a E Af.
Thus, in the following theorem, which is irmnediate
from the above lemma, we are able only to give sufficient conditions on an ideal A in order that A[[X]] ~ A R [ [ X ] ] .
THEOREM 5.
Let A be an ideal of R and suppose that for each countably senerated
ideal C c A there exists a positive integer k and a finitely generated ideal B c A such that ck E B for each c E C.
Then A[[X]] ~
JA-R[[X]].
We shall say that an ideal A is an ideal of strong finite type (or, an SFTideal) provided there exists k E w and a finitely generated ideal B c A such that a
k
E B for each a E A.
We say that the ring R satisfies the SFT-property if each
ideal of R is an SFT-ideal [2]. property,
then R satisfies
It is easy to see that if R satisfies
the SFT-
the ascending chain condition for radical ideals,
R has Noetherian prime spectrum.
The converse does not hold.
that i%
For example, a rank
one nondiscrete valuation ring does not satisfy the SFT-property. Following Theorem 3, we referred to an example of a valuation ring V with maximal ideal M such that M is not finitely generated, yet My[x]] then, My[x]] = ~ [ [ X ] ] .
But M is not an SFT-ideal since
of a finitely generated ideal. A[[X]] ~ A R [ [ X ] ] ,
THEOREM 6.
= MV[[X]].
Clearly
it is not even the radical
However, if we impose globally the condition that
then we obtain the following result [I, Theorem i].
Th__£efollowin$ conditions are equivalent.
(i)
R satisfies the SFT-property.
(2)
A[[X]] ~
(3)
P[[X]] = ~
~
for each ideal A of R. for each prime ideal P of R.
In view of the somewhat analogous nature of Theorems 4 and 6, it seems reasonable to ask whether the condition that P[[X]] = PRy[X]]
20
for each prime ideal
P of R implies that R is Noetherian.
3.
Krull dimension in R[[X]].
The author does not know the answer.
We say that R has dimension n, and write dim R = n,
provided there exists a chain Po c PI c •
... c pn of n + I prime ideals of R, where
Pn # R, but no such chain of n + 2 prime ideals. if dim R = n, then n + 1 ~ dim R[X] ~ 2n + i. n + 1 ~ dim R[[X]], however,
Seidenberg has shown in [II] that
It is easy to see that one also h a ~
the following result shows that dim R[[X]] has no upper
bound [I, Theorem i].
THEOREM 6.
If R does not satisfy the SFT-property,
then dim R[[X]] = =.
In order that a PrHfer domain D satisfy the SFT-property,
it is necessary and
sufficient that for each nonzero prime ideal P of D, there exists a finitely generated ideal A such that p2 c A c p [2, Proposition 3.1].
In particular,
a valuation
ring V satisfies the SFT-property if and only if it contains no idempotent prime ideals.
A rank one nondiscrete valuation ring,
therefore, provides a simple example
of a Prefer domain which does not satisfy the SFT-property.
An integral domain D is
called an almost Dedekind domain provided DM is a rank one discrete valuation ring for each maximal ideal M of D.
An almost Dedekind domain which is not Dedekind
[8, p. 586] provides another example of a Prufer domain which does not satisfy the SFT-property. If D is a Prefer domain with dim D = n, then Seidenberg has shown in [12, Theorem 4] that dim D[X] = n + I.
The analogous statement for power series rings
does not hold, for, as we have just seen, D need not satisfy the SFT-property. we may have dim D[[X]] = ~. satisfy the SFT-property, rings.
However,
Thus,
if we restrict our attention to rings which
then we can extend Seidenberg's result to power series
Namely, we get the following result [2, Theorem 3.8].
THEOREM 7.
Let D be a Prefer domain with dim D = n.
are equivalent. (i)
D satisfies the SFT-property
(2)
Dim D[Ex]]
(3)
Dim D[[X]] < ~.
= n + 1.
21
The followin$ statements
It can be easily shown that if R is any ring with dim R = 0, then statements (i) - (3) of Theorem 7 are equivalent which
(i) - (3) are not equivalent.
for R.
The author knows of no example for
In an attempt to find such an example, one
might be lead to consider rings R which satisfy the SFT-property but for which dim R[X] > 1 + dim R.
To construct one such ring,
simple transcendental extension of k.
let k be a field and K = k(t) a
Suppose that V = K + M is a rank one discrete
valuation ring with maximal ideal M (e.g., take V = K[[Y]]).
If we set D = k + M,
then D satisfies the SFT-property and dim D = I, yet, dim D[X] = 3 [8, Appendix 2]. In this case, however, we get that dim D[[X]] = 2 [9]. From the statement of Theorem 6, one can conclude only that there is no bound on the lengths of chains of prime ideals of R[[X]].
The proof given in [I, Theorem
i] shows, more strongly, that R[[X]] contains an infinite ascending chain of prime ideals.
It is easy to see, then, that when statements
(I) - (3) of Theorem 7 are
equivalent
for a rin~ R, then they are equivalent to the statement "R[[X]] satisfies
the a.c.c,
for prime ideals".
The proof given in [I] for Theorem 6 shows, in fact,
that if R does not satisfy the SFT-property, P of R and an infinite chain Q1 c Q2 c Qi n R = P for each i E w.
then there exists a nonzero prime ideal
... of prime ideals of R[[X]]
such that
Foll~qing a proof given by Fields in [6, Len~na 2.4],
Arnold and Brewer show in [3, Proposition i] that if Q is a prime ideal of R[[X]] such that Q N R = M is a maximal ideal of R, then either Q = M + (X) or Q ~ My[x]]. Thus, if D is a one dimensional integral domain which does not satisfy the SFTproperty,
then there exists a maximal ideal M of D and an infinite chain
Q1 c Q2 c
... of prime ideals of D[[X]] such that MD[[X]] c Q1 c Q2 c
... c M[[X]].
It follows from Theorem 16.10 of [8] that if P is a prime ideal of the Prefer domain D, then each prime ideal of D[X] contained in P[X] is the extension of a prime ideal of D.
Our remarks in the preceding paragraph show that this need not
occur in power series rings.
THEOREM 8. SFT-property,
We do have the following somewhat analogous result.
Let D be a finite dimensional Prefer domain which satisfies the
and let P be a prime ideal of D.
contained in P[[X]] have the form plY[X]]
The only prime ideals of D[[X]]
for some prime ideal PI o f D.
22
4.
Valuation overrinss of D[[XI].
The domain (D[X])p[x]
Let P be a prime ideal of the integral domain D.
is a valuation ring if and only if Dp is a valuation ring.
For
Krull domains, the analogous statement for power series rings holds [9, Theorem 2]. In [3] Arnold and Brewer have considered the following question:
If P is a prime
ideal of the domain D, when is (D[[X]])p[[x] ] a valuation ring?
Necessary conditions
are given in the following theorem [3, Theorem I].
THEOREM 9.
Let P be a prime ideal of the intesral domain D.
is ~ valuation rin$, then Dp is a rank one discrete valuation rin$. (D[[X]])p[[x]]
If (D[[X]])p[[X]] Moreover,
is rank one discrete.
The proof given in [3] for Theorem 9 shows the following:
Let (0) c Q c p
= I anDp be prime ideals of D and assume that Dp is a valuation ring with QDp = N n= for some a E P - Q.
Then (D[[X]])Q[[X] ] is not a valuation ring.
Thus, if P is a
prime ideal of any valuation ring V, with dim V = n > i, then (V[[X]])p[[x]] a valuation ring.
is not
This shows that the condition that Dp be a rank one discrete
valuation ring is not sufficient to insure that (D[[X]])p[[x] ] is a valuation ring. Our observations in the previous section yield another simple example. almost Dedekind domain which is not Dedekind, of D and an infinite chain QI c Q2 c MD[[X]] c QI c Q2 c ... c M[[X]].
If D is an
then there exists a maximal ideal M
... of prime ideals of D[[X]] such that
In view of Theorem 9, (D[[X]])M[[X]]
is not a
valuation ring. The following theorem gives sufficient conditions
for (D[[X]])p[[x] ] to be a
valuation ring [3, Theorem 2].
THEOREM I0.
Suppose that P is a prime ideal of the intesral domain D and that
Dp is a rank one discrete valuation Tins. then (D[[X]])pD[[X]]
If PD[[X]] is ~ prime ideal of D[[X]],
is a rank one discrete valuation Tins.
In particular,
if
PD[[X]] = P[[X]], then (D[[X]])p[[x] ] is a valuation ring.
While the assumption that PD[[X]] is assumption that PD[[X]] = P[[X]],
prime
is ostensibly more general than the
the authors in [3] are unable to provide an
example of a prime ideal P which satisfies the hypotheses of Theorem 10 and which has
23
the property that DP[[X]] is prime but PD[[X]] # P[[X]].
If V is a rank one non-
discrete valuation ring with maximal ideal M, it is the case that MV[[X]] is prime, yet, MV[[X]] # M[[X]] (see eerana I of [3] for a proof due to M. Van der Put). The conditions given in Theorem I0 are not necessary, for one can construct a Krull domain D with a minimal prime ideal P such that PD[[X]] is not prime (in particular, PD[[X]] # P[[X]]), yet, (D[[X]])p[[x]] is a valuation ring [3]. Although the following theorem gives necessary and sufficient conditions in order that (D[[X]])p[[x] ] be a valuation ring, these conditions are not entirely satisfactory since they are not given in terms of the ideal structure of D[3, Theorem 3].
THEOREM ii.
Let P be a prime ideal of the intesral domain D and suppose that
Dp is a rank one discrete valuation rin$.
The followin~ conditions are equivalent.
(i)
(D[[X]])p[[X] ] is a valuation rin$.
(2)
P(D[[X]])p[[x]] = P[[X]](D[[X]])p[[x] ].
(3)
Dp[[X]] n ~.~.(D[[X]]) ~ (D[[X]])p[[x] ].
REFERENCES
i.
Arnold J., Krull dimension in power series rings, Trans. Amer. Math. Soc. (to appear).
2.
, Power series rings over Prefer domains, Pacific J. Math. (to appear).
3.
Arnold, J. and Brewer, J., On when (D[[X]])p[[x] ] is a valuation ring (submitted).
4.
.. . . . . . . Proc. Amer. Math. Soe
5.
, Kronecker function rings and flat D[X] - modules, 27 (1971), 483-485.
Fields, D., Zero divisors and nilpotent elements in power series rings, Proc. Amer. Math. Soc., 27 (1971), 427-433.
6.
, Dimension theory in power series rings, Pacific ~. Math 35 (1970), 601-611.
24
7.
Gilmer, R., A note on the quotient field of the domain D[[X]], Proc. Amer. Math. Soc. 18 (1967), 1138-1140.
8.
, '~ultiplicative Ideal Theory", Queen's Papers on Pure and Applied Mathematics, No. 12, Kingston, Ontario, 1968.
9.
Gilmer, R. and Heinzer, W., Rings of formal power series over a Krull domain, Math. Zeit. 106 (1968), 379-387.
i0.
Kaplansky, I., "Commutative Rings", Allyn and Bacon, Boston, 1970.
Ii.
Seidenberg, A., A note on the dimension theory of rings, Pacific J. Math. 3 (1953), 505-512.
12.
,
A note on the dimension theory of rings II, Pacific J. Math.
4 (1954), 603-614.
13.
Sheldon, P., How changing D[[X]] changes its quotient field, Trans. Amer. Math. Soc. 159 (1971), 223-244.
25
KRULL DIMENSION OF POLYNOMIAL RINGS
J.W° Brewer t P.R. Montgomery~ University of Kansas,
E.A. Rutter
Lawrence,
Kansas
and W.J. Heinzer Purdue University,
Lafayette,
Indiana
ABSTRACT. Let Q be a prime ideal of R[X], ..., X ] and let P = Q ~ R. This paper investigates~the relationship between the ranks of Q and P[X], ..., Xn]. The results are used to recover some weIl-known results concerning the Krull dimension of R[XI, ..., X ]. The paper also contains a number of examples r~lated to questions w h i c h arose in connection with the above investigation.
Let R be a commutative unitary ring with XI, minates over R. Prufer domain,
..., X n indeter-
It is well-known that if R is a Noetherian ring or a then dim (R[X 1 . . . . .
Xn]) = n + dim (R), where dim (S)
denotes the Krull dimension of the ring S.
It is customary [3] and [9]
to treat these two classes of rings in a different Kaplansky [7], for the case of a single variable,
fashion.
However,
treats the two cases
in a unified manner by introducing the notion of a strong S-ring. strong S-rings respected polynomial
extensions,
If
then this approach
could be extended to several variables by means of an induction argument.
But strong S-rings do not respect polynomial
extension as we
shall see in Section 2 and in fact, we shall show by examples that it is not possible to give an n-variable definition of strong S-ring in 26
such a way that the approach can be extended. ment is to be given for several variables, found.
So, if a unified treat-
another approach must be
The properties of strong S-rings which are crucial for proving
that dim (R[XI]) = 1 + dim (R) are:
(i) rank (P) = rank (PtXI]) for
each prime ideal P of R and (ii) if Q is a prime ideal of ~ X I] with Q D (Q n R)[XI] , then rank (Q) = 1 + rank (Q n R)[XI]. extend these ideas to n variables, rank (P[X 1 .....
In trying to
one sees readily that rank (P) =
Xn] ) in the classical cases mentioned above and there-
fore the problem is to relate the rank of a prime Q of R[X I, ..., X n] with that of (Q n R)[X 1 .....
Xn].
The principal positive result of
this paper is Theorem i, which states that for an arbitrary commutative ring R and an arbitrary prime ideal Q of R[X 1 . . . . .
Xn] , rank (Q) =
rank ((Q n R)[X I . . . . .
X n]) + rank (Q/(Q N R)[X I .....
rank ((Q n R)[X 1 .....
Xn] ) + n.
X n])
An appealing feature of Theorem i is
that its proof is not only brief but also elementary and therefore it seems to make available
from scratch, more readily than ever before,
information about the ranks of prime ideals of R[X 1 ..... Moreover,
Xn].
not only can Theorem 1 be used to prove the classical dimen-
sion theorems, but it also yields immediately that dim (K[XI,
.... Xn]) = n, when K is a field.
Some other applications
include a greatly simplified proof of the "Special Chain Theorem" of Jaffard and the fact that if all maximal ideals of R[XI,
..., X n] have
the same rank, then R is a Hilbert ring. Perhaps the most interesting part of the paper is Section 2 which is given over to the construction of counterexamples which arise in Section i.
In particular,
to questions
it is shown that strong
S-rings do not respect polynomial extension and that if all maximal
27
ideals of R[X I] have the same rank, it need not be true that all maximal ideals of R have the same rank. Our notation is essentially that of [7].
However, we shall
write "r(P)" in place of "rank (P)" for the rank of a prime ideal P of the ring R.
I.
The Main Theorem and Some Applications
We proceed to the proof of our main theorem pausing only to prove a preliminary result.
LEMM
i.
Let Q be a prime ideal of R[X I] and let P = Q A R.
If Q D P[Xl] , then r(Q) = r(P[Xl] ) + i, and for each integer n > i, r(Q[X2,
.... Xn]) = r(P[X I . . . . . Proof~
Xn]) + i.
Both assertions are obvious if r(Q) = =
In the finite
case to prove that r(Q) = r(P[ X I] ) +I, we induct on r(P). then r(P[ X I] ) = 0.
If r(P) = O,
Since three distinct primes of R[ X I] cannot have
the same contraction to R, P[ X I] is the unique prime ideal properly contained in Q and r(Q) = i = i + r(P[ Xl] ).
Assume the result to be
true for all k < m, where m > 0 and r(P) = m.
To prove that r(Q) =
r(P[X I]) + i it suffices to prove that r(Ql) ~< r(P[X I] ) for each prime ideal QI c Q.
Thus, we let PI = QI n R.
If PI -- P' then
P[XI] C QI c Q, QI = e[xl] and r(Ql) ~< r(P[X I]).
If el c p, the
induction hypothesis implies that r(Ql) ~< r(P IX I] ) + i ~< r(P[ X I] ). i As for the second assertion considering R[X I, ..., X n] as R[X I][X 2 . . . . , X n] we have that NIX 2 . . . . . (R N Q)[X2, as R[X2,
..., X n]) = P[X 2 . . . . .
x n] .
Xn] N Q[X2,
..., Xn] =
Then regarding R[XI,
..., X n]
..., X n][X I] , it follows from the first part of the lemma that
28
4
r(Q[X 2, .... Xn]) = r(P[Xl,
THEOREM i.
..., Xn] ) + i.
If Q is a prime ideal of R[XI,
Q n R = P, then r(Q) = r(e[Xl, r(P[Xl,
.... X n] with
.... x n]) + r(Q/P[Xl,
..., x n]) ~<
.... x n] ) + n. Proof.
We use induction on n, the case n = i being immediate
from Lemma i.
Therefore, we assume the result for all k < n and set
QI = Q n R[XI] .
If QI = P[XI]' the result is immediate from the
induction hypothesis upon regarding R[XI, R[X I][X 2, .... X n]. r(Ql[X2,
.... Xn] ).
assumption implies that r(Q) = r(Ql[X2,
X n]) ~< (n-l) + r(Ql[X 2 . . . . .
X n]) ~< r(P[X I . . . . .
Since i + r(Q/QI[X2, r(P[X I . . . . .
as
Moreover,
x n]).
Combining these
Xn] ) +
x n]) + n.
.... x n]) ~< r(Q/p[x I . . . . .
x n]) + r(Q/P[X I . . . . .
the induction
.... Xn] ) +
results we have that r(Q) = i + r(P[X I . . . . . r(Q/QI[X 2 . . . . .
Xn ]
If P[X I] c QI' then Lemma i implies that
.... Xn] ) = i + r(P[Xl,
r(Q/QI[X 2 . . . . .
• -.9
x n]).
x n] ), r(Q) ~<
But the reverse of this last
inequality is clear, so r(Q) = r(P[X I . . . . , x ]) + r(Q/P[Xl,
..., x n] ).
n
We remark that Theorem i is of no greater depth than the fact that there cannot exist in R[X I] a chain of three distinct prime ideals having the same contraction to R.
This fact in turn is really no
deeper than the fact that KtX I] is a PID if K is a field. obtain an elementary proof of the next result.
Thus, we
The customary approach
is to make use of the fact that fields are Noetherian rings and then appeal to the Hilbert Basis Theorem and to Krull's Principal Ideal Theorem.
29
COROLLARY i.
Proof. dim (K[X 1 . . . . .
If K is a field,
then dim (K[XI,
.... Xn] ) = n.
The chain (0) c (XI) c ...c(xI, X2 . . . . , Xn) shows that Xn]) > n and that dim (K[X 1 . . . . , Xn]) < n is clear
from Theorem i.
In order to illustrate how Theorem i can be used to prove the other classical dimension theory results, we generalize a little to bring the argument
into focus.
COROLLARY 2. localizations.
Let S be a class of rings w h i c h is closed under
Suppose further than whenever
member of S, r(M) = r(M[X 1 . . . . . dim (SIX 1 . . . . .
Xn] ).
X ]) = n + dim (S). n
or semi-hereditary,
dim (R[XI,
the ring R is semi-hereditary
(R,M) is a quasi-local
Then if S E S,
In particular,
if R is Noetherian
.. ., X n ]) = n + dim (R) if finitely generated
(Recall that
ideals of R are
projective.)
Proof. R[X I, .... Xn]. and since
Let R E S and let Q be a maximal
ideal of
If P = Q N R, then since r(Q) = r(Q(R[XI,
S is closed under localizations,
.... Xn])R\p)
in order to see that
r(Q) ~ n + dim (R), it suffices to treat the quasi-local case (R,P) w i t h Q N R = P.
By Theorem i, r(Q) ~< n + r(P[X I . . . . .
XnJ) =
n + r(P) = n + dim (R). If R is Noetherian and if P is a prime ideal of R, then r(P) = r(P[X 1 . . . . , X n]) by the Principal converse
Ideal Theorem [7, p. ii0]
and its
[7, p. 112].
If R is semi-hereditary and if P is a prime ideal of R, then Rp is a valuation domain [ 2, p. 113].
Thus, we need only see that if V is
30
a valuation domain having maximal r(P[X 1 . . . . . V[XI,
Xn] ).
ideal P, then r(P) =
We prove that if Q is a nonzero prime
..., X n] with Q c P[Xl . . . . .
It clearly suffices
Xn] , then Q = (Q n v)[x I . . . . , Xn].
to prove that Q ! (Q n V)[X 1 . . . . .
f ~ 0, then f = ag where a E V and some coefficient Then g ~ PIXI,
..., Xnl D Q and thus a c Q n V.
f = ag ~ (Q n V)[X 1 . . . . . of V[XI,
Xn] .
..., X n] contained
that r(P[X I . . . . .
ideal of
Xn].
If f E Q,
of g is a unit.
We conclude
that
It now follows that the prime
in P[XI,
..., Xn] are extended
ideals
from V and
X n]) = r(P).
Kaplansky called an integral domain an S-domain [7, p. 26] for
each
rank
ideal of R[XI]. prime
one
prime
ideal
if
P of R, P[X I] is a rank one prime
A ring R is then called a strong S-ring if for each
ideal N of R, R/N is an S-domain.
easily seen to be equivalent
This latter condition
to the requirement
of R extend to adjacent primes of R[XI].
is
that adjacent primes
The class of strong S-rings
is then clearly closed under localizations
and homomorphic
images.
Since in computing
the ranks of extended primes of R[XI] , there is no
loss of generality
in first localizing,
strong S-ring is equivalent for each one-dimensional image of a localization
the condition that R be a
to the requirement
quasi-local of R.
that dim (D[ X I] ) = 2
domain D which is the homomorphic
This in turn is the same as requiring
that the integral closure of each one-dimensional which is the homomorphie domain [8, p. 511]. S-ring,
image of a localization
Moreover,
quasi-local
domain D
of R be a Prefer
Kaplansky proved that if R is a strong
r(P) = r(P[XI] ) for each prime
ideal P of R and it follows
easily from this that dim (R[XI]) = 1 + dim (R). If C is a class of rings closed under localizations
31
and quotients
by prime ideals and having the additional property that for each R dim (R[XI]) = 1 + dim (R), it is clear that dim (D[XI]) = 2 for each one-dimensional
quasi-local domain in C.
Thus, the strong S-rings form
the largest class of rings having all three of these properties.
How-
ever, this is a strictly smaller class than the class of all rings satisfying the "dimension formula" for we show in Example 5 of Section 2 that there exists a domain D which is not a strong S-ring but is such that r(P[XI, integer n.
.... Xn] ) = r(P) for each prime P of D and each positive It then follows from Theorem 1 that dim (D[ XI,
.... X n] ) =
n + dim (D). Using the fact that the integral closure of a one-dimensional quasi-local S-domain is a Prufer domain,
it can be shown that if P is
a rank one prime ideal of a strong S-ring R, then P[XI, rank one prime ideal of R[XI,
..., X n] for each positive integer n.
In view of this one might hope to show that r(P{ X 1 ..... for any prime P of a strong S-ring. Theorem 1 that dim (R[X 1 ..... however,
..., X n] is a
X n] ) = r(P)
It would then follow from
X n] ) = n + dim (R).
that this is not the case.
We shall show,
In fact, we exhibit in Example 3
of Section 2 a strong S-ring R such that dim (R[XI, X 2] ) > 2 + dim (R). As Robert Gilmer pointed out to us, the first assertion of Theorem 1 can be deduced from the "Special Chain Theorem" of Jaffard 15, p. 35].
In fact, the two results are equivalent.
A chain C = {Q0 c QI c ... c Qm } of prime ideals of R|XI,
..., X n] is called a special chain if for each Qi' the ideal
(Qi N R)[Xl,
..., X n] belongs to C.
With this notation we prove
32
COROLLARY of finite rank,
3.
(Jaffard)
If Q is a prime
ideal of R[ X 1 , .. .,
then r(Q) can be realized as the length of a special
chain of primes of R[XI,
..., X n] with terminal
lar, if R is finite dimensional,
element Q.
then dim (R[XI,
The second assertion
is immediate
In particu-
.... Xn]) can be
realized as the length of a special chain of primes of R[XI,
Proof.
Xn ]
.... Xn].
from the first and for
the proof of the first, use Theorem i and induction on r(Q)0 If Q is a prime ideal of R[X I] with r(Q) < =, then by Corollary there exists a special chain in R[X I] terminating length r(Q).
In fact,
3
in Q and having
it is not hard to show that there exists a
special chain ¢ = {Qo c QI c ... c Q} of length r(Q) and such that the corresponding
chain (Q0 N R) ! QI n R ! ... ! Q N R of contracted
is saturated--that
is, between two distinct members
prime ideal can be inserted. as we shall see in Example variable,
of the chain no
This fact is not true for two variables
3 of Section 2.
Furthermore,
even for one
we shall see in Example 4 that it is not always possible
find a special chain of length r(Q) such that the corresponding of contracted
to
chain
ideals has length r(Q n R).
Our next result of [7].
ideals
is in the spirit of Exercises
3 and 4, p. 114
See also p. 126 of 16].
COROLLARY the same rank,
4.
(a) If all maximal
then R is a Hilbert ring.
(b) Suppose that R is a Hilbert of R have the same rank. r(M]X 1 . . . . .
ideals of R[X 1 . . . . , X n] have
If n is a positive
Xn]) for each maximal
ideals of R[X 1 , ... , X n ]
ring and that all maximal
ideals
integer such that r(M) =
ideal M of R, then all maximal
have the same rank.
33
In particular,
if R is
either Noetherian or semi-hereditary,
then all maximal
ideals of
R[X I, ..., X n] have the same rank.
Proof.
(a):
Since R is a Hilbert ring if and only if
R[X I, ..., X m] is a Hilbert ring for each positive suffices to prove (a) for n = I. R[X I] and set M = M' n R.
Thus,
integer m, it
let M' be a maximal
If M is not maximal,
ideal of
say M C p, then
r(M') = 1 + r(M[XI] ) < 1 + r(P[XI] ) and P[X I] is not a maximal
ideal.
It follows that M is maximal and that R is a Hilbert ring. (b):
We first make a couple of easy observations.
integral domain,
then there is a one-one correspondence between prime
ideals P' of D[XI,
.... X m ] with P' n D = (0) and all prime ideals
of K[X I, .... Xm], K the quotient maximal
If D is an
ideals of KIXI,
field of D.
Moreover,
since all
..., X m] have rank m, (a well-known fact
easily proved by induction using Corollary 1 and elementary properties of Hilbert rings) p' n D = (0)
if P' is a maximal
then r(P') = m.
ideal of D[X 1 . . . . .
To prove (b)
'
maximal
ideals of RIX 1 . . . .
R[X 1 . . . . .
Xn]/(Mi[X 1 . . . . .
'
and M' 1
X n] and set M i = M'
i
~ R.
be 2
By forming
Xn] ) and applying the above observation,
we see that r(M'i/(Mi[X 1 . . . . . r(MI[X 1 . . . . .
let M' '
X m] such that
Xn])) = n.
Thus, r(M'l) =
Xn]) + n = r(Ml) + n = r(M2) + n = r(M21X 1 . . . . .
Xn]) +
n = r(M'2).
The first two examples of Section 2 show that neither part of Corollary 4 can be strengthened.
The first of these examples
is an
integrally closed Hilbert domain D of dimension one such that not all maximal
ideals of D[X I] have the same rank.
closed Hilbert domain D such that all maximal
34
The second is an integrally ideals of D[X I] have the
i0
same rank but such that this is not true of D.
2.
Examples
We present here the examples promised
in the preceding section.
The examples are rather technical but very illuminating and the reader wishing to gain insight into the relationship between ranks of primes of R and ranks of primes of R[XI,
..., Xn] would profit from studying
them. Despite the fact that our first construction examples of least importance to this paper, generality and for its possible usefulness the construction
is used to give the
it does possess a certain in other contexts,
we give
in that generality.
Let R be a Prufer domain w i t h quotient field K. to find a domain D whose local behavior is identical
We would like
to that of R
except at a single prime ideal P of D and at P we w i s h certain pathology--namely,
we want Dp to be a one-dimensional,
quasi-local,
integrally closed domain which is not a valuation domain. able to do as we shall see from Theorem 2.
This we are
We require some
preliminaries. If S is a commutative
ring and if Y = {Y~}
is a collection of
indeterminates over S, recall that S(Y) denotes the ring where T is the set of all f c S[Y] generated by the coefficients for the pertinent
PROPOSITION of indeterminates i)
such that c(f),
of f, is equal to S.
(S[Y]) T,
the ideal of S See [3, p. 379]
facts about S(Y).
i.
Let R be a Pr{ifer domain,
X = {X } a collection
over R and let P be a prime ideal of R.
If Q' is a prime ideal of R[X] with Q' ! P[X], 35
then
ii
Q' = (Q' n R ) t X ] . 2)
Q ~ QR(X) i s a o n e - o n e
between the of R(X).
set
of all
lattice
prime ideals
preserving
o f R and t h e s e t
correspondence of all
prime ideals
Moreover, RQ(X) = (R(X))QR(X) = (R[X])Q[X ] .
3)
If R is a Hilbert ring, then R(X) is a Hilbert ring.
Proof.
i):
There exists a one-one correspondence between all
prime ideals of R[X] contained in P[X] and all prime ideals of (RIX])p[x]
D (R[X])R\P = Rp[X] .
Thus,
(R[X])p[x]
is a ring of quotients
of Rp[X] and, in fact, (R[X])p[x] = (RpIX])pRp[X].
Thus, we may start
over with R a valuation domain and P its maximal ideal. and if f E Q', then e(f) = (a I . . . . . a'. = 1 for some i. 1 that a E Q' n R. 2):
an) = a(a'l,
If Q' _c p[ X]
..., a'n) where
Thus, f = ag where g ~ P[X] o Q' and it follows
Hence, f E (Q' n R)[X] and Q' c (Q' N R)[X].
That Q ~ QR(X) is a one-one correspondence between the set
of all prime ideals of R and the set of all prime ideals of R(X) now follows from i) and [3, p. 380].
As for the second assertion of 2),
if S = {f c R[X] I c(f) = R}, then Q[X] N S = ~ implies that (R[ X] )Q[ X] = (R(X))QR(X). T = {g • RQ[X]
On the other hand, RQ(X) = (RQ[XI)T , where
I c(g) = RQ} = RQ[X] \ QRQ[X].
Therefore, RQ(X) =
(R[ X] )Q[ x] " 3):
We n e e d o n l y p r o v e
QR(X) of R(X), QR(X) =
that
for each non-maximal prime ideal
HOHR(X), where {MR(X)} is the set of all prime
ideals of R(X) properly containing QR(X). if H o Q if and only if H[X] o Q[X]. HAHIX] = Q[X] = QR(X) D RIX].
HR(X) D QR(X) if and only
Thus,
Therefore
(HNHR(X)) D R[X] =
since '
DHR(X) and QR(X) are H
extended ideals of R(X) having the same contraction to R[X], they are equal. 36
12
We remark that in the above proposition,
the valuation domains
RQ and RQ(X) have the same value group and consequently,
they have
identical prime ideal structures.
THEOREM 2.
Let R be a Prufer domain with quotient
let YI' Y2 be indeterminates
over K.
Consider T = R(YI, Y2 ),
U = (KIYI, Y2])(yl ) = K(Y2) + M, and V = K + M. P = M e D.
field K and
Put D = T n V and
Then
i)
The quotient
field of D is K(YI, Y2).
2)
Dp = V is one-dimensional,
quasi-local,
integrally closed
and not a valuation domain. 3)
P is a maximal
4)
D is integrally closed.
5)
D[I/Y I] = T.
6)
Q ~QT
of D distinct
ideal of D.
is a one-one correspondence
between all prime
from P and all prime ideals of T.
Moreover,
ideals
DQ = TQT =
R ( Q T n R ) ( Y I ' Y2 )7)
If R is a Hilbert
Proof. 2):
then D is a Hilbert ring.
RIYI, YIY2 ] ! D.
If S is an arbitrary multiplicative
DS = TS n VS . Proposition
i):
ring,
In particular,
if S = D\P,
then V S = V.
By
i, 2), since R\(0) ~ K\(0) ~ S, each nonzero prime ideal
of T = R(YI, Y2) meets S and it follows second assertion, 3):
system in D, then
that T S = K(YI, Y2 ).
For the
see [3, p. 560].
Let d • D\P.
k = a/b, a, b • R\(0).
Then d = k + m, k • K\(O), m • M. Since R ~ D, bd = a + bm • dD.
bm = bd - a • M n D = P and hence,
a = bd - bm • dD + P. 37
Thus,
Write
13
a + Y1 E dD + P and a + Y1 is a unit in V since a @ R\(O). is a unit in T by definition.
But a + Y1
It follows that dD + P = D.
4):
T and V are integrally closed.
5):
D[i/Yl ] = T[i/Yl ] n V[i/Yl ] and V[I/Y I] = K(YI, Y2 ).
6):
By 5), we need only see that P is the lone prime ideal of D
which contains YI"
Suppose not and let Q be a prime ideal of D
containing YI and distinct from P. and DQ = TD\ Q N VD\Q TD\ Q = DQ.
= TD\Q"
But Y1 c QDQ.
Since P is maximal, P n (D\Q) #
Y1 is a unit in T and hence in
That TQT = R(Q T nR)(YI ' Y2 ) follows from
Proposition i. 7):
Since P is maximal, if Q is a non-maximal prime ideal of D,
then QT is a non-maximal prime ideal of T.
Thus, by Proposition i, 3),
QT is the intersection of the prime ideals of T properly containing it. It now follows easily that D is a Hilbert ring.
Using Theorem 2, we can now show that neither part of Corollary 4 can be strengthened.
Throughout, Z denotes an indeterminate over the
integral domain D.
EXAMPLE i°
This is an example of an integrally closed Hilbert
domain D of dimension one having the property that dim (D[Z]) = 3. Moreover, some maximal ideals of D[Z] have rank two and some have rank three. Let R be a Dedekind domain and let K denote the quotient field of R.
The notation being as in Theorem 2, since dim (D[Z]) < 3, it
suffices to prove the "moreover" assertion.
If Q is a prime ideal of
D distinct from P, DQ is a DVR and rank (Q[Z]) = i. ideals of D[Z] contracting to Q have rank two.
38
Thus, maximal
Since D is integrally
14
closed and Dp is not a valuation domain, ~ank (P[Z]) = 2 [3, p. 218]. A maximal ideal of D[Z] contracting to P must have rank three.
EXAMPLE 2.
This is an example of an integrally closed Hilbert
domain D with the following two properties: has a maximal ideal of rank one. ideal of D[Z] has rank three. R and then apply Theorem 2.
(i) dim (D) = 2, bu___ttD
(2) dim (D[Z]) = 3 and each maximal
Our problem is to choose the appropriate Moreover, since the D constructed will be
a Hilbert ring, a maximal ideal of D[Z] will contract to D in a maximal ideal of D, and because the rank of a maximal ideal of D[Z]
is
one more than the rank of its contraction, we will be finished as soon as each maximal ideal of R has rank two. the fact that rank (P[Z]) = 2.
Of course, we are using here
To choose such an R, let R be a Bezout
domain having divisibility group H as given in |4, p. 1371] and such that H has infinitely many prime filters.
(The terminology of [4] is
"minimal primes", but "prime filters" seems to be more standard.) This R will suffice.
EXAMPLE 3.
We are going to construct a two-dimensional
quasi-local domain (R, M) having the following properties: (i)
R has a unique rank one prime ideal P,
(2)
P has the property that both Rp and R/P are DVR's, and
(3)
R has valuative dimension three.
(Recall that the integral
domain D has valuative dimension m < ~ if there exists a rank m valuation domain between D and its quotient field but no valuation domain of larger rank.) Supposing for the moment that we have done this, let's consider the implications.
39
15
R is a strong S-ring.
To see this, notice
one since Rp is a DVR and that PIX I] c MIX1]
that P[X I] has rank
are adjacent
since R/P
is a DVR. R[X I] is not a strong S-ring. following
theorem of Arnold
of finite valuative m ~ k-l,
[i, p. 323]:
dimension k.
then dim (D[XI,
This is a consequence
of the
Let D be an integral domain
If m is a positive
.... Xm] ) = m + k.
integer such that
In our case, k = 3 and so
dim (R[XI, X 2]) = 5 > i + dim (R[X I]). I f Q is a rank five maximal chain of length five with terminal Thus,
in contrast
ideal of RIX I, X 2] , then no special element Q can contain P[ Xl, X2].
to the single variable
need not be tru____~ethat all ranks of ~rime lengths
case,
ideals are realizable
of special chains where the contracted
This assertion
for two variables
it
as the
chain is saturated.
follows again from the fact that Rp and R/P are DVR's.
Now to the construction. achieved by casting the example
In the hope that some clarity may be in a more general
setting, we isolate
the following portion of the argument.
LEMMA 2.
Let (V, M) and W, N) be quasi-local
a common subfield K and set R = K + (M A N). domain with maximal
ideal M n N.
Moreover,
domains containing
Then R is a quasi-local if W is a rank one
valuation domain and if V and W have the same quotient field, each nonzero nonmaximal
then for
prime ideal Q of R, RQ = VQ, for some prime
ideal Q' of v.
Proof. consisting
That R is a domain and that M A N is an ideal of R
entirely of nonunits
with a c K, m
is obvious.
e (M \ N) then a # 0 and so
40
If r = a + m E R \(M n N),
16
i/r = (i/a) + (-m)/(a(a + m)) C R since (-m)/(a(a + m)) E M ~ N. To prove the second assertion, suppose that Q is a nonzero prime ideal of R with Q c (M N N) and choose t E (M n N) \ Q.
If s E V,
then since W has rank one, there exists a positive integer e such that tCs E N.
Therefore, tCs ~ M n N and s E RQ.
Since t is a unit in RQ
but not in V, V c RQ.
So, to construct the promised R, we begin by building valuation domains V and W so as to enable us to apply Lermna 2. Let K be a field with YI' Y2 and Y3 indeterminates over K.
Let
V I be the valuation domain of the Y3-adic valuation on K(YI, Y2)[Y3]. V I is a DVR with residue field K(YI, Y2).
On K(Y I, Y2 ), build a rank
one valuation domain V* by mapping YI and Y2 to two rationally independent real numbers, say YI ~ i and Y2 ~ ~"
If f denotes the
canonical map from V I onto K(YI, Y2) , then f-l(v*) = V is a rank two valuation domain of the form K + M I. prime of V, V I = VPl.
Moreover, if PI is the rank one
(We refer the reader to [3, p. 190] for a more
general treatment of this "pull-back" construction of valuation domains.)
Now, if W is the valuation domain of the (Y3 + l)-adic
valuation on K(YI, Y2)[Y3], then W is a DVR of the form K(YI, Y2 ) + M 2 and W # VPl.
Set R = K + M I N M2.
By Nagata's Theorem [3, p. 262],
D = V n W is a two-dimensional Prufer domain with quotient field K(YI' Y2' Y3 ) and the maximal ideals of D are M I n D and M 2 n D.
Thus
MI n M2 is a nonzero ideal of D and it follows that the quotient field of M I n M 2 is K(YI, Y2' Y3 )"
From this we see also that the quotient
field of R is K(YI, Y2' Y3 )"
Since R ! K + M 2 and since the valuative
dimension of K + M 2 is three [3, p. 561], the valuative dimension of R
41
17
is three. M1 n M2"
Moreoever, by Lemma 2, R is quasi-local with maximal ideal But P = P1 n R is a nonzero nonmaximal prime of R for P # (0)
and if P ~ M 1 n M2 , then P1 ~ (MI N D)(M 2 n D) which is impossible. So again by Lemma 2, P is the unique rank one prime of R and Rp = V I, a DVR.
It remains only to show that R/P is a DVR.
that R/P = D/(P 1 n D).
It suffices to show
Now R/P ! D/(P 1 n D) and since for d • D there
exists a • K with d - a • MI, to show that D/(P 1 n D) ! R/P, it suffices to show that for d • M 1 N D, there exists b • R such that ~ • D/(P 1 n D) equals b • R/P.
But P1 n D and M 2 n D are comaximal and so such a
b does exist.
As a variant on the preceding construction we have
EX~PLE
4.
This is an e x ~ p l e
of a domain R such that there
exists a prime ideal Q o~ R[X I] h~ving the ~roperty that for each special chain of primes o~ RIX I] t e ~ i n a t i n g
i~ Q and having length
r(Q), the corresponding chain of contracted primes fails to have length r(Q n R). It will suffice to find a quasi-local domain (R, M) having the following properties: i)
The prime ideal s t ~ c t u r e of R is given by the diagram
/\
M
2)
dim (R [X I ] ) = 5.
3)
(0) c Ql[Xll
c Q2[X1] c M[Xll i s a s a t u r a t e d
42
chain.
18
To build such a domain, let K be a field with YI' Y2' Y3 and Y4 indeterminates over K.
Consider (K(YI, Y2' Y3)IY4])(Y4) =
K(YI' Y2' Y3 ) + Ml and (K(YI)[Y2])(y2) = K(YI) + M 2. and set V = K + M V.
Let M V = M 2 + M I
From Lemma 2 and standard facts about the "D + M-
construction", V is a two-dimensional quasi-local domain with maximal ideal M V and with M I as unique rank one prime.
Note that V[X I] has
dimension five with primes properly between (0) and MI[X I] and between MI[XI ] and Mv[XI ]. Now consider (K(YI, Y2' Y3)[Y4])(Y4+I ) = K(YI' Y2' Y3 ) + NI' (K(Y I, Y2)[Y3])(Y3+I) = K(Y I, Y2 ) + N 2 and define a rank one valuation domain on K(Y I, Y2 ) of the form K + N 3 by giving Y1 and Y2 values 1 and ~.
Let W = K + N where N = N 3 + N 2 + N I.
Then W is a rank
three valuation domain of K(Y I, Y2' Y3' Y4 )" Let M = M V n N and R = K + M.
Note that YIY2 , Y2' Y3Y4 ' Y4 E M
so R has quotient field K(YI, Y2' Y3' Y4 )" a E M, R[i/a] = VIi/a] n W[i/a].
Moreover, for each nonzero
Thus, if a = Y2(Y4 + i), then a c M
and VIi/a] = VMI = K(YI, Y2) + M 1 and W[i/a] = K(¥1, Y2' YB' Y4 )" So, letting P = M 1 A R, we have Rp = VMI = K(YI, Y2 ) + M I. Taking a = Y4Y2 E M, VIi/a] = K(YI, Y2' Y3' Y4 ) while W[i/a] = W(N2+NI ) = K(YI, Y2) + (N 2 + NI).
Taking a = Y4(Y3 + i) E M, VIl/a] =
K(YI' Y2' Y3' Y4 ) while W[i/a] = WNI = K(YI, Y2' Y3 ) + NI"
It follows
that if Q1 = N1 n R and Q2 = (N2 + NI) n R, then RQI = WNI and RQ2 = W(N2+NI )This R has the desired properties.
43
19
EXAMPLE 5.
This is an integral domain R which is not a strong
S-ring but which does have the property that for each prime P of R and each positive integer n, r(P) = r(P[XI,
.... Xn] ).
We are indebted to
J.T. Arnold for the idea. Let K be a field with ZI, Z 2 indeterminates over K.
Let
V = K(ZI) + M be the valuation domain of the Z2-adic valuation on K(Z I, Z 2) and set D = K + M.
D is a one-dimensional
quasi-local domain of valuative dimension two. Theorem [i, p. 323], dim (D[XI, integer m.
If P = ~ X I ] ,
.... Xm] ) = m + 2 for each positive
Xn]), we merely handle the few cases which
Thus, r(M[XI, YI' "''' Yn ]) > 2.
dim (D[XI, YI' "''' Yn]/M[XI' YI . . . . .
r(M[XI' YI' "''' Yn ]) = 2.
But
Yn ]) = dim ((D/M)[X I .... ,Yn ]) =
n + i and dim (D[XI, YI' "''' Yn ]) = n + 3.
r(e) = i = r(P[YI,
To show
then r(P) = 2 since R is integrally closed but
not a valuation domain.
r(P) = 3.
Thus, by Arnold's
Put R = D[X I] and let P be a prime ideal of R.
that r(P) = r(P[X I . . . . . arise.
integrally closed
Hence, we must have that
If P n D = (0), then Rp is a DVR and
.... Yn]).
By Lemma i, r(P[YI,
If e n D # (0) and if P # M[XI] , then .... Yn ]) = r(M[XI' YI . . . . .
Yn ]) + i = 3.
REFERENCES
[i]
Arnold, J.T., On the dimension theory of overrings of an integral domain, Trans. Amer. Math. Soc. 138 (1969), 313-326.
[2]
Endo, S., On semi-hereditary rings, J. Math. Soc. Jap. 13 (1961), 109-119.
[3]
Gilmer, R.W., Multiplicative Ideal Theory, Queen's Papers on Pure and Applied Mathematics,
Kingston, Ontario,
44
1968.
20
[4]
Heinzer, W. J., J-Noetherian integral domains with i in the stable range, Proc. Amer. Math. Soc. 19 (1968), 1369-1372.
[S]
Jaffard, P., Theorie de la Dimension dans les Anneaux de Polynomes, Gauthier-Villars,
[6]
Kaplansky,
Paris, 1960.
I., Commutative Rings, Queen Mary College Mathematics
Notes, London, 1966. [7] [8]
, Commutative Rings, Allyn and Bacon, Boston, 1971. Seidenberg, A., A note on the dimension theory of rings, Pac. J. Math. 3 (1953), 505-512.
[9]
, On the dimension theory of rings, Pac. J. Math. 4 (1954), 603-614.
45
A NOTE ON THE FAITHFULNESS OF THE FUNCTOR S ~R(-)
J.W. Brewer and E.A. Rutter University of Kansas, Lawrence,
Kansas
ABSTRACT. If R and S are rings with S a unitary extension of R, the faithfulness of the funetor S ~ ( - ) is studied.
Let R and S be unitary rings with S an extension of R. the functor S ~R (-) is additive,
Since
it is faithful if and only if it
reflects zero maps [ 2, pp. 52 and 56].
It is natural to ask whether
or not this is equivalent to the apparently weaker condition that S ~R (-) reflects zero objects.
Of course,
flat extension of R [2, Prop. 7.2, p. 57]. is to show that in general, however,
this is the case if S is a The purpose of this note
this is not so.
In fact, we prove
that S ~R (-) is faithful if and only if S is a pure extension of R in the sense of Cohn [I] and we give an example of integral domains R and S having the following two properties:
Let 0
(I)
S is not a pure extension of R and
(2)
S ~R(- ) reflects zero objects.
>E
> F be an exact sequence of right R-modules.
The
sequence is said to be pure if and only if the sequence 0
> E ~R M
> F ~R M is exact for each R-module M.
46
When R is a subring
of S and the sequence O--->R--->S is pure, we say S is a pure extension of R.
Theorem.
Let R C S be rings, R a subring of S.
the functor S RR(-) be faithful,
In order that
it is necessary and sufficient
that S
be a pure extension of R.
Proof.
Since the functor S RR(-)
and only if it reflects
is additive,
the zero mapping;
each nonzero R-homomorphism
Assuming
the mapping
that S is a pure exten-
sion of R, let f : M'--->M be a nonzero R-homomorphism following commutative
Since the vertical
if
that is, if and only if for
f : M'-->M of R-modules,
1 S R f : S RRM'--->S RR M is nonzero.
it is faithful
and consider the
diagram.
S i R M~
ls~f>s iRM
R ~R M'
iR~f> R ~R M
arrows are monomorphisms and since 1 R ~ f is nonzero,
i S N f is nonzero and it follows that S RR(-)
is faithful.
Suppose now that S is not a pure extension of R and let M be an R-module
such that 0-->R ~RM-->S ~R M is not exact.
with M, we see that this is equivalent
and denote by f the injection of M' into M.
f
commutes
>
M
M
and by the definition of f, eMo f = O.
47
Let M' = ker eM
The diagram
is f>s
M'
R ~R M
to the condition that the
canonical map eM : M-->S ~R M is not a monomorphism.
s RM'
Identifying
But eM,(M' ) generates
S ~R M' as an S-module and therefore i S M f = 0.
We conclude that
S MR(-) is not faithful.
Before providing the example promised in the introduction, we make some preliminary observations. R a subring of S.
Let R ! S be unitary rings with
Theorem 2.4 of [i]
shows that S is a pure extension
of R if and only if each system of linear equations with coefficients in R which is solvable in S is solvable
in R.
Thus,
if S is a pure
extension of R, then each linear equation which is solvable in S is solvable in R.
We observe that this latter condition is equivalent to
the condition that AS N R = A for each right ideal A of R. the desired example, of K.
To construct
let K and F be fields with F a proper subfield
If X is an indeterminate over K denote by S the ring of formal
power series in X over K.
Then S = K + M, where M is the maximal
of S, and if We set R = F + M, then M is a common maximal and S.
AS N R ~ A.
that aS A R
Let s E S ,
PaR.
By the
it suffices to find an ideal A of R such that s ~ R and let a e M ,
aS n R = aR, then as = at for some t E R
a ~ 0.
Then a s e R and if
which is absurd.
It follows
It remains to show that S ~R (-) reflects
Suppose N is an R-module such that S ~R N = 0. S-module
ideal of R
We verify first that S is not a pure extension of R.
above observations,
ideal
Then E ~R N = 0, for any
E since E ~R N = (E ~S S) ~R N = E ~S (S ~R N) = 0.
lar, S/M ~R N = 0 and M ~R N = 0. MN via the mapping m ~ n-->mn.
zero objects.
In particu-
Thus, MN = 0 since M ~R N maps onto
Since M c R, there is a natural
R-monomorphism 0-->R/M-->S/M.
As M is a common ideal of R and S, it
annihilates both R/M and S/M.
Hence,
the usual way as R/M-modules.
If this is done, then the R-homomorphisms
and the R/M-homomorphisms
these modules may be regarded in
between them are identical. 48
Thus, the above
monomorphism splits since R/M is a field.
Therefore,
the sequence
0-->R/M ~RN--->S/M HRN is exact and hence R/M ~R N = 0.
But R/M ~R N =
N/MN and since MN = 0, we conclude that N = 0.
REFERENCES
i.
P. Cohn, On the free product of associative rinss, I.
Math. Zeit. 71 (1959), 380-398. 2.
B. Mitchell, Theory of Categories, New York:
Press, 1965.
49
Academic
On A PROBLEM IN LINEAR ALGEBRA David A. Buchsbaum and David Eisenbud
In its vaguest and most tantalizing
form~ the problem referred to in the
title of this paper is to say something about the solution of systems of linear equations over a polynomial ring
R = K[XI,...,Xn]
, where
K
is a field.
difficulty is~ of course, that a system of linear equations over
R
The
may not
possess a set of linearly independent solutions from which all solutions may be obtained by forming linear combinations with coefficients
in
R .
Hilbert's Syzygy Theorem suggests a promising approach to this problem: ~i
Let
be a system of linear equations~ which we will regard as a map of free
R-modules~
say
~i: F l - - >
Fo
If we form a free resolution of the cokernel of
~i ' that is an exact sequence of free R-module of the form
°n- ~
~n (i)
Fn
>Fn_ I
~2
~Fn_ 2
>: ..
>F 2
>F I
~>Fo
'
then the Syzygy Theorem tells us that the system of linear equations represented by
~n-1
cisely,
does have a full set of linearly
independent solutions.
More pre-
it tells us this in the graded or local cases; in general~ Ker ~n-i
merely stably free. may replace
Thus, replacing
Fn
by
Ker ~n-i
is
in the sequence (i), we
(i) by a finite free (or stably free) resolution of the form
(2)
o
>F n
~n >Fn-i
>...
>FI
>F o
In order to use the approach to linear equations that Hilbert's Theorem suggests~
it seems to be necessary to study the relations that must hold between
the maps
~i
in a finite free resolution of the form of
establishing a relationship between the maps [6].
Letting
Theorem i.
(3)
nAX
(2).
The first result,
~i ' was also obtained by Hilbert
denote the n th exterior power of a module X, we state it as;
Suppose that
R
0
is a noetherian ring.
>R n
~2
>R n+l
~i
If
>R
is an exact sequence of free R-modules, then there exists a non-zerodivisor
50
a c R
n
such that~ after making the canonical identification ~Rn+l*
Rn+l
, we have
ARn* ~ R
and
n ~i = aA ~2" "
(Hilbert actually proved this only when
R
is the graded polynomial ring in
two variables over a field; he did it to illustrate the application of the Hilbert function.
The first proof that works in the generality in which we have stated
the theorem is due to Butch [5].
The most elementary proof is contained in [7].
We shortly give a new proof of this result.) Theorem i can be applied to a special case of Grothendieck's Theorem in a way that we will now describe. polynomial rings to regular
To do so, we w i l l shift our point of view from
local rings; just as in the polynomial case, every
modnle over a regular local ring has a finite free resolution. Grothendieck's ring, and let
x ¢ S
erated R-module
lifting problem is the following: be such %hat
R = S/(x)
let
is regular.
S
be a regular local
Given a finitely gen-
M , does there exist a finitely generated S-module
M~
such
that
(2)
x
If a module
is a non-zerodivisor M # satisfying
on
M# ?
(i) and (2) exists, it is called a lifting of
M
That the lifting problem is closely related to the structure of free resolutions is shown by the next lemma. Lemma 2.
Let
S
be any ring, and let
x e S
be a non-zerodivisor.
Let
R = S/(x) , and let
~2
F: F 2
~l > FI
be an exact sequence of free R-modules.
> F°
Suppose that
# # is a sequence of free S-module which reduces modules then Coker ( 4 ) Proof:
is a lifting of
(x) to
If
~01 ~02 =
0 ,
Coker(~l).
The hypothesis allows us to construct a short exact sequence of complexes
o
>~F #
x
>~#
>~
> 0
The associated exact sequence in homology contains
(4)
Hl(F)
> H o ( ~ #)
x
> Ho ( ~ # )
51
>Ho(IF)
> 0
On the other hand, HI(IF) = 0
Ho( IF#) = Coker(~#l) ~ Ho(IF) = Coker(~l) and, by hypothesis
Thus
(4) becomes
0--.>
Coker(~l#)
x
> Coker(~)
> Coker({01)
> 0 ,
and the lemma is proved. Lemma 2 implies the liftability of various classes of modules. in the situation of the lifting problem, any module one can be lifted:
If
0
> FI
>F
For example,
M of homological dimension
is a free presentation of
M ~ then
O
there exists a homomorphism ~#i
of free
(for example, to construct
one can choose bases of
~I
each element of the matrix of
~i).
S-modules which reduces to FI
By lemma 2, Coker (~#i)
and
~i modulo
F2
x
and lift
is a lifting of
M
Theorem i can be applied to the lifting problem through the use of lemma 2. Together they imply the liftability of cyclic modules of homological dimension 2.
For, if (3) is the free resolution of such a module, we may choose a map
# ~2
of free S-modules and an element
a # e S which reduce, modulo (x), to
# and a respectively.
Letting
# #
q01
~02
# =
, it is easy to see that
a
#
~i q02 = 0
By theorem i,
~i
reduces to
and shows that Coker(~l) is a lifting of
~i
modulo
Coker(~l)
(x) , so lemma 2 applies
.
We now return to the problem with which we began: what can one say about the relations between the maps
q0i in (2) ? If the F. were vector spaces, the l exactness of (2) would be equivalent to a condition on the ranks of the maps q0i . In the case of, say, a local ring
R , what additional condition or condi-
tions must be imposed in order to ensure exactness of a sequence of free R-modules? Let
In order to answer this, we introduce some terminology and notation. R
be a commutative ring and
free R-modules. r
that
r
~
> G
a map of finitely generated
to be the largest integer
r
such
r
A~: AF
ideal of
~: F
We define the rank of
R
>
AG
is not zero.
If
generated by the minors of
The definition of
I(~)
of order
~ .
I(~)
F
and
It can also be shown that the map
k , a map kAG* ® kAF
~
52
the
r .
makes sense if one chooses bases of
writes the matrix associated to induces, for every
r = rank (~)~ we denote by ~
>R (where G* = HomR(G,R))
G
and
~:F-->G and
With this notation, we can state Theorem 3 [2]. Let
R
be a noetherian commutative ring, and ~n
~: 0
~i
-> Fn
>Fn_ I
a complex of free R-modules.
i)
ii)
Then
> ---
~
is an exact sequence if and only if
rank(~k+l) + rank(~ k) = rank(Fk) l(~k)
>F o
>F I
for
k = i, .... n
contains an R-sequence of length
k
for
and
k = l,...,n
We make the convention that condition ii) is automatically satisfied if l(~n) = R. As a result, when
R
is a field, condition i) becomes the only (and the usual)
condition for the exactness of a complex of vector spaces. Theorem 3 provides one very quick proof of Theorem i. • : 0
>R n-I (aij)>R n (Yi) >R
~2 = (aij) and
>R/I
is exact.
~i = (Yi) ' we know by Theorem 3 that
I(~2) contains an R-sequence of length two. the minors of order 0
>0
For suppose
>R (Ai) >B n
n-i ~
But
of the matrix (aij) . >R n-I
Letting
rank(~2) = n-i
I(~ ) = (Ai) where 2 Thus, the sequence
and [Ai}
are
is exact since the composition is clearly zero
and Theorem 3 applies. - 0
Because the sequence
>R
(Yi) >R n
~
>R n-I
is of order two,
we get a map of complexes
0
>R (Yi)
0-->R and thus
Yi = a~.l for
>R n
q0~
>Rn_ 1
(~i) >Rn i = l,...,n
Although Theorem 3 provides us with a nice proof of Theorem i, it is not clear how that gets us any further into the problem of determining relations among the maps of an arbitrary finite free resolution.
In [4], we have exploited
Theorem 3 to obtain the following two results: Theorem 4 [4~ Th. 3-1]. Let
R
be a noetherian ring, let (2) be an exact se-
quence of free R-modules, and let r k
= rank (~k).
53
Then for each
k, 1 < k < n,
there exists an unique homomorphism
i)
an =
ii)
rn rn A ~n: R = AFn
for each
rk > A Fk_l~
ak: R
rk_l A F{_ I
such that
rn > AFn-1
k < n , the diagram rk A ~k
rk A Fk
rk > A Fk_ I
R/ a commutes. Using the maps
a kl :
a k , we may define maps
rki~ A FR_ I
to be
> F~_ I
the composite:
rk-1
-i
rk+l
A Fk_l
r n: ~ Fk_ I @
where
a k @1 rk
= R e A F~_I - - ~ r k +i A F~_ I
rk+l
AF~_le
>F[_ I
AF~_I
n
>F~_I
is the usual action of
AFk_ I
on
AF{_ I
(see Ill). Theorem
5.[4, Th. 6.1].
Let notation and hypothesis be as in Theorem 4. rk-I k ~ 2 , there exist maps bk: F{ > A Fk_ I making the diagram
rk-i A Fk
rk-i A ~k >
Then for
r -I k A Fk_l
ak- I
bk F* k
commute. Theorems minors
4 and 5 give us fairly strong information about the relations
of orders
rk
and
rk-i
of the maps
If we have the exact sequence
54
~k
of the
in a finite free resolution
(2).
F;
then
O
.> Rm-2
O3
> Rm
~2
> R3
O1
> R
,
r I = 2 , r I -i = i , and it is possible to use Theorems 4 and 5 to express
the maps
~i
and
O2
completely
in terms of the minors of
O3
of order
m-2
(see [4]). Consequently just as Theorem i was applied to the lifting problem for cyclic modules of homological dimension 2, Theorems 4 and 5 may be applied to show that cyclic modules
R/I
of homological dimension 3 may be lifted, provided
I
is generated by 3 elements. In the general case of a cyclic module results applied to a resolution of
0
> R p - -O3 ~
R/I
of homological dimension 3, our
R/I:
R m - -O2 ~
Rn
~i
> R
only give us information about the minors of order
> R/I
n-I
and
> 0
n-2
of
~i
One
might therefore hope that further information about the lower order minors of and
O2
might be obtained from information about the lower order minors of
This idea, of course,
is completely demolished when one finds resolutions
~i O3 .
of the
form (5)
0
particularly
> R
if
I
O3
> Rn
On
> R n - -O1>
R
contains an R-sequence of length three.
local ring, this turns out to be the case precisely when ring.
It is
> R/I
If
R/I
R
> O,
is a regular
is a Gorenstein
possible to show, however, that the lifting problem for 5) in the
Gorenstein case reduces to a problem of lifting
0
where
> Rp
J
¢3
> R m - ~2 - ~
~i
> R
>R/J
> 0
is an ideal generated by four elements and contains an R-sequence of
length three.
Since
p = m-4+l = m-3 , the map
than those of order p,
provided m -3 > i
it possible to have an ideal i)
R4
R/I
ii)
h~
iii)
I
i.e.
¢3
will have lower order minors
m > 4 .
I in a regular local ring
is a Gorenstein ring R/I = n
and
is minimally generated by
n + i elements?
55
The question arises: is R
such that
In [3], we show that this situation cannot arise.
In fact, the question arises
as to what restrictions there are on the number of generators a regular local ring
R
when
R/I
is
Gorenstein.
dim R/I = 0 , we have found such ideals
If
of an ideal
dim R = 3
I
in
and
I generated by five, seven, and nine
elements, but none that are generated by an even number of elements. This appears to be R e d up with questions about the possible skew-symmetry of the map
0
> R
~3
~2
> Rn
(or probable)
in the resolution of such an ideal:
~2
> Rn
~i
> R
R / I ~
0
A computer program worked out by R. Zibman for the Brandeis PDP-10 computer has provided numerous examples of such ideals information on the problem.
I
and may yield same helpful
In any event, it seems evident that our problem in
linear algebra has many ramifications,
the nature of which we are just beginning
to discover.
Bibliosraphy
i.
N.
Bourbaki, Elements de Mathematiques, Algebre, Ch. III, Hermann,
2.
D.A. Buchsbaum and D. Eisenbud, What makes a complex exact, J. Alg. to appea~
3.
, Remarks on ideals and resolutions,
1970.
Symposia Math.
to appear. , Some structure theorems for finite free resolutions,
4. To appear. 5.
L. Butch, On ideals of finite homolo~ical dimension in local rings, Proc. Cam. Phil. Soc. 64 (1968) 941-946.
6.
D. Hilbert, Uber die Theorie der al~ebraischen Formen, Math. Ann. 473-534.
7.
I. Kaplansky, Commutative Rin~s, Allyn and Bacon, 1970.
56
(1890),
MAXIMAL
IDEALS
IN P O L Y N O M I A L
RINGS
E d w a r d D. D a v i s and A n t h o n y V. G e r a m i t a Dept. of M a t h . r SUNYA, A l b a n y , N. Y. 12222, U S A Dept. of M a t h . , Q u e e n ' s U n i v . , K i n g s t o n , Ont., C a n a d a
T h i s n o t e is c o n c e r n e d w i t h e s t i m a t e s of the m i n i m a l n u m b e r of g e n e r a t o r s for m a x i m a l ideals in p o l y n o m i a l r i n g s -- e s t i m a t e s e x t e n d i n g the w e l l k n o w n facts of the case w h e r e i n the c o e f f i c i e n t r i n g is a field.
As ever,
all r i n g s
are c o m m u t a t i v e w i t h
the m i n i m a l n u m b e r of g e n e r a t o r s c o u r s e be i n f i n i t e . ) b u t in m a n y
cases
the p o l y n o m i a l Hilbert's
shall obtain
sharp estimates
a generalization result
m e n t of the N u l l s t e l l e n s a t z This m e a n s
that there exists
reader who
cares
ideals
only
in some e a s i l y o b t a i n e d ,
consequence
= n if R is a f i e l d
Noetherian
of
[8].
R and as c o r o l l a r y
We the
H i l b e r t ring,
for
ideals.
as above, and P = MnR.
We n e e d one
rendering
of the K r u l l - G o l d m a n
[4, 6]: R/P is a G - d o m a i n
[5, Thm.
t in R s u c h t h a t P R rings
is m a x i m a l . t can m a n a g e w i t h o u t
We n e e d a l s o the u p p e r s e m i c o n t i n u i t ~ I finitely
ists t in A - Q s u c h t h a t ~(IAt)
i d e a l of
infinitely many maximal
for N o e t h e r i a n
of a r i n g A, w i t h
denote
(This n u m b e r m a y of
A well known
for R a r e g u l a r
polished
I.
L e t v(I)
for M a m a x i m a l
for a r b i t r a r y
a Dedekind domain with
from Kaplansky's
fact.)
of ~(M)
is that ~(M)
F r o m n o w on R, S and M are fact
interested
ring S = R[Xl,...,Xn].
sharpest possible example,
of the i d e a l
We are h e r e
Nullstellensatz
1 ~ 0.
generated
= ~(IAQ).
57
of ~:
treat23]. (The this
For I and Q
and Q p r i m e ,
t h e r e ex-
THEOREM. (I)
For P maximal,
(2)
In any case,
(3)
For P maximal
Proof. R/P;
W i t h R, S, P, M as above, ~(M)
~(M)
semicontinuity
ring
Then:
__< ~ ( P R p ) + n + l . ~(M)
< u(PRp)+n.
S/PS is a p o l y n o m i a l
(I) is t h e n an i m m e d i a t e
the c o e f f i c i e n t
finite.
< ~(P)+n.
and n p o s i t i v e ,
For P maximal
a s s u m e ~(P)
consequence
is a field.
As
for
ring over
of the k n o w n (2), o b s e r v e
case w h e r e i n t h a t the u p p e r
of ~ and the fact t h a t R/P is a G - d o m a i n
existence
of t in R such t h a t P R t is m a x i m a l
(i) then,
~(MSt)
~ ~(P~p)+n.
the f i e l d
guarantee
the
and 9 ( P R t) = ~(PRp).
By
T a k e m in M c o m a x i m a l w i t h
t and let I
be the i d e a l of S g e n e r a t e d by m and ~ ( P R g e n e r a t e MS t. of S.
Then I = M because
As for
(3), n o t e
n o m i a l r i n g S/PS shows s h o w t h a t it s u f f i c e s say
t h a t the N u l l s t e l l e n s a t z that M ~ R [ X I ]
~ ~(PRp)
tf-p, w h e r e Hence
tr+tf*,
where
ated maximal
(reR, pcP). it s u f f i c e s
f* d e n o t e s
REMARKS.
of
apply
in the p r o o f of
(I) to c o n c l u d e
shows
domain
of
image
is p r i n c i p a l ; (2) we h a v e
2. one
M = gS+PS.
1 = tr+p =
gener-
(Observe t h a t this alone.
a n d A/I
And somewhat
semilocal,
~(I)
can a v o i d use of the f a c t a b o u t
(2) by a s s u m i n g ~(M)
t h a t ~(MSM)
Let g =
For Q a f i n i t e l y
~ ~(QAQ)+I.
generated
(i)
of f in A.
"upper semicontinuity"
~ ~(P~)+n; than
and M a n o n p r i n c i p a l
that one cannot
t h a t tA = A:
(2) gives:
~(Q)
3. In g e n e r a l no b e t t e r e s t i m a t e Dedekind
t as in the p r o o f
the c a n o n i c a l
For I finitely
l + m a x { ~ ( I A Q ) IQ m a x i m a l } . ) G-domains
f a c t and
In this case M / P S
to c h e c k
i d e a l of a r i n g A,
generally:
This
to the p o l y -
T h e n s i n c e rg = f - p f - p r ,
i. T a k i n g n = 0 in
f a c t is a c o n s e q u e n c e more
Taking
applied
for any R - a l g e b r a A s u c h t h a t tA = A.
1 = tr+p
for A = S/gS,
is m a x i m a l .
to take n = i.
f g e n e r a t e s M m o d PS.
t h a t ~(PA)
)+n e l e m e n t s of M w h i c h P ISQ = M S Q for e v e r y m a x i m a l i d e a l Q
then
maximal
ideal.
Localize
at P;
apply Remark
(2) is p o s s i b l e :
in g e n e r a l w e a k e n
58
finite:
Take R = S a
4. R e m a r k
the h y p o t h e s i s
i.
on n in
3 (3).
-
Nor
can
one
for
n = 1 wherein
principal
weaken
regular is
Rp
ideal
generated
nonmaximal
is), by
ring,
COROLLARY. tive is
number
generated
Question:
dim(R)
by
Yes =
2
Dedekind
in
domain
proof
Suppose
S with
show
that
~(Q)
the
product
I = HJ,
ponents From
of
of
this
I and
with
J.
~(I) ~ ideals. for
It 2 if,
to an follows for
I a radical
ideal that
ideal.
~(M)
M is
a non-
If S M is
= dim(SM) ; whence
necessarily
maximality known
for
so
by
for
taking
P
R a
R a field:
a polynomial
a regular
ring
Noetherian
the
cases:
in a p o s i -
Hilbert
ring
one
deduces
Q = JK,
and
routinely
--
where
For of
the
"divide"
of For
is
this
the has
R a PID
these
59
n =
--
of
i.
remarks
R a
discovered
For the
Q any proof
of
if Q n R
is
proper
are
ideal
1 primary
com-
component
is
that
of
I.
I is
comaximal
J / ( J ~ R ) S is.
by
and
treatment.
e.g.,
distinct
observed
n = i,
Heinzer
b y H --
(K6R) S a n d
product been
by W.
height
if
--
-- w a s
0-dimensional
K =
[i]);
corollary
I a nonzero
Q / ( Q n R ) S is p r i n c i p a l J~R
(Endo
adapts
is p r i n c i p a l
"the"
proviso?
Heinzer-Davis
domain
ideals.
J is
and
ring"
1
ideals
the
one
= the
Geramita
intersection
--
of
maximal
a Dedekind
the
"Hilbert
dim(R)
case
by
nonzero,
(I:H)
case
of
many
maximal
examplep
A special
P's
well
2 if Q / ( Q n R ) S
decomposition
S-isomorphic
-- n o t
without
Gilmer,
Q~R
H is
J =
[2]
is e s s e n t i a l l y
~
and
examples
= dim(Rp)+n.
then
A special
R is
distinct where
over
valid
by
that
of
to
S,
now
domain
> 0,
insure
semilocal
(3)
ideal
(3)
of
of
n
ideal
infinitely
Swan,
creates
sequence.
[3]).
with
by
The
proper
certain
d i m ( S M)
a theorem
maximal
result
(Geramita
independently Davis.
results
easily
P = 0.
and
if w e
one local
sequence
So
a regular
this
Pz
that
indeterminates
Is
Answer:
so
P maximal
Every
of
and
4).
there
on
for which
a regular
(Remark
-
a 1-dimensional
R Notherian
(iff
Hilbert
hypothesis
R is
maximal
Assume
M
the
3
Hence
maximal
Gilmer: special
~(I)
~
cases
2
-
of
a t h e o r e m of Serre
Question:
How much of Serre's
domains? setting
[7]: v(I)
Observe
S-module
-
~ 2 if ~(IS N) ~
2 for every m a x i m a l N.
theorem survives
for arbitrary
Dedekind
that the validity of the theorem in the more general
is an immediate
sharpening
4
corollary
of the F o r s t e r - S w a n
of the Eisenbud-Evans
conjectured
bound on the number of generators
of an
[0].
REFERENCES
[0] Eisenbud,
D. and Evans,
over p o l y n o m i a l [i] Endo,
S.
15
E. D. Pac.
[3] Geramita,
[4] Goldman,
(1963)
A. V.
O.
20
(1967)
(1951)
I.
[6] Krull, W.
Jacobsonscher
No.
[8] Zariski,
J. Math.
class,
R-sequences
correspondences,
...,
rings,
to appear.
and the Hilbert Nullstellensatz,
Rin~s,
Ringe, Math.
Sur les modules
Foundation
and
197-209.
Allyn and Bacon,
Hilbertscher
Z. 54
(1951)
projectifs,
Boston
354-387.
Seminaire
Dubreil-Pisot
of a general theory of b i r a t i o n a l Trans.
A.M.S.
60
53
(1970).
Nullstellensatz,
2 (1960/61).
O.
Soc.
136-140.
Commutative
Dimensiontheorie,
J-P.
rings,
ideals in p o l y n o m i a l
Hilbert rings
[5] Kaplansky,
[7] Serre,
over p o l y n o m i a l
339-352.
Maximal
Z. 54
efficiently
these proceedings.
Ideals of the p r i n c i p a l
J. Math.
Math.
G e n e r a t i n g modules
rings,
Projective modules
Japan
[2] Davis,
E. G.
(1943)
490-542.
A CANCELLATION PROBLEM FOR RINGS PAUL EAKIN and WILLIAM HEINZER 0.
Introduction. Suppose A and B are rings. Let us say that A and B are stably equivalent if there is an
integer n such that the polynomial rings A [ X 1. . . . . X n] and B[Y 1. . . . . Yn ] are isomorphic. A number of recent investigations [CE], [AEH], [EK], [BR], [H] are concerned with the study of this equivalence. The most obvious question one might ask is: (0.1)Cancellation Problem. Suppose A is a ring. What conditions on A guarantee that for any ring B, A is stably equivalent to B only if A is isomorphic to B?
This cancellation problem was first investigated in [CE] and later in [AEH], [EK], [BR] and [H]. In [H], Hochster provides an example of two stably equivalent rings which are not isomorphic. In addition to providing an example, [HI also draws our attention to the connection between the cancellation problem for rings and the well-studied cancellation problem for modules. In the study of the cancellation problem for rings the following concepts have proved useful. (0.2) A ring A is said to be invariant provided that whenever A is stably equivalent to some ring B, then A is isomorphic to B. (0.3) A ring A is said to be strongly invariant provided that whenever o: A [ X 1 . . . . . X n] ~ B[Y 1 . . . . ,Yn ] is an isomorphism o f polynomial rings, then o(A) = B.
Obviously, strongly invariant rings are invariant. The most obvious example of a strongly invariant ring is the integers Z. On the other hand there are invariant rings which are not strongly invariant. The following isa result from [AEH]. (0.4) Suppose A is a domain o f transcendence degree one over a field, then A is invariant. Moreover A is strongly invariant unless there is a field k such that A = k [ X ] . This result tells us that if k is a field then A = k [X] is an invariant ring. However A is n o t strongly invariant, for if A = k [X] and B = k[Y] then the identity is an isomorphism of A [Y] o n t o B [X] which does not take A o n t o B. A fair a m o u n t can be said about rings of Krull dimension one which are not strongly invariant [AEH], but this case is not completely settled. In particular there is a gap in our knowledge o f the case when A is a Dedekind domain. In section five we have a discussion of the cancellation problem for Dedekind domains and present some questions whose affirmative answer would affirmatively settle the cancellation problem for all one dimensional noetherian domains. One way t o approach t h e cancellation problem for a particular ring A is to write
A [ X 1. . . . . X n] = B[Y 1. . . . . Yn ] for,some ring B and look at D = ANB. In the case of domains, if the transcendence degree of A over D is at
61
most one and S denotes the non zero elements of D, t h e n e i t h e r A = B o r A
s -~ B s ~ k [ X ]
where k is the
quotient field of D. This follows from the result on domains of transcendence one over a field mentioned above, with information like A s = k [ t ] , a bit of additional hypothesis will often yield results like A ~ B-,~ D[X].
When the transcendence of A over D is greater than one, the problem becomes considerably more
complex. In particular, the cancellation problem for C [ X , Y ] , the polynomials in t w o variables over the complex numbers is not settled. C.P. Ramanujam has taken a particularly sophisticated topological approach to this question JR]. He characterizes the complex 2-space as a non singular, contractible algebraic surface which is simply connected at infinity. To our knowledge, this has not yielded a solution to the cancellation problem for C [ X , Y ] ,
C
the complex numbers.
We begin this article by examining Hochster's example from [ H ] . We observe that this provides a nice introduction to an interesting class of rings, called locally polynomial rings in [ES]. (Given a ground ring R and an algebra A over R, we say that A is locally a polynomial ring over R if Ap = A®RR p is a polynomial ring over Rp for every prime p of R). Over a fairly substantial collection of rings we give a number of characterizations of locally polynomial rings "in one variable." In particular we show that if D is a domain, the affine (ie. finitely generatedring extension) locally polynomial rings in one variable over D are precisely the symmetric algebras of the invertible ideals of D. We then use this result to indicate an approach to the cancellation problem for k [ X , Y ] ,
k an algebraically closed field.
Before proceeding let us make the usual stipulations: all rings are commutative with identity and all modules are unital.
1.
Symmetric Algebras and Non Invariant Rings. In this section we give Hochster's example of a non
invariant ring. This example, in addition to showing that stable equivalence of rings is not a trivial relationship (i.e. isomorphism) also serves to remind "commutative algebraists" that its nice to know some topologyespecially the theory of vector bundles. Let us recall the definition and basic properties of the symmetric algebra of a module.
(1.1) Definition. Let R be a ring and M an R module. A symmetric algebra of M over R is a pair (~0,S(M)) consisting of an R isomorphism ~ and an R algebra S(M) such that ~: M ~ S(M) and such that if a is any Rhomomorphism of M into an R algebra A, then there is a unique R-homomorphism h such that the following diagram commutes M
~ S(M)
Every module has a symmetric algebra which is unique up to an R-isomorphism [C]. In fact, if M is an R module, S(M) is R isomorphic
to the "'abelianization'" of the tensor algebra of M over R [C]. Another
realization of the symmetric algebra of a module which is often quite computable is the following.
62
Let M be a module and write 0
-~
K
~
F
~
M
-*
0
exact
where F is a free module, say F = ~ RZi. Let {Xi)ie [ be indeterminates over R and consider J(K) = ie[ {~; riX i e R [ { X i ) I ~ riZi ¢ K). Let 6L denote the ideal in R [ { X i } i d ] generated by J. Then R[M] = R [ {X i) ie]]/6"(
is the symmetric algebra of M over R. To see this one observes that R [M] is a graded,
homogeneous ring over R whose module of 1-forms is isomorphic to M. Thus we have an R-homomorphism h such that M
i
a [M]
S(M)
commutes.
Let z i = rt(Zi). Then we have a mapping R [ { X i } i ¢ l]
~
S(M)
given by
xi~0(zi).
Since J C Ker £ we have a well defined mapping R [M] ~ S(M) which is surjective. Since ho£ = id we have R [M] is R-isomorphic to S(M). Using the above, or referring to [C] one can easily see the following well known facts. (1.2) Lemma. Let R be a ring, then (i)
If F is a free R module on a set of cardinality ], S(F) is R isomorphic to the polynomials over R in
a set of indeterminates having cardinality I. (ii)
If A and B are R modules then S(A~B) is R isomorphic to S(A) ORS(B).
Recall that t w o R-modules M and N are stably equivalent if there is a free module F such that M(DF ~ N@F. In this case, if F is free on n generators then, taking symmetric algebras we have S(M)®RS(F) ~ S(MI~F) ~ S(N(~F) ~ S(N)®RS(F ). S(F) ~ R[X 1 . . . . . X n] . S ( M ) [ X 1 . . . . . X n]
But Hence we have
~ S(M)®R R[X1 . . . . . Xn] ~ S(N)®R R[X1 . . . . . Xn] ~ S(N)[X1 . . . . . Xn]"
Thus stably equivalent modules give rise to stably equivalent rings. This provides the basic idea of Hochster's example: find two stably equivalent modules which are not isomorphic, then prove that their respective symmetric algebras are not isomorphic. (1.3) Lemma. Let R be a ring and M and N two finitely generated R modules. If S(M) and S(N) denote the respective symmetric algebras then S(M) is R isomorphic to S(N) if and only if M is isomorphic to N.
63
4
Proof. Clearly M ~- N : ~ : S(M) is R-isomorphic to S(N). Conversely, suppose S(M) is R-isomorphic to S(N), say by an R-isomorphism ~o. Now S(M) is a homogenous graded ring whose module of 1-forms is isomorphic to M. Identify M with these 1-forms and let {r/i)iei be a basis for M. Then write ¢(rt i) = j~OCXij e S(N). Here aij is a homogenous element of S(N) of degree j. Since ¢ is an R-isomorphism, we have ~°(rti-a°i) = j~=l aij • Denote this element by 3,i. Then since S(M) = R [ (r/i} i e [ ] = R [ {r/i-C~oi} ] we have S(N) = R [ ~3,i}iei ] . By grading it follows that (c~1 i}ie[ generates the 1-forms of S(N), which is a module isomorphic to N. There is then a well defined surjective R homomorphism a: M -* N
given by
o(r/i) = a i l
which takes M onto N. By symmetry there is a surjection o': N ~ M. But o'oa is a surjective mapping of M onto itself. Thus o'oa is an isomorphism [V] and so is o. Thus, by this lemma we see that if M and N are finitely generated, stably equivalent, non isomorphic modules, then S(M) and S(N) cannot be R-isomorphic. The idea now is to find two such modules and prove that any supposed isomorphism of the symmetric algebras would (essentially) have to be an R isomorphism. To this end we recall the following from [S]. n Let k d e n o t e t h e real n u m b e r s and R = k [ X o , . . •, X n ] / i ~=0 X i 2 - 1 = k [ x 0 , x 1 . . . . ,Xn]. Let F denote the free m o d u l e on generators e o , . . . , e n and define o: F ~ F b y
o(ei) = xi(E xie i) . T h e n one has
o 0 ~ Hence
PI9 R~
P
~
F
~
R n+l = Rn~
(Zxiei)R ~ R.
0
exact.
Moreover P ~ R n (ie
P is free) if and o n l y if n=1,3,7.
The notation of this statement holds for the next two lemmas.
(1.4) Lemma. The ring R above is strongly invariant. Proof. Suppose 7': R IX 1. . . . ,X n] -~ B [Y1 . . . . . Yn ] . We may assume R [X I . . . . . X n] = B [Y1 . . . . . Yn ] and show R = t3. Suppose xi=boi+bliYn
+'" ' + b k i Y k
w h e r e b i j e B [ Y 1. . . . . Y n _ l ].
Then ~ x i 2 = 1 implies ~b2,i = 0 if k is greater than zero. But R[X 1 . . . . . X n] is formally real (ie. 0 cannot be written non trivially as a sum of squares there. Thus x i = bol for each i and x i e B [ Y 1 . . . . . Yn--1 ] • By further reduction, x i e B for each i and hence R C B. But it now follows easily that R = B. (1.5) Lemma. If ¢: S(P) ~ R [X o . . . . . X n _ 1 ] then up to an automorphism of R, ~0 is an R isomorphism. Proof. By adding an indeterminate and extending trivially we have
64
~0:S(P)[Y] ~ R[X o ..... Xn_ I ] [ Z ] . Since S(P)[Y] = R [Z o ..... Zn], one concludes that, up to an automorphism a of R, ~ois an R isomorphism. However one then has S(P) ~
a [ x o ..... X n - 1 ]
~1
R[Xo ..... X n - l ]
and a-lo~0 is an R isomorphism. Thus, with the notation of the previous lemmas, if R IX 1..... X n] is isomorphic to S(P), then P ~ Rn by (1.3). So for n # 1,3,7 we see that the rings S(P) and R [X 1..... X n] are non isomorphic, stably equivalent rings. (1.6) Remark. In case n = 2 in the above, the integral domain R has the following properties (i)
R is strongly invariant
(ii)
R [X] is invariant, but not strongly invariant
(iii) R[X,Y] is not invariant.
Proof. We have already seen (i) and (iii). To see (ii) we observe that the proof of (1.4) actually shows somewhat more. We have seen that if R IX 1..... X n] = B [Y]. Then R C B. We can thus appeal to the following result from [AEH]. Let D be a unique factorization domain (UFD) and let (X i) be indeterminates over A. If R is a UFD such that D C R C D[ {Xi} ] and such that R has transcendence degree one over D, then R is a polynomial ring over D.
Remark ( 1.6) has to be considered when one gets overly optimistic about the prospects of proving k[X,Y] invariant, k a field.
2.
Locally Polynomial Rings. Let D be a ring and A an algebra over D. If for every prime p C D, Ap =
Dp ~D A is a polynomial ring over D, we say that A is locally a polynomial ring over D. These rings are investigated in [ES]. Since the formation of symmetric algebras respects change of ring, we find that the symmetric algebras of locally free modules provide plenty of examples of locally polynomial rings. (2.1) Lemma. If M 1 and M 2 are modules over a ring R such that S(M1)[X 1..... X n] ~ S(M2)[Y 1..... Yn ] R (ie. the algebras are R-isomorphic]. Then for every prime p C R, S(M 1) O Rp Rp S(M2) ® Rp.
Proof. By lemma (1.3) we have that M 1 and M 2 are stably equivalent modules. Thus MI(~RR p and M2®RR p are stably equivalent Rp modules. Thus MI®RR p is Rp isomorphic to M2~RR p by a theorem of Bass which assures us that stably equivalent modules over a local ring are isomorphic [B]. Thus S(MI®RR p) = S(M1)®RR p is Rp isomorphic to S(M2)®RR p = S(M2®R Rp).
65
With reference to the preceeding lemma we see that if S(M 1) is a domain and S(M 1 ) is stably equivalent to S(M2), then even though S(M 1) may not be isomorphic to S(M2), they still have isomorphic quotient fields. Referring to Hochster's example then, we see that this is not a counterexample to the following "bi-rational cancellation problem". Suppose A and B are domains and A[X] ~ B[Y] is an isomorphism o f polynomial rings. Are the quotient fields of A and B isomorphic?
This question is considered in [ A E H ] .
It is certainly related to the following, well studied question of
Zariski: Zariski Problem, Let K and K' be finitely generated fields over a field k. Assume that simple transcendental extensions of K and K' are k-isomorphic to each other. Does it follow that K and K' are kisomorphic to each other? 1
3.
Some Interesting Types of Ring Extensions. In studying the cancellation problem we have been led to
consider the following possible relationships between a ring R and its extension A. (l)
A is the symmetric algebra of a projective R module
(2)
A is projective in the category of R algebras. That is, given a diagram of R-algebras and R
homomorphisms. A
B
~
C
~
0
with the bottom row exact, there is a R homomorphism 0 : A -+ B which makes the diagram
A
B~C~O
(3)
commute.
A is a retract of a polynomial ring over R. That is, there is an idempotent endomorphism of some
polynomial ring R[ (Xi}ie I] which has A as its range. (4)
There exists an R-algebra A* such that A®RA* is a polynomial ring over R.
(5)
There are indeterminates (Xc~) and {Yi3} such that R[ {Xc~}] = A[ (YI3}].
(6)
There exist indeterminates {Xc~} such that R C A C D[ {Xc~} ] and A is an inert subring o f
R[ {Xa) ]. That is, if a,b are non zero elements of R[ {Xc~}] and ab e A then a e A and b e A.
1
This question is studied by Nagata in [ T V R ] .
He notes there that it was raised by Zariski at a 1949
Paris Colloquium on algebra and the theory of numbers.
66
7 (7)
A is locally a polynomial ring over R.
A t present the relationships among these properties is not clear. However, some connections do f o l l o w readily. (3.1) Lemma. With regard to the above conditions we have the following implications: (7)
~
(1)
=
(2)
4
~
5
¢=~ (3)
f
In case R is a domain we have 15) ~ (6), and if A is an affine ring of transcendence degree one over the domain R then (7) ~ (1). Proof. (1) ~ (7)
Let P be a projective R module and Q a prime of R. Then SR(P)®RR Q = SRQ[P®RRQ].
But P®RRQ is free, hence SRQ[P®RRQ] is a polynomial ring. (2) <==~(3) The proof is the same as showing that projective modules are retractions of free modules. (1) ~ (4)
Let A = S(P) where P is a projective R module and let P* be an R module such that P(3P* is
free. Then S(P)®RS(P*) = S(P~P*) is a polynomial ring over R. (4) = (5)
Suppose A ® R A ' is a polynomial ring over R. Then look at A® A'®
A®
A'®
A®
.....
associated as (A ® A') ® (A ® A ' ) ® (A ® A ' ) ® . . . . .
S
and as A ® ( A ' ® A ) ® (A' ® A) ® (A' ® A ) ® . . . .
ADS.
We observe that S is the polynomials in some (large) numbers of variables over R. Hence A ® S is a ring of polynomials over A which is what was to be shown (5) ~ (3)
Obvious since A is a retract of A [ {Xcx} ].
In the case of a domain (5) ~ (6) is obvious since then (and only then), A is an inert subring of A [ {Xc~)]. We will be done if we show (7) =~ ( 1 ) under the hypothesis that A is a domain of transcendence degree one and
an affine ring over R, For convenience, this is decomposed into a few lemmas.
(3.2) Lemma. Let R be a domain and A an affine ring over R which is locally a polynomial ring in one variable over R. Then there exists a set { x I . . . . . Xn) C A such that A = R [x 1 . . . . . x n] and given any prime
P C R, Ap = R p [ x i ] for some i, 1 ~< i ~< n (i may depend upon the choice of P). Proof. Let {Pc~}c~e[ denote the primes of R and Xc~ an element of A such that Apex = Rp~[ x ~ ] . Then one has R [ { xc~}(xe] ] = R. This follows since we can show that, as R modules they are locally equal, and hence
67
equal. SinceA is affine over R we may assume A = R[ x 1. . . . . xn] where x i e {x~e}~e I . Moreover, for each prime P C R, Ap = Rp [ x i] for some x i e { x 1. . . . . Xn}. For let Xp be a local generator at the prime P. Then for each i xi = ~i Xp +/3 i
el,/~i e K = quotient field of R .
By uniqueness of representation, ~xi,/~i e Rp. Residually mod P we have A p / p A p = Rp/pRp I x 1. . . . . x n] = k [ X p ] = k [ x 1. . . . . Xn] where k is the field Rp/pRp. Since x i = cxi Xp + ~i one of the Ei is not zero, (ie one of the ~i is a unit in Rp, say cx1. Then one has R p [ × p ] = R p [ X l ] .
(3.3) Lemma. Let R be a domain and A = R [ x 1. . . . . x n] a transcendental extension of R such that for any prime P C R, A p = R p [ X i] for some i, 1 ~< i ~< n. Let K denote the quotient field of R and fix some Y such that R(o ) = K [ Y ] . Then one can write x i--a iY+b i The
with
ai , b i e K .
n R module ~ R a i is an invertible fractionary ideal of R. i=1
Proof. L e t ~
be a maximal ideal of R. Then RT~[X 1. . . . . X n] = R7~[Xi] for some i. Hence a1 Xl = aT xi + ~1 _ an Xn-a-~-- x i + ~ n .
By uniqueness of representation
#
~j
e R-F/L. Thus (a 1. . . . . a n) Rnt =a i RZzL. So (a 1. . . . . an) is locally
=
principal and thus invertible. (3.4) Lemma. With the notation of (3.3) let ~
denote the invertible ideal (a I . . . . . an). Then A is R
isomorphic to S ( ~ ) , the symmetric algebra of the projective R-module (~. Proof. Consider M, the R submodule of A generated by { x I . . . . . Xn}. Represent M as ((alY+b 1) . . . . . (anY+bn)). Then there is an obvious natural mapping M
o_, ~
xi
(7 -*
~
0
given by
a i.
Since(3~f, is a projective R module, there is a mapping o': (3t~ -+ M such that oo' = id(~L. Then o' induces a unique R mapping h such that
68
0
o~
1 s(6~)"
A /
/
/
commutes. h
In fact, h is an isomorphism. For let p be any prime of R and let apY + bp be a local generator for R at P. Then let
n n Zp = o'(ap) = i~ 1 ;ki x i = i~ 1 ;ki (aiY + b i) = (~; Xiai)Y + (i~ 1 ;kibi).
It turns out that Zp is a local generator for A at p. For applying o we have ap = o ( o ' ( a p ) ) =
o(~;X i x i) = ~;Xia i .
Thus Zp = apY +/3. Since Zp ~ Ap = R p [ a p Y + b p ] , Zp = e(apY+bp) + $ where ~,$ e Rp. Thus e = 1 and Z p - $ = apY+bp so A p = Rp[apY+bp] C Rp[Zp] _C Ap, and we have equality. Thus the mapping:
h (S(0-C))p ' ~
Ap
is surjective.
But S(l~)p = Rp[T] and any Rp surjection of Rp[T] onto Rp[Zp] would necessarily be an isomorphism. Thus the mapping h is locally an isomorphism therefore its an isomorphism. This completes the proof of the lemma, the theorem now follows immediately,
o
(3.5) Theorem. Let A be an affine ring of transcendence degree one over the domain D. Suppose D is locally a UFD, then the conditions 1--7 above are equivalent. Proof. (6) ~ (1) follows in view of the result from [AEH] quoted in (1.6), and since a retract of a UFD is a UFD 1 (2) =~ (7) follows from the same result.
1
[]
Ed Enochs called this charming little result to our attention. His argument goes as follows: Let R C S
be rings, R a retract of the UFD S under the mapping o. It is easily seen that each element in R is a product of irreducibles, thus one need only show that irreducibles in R are prime. Let ~r e R be irreducible then factor 7r in S, 7r = Pl " " • Pk ' each Pi prime. Apply o to both sides and use the fact that lr is irreducible to conclude that for a suitable numbering o(pi) is a unit, i /> 2. Thus ~ = # o ( p l ) , /z a unit in R. Now if lrlab in R. Then one can assume p l q = a. Since /z is also a unit in S, ( p l p . ) ( / z - l q ) = a. Then applying o wesee ~ r ( # - l o ( q ) ) = a . Thus a e ~ R
and 7r isprime.
69
10
(3.6) Remark. Theorem (3.5) should generalize (at least in the one variable case) to a much larger class of domains than those mentioned here. One would expect most of it to go over to domains D such that D [X] is D invariant. A ring A is D-invariant provided that whenever A [ X 1 . . . . . X n]
-~ B[Y1 . . . . . Yn ] is a
D--isomorphism of polynomial rings, then A is D-isomorphic to E3. In [AEH] it is shown that for'D a locally HCF 1 ring D [ X ] is D, invariant. It is easy to see that for such rings, all of the implications of (3.5) hold except possibly (2) =~ (7). (3.7) Corollary.
Let D be a domain which is locally a UFD and let ~'L be an invertible ideal of D. Then
S(~'(), the symmetric algebra o f ~ ( is D-invariant. 2 Proof. Suppose S(0/,) IX 1 . . . . . X n] = B[Y1 . . . . . Yn ] with D C S(<3-~) N B. Then localizing everything at a prime p of D, S((3~,)p = D p [ X p ] , thus Bp = D p [ Z p ] , by the result from [AEH] quoted in the proof of (1.6). Thus B is locally a polynomial ring over D, and since S(O'L) is affine over D, so is B. Thus by Theorem (3.5) B = S(G) where G is an invertible ideal of D. If F n denotes the free D module of rank n, then our hypotheses implies S(O-L(~ Fn) is D isomorphic to S(GI~Fn). Thus ~
~F n ~ G I~ F n and £ ~
G [K, p.328] and hence
s ( o ~ l ~ S(G)) = B.
4.
An Approach to Cancellation for G [ X , Y ] .
In this section, we return to what must be regarded as the
next fundamental problem in our study; whether A [ T 1 . . . . . T n] = ~ [X 1 . . . . . Xn+ 2] implies A ~ ~ [ X , Y ] . We offer the following observation. (4.1) Theorem. Let k be an algebraically closed field and A [T] = k [ X , Y , Z ] (T and Z represent the same
finite number of indeterminates). If k is properly contained in A N k [ X , Y ] then either A = k [ X , Y ] or there is an f such that A n k [ X , Y ] = k [ f ] .
Proof. Let R = A n k [ X , Y ] .
In the latter case, there is a g such that A = k [ f , g ] .
The t w o cases obviously depend upon the transcendence degree of R over k.
Since R is algebraically closed in A, if the transcendence is t w o we clearly have equality. If R has transcendence degree one over k then it is known [ A E H ] that R has the form k [ f ] .
Now k [ f ] is an inert subring of A,
hence any prime element of k [ f ] already generates a prime ideal in A. Let S denote the non zero elements of k [ f ] , then we have A S = k [ X , Y ] s [ Z ] .
Since A ~=k[X,Y], A S # k [ X , Y ] S' Moreover, k(f) is algebraically
closed in A S and A S is of transcendence degree one over k(f). Thus by (0.4), A S = k(f) [t] for some t e A. N o w let V(f) denote the rank one discrete valuation ring k [ f ] (f). Then we have the following situation:
1.
HCF stands for highest common factor. These are the rings whose groups of divisibility art lattice
ordered. A m o n g the rings w h i c h locally have this property are the Prufer domains. A student of Hochster has considerably extended the list of domains D such that D [X] is D-irwariant. 2.
Using (4.9) in [ A E H ] one can see that HCF can be substituted for UFD.
70
11 V(f) C V ( f ) [ t ]
_C A S , C k ( f ) [ t ] ,
where
S' = k [f] - f k [ f ] . We claim that t can be chosen so that A S, = V(f) [ t ] .
First we show that we can choose t
so that in A/f, t is transcendental over k. Let t = b o + b l Z + ' • "+ bnZn
with
bi e k [ X , Y ] .
Remember that k is algebraically closed and observe that the element t fails to be residually transcendental over k in A / f only if f divides b 1. . . . .
b n and b o is residually equal to some (xo e k (if there is more than one
Z you've got to use forms of degree n in place of zn). But then f divides t - ( x o in k [ X , Y , Z ] , say fh = t - ( x o. t-c~ o Since t--cz o c A, h c A and h - - 7 will serve equally well for t. We can repeat this only finitely many _
times before we come to the point where one of the b i, i ~> 1 is not divisible by f and hence this choice of t is residually transcendental over k mod f. But now we have A S, = V f [ t ] .
For let a e AS,. Then
a = ' Y o + 3 ' l t + ' ' ' + 3 , k tk where ~/ie k(f). But each ")'i e V(f), for if - £ were the least value of any 7i in the f - a d i c valuation then f£a = (f£3,o) + (f£,),l)t + ' . . + (f £-),k)t k.
Then
each f£"fi is an element of k[f] (f) and at least one of them is a unit there. Taking residues rood f, the left side vanishes, and hence the right side becomes an equation of algebraic dependence for t over k mod f, a contradiction. We have thus shown that if p is any prime of R = k [ f ] , then A p = ASRR p is a polynomial ring in one variable over Rp. Thus by Theorem (3.2), A is k [ f ] isomorphic to the symmetric algebra of an invertible ideal of k [ f ] .
Since k [ f ] is a PID, this ideal is principal, and the symmetric algebra is k [ f ] [ U ] .
The image of U under the now extant k [ f ] --isomorphism provides a g such that A = k [ f , g ] .
[3
An interesting corollary is the following: (4.2) Corollary. Suppose k is an algebraically closed field and k [ X , Y , Z ] = k [ X ' , Y ' , Z ' ] .
The possibilities
for R = k [ X , Y ] n k [ X ' , Y ' ] are the following (i)
R =k
(ii)
R = k[X,Y]
(iii) R = k [ f ] .
In this case there exist a g and an h such that k [X,Y] = k[f,g] and k [ X ' , Y ' ] = k[f,h].
Thus the only non-trivial possibility for R is essentially when X = X' and the intersection is k [ X ] . With regard to the cancellation problem for ~ [ X , Y ]
we now see the following.
(4.3) Corollary. Let (~ denote the complex numbers and suppose ~ [ X , Y , Z ] = A [ T ] .
Then A ~ ~ [ X , Y ]
and only if then is an automorphism o of ~ [ X , Y , Z ] , say a(X) = X', o(Y) = Y', o(Z) = Z' such that
~ [X',Y'] nA~
k.
The steps in the proof of (4.2) are
71
if
12 (1)
Observe that A N k [ X , Y ] = k [f] is an inert subring of A (this is equivalent to the assertion that
(2)
Argue that A [ k ( f ) ] = k(f) [t] for some t ~ A.
(3)
Show that A is locally a polynomial ring in one variable over k [f].
(4)
Appeal t o (3.5).
f-X
is prime in A for each X e k)
In step (3) we use the fact that all operations take place in k [ X , Y , Z ] to argue that, for each X, only a finite number of modifications of t are necessary to arrive at a local generator at k [ f ] ( f - h ) "
If we are willing
to use a bit more machinery, then we can follow steps (1) and (2) by a global version of step (3), thus eliminating the need for (3.5). In fact, the following is true: (4.4) Theorem. Let k be an algebraically closed field and A a t w o dimensional affine ring over k which is a UFD. Suppose there exists an f E A such that
(i) k[f] is an inert subring of A. Equivalently; ( f - X ) A is a prime ideal for every X e k. (ii)
A[k(f)] =k(f)[t]
Then there exists g e A such that A = k [f,g]. Before giving a proof of (4.4) we need to recall some facts about quadratic transformations. Let RT,f/ be a local ring and V a valuation ring which dominates R. Let x 1 . . . . . x k be a minimal generating set for ~7~,and assume x k has minimal value in the valuation associated with V. Let ~ t
denote
the center of V on R I = R [ x l , . . . , X k - 1 ] and consider the local ring R 1 7 ~ f This is a quadratic transform xk xk (also called a quadratic diJitation) of R along V. Obviously such a procedure can be iterated. We then speak of a sequence of quadratic transformations of R a l o n g V .
LetR < R 1 < R2<'"
n <
be a sequence
of quadratic transformations of R along V. If UR i = V we say that the sequence "reaches" V. If R n = V for some n, we say that the sequence reaches V in a finite number of steps. In some situations, it is possible to assert that every sequence of quadratic transforms of R along V will reach V in finitely many steps. We need the following special case of a general theorem of Abhyankar [A, p.336].
(4.5) Lemma. Let X and Y be indeterminates over the field k and V a rank one discrete valuation ring such that k[X,Y] C V C k(X,Y). Suppose that the residue field of V is transcendental over k. 1 Then if p denotes the center of V on k [ X , Y ] , any sequence of quadratic transformations of k [X,Y] p along V must reach V after finitely many steps. We can now give a fairly straightforward proof of (4.4).
1.
In the language of Zariski-Samuel, Vol. II, p.95, V is a prime divisor.
72
13 Proof of (4.4). The idea is to choose t as a first approximation to g and argue that only finitely many modifications are necessary to arrive at a "perfect f i t . " We start out then with gl = t and k[f,g 1] C A C k(f) [gl ] . Since A is affine over k, it is easily seen that there is some ~0e k[f] such that k[f,g 1] C A c k[f, gl, 1/~o]. Let h 1. . . . . h s denote the prime factors of ~o. With the possible exception of the h 1. . . . . h s adic valuations of A, every essential valuation of A is an essential valuation of k[f,g 1] . We claim that we can modify gl to get g2 so that the h 1. . . . h s _ l - a d i c will be the only possible essential valuations of A which are not essential for k[f, g2]. When we have done this we will be through, for it would follow by reduction that A = k[f,gs+ 1] . Since k is algebraically closed, there is no loss in assuming h s = f. Denote by Af the localization of A at the prime fA. Then Af is a rank one discrete valuation ring which has residue field transcendental over k. If Af is centered on a minimal prime of k [ f , g l ] , then it is essential for k[f,g 1] and we take gl = g2" Otherwise A f is centered on a maximal ideal of k [ f , g ] , say (f,g-;k). We replace gl by g] - g l - X f If A f is not centered on a height one prime of k[f, g~], we repeat this to get g2. There must exist an £ such that Af is centered upon a height one prime of k[f, gl~]. This is because this selection procedure constitutes taking a sequence of quadratic transforms of k [f,gl ] ( f ' g l - X ) along Af. This sequence must reach Af by (4.5). Set g2 = g~" Then Af is an essential valuation of k[f,g2]. Since the formation of g] . . . . . g~ has taken place within k[f,gl,1/~0] = k[f, g2,1/~] and we know that the hs-adic valuation is essential for the rings A and k[f,g2], we are reduced to k[f, g2] C A C k[f,g2,h 1. . . . . h s _ l ] . Thus A = k[f,gs+ 1] by reduction.
[]
(4.6) Remark. Professor Abhyankar has shown us how to remove the assumption that k is algebraically closed from (4.2). His argument uses the fact that for k" algebraically closed, every k--automorphism of k [ X , Y ] is a product of elementary automorphisms. We should also remark that (4.4) generalizes the results announced in the Notices, Nov. 1971, abstract no. 6 8 9 - A 2 7 .
5.
Problems. The investigation of the cancellation problem provides numerous simply stated problems
whose solutions appear to be non trivial. We take the opportunity to discuss a few of these in this closing section. Some of these questions are also posed in [ A E H ] . (5.1) Question. If A and B are domains and A [ X 1 . . . . . X n] = B [ Y 1 . . . . . Yn ] , are the quotient fields of A and B isomorphic? This is the "bi-rational cancellation problem" mentioned after (2.1), (5.2) Question. If A [ X 1 . . . . . X n] = B[Y 1 . . . . . Yn ] does there exist an isomorphism of A into B such that B is a finitely generated ring extension of the image of A? This is easily seen to be the case for domains with n = 1. In considering invariance and strong invariance when A [ X 1. . . . . X n] = B [ Y 1. . . . .
Yn ] , it would perhaps
have been more precise to define n-invariant and n-strongly invariant. (5.3) Question. Is it possible for a ring to be n-invariant (or n-strongly invariant) but not m invariant ( m -
73
14 strongly invariant)? (5.4) Question. If A is a strongly invariant integral domain and A [ X 1 . . . . . X n] = B [ Y 1 . . . . . Y m ] must it f o l l o w that A C B? Of course we would like to know which ring-theoretic invariants are also invariants of the stable equivalence relation. While global and valuative dimensions offer no problems, Krull dimension does. (5.5) Question. Suppose A is an integral domain of Krull dimension d and A [ X 1 . . . . . X n] = B [ Y 1 . . . . . Y n ] . Does B have Krull dimension d? We can offer one little result which may provide an approach to this problem (5.6) Theorem. Let A be an integral domain of Krull dimension one and suppose A [ X ] = B [ Y ] (one X; one Y), then B has Krull dimension one. Proof. It is easy to see that an integral domain with a non zero Jacobson radical is strongly invariant. Thus we may assume that A has Jacobson radical zero. Thus every prime of A is an intersection of maximal ideals and it follows that A is a Hilbert ring. Since it is a homomorphic image of the Hilbert ring A [ X ] , is a Hilber~ ring. Let P be a depth t w o prime of B, and P = N ~
a representation of P as an intersection
of maximal ideals of B. Then P[Y] = n ~ c ~ [ Y ] and each T/~c~[Y] is a depth one prime of B [ Y ] . then each 7 ~ [ Y ]
B
If P 5 0
has height two and thUS~c~[Y ] N A = Nc~ # 0. This is because if 7~c~[Y] n A = 0 then
we could localize at the non-zero elements of A where (~r/Tc~[Y])e could have height at most one. Therefore N e [ X ] C ~ I ~ [ Y ] , and since depth Nc~[X] = 1, we must have N e [ X ] = ~ 7 ~ [ Y ] where Ne is a maximal ideal of A. Then we must have ( n N e ) [ X ] = N (Ncx[X ] ) = n ( 7 7 ~ [ Y ] ) = P [ Y ] .
It then follows that (N N~) is a
non maximal prime of A, thus it must be zero. Therefore P = 0 and B is one dimensional.
[]
There is evidence to the effect that a Dedekind domain is either strongly invariant or a polynomial ring. If this were true, it would completely solve the cancellation problem for one dimensional noetherian domains. From [ A E H ] we have: (5.7) Theorem. Let A be a Dedekind domain which is not strongly invariant, Then there is an element s in A such that A [ 1 / s ] is the polynomials in one variable over some field k. If A contains the rational numbers then A is the polynomials in one variable over some field. Using this result, the cancellation problem for Dedekind domains reduces to the following (5.8) Question. Let V be a rank one discrete valuation ring with quotient field k and k(t) a simple transcendental extension of k. Let V * be an extension of V to k(t) such that t e V * and the residue field of V * is algebraic over that of V. Let A = V * N k i t ] .
Then A is a Dedekind domain [ A E H ] .
Is A strongly
invariant? Invariant? The solution to this problem will finish the cancellation problem for Dedekind domains. We observe in [EH]
74
15 that a domain of the above type cannot be euclidian. This, of course solves the cancellation problem for euclidian domains. In this context it is also w o r t h noting that if the valuation V * above is an unramified extension of V if and only if resulting ring is a principal ideal domain. Since this construction is very easily implemented, it provides a large and quite tractable family of non euclidian principal ideal domains. There is an interesting question on extending valuations which, if answered affirmatively would completely solve the cancellation problem for one dimensional noetherian domains by providing an affirmative answer to (5.8). It goes as follows: (5.9) Question. Suppose V is a rank one discrete valuation ring with quotient field K. Let K(u,w) be a pure transcendental extension of degree two and V * a rank one discrete extension of V to K(u,w). Suppose Vu =
V * N K(u) and Vw = V * n K(w) have residue fields which are algebraic over the residue field of V. Must the residue field of V * be algebraic over that of V?
If the residue field of V is of characteristic zero, orperfect, the answer is yes. To see this, extend V * to a larger complete, rank one discrete valuation ring V * whose coefficient field contains the algebraic closure of that of V. Now complete V, V u and V w within V ' t o obtain V, V u and V w respectively. Since the ~esidue field of Vw
V
is perfect we may assume that the coefficient fields of V,
are all contained with that of
V*
[N, (31.9), p.110],
Let
W
Vu
and
denote the completion
of the integral closure of~) in ~ * . Then the residue field ~f W is in fact the algebraic closure of that of V. Moreover V u and V w are integral over W. For if f e Vu, then since f is residually algebraic over V, if77L is the maximal ideal of W, then W[f]/~R.W[f]
is a finite module over W/TfL. Hence W[f]
is a finite W-module
by [N, (30.6) p.103]. Now let W* denote the integral closure of W in ~ * . Then W* is an extension of both V u and V w with residue field algebraic over that of V. But W* N K(u,w) is V*. One can see from the above argument that an affirmative answer to (5.9) would follow from a negative answer to the following. (5.10) Question. Suppose V C V * are complete rank one discrete valuation rings with residue fields k and k'(t). Suppose k" is an algebraic extension of k and t is transcendental over k. Can there exist distinct complete rank one discrete valuation rings V u and V w such that each has residue field k' and V _C_V u n V w
C V*? In considering this question one may assume that k' is an infinite, purely inseparable extension of k, that the quotient field of V is algebraically closed in those of V u and Vw, and that both V u and V w are algebraically closed in V*. Abhyankar and Moh have shown the following to be true. (5.10)
Let k be a field of characteristic zero and f c k [ X , Y ] .
If k [ X , Y ] / ( f ) = k It] then there exists
ge k [ X , Y ] such that k [ X , Y ] = k [ f , g ] . This raises the following possibility for solving the cancellation problem for k [ X , Y ] . affirmatively answer the following t w o questions.
75
Suppose we could
16 (5.11) Question. Does (5.10) generalize to three variables? That is, suppose there exists f e k [X,Y,Z] such that k [ X , Y , Z ] / ( f ) = k [u,v]. Can one always find a g and an h such that k [X,Y,Z] = k [f,g,h] ? (5.12) Question. Suppose A [ T ] = k [ X , Y , Z ] , where A is a ring, T is transcendental over A and X,Y and Z are indeterminates. Does there exist a surjective homomorphism 3,: A --* k[t] ? Were one to answer both of these questions affirmatively, then the kernel of 1' would become the " f " in (4.1) and the cancellation problem would be solved for k [ X , Y ] , k a field of characteristic zero.
(5.13) Remark. Obviously, a problem similar to our is the following: Suppose A and B are rings such that A [ [X] ] o_, B [ [ Y ] ] is an isomorphism of rings of formal power series. Are A and B isomorphic? M. J. O'Malley has begun this investigation in [ 0 ' M ] . He has proved the analogs of some of the results of [CE]. In particular he shows that with the above notation, if o(A) C B then o(A) = B. He also shows that if the Jacobson radical of A is zero, then the above conditions imply o(A) = B.
Acknowledgements: We are grateful to professor Abhyankar for the benefit of several stimulating conversations on this material. We also express our gratitude to Wanda Jones for her patience in typing the manuscript.
76
17 References [A]
S. Abhyankar, On the valuations centered on a local domain, Amer. J. Math. 78(1955) 321-348.
[AEH]
S. Abhyankar, P. Eakin and W. Heinzer, On the uniqueness of the ring of coefficients in a p o l y nomial ring, to appear J. Algebra.
[B]
H. Bass, K-theory and stable algebra, I HES Publications Mathematiques 22(1964) 5--58.
[BR]
J. Brewer and E. Rutter, Isomorphic polynomial rings, to appear Archiv der Math.
[C]
C. Chevelley, Fundemental Concepts o_f Algebra, Academic Press, New York (1956).
[CE]
D. Coleman and E. Enochs, Polynomial invariance of rings, Proc. AMS 27(1971) 247-262.
[EK]
P. Eakin and K.K. Kubota, A note on the uniqueness of rings of coefficients in polynomial rings, Proc. AMS 23(1972) 333-341.
[EH]
P. Eakin and W. Heinzer, Some Dedekind domains with specified class group and more non euclidian PID's, to appear.
[ES]
P. Eakin and J. Silver, Rings which are almost polynomial rings, to appear Trans. AMS.
[H]
M. Hochster, Non-uniqueness of the ring of coefficients in a polynomial ring, to appear Proc. AMS.
[K]
I. Kaplansky, Modules over Dedekind rings and valuation rings, Trans. AMS 72(1952) 372-340.
[N]
M. Nagata, Local Rinqs, Interscience, New York (1962).
[N,tvr]
M. Nagata, A theorem on valuation rings and its applications, Nagoya Math J. 29(1967) 85-91.
[O'M]
M. O'Malley, Power invariance of rings, to appear.
[B]
C.P. Ramanujam, A topological characterization of the affine plane as an algebraic variety, Annals of Math. 94(1971) 69-88.
[v]
W. VasconcelosOn finitely generated flat modules, Trans. AMS 138(1969) 505--512.
77
THREE CONJECTURES ABOUT MODULES OVER P O L Y N O M I A L RINGS by David Eisenbud and E. Graham Evansp
I)
Introduction.
generating K-theory,
In [E-E,I]
modules
and ideals,
all depended
we observed that several results and several results
on essentially
we showed that one of these results every ideal in a noetherian
ground ring is a p o l y n o m i a l d-dimensional
R can be generated,
:
we
by
i] could also be s t r e n g t h e n e d
polynomial examples
ring.
A brief
of results
provements
of this type.
elements. in
case the ground ring is a
We were
led to conjecture
of
that im-
in [E-E,I]
should be possible
in
our conjectures,
with some evidence
for
In section 2 we state the conjectures
ber of the known theorems w h i c h are special cases. prove that all our conjectures local rings
discussed
ring case.
In this paper we present their validity.
is a
look at the literature p r o d u c e d a number
in most of the theorems
the p o l y n o m i a l
in
R
then every ideal of
This result made us ask whether the other results [E-E,
,
in case the
showed that if
d
that
ring can be generated
ring of the form Six],
up to radical,
In [E-E,2]
- the theorem of K r o n e c k e r
can be s t r e n g t h e n e d
ring
on
about algebraic
the same arguments.
d-dimensional
up to radical, b y d+l elements-
noetherian
Jr.
of positive
number of generators
hold for p o l y n o m i a l
dimension.
and list a num-
In section
rings over semi-
We also establish a theorem
of a p r o j e c t i v e
3 , we
on the
module of rank i which is another
special case of our conjectures. We recall from [E-E,I]
some of the t e r m i n o l o g y
will use:
78
and results that we
Let Then
R
be a ring and let
~(R,M) If
p is a p r i m e
of R/p.
in M at p if
if it is b a s i c The h e i g h t
sion
of
Rp
ht(p) ~ t .
If
of the longest
fn
chain
cial
ideal of
convenience,
of b a s i c
of
we
in section
elements.
ideals,
m
is
a basic
of
R
p , of
R , dimt(p)
@t
which
R
is the length p .
two of the r e s u l t s
state
dimen-
such that
contain
The first
ele-
.
is the K r u l l
of
We w i l l
.
i d e a l of R , then
ideal p
R , ht(p)
3 •
M over R
is the K r u l l d i m e n s i o n
(M/Rm)p).
state
R-module.
result
from
concerns
it only for the
spe-
case w h i c h we w i l l use here.
Theorem Let
A:
[EE-I]
M' ~ M
every p r i m e p
p
of
a prime
> ~(Rp,
of e l e m e n t s
that we w i l l use
the e x i s t e n c e
and p
set of p r i m e
is a p r i m e
For the r e a d e r ' s [E-E,1]
m cM,
ideal
generated
of g e n e r a t o r s
M at p for every p r i m e
is the p
number
~(~,~)
of a p r i m e @t
be a f i n i t e l y
i d e a l of R , then dim(p)
If M is an R - m o d u l e ,
m is b a s i c ment
is the m i n i m a l
M
•
If
that
: Let
R
be f i n i t e l y ideal
p
generated
of
R ,
m l , . . . , m u ~ M'
(r,ml)
c R @ M
of the f o r m ml+rm~, The
other
result
be a n o e t h e r i a n R-modules,
and
w i t h dim R = d < ~ . suppose
M'
is
(dim(p)+l)-fold
generate
M'
, and if
is basic, where
ring,
then there u m' ¢ i~2 Rmi"
is that b a s i c
basic
r c R
is a b a s i c
elements
that for in
M
is g i v e n element
in p r o j e c t i v e
at
such
of
modules
M
are
unimodular:
Lemma
i [E-E,I].
generated and only
Section The Let ring.
S
If
projective
R
is a c o m m u t a t i v e R-module,
if it g e n e r a t e s
2.
an
a free direct
we w i s h to c o n s i d e r
be a n o e t h e r i a n
ring,
and
let
d = dim R .
79
and P
element
summand
The Conjectures, and a S u r v e y
conjectures
Set
then
ring,
of
of K n o w n
is a f i n i t e l y
m ~ P
is b a s i c
P .
Special
Cases.
are the following: R = S[x]
be the p o l y n o m i a l
if
i)
If
M
is a finitely
for every prime
ideal
In particular, P 2)
generated
if
p
P
such that
R , then
M
is a projective
~(Rp,~)
has a basic
R-module
~ d
element.
of rank
d , then
has a free summand.
If
P
is a finitely
and if
Q
generated
is a finitely
Q • P m Q e P' , then 3)
of
R-module
Let
M
be a finitely
of primes
of
M
R-module
generated projective
of rank > d
R-module
such that
P m P,. generated
R-module,
R such that dim(p)
n = Then
projective
max(~(R pe@ v'Mp)
can be generated
< d .
and let
@
be the set
Set
+ dim(p))
by
n elements.
All three statements become true [E-E,I] if d is replaced b y d+l. It is easy to see that given their generality, these conjectures are the strongest suppose
that
J
possible.
To see that
is a maximal
M = J ~...~ J (d-i times). for if
(rl,...,rd_l)
radical
of the ideal
ideal of height
Then
M
d
and
element,
contradicting
then
J
Krull's
i ,
let
cannot have a basic
were a basic ZRr i
this is so for conjecture
element
,
would be the
principal
ideal
theorem. To see that conjecture 2 fails be the coordinate
S3 .
it is known that ~
2 x 3 - I), and let
Then
rank P = 2 = dim R-I
then
~ m P/XP ~ S 2
which
As for conjecture
where
ideal K
elements
I
~
free. and
2
d-l,
3 , Murthy
, and
~ ~ S
R = S[x]
R ~ P m R 3.
However,
[Mur,2]
let
S
of
is free, but and if
P = ~®S R P -~ R 2 ,
has given an example
2, in a ring of the form
such that
implies
be the cokernel
If we set
but r e q u i r ~ 3 generators
Conjecture
by
is a contradiction.
of height
is a field,
~
is projective
is not
then
unmixed
d
ring of the real 2-sphere,
2 2 S ~IR[Xl,X2,X3]/(Xl+X2+ sJxI'x2'x3)-~
if we replace
I
can be generated
R =
of an K[X,Y,Z]
locally by 2
globally.
that if
R = K[XI,...,X d]
80
with
K
a field,
then every projective
of rank > d
is free.
This is a weak form of
Serre's problem. The following Corollary illustrates
the application
of conjecture
3: Corollary to Conjecture ring, and let
I
rated locally by
3 : Let
R = S[X]
be an ideal of g
elements.
R . Then
be a noetherian polynomial
Suppose that I
I
can be gene-
can be generated b y
max( d , g+dim R/I) elements. The Forster-Swan Theorem [E-E,I] that
I
implies in the above situation
can be generated by max(d+l,
g + dim R/I)
elements. It would be tempting to formulate a stronger version of conjecture 2, to parallel
[E-E,I,Thm.Aiib].But
this stronger form is false, as is
any form strong enough to imply the Stable Range Theorem for the ring K[XI,...,~], [Vas]
where
shows that
K
d+l
is the field of real numbers.
is best possible value in this case).
We do not know whether, of the conjectures,
(An example in
E(d,R)
for a ring R
satisfying this hypothesis
is transitive
on unimodular
elements
(See [ B a s s ~ ] ) . We will now ennumerate the special cases of our conjectures that we have found in the literature. The best known of these special cases is Seshadri's if
S
is a principal ideal ring, then every projective
is free.
This has been generalized by
and Murthy [Mur,I]
Serre,
to the case in which
S
a free module and an ideal.
S[X]-module
[Set], Bass [Bass,l]
is any 1-dimensional
ring with only finitely many non-regular maximal ideals; the theorem says that ar
theorem that
S[X]-module
in this case
is the direct sum of
(The freeness of all projectives
81
in case
S
is a p r i n c i p a l
ment.)
This
is p r e c i s e l y
to p r o j e c t i v e
modules.
from c o n j e c t u r e module case
ideal d o m a i n
immediately
the c o n c l u s i o n
Moreover,
i , since
f r o m an ideal
follows
[Kap,
p.76].
The
state
of our c o n j e c t u r e
in this
it is a l w a y s
from this
case,
conjecture
possible truth
i
applied
2 follows
to c a n c e l a p r o j e c t i v e
of c o n j e c t u r e
3 in this
is open. Another
is k n o w n
situation
for w h i c h
is that in w h i c h
variables in this
over a field.
case
R
the truth
of a part
is a p o l y n o m i a l
Bass
[Bass,
every p r o j e c t i v e
3]
of c o n j e c t u r e
i,
ring in an odd n u m b e r
, C o r o l l a r y 4.3,
proves
of
that
P , w i t h rank P = dim R , has a free
summand. On the n o n - p r o j e c t i v e slightly M
weaker
is a d i r e c t
(The c o n n e c t i o n is d e s c r i b e d
result,
side,
actually
sum of ideals between
in
Cot.
sional
semi-local
over
2 holds.
Bass and S ~ h a n u e l
where
S
is a p o l y n o m i a l
main.
Bass
polynomial
This
enables
it follows,
ring.
Endo
in a ring of the f o r m cipal i d e a l maximal
prove
that
domain.
ideals
M
and b a s i c
[Endo]
discussed
give
where
3 , the
cases
element. elements
statement
proved
R = S[X]
principal
ideal do-
, where
, where
S
82
SO
proves
is a and
if
if
n > d .
is free.
theorem
, where
Geramita
rows
is i m m e d i a t e
3
S
dimension,
P
of
conjecture
for
on u n i m o d u l a r
conjecture
the p r o b l e m
in w h i c h
R = S[X]
2 in case
So[XI,...,Xn]
S[X]
1 ,
is a o n e - d l m e n -
our c o n j e c t u r e
from the structure
[Endo]
S
ring of p o s i t i v e
conjecture
In [Ger], of
if
is t r a n s i t i v e
for i n s t a n c e
has a b a s i c
ring over a s e m i - l o c a l
h i m to p r o v e
As for c o n j e c t u r e
euclidean
[B-S]
ring over a s e m i - l o c a l
d = dim R , then E(n,R)
M
is free,
results
2] p r o v e s
for a
.]
2 . Endo
His
in [E-E,2]
if , in c o n j e c t u r e
in [E-E,2]
S[X,Y]
domain.
in [Bass,
7
that
R , then
the result
[E-E,1,
every projective
implies
of
We now turn to c o n j e c t u r e when
the p r o o f we gave
in case
dim S=O;
for m o d u l e s for m a x i m a l
over a ideals
is a s e m i l o c a l conjecture
is a D e d e k i n d
domain
3 .
prinfor
Davis and Geramita form
[D-G] prove it for m a x i m a l
R = S o [ X I , . . . , X n]
positive
, where
interesting
implied b y a theorem of Murthy holds if
field,
is an a r b i t r a r y
over rings of the semilocal ring of
dimension.
A particularly
3
So
ideals
R
and
is
M
Section 3 • In this
is any ideal of
2 Special
Cases
section we will R
cludes a number were m e n t i o n e d
form
3 always holds
Theorem
i .
S
conjectures
Remarks.
The hypothesis
only finitely many prime tion in the proof
section.
S
maximal height.
A version
S
is
We will also prove that con-
modules
of rank 1 .
ring with a noetherian
R = S[XI,..,Xn].
S
Set
d = dim R .
to be that
may be p r o v e d
of the primes
I using j-primes,
version
has
The only m o d i f i c a -
of Theorem
of the
of
S
etc.,
just as we will prove Theorem
some n o n - c o m m u t a t i v e
Then
S
is the replacement
by the i n t e r s e c t i o n of Theorem
spectrum
R •
can be weakened
that would be n e c e s s a r y of
, where
in
This result in-
ideals of m a x i m a l height.
Jacob saa radical
[E-E,I]
dimension.
of section 2 hold for
on
over a
special cases of the conjectures w h i c h
be a semi-local
the three
in 3 variables
R = S [ X I, .... X n]
for p r o j e c t i v e
of dimension > 0 , and let
shows that conjecture
all three of our conjectures
ring of positive
in the previous
Jecture
is
R .
establish
of the known
Let
, which
3
of the Conjectures
has the
a semi-local n o e t h e r i a n
Presumably,
[Mur,2]
the ring of p o l y n o m i a l s
the case in which
style of
special case of conjecture
of in the
I .
i , as in [E-E,I],
is also true. All three parts
Lemma 2: Let Let K ~ M
R
of Theorem
I depend on the following
be a n o e t h e r i a n
be finitely g e n e r a t e d
ring,
and let
R-modules, 83
I
lemma.
be an ideal of
and let
t
R
.
be an integer
such that for every p r i m e at
p .
Suppose
a)
(r,k)
b)
The
Then
image
in
M
Proof
of T h e o r e m
of
k
p
The
element
M/IM
k'
Let
J on
R
in
M
such that
such that
statement
S
follows
R-module
k + rk'
is
"
and
radical
shows from
that
of
S , and
dim(R/I)
the first b e c a u s e
generates
a free
summand
set
< d . a basic
[E-E,1,
1]. the f i r s t
statement,
R
contains
I , we h a v e
which
,
(M/IM)~)
w h e r e we h a v e w r i t t e n (R/I)
ing the
~
, with
M
We w i l l is a p r i m e i d e a l of
basic
of
Thus
ideal
m + m' R
of
m' = m ( m o d
M/IM
m'
c IM
This
p
by
Since
[E-E,
element.
I].
Apply-
element
such that
at all p r i m e s
shows that
I •
m + m'
"
q N S
On the other hand,
I).
modulo
to the b a s i c
w i t h ht q = d , then
q m I .
p in
P e @d-i
is b a s i c
every p r i m e
to this b a s i c
t = d-l, exists
for
~ d ,
element
reduces
there
at every p r i m e
ideal
lo I since ~ +
a
K = M and
show that
S .
p(Rp,Mp)
which
c R e M , we see that in
we n o t e that,
for the r e d u c t i o n
b e an e l e m e n t
Lemma
is b a s i c
=
< d , tl.ere exists
m c M
(l,m)
c IK
be the J a c o b s o n
To p r o v e
Let
are e l e m e n t s
basic
is basic.
an e l e m e n t
in a p r o j e c t i v e
~((R/I)~
dim
(dimt(P)+l)-fold
k ¢ K
for a l l p c @t
l:
second
and
in
The h y p o t h e s i s
l)
of
at
' K is
is b a s i c , and
exists
basic
Lemma
r ¢ R
~ R ~ M
there
I = JR.
that
P ~ @t
of
R .
If
q
is a m a x i m a l
m + m' is b a s i c
m + m'
is b a s i c
moduat
q,
as required. 2.) there
We b e g i n b y m a k i n g
exists
is e n o u g h at a time.
a projective
to be able
the f a m i l i a r
module
to c a n c e l
Thus we may
suppose
Q'
reduction
such that
the r a n k
1 free
R S P m R~P'.
84
to the
QeQ'
is free,
summands Let
case
of
Q = R: so it
Q~Q'
one
f: ReP' + R e P
be
8 the isomorphism.
(l,O)eR@~
is clearly basic,
= (r,m o) e R ~ P , then
(r,mo)
is basic.
morphism
carries
(r,mo)
a
of
a commutative
This
R @ P
diagram with exact R
~
R @ P'
~ P' ~ 0
0
~
R
-~
R ~ P
~ P
a , we again
use
[E-E,
to show that there
such that
E o + r~l
is basic
to denote
reduction
modulo
We can now apply the (r, mo+ rml)
in part basic
map
, it follows
If
~
is basic
in
P
, we have written
-
and
t = d-I
there exists
to the ele-
an element
at all primes
that the element
a map
follows
denotes P ~ R
Lemma 1 ,
m
R ~ P
taking
the map
carrying
generates 1
a =
the pattern
p e @d-I
m = mo +
m to
a free
in [Bass,
R ~ P carrying
1 to
l-r
(such maps
summand
of P), and
" As
rml+ rm 2 is
2
~ and [E-E, ml+m 2 ,
exist because, y
denotes
the
to -m , then we may take
1 + ~ ?+~+?~
3)
mI
P .
l,Cor.4]:
by
K = M
By the Lemma,
mo+rml+rm 2
I , above
and the fact
exists an element
P/IP , where
Lemma with
The rest of the proof
denotes
in
I Theorem A]
I .
e R ¢ P.
such that
in
form:
P ~ P.'
(R/I) < d
m 2 ¢ IP
We will thus obtain
~ 0
dim
ment
(1,O).
rows of the following
~
To construct that
f((l,O))
We will show that some auto-
to
0
shows that
so if
Unless n < max(d + ~(Rp,M) ) , ~ dimp=d
~
1
i +?O
/
this is the conclusion
of the
%4o
usual Forster-Swan p of dimension primes
p
theorem
d such that
of dimension
[E-E,I]. %4
< d -i
Therefore
there must
0 , and so, a f o r t i o r i , such that 85
~
@ 0 .
exist primes
there
Thus n
exist
~ d .
Now suppose exact s e q u e n c e (* )
~(R,M)
It f o l l o w s
the
pattern
is b a s i c
which
Rt =
, Lemma
i]
Rm @ P , by
[E-E,I
We w i l l t h e n h a v e
rank
t-l.
contradicts
P -~ M , so
and
K
)-fold b a s i c is
in
It is easy to see that
Rt .
Thus
contains which
of
[E-E,
a basic
r e d u c e s to
to the b a s i c k'
c IK
P
"
Since
~ @d-i
the p r o o f
tern
K
i, Thm. element [
contains
such that
k + k'
k+k' = m
is b a s i c
The main of
Rt
Instead,
at w h i c h
difference between with
p
k ¢ K
in
dim p < d ,
the i d e a l
I ,
as is
K
in
K
Rt/IR t
in
be an e l e m e n t
e x i s t s an ideals
I , it f o l l o w s as in
R t , as r e q u i r e d .
l]
Lemma follows
the p a t -
so c l o s e l y that we
we w i l l r e m a r k on the p o i n t s
before
We a s s u m e
in that
him.
L e m m a 2 of this paper,
A = R
t
dim P = d .
at a l l p r i m e
of this
in [E-E,
R
L e m m a w i t h t = d-I
c h a n g e s m u s t b e made.
a c o p y of [E-E,1]
[E-E, 1]
the
modulo
The p r o o f
of T h e o r e m A g i v e n
of T h e o r e m A
Let
is b a s i c
of i~ that
it in detail.
at
the i m a g e of
Applying
in
the r e a d e r has
A ii) b)
p
e R ~ R t , we see that t h e r e
of P r o o f of L e m m a 2 :
the p r o o f
ideal
is Just as b a s i c
IR t .
indu-
is
s u c h that
is b a s i c
w i l l not g i v e
p
K
p
R t / IR t .
is
elements.
of
primes
k
of the p r o o f
element
such that
of
(l,k)
a basic
that
P .
= t , p r o v i n g the theorem.
at e v e r y p r i m e
implies
modulo
t-I
that
at
P
Rt ~ M
t > n , it f o l l o w s
if a p r i m e
~
epimorphism
~(R,M)
contains
Rt
A]
the
Rt
module
above,
can b e g e n e r a t e d b y
that
in R t / I R t
element
element
Sketch
K
in R t
d-fold basic
then the i m a g e
M
our a s s u m p t i o n
a n d the a s s u m p t i o n
(dim p + l
R t , so that
P -- t - I > d , so b y p a r t 2),
It r e m a i n s to s h o w t h a t (*)
in
, for some p r o j e c t i v e
On the o t h e r hand,
ces an e p i m o r p h i s m
From
show that
Cor. 5
e K
This
], we w i l l
of [E-E,I,
m
of r a n k
is a short
~ M ~ 0
there e x i s t s an e l e m e n t
free
that t h e r e
of the f o r m
0 ~ K ~ Rt
Following
= t > n.
and Theorem
, is that in L e m m a 2, m I is a s s u m e d
86
i0
basic mod I , and we w i s h to produce T h e o r e m A~) w h i c h essentZally
lie in
elements
a i (in the notation
I . (The appearance
trivial change).
b y a number of a p p l i c a t i o n s
The elements of Lemma 3
of the sets
a. 1
of
are actually
[E-E,1].
of Lemma 3 , it suffices to prove that we may take elements
aj
in Lemma 3
don't matter).
Lemma 3 , in section 5 of [E-E,I] in the notation
of
@t
is an obtained
In the notation ale I
(the other
Turning to the proof
, we see that
of
aI
of
is chosen
so that,
[E-E,I], m I + aalm I
is basic
in
M
Suppose that of this
at a certain finite mI
is basic mod I, and that
list that contain
ml+aalml basic mod Thus it suffices contain
I .
list of primes
I .
pl,...,pv
Pl '''''Pv
Then any choice of
.
are the primes
ale I
makes
I, and therefore basic at each of the primes ~Y"Pv''
to deal with the primes
To do this, we first
pv,+l,...,pv
choose
aI
, w h i c h do not
in the way that the
proof
of Lemma 3 instructs u s - not n e c e s s a r i l y in I. Then we p i ~ k s¢l v such thac s ~ u . P i ; this is possible, since otherwise I w o u l d b e conl=V ' + 1 tained in one of t~e primes in the union. We can now replace a I by
sale I
and continue with the proof as given in [E-E,1].
The next theorem proof
covers a special case of conjecture
is similar to that in
T h e o r e m 2.
Let
polynomial
ring,
S
P
ring,
then
We p r o c e e d b y induction S
is artinian.
lemma,
in this case
N
on
87
of rank i .
If
If
dim S = 0 ,
radical.
By Nakayama's
P/NP).
is a direct p r o d u c t
is a direct product
be the
elements.
be its nilpotent
N = O, so that S R
d
R-module
d-1 = dim S .
~(R,P) = ~(R/NR,
Thus we may assume But
Let
and let R = S[X]
be a projective
dim R = d , then P can b e g e n e r a t e d b y
Proof.
The
[E-E,2].
be a n o e t h e r i a n
and let
3).
of p r i n c i p a l
of fields.
II
ideal domains,
so
Now suppose subset which S
P
is cyclic.
that dim S > 0 .
is the complement
Let
~ g S
of the union
be the multiplicative of the minimal primes
of
.
Then rated
SM
over
exists
has dimension S [X] = R
an element
by one element
u ¢ M
(3)
0 , so by induction, Pl¢
P~
can he gener-
P " It follows
that there
such that
uP g Rpl Since
u
dim S/(u)
is not in any of the minimal primes
< dim S •
R/(u)-module
P/uP
Thus,
by induction,
can be generated
by
of
S ,
the rank 1 projective d-I element
~2,...,~d
,
that is , d
(4) We claim that if modulo
(h) to
P2'''''Pd
are any elements
of
P
which reduce
~ 2 , . . . , ~ d , then d
p
V =
2
RPi
i=l It is enough to prove ]prime ideal
q
of
that this holds after R .
If
u ¢ q
localizing
at an arbitrary
then
d i=l follows
immediately
(5) follows
from
q
(3).
from Nakayama's
If, on the other hand,
lemma and
$8
(4).
This
u c q , then
concludes
the proof.
12
REFERENCES Bass, H., [Bass, I] "Torsion Free and Projective Modules," Trans. A.M.S. 102 (1962), 319-327. [Bass, 2] "K-Theory and Stable Algebra " Publ. Math. IHES no. 22 (196~), 5-60. [Bass, 3] "Modules Whic h Support Nonsingular Forms," J. Alg. 13 (1969), 246-252. Bass, H., and Schanuel, S. [B-S] "The Homotop[ Theory of Projective Modules," Bull. A.M.S. 68 (1962), 425-428. Davis, E., and Geramita, A., [D-G] "Maximal Ideals in Polynomial Rings." these Proceedings. Eisenbud, D., and Evans, E. [E-E,1] "Generating Modules Efficiently: Theorems from Algebraic K-Theory." j. Alg., To appear. [E-E, 2] "Every Algebraic Set in n-space is the Intersection of n Hypersurfaces." To appear. Endo, S. [Endo] "Projective Modules Over Polynomial Rings." Soc. Japan 15 (1963), 339- 352.
J. Math
Geramita, A. [Ger], "Maximal Ideals in Polynomial Rings ," Queen's University Mathematical Preprint no. 1971-56. KapLansky, I., [Kap] Infinite Abelian Groups, Revised edition, The University of Michigan, Ann Arbor, (19~9). Kronecker, L., [Kro] "Grundz~ge eine arithmetischen Theorie der algebraischen Grossen," J. Reine. Angew. Math. 92(1882), 1-123. Murthy, P. [Mur,1] "Projective Modules Over a Class of Polynomial ;Rings," Math. Z. 88 (1965), 184-189. [Mur, 2] "Generators for Certain Ideals in Regular Rings of Dimension three" To appear. Serre, J.-P. [Ser] "Sur les Modules ProJectifs," Sem. Dubriel-Pisot. t. 14 , 1960-1961, no. 2 . Seshadri, C.~[Ses] "Triviality of Vector Bundles over the Affine Space K~ '' Proc. Nat. Acad. Sci. 44(1958), 456-458. Vaserstein, L. [Vas] "Stable Rank of Rings and Dimensionality of Topological Spaces~ Functional Analysis and its Applications 5 , (1971), I02-110.
Brandeis University Waltham, Massachusetts M.I.T. Cambridge, Massachusetts
89
PRUFER-LIKE CONDITIONS ON THE SET OF OVERRINGS OF AN INTEGRAL DOMAIN
Robert Gilmer, Department of Mathematics Florida State University, Tallahassee, Florida, 32306
This paper considers ten conditions on the set of overrings of an integral domain D with identity. Each of these conditions is satisfied if D is a Prufer domain. Relations among the conditions are discussed, and several related questions are mentioned.
Let
D
be an integral domain with identity with quotient field
--Overring of
D,
we mean a subring of
of overrings of (0)
D
D,
K
containing
D.
K.
We denote by
By an ~
the set
and we consider the following eleven conditions.
is a Prufer domain; that is, finitely generated nonzero ideals of
are invertible. (i)
Each valuation overring of
D
valuation ring) is a quotient ring of
(that is, an overring of
D
D.
(2)
Each overring of
D
is a Prufer domain.
(3)
Each overring of
D
is integrally closed.
(4)
Each overring of
D
is flat as a D-module.
(5)
Each overring of
D
is an intersection of quotient rings of
(6)
If
J • Z,
If
J • Z
and if
prime ideal of
J
lying over
(8)
J • Z
and if
of
that is a
then the prime ideals of
J
D.
are extensions of prime ideals
D. (7)
ideals of (9)
If J
lying over
The set
Z
P
is a prime ideal of
dim J _< dim D
D,
then there is at most one
D,
then a chain of prime
P. P
is a prime ideal of
has at most one member.
is closed under addition--that
Jl + J2 = {Xl + x2 I x i • Ji } (i0)
P
is in
for each
Z. J
in
~.
90
is, if
Jl' J2 • Z'
then
In the sequel we discuss relations among the preceding conditions. lar, conditions D
(0) - (4)
are equivalent and they imply conditions
is integrally closed, then conditions
Conditions
(i) - (3)
(0) - (9)
and Some Related Questions
(0)
is a Prufer domain I if and only if each valuation overring of
ring of
D.
J, V • E
It then follows easily that
with
quotient ring of
J ~ V,
(5) - (i0); if
are equivalent.
W. Krull in [42, p. 554] established the equivalence of D
In particu-
and if
V
(0)
and
(2)
is a quotient ring of
and D
(1)--that is,
is a quotient
are equivalent, for if D,
then
V
is also a
J.
Since a Prefer domain is integrally closed, the implication from the equivalence of
(0)
and
(2).
(0) ÷ (3)
follows
E. D. Davis proved the reverse implication
in [12, p. 198]; he also extended his results to commutative rings
R
with identity
with few zero divisors, the definition of such a ring being that the set of zero divisors of
R
is a finite union of prime ideals of
Several questions related to the equivalence of sidered in the literature.
R. (0)
and
(3)
have been con-
A dual question that is easy to answer is:
conditions is each subring of
D
with identity integrally closed?
Under what
The answer to
this question is contained in [20, Theorem i].
THEOREM A.
field
The following conditions are equivalent.
(i)
Each subring of
(2)
D
Q,
D
with identity is integrally closed.
is an algebraic extension field of a finite field or
D
has quotient
the field of rational numbers.
iKrull uses the term "Multiplikation ring" in [42] instead of "Prufer domain". As Krull noted, his use of this term was in conflict with its meaning in older literature. Current usage of the term multiplication ring has reverted to that of the older literature--a commutative ring R in which the containment relation A ! B for ideals implies the existence of an ideal C such that A = BC (that is, "every divisor is a factor"); see [27] for more on multiplication rings. The first use of the term "Prefer domain" (or rather, "Prefer ring") that I have found in the literature appears in H. Cartan and S. Eilenberg [ii, p. 133]; I. Kaplansky repeats the terminology in [40]. The term "Prefer ring", as it is currently used, includes commutative rings with zero divisors; see [32] and [46, Chapter i0]. 9~
It follows from Theorem A that if each subring integrally closed, then each such J/A
J
J
of
D
with identity is
is a Euclidean domain with the property that
is finite for each nonzero ideal
A
of
J.
In [19, Theorems 3, 4], R. Gilmer gives necessary and sufficient conditions, reminiscent of those in
(2)
of Theorem A, in order that each subring of
D
should
be Noetherian; W. Borho [8] has investigated this question for rings (not necessarily commutative) with zero divisors, and the "Noetherian pairs" of A. Wadsworth [54] are another variant of the same question 2. In [27], Gilmer and J. Mott considered several questions of the following type: Determine necessary and sufficient conditions in order that each integrally closed subring of
D
should have property
P.
To give an indication of the results of [27],
we cite a portion of its Theorem 3.1.
THEOREM B.
The following conditions are equivalent.
(i)
Each integrally closed subring of
(2)
Either
characteristic (3)
K p ~ 0
has characteristic and
0
D and
is completely integrally closed. K/Q
is algebraic, or
K
has
tr.d. K/GF(p) S i.
Each integrally closed subring of
K
is a PrSfer domain.
In analogy with Noetherian pairs, we could, of course, define an integrally closed pair to be a pair
(R, S)
that
S
R
is a subring of
of commutative rings with a common identity such
and each subring of
S
containing
R
is integrally
closed 3 .
Flat Overrings
In [50], F. Richman proved that ring of
D
is flat as a D-module.
the structure of
D
D
is a PrHfer domain if and only if each over-
This particular equivalence is useful in relating
to that of an overring
J
of
D,
primarily because of Theorem C,
2For more on questions of the form "characterize rings R such that each subring of R has property P", see [19, Theorem 5], [25], and [30]. 3Several persons, including E. D. Davis, A. Grams, I. Kaplansky, T. Parker, and A. Wadsworth, have shared with me some observations about such pairs after the conclusion of my talk; see, for example, the next paper (by Davis) in these Proceedings. 92
which we cite later. and
(4)
At this point, we remark that the equivalence of
(0), (3),
generalizes to the case of rings with zero divisors; see [32] and [46,
Theorem 10.18]. There are several papers in the literature that touch on the subject of flat overrings.
The papers [i, 2, 3] of T. Akiba dwell on the topic, while [50], [45],
[35], [32], and [5] also contain some of the general theory of flat overrings.
As a
nice summary statement, we cite a result that appeared in a preprint of [5].
THEOREM C. T,
and let
Let
R'
R
be a commutative ring with identity with total quotient ring
be a subring of
T
containing
R
such that
R'
is flat as an
R-module. (a) and
There is a generalized multiplicative
AR' = R'
PR'
for each
Let
A
and
(b)
As : A R ' .
(c)
AR' = R'
(d)
Let
Q
is prime in
R,
and let
be a P-primary ideal of R',
QR'
(f)
If
S
in
R
such that
R' : RS
S. A'
if and only if there exists
A' = (A' n R)R'.
m
in
B be ideals of
(e)
is
R
PR'-primary,
A' = (a~, ..., a') m
bl, ..., b n • R A'
A
system 4
B ~ S
R'.
such that
such that
A ~_ B.
QR' c R'.
P = PR' n R,
and
Then
PR' c R',
Q = QR' n R.
is finitely generated, then there exist i -< i -< m,
such that for
be an ideal of
i -< j < n,
a'b. • R i ]
and
n
:
i=l j=l
a'b.R'. m ]
4By a generalized multiplicative system in a commutative ring T, we mean a nonempty family S = {I } of nonempty subsets of T (in forming generalized quotient rings, there is no loss of generality in assuming that each such that
I I 8 = {El xiYi I x i • I , Yi • 18} • S
alized quotient ring
TS
of
in the total quotient ring of
T
with respect t o T
such that
xl
for all S, c T
I ~
is an ideal of and
B.
T)
By the gener-
we mean the set of elements for some
I
in
S.
x
The no-
tion of a generalized quotient ring, which obviously generalizes the concept of a quotient ring with respect to a regular multiplicative system, seems to have originated with W. Krull [44, Section 2]. More recently, D. Kirby [41], L. Budach [9], M. Griffin [32], W. Heinzer, J. Ohm, and R. Pendleton [35], J. Arnold and J. Brewer [5], and H. S. Butts and C. Spaht [i0] have used the concept. It is interesting to note that the set of generalized quotient rings of D i n c l u d e s the set of overrings J of D that can be written as an intersection of looalizations of D [35, Prop. 4.3 and Cot. 4.4]. 93
(g)
(A n B)R'
(h)
](AR') = (/A)R'.
(i)
If
B
: AR'
n BR'.
[AR':BR'] = [A:B]R'.
is finitely generated, then
Of several homological characterizations terms of overrings, here.
H. Storrer [53] has proved that
and only if the injection map of commutative rings for each
J
D
in
into ~.
domain if and only if the D-module
Conditions
(5) - (8)
of Prefer domains, we mention one, in
J
Storrer also proved that
D[s] @D D[s]
implies
D.
(7)
are satisfied if
Moreover, part
and that D
D
is a Prefer
is torsion-free for each
s
in
K 5.
and Some Related Results
satisfied for each flat overring (6)
is a Prefer domain if
is an epimorphism in the category of
As we have previously observed, a flat overring of localizations of
D
J
(7)
(e) of
D
is an intersection of
of Theorem C shows that condition D.
implies
(5)
is
It is clear that (8).
Consequently,
conditions
(5) - (8)
is a Prefer domain.
E. D. Davis asked in [12, p. 200] if a domain intersection of localizations of alized quotient ring of
D)
D
( and
hence
D
for which each overring is an
each overring of
is, in fact, a Prefer domain.
to this question is affirmative if each prime ideal of
D
D
is a gener-
He proved that the answer has finite height.
In
[21], Gilmer and Heinzer conducted a thorough investigation into the theory of a domain
D
for which each overring is an intersection of localizations;
such a domain a QQR-domain or a domain with the QQR-property.
they called
Gilmer and Heinzer
proved that the integral closure of a QQR-domain is a Prefer domain; moreover, valuation overring of a VQR-domain) D
D
is an intersection of localizations
of
D
(that is,
and if the ascending chain condition for prime ideals holds in
is a Prefer domain.
Finally,
if each D D,
is then
[21] contains examples that show that a VQR-domain
need not have the QQR-property, and that a QQR-domain need not be a Pr[fer domain.
5A. Hattori [33] proved that D is a Prefer domain if and only if for all torsion-free D-modules M and N, M 8 D N is torsion-free; some of Hattori's results were generalized by S. Endo in [13]. 94
It seems appropriate
at this point to mention
and Heinzer
in [21] to obtain their examples.
imal ideal
M,
D
if
¢
is a subring of
structure
of
DI
ring extension such domains to provide field of
D,
then
the domains
one frequently
is realized
this reason,
D I : ~-I(D)
assumes
as a subring
of
consider
over
ring.
V;
such that Krull's
the closely related ring
({Xl}),
S
have property
example
is in a paper
and
B
amounts
arithmetisch
The
A
as a
construc-
f(X, Y)/g(X,
to taking
Y) e k.
J = k + M, k[X, Y](X)
where
R
Let Y)
k
J be
in two
In terms of a where
M
= k(Y) + M.
is a subring S[{X~}]
[A
Krull
closed domain
is the following:
f(0, Y)/g(0,
closed),
D + M
D + M
integrally
functions
ring
and for
[43, p. 670] by Krull.
this is an example
is integrally
is not endlich
use of the
of a domain with property
Krull uses the same domain as an example
M.
is, the residue
can be found in the classical
D + (X, Y)K[X,
D
and if
can be realized by
D I = D + M,
is an ideal of the polynomial ring
integrally
v-operation
J
R + B,
(in modern terminology,
where
of
valuation
domain that is not completely Y],
of
2 of [16] as the
quasi-local
and
construction
To obtain an example BA
containing
V = £ + M--that
the earliest
g(0, Y) ~ 0
then such a construction
of H. Prefer 6.
A = V/M,
and the structure
in Appendix
be the set of all rational
k
V
in this case,
His description
ideal of the rank one discrete
commutative in
J
construction,
maximal
that
of a one-dimensional
that is not a valuation a field and let
V
onto
ring with max-
have usually been used in the literature
In the form just described,
gives there an example
D + M
of
DI
tion that I have found in the literature
variables
is a subring
I refer to such a construction
construction.
is a valuation V
of
used by Gilmer
and hence a rather wide range of properties
Because
examples,
V
of
reflects both the structure
D I.
V
If
is the canonical homomorphism
A,
of
the construction
is the If we
of the contained
paper [49, p. 19] that does not
of an integrally
closed
Prufer gave the example
closed and
D ~ K.
of an integrally
In [42, p. 569],
closed domain on which the
brauchbar.
6prufer's paper [49] represents the first study of Prufer domains, per se, in the literature. With proper modesty, Pr~fer's term for an integral domain with identity in which nonzero finitely generated ideals are invertible was a domain with property I_BB. 95
Another important use of the
D + M
construction is in the paper [52, p. 604]
of A. Seidenberg; Seidenberg used such domains to prove that if tive integers such that domain
J
such that
n + i ~ k ~ 2n + i,
dim J = n
and
n
and
k
are posi-
then there is an integrally closed
dim J[X] = k.
In recent years, the
D + M
construction and related construction for polynomial rings have appeared frequently in the literature 7, and some recent papers have contained theorems concerning the structure of such constructions; see [29], [22], [16, Appendix 2], [7]. Gilmer has generalized some of these theorems to finite intersections We return to our discussion of relations between conditions
(0), (5) - (8).
One of the examples of Gilmer and Heinzer in [21] shows that conditions and
(9)
do not imply
(0).
On the other hand, conditions
(7)
In [18], n hi= I (D.I + M.).
and
(5), (6), (8)
are equiv-
alent and are, in fact, equivalent to the condition that the integral closure of is Prefer [16, Th. 16.10 and Th. 22.2].
Consequently, conditions
(0) - (8)
D
are
equivalent for an integrally closed domain. If each overring of equivalence of
(0)
and
D (2)
is a quotient ring of that
D
D,
then it follows from the
is a Prefer domain.
Such domains were con-
sidered independently by Davis [12] and by Gilmer and Ohm [28]. say that
D
has the QR-property if each overring of
A Prefer domain need not have the QR-property.
D
Following [28], we
is a quotient ring of
D.
In fact, the following theorem was
proved independently by Davis, Gilmer and Ohm, and by O. Goldman [31].
THEOREM D.
If
D
is Noetherian, then
D
has the QR-property if and only if
is a Dedekind domain with torsion class group.
In [28], Gilmer and Ohm prove that a Prefer domain power of each finitely generated ideal of strengthened this result to:
D.
A
of
has the QR-property if a
is principal; in [48], Pendeleton
The Prufer domain
if for each finitely generated ideal ideal of
D
D
D,
D /A
has the QR-property if and only is the radical of a principal
Must a domain with the QR-property have torsion class group?
This
7in fact, I have heard the statement made in jest that if an example exists, it can be realized in the form D + M. Since each domain with identity is trivially of the form, the statement is literally true, but in making this observation, I avoid the spirit of the speaker. 96
D
question remained open for several years; it was answered in the negative by Heinzer in [34].
The example that Heinzer gave (not a
D + M
construction)
is related to a
construction of Goldman in [31]. The search for a characterization quotient rings is not lost, however.
of Prufer domains in terms of overrings being Both Arnold and Brewer [5] and Butts and Spaht
[i0] have obtained such a characterization. that
D
Specifically, Theorem 1.5 of [5] states
is a Prufer domain if and only if each overring of
for some generalized multiplicative
system
S = {I }
Arithmetic Relations on the Set of Overrings of
In the terminology of [24], Jl + J2 = {al + a2 I ai ~ Ji }
D
D
is of the form
of invertible ideals of
D.
D
is a A-domain if for all
is a subring of
DS
K.
Jl" J2 e Z,
A Prefer domain is a A-domain;
this follows since a valuation ring is obviously a A-domain and since a domain
D
a A-domain if and only if
More
generally,
DM
is a A-domain for each maximal ideal
M
of
D.
is
a domain with the QQR-property is a A-domain [24, Theorem 5], and hence a
A-domain need not be a Prefer domain.
It is true, however, that the integral closure
of a A-domaln is a PrHfer domain. Condition
(9)
is distinguished
it is primarily arithmetic in nature. and only if
xy e D[x] + D[y]
for all
from the other conditions on our list because In fact, it is true that x, y e K.
D
is a A-domain if
Several characterizations
of
Prufer domains, in terms of arithmetic relations on the set of ideals, are known. As a fair summary of these results, we state Theorem E (see [38] and [16, Section 21]).
THEOREM E.
The following conditions are equivalent.
(a)
D
is a Prefer domain.
(b)
A n (B + C) = (A n B) + (A n C)
(c)
(A + B)(A n B) : AB
(d)
A(B n C) = AB n AC
(e)
A:B + B:A = D
(f)
(A + B):C : A:C + B:C
for all ideals
A, B, C
of
D
with
C
finitely
C:(A n B) : C:A + C:B
for all ideals
A, B, C
of
D
with
A
and
for all ideals
for all ideals for all ideals
for all
A, B
A, B A, B, C
of
A, B, C
of" D.
D. of
D.
finitely generated ideals of
D.
generated. (g)
97
B
finitely generated. (h)
AB = AC,
zero, implies that
for ideals
A, B, C
of
D
with
A
finitely generated and non-
B = C.
Several of the conditions of Theorem E have been considered for commutative rings
R
with identity [46, pp. 150,151].
This is especially true of condition
--the condition that the lattice of ideals of tributive.
R,
under
+
and
n,
(b)
should be dis-
L. Fuchs in [14] refers to such rings as arithmetischer Ringe, and C. U.
Jensen has investigated these rings (arithmetical rings, in Jensen's terminology) in a series of papers (see, in particular, [39]). Besides [24], a few other papers that have touched on the question of characterizing Prefer domains of
D
D
in terms of arithmetic relations on the set of overrings
are [47], [17], and [23].
Dimension Theory of Overrings
It follows from part overring of
D.
(e)
of Theorem C that
Thus condition
generally, condition
(i0)
(i0)
dim J N dim D
is satisfied if
follows from condition
D
(8).
D',
J
is a flat
is a Prufer domain; more Since dimension is pre(i0)
served by integral extensions, it is clear that condition if and only if it is satisfied for
if
the integral closure of
is satisfied for D.
But unlike the
other conditions we have considered, an integrally closed domain satisfying need not be a Prefer domain.
If
D
(i0)
For finite dimensional domains, the following theorem
gives a characterization of domains satisfying
THEOREM F.
D
has finite dimension
(i0).
n,
then the following conditions are
equivalent. (a)
dim J -< n
for each
J
in
E.
(b)
The valuative dimension of
D,
(c)
dim D[XI, ...
dim
V
D,
is
n°
X ] = 2n. n
A proof of Theorem F is given in [16, pp. 346-9]; Arnold proves a generalization of Theorem F in Theorem 6 of [4].
The notion of valuative dimension was introduced
by P. Jaffard in [36] (see also [37, Chapitre IV]);
98
dim
V
D
is defined to be
i0
sup {dim V I V
is a valuation
always holds,
and it follows
domain
D
or if
integrally
domain if and only if
where D
(i0)
dim D I
is a positive
than or equal to dim D k = k struction
and
k,
F that
condition
closed Noetherian is at most
i.
dim
domain There
closure D = i
v
proved by Seidenberg integer
and if
dim D = dim
m
dim v D k = m.
[16, p. 572].
In general,
dim v D = r < ~,
then for
but
if
D
D
is a Prefer
is satisfied DI
for each
is a Prefer case in which
is a Prufer domain--the that the integral
case
closure of
in [51, p. 511]; see also [15]. ~
or is a positive
Dk
closed domain
integer greater Dk
can be realized by a
such that D + M
con-
{dim D, dim D[XI] ,
for example,
s a r,
DI,
of
the sequence
can be wild indeed;
D
(i0)
implies
is
Such a domain
v
dim D ~ dim v D
is one nontrivial
then there exists an integrally
dim D[XI, X2] , ...} But if
The inequality
In particular,
This result--that
is a Prufer domain--was k
D}.
implies that the integral
dim D = i.
If
of
from Theorem
is Noetherian.
finite-dimensional
condition
overring
dim D[XI,
see [7, Section
5] and [6].
..., X s] = r + s
[36, Th~oreme
Resume
In summary, conditions ditions
conditions
(5) - (i0).
(5) - (9)
Finally,
integrally D
n
The integral
are equivalent, closure
and these conditions
of a domain satisfying
is a Pr{ifer domain, but the domain of Example
that the combination closed.
(0) - (4)
of conditions
for each positive
closed domain
Dn
(5) - (i0) integer
such that condition
one of the con-
4.1 of [21] shows
does not imply that
n > i,
imply
D
is integrally
there is an n-dimensional (i0)
is satisfied
for
Dn,
but
is not a Pr[fer domain.
.REFERENCES
[i]
Akiba,
T. Remarks
on generalized
quotient rings,
Prec.
Japan Acad.
40, 801-806
(1964). [2]
Remarks
on generalized
rings
of quotients
II, J. Math. Kyoto Univ.
5,
39-44 (1965). [3]
Remarks 205-212
on generalized
rings of quotients
(1969). 99
III, J. Math.
Kyoto Univ.
9,
3].
ii
[4]
Arnold, J. T. On the dimension theory of overrings ef an integral domain, Trans. Amer. Math. Soc. 138, 313-326 (1969).
[5]
Arnold, J. T., and Brewer, J. W. On flat overrings, ideal transforms, and generalized transforms of a commutative ring, J. Algebra 18, 254-263 (1971).
[6]
Arnold, J. T., and Gilmer, R. The dimension sequence of a commutative ring, in preparation.
[7]
Bastida, E., and Gilmer, R. Overrings and divisorial ideals of rings of the form D + M,
preprint.
[8]
Borho, W. Die torslonsfreien Ringe mit lauter Noethersehen Unterringen, preprint.
[9]
Budach, L. Quotientenfunktoren und Erweiterungstheorie, Math. Forschungsberichte
22, VEB Deutscher Verlag der Wiss. Berlin, 1967. [i0]
Butts, H. S., and Spaht, C. G. Generalized quotient rings, to appear in Math. Nach.
[ii]
Cartan, H., and Eilenberg, S. Homological Algebra, Princeton Univ. Press, Princeton, N. J. (1956).
[12]
Davis, E. D. Overrings of commutative rings. II. Integrally closed overrings, Trans. Amer. Math. Soe. ii0, 196-212 (1964).
[13]
Endo, S. On semi-hereditary rings, J. Math. Soc. Japan 13, 109-119 (1961).
[14]
Fuchs, L. 0ber die Ideale arithmetischer Ringe, Comment. Math. Helv. 23, 334-341
(1949). [15]
Gilmer, R. Domains in which valuation ideals are prime powers, Arch. Math. 17, 210-215 (1966).
[16]
Multiplicative
Ideal Theory, Queen's University, Kingston, Ontario
(1968). [17]
On a condition of J. Ohm for integral domains, Canad. J. Math. 20, 970-983 (1968).
[18]
Two constructions of Prufer domains, J. Reine Angew. Math. 239/240, 153-162 (1969).
[19]
Integral domains with Noetherian subrings, Comment. Math. Helv. 45, 129-134 (1970).
[20]
Domains with integrally closed subrings, Math. Jap. 16, 9-11 (1971).
100
12
[21]
Gilmer, R., and Heinzer, W. Intersections of quotient rings of an integral domain, J. Math. Kyoto Univ. 7, 133-150 (1967).
[22]
On the number of generators of an invertible ideal, J. Algebra 14, 139-151 (1970).
[23]
Gilmer, R., and Huckaba, J. A. The transform formula for ideals, J. Algebra 21, 191-215 (1972).
[24] [25]
A-rings, preprint. Gilmer, R., Lea, R., and O'Malley, M. Rings whose proper subrings have property P,
[26]
to appear in Acta Sci. Math. (Szeged).
Gilmer, R., and Mott, J. L. Multiplication rings as rings in which ideals with prime radical are primary, Trans. Amer. Math. Soc. 114, 40-52 (1965).
[27]
Integrally closed subrings of an integral domain, Trans. Amer. Math. Soc. 154, 239-250 (1971).
[28]
Gilmer, R., and Ohm, J. Integral domains with quotient overrings, Math. Ann. 53, 97-103 (1964).
[29]
Primary ideals and valuation ideals, Trans. Amer. Math. Soc. 117, 237-250 (1965).
[30]
Gilmer, R., and O'Malley, M. Non-Noetherian rings for which each proper subring is Noetherian, to appear in Math. Scand.
[31]
Goldman, O. On a special class of Dedekind domains, Topology 3, 113-118 (1964).
[32]
Griffin, M. Prefer rings with zero divisors, J. Reine Angew. Math. 239/240,
55-67 (1969). [33]
Hattori, A. On Prufer rings, J. Math. Soc. Japan 9, 381-385 (1957).
[34]
Heinzer, W. Quotient overrings of an integral domain, Mathematika 17, 139-148 (1970).
[35]
Heinzer, W., Ohm, J., and Pendleton, R. On integral domains of the form P
[36]
minimal, J. Reine Angew. Math. 241, 147-159 (1970).
Jaffard, P. Dimension des anneaux de polynomes.
La notion de dimension
valuative, C. R. Acad. Sci. Paris Ser. A-B 246, 3305-3307 (1958). [37]
Theorie de la Dimension dans les Anneaux de Polynomes, GauthierVillars, Paris, 1960.
101
n Dp,
13
[38] [39] [40]
Jensen, C. U. On characterizations of Prefer rings, Math. Scand. 13, 90-98 (1963). Arithmetical rings, Acta Math. Acad. Sci. Hungar. 17, 115-123 (1966). Kaplansky, I. A characterization of Prufer rings, J. Indian Math. Soc. (N.S.) 24, 279-281 (1960).
[41]
Kirby, D. Components of ideals in a commutative ring, Ann. Mat. Pura Appl. (4) 71, 109-125 (1966).
[42]
Krull, W. Beitrage zur Arithmetik kommutativer Integrit~tsbereiche, Math. Z. 41, 545-577 (1936).
[43]
Beitr~ge zur Arithmetik kommutariver Integritatsbereiche. II. v-ldeale und vollstandig ganz abegeschlossene Integrit~tsbereiche, Math. Z. 41, 665-679 (1936).
[44]
Beitr~ge zur Arithmetik kommutativer Integrit~tsbereiche. VIII. Multiplikativ abgeschlossene Systeme yon endlichen Idealen, Math. Z. 48, 533-552 (1943).
[45]
Larsen, M. D. Equivalent conditions for a ring to be a P-ring and a note on flat overrings, Duke Math. J. 34, 273-280 (1967).
[46]
Larsen, M. D., and McCarthy, P. J. Multiplicative Theory of Ideals, Aaademic Press, New York, 1971.
[47]
Ohm, J. Integral closure and
(x, y)n = (xn, yn), Monatsh. Math. 71, 32-39 (1967).
[48]
Pendleton, R. L. A characterization of Q-domains, Bull. Amer. Math. Soc. 72, 499-500 (1966).
[49]
Prefer, H. Untersuchungen ~ber die Teilbarkeitseigenschaften in Korpern,
J.
Reine Angew. Math. 168, 1-36 (1932). [50]
Richman, F. Generalized quotient rings, Proc. Amer. Math. Soc. 16, 794-799 (1965).
[51]
Seidenberg, A. A note on the dimension theory of rings, Pacific J. Math. 3, 505-512 (1953).
[52] [53]
On the dimension theory of rings II. Pacific J. Math. 4, 603-614 (1954). Storrer, H. H. A characterization of Prefer rings, Canad. Math. Bull. 12, 809812 ( 1 9 6 9 ) .
[54]
Wadsworth, A. Noetherian pairs and form.,
Thesis,
Chicago,
the 1972.
102
function
field
of a q u a d r a t i c
INTEGRALLY
CLOSED PAIRS
E d w a r d D. Davis D e p a r t m e n t of M a t h e m a t i c s S t a t e U n i v e r s i t y of New Y o r k at A l b a n y A l b a n y , N. Y. 12222 U.S.A.
In the d i s c u s s i o n f o l l o w i n g G i l m e r ' s a d d r e s s K a p l a n s k y i n t r o d u c e d the n o t i o n " i n t e g r a l l y c l o s e d pair", r a i s e d the q u e s t i o n of c h a r a c t e r i z i n g t h e s e p a i r s and c o n j e c t u r e d an a p p r o x i m a t e answer. Mott wisely conjectured t h a t the m e t h o d s of Davis' t h e s i s [i] m i g h t w e l l p r o v i d e an answer. This n o t e e s t a b l i s h e s the v a l i d i t y of b o t h c o n j e c t u r e s , at l e a s t in the N o e t h e r i a n case, and s o m e w h a t m o r e g e n e r a l l y , for K r u l l d o m a i n s .
INTEGRALLY
C L O S E D P A I R S OF K R U L L D O M A I N S .
cellent discussion
of " K r u l l d o m a i n s "
The r e a d e r w i l l
in G i l m e r ' s
r e g a r d the n o t i o n
as so m u c h e s o t e r i c a ,
closed Noetherian
domains"
book
(A,B)
closed
if A is a s u b r i n g of B and all i n t e r m e d i a t e
and B)
are i n t e g r a l l y
closed.
ideals
of A, Ap m e a n s
the l o c a l i z a t i o n
the i n t e r s e c t i o n parenthetical theorem
c l o s e d if, B algebraic height
observation
is this w r i t e r ' s
THEOREM
i.
A b i t of n o t a t i o n :
of the l o c a l i z a t i o n s concluding
rings
of A w i t h
respect
(2) is unique, (Consequently:
consisting
the p a i r
is i n t e g r a l l y
(A,B)
of A w i t h such.
closed,
of m a x i m a l
ideals w h i c h
t h e n B is a l o c a l i z a t i o n
of i n v e r t i b l e m a x i m a l
is c o m p l e m e n t a r y
ideals.)
10.3
(i) A a f i e l d w i t h respect
to a set of
The e x c l u d e d set in
t h o s e m a x i m a l M for w h i c h MB = B.
For A N o e t h e r i a n b u t not a field,
integrally
The
conjecture.
only maximal
of e x a c t l y
of P.
of K a p l a n s k y ' s
(2) B a l o c a l i z a t i o n
that excludes
to P -- i.e.,
of the f o l l o w i n g
and o n l y if, one of the f o l l o w i n g holds:
1 primes
(including A
for P a set of p r i m e
the s t a t e m e n t
understanding
"integrally
is i n t e g r a l l y
of A at the p r i m e s
For A a K r u l l domain,
o v e r A;
[2]; s h o u l d he
he may s u b s t i t u t e
The p a i r of d o m a i n s
f i n d an ex-
if the p a i r
of A w i t h
to a u n i q u e l y
respect
(A,B)
is
to a set
determined
set
-
proof
The
progresses
via
blanket
assumption:
algebra
over
closed,
B is n e c e s s a r i l y
case
(i) ; h e r e a f t e r
system
a sequence
(A,B)
a domain
-
of
lemmas
is i n t e g r a l l y
abounds
we
2
closed.
in s u b a l g e b r a s
algebraic
assume
over
A not
for w h i c h Since
which
A.
This
a field.
ring between
A S and B S is of the
ring between
A and
B;
therefore
the p a i r
this
freely
We
LEMMA
shall
i.
We
A,
element
o v e r A; we
lose n o t h i n g
A[x]
is for
a ring
we
following
nonunit height
whence
A[x]
If A is
assume points:
of A lies 1 primes;
if,
and o n l y
The
existence
from these
if, and
points
a consequence
the
local
closed,
local
N for w h i c h
proof
every
A = Ap,
and
N contains unicity
from
of
lemma
domain
for M; Ap
A[x]
M and
over
observe
Consequently
Lemma.
i/x is in A. and
and
[i]
in any
for
Thus:
case
104
further
B
B = AN
1 prime;
N subsets
M of
P such
P is of
details. comment
domain;
the
in
set
each
of A's
P, A M c o n t a i n s
ring that
only
x lies
explicit
of A is a K r u l l
M excludes
of:
that
use w i t h o u t
is a v a l u a t i o n
that
of A,
= A[zx].
then
case,
1 of
height
a subset 2;
B is a l g e b r a i c
the o b s e r v a t i o n
localization one
of A.
NB ~ B.
of t h e o r e m
A a Krull
comment.
y in A and x i n t e g r a l
Nakayama's
of A in the
integrally
explicit
z ~ 0 a nonunit
and x in B-A,
in at l e a s t
and
For
= A by
i/x is in A f o l l o w s
consult
Henceforth
in A.
of
f o r m C S for C a
field
Since
f o r m x/y w i t h
is i n t e g r a l l y
s e t of p r i m e s
That
A [ x 2] --
the
A[zx]
of q u o t i e n t s
N the
Proof.
2.
the
A local.
disposes
is also
and w i t h o u t
in the q u o t i e n t
taking
t h a t x lies
= A+zA[x];
LEMMA
by
of B is of
show
that because
fact
B is c o n t a i n e d
Proof. each
use
integrally
S a multiplicative
every
(As,B S)
a
the p o l y n o m i a l
remark
For
now make
are n o t
in A,
closed.
we
for e a c h
AN
P in
P.
B = A M follow
maximal
ideals
is
-
3.
LEMMA
Proof. of
-
If A is l o c a l and B ~ A,
S i n c e B ~ A, we
P excluding
of A i m p l i e s
3
t h e n A is of d i m e n s i o n
i.
lose n o t h i n g by a s s u m i n g B = A M for M a s u b s e t
a s i n g l e p r i m e P.
Then
that B is a r i n g of q u o t i e n t s
that B = A x for x in P but in no m e m b e r of M.
Were
there
a n o n u n i t y n o t in P, t h e n y / x w o u l d be in B-A, b u t x/y n o t in A -- a contradiction
LEMMA pair
4.
(A,AN)
Proof.
of l e m m a 2.
For
Finally
the s u f f i c i e n c y
N the set of n o n m a x i m a l
is i n t e g r a l l y
ideals
Careful examination
the p r o o f of l e m m a
s e l e c t a here,
elsewhere.
Call
pair
Except (A,B)
grally
is m e r e l y
closed.
c l o s e d if, valuation and o n l y
2.
if,
ring,
For A local,
and o n l y ring;
general
setting;
closed pairs
A a valuation
THEOREM
relatively
integrally
if, B = Ap
if
rings
-- m o s t
results;
we
for p u b l i c a t i o n
integrally integrally
closed closed
an i n t e g r a l l y
closed
c l o s e d p a i r w i t h B inte-
The c e n t r a l p o i n t of the g e n e r a l i z a t i o n
integrally
examples:
more general
case of A a field,
a relatively
2 is v a l i d in the m o r e atively
(A,B)
all i n t e r m e d i a t e
for the t r i v i a l
in M~A;
of the a b o v e a r g u m e n t s
2 -- r e v e a l s
a p a i r of d o m a i n s
If M ~ A is
In e i t h e r case C M = A M ~ A.
l e a v i n g the d e t a i l e d e x p o s i t i o n
if A is a s u b r i n g of B w i t h in B.
to P, the
r i n g C.
of N c o n t a i n e d
M ~ A is in P, then AM~ A is a v a l u a t i o n ring.
present
belonging
i d e a l of an i n t e r m e d i a t e
n o t in P, then A M ~ A = A M for M the m e m b e r s
importantly
(2):
closed.
Let M be a m a x i m a l
GENERALIZATIONS.
of
whence
in the
is t h a t l e m m a
local case,
rel-
are c o m p o u n d e d o u t of the two o b v i o u s
a n d the t r i v i a l e x a m p l e
the p a i r
(A,B)
of A = B.
is r e l a t i v e l y
for P a p r i m e w i t h P = PAp
for A not a field,
the p a i r is i n t e g r a l l y
in a d d i t i o n A is i n t e g r a l l y
105
closed.
integrally and A/P
a
c l o s e d if,
-
We
now p l a y a b i t l o o s e w i t h
fast-Noethersche composition of h e i g h t primes
and a l s o
associated
theorem
3.
in the s e n s e
For
closed
s i o n i).
i.
Since
(A,B)
if, and only if,
decomposition
domain
de-
to p r i n c i -
is e i n b e t t u n g s f r e i ,
A,
the p a i r
(A,B)
of p r i n c i p a l
ideals
is u n i q u e
if,
and o n l y
to a set of m a x i m a l
determined
ring
for A e i n b e t t u n g s f r e i .
closed
re-
that excludes (of d i m e n -
ideals M for w h i c h MB = B can serve
integrally
respect
is r e l a t i v e l y
B is a l o c a l i z a t i o n of A w i t h
as
(Therefore:
if, B is a l o c a l i -
ideals
ideals M for w h i c h A M is a v a l u a t i o n
set is u n i q u e l y
a primary
if its only
ideals M for w h i c h A M is a v a l u a t i o n
is r e l a t i v e l y
set of m a x i m a l
admits
a domain
and at m o s t a f i n i t e n u m b e r
a Krull
fast-Noethersche
set, w h i c h
z a t i o n of A w i t h
Call
case Of:
The set of m a x i m a l
the e x c l u d e d
[3]:
nonunits
of the p r i m a r y
s p e c t to a set of a s s o c i a t e d p r i m e s only such maximal
of K r u l l
call s u c h a d o m a i n e i n b e t t u n g s f r e i
1 is a s p e c i a l
integrally
two n o t i o n s
lies in at l e a s t one
are of h e i g h t
THEOREM
-
if e a c h of its n o n z e r o
1 primes;
pal i d e a l s
4
complementary ring;
this
to a
latter
for A e i n b e t t u n g s f r e i . )
REFERENCES
[1] Davis,
E. D.
Overrings
closed overrings,
[2] Gilmer,
R. W.
of c o m m u t a t i v e Trans.
Multiplicative
[3] Krull,
W.
Ontario
Math.
-- No.
II.
(1964)
Integrally 196-212.
Queen's
12, Q u e e n ' s
Papers
on P u r e
University,
(1968).
Einbettungsfreie,
Oberringe,
ii0
I d e a l Theory,
and A p p l i e d M a t h e m a t i c s Kingston,
A.M.S.
rings.
fast-Noethersche
Nachr.
21
106
(1960)
Ringe
319-338.
und ihre
NOETHERIAN
INTERSECTIONS
OF INTEGRAL
DOMAINS
II
William Heinzer* Purdue
If
R
and
would
like
that
D = R~V
that
R
are integral
to c o n s i d e r
or
V
A sample
If
when
R
R
ring and
if
K?
(ii) when
that
V
that
are
R
V
of
as in (i),
quasi-local,
and was
extended
to the p r e s e n t
and myself.
Access
Section in S e c t i o n
of w h e t h e r
i.
Some
Throughout quotient called *This
our results
remarks
this s e c t i o n K.
R
The
R
radical
research
supported
was
of
and
is a rank
the case w h e n form
V
one
then
of
if
denote
out
(i)
and
to us in our
Ohm.
(ii);
while
(but do not resolve)
noetherian
domain,
R} again be n o e t h e r i a n . of integral
integral
domains ideals
R.
107
by Jack Ohm
jointly with
of the m a x i m a l
by NSF Grant
is
to Jack Ohm in that many
on i n t e r s e c t i o n s V
R
jointly
was u s e f u l
and also discuss
intersection
the J a c o b s o n
R
is n o e t h e r i a n
on q u e s t i o n s
one prime
and
if
D.
a 2-dimensional
is a h e i g h t
general
field
(iii),
for
conclude
is the following:
V
D
manuscript
given here were w o r k e d
1 contains
R* = N { R p I P
for
I am also i n d e b t e d
2 we deal with
the q u e s t i o n must
to P r e k o w i t z ' s
results
say
are n o e t h e r i a n .
Prekowitz
of the other
we
can one
is i r r e d u n d a n t ,
to Bruce
of the result.
K,
is also n o e t h e r i a n ?
and
D = R~V
1 is due
extension
can one
questions
radical
field when
(iii)
D = R~V
localizations
and
(i)
D = RNV?
Jacobson
such
quotient
questions:
of
the same h y p o t h e s i s
and only
with
can say on these
has n o n z e r o
both
Result
can one say
of w h a t we
With
field
is a l o c a l i z a t i o n
valuation
2)
domains
the f o l l o w i n g
has q u o t i e n t
are n o e t h e r i a n ,
i)
V
University
GP-29326AI
domains. with of
R
is
2 i.i Lemma. Jacobson Proof:
If
V
is a v a l u a t i o n
radical It will
field of
D.
a ~ V,
then
I/(l+a)
e
suffice
If
1.2 Remark.
ring
V
k
clear
{V }
as
{a e All + ra
that
any e l e m e n t
implies in
radical
radical
is not
radical
zero,
that
D = RNV
X
field
K.
in the q u o t i e n t
If
a e J
in
V,
of
T
and
so
and
that n o n z e r o
radical. ideals
in
V
a rank
has q u o t i e n t
of subrings
then
T =
~ V
one
field K.
R = k[X],
in
A
for all
is a unit
again has For R
D or
T.
r e A}, V
and
This
is
of a ring
A
and the fact is a unit
in
T.
R and V
b o t h have n o n z e r o
quotient
field
having V
of
ring
the p r o p e r t y
radical
in each
if
of some has
radical
of the J a c o b s o n
from 1.3 that
D = RNV
for
= k.
in the J a c o b s o n
which
even
an i n d e t e r m i n a t e ,
J ,
is a unit
Jacobson
quotient
contract
K, then
field
D
K
to n o n z e r o
ideals
D. This yields
the f o l l o w i n g
1.5 P r o p o s i t i o n . K
which
is a c o l l e c t i o n
It follows
has n o n z e r o
has n o n z e r o
is c o n t a i n e d
in
from the c h a r a c t e r i z a t i o n
Jacobson
J
R
R
quotient
a e D.
D = RAV
is c o n t a i n e d
1.4 Remark.
and
then
is a field, then
If
~ J
K
a e V,
has J a c o b s o n
has J a c o b s o n
that
has
that
it need not follow
V = k[I/X](I/X),
if
is a unit
R
if
1.3 Remark.
and
on
= D.
If
For example,
D = RAV
to show
a e J
1 + a RAV
valuation
J, then
ring
and if
has
R
quotient
If
n 1 {Vi}i=
has n o n z e r o field
extension
of I.i.
is a finite
Jacobson
radical,
set of v a l u a t i o n then
D = RN(
rings
on
A ni=l Vi)
K.
In c o n n e c t i o n w i t h
i.i and 1.5, we w o u l d
like
to raise
the
following: 1.6 question. with
quotient
If field
R
and
V
K, must
are
1-dimensional
D = RNV
108
quasi-local
have q u o t i e n t
field
domains K?
3 The following l-dimensionality 1.7 Example. Let
simple
example
assumption
Let
k
shows
the necessity
in 1.6.
be a field and let
R = k[X, X2y]
and
2-dimensional
and are subrings element
of
R
the property Y-degree,
of the formal power
Given that conditions natural
on
ideals
R V
of
quotient
V
of
V
is a G-domain,
follow
to be almost V-module
that an element
element
of K
Proof.
V* if
to
D*
D
D*
radical
V
field k(X, Y) = K,
k[[X,
Any
Y]]
is greater
has
than
V
generated
has smaller
if
V
V[e]
field
quotient
[7, p. 1210
radical
field.
may be assumed
of
V
and
It would
to be an
in
K
and let
of the nonzero N
V[~]
is said
is nonzero.
with
every
has quotient
has quotient N
is a nonzero
field K.
be the con-
primes
K, it will suffice
109
V
in a finite
D = RAV
D* = R A V *
field
so
field of
VCV*CK
V, then
of the intersection
has quotient
Jacobson
prime
if the
ring extension
is contained
and
only if)
V * is a G-domain,
a
is a G-domain
if the nonzero
of the quotient
over
has quotient
K,
rings.
is a G-domain integral
V
for
field
or equivalently
has nonzero
~
and looking
has quotient
if the conductor
(and obviously
VCV*CK,
To see that
If almost
Suppose
traction
over
or equivalently
1.8 Proposition.
Since
R
D = RNV
integral
are
is greater than the
to be that
intersection,
of rank one valuation
We recall
field
seems
from 1.8 that in such an example
intersection
in
Y-degree
D = RAV
is a finitely
but
V
k[[X, Y]].
series
the
Jacobson
V
I am aware of no example where V
ring
is said to be a G-domain
have nonzero
field of
series
V
that
to put on V
R and
= k.
in order
domain
.
(xyZ,y)
of each monomial
has nonzero
hypothesis
(an integral
R nv
be indeterminates.
Y]
as a formal power
and for any element Thus
Y
local rings having quotient
that the X-degree
the X-degrse.
and
'
regular
regarded
X
V = k[XY 2
(x,xZy) both
of the
of
ideal
V*. of
D*.
to show that
N
4
is contained
in the quotient
of
n,
n
~,
say
and
a
field of
n+l
field of
is contained are in
RAV
precisely
If
V
= D, so
G-domains
R
valuation
R
V
constructed
V
in
V[~].
Thus
is in the quotient
Let
obvious
V~z
has quotient VI,...,V n
Vi
is contained
valuation
of
K.
R
has quotient
has quotient
K.
Apply
if there
in
[14].
extension
almost field
field
of rank one quasi-local
rings have been Perhaps
these
of 1.8. with quotient
integral K
field
over
field
V..z
if (and obviously
K
Then only if)
K. with quotient
in only a finite number
field
exist
Jacobson
K.
field
K
of rank one
radical,
In particular,
then
V 1 A...NV
n
1.5 and i.i0.
1.12 Remark. ideals
containing
of 1.8,
is
1.6.
has nonzero
D = RAV In... AV n field
rings
of 1-dimensional
be G-domains
such that each
If
V
are intersections
be G-domains
has quotient
rings
over
does not have quotient
and Ribenboim
VI,.. , V n
Let
then it is easily
of rank one valuation
nv n
I.ii Corollary.
integral
examples
with
nV*n
almost
RAV
V iCV~CK
D* = R A V ~ A . . .
G-domain,
in answering
the following
D = RAV In...
A 1-dimensional is a G-domain,
VI,...,V n
quotient
then some power
V
closed
application
exist which
in [i0]
would be useful
and suppose
K
by repeated
are intersections
i.i0 Corollary.
if
of
n+i/n
of the rank one valuation
Interesting
by Nagata
We note
of
such that
and
rings.
which
examples
Hence,
and
K, then such
prime
a = ~
is an integrally
the intersection
[4, p. 359].
Proof.
~ e N,
in the conductor
seen that the set of elements
domains
If
D.
1.9 Remark.
V
D.
field
domain with
so for example,
are 1-dimensional K, then
only a finite number
semi-local
Vl n . . . N V n
110
it follows
from i.ii
(noetherian)
has quotient
of
field
domains K
that with
and in
5 fact
RAVIN
Jacobson
... N V n
to the q u e s t i o n
field
K
whenever
of w h e n
R
or
R
has n o n z e r o
V
is a l o c a l i z a t i o n
D = R ~V.
1.13 P r o p o s i t i o n . a valuation D = RNV
Let
1 + x
V
R
P
denote
then
y / ( l + xy)
of
R,
are in
D
of
e P + yD P
R
rank one v a l u a t i o n x e RkV
and
and
R
ideal Proof.
of
ideal of
ring such
If
R
empty.
Let
b ~ 0.
Since
sufficiently Moreover,
xn
of
then
I/(I + xy)
for
and
x e J\V,
V
then
e (DNJ)\P.
is a u n i t
R
D.
If
and x/(l+x)
Jacobson
in
P
If
and
R.
Hence
R
is m a x i m a l .
is
If
x ~ J\V,
then
and h e n c e
a unit
radical
R ~ V,
then
xny ~ V
has n o n z e r o D = RNV
J
and
J ~ V.
in
D.
for some p o s i t i v e
Jacobson
D, then
V
V
r a d i c a l J,
is c e n t e r e d
is a
For if
is an i r r e d u n d a n t
integer
V
n.
is
intersection, on a m a x i m a l
for some m u l t i p l i c a t i v e
of the i n t e r s e c t i o n If
y e V,
large p o s i t i v e in
Hence
~ = a/(x n + b)
Since
y = a/B, y e Dp,
R
and so
and
b E J
B = b / ( x n + b) Dp = V.
111
xnv
It r e m a i n s
a, b e J A D , n
is a u n i t
D
a in
is a u n i t and
to s h o w
in
is non-
for
xn + b
are in
S
SAP
where
b / ( x n + b) imply
system
implies
y = a/b
quasi-local,
integer and
D = RNV then
is 1 - d i m e n s i o n a l
a unit
is a l o c a l i z a t i o n
is
D.
then
R = DS
x ~ SAP. V
V
D.
J
Dp = V.
By a s s u m p t i o n
Irredundance
and
to s h o w that
that
of
radical
ideal of
on
x c JXV,
has n o n z e r o
and
V
e P
in b o t h
quasi-local,
D
R
and a = i/(i + x)
is a l o c a l i z a t i o n P
then
of
It remains
y e J, y ~ 0,
1.15 P r o p o s i t i o n . 1-dimensional
Jacobson
I/(I + xy)
is a u n i t
If
if
and
D.
is a m a x i m a l
1.14 Remark.
I/(l+x) and
z ~ D\P, y e ( D N J ) \ P ,
Thus
J ~ V,
the c e n t e r
y ~ V,
a localization
a + yz
has n o n z e r o
is c e n t e r e d on a m a x i m a l
is a u n i t
y e R\D,
If
ring s u c h that
and
Proof.
D.
quotient
radical.
We turn now of
has
that
V. in R.
B ~ P. P
is
6 maximal. then
We note ~z + x
P + zD = D,
1.16
~ = b/(x n + b)
is a unit so
We note
that
P
in
and
is a m a x i m a l
the f o l l o w i n g
Cor.ollary.
R
If
R
ideal
on
N V n.
ideals
and are l o c a l i z a t i o n s
D
We w o u l d
like next
1.16 by w e a k e n i n g rings.
The
Moreover,
and help
1.17 E x a m p l e
where
of c h a r a c t e r i s t i c
of
zero
T = k[X, Y](y)
= k(X)
R = k(X 2) + M
and
and
easily R
V
and
V
which
1-dimensional are not
Let
X2)
R = k(Xl)
+ M
X1
and
and
localizations
of
Also
the
quotient
R
and
yields V
a
of
One
+ M,
closed
are d o m i n a t e d
having
Let of
T.
= k + M,
can also
Let and
in this way
closed
domains
Again begin with
where
F
is a field,
independent R
and
domains
that by
and
be a field
ideal
integrally
then
1.19.
domains
k
D = RAV
algebraically
D = RAV
local Let
RNV.
F + M
Note
on such
of P r o p o s i t i o n
282].
k + M.
integrally
= k + M.
some bounds
Then
of
V
a say
over
k.
are
and are not
1.8 a c o n s t r u c t i o n
the same q u o t i e n t
by a common v a l u a t i o n
of
field.
ring of their
field.
1.18 E x a m p l e localization Let
RNV
X2
on m a x i m a l
are rank one v a l u a t i o n
indeterminates.
quasi-local
V = k(X2)
quasi-local
form always
of
of
for g e n e r a l i z a t i o n
is the m a x i m a l
localizations
l-dimensional
this
be
M
ring of the form
with
Vi
[2, p.
where
Thus
VI,...,V n
are c e n t e r e d
the h y p o t h e s e s
X, Y
localizations
rank one v a l u a t i o n F = k(Xl,
+ M,
Vi
are 1 - d i m e n s i o n a l
and let
and
is a l o c a l i z a t i o n
indicate
D = RNV
D.
and 1.15.
radical
R
that the
V
in
D.
V = k(X 2 + X) + M.
are not
construct
and
a unit
1.14
possibilities
to justify
R
are not l o c a l i z a t i o n s
R
to examine
two examples
generalization
Jacobson
irredundant
the h y p o t h e s i s
following
of 1.13,
of
If z e D\P,
D.
K, then
D = RNV IN... of
in
consequence
rings
(JND)\P.
V, and hence
has n o n z e r o
are rank one v a l u a t i o n
is in
r = k[X,
to show that of
D.
Let
Y] (y) = k(X)
k
in 1.13,
if
be a field + M,
where
112
JCV, and let M
then X, Y
R
n e e d not be a be i n d e t e r m i n a t e s .
is the m a x i m a l
ideal
of
T.
7
Let
R = k[X]
radical
of
+ M
R,
and
V
is a valuation
is not a localization 1.19 Proposition. a G-domain
for each
Proof:
If
rings
i, then
Since
the maximal maximal
R
K, say
ring of
K
valuation
ring of
V.
is integral
the conductor
Let
V i.
of
= k + M,
so
of
R
of
It follows
such that
s
K
s
and some power of
V
in
Thus
si
every nontrivial
V.
sn
sn e V A R
is in
is in the
is in the maximal
s, say
say
in a rank one
containing
V
V[s].
Then
is contained
that
ring of
is
R ~ Vi
D' = R A V I A . . . A V n ,
is a G-domain,
V
V
D = RAV.
si ~ S
V
containing
J, and
and if m o r e o v e r
s = Sl...s n.
Since
valuation
over
is the Jacobson
D = RAV
VI,...,Vn,
there exists
V i.
ideal of each
M
in only a finite number of rank
is a localization
valuation
and
is a localization
ideal of
Then
Jacobson radical
is contained
R ~ Vi,
every nontrivial
ring,
has nonzero
V of
+ M.
D.
R
R
By 1.16,
R = D~.
of
such that
one valuation
V = k[I/X](I/X)
ideal of
Therefore
s
is contained
in
and
i/s n e R.
Hence
D[I/s n] = R A V [ I / s n] = R. 1.20
Corollary.
1-dimensional number each
i
Proof:
implies
Apply
however,
such that
V
R
and
V
radical
is contained rings
of
are both localizations
ideal of
and
V
is
in only a finite
K, then of
D
R ~ Vi and
V
for is
D.
1.19 and 1.15. the assumption
can happen for group that
has nonzero Jacobson
of rank one v a l u a t i o n
on a maximal
Without
V
R
quasi-local
VI,...,Vn,
centered
but
If
RAV
V
that
R
has nonzero Jacobson
a rank one v a l u a t i o n = D
is irredundant,
is not a localization the following
of
corollary
D
it
ring not having rational value D
again has quotient
[12, p. 330].
to 1.13 and 1.14.
113
radical,
field
We can give,
K,
8 1.21 Corollary. irredundant centered Proof. RAV
Dp
on a height
that
would
would
imply
V
of
so
one v a l u a t i o n
again has P
of
ring,
quotient D, then
is e q u i v a l e n t
RS~V RS
= Dp.
is also
have n o n z e r o
Since
Thus
Dp.
is an
K, and
V
is
irredundance
R S = K; P
and
Hence,
if
one prime, R S ~ K,
by 1.13
It follows
of
RS ~ K
is a h e i g h t
radical which
of
= D
Dp = V.
a G-domain.
Jacobson
is a l o c a l i z a t i o n
to
RAV
field
Dp = R S ~ V S = R S A V .
Dp = V
is a G-domain, RS
D
then
irredundance
then
is a rank
one prime
S = D\P,
implies
and
V
intersection,
If
implies
If
that
and 1.14 RS = K
Dp = V.
2.
Noetherian
sections.
We
on n o e t h e r i a n to denote
can now q u i c k l y properties
integral
2.1 Theorem.
If
of
domains
R
rank one v a l u a t i o n NV n
corollaries
and a q u e s t i o n
give
rings w i t h
We
quotient
has n o n z e r o
is n o e t h e r i a n
some results
D = R~V. with
if
R
continue
to use
of
R
[5]
and
V
K.
radical
field
inter-
in the spirit
field
Jacobson
quotient
if and only
on n o e t h e r i a n
K,
and the
and
VI,...,V n
then
are
D = RAVIN
irredundant
Vi
...
are
noetherian.
Proof. so
D
By
1.16,
noetherian
Also by 1.16, D.
R
Thus
and the
implies
each
I.i0]
then R
D
2.2 Theorem.
If
1-dimensional
quasi-local
VI,...,V n
R ~ Vi
for each
V. z
and
are n o e t h e r i a n .
1
are
yields
is c e n t e r e d that
if
localizations
R
Vi
D,
are n o e t h e r i a n .
on a m a x i m a l
and the
of
ideal
irredundant
of Vi
is n o e t h e r i a n .
such
Jacobson
that
V
then
D = RAV
114
radical
and
is c o n t a i n e d
of rank one v a l u a t i o n i,
V.
and the i r r e d u n d a n t
has n o n z e r o
number
V
R
irredundant
[S, T h e o r e m
are n o e t h e r i a n ,
irredundant
rings
of
is n o e t h e r i a n
V
is
in only
K,
a finite
and if m o r e o v e r
if and only
if
R
9 Proof:
By 1.20,
noetherian centered system Q
implies
of
2.3 Remark. first
Example
local
local
and
overring
(noetherian)
2.4 Corollary.
Proof.
overring
If
D = RAVIA is centered
ideal of
R'
quasi-local Therefore overring D = RAVIN
... n v n
D
is noetherian;
for
2.2, we
and
being
D
V
being
R
the hypothesis that
R
1.19]
1-dimensional an
are semi-local
that
Finally, R
and
V
and
V
have
are 1-dimensional
is not noetherian. to Theorem
with nonzero
(noetherian) D = R~V
such
ideal
P
P, then R'CW, that
induction
is noetherian.
115
2.2.
Jacobson that
R
is a noetherian ideals
By 2.2 and 1.20,
insure
of Theorem
ring of rank > i.
are the maximal
< K, and a simple
if
It now follows
[5, Example
and
corollary
semi-local
such that
is
i.i]),
such as
and
D
is noetherian
from
D V
it is easy to construct
where
D = RNV
our hypotheses
R
domains
< K, it can happen
on a maximal
domain
with
Also
that without
distinct
RCDQ.
then
example
D = RAV
. . . N V n.
[5, Lemma
generalizations
is a valuation
Pl,...,Pn
so
1.8].
A simple
< K, then
D,
for any multiplicative
then
are noetherian,
D = RAV
R
Since
Q ~ P,
the following If
D.
of
Also by 1.20,
(see for example
possible
and
is 1-dimensional
no common
of
(noetherian)
V
1.17 shows
We can give
Then
V
[5, Corollary
of the form
no common
P
but not noetherian.
(noetherian)
V1
and
localizations
are noetherian.
such that
that one can have
example
V
ideal
the following.
quasi-local
are both
and
Concerning
2-dimensional
V
D R
apply
V
D S = RSAV S
of
that if
example,
shows
R
D,
is a prime
note
and
on a maximal S
easily
R
of
of
RCR~.
Hence
and
V2
RCW
is a prime W or
is any VICW.
have no common
argument yields
have
V i ~ Vp.. i is noetherian,
Q
if
V
and
domain.
V, let
and if
then either R'
and
R' = R N V 1 R'
radical
that
i0 2.5 Remark. Jacobson maximal then
The simplest
radical ideals.
are,
has nonzero
However,
if
~
radical
of
RS,
radical
rings.
Thus
noetherian
domain
T
can be realized
noetherian
domain
R
with nonzero
R
ideal of
R
easily yields
in
and
R,
~R S
noetherian spectrum
out to me,
Jacobson
radical
the maximal
RS
containing
the maximal
Also as Graham Evans pointed
formal power series
ideal
so the localization
for w h i c h
simple way to obtain domains with nonzero
R
subset of
Since no prime
Jacobson
large dimension.
taking
is a nonzero
this sort of localization
domains with nonzero
domains with nonzero
those with only a finite number of
is a m u l t i p l i c a t i v e
Jacobson radical. RS,
of integral
in general,
in the Jacobson
is lost in
has
of course,
S = {i + ala e ~}
is contained
examples
a very
is by
spectrum of any
as the maximal
spectrum of a
Jacobson radical by setting
T[[X]].
=
We should like to conclude with some question,
cf. Krull
2.6 Question. R~ =
A{RpIP
If
comments
on the following
[8]. R
is a 2-dimensional
is a height
one prime
of
noetherian R},
domain
then must
and
R~
again be
noetherian? Ferrand R
and Raynaud prove
is a 2-dimensional
R~ =
~
{Rplht(P)
It follows
= i}
over
R
then height
In fact,
closure
118 and p. R ~CA,
so
of
120].
x
This A
R~
R~
argument
is integral that
x e A
are precisely
that if over
R~
is then a 2-dimensional
If
ideals A
of
116
R,
is a then
is integral let
A
be the domain
have height
has some maximal
ideals
R
semi-local A
such that the maximal the maximal
that if
is noetherian.
can be seen as follows:
If all maximal
298]
and
R, then
the assumption
R ~ is noetherian.
I, then choose
that contain
R.
over
and
1.4, p.
domain
localization
domain
may be deleted.
integral [9, p.
is integral
semi-local
is noetherian.
[3, Corollary
(noetherian)
readily by a simple
2-dimensional R~
local
in
ideals
of height
ideals of i.
A It
2, of
Ii follows
from
closure
of
of
R
[6, R *.
Let
obtained
integral
Theorem 2 . 7 , B
by a d j o i n i n g
integral
one p r i m e o f
Thus
for
true that
R*
R
that
proposition
that
must be noetherian
R*
integral
closure
of
2.7 Proposition. R* = N { R p I P integral maximal
TCA,
of
of
R
extension
o f an e q u a t i o n is
over
is
a height
B, t h e
ideal
that
R*
domain,
Ferrand-
it is at least
is noetherian.
has noetherian
all domains
between
The spectrum
R
one prime If
which
of
R},
noetherian
and
and the
and let
R* ~ A, and if
{Pi }
are also of height
domain,
A
denote
denotes
one,
then
T = R* A { A p . } is noetherian. i is a flat T-module. In particular, R*
R*
of
a semi-local
let the
the set of R* is
if and only if
and
integral
is noetherian.
be a 2-dimensional
R. A
ring
B* = A { B p I P
noetherian
when
B
the
are noetherian.
Let
closure
noetherian
R
Then
integral
B* = R*
will show
is a height
ideals
is
is
generated
and
at any prime
following
A[i/x]
coefficients
R*.
B*
a 2-dimensional
localized
the
A[1/x],
Since
implies
R
over
closure
that
a finitely
to
1/x
B} = R*.
Raynaud result
150]
denote
dependence for
domain w i t h
p.
Moreover, has noetherian
spectrum. Proof.
If
M
A M = N{AQI Q is a height follows PinT
is a maximal is a height
one prime
that
TEA.
of If
is also maximal,
centered Theorem
on a maximal i.i0]
that
if and only if
T
since
field of
T,
R~ T
of
one prime R Pi
A
of
of height A
such that
RpCAQ.
T.
valuation
It therefore
is a flat T-module,
extension
of
Since T,
T
is a flat extension
having noetherian
117
M},
ideal ring
follows R~
of
T
R*
It A, then
is noetherian domain
spectrum.
contained
implies
= P
Ap. is z from [5,
is a noetherian
has noetherian of
and Q A R
R ~CA M .
and that
A
spectrum
spectrum.
in
Thus
one maximal
so the noetherian
ideal of
2, then
contained
is a height
is noetherian.
and is an integral Finally,
R~
ideal
in the quotient has noetherian
12 We close with the following example which shows that between a 2-dimensional noetherian domain
R
and the integral closure of
there can exist a non-noetherian
domain
B
such that
B
R
localized
at any prime ideal is noetherian. 2.8 Example. the integral
Let
T
be a 1-dimensional noetherian domain such that
closure of
for each prime ideal finite
TQ-mOdule.
example,
Q
of
is not a finite T-module, but such that T, the integral closure of
(Such a domain
as in Nagata
indeterminate
T
T
may be constructed,
[Ii, p. 35] or Dade
and let
R = T[X].
Then,
[i].)
Let
and the integral
ascending and let
B
B = R
TQ[X].
[{TiX1}~ " <...
1] "
then
Bp
Then
T
for
be an
{Ti}~= 1
domain
B
between
be a strictly
and the integral
closure of
T,
(T1X) < (T1X, T 2X 2) <... <
is a strictly ascending chain of ideals of
is not noetherian.
Q = PAT,
R: let
chain of domains between
(TIX,...,TnXn) so
closure of
X
is a
as Kaplansky pointed out to me,
it is easy to see the existence of a non-noetherian R
TQ
However,
for any prime ideal
P
of
B, B, if
is a localization of an integral extension of
Since the integral closure of
must be noetherian.
118
TQ[X]
is a finite module,
Bp
13 REFERENCES i.
E. C. Dade, Rings in which no fixed power of ideal classes becomes invertible, Math Annalen 148(1962), 65-66.
2.
P. Eakin, The converse to a well known theorem on noetherian rings, Math Annalen 177(1968), 278-282.
3.
D. Ferrand and M. Raynaud, Fibres formelles d'un anneau local noeth6rien, Ann. scient. Ec. Norm. Sup. 3 (1970), 295-311.
4.
R. Gilmer and W. Heinzer, On the complete integral closure of an integral domain, J. Australian Math. Soc. 6(1966), 351-361.
5.
W. Heinzer and J. Ohm, Noetherian intersections of integral domains, Trans. Amer. Math. Soc. 167(1972).
6.
W. Heinzer, J. Ohm, and R. Pendleton, On integral domains of the form
ADp, P minimal, J. Reine Angew. Math.
241(1970), 147-159.
7.
I. Kaplansky, Commutative Rings, Allyn and Bacon, Boston (1970).
8.
W. Krull, Einbettungsfreie fast-Noetherche Ringe und ihre Oberringe, Math. Nachr.
9. i0.
21(1960), 319-338.
M. Nagata, Local Rings, Interscience, New York (1962). M. Nagata, On Krull's conjecture concerning valuation rings, Nagoya Math. J. 4(1952), 29-33.
ii.
M. Nagata, On the closedness of singular loci, Publs. Math. Inst. Hautes Etudes Sci. 2(1959), 29-36.
12.
J. Ohm, Some counterexamples related to integral closure in D[[X]], Trans. Amer. Math. Soc. 122(1966), 321-333.
13.
B. Prekowitz,
Intersections of quasi-local domains, Chicago Ph.D.
thesis, 1971. 14.
P. Ribenboim, Sur une note de Nagata relative 8 un probl~me de Krull, Math. Zeit. 64 (1956), 159-168.
119
COHEN-MACAULAY MODULES
Melvin Hochster I University of Minnesota Minneapolis, Minnesota 55455
The object of this paper is to discuss the conjecture, which will be abbreviated (E) , that every complete local ring of dimension n possesses a finitely generated module of depth n . It is noted that several conjectures which have been open for some time follow from (E) , and the connection of (E) with Serre's conjecture on multiplicities over regular local rings is discussed. In fact Serre's conjecture is proved for dimension ~ 4 using the ideas under consideration. A number of proofs of (E) for the twodimensional case are given, and some possible methods for handling the general case are discussed. One of these is proposed as particularly worthy of study and is applied to an interesting class of examples in dimension 3 to obtain modules of depth These examples do not yield easily to other techniques.
I. Throughout, homomorphisms unital. If
M
THE BASIC FACTS AND THE BASIC CONJECTURE
all rings are assumed to be commutative, with identity, ring
are assumed to preserve the identity,
is an
R-module, we say that
..., ri_l)M ,
R-module, depth M-sequences
M
if
I < i < k .
(I,M)
or
contained in
finitely generated, then
then
R
we write
G(I)
or
rl, ..., r k
in
(rl, ..., rk)M / M
D(I,M) I ,
If
form an r. 1
is an ideal of
IM/
M .
D(I,M) = =
D(M) = D(P,M) grade I
I
and
R
M-sequence or
is not a zerodivisor on R
and
M
is an
denotes the supremum of lengths of
provided
M = 0) we make the convention that
of
and modules are assumed to be
By a local ring we mean a Noetherian ring with a unique maximal ideal.
a regular sequence on M/(rl,
3.
for
If If
IM = M (e.g. (R,P)
M / 0 .
D(I,R)
R
If
I = R
is local and
is finite if .
if
If
I
M
or is
is an ideal
is Noetherian and
I # R ,
G(I) < = .
IThis research was supported in part by National Science Foundation Grant GP-29224X.
120
2 Let
M
g(M) < =
be a finitely generated module over a local ring
if and only if
generally,
if
dimM=
AnnRM
is
P-primary,
Krull dim(R/AnnRM ) ,
where
g
(R,P) .
Then
denotes length.
we have that the following
More
conditions
are equivalent: i)
k
is the least possible integer such that
2)
The images of
rl,
3)
For each
1 < i < k ,
A sequence
i ,
rl, ..., r k
system of parameters
..., r k
for
M ,
or C-M, if
dim M/(rl,
and then M
condition is preserved by localization M
is
C-M
over
R
conditions
(R,P)
for
C-M If
over R
~
for every
M
is an
is local and
M
(in fact,
D(M) _< dim M) .
for
can be achieved.
D(M)
is a maximal
C-M
C-M M
over
to the
The
at a smaller prime in Supp M .
If
R/I .
When
is Cohen-Macaulay
R-module,
then
if
Mp
D(M) _< dim R
It is natural to ask whether the upper bound If it is achieved for a certain R .
C-M
R
Supp M .
is a finitely generated
module over
is the same as a faithful
in
is called
M-sequence.
if and only if it is
P
•
is called a
The condition is equivalent
is not local, we say that a finitely generated module is
R/AnnRM
See Serre [16], Ch. IIIA.
over a local ring
D(M) = dim M .
for
..., rk)M = dim M - i .
k = dim M .
statement that every system of parameters
I ~ AnnRM ,
..., rk)M) < = .
form a system of parameters
satisfying these equivalent
A finitely generated module Cohen-Macaulay,
g(M/(rl,
In case
module.
R
is a domain,
In general,
M ,
we say that
a maximal
a maximal
dim R
C-M
C-M
M module
module is
torsion-free. If
dim R _< 1 ,
one can produce a maximal
prime of maximal coheight. domain case.
In general,
But in general,
To see this, suppose
R
so that if Spec(S). C-M
over
S
possesses
chain condition,
is the ring,
In fact, if S
this trick reduces the problem to the
a
C-M
and
M
2) Grothendieck's
..., Xn]
Spec(S)
module
with
Supp M = Spec(R)
or formal power series ring over
[Q E Spec(S):
S = R[Xl,
with support
module by dividing out by a
even in dimension 2, the answer is negative.
Then every quotient of a polynomial I) the saturated
C-M
,
SQ
is
or
R[[Xl,
and facts
121
C-M]
l)
(CMU) condition,
satisfies [6] p.162,
is Zariski open in
..., Xn] ] , and
R
.
2)
then
M ~ RS
are immediate from
is
Serre [16], p. IV-2A, Corollaire 3, and Grothendieck [6], p. 163, Remarque (6.11.9) (ii), respectively.
The two-dimensional local domain
R0
discussed by Kaplansky
in [ii], §§ A and 23 (of course, the example is originally due to Nagata), fails to satisfy 1), while the two-dimensional local domsin discussed by Ferrand and Raynaud [5], Proposition 3.5, fails to satisfy 2). However, if
R
is complete both I) and 2) are automatic because
quotient of a regular local ring.
Conjecture (E).
If
R
R
is a
This author conjectures:
i__ssg complete local ring,
R
possesses a maximal
Cohen-Macaulay module.
Equivalently:
Conjecture (E').
If
R
is a complete local domain,
R
possesses a faithful
Cohen-Macaulay module. If
S
is a complete local domain, we know (see Nagata [13], p. 109,
Corollary (31.6)) that regular local ring
S
R .
also a finitely generated
is a module-finite domain extension of a complete If
M
is a finitely generated
R-module ,
and
S-module ,
Ds(M) = DR(M ) ,
then
M
either from Serre [16],
p. IV-18, Proposition 12 or from Levin and Vasconcelos [12], p. 317, Lemma. R ,
every module has finite projective dimension and if
dimension, then for every finitely generated DR(E)+PdR E = D(R) = dim R . DR(M ) = D(R) ,
and
R-module.
PdRM = 0 .
denotes projective
if and only if
This says that
M
is an
S-module
It follows that an equivalent conjecture to
( E ' ) is:
Conjecture (E").
l__ff R
is a complete regular local ring an___dd S
module-finite domain extension, then for some integer of an
Over
E ,
Ds(M) = dim S = dim R
i.e. if and only if
which is a nonzero free (E)
Hence,
R-module
pd
is
n ~ 1
Rn
is a
has the structure
S-module.
To give
Rn
an
S-module structure (by which we mean, of course, an
structure which extends its given
S-module
R-module structure) is equivalent to giving a
122
4 ring homomorphism of
S
into the ring
extends the usual identification of
~n(R)
R
of
n by n
matrices over
with the scalar matrices.
R
which
This enables us
to reformulate the conjecture thus:
Conjecture
(E'").
If
R
i_~s ~ complete regular local ring and
module-finite extension domain, then for some integer homomorphism
R -~(R)
extends to a ring homomorphism
A finitely generated module if
M ~ 0
and
G(AnnRM)
in general, if p. 30. S
If
R
pdM
n ~ ] ,
M
: pd M (so that, in particular,
is finite and
M ~ O ,
is a complete local domain,
is a complete regular local ring and
then precisely the same as a perfect
R
Q
then
R
pd M
.
is called perfect < ~ ).
Recall that,
G(AnnRM) ~ p d M ;
see Rees [15],
can be represented as
is prime.
is a
the usual ring
S ~(R)
over a Noetherian ring
S
A maximal
S-module with annihilator
S/Q ,
C-M
where
R-module is
Q .
Hence, we can
state one last form of the basic conjecture:
Conjecture ideal of
S ,
(E"").
If
S
is a complete regular local ring and
then there exists a perfect
2.
S-module
CONSEQUENCES OF CONJECTURE
M
such that
Q
i_~s ~ prime
AnnsM : Q .
(E)
We now discuss a number of open questions which are settled affirmatively by (E) .
The first of these can be regarded as a homological version of Krull's
principal ideal theorem.
Conjecture be an
(A).
R-module,
Let
and let
R - S
be a homomorphism of Noetheri~n ring~, let
I = AnnRM .
Let
Q
be a minimal prime of
IS .
M Then
ht Q S PdRM .
We shall use ht J
for
ht Q
for
inf[ht Q : J < Q ,
To understand conjecture R : Z[Xl,
..., Xn] ,
To give a homomorphism
dim RQ , Q prime]
and for an arbitrary ideal
we use
.
(A) better, consider the example where
I : (Xl, ..., Xn) , R ~ S
J ,
and
M : R/I .
is the same as to specify
123
In this case, n
elements
PdRM : n .
Sl, ..., sn
of
S
to serve as the images of
Xl, ..., x n
Then
IS = (Sl, ..., Sn) ,
and
the conjecture asserts that the height of any minimal prime of an ideal generated by
n
elements is at most
n .
On the other hand, if we take minimal prime
P
of
theorem [15], p. 30
I ,
R = S
and note that
grade I ~ ht P
for any
we can conclude another known result from (A):
that
G(AnnRM) ~ PdRM
if
M ~ 0 .
Rees'
One interesting point here
is that the generalization of Rees' theorem which allows the change of ring is true (Theorem 1 of [8]) even in a non-Noetherian version, (E) ~
(A) ;
in fact, a very weak form of
(E)
and this permits us to conclude
suffices.
We return to this
point soon. The so-called "intersection theorem" is an easy corollary of converse is less trivial but also true.
Conjecture
(A').
R-modules such that
l_~f (R,P)
then
the
We first recall the intersection theorem:
is a local ring an__dd M, N
&(M~N) < ~ ,
(A) :
are finitely generated
dim N _< pd M .
This result is proved in Peskine and Szpiro [14] in many interesting special cases:
if
R
has characteristic
p > 0 ,
finitely generated over a field, if
if
pd M _< 2 ,
R
is a localization of a ring and, of course, if
the result was known, Serre [16], p. V-18, Th~oreme 3. the case where
N
is prime cyclic, i.e.
Proposition 2.1. Proof.
Assume (A).
The hypothesis P-primary.
Let
Now assume
Rp ~ SQ
If
P
S .
Hence,
(A') .
Let
where
( A ' ) when
is equivalent to the hypothesis
I -- Ann M .
Then
I+Q
is
P-primary,
and
S , M , I
is the inverse image of
Q ,
and
Q
primary to
is prime.
(A') holds.
N = R/Q .
Ann M + Ann N IS
and we must show that
is
by
(A) . (A)
be
we can consider the local homomorphism
ht Q _< PdRM .
124
S = R/Q.
is primary to the
Thus, we can assume that
is a local homomorphism of complete local rings and that
Q ,
Let
as in the statement of
instead of the original one, and we can complete.
(R,P) ~ (S,Q)
Q
dim N = dim S = ht P/Q _< pdRM , R-
is regular
It is trivial to reduce to
(A) holds if and only i__ffConjecture
It suffices to prove
&(M~N) < ~
maximal ideal of
given.
Conjecture
N ~ R/Q ,
R
IS
is
By making faithfully flat
6 local extensions of
R
and
coefficient ring of
S ,
S
we can get a coefficient ring of
( A ' ) with
R ~ S
is onto, i.e. that
p > 0 ,
if
(A')
R , S
is regular, or if
are finitely generated
we can
S = R/J .
We can
holds in certain R
has characteristic
k-algebras for some field
k ,
if
R
PdRM ~ 2 .
(E)
show that the weaker conjecture
(E*).
If
(possibly non-Noetherian)
Proposition 2.2. Proof.
(A)
is known to hold locally, e.g. if
We next want to show that
Conjecture
R
N = R/J .
By modifying this argument, we can deduce that situations where
to map onto a
and by adjoining analytic indeterminates to
then further guarantee that the map then apply
R
(E*)
(R',P')
Conjecture
(E*) for
below implies
(A) .
We shall even
(A) .
is a complete local domain, there is a
R-module
It suffices to prove
be as in Conjecture
does, in fact, imply
T
such that
P'T ~ T
(E*) implies Conjecture ( A ' ) when
R' = R/Q .
N = R/Q
and
D(T) = dim R'
(A) . Let
is prime cyclic.
T
Then by Theorem i of [8],
dim N = D(P/Q, T) = D(IR', T) ~ pd M .
Let us say that a local ring of
R
there is a
Corollary 1.3.
C-M
R
module over
Conjecture
has abundant R
C-M
modules if for every prime
whose annihilator is
(A') holds for a local ring
R
Q .
if
R
has
abundant Cohen-Macaulay modules.
Note the following corollary of Conjecture
Conjecture
(AI).
If
R
(A) :
is a local ring which possesses a module
finite length and finite projective dimension, then
(A') ~
(AI)
because we may take
N = R
R
and then
M
i__ssCohen-Macaulay.
dim R _< pd M _< D(R) , as
required.
R. Y. Sharp [17] has recently proved a formal duality between finitely
125
of
generated modules of finite injective dimension and finitely generated modules of finite projective dimension over any
C-M
local ring
Gorenstein local ring~ and, in particular,
R
which is a quotient of a
over any complete
C-M
local ring.
On
the other hand, Bass has conjectured [1]:
Conjecture
(B).
If a local rink
of finite injective dimension, then
(R,£) possesses a finitely generated module R
i_As C-M .
This can be reduced to the complete case and, to some extent, is dual to Conjecture
(A1).
Peskine and Szpiro [IA] have proved
that they have proved implies of
R
(B) :
(A') .
They also note in
dim R' = dim R(=n)
finitely generated as an
and
finitely generated Ann M + Ann N
Ann M + Ann N
R-modules with is
for each
n
in roughly the same cases
that a weak form of
(E)
over a local extension
dim R'/PR' = 0 .
~-
If
pd M < = ,
P-primary), i,
R'
(The module is to be
R'-module.)
We now want to discuss multiplicities.
(i.e.
[14]
one needs a module of maximal depth
such that
(B)
(R,P) then if
we have that
and each
is local and
&(Tori(M,N))
M, N
are
&(M~N) <
Tori(M,N) < ~ ,
is annihilated by
so that
eR(M,N) : e(M,N) : Z i (-1)i%(Tori(M,N))
is a well-defined integer, which we refer to as the intersection multiplicity of M, N .
If
R
is regular, one knows that under these hypotheses
dim M + dim N ~ dim R ,
Serre [16], p. V-18 (and it is conjectured that this holds
without the regularity:
pd M < ~
knows that if the completion of
and R
valuation ring (in particular, if e(M,N) > 0
if and only if
&(M~N) < ~
should be enough), and one also
is a formal power series ring over a discrete R
is unramified) then
dim M + dim N = dim R .
holds for all regular local rings [16], p. V-14. shall prove below that if both
(E)
e(M,N) k O
Serre has conjectured that this
This question is still open.
and an additional conjecture
(F)
2At the time of this talk, I mistakenly thought that Icould prove that Serre's conjecture without assuming (F) .
126
and
We
hold 2 ,
(E) implies
then Serre's conjecture holds.
Conjecture generated
(C).
Let
We first state Serre's conjecture formally:
(R,P)
R-modules such that
and only if
be a regular local ring and let g(M®N)
< =
Then
e(M,N) ~ 0
M, N and
be finitely
e(M,N) > 0
if
dim M + dim N = dim R .
We next want to discuss Conjecture
(F):
afterwards we shall prove
(E) + (F) ----> (C) . Let
E
be a perfect module over
such that
G(IR) ~ G(I) ,
perfect
where
A = Z[Xl,
I = AnnAE ,
..., Xn] .
but
IR # R ,
R-module of the same projective dimension as
projective resolution of E' = R ® A E say that
(i.e. E'
is
E ,
then
R®A ~
E .
R
is an A-algebra
then
R®AE
is a
In fact, if
~
is a
is a projective resolution of
Tor~(R,E) = O, i ~ I) . g-perfect.
If
See [8].
In this situation, we shall
The significance of this notion for the study of
multiplicities is apparent from the following:
Lemma 2.4. R-module.
Let
Let N
(R,P)
b_!e _a C.ohen-Macaulay ring and let
b_~e a_ finitely ~enerated
M
R-module such that
be a
g-perfect
g ( M ® N) <
Then: a)
dim M + dim N ~ dim R ,
b)
e(M,N) ~ 0 ,
c)
e(M,N) > 0
Proof.
and
if and only if
dim M + dim N = dim R .
As usual, we can assume that
is an unramified regular local ring. over
A : Z[Xl,
A - S .
Let
I = AnnAE .
..., Xn] .
M * = S®A E .
R
is complete.
Suppose
The homomorphism
A - R
We shall show that
We need to show that
M = R®AE
G(IS ~ G(IR)
M* .
Write ,
R = S/J ,
where
of
M*
over Thus,
ever
R .
S ,
Regard
Tor~(M*,N)
and
N
E
over
A ,
R® s(M*GAg )
as sn
: Tor~(M,N)
S-module. ,
and
so that
is also But
g-perfect.
M* ® A g
IN : 0 ,
.
=
Let
g
be
is a projective resolution R®sM*
= M
(R®~M*®Ag))~RN:(M*®A~)®sN.
eR(M,N ) : es(M*,N ) .
127
Let
G(IR) = G((J+IS)/J)
is a projective resolution of
Since
is perfect
lifts to a homomorphism
ht (J+IS)/J = ht (J+IS) - ht J ! ht J + ht IS - ht J = ht IS = G(IS) a projective resolution of
E
where
Since
Conjecture
(C)
holds for unramified regular local rings, all
three facts will follow if
dim R - dim M - dim N = dim S - dim M* - dim N .
But
dim R - dim M
= D(R) - depth M = pd M ~ pd M* = D(S) - depth M* = dim S - dim M* ,
This motivates the following definitions. Let
_G(R)
be the Grothendieck group of
homological dimension. classes Let
[E]
of the
Let
g(R)
be the subgroup of
(= R[tl,
E
m _< n ,
for each
and we therefore have maps
F(R) = lira ~(Rn) n
Conjecture
be a local
C-M
ring.
_G(R) generated by the
of finite length.
..., tn]PR[tl,
G(Rm) ~ G(Rn) , m, n,
by
(R,P)
R-modules of finite length and finite
g-perfect modules
R n = R(t I . . . . , tn)
Let
as required.
[E] ~ [ E ® R R n ] m
.
.-., tn])
.
Let
We have maps
The maps take
G(Rm) ~ G(Rn)
,
~(R) = _G(R)/g(R).
g(Rm)
into
m _< n .
g(Rn)
Let
-
(F).
For each Cohen-Macaulay local ring
For applications to multiplicities,
R ,
F(R) = 0 .
we only need this conjecture when
R
is a
local complete intersection. We shall prove If
(F)
dim R = 0 ,
for the case
dim R ! 2 .
every module of finite projective dimension is free and
there is nothing to prove. If
dim R = i ,
and
M
M = Coker (Rn ~ R n)
for some
is then
let
g-perfect:
the entries of f*: A n - A n
(xij)
is a perfect module with n ,
where
det f
pd M = I ,
is not a zerodivisor.
A = Z[xij: 1 ~ i,j ~ n] ,
and map
A
to
to the corresponding entries of the matrix of
have matrix
(xij) ,
and let
then
E = Coker f* .
Then
M
R
M
itself
by taking f .
Let
arises from
E
in the required manner. To obtain the result for dimension 2 we use a result of Buchsbaum and Eisenbud [3] • Let
Let
(xij)
be a
A = Z[xij]i j .
Let
bI
by
b0
matrix of indeterminates
f: A bl - A bO
over
be the map whose matrix is
E(bl, bo) = Coker f .
128
Z, b I ~ b 0 . (xij) .
Let
I0 Lemma 2.5 (Buchsbaum-Eisenbud). dimension
bl-b0+l .
E(bl, b 0)
i_Asperfect of projective
Its annihilator is the ideal of maximal minors of
and that ideal has grade
(xij) ,
bl-b0+l .
If
bl-b 0 = 1 ,
~
Betti numbers of
E
are
b0, bl, l, i.e.
b0, b0+l , 1 .
If
bl-b 0 = 2 ,
the Betti numbers of
E
are
b0, b0+2 , b0+2, b O.
In fact, in [3] Buchsbaum and Eisenbud give explicit minimal resolutions of these modules, which they refer to as "generic torsion modules". if
(xij)
has entries in
matrix,
bI ~ b0 ,
bl-b0+l
then
P ,
where
(R,P)
It follows that
is local, and is a
bI
by
then if the grade of the ideal of maximal minors of
Coker (xij)
is perfect of projective dimension
b0 (xij)
bl-b0+l
is
and has
the same Betti numbers (and, in fact, essentially the same resolution) as a matrix of indeterminates does in the generic case.
In the sequel we want to consider what we shall refer to as resolutions with matrices in "general position". which is a resolution of H0(~) ,
which is
M .
M Let
Consider a free complex
over (t)
~ = 0 - R n ~ ... ~
04 0
R , i.e. the homology vanishes except for denote
Z i b~
indeterminates.
Then
1
%(t) = ~®RR(t)
gives a resolution of M(t) = M®RR(t) over R(t) . We can make b. a change of basis in each R(t) i in which the change of basis matrix has as entries the appropriate
b~
indeterminates (so that the sets of entries of the
n
1
change of basis matrices are mutually disjoint).
The matrix of each map is then
multiplied on each side by a square matrix of indeterminates of appropriate size, and the two matrices of indeterminates used for a given map are disjoint (note that R(tij) ,
1 ~ i ,
j ~ n ,
has an
R-automorphism which takes the matrix
(tij)
to
its inverse, so that the inverse of a matrix of "round brackets indeterminates" is just as "good" as the original matrix of indeterminates). Generally speaking, if represents
p2+q2
U
is a
p
indeterminates, say
faithfully flat over
R(t)
we shall say
by (tij)
q
matrix over and
(t~j) ,
(tij)U(tlj)
129
R ,
and
then if
(t) S
is
is in general position in
S.
II Lemma 2.6.
Let
general position, ha___~sgrade of
g .
U*
U~
s ~ r ,
g .
by
flat extension
Multiplying
r
minors
(Rings are assumed
matrix of indeterminates):
grade.
r
The grade will be at most
Faithfully
s
bv
an_dd suppose that the ideal of
Then the ideal of
ha__~s grade
Proof.
be the result of putting an
r
of any
Noetherian
(r+g-l)-r+l
r
matrix
by
r
r+g-i
U
minors by
r
of
here.)
= g
(which is what it is for a
and localization
on the right by the matrix of round brackets
only increases indeterminates
of a given size.
assume that we are only multiplying
on the left by a matrix of square brackets
r+g-I
by
r
The submatrix of
U~
then has the form
matrix of indeterminates,
and if
R
the entries
be the ideal generated
by the
minors
can choose a prime
Q
of
replace
without
ideal
R
by
J0
~
of
r
by
G(J 0) = G(JoR[t]) that
(R,P)
r
R[t]
such that
miners
of
J G(Q) J D(R[t]Q)
is local and that some i.e. is invertible
whose first
r
rows form an
Then
U = AV
r
altering
determinant,
are zero.
by
where
r
by A
r
of
T').
= T'V ,
which is an
It follows
r+g-I
G(JRp)
< g .
P ,
minor of Let
V
s
by
over by
r
s
U
ring, we can R .
we can
Now, if the JoR[t] ~
s
by
r
Q
and
matrix
and whose last
A
s-r
matrix.
r
rows
Now,
is invertible,
matrix of indeterminates by
then we
has invertible
because
r
J
Hence, we can assume
be the
that the grade of the ideal of
Let
Q N R = P ,
then
invertible
R ,
is an
G(J) < g , If
in
T
to
< g .
identity matrix
is an
is simply a new matrix of indeterminates TU = T(AV)
r
T
If
a contradiction.
by
where
of
TU .
is contained
as a matrix.
r
of
D(R[t]Q)
< g ,
TU ,
t..
the hypothesis U
we might as well
was the original
work over the ring obtained by adjoining r
Hence,
g .
does not change the ideal of minors
indeterminates.
U
submatrix
the problem is to show that it is at least does not change grade,
in
minors
TA = T' and
(a submatrix
of
TU
is
g ,
as required.
We now continue
the discussion
Suppose that we are working and finite projective Pass from
R
to
R(t)
of Conjecture
in dimension
dimension
(F) in dimension
2 and higher.
n .
Let
M
be a module of finite length
(equivalently,
let
M
he perfect,
for some large family of indeterminates
130
with (t)
pd M = n).
and assume
12
that
0 ~R
is a free minimal b0, bl,
..., b
b0+b2+... ~(M)
Then
of
M
Let
bI
by
b0
U0
M* = Coker U 0 generic
follows
of
U0
M
and
n .
Thus,
Then
MI
of
U
b0
consisting
pd M I ~ n
~
n ,
while
rows.
generators
for the row space.
These are minimal
as to give a relation hl-(bo+l)
We call
U
Hence,
the rows of the submatrix
of
generate
minimally)
(not necessarily
V
if
V
U
rows.
n
of
.
and
G(Ann MI) ~ n .
as well. M*
for It
It is worth
. bl-(b0+l)
rows of
spanned by the first U
are minimal
on these generators
is the same
and then project
on the last
is the second matrix of the resolution
consisting
then
The ideal
0 ~ M I ~ M* ~ M ~ 0
n
and
which comes from a
since the rows of
To give a relation
U ,
and it follows that
by the submodule
generators,
of
b I > b0+n-i
b0+n-I
by the images of the last
on all the rows of
cocrdinates.
minors
If equality holds,
dimension
dimension
M = Coker U .
If not,
Ann M I D Ann M*
of the row space of
bl-(b0+l )
b0
.
of the first
by Lemma 2.6,
of projective
is generated
so that
by
We get a short exact sequence
is perfect
Also note that in the quotient
= bl+b3+ . . . .
and we are done.
noting that from [3], we know the entire resolution
U
Then
E. (-l)ib. = 0 , i.e. 1 1
b I ~ b0+n-I
is a perfect module of projective
MI .
that
of
in general position.
with the ideal of
has grade
torsion module.
a certain
with matrices
torsion module,
be the submatrix
of maximal minors
040
matrix of the first map,
is power associated
comes from a generic
Let
-
~(M) = (E i bi)/2 = b0+b2+...
the grade of this ideal must be M
h
M .
be the
Ann M
... ~ R
are the Betti numbers
n
the index of U
n~
resolution
= bl+b3+ . . . .
Let
h
of the last
the relations
bl-(b0+l)
cf
columns
on the specified
generators
of
MI • Now, let residue
r i = rank
class field of
(Tori(M~,k)
~ Tori(M,k))
R .
131
,
0 < i < n ,
where
k
is the
M,
13 Lemma 2.7. Proof.
The index
Let
i(~)
is
T i = Tot i ( , k )
i(M) + (i_(M*) - Z i ri) .
and consider the long exact sequence: •
- Ti+l(M) ~ T i ( ~ ) ~ Ti(M* ) ~ Ti(M ) ~ Ti_l(~) ~ .... Betti numbers of
MI, M, M*,
forth.
Since the rank of
map is
ci-r. ,
respectively.
Ti(M*) ~ Ti(M)
Then is
ri ,
the rank of the following map is
mind that
b 0 = r0 )
we get
If it were true that
b.-r. ,
M
in
trivial, and
F(R)
i(~)
i
th
and so
and it follows that
i
Summing these equalities (and keeping in
2i_(M1) = 2i_(M)+2i_(M*)-22i r i ,
E. r. > i(M*) l
class of
i .
be the
the rank of the preceding i
for each
ai, bi, ci
ai = dimkTori(Ml,k ) ,
i
a i = (bi+l-ri+l)+(ci-ri)
Let
as required.
always, we would be able to show that the
i
is trivial by induction on would be less than
i_(M) .
i_(M): the class of The inequality
M*
is
E. r. > i(M*) i
l
does hold in dimension 2 (we shall prove something quite a bit stronger), but in dimension 3 equality is possible. In dimension 2,
b I = b2+b 0 ,
and the long exact sequence for
Tor
takes the
form:
0 - T2(MI) - k f k b2 - TI(M1) ~ k b0+l
k
Since
T2(MI) / 0 ,
T2(~) ~ k ,
k bl (bo+l)
and
right hand side, it is easy to see that bl-(b0+l) = b2-1 , b2,
and
1 ,
f
is g
0 .
is
0 .
columns.
0
Working backward from the Thus,
~
has Betti numbers
V
consisting of the last
We have shown:
Corollary 2 . 8 . a local ring
k bO
and comes from a generic torsion module.
Moreover, its matrix is exactly the submatrix of b2-1
k bO
R .
Let
E
be ~ perfect module of projective dimension 2 over
Then for some
(t) ,
E(t)
is a quotient of a perfect module
arising from ~ generic torsion module by ~ perfect submodule arising from ~ generic torsion module, both of projective dimension 2. In dimension 3, this technique seems to fail.
132
However, one can still show
14 that
Ei ri ~ ~(M*)
index.
,
so that passing from
For the long exact sequence
0~k3~ a
that
r0 = b0
which
case we get equality.
a3 = b 0 ,
and
kb0 ~ kb3 ~ k 2 ~a
f
the new Betti numbers
,
.
Thus,
Now,
~(M*)
< ~(M)
kal
= 2b0+2
unless
r3 = r2 = 0 ,
in order to generate
and we can see
r3 = r 2 = 0 ,
a 0 = bl-b0-2
in this case.
g-modules
,
in
it is easy to see that
in addition to
determined
the need for more
generic torsion modules
can never increase the
kb0+2 ~ kb2
~(M1)
aI = b 2 ,
are completely
occur, which indicates
0 .
In the case and
MI
k bl ~ kbl-bo -2 ~ k b0 ~ k b0 ~ 0
must be
r I = bo+2
a 2 = b3+b0+2
to
takes the form:
kb0 +2
and it is easy to see that
M
besides
,
so that
This situation
can
those coming from
the stable Grothendieck
group.
This
is not unexpected.
We next want to prove a result which yields 4 •
However,
modules
over a
Serre's
we first want to note the following: C-M
the biggest integer
local ring such that
(R,P)
such that
Tori(M,N ) # 0 .
let
g(M®N)
Then
conjecture M
and
< ~
in dimension N
be perfect
Let
i
i = dim R - dim M - dim N .
To see this, we first note that either from the main result of Buchsbaum Eisenbud But
[2] or from [8],
be
and
i = pd M - D(Ann M, N) .
pd M - D(Ann M, N) = D(R) - D(M) - D(Ann M + Ann N, N) =
D(R) - dim M - D(P,N)
= D(R) - dim M - dim N ,
dim M + dim N = dim R , while if e(M,N)
Theorem 2.9. following I)
- ~(TorI(M,N))
Let
two conditions
(R,P)
and
Tor
and
there are only two nonvanishing
be a regular local ring of dimension
are equivalent:
%(M®N)
Hence, e(M,N)
if = ~(M®N),
Tor 's
and
-
For an__nc f two finitely generated modules
dim M + dim N < n
D(R) = dim R .
there is only one non-vanishing
dim M + dim N = dim R - 1
= $(M®N)
and
< ~ ,
e(M,N)
133
M, N
= 0 .
such that
n .
Then the
2)
For any two perfect
$(M®N)
< ~ ,
e(M,N)
I f Conjecture then
I)
and
abundant
2)
do hold
t = (pd M - O(Ann M))
is the hypothesis
assume
that
maximal
t > I .
dim N : (D(N)) appropriate
G(1)
= m
and that
by avoiding But
~ ht I = G(1)
construction,
N ®
holds
(R/(rl,
N ®
..., rm)
short
sequence:
since
R rl,
..., rk) )
..., rm) ) . onto
M ,
since
and
e(M,N)
is annihilated
we can assume similar
reasoning
the case where
k _< m-I
by
then
vanishes
(rl,
than what
pd M = m = G(Ann M) we can also assume
,
that
we can let
M *s
m > dim N ,
are
M* = R/(rl,
Let
sequence.
certainly
(M,N)
to a
.
Hence~ Assume,
Note
G(Ann M) ~ m ,
is perfect,
t > i .
of
vanishes.
the integer
M
By
Therefore,
so that
so that
..., rm_l),
134
C-M.
n - m .
,
JN = 0 ,
g-perfect,
e(~,N)
since
since
and this leads
is
is perfect.
and
,
an
that must be
so that this
I ,
it was for
N
that is a
construct
sum of copies
..., rm) c
so that
claim
to show that
length,
..., rm)
and
segment
= D(I,N)
if and only if
on
We first
since we can then
Now,
what we
The case
t ,
respectively.
and therefore
since both
.
,
has
by induction
has an initial
,
R
Evidently,
be the required
m
pd M I = pd M - I , one less
.
< [(n+l)/2]
and
for smaller
= D(I+J,N)
Ann M *s = (rl,
We must have that ,
2)
of two sets of primes
..., r
t = dim R - dim M - dim N
dim M < dim R . Since
,
is at least that
which
0 - M I ~ M *s - M - 0 .
= 0 ,
(MI,N)
and
We can map a direct
e(M*S,N)
pair
R .
Ann N ,
has finite
dim M *s = dim M ,
pd M ~ m+l
for
We proceed
is regular
and
and if
I
: D(P,N)
< = ,
MI
(C) holds
the result
the union
6(M*S®N)
that
R
N)) + (dim R - dim M - dim N)
in
and let
(R/(rl,
M* = R/(rl, exact
for
holds.
G(1) ~ D(I,N)
D(N)
,
and dim N = k ,
for
i)
hold
I)
Ann M ,
R-sequence
: D(I,N)
at each stage.
D(I,N)
be
2)
an___dd
. in dimension
To show this it is only necessary
sequence
and
and
of
Now assume
I, J
is a maximal
N-sequence.
avoided
2)
Let
])
the equivalence
(pd N - G(Ann
t = 1
intersections
Conjecture
then
dim M + dim N = n-I
= 6(TorI(M,N))
If
then
holds,
+
such that
6(M®N)
R .
modules,
2)
M, N
for complete
for
We first prove
show is that if
that there
i.e.
(F) holds
Cohen-Macaulay
Proof. must
: 0 ,
modules
t
This
for the shows
that
and by we must be in say,
that
dim N < dim R - d i m M = m . map
M ~-s
onto
M
for a
16 suitable
s ,
and we get a short exact sequence:
pd M = m
and
pd M *s = m-I ,
and it follows that dim M + 1 .
M1
pd ~
From the fact that
t > I ,
conclude that vanishes.
~ M *s - M ~ 0 .
Moreover,
G(Ann ~ )
is perfect of projective dimension k ~ m-I
involved have finite length, and because
= m-I .
O -~
what it was for the pair
t
M *s
e(M,N)
for the pair
(M,N) ,
is
g-perfect we can still
vanishes if and only if (MI,N)
is precisely one less than
I)
and
2) .
holds for complete intersections of dimension J [(n+l)/2] R
of dimension
n .
one of these two modules, say before, there is an call this
d 2 (n-l)/2 ,
dim S ~ (n+l)/2 , and
g(S(t))
es(t)(M'(t),N(t) )
But
e(M,N)
holds.
> 0 .
M = R/Q ,
Q'
S = R/(rl,
since
[M'(t)]
g(S(t)) I)
Q, Q'
Let
T, T'
respectively.
We must show that
..., rd) .
holds for
S .
show that
e(T,T')
> 0 .
,
holds and
g(S(t))
and
R
has abundant
dim M + dim N = dim R , M, N
[M'(t)]
es(t) ( , N ( t ) )
In a prime filtration for b
C-M
and
135
T'
modules,
then
are prime cyclic, as usual, say
be Cohen-Macaulay modules over
T
and
and we know that
or
h' ,
e(T,T') = bb'e(R/Q,R/Q')
But since
But
so large that
T
R or
,
having T' ,
Q
or
and the other primes
will be of strictly higher height, and their factors will contribute It follows that
Since
.
will occur a positive number of times, say
multiplicity.
M-sequence;
S-module of finite length.
(t)
lies in
2)
we know that for
dim M' = dim M - d
es(t)(M'(t),N(t)) = es(M',N)
We must show that if
.
(F)
is a perfect
This reduces to the case where
N = R/Q'
annihilators
and let
(F)
By the same reasoning as
(F) we can choose
vanishes because
.
which is also a maximal
and conjecture
It remains to prove that if (C)
J = Ann N
By Conjecture
vanishes on the generators of
then
dim M ~ (n-l)/2 .
M' = M ® R S
Now assume that
dim M + dim N = n-i ,
dim M' + dim N = dim S - I ,
dim S = dim R - d . lies in
M ,
rl, ..., r d ~
eR(M,N) = e s ( M ® R S, N) , We still have
But since
R-sequence in
R-sequence
e(MI,N)
we are done.
We have demonstrated the equivalence of
holds for
and dimension
dim M *s + dim N = dim M + dim N + 1 < dim R
and that
Since the integer
~ G(Ann M *s) =m-l,
we still know that the tensor products
so that from the fact that
e(M*S,N) = 0
m-I
Since
0
to the
and it suffices to
are perfect and
17 dim T + dim T' = dim R ,
there is only one non-vanishing
e(T,T')
as required.
= %(T®T')
> 0 ,
Corollary 2.10.
If Conjecture
holds fo___Krcomplete intersections holds
(in dimension Proof.
abundant
(in dimension
(in dimension
< [(n+l)/2]
and
n) an__~dConjecture ) ,
(F)
then Conjecture
(C)
n) .
Conjecture
C-M
(E) holds
Tor ,
modules,
(C) can be reduced to the complete case. by Conjecture
But then
R
has
(E), and the result is immediate from
Theorem 2.9.
Corollary 2.11. Proof. 2 ,
Conjecture
We can reduce to the complete case.
and since
2 ~ [(4+1)/2]
rings in dimension
~ 4
height
itself is
~ I ,
is true again. Conjecture
(C) holds for regular local rings of dimension < 4 •
R/Q
,
(F)
in dimension
it suffices to show that complete regular local
have abundant C-M .
The only remaining
Since we know
C-M
modules.
If the prime
case is where
Q
If the prime
Q
has coheight
~ i ,
dim R = 4
and
has this
ht Q = 2 .
But
(E) is known for complete local domains of dimension 2 (see the next
section). We conclude this section by noting one further consequence in fact, proved to he equivalent known precisely when
Conjecture
(A')
(D).
If
is an
(A')
is known:
M
(E)
which is,
in Peskine and Szpiro [14] , and is
the zerodivisor
conjecture.
is a finitely generated module of finite projective
dimension over a local ring r I, ..., r k
to
of
R
and
rl, ..., r k
is an
M-sequence
,
then
R-sequence.
9-
THE TWO-DIMENSIONAL
We want to give several proofs of
(E)
CASE
when the dimension of the ring is
2.
We can assume that we are dealing with a complete local domain.
First proof. integrally
Let
M = R'
be the integral
closed local domain of dimension
136
2 ,
closure of DR,(R')
R . = 2 .
Since
R'
But then
is an
18
DR(R' ) : DR,(R')
= 2 .
[12] for proving Bass'
This is essentially conjecture in the
conjecture is the same as the conjecture question
(not conjectured)
Second proof.
the technique of Levin and Vasconcelos
D(R) = 1
case.
"IV implies III" which is raised as a
on p. 162 of [lO].
Since the ring
S
is a complete local domain,
finite extension of a complete regular local ring S-module structure and is free as an PdRS ~ i ,
and if
Note also that Bass'
R-module.
0 ~ Rn ~R m ~ S ~ 0
R . For
Then
S
is a module-
HomR(S,R )
DR(S ) ~ I
has an
implies that
is exact, then we get:
(*) 0 ~ HomR(S,R ) - R n ~ R m ~ Ext~(S,R) ~ 0
is exact, and since a 2-dimensional
HomR(S,R )
is a second module of syzygies of
r2g(s)
and
r2g = rlf,
= rlf(s)
divides
,
applying a functor to an
D = Homs(
Let
, E(S/Q))
E
f(s)
then for each for every
s
s ;
rl, r 2 in
is an
S ,
but then
(E(S/Q)
r2
n = dim E ,
parameters
for
dim S = 2
is a maximal
m ,
however,
and C-M
Then E
where
HQi
s
then
(Note:
S ,
where
S
of
if we write
T i(E) -~ ExtR-i(E,R)
;
S/Q).
(respectively,
(respectively,
is any 2-dimensional
module over
To see that the statements
Sl, s2)
Sl, s2)
and
See
S = R/I ,
where
one does not want
because it seems to depend on the choice of s
(S,Q)
is local cohomology
is the injective hull over
and suppose E .
The functor used here is similar to
T i = D(H ) ,
[7], pp. 87 and 88.
is regular of dimension
this as a definition,
S-module.
be a finitely generated module over
and let
Grothendieck-Hartshorne
if
divides
for if
This proof is a variation on the same basic idea as the second
is local and complete,
Le~
r2
,
~ 2 ,
f .
the one used above.
R
D(HOmR(S,R))
f, g E Hom(S,R)
and so
Third proof. proof:
over
regular local ring it must be free.
One can also see directly that R-sequence
Ext~(S,R)
R .)
is part of a system of
is a
Tn(E)-sequence.
finitely generated
S-module,
T2(E)
S . above hold, note that since
137
Thus,
dim E/sE = n-1 ,
19 Tn(E/sE) = 0 .
Consider the exact sequence
E" = UnAnnEsn
Since
dim E" < dim E ,
long exact sequence for E' ,
T
that
and the exact sequence
s
Tn(E ') ~ Tn(E)
s2
is not a zerodivisor on
we have an exact sequence have an injection
yields
,
as required. E .
~(E)
E
0 ~ Tn(E)/sTn(E) ~ Tn-l(E'/sE ') . E'/sE'
and hence not a zerodivisor on
is not free and has depth
Fourth proof. dimension
E'
as above.
We
~ Tn(E')/sTn(E ') .
But
,
Since s2
~ 2 .
Let
Q Q(n)
Let
for all sufficiently large shall prove this in
§5 •
and so we
dim E'/sE' = n-1
and
is not a zerodivisor on
Tn(E)/sTn(E) let
(This result does
S
be a regular local
a torsion-free reflexive module which is not free.
: HOms(E,S )
Tn(E'/sE ') = 0 ,
Now suppose
Define
Tn(E)/sTn(E)
not extend to systems of parameters of length three: ring, and
is not a zerodivisor on
0 - Tn(E ') Z Tn(E ,) ~ Tn-I(E,/sE ,) . ... ,
is part of a system of parameters for
Tn-l(E'/sE ') ,
s
dim E'/sE' < n ,
is part of a system of parameters for
must show that
s2
Since
where
and it follows from the
But
0 ~ E' ~s E'/sE' ~ 0
is a nonzerodivisor on
s = sl, s 2
Tn(E '') = 0 ,
Tn(E ') ~ Tn(E)
0 - Tn(E'/sE ') - Tn(E ') ~ Tn(E ') - . . . . and
0 ~ E" - E - E' - 0 ,
Then
< 3 .)
be a height one prime of a complete local domain
S
n = Q SQn S ,
symbolic power.
Then
is at least
We
n ,
D(Q (n))
as usual, the (not
n
th
G(Q (n)) !~)
2.
of
It is the circle of ideas involved in this proof which
we feel is most likely to generalize to higher dimensions.
4.
APPROACHES TO HIGHER DIMENSION
In this section we consider several techniques which might be used to prove Conjecture
Let
(E) in higher dimension.
R
be a complete regular local ring.
module-finite extension domains of and
One technique is reserved for
RCSoCS
,
SO
Consider the smallest class
such that
is in
I)
R
in
G
G ,
and
3)
an
S-free module finite extension domain of
S ,
then
We shall call the
then
R
R-algebras in
G
is in if
the accessible
138
§5 , however.
S T
C , is in
is in
C
of
2)
if
S
is
G
and
T
is
G .
R-algebras.
It is quite
20
trivial that the class
8
of module-finite extension domains of
maximal
C-M
modules satisfies
maximal
C-M
module over
case C-M
3) ,
if
M
T .
In dimensions
G ;
if
C-M
module over
Thus, accessible
1
and
R • S ,
2
as the first proof of
(E)
2) ,
C-M
S ,
which possess M
is a
module over
SO .
then
if
M®sT
R-algebras have maximal
S
is in
~ .
S'
of
R
In
is a maximal C-M
every module-finite extension domain of
the integral closure
but then its subalgebra
In case
it is also a maximal
is a maximal
module over
accessible:
S ,
I), 2), and 3) •
R
modules. R
is
is R-free and hence in
This gives essentially the same proof
in dimension 2 of the preceding section.
Thus, it is natural to ask in higher dimensions:
Question 4.1.
l_~s every module-finite
extension domain of a complete regular
local ring accessible?
An affirmative answer would prove are all
R-split
the class of
(i.e. will contain
(E) .
R
Note that the accessible
as a direct summand a_~s R-modules),
R-split extensions satisfies the conditions
cannot have an affirmative answer to extension domain of This i_gstrue:
R
is
Question 4.1
R-split when
R
C-M
dim R = 3
and
R
completion
S
of
Because
S
extension algebra.
Thus, we
unless every module-finite
complete local rings which have no
In fact, let
R
be an integrally closed
local domain over a field of characteristic is not R
C-M .
C-M .
We give such an example in
Suppose
(cf. [9]).
T
were a module-finite
Since
nT(T) = Os(T) _
S
§5 •
(E"') .
Let
T
C-M
and suppose Then the
onto
extension algebra of S
is a direct summand of
by means of a T
as
S-modules,
.
A different approach to the higher dimensional of view of Conjecture
0 ,
is an integrally closed 3-dimensional complete local ring
contains the rationals we can retract
modified trace map
we have
3) •
see [9].
algebro-geometric
which is not
2) ,
since
is a complete regular local ring.
Note, however, that there are 3-dimensional module-finite
R-algebras
S
case involves taking the point
be a module-finite extension of
139
R ,
let
S.
21 Sl, ..., sm assume
be a set of generators for
s I = I) ,
relations on
and let
mn
S
indeterminates
1 < k < m ,
over
R .
I -< i,j -< m
In order to map 2
as an
(ril , ..., rim) ,
Sl, ..., sm
s.s. i j = C =I r.. l~k sk ,
S
For each
i, j
generate the module of
we can write
"
such that if
~
~(R)
we must find values in
denotes the
then the values for the indeterminates
X I = the
(Cn)
(for convenience,
1 < i < t ,
homomorphieally into
xij k
R-module
n
by
E.j r..X. i j j = O,
n
n
by
Let
rl, ..., r d
for fixed lq
n
for some
n ~
i < i < t . -
-
(r , ..., r
the equations
.
q ,
(~n)
we can solve
In certain good cases
(Xijk)
identity matrix.
(e.g. if
R .
(~n) R
.
have a solution in
be a regular system of parameters for
and for every integer
matrix
for
satisfy the matrix equations:
xixj : ~:I rijkX~, 1 _< i,j _< m We want to know whether,
n
R
R .
Suppose that
modulo
has an uncountable
algebraically closed residue class field) it is possible to show that a system of equations has a solution in
R
if and only if it has a solution modulo every
power of the maximal ideal or, equivalently, modulo every (~n) n
modulo
as an
lq
is the same as to find an
(R/lq)-module.
Henc%
if
R
R ,
for some integer
and every integer
n
then
module which is free of rank
Since
R/lq
is a
S
n
is a complete regular local ring with an
possesses a maximal
as an
q ,
S/IqS
is a field and
X
is a module-finite
C-M
module if and only if
(R/lq)-module.
0-dimensional Gorenstein local ring, i.e. a self-injeetive
Gorenstein local ring and R-fret
S
possesses a finitely generated
local ring, it might be tempting to conjecture that if
S-module is
But to solve
q
(S/lqS)-module which is free of rank
uncountable algebraically closed residue class field and extension domain of
I
S
R
is a
0-dimensional
is a module-finite local extension, then some
but this is false. E.g. let is an indeterminate,
and let
140
S = k[x] : k[X]/(X 3) , R : k[x 2] .
where
It is quite easy to
k
22 see that no
S-module is
R-free.
Quite generally, if
(R,P) ~
same socle nonzero and
yR = yS
R-free.
then
For if
R/P ~ S/Q) ,
E
is an
E
as an
yu = 0 ,
S-module:
and if
2~/ = ~i (Yri)ei = O
Ann S e I : 0 , as
S-modules
and
S
(since
We also know that
gives an
Se I S
R/P = S/Q ,
then no
R-free,
E / 0 ,
R-module, then they are QE = PE .
and not all the
For if r.
u
is in
are in
P ,
i
R-relation on the
S-injeetive).
R / S ,
1
is a submodule of
is
as an
in fact,
i
and
S-module which is
E
u = Z. r.e. I
then
0-dimensional local rings with the
are minimal generators for
minimal generators for QE - PE ,
are
(which implies that
S-module is
el, ..., e t
(S,Q)
E
ei
Now, since
ye I ~ 0 ,
and therefore a direct summand
But then
PS = QS = Q ,
and it follows that
since
PE = QE .
R = S .
This argument yields the following:
Proposition 4.2.
Let
(R,P)
algebraically closed and let Suppose that
S
P-primary ideal such that
J
Proof. S/IS
S
be a complete regular local ring with b_ee a module-finite domain extension,
possesses ~ maximal I
o__ff R ,
lies over S/IS
C-M
module.
there is an ideal
S ~ R .
Then for each irreducible o__ff S
IS
properly larger than
I .
is a local extension of
cannot have the same socle as
disjoint from the socle of
Thus, if Conjecture extra assumption on
J
R/P
S .
R/I .
R/I , J
R/I
and
R/I
is Gorenstein.
and there is an ideal
J/IS
Hence, of
S/IS
is the required ideal.
(E) is true, the conclusion of (4.2) holds without
the
It would be interesting to know whether this is really
true.
Another idea which might work in the higher dimensional case is to show that if
S
is a module-finite
then for some functor
F
some finitely generated still an
S-module:
extension domain of a complete regular local ring from finitely generated S-module
E ,
F(E)
is
R-modules to themselves and for R-free and
in fact, it is a HOmR(E,E)-module.
utilized in the second proof of
(E)
in dimension
141
R ,
2
/ 0 .
For
F(E)
is
This general idea is in the preceding section, and
23 a variation generated
(in which the functor is defined only on a subclass of finitely R-modules, namely,
S-modules)
for example, that in dimension S-module
E
3
1
For then, by the same exact sequence HomR(E,R )
We note,
it would be sufficient to find a torsion-free
of homological dimension
can conclude that
is used in the third proof.
is
over
(*)
R
such that
DR(EXt~(E,R))
given in the second proof of
> 0 .
(E) ,
we
R-free.
We devote the next section to our remaining suggested approach to the higherdimensional case. purposes,
We conclude this section with the remark that, for certain
adjoining round brackets indeterminates is harmless in looking for
maximal
C-M
modules over a complete local domain.
In other words, if several indeterminates, S .
R .
homomorphism
Then
is a complete local domain and
then if
To see this, represent
local ring
R[t]
(S,P)
S(t)
S
S(t)
is module-finite over for some
n .
R(t) ,
(R',P')
of
R
We can choose an
such that
h
R' .
homomorphism m ,
.
But since
in
R'
denote generators for
P .
R[%l/h]
R'
(R,P)
If
is
0 th
integer
such that
h
extension domain R'/P'
and then we
specializes to a
R-free we can view
S ~ T~m(R)
,
which induces a R' ~ ~m(R)
for
.
deficiency
is a finitely generated
M
definition depends on the choice of The
~ R'
in
such that
denotes a local ring and
M / 0
to associate certain integers with
notation).
such that
h
DEFICIENCIES AND SYMBOLIC POWERS
Throughout this section,
d
t
and then we have a homomorphism
5.
R
R-free module-finite
This gives a homomorphism S -~(R')
and we have a
does not vanish identically on
can choose values for the variables unit of
module, so does
We can choose a polynomial
with some coefficient outside the maximal ideal of 1/hi).
C-M
denotes
as a module-finite domain extension of a complete
f: S(t) ~ n ( R ( t ) )
f(S) ~ n ( R [ t ,
some
possesses a maximal
(t)
PI' "''' Pw R-module, we want
which we call the deficiencies Pl' " ' "
Pw
defo~ = def0M
L(M) A p d M = 0 ,
where
142
of
M
(the
but we shall ignore this in our we define to be the smallest
L(M) = H~(M) = [m E M: ptm = 0
for
24 some integer
t] = U t AnnMPt .
indeterminates,
To define the
w s i = Zj= 1 tijP~~ ,
let
defkM : def~(t)M(t)/(s! , ..., Sk)M(t)
k th
deficiency let
I < i < k ,
M
M
occurs when
k = D(M) ,
kw
and let
.
It is trivial to see that the first nonvanishing module
tij. be
deficiency of a nonzero
and we regard it as a measure of how far away
is from having depth one more than it actually does. The significance
Theorem 5.1. D(M) ~ k ,
R(t)
.
Le__~t M
N be a nonzero submodule of defkM P M . Then D(N) ~ k+l .
Nc
The question of whether
Thus, we may adjoin
deficiency,
Sl, ..., sk
(M/N)(t) ~ M(t)/N(t) '
kw
denote
,
and hence on
not an associated prime of
N(t)
N(t)'
have an element
N(t)
(t) in
M(t)
u 6 (s I, ..., Sk)M(t)
P(t)
Working modulo
(Sl, ..., Sk)M(t )
defoM (t)
is - 0
Suppose we
We must show
so it suffices to show
u' E L(M(t)') , since P ( t ) u ~ (s I, . ., . Sk)N(t) . .~ (Sl, . defkM u' 6 P(t) M(t)' , since u E N c pdefkMM . Hence,
we see that
, Sk)M(t)
,
and
,
M(t)' = 0 ,
as required.
We can now complete the argument in the fourth proof of given in
D(N(t)) 2 k .
(Sl, ..., Sk)N(t ) .
that
u 6 (Sl, ..., Sk)M(t ) ,
It
0 ~ N(t)' ~ M(t)' - (M(t)/N(t))'
P(t)u~
k th
and
Then it remains to show that
N P(t)
of
"general position".
u E (Sl, ..., Sk)N(t ) = N(t) ~ (Sl, ..., Sk)M(t ) ,
i.e.
D(M/N) ~ k .
as in the definition
that
u E L(M(t)')
with
..., Sk)M(t ) = (Sl, ..., Sk)N(t ) .
such that
.
(R,P)
such that
as well, so that
Since
N(t) N (sl,
of the following:
is unaffected by tensoring with
Sl, ..., sk
...., Sk).
is exact, we have that of
M
is a regular sequence on both
®R(t) R(t)/(Sl'
u
D(N) ~ k+l
indeterminates
and we obtain a sequence
follows that
Let
be a finitely generated module over
and let
Suppose that Proof.
of this notion for us here is a consequence
§3 •
143
(E)
in dimension
2
25 Corollary 5.2.
Let
(R,P)
be ~ Complete local domain ~n~
Drime of coheight at least one. Proof. all m
m ,
Let
d = deflR .
Clearly,
since any element of
is so large that
intersection
P - Q
Q(m) ~ pd ,
D(R)
Q(m)
and
we are done.
D(R/Q (m))
Q
is
0 ,
R
of
nopzero
m ,
D(Q (m)) ~ 2 .
R/Q (m)
Hence,
if
is a domain the
and since pt
~
are at least one for
on
But since
is contained in a given power
R
P
is complete it
for all sufficiently
m .
In fact, suppose that whenever define
Q(m)[M]
Lemma 5.3.
for some
Le___tt (R,P)
b__eeprimes o__ff R . integer.
Let
M
a
localizing,
I) implies since
R
in
R-P ,
i) ~
M
au E pmM]
F = Rk .
By the Artin-Rees
large integer
m ,
and
~(t)[M]
R
we can embed
Q(m) c p t *
Q(m) c p t *
we only need to show that
Q~
Q(m) c Q.(m)
is any prime of ,
3)
(37.8)
M
lying over
irreducible
t*
and
R .
m ,
since
To show that
we have
~
Since
144
S
Q~(m) = O ,
and we are done.
At this point we want to propose that a systematic
By
such that
~m Q(m)S = O . Q ,
of
in a free module
for large
be the completion of
S
by Corollary
is analytically
Let
if
be a given
Q(m)[M] c i l
Q(m)[M] c Q(m)[F] c pt*F .
is a domain,
t
Q ~ QI
over a complete regular local ring.
Then it suffices to show that
m ,
Q c QI '
irreducible.
lemma we can choose an integer
S
so that
closed, o__rr
2) implies
(R,P)
over
we
o__rr
is analytically --
RQI
,
R
Then we have:
3)
is torsion-free
for large
•
is integrally
is module-finite
Since
a prime of
M ~ (M/QmM) ® R Q
2) RQI
3) obviously,
QI = P "
Q
R-module and let
= P ,
we can assume instead that
ptM c pt*F .
R-module and
be a torsion-free
Then for every sufficiently
Nagata [13] ,
is an
be a complete local domain an__~dlet
Suppose either that:
Proof.
M
to be the kernel of the natural map
Q(m)[M] = [u E M:
while
large
is a nonzerodivisor
of the symbolic powers of
follows that large
Then for all sufficiently
Q
study of deficiencies
26 (defiQ(m)[M])
and residual deficiencies
(defiM/Q(m)[M])
of symbolic powers of
primes on modules be made, because understanding the behavior of these functions as m - ~
could well lead to a proof of Conjecture
(E) .
The general idea is to look for filtrations where the inclusions are proper, and complete local domain I)
(R,P)
D(M/Mi) ~ I ,
M
0 = M 0 c ~ - ~ ... c M n : M ,
is a module of dimension
n
over a
such that the filtration satisfies the conditions:
0 < i < n-I (so that
D(Mi+I/Mi) ~ I
for each
i
as well),
and defn_i_l(Mi+2/M i )
2) ~i+i/Mi ~ P
Mi+l/Mi , O < i
If one has such a filtration one can show by induction on D(Mi+I/Mi) ~ n-i ,
0 < i < n-I ,
so that, in particular,
n-i
that
D(MI) ~ n .
We shall now restrict attention to dimension 3 for a while to see what might be enough.
Let
torsion free
(R,P)
be a complete local domain of dimension 3 and let
R-module of finite type.
If for each
i
we can find
M
he a
N. c PIM i
such that
D(M/Ni)
> 2 ,
then eventually the
N.
--
Theorem 5.1. D(M/Ni) ~ I and
D(Ni)
~ 2 ,
we have that
D(M) ~ 2 ,
D(M) ~ 2 .
D(M/N i) ~ 2 ,
is a height one prime of
instead of requiring require that that
RQ
and
We can then apply
i ~ def2M
taken together
D(N i) ~ 3 •
However, if we are taking the sequence Q
by
In fact, by a first application of Theorem 5.1, the facts that deflM M and N i c P imply that D(Ni) ~ 2 , and then since D(M/N i)
Theorem 5.1 again:
where
3 ,
i
are both
imply that
have depth
R ,
D(M/Ni) ~ 2 ,
defl(M/Ni)
N. l
is a discrete valuation ring.
Q(m)[M] ,
we do not need to know quite as much:
so that
be bounded as
to be a subsequence of
defl(M/Ni)
i -- =
: 0 ,
we only need to
However, we shall need to assume
This is true for all but finitely many
height one primes in a complete local domain since the singular locus is closed.
Proposition 5.4.
Le___tt (R,P)
be a complete local domain of dimension 3 •
Let
Q
be a height on___eeDrime of
Let
M
be a finitely generated torsion-free module over
defl(M/Q(i)[M] )
is bounded as
R
such that
i ~ ~ .
RQ
is a discrete valuation ring. R .
Suppose that
Then for all sufficiently large
145
i ,
27
D(Q(i)[M]) = 3 . Proof.
Since
height two prime bound
QI
is complete and the singular locus is closed we can choose a of
defl(M/Q(i)[M])
Lemma 5.3)
lemma).
and choose
and
M I = Q~J)[M] c PJ'M .
RQI
is regular.
Let
Choose
Let
j
d = def2M I .
k
(by Choose
i
(which is possible from Lemma 5.3 and the Artin-Rees
D(MI/Q(i)[M]) ~ 2 .
Remark 5.5.
Q c QI
J' 2 max [k, deflM ] .
defl(M/Q(i)[M]) ! k ,
D(Q(i)[M]) 2 3 ,
M
such that
Q(i)[M] c pdM I
Since
follows that
R
so large that
so large that
Let
R
as required.
and
Since
MI/Q(i)[M ] c Pk(M/Q(i)[M]) Q(i)[M] c pdM 1 ,
,
it
we have that
(We have found a good filtration.)
It is possible to bound deficiencies in a more "functorial" way.
be a finitely generated
R-module,
T = R + Pz + p2z2 + ... = R[Pz] c R[z] , Then there is a functor
F
F(M) = M + PMz + p2Mz2 +
...
from c
(R,P) local. where
z
is an indeterminate over
R-modules to graded
M[z] = M ® R R [ Z ]
,
Let R .
T-modules: F(M) = Im(M®RT ~ M®RR[Z])-
i.e.
Then
defo(M) S since
~T(F(M))
~(H~T(F(M))
= Z i (H~(M) N piM)zi
and
,
H~(M) 0 piM
contribution to the described length for each
i ,
makes a nonzero
0 < i < defoM .
It follows at
once that
0 def i(M) _< g(HpT(t ) (F(M(t) ®R(t)(R(t)/(Sl,
We next want to make estimates of the behavior of
..., si))) ) .
defl(M/Q(i)[M])
under good
conditions.
Proposition 5.6. M
Let
(R,P)
be ! complete local domain of dimension 3, let
be a finitely generated torsion-free
prime of
R
has depth 2.
such that
RQ
R-module, and let
Q
be a height one
is discrete valuation ring and such that
Then there is a constant
c
146
such that
M/Q(1)[M]
defl(M/Q(i)[M]) J ci
fo__Er
28
all i . Proof. that in
Choose
yQ(1)[M] R(t),
for
x E Q
= xM .
such that
Let
s
y
of
def I .
Then
y , s
We shall begin by proving by induction on
w E Hi = ~ ( t ) ( M ( t ) / ( Q ( t ) ( i ) [ M ( t ) ] + s M ( t ) ) ) y i-I w E E i = Q(t) (i) [M(t) ]+sM(t)
Ei ,
and
in
P - Q
such
.
P
.
that
i f and o n l y i f y i-1 w
On the one hand, if
Rad ( y i - l , Q ( i )
On the other hand, assume
i
of
is a system of parameters
(M/Q(1)[M]) ®R(t) = M ( t ) / Q ( t ) ( 1 ) [ M ( t ) ]
and
( y i - 1 Q(i), s ) w c
and choose
be a general linear combination of generators
as in the definition
R(t)/QR(t)
Rq = XRQ
w E H.
is in
s) D Rad (Q, y, s) = P ( t ) If
i = 1 ,
Ei
then
.
we have that
1
y~w + sv E E 1 •
But since
and since
is a system of parameters
y, s
• + sv 6 Q(t)(1)[ M(t)] y~w Now suppose
M(t)/Q(t)(1)[M(t)]
and
Q(t)(i)[M(t)] c Q(t)(1)[M(t)] Suppose
w = w' + sv' ,
We then obtain
yw' = xw* ,
that
yi-2w*
is in
is in
It follows that
Ei_ I .
w = y l-1 w
Then
w
,
,
where
•
is in
Q(t I l)[M(t)]+sM(t),
Since
is in
Q(t)(1)[M(t)]+sM(t) and
v"
v'
show that
d
such that
I
w
is in
yQ(t)(i)[M(t)]
By the induction hypothesis,
say
(Q(t), s, y) •
Suppose that
w
w' = yiw" .
yi_lw,, E E i
Let
w' 6 yiM(t) Choose
But then
j
and
we then have
is in
as required.
c = 2d .
We need only
H. N P(t)2diM(t)
i
where
.
and
y(yJw' + sv") = x(yJw * + sv*)
.
Then
i
(Q, s, y)2iM(t) c QiM(t) + sM(t) + yiM(t)
w = w' + h
M(t)
yi-lw = yi-l(w' + sv') =
P(t) d c
H. A p(t)ciM(t) c E.
is in
.
then must be in
yi-lw' + s(yi-lv ') = yi-2xw* + s(yi-lv ') E xEi_ I + sM(t) c E i , Now choose
R(t)/Q(t)
from the fact that
is in
Q(t)(1)[M(t)]
yJ(yw') + s(yv")
Q(t)(i-l>[M(t) ] .
is in
of depth 2 over
R(t)/Q(t)
it follows that w'
Then
yv" = xv* .
" y3w* + sv*
so
,
where
as well.
C-M
yJw + sv E Q(t)(i)[M(t)]
yJw' + sv" E Q(t)(i)[M(t)]
Q(t)(1)[M(t)]
for
we can deduce that
i > I ,
is
and
h E E i c H.m '
such that
yJw' E E i •
w' = ylw" E E.I '
so that
.
Since
so that Then
QiM(t) + sM(t) = EicHi, w' E H i N yiM(t)
yj+iw"
E Ei ,
so that
w = w' + h E E i + E i = E i ,
as required.
This result falls short of what we would like to have:
147
,
some additional
29
condition is n e e d e d to guarantee boundedness of the residual deficiency.
But the
conditions of the hypothesis are very easy to satisfy, so that it might well be possible to meet extra conditions. The following lemma shows how to construct situations which satisfy the hypothesis of Proposition 5.6.
D. Eisenbud suggested trying this type of
construction to me.
Lemma 5.7. such that
Let
(R,P)
be a local domain, and let
RQ is a discrete valuation ring.
torsion-free module over R-module
M
Proof.
such that
R/Q .
Le___~t E
E
Then there is a finitely generated torsion-free
M/Q(1)[M] m E .
The localization of
in a free
0 ~ E ~ F
E
(R/Q)-module
at the prime
0
F = (R/Q) t
of K
R/Q of
is a finiteR/Q .
way that
(R/Q) p
E
as
Im f ,
where
maps onto the image of g: R p ~ R t .
R-modules:
Let
E
We can
in such a way that the inclusion
becomes an isomorphism when we localize at the prime
can therefore represent
of free
be a height one prime
be a finitely generated
dimensional vector space over the field of fractions embed
Q
0
of
R/Q .
f: (R/Q) p ~ (R/Q) t = F in
F .
M = g(R p)
We can lift M c Rt
f
We
in such a to a map
g
is torsion-free.
Since the diagram
Rp _onto > M ~-__> R t
o I
n t
O
commutes onto
E .
E = f(u(RP))
u
¢~r
\o
, ~
(R/Q)p onto,>~, ~
= v(g(RP)) : v(M)
,
v
n t O + (R/Q)o
and there is an induced map
It remains to see that the kernel of this map is precisely
¢
of
Q(1)[M]
and for this purpose it suffices to compute the kernel after localizing at may therefore assume that so that the inclusion of free and consequently f r e % It is clear that to
E
induced by
(R,Q) E
in
is a discrete valuation ring and that (R/Q) t
is an isomorphism.
Then
M
Q .
is torsion-
and we are trying to show that the kernel of
is surjective.
It suffices to show
148
We
R/Q : K ,
¢
Q~i is contained in the kernel, and we know that the map of ¢
M
is
QM. M/QM
3O that this map is an isomorphism, or, equivalently, that Since
¢
is surjective, dimK(M/QM ) ~ t ,
dimK(M/QM ) = rank M ~ t ,
a maximal
C-M
module over
Q
such that
module
R
E
such that
M~
Rt ,
and we are done.
Now, given a complete local domain height one prime
and since
dimK(M/~ ) = dimKE = t .
RQ
over
(R,P)
of dimension 3 we can choose a
is a discrete valuation ring.
R/Q ,
We can construct
and we can then construct a torsion-free
M/Q(1)[M] e E .
This verifies the statement that the
hypothesis of Proposition 5.6 is easily satisfied. To see that some extra condition really ~ needed to guarantee that defl(M/Q(i)[M] )
is bounded, not merely
Example 5.8. ideal
Let
(x,yz) c R ,
and let
Q = zR .
, where Then
Q(i)[M ] = (xz i ,yz i ) ~ (x,y)
depth 2, but that
R = k[[x,y,z]]
O(i) ,
defl(M/Q(i)[M])
we consider the following:
k
is a field.
Let
M
M/Q(1)[M] ~ x(R/zR) ~ R/Q
has depth 2 for all
i ,
be the has
which shows
is not bounded.
Despite this, there are interesting cases where the idea of Proposition 5-4 succeeds.
We conclude this paper with one class of such examples.
Example 5.9. Let
f
Let
k
be an algebraically closed field of characteristic
be an irreducible homogeneous polynomial in
XI, X2, f
k[Xl, X2, X3]
is a homogeneous system of parameters but
that the partial derivatives of
f
simultaneously only at
(One example is
(0,0,0).
conditions guarantee that
such that
X 2 ~ (XI, X2, f) 3
with respect to
X1, X2, X3
f = ~
and such
vanish
+ X~ + X~ .)
R 1 = k[Xl, X2, X3]/(f ) = k[Xl, x2, x3]
0 .
These
is integrally
closed but not proper in the sense of Chow [4], P. 817, and it follows from the Theorem on p. 818 of [4] that the Segre product of R 2 = k[Yl, y2 ] R = k[xiYj:
is not Cohen-Macaulay.
i = i, 2, 3
and
R1
with the polynomial ring
This Segre product
j = I, 2] c Rl[Yl, y2]
closed ring of dimension 3 which is not Cohen-Macaulay. depth 3 module using the ideas of this section.
is then an integrally We shall construct a
(We could localize at the
homogeneous maximal ideal and complete without essentially changing the situation,
149
31 but it will be simpler to continue in the graded case.) Let
Q = yl~[yl,Y2] N R .
R/Q ~ k[XlY2,x2Y2,x3Y2] ~ R I R .
We shall show that
Then
Q(i)
is a
note that
Q(i) = Yli R I[YI'Y2] A S .
that proof.
xlY2~ x2Y 2
for all
C-M
D(R/Q (i)) = 2
where
y~
i ,
Then
Then
f0x2 = g0xl
in
~
,
f0y]'Pyq
and since
i .
Xl, x 2
150
modulo
In fact, we shall To see this,
Yi' Y2 g
=
(yl) .
i = i, 2, 3] We shall show
which will complete the
fx 2 = gx I ,
and
follows easily.
Yl
R/Q (i) ,
bihomogeneous pieces of the same bidegree in =
is a height one prime of
which certainly suffices.
is the image of
f
in fact,
Hence, R/Q (i) = k[xiY7', xiY2:
fx2Y 2 = gxlY 2 .
we might as well assume that
Q
module for large
is a regular sequence in
Suppose that
is a domain;
is two-dimensional, and
show that
= Sl[Yl, y2]/(yl) ,
R/Q c ~ [ y 2 ]
and
f, g
break into
for which this is true, and g0ylPy q ,
is an
where
f0' go E R I .
Rl-sequence , the result
32
BIBLIOGRAPHY
I.
H. Bass,
2.
D. Buchsbaum and D. Eisenbud,
On the ubiquity of Gorenstein rings,
Math. Z. 82 (1963), 8-28.
What makes a complex exact?,
to appear in
J. of Alg. 3.
,
Remarks on ideals and resolutions, preprint.
4.
W. L. Chow,
5.
D. Ferrand and M. Raynaud,
On unmixedness theorem,
Amer. J. Math.
86 (1964), 799-822.
Fibres formelles d'un anneau local
Noeth6rien, Annales Sci. d e l'Ecole Normale Superieure 3 (1970), 295-312. 6. IV.
A. Grothendieck (with J. Dieudonn@),
(Seconde partie),
El@ments de g$om@trie alg@brique,
Publications math@matiques de l'I. H. E. S. n ° 24, Paris,
1965. 7.
A. Grothendieck (notes by R. Hartshorne),
Verlag Lecture Notes in Mathematics 8.
M. Hochster,
9.
,
I0.
,
Springer-
No. 41, 1967.
Grade-sensitive modules and perfect modules, Contracted ideals from integral extensions,
I. Kaplansky,
II.
Local cohomology,
Commutative rings,
Allyn and Bacon,
preprint. preprint.
Boston, 1971.
Topics in commutative ring theory, I. and IIl.,
duplicated
notes. 12.
G. Levin and W. Vasconcelos,
Pacific J. Math.
Homological dimensions and Macaulay rings,
25 (1968), 315-323.
13.
M. Nagata,
Local rings,
14.
C. Peskine and L. Szpiro,
Interscience Tracts 13,
Dimension projective finie et cohomologie locale,
Thesis (Orsay, Serie A, N ° d'Ordre 781), 15.
D. Rees,
16.
J. P. Serre,
in Mathematics 17.
to appear in Publ. I. H. E. S.
The grade of an ideal or module,
Philosophical Society
New York, 1962.
Proc. of the Cambridge
53 (1957), 28-42. Alg~bre locale.
Multiplicit@s.
Springer-Verlag Lecture Notes
No. II, 1965.
R. Y. Sharp,
Application of dualizing complexes to finitely generated
modules of finite injective dimension,
preprint.
151
33
Comments I.
There
is a serious
theorem
in [9]
Question
4.1).
if
R
the result
a field,
and the general
of a weak
It can be shown classes
get a version
where ,
Sl,S 2
of
case
is
(E) [9].
that
the
is an Rn-sequence ,
so that we need only divide
of all g-perfect
of Conjecture
152
case
form of Conjecture
in the 2-dimensional
lim G(Rn) -+ n out by these instead
after
still holds
in a revised version
[Rn/(Sl,S2)],
generate
of the main
However,
This will be shown
2.
gap in the proof
(see the first paragraph
contains
a consequence
and Corrections
(F)
modules
to hold.
to
C O M M U T A T I V E
RINGS
Irvin~ Kaplansky
i. I N T R O D U C T I O N I have
c h o s e n to s p e a k on the subject
a t o p i c w h i c h has maturity
opportunity
to o b s e r v e that
a historical touch time
f a s c i n a t e d me for years,
as one of the b a s i c
to take the hasten
on the recent to r e c o r d
The
to s u r v e y
subject
lightly
field begins,
number
The
each
defined
by
therefore
pen has
eras,
signalled
quadratic
of the i n t e g e r s
functions
or r a t h e r
by the a p p e a r -
established,
to
painstaking
an i n s i g h t
w i t h the
field.
to a
At the
of any
and so did the p a r a l l e l This
is the
"one-dimensional"
case we have to look to the
d u r i n g the
in the i n t r o d u c t i o n
they g a i n e d
of a q u a d r a t i c
of one v a r i a b l e .
geometry
did,
forms a m o u n t e d
the r i n g of i n t e g e r s
For the n - d i m e n s i o n a l
" r e s t e d on a most
3- THE F I R S T
Let me
and spend m o r e
century mathematics
binary
field became well
has a p t l y w r i t t e n ,
of the a x i o m a t i c
of
and K r o n e c k e r ,
of w o r k on a l g e b r a i c
from which
of 19th
theory
account
of a l g e b r a i c
geometers
like
paper.
Dedekind,
case of our subject. stream
I will
authoritative
into five well
four p e r i o d s ,
as so m u c h
complete
of Kummer,
algebraic
perspective.
background,
N. B o u r b a k i ' s
g r o w n to
I would
[6] are a c c o m p a n i e d
standard.
classical
rings,
ERA
Disquisitiones. G a u s s ' s
theory
which
era and then
2. THE P R E H I S T O R I C
virtually
chapters
erudite
seems to me to d i v i d e
ance of a v e r y i m p o r t a n t
hands
seven
on the
history,
and one w h i c h has of m a t h e m a t i c s .
it in h i s t o r i c a l
Bourbaki's
Noetherian
in detail.
a "prehistoric"
The
"structures"
note up to his u s u a l
comparatively
yet
of c o m m u t a t i v e
~OJ,
study
not always
19th century. that
A. W e i l
the a l g e b r a i c
of n u m e r o u s
special
found a m o n g m o d e r n
examples,
exponents
creed".
PERIOD
19th c e n t u r y b e g a n w i t h a m o n u m e n t a l
piece
of work;
it also ended
w i t h one. Hilbert's memoir (i) He p r o v e d nomials
[14] m a d e
that
three
extremely
if K is a f i e l d and
in n v a r i a b l e s ,
then
every
important
contributions.
R=K[Xl,...,Xn] the
ideal of R is f i n i t e l y
153
r i n g of p o l y generated.
-
L a t e r this was recast the Hilbert finitely theorem
and s l i g h t l y
basis theorem:
generated,
u n i t e d at one stroke
expressible
are a g a i n p o l y n o m i a l s . The
combination This
"syzygies"
investigated
generated
is to say,
observations
a finite
any p o l y n o m i a l
not,
set of in I is
coefficients
however,
that
ordinarily
be
by r e l a t i o n s
geometers.
For d e e p e r i n s i g h t
he next
ul,...,u r. He r e c o g n i z e d that he was now d e a l i n g
rather than
an ideal,
and p r o c e e d e d to the
This
is
decisive
... + Urf r = 0
but s h o w e d
"second
you reach O, or to state it otherwise, reached.
call
ideal
Hilbert's
of fl,...,f r w i t h
is m e a s u r e d
by the a l g e b r a i c
the r-ples
with a module
ring R every
t h e r e is t h e r e f o r e
expression will
lack of u n i q u e n e s s
ulf1+ called
form we now
case by case.
in K[xl,...,Xn]
as a l i n e a r
to the
in R[x].
is true
for I, say fl,...,f r. That
generators
unique.
generalized
a large n u m b e r of f i n i t e n e s s
been m a d e
(2) If I is an i d e a l
-
if in a c o m m u t a t i v e
t h e n the same
that had p r e v i o u s l y
2
tour de force was
it was
syzygies".
again
finitely
after n steps
Theorem:
after n-1 steps a free module
a farsighted
anticipation
is
of h o m o l o g i c a l
algebra. I in K[x1,...,Xn]
(3) G i v e n an i d e a l homogeneous
ideals,
s t u d i e d the
asymptotic number
degree
contained
Subsequently treatment Three
in I, and was
this t u r n e d
later,
Nullsteliensatz. of f u n c t i o n a l
of l i n e a r l y
I would
another
forms
at some point.
ideal
fundamental
Analogously,
in n v a r i a b l e s
t h i n k of R as c o n s i s t i n g
o v e r the
consists
of f u n c t i o n s
generation
complex
be the r i n g of
c l o s e d f i e l d K, and
space of n - t u p l e s
over K.
ideal in R is the set of all polynomials
The p o l y n o m i a l
is a good deal h a r d e r
theorem.
theorem
But here
i~ u n c o u n t a b l e .
is a curious remark:
(I say " c u r i o u s "
154
t h a n the
continuous
the N u l l s t e l l e n s a t z
because,
functions
functions
every maximal at some point.
Theorem:
the
if X is a
of all the
let R=K[x~ .... , x ~
on the
rigorous
theorem:
the y o u n g e r
algebraically
polynomial.
varieties.
the r i n g of c o n t i n u o u s
in C(X)
Hilbert
of a g i v e n
for the m o d e r n
of a l g e b r a i c
like to try to i n t e r e s t
space and C(X)
only w i t h
characteristic
so let me b e g i n by r e c a l l i n g that
every maximal
polynomials
independent
led to a c e r t a i n
proved
dealt
can be g e n e r a t e d by forms)
out to be a key t o o l
Hilbert[15~
analysts,
Hausdorff
on X, t h e n vanishing
(actually Hilbert
ideals w h i c h
of i n t e r s e c t i o n m u l t i p l i c i t i e s
years
compact
i.e.,
vanishing function
is easy if K
a m o n g o t h e r things,
in H i l b e r t ' s
-
day one did not
specify
of c o m p l e x n u m b e r s . ) me b r i e f l y recast
repeat
K,
it.
Given
element
that the e l e m e n t s This
The N u l l s t e l l e n s a t z ideals
and points.
between prime course
not just
the p r i m e
he p r o v e d primary
that
This
additions
were m a d e
up to his
Cambridge
tract
of his time, recently. himself
~7,
cannot p.68~
linearly
inde-
between
require
algebra
[21] w h e n
as an i n t e r s e c t i o n
powers.
in a n u m b e r of p a p e r s Macaulay
of his w o r k won full
was
leading
far a h e a d
appreciation
w i t h the
of
as we can expect
of prime
respects
fail to be i m p r e s s e d
is of
and geometry.
us to study all ideals,
as a p r o d u c t
In m a n y
found M a c a u l a y ' s
correspondence This
step was t a k e n by L a s k e r
by M a c a u l a y
[23].
between maximal
varieties.
is as good an a n a l o g u e
and some a s p e c t s
One
in l i n e a r
for R is a v e c t o r
to a o n e - t o - o n e
an e x p r e s s i o n
of an i n t e g e r
Important
over K, are
algebraic
the b r i d g e
The next
admits
theorem
of the r e p r e s e n t a t i o n
can be
contain a
exercise
correspondence
can pass
and g e o m e t r y
ideals.
field let
over K.
in b u i l d i n g
any ideal
ideals.
a ranging
and i r r e d u c i b l e
of b o t h a l g e b r a
closed)
(and a nice
if K is u n c o u n t a b l e ,
F r o m this one
step
[20], but
i d e a l M in R, our p r o b l e m
sets up a o n e - t o - o n e
ideals
a vital
The needs
i/(u-a),
is a c o n t r a d i c t i o n dimension
K was the
[19] and
R/M c o i n c i d e s w i t h K. Now if the f i e l d R/M
It is a fact
algebra)
of c o u n t a b l e
in
(since K is a l g e b r a i c a l l y
u.
pendent. space
a s s u m e d that
is c o n t a i n e d
a maximal
that
is l a r g e r t h a n K it m u s t transcendental
-
it b e i n g t a c i t l y
The p r o o f
to the s t a t e m e n t
3
fact that
w o r k to be " t e i l w e i s e
only
Krull
recht m ~ h s a m " .
4. THE S E C O N D P E R I O D The
s e c o n d era was
tance
inaugurated
of this p a p e r
exaggeration
a x i o m the
ideals.
conclusion
generated, A very
intersection expressible
of H i l b e r t ' s
simple,
general
of a finite
number
as an i n t e r s e c t i o n that u s h e r e d
irreducible
is p r i m a r y .
ideal
It is e n t i r e l y a p p r o p r i a t e later
called
paper
[28]. The i m p o r -
it is s u r e l y not m u c h of m o d e r n
basis
she takes
theorem:
of two
larger
that rings
155
ideal
is
condition
on
any ideal
ideals
ones).
in the new era:
"Noetherian".
chain
shows that
of i r r e d u c i b l e
as a sole
that every
the ascending
argument
of an
algebra.
of the b a c k g r o u n d
or equivalently
charming argument
were
that
to call her the m o t h e r
After a leisurely description
finitely
by Emmy N o e t h e r ' s
is so great
(i.e.
And t h e n
is the ideals not comes the
a short p r o o f that
w i t h the a s c e n d i n g
chain
any
condition
-
Let me take the normalization ring
in w h i c h
is n i l p o t e n t Let
4
-
space to r e c o r d the proof.
that
the i r r e d u c i b l e
0 is i r r e d u c i b l e (for this
is what
ideal
I do it w i t h the h a r m l e s s
is O.
So: we have
and have to p r o v e that it m e a n s
to say that
a Noetherian
every
zero-divisor
0 is a p r i m a r y
ideal).
ab=O w i t h b~O. Let I
ascend,
be the a n n i h i l a t o r of a n . The ideals ( I ) n u l t i m a t e l y b e c o m e stable, say at I k. We claim
and t h e r e f o r e
Ik~Rak:o.
x=ya k lies in the i n t e r s e c t i o n .
For s u p p o s e
O:xak:ya 2k,
Then
Y c I2k. But I2k :Ik, so yak:o, x:O. By the i r r e d u c i b i l i t y of O, e i t h e r I k or Ra k is O. But I k c o n t a i n s b. Hence Rak:o and a is n i l p o t e n t as required. In a s u b s e q u e n t treatment
of rings
(subsequently book
D~
[29], E m m y N o e t h e r
w i t h the p r o p e r t i e s
called Dedekind
taught
I insert Artin
paper
all these
at this point
descending
things
as the n a t u r a l algebras
it.
Natural
dimensional rings w h i c h and r e m a i n rings
are not
the next n a t u r a l
rings w h i c h rings.
are f i n i t e l y
Among
PERIOD
In the e a r l y
1920's
commutative
at the t i m e
[17] gave
and m a k e s
The t h i r d era was the
day K r u l l has
rings w i t h p e n e t r a t i o n
of 1935
principal
inaugurated
ideal theorem.
37 of
worse
~7].
Krull
The
one on p a g e
common.
Infinite-
Noetherian
substantially
under
(possibly non-commutative)
over
commutative class
Noetherian
of rings!
to carry on the t r a d i t i o n .
investigated
every
and d i s c e r n m e n t .
aspect
of
His E r g e b n i s s e
subject
as he saw it
r e a d i n g today.
by the p a p e r
gives this
of
of the A r t i n i a n
commutative
for this
arose
rings
I am b e g i n n i n g
~167
in w h i c h
Krull proved
( W a r n i n g to the g r o u p - t h e o r i s t s :
n o t h i n g to do w i t h the p r i n c i p a l matters
of
those
Artinian
are hard to w o r k w i t h
with
over fields class
In ~927
subject But
are not
a b r o a d view of the
stimulating
note.
of rings;
the
tangible
algebras,
a name
a s t r o n g new voice
up to the p r e s e n t
monograph
is the
we n e e d
rings
that,
generated modules
o t h e r things,
5. THE T H I R D
I argue
algebras
step
number theory
of a l g e b r a i s t s .
class
of r i n g theory.
the most
finite-dlmensional
largely mysterious.
generation
and c o n t r o v e r s i a l
of A r t i n i a n
axiomatic
van der W a e r d e n ' s
way to g e n e r a l i z e
perhaps
and f i n i t e - d i m e n s i o n a l
control,
Right
rings,
long,
For forty years t h e s e
to a p i e c e
examples
division
in a l g e b r a i c
call the dual
chain c o n d i t i o n .
finite-dimensional to doubt
to a whole
one m i g h t
have been r e g a r d e d
found Before
a parenthetical
[i] i n t r o d u c e d what
w i t h the
rings).
gave her c e l e b r a t e d
ideal t h e o r e m
of group theory.
name to two t h e o r e m s :
25 is a c o m p a r a t i v e 156
this has
see p a g e s
triviality.)
To m a k e 25 and
-
As N o r t h c o t t theorem rings.
justly
arrived,
remarks
to B o u r b a k i ' s
ideals
one p a s s e s
theorem;
is minimal
relatively
a prime
that
of the
his s e v e n c h a p t e r s
rank of a p r i m e
chain.
over an i d e a l
saying
in a Noetherian
Theorem:
over a principal
ideal,
. . . 3 Pn of prime
ring,
ideal has rank ~ i. F r o m
e a s i l y to the k i n g - s i z e p r i n c i p a l
ideal m i n i m a l
subject.
ideal theorem.
a c h a i n P = P o ~ PI
of l e n g t h n, but no l o n g e r
a prime ideal which this
exists
ideal
s h a d o w of p o l y n o m i a l
asinorum
thoroughness
I i n t r o d u c e the
that P has r a n k n if t h e r e
u n t i l the p r i n c i p a l
a pale
a kind of pons
have not yet r e a c h e d the p r i n c i p a l To state the t h e o r e m ,
i05-i06],
rings were but
it r e m a i n s
It is a t e s t i m o n i a l
-
E30, PP.
Noetherian
To this day
5
ideal
g e n e r a t e d by n e l e m e n t s
has rank < n. K r u l l was
of course
inspired
and t h e i r h o m o m o r p h i c is in that
context
I note that Rees
Rees
in
variety
about p o l y n o m i a l
rings
the p r i n c i p a l
ideal t h e o r e m
of c l a s s i c a l p r o p e r t i e s
of the d i m e n -
and of the d i m e n s i o n
gave a s p a r k l i n g new p r o o f
[31~, a p a p e r that
of an i n t e r s e c t i o n .
of the p r i n c i p a l
also m a r k e d the a p p e a r a n c e
ideal
of the A r t i n -
lemma.
The p r i n c i p a l
ideal theorem yields
in a N o e t h e r i a n strengthened
ideals
r i n g s a t i s f y the
form,
a fixed prime.
Of course,
and a f o r t i o r i
I see the next technique
paper
[3~
was
n o t e d that
"local"
nation
can be f o u n d
"quasi-local"
be a r e a s o n a b l e The two m a j o r
have
Krull
on local rings. although
need
The
it was not
in c o m p l e t e
has stuck.
generality. to
It is to be
Noetherian.
concept
It is r e a s o n -
s h o u l d h a v e the
convention
I suggest
short
is not u n i v e r s a l l y
"not n e c e s s a r i l y
literature.
for a c o m m u t a t i v e
choice
the
circumlocution
in the
c h a i n on all
enough there
c a l l e d such a r i n g a " S t e l l e n ring"
important
arisen:
from
ideal r e d u c e s m a n y p r o b l e m s
to i n c o r p o r a t e
that the most
and the d r e a d f u l
local ring"
~8~
it a r r i v e d
at a m a x i m a l ideal.
in a
from a fixed prime.
~938 p a p e r
r e n a m i n g to "local
But two d i f f i c u l t i e s
accepted,
but p e r v e r s e l y
getting established,
is here m e a n t
able in m a t h e m a t i c s name.
chain condition
ideals
b o u n d on chain d e s c e n d i n g
ascending
in 1946 that
localization
but C h e v a l l e y ' s
ideals,
as K r u l l ' s
case of a single m a x i m a l
ring",
descending
on chains
landmark
In p a r t i c u l a r ,
i n f o r m a t i o n that the p r i m e
by a x i o m we have the a s c e n d i n g
on p r i m e
of l o c a l i z a t i o n
until U z k o v ' s
the
to wit w i t h a u n i f o r m
not be a u n i f o r m b o u n d
the
facts
B r i e f l y put,
a consequence
sion of an a l g e b r a i c
theorem
by the k n o w n
images.
Noetherian
that the
r i n g w i t h one m a x i m a l
desig-
ideal w o u l d
on w h i c h to agree.
accomplishments
in
~8]
157
are the i n t r o d u c t i o n
of the t e c h -
-
nique of completing special
a local ring,
class of local rings.
local ring R can be generated
if this m a x i m u m
is attained.
[7] "regular"
not the happiest meanings
choice
and the recognition
time a definition
of regularity
REMARK.
This account
alternative restricted
at the latest
for regularity,
corresponded
of a nonsingular
(i) If R is regular,
(2) Is a regular
sequence rings,
raised
local rings that
effort,
Cohen's
for geometric
thesis
stubbornly
complete
regular
local rings.
a kind of climax theorems
local rings,
now had a m e t h o d localize
easily.
in answering But the
then
local
which came close to
for launching
(hopefully)
a systematic ideal,
work your way
The method works with gratifying should
for this
for complete
at a general maximal
theory,
frequency,
learn how to use it. But it does
e.g. the two problems m e n t i o n e d
above
continued
to hold out.
A pat.tern began to appear repeatedly. for a special geometric
rather
also succeeded
the geo-
remained open.
[9] provided
structure ring.
29
He gave structure
and every young ringtheorist not always work,
geometry,
[41, p.
regular
back to the original
domain?
(1) in the affirmative
The subject
use Cohen's
factorization
Zariski
attack on a proposed problem: complete,
ideal in R, is Rp (the localiza-
settles
of developments.
especially
original paper.
arise in algebraic
for both problems,
being decisive.
in Krull's
local ring a unique
the second affirmatively
The late I.S.
concept
This was fully set
to P) regular?
interpretation
case,
to a very classical
on a variety.
and P is a prime
tion of R relative
general
used
Chevalley
[4~.
were promptly
With additional
2,947
somewhat.
point
metrical
has
different
Krull and Chevalley and furthermore
It turned out that regularity
For the regular
"regular"
[39] of an entirely
indeed,
Two questions
count,
in order and It was perhaps
of all, there was already at that
is over-simplified:
his local rings
forth by Zariski
accepted.
for rings.
definitions
that
called R a "p-Reihenring"
Another name was doubtless
is now universally
and, worst
ideal M of a
then n is the maximal
in R. Krull
due to von Neumann
concept
of a certain
if the maximal
by n elements,
since,
in mathematics;
-
Recall that
length of a chain of prime ideals
Chevalley's
6
class of Noetherian
case. Workers
A certain theorem would get proved
rings,
usually
in the field relentlessly
to remove the restriction.
This was partly
158
something accepted
like the the challenge
for the usual reason:
the
-
elegance
and c l a r i t y
a n s w e r to current diophantine integers
and a n t i c i p a t e d
geometry
calls
since
in this
local
no n i l p o t e n t
elements.
Chevalley This
the exact
its q u o t i e n t that
the
course
integral
is p e r h a p s did not
is f i n i t e l y
closure
the
in
subject
of
o v e r the
the
inter-
Let R be a
c o m p l e t i o n R ~ has local domain,
in a s p l e n d i d p a p e r
not [3~,
for e v e r y r i n g f b e t w e e n R and
generated
of T is a f i n i t e l y
that this
a little m o r e
for a g e n e r a l
[26]. But Rees,
c o n d i t i o n needed:
it is s t a n d a r d
it was
rings
generalize.
[8] p r o v e d that
is not true
closed
field which
for i n s t a n c e ,
of p o l y n o m i a l
which
case the t h e o r e m
domain.
But p a r t l y
images.
instance
e v e n if R is i n t e g r a l l y pinpointed
needs;
for a study
and t h e i r h o m o m o r p h i c
geometric
-
of a good g e n e r a l i z a t i o n .
I w i l l give an e x p l i c i t esting
7
condition
o v e r R, we m u s t
generated
is f u l f i l l e d
have
T-module.
Of
in the g e o m e t r i c
case. Incidentally,
there
an integrally
closed geometric
again
is a c o m p a n i o n t h e o r e m
local
a domain.
Nagatats
example
in
generalization.
It w o u l d
be n i c e
to have
showing exactly
what
6. THE F O U R T H
is n e e d e d
by the a r r i v a l Let us r e c a l l (i.e.
k e r n e l KI, If the gical
a direct
summand
finished
N o w it turns behaves
Krull's
theorem,
was h e r a l d e d
was a n n o u n c e d
the job
in
E34].)
that
a
a k e r n e l K2,
etc.
is K
we say that the h o m o l o n =n; the value of n does not
out.
If no K
Buchsbaum in
is p r o j e c t i v e n chains of syzygies.
and Serre.
[2], w i t h
(The
full d e t a i l s
Let R be local w i t h m a x i m a l
if and only if d(M)<~. further
onto A, g e t t i n g
to H i l b e r t ' s
by A u s l a n d e r ,
a projective
For r e f e r e n c e
w h e n R is r e g u l a r ,
a little
d(A)<~
in
i d e a l M. later
for
A.
out,
very w e l l
regularity
carried
portion
it f o l l o w s
every R - m o d u l e
was
A map
onto K I g e t t i n g
d(A)
Note the r e s e m b l a n c e
R is r e g u l a r
and its a d v e n t
of a free m o d u l e )
of A is n, and w r i t e
Auslander-Buchsbaum
we note that
a complete
t h e o r e m to hold.
G i v e n an R - m o d u l e
of the K's to be p r o j e c t i v e
d(A)=~.
[31; Serre
out
algebra.
a projective module
The big t h e o r e m was p r o v e d
Then:
for Z a r i s k i ' s
as we need.
on the way the p r o c e s s
we w r i t e
again rules
R ~ is
a c o m p a n i o n to R e e s ' s
era right now,
of h o m o l o g i c a l
dimension
depend
fourth
as m u c h
t h e n map
first
[26]
if R is
[42]:
then the completion
PERIOD
We are l i v i n g in the
module
due to Z a r i s k i
domain
for the t e c h n i c a l under
localization,
is e a s i l y p r o v e d
problems
was
r e a s o n that
solved.
that this
to be i n h e r i t e d This r e s o u n d i n g
159
homological
dimension
characterization
by Rp.
Thus the
triumph
first
of of
of the new h o m o l o -
-
gical method marked Noetherian Krull's
a t u r n i n g point
p r o b l e m on u n i q u e
of the
factorization
to A u s l a n d e r
this
solution took
into
the f o l l o w i n g :
subject
of c o m m u t a t i v e
map
K is free,
common divisor
One t r a n s l a t e s
generated
a free m o d u l e
the
till form
the p r o b l e m prove
onto the two g e n e r a t o r s
on one g e n e r a t o r .
to p r o v e u n i q u e
years
by two e l e m e n t s ,
d(I) ~ i, K is p r o j e c t i v e ;
necessarily
needed
four m o r e
[4]. Let me d e s c r i b e
of MacRae.
if I is an ideal
a k e r n e l K. Since
had to wait
and B u c h s b a u m
at the hands
d(I) ~ I. ( E x p l a n a t i o n : local,
-
rings.
it too y i e l d e d
getting
8
of I,
and since R is
This
gives
factorization.)
us the
greatest
Now s u p p o s e
one
d(I) = O, i, or ~. The r e g u l a r i t y h y p o t h e s i s w o u l d rule
could prove
out ~, and all w o u l d be done.
MaeRae
has p r o v e d
the d e s i r e d
theorem,
if I is any ideal generated by t~o elements in a commutative Noetherian ring, then d(I)=O, ~, or ~. He did it first for
w h i c h we restate:
domains
in
[24];
and D. M u m f o r d , Lamentably theorem
in
[25], w i t h the aid of s u g g e s t i o n s
he c o m p l e t e d
(or not,
depending
the proof. on your point
of this kind for ideals
Many
of the
subsequent
that
is, the h o m o l o g i c a l
I have no doubt arbitrarily,
I have
generated
developments
that m o r e
tools were
of view)
by three
might
there
or m o r e
be d e s c r i b e d
sharpened
applications
selected
from M. A u s l a n d e r
await
five of t h e s e
is no elements.
as " i n t r o s p e c t i v e " ,
for t h e i r
own sake.
future d i s c o v e r y .
developments
But
Somewhat
for a b r i e f
discussion. i. Tate
[3~
gave a q u a n t i t a t i v e
Serre t h e o r e m
(ABS).
sharpening
W i t h M the m a x i m a l
(the kth "Betti n u m b e r " )
denote
the
successive
(k-1)st
above.
kernel,
when
says
that
the n u m b e r kernels
of a local r i n g R,
of e l e m e n t s
needed
are f o r m e d
let B k
to g e n e r a t e
as d e s c r i b e d
Bl(=n , say) is the m i n i m a l n u m b e r of g e n e r a t o r s
In p a r t i c u l a r ,
of M. ABS
of the A u s l a n d e r - B u c h s b a u m -
ideal
if some Bk=O t h e n R is r e g u l a r .
Now one knows
that
B k is at least as large as the b i n o m i a l c o e f f i c i e n t nCk. Tate proves: if Bk=nC k for a s i n g l e sharpened
this
the P o i n c a r @
considerably.
series
technique
was to form an a l g e b r a
Gulliksen
function
have
is to s e t t l e w h e t h e r of Z; this
is k n o w n
cases.
if it could be chosen to be m i n i m a l recently,
and he and o t h e r s
A challenging problem
~ BkZk is a r a t i o n a l
in a n u m b e r of s p e c i a l Tate's
k ~ 2 t h e n R is r e g u l a r ,
[13] a n s w e r e d
2. There
is a dual t h e o r y
modules.
Bass
resolution
in an a p p r o p r i a t e
this
class
dimension,
of N o e t h e r i a n
160
and he a s k e d
sense.
Just
in the a f f i r m a t i v e .
of h o m o l o g i c a l
[5] s t u d i e d the
of M,
b a s e d on i n j e c t i v e
rings
R w i t h the
_
property
that the m o d u l e
does not
do this
has p r o j e c t i v e
R i t s e l f has
on the p r o j e c t i v e
d i m e n s i o n 0.)
w i t h one that G o r e n s t e i n purposes,
and so these
a surprisingly
finite
injective
side of the ledger,
It t u r n e d
out that
[~i] had i n t r o d u c e d
rings
large n u m b e r
are
dimension.
the
of c o n n e c t i o n s
for of course R
condition
for a l g e b r a i c
called G o r e n s t e i n
(One
rings.
coincided
geometric Bass
with other topics
surveyed
in N o e t h e -
rian rings. 3. Like
any good t h e o r e m ,
t h e o r e m due to L e v i n [2~.
Suppose
ABS d e s e r v e s
is p u b l i s h e d
for the m a x i m a l
for some p o w e r M n. We cannot equal 0 in a n o n r e g u l a r 4. The
theorem proved
R-sequence the
is,
in his joint
conclude this
~n
divisor
in R/(al,a2) , etc.
R-sequences
natural
way in the i n i t i a l
stages
[~
and of Rees
I shall m e r e l y still
[3~.
another
be c a p a b l e
generated
of b e i n g
Vasconcelos's
theorem:
by an R - s e q u e n c e The a p p l i c a t i o n
to ABS
fifth r e m a r k
and
of A u s l a n d e r report
of t h e i r own,
ring,
for w h e n I=M,
school:
d(A)<~,
and x is a z e r o - d i v i s o r
divisor
on A?
(R/I)-module.
M/M 2 is a v e c t o r
free.
if R is local,
in R, does
for
I can be g e n e r a t e d
and I/I 2 is a free
is an open q u e s t i o n w h i c h has b e e n
to the C h i c a g o
of a local
is that M should
N o w we are ready
I in a local
is c e r t a i n l y
in a v e r y
and B u c h s b a u m
are here to stay.
condition
by an R - s e q u e n c e .
if d(I)<~
An
exactly,
into a n o n z e r o -
of the r e g u l a r i t y
time the
is i m m e d i a t e ,
space over the f i e l d R/M 5. This
this
More
their appearance
an e x t e n s i v e
for an ideal
if and only
made
say that R - s e q u e n c e s
characterization
ideal M;
R-sequences.
if a I is a n o n z e r o - d i v i s o r
of the w o r k
They d e s e r v e
emphatically
ring R with maximal
a challenge
concerns
in R/(al) , a 3 m a p s
into a n o n z e r o - d i v i s o r
but
exception.
of n o n z e r o - d i v i s o r s .
al,...,a n in R is an R - s e q u e n c e
sequence
in R, a 2 maps
I quote
L38]
d(Mn)< ~
for M n can
that R is r e g u l a r ,
a sequence
following
r i n g R we have
is the only
by V a s c o n c e l o s
The
paper with Vasconcelos
ideal M of a local
R. But:
so to speak,
generalizations.
s o m e t h i n g of
A an R - m o d u l e
with
it f o l l o w that x is a zero-
7. C O N C L U S I O N There
are m a n y m o r e
and I hope main
topics
that
an e x p e r t w i l l w r i t e
t h e o r e m on a l g e b r a i c
fourteenth problem, classical problems, for c h a r a c t e r i s t i c
s h o u l d be s u r v e y e d one soon.
correspondences,
Nagata's
ingenious
the r e s o l u t i o n O, the
formula
Nagata's
I mention
solution
counter-examples
of s i n g u l a r i t i e s for i n t e r s e c t i o n
161
in a full report,
As e x a m p l e s
Zariski's
of H i l b e r t ' s
to a flock of
and H i r o n a k a ' s
proof
multiplicities
as an
-
alternating problem
sum of dimensions
of the rigidity
factorization series
domains
iO
-
of Tor's,
of Tor,
Samuel's
and his notable
Lichtenbaum's investigations
examples
rings,
and finally
the generalization
of everything
to schemes,
as required
braic geometry appropriately picture
(a scheme pasted
being,
together).
in the contemporary
mathematical
of power
by the Grothendieck
speaking,
But I hope
of the
on unique
of failure
by the new foundations
crudely
of the field of commutative
solution
a number
school
of alge-
of rings
I have given a useful
Noetherlan
rings and its status
scene.
BIBLIOGRAPHY The bibliography ference, ~0~,
except
[12~,
is confined that
[2~,
and
Hamburg ~ (1927), 2. M. Auslander
Nat.
is an actual
Zahlen,
Abh. Math.
(1957),
4.
Acad.
Sci.
Sci.
5. H. Bass,
Univ.
Homological dimension in Noetherian
USA 42 (1956),
36-38. Amer.
Math.
Proc.
Nat.
390-405.
, Unique factorization Acad.
Sem.
251-260.
, Homological dimension in local rings, Trans. Soc. 8 5
re-
a reader may wish to consult:
[4~.
and D.A. Buchsbaum,
Proc.
3.
that
Zur fheorie der hypercomplexen
I. E. Artin,
rings,
to items to which there
I have added texts
USA 45 (1959),
in regular
local rings,
733-734.
On the ubiquity of Gorenstein rings, Math.
Z. 8 2
(1963),
8-28, Alg~bre commutative,
6. N. Bourbaki, 7. C. Chevalley,
Ch. I-VII,
Hermann,
On the theory of local rings, Ann.
Paris,
of Math. 4 4
1961-1965. (i943),
690-708.
8.
, Intersections Amer.
9- I.S.
Math. Cohen,
Trans.
Amer.
Soc. 5 7
of algebraic and algebroid varieties,
(1945),
On the structure and ideal theory of complete Math.
Soc. 59
Trans.
i-85.
(1946),
162
54-iO6.
local rings,
-
11
-
I0. J. Dieudonn@, Topics in Local Algebra,
Notre Dame Mathematical
Lectures no. i0, Notre Dame Univ. Press, Notre Dame, Indiana, 1967. ii. D. Gorenstein, An arithmetic theory of adjoint plane curves, Trans. Amer. Math. Soc. 72 (1952), 414-436. 12. A. Grothendieck (avec J. Dieudonn@), Elements de G~om~trie Alg~brique, Ch. IV, Inst. Hautes Etudes Sci. Publ. Math., 20, 24, 28 (1964-5-6). 13. T.H. Gulliksen, A proof of the existence of minimal R-algebra resolutions, Acta Math. 120 (1968), 53-58. 14
D. Hilbert, Uber die Theorie der algebraischen Formen, Math. Ann. 36 (1890), 473-534.
15
Math. Ann. 42 (1893),
, Yber die vollen Invariantensysteme,
313-373. 16
W. Krull, Primidealketten
in allgemeinen Ringbereichen,
S.-B. Heidel-
berger Akad. Wiss. Math.-Natur. KI. (1928), 7. 17
, Idealtheorie, gebiete,
18
Ergebnisse
der Mathematik und ihrer Grenz-
Springer-Verlag, Berlin, 1935. , Dimensionstheorie
in Stellenringen,
J. heine Angew. Math.
179 (1938), 204-226. 19
, Jacobsonsche hinge, sionstheorie,
20
Hilbertscher Nullstellensatz,
Dimen-
Math. Z. 54 (1951), 354-387.
S. Lang, Hilbert's Nullstellensatz
in infinite-dimensional
space,
Proc. Amer. Math. Soc. ~ (1952), 407-410. 21
E. Lasker, Zur Theorie der Modulen und Ideale, Math. Ann. 60 (1905), 20-116.
22
G. Levin and W. Vasconcelos, Homological rings,
dimensions and Macaulay
Pacific J. Math. 25 (1968), 315-323.
23
F.S. Macaulay, Algebraic theory of Imodularl Tracts no. 19, Cambridge, 1916.
24
R. MacRae, On the homological dimension of certain ideals, Proc. Amer. Math. Soc. 14 (1963), 746-750.
25
systems,
Cambridge
, On an application of the Fitting invariants,
J. Algebra
(1965), 153-169. 26
M. Nagata, An example of normal local ring which is analytically ramified,
27
Nagoya Math. J. ~ (1955), 111-113. , Local Rings,
Interscience,
163
New York, 1962.
-
28. E. Noether, 24-66. 29.
Idealtheorie
1 2
-
in Ringbereichen,
Math. Ann. 83 (1921),
, Abstrakter Aufbau der Idealtheorie in algebraischen Zahl- und Funktionenk~rpern, Math. Ann. 96 (1927), 26-61.
30. D.G. Northcott,
Ideal Theory,
Cambridge Univ. Press, New York 1953.
31. D. Rees, Two classical theorems of ideal theory, Philos. 32.
, A theorem of homological
Soc.
5__22 ( 1 9 5 6 ) ,
33.
algebra,
Proc. Cambridge Philos.
605-610.
, A note on analytically Math.
Proc. Cambridge
Soc. 52 (1956), 12-16.
unramified
local rings, J. London
Soc. 36 (1961), 24-28.
34. J.-P. Serre, Sur la dimension homologique Noeth~riens, Tokyo,
Proc.
Internat.
des anneaux et des modules
Sympos. Algebraic Number Theory,
1955, pp. 175-189.
35. J.F. Tate, Homology of Noetherian J. Math.
rings and local rings,
Illinois
i (1957), 14-27.
36. A.I. Uzkov, On rings of quotients
of commutative
rings, Mat. Sbornik
N.S. 13 (1943), 71-78. 37
B.L. van der Waerden, Moderne Algebra, Berlin,
38
1931.
vol.
II, Springer-Verlag
(Also later editions and English translation.)
W. Vasconcelos,
Ideals generated by R-sequences,
J. Algebra ~ (1967),
309-316. 39
J. von Neumann,
On regular rings,
Proc. Nat. Acad. Sci. USA 22
(1936), 707-713. 40
A. Weil, Foundations Publ. vol.
41
Amer. Math. Soc. Colloq. R.I., 1946, Rev. ed. 1962.
O. Zariski, The concept of a simple point of an abstract algebraic variety,
42
of Algebraic Geometry,
29, Amer. Math. Soc. Providence,
Trans. Amer. Math. Soc. 62 (1947), 1-52. , Analytical
irreducibility
of normal varieties,
Ann. of
Math. 49 (1948), 352-361. 43. O. Zariski and P. Samuel, Commutative Algebra, Princeton,
Van Nostrand,
N.J. vol. I, 1958, vol.j II 1962.
University of Chicago Chicago,
164
lllinois
Addendum
13
-
(July,
1972)
This lecture w a s originally presented in June, 1968 as the first JefferyWilliams lecture of the Canadian Mathematical Congress.
It w a s published in 1972
by the Canadian Mathematical Congress, along with other lectures of this series. Thanks are expressed to the Congress, and to John J. M c INarnee, their Executive Director, for permission to reprint it here. I shall take the opportunity to update the lecture a little, beginning with the fact that Krull died in 1971.
His m o n o g r a p h [17] appeared in a second edition, re-
vised and expanded, in 1968. In E G A ,
the Grothendieck-Dieudonnei Elements
Chapter IV appeared as no. 32 in the I H E S series.
[12], a fourth part of
Springer's reprinting of E G A ,
in book f o r m and slightly revised, has begun with the appearance in 1971 of Chapters 0 and I. It should be added that §6 of Chapter 0 consists of significant "CompleSments '' concerning c o m m u t a t i v e algebra. H e r e is an indication of the order of magnitude faced by a person planning to update himself in commutative ring theory since 1968.
With the Jan. 1968 issue,
Mathematical Reviews began to have a section titled " C o m m u t a t i v e associative rings and algebras".
Since then, to the middle of 1972, approximately 750 items
have appeared under this heading.
I shall content myself with listing the books that
have appeared (in addition to m y C o m m u t a t i v e Rings,
Allyn and Bacon, 1970).
In
this list I a m interpreting c o m m u t a t i v e ring theory rather narrowly. M.F.
Atiyah and I.G. MacDonald,
Addison-Wesley, R.W.
Introduction to C o m m u t a t i v e Algebra,
1969.
Gilmer, Multiplicative Ideal Theory,
Applied Mathematics, no. 12, 1968.
Queens Papers on Pure and
In revised and expanded form, this will
shortly be published by Dekker.
165
-
M.D.
14
-
Larsen and P.J. McCarthy,
Multiplicative Theory of Ideals,
Academic Press, 1971. H. Matsumura,
Commutative Algebra, Benjamin,
1970.
D.G. Northcott, Lessons on Rings, Modules, and Multiplicities, Cambridge, J.T. Note
Series
1968. Knight, No.
Commutative
5, Cambridge,
Algebra,
1971.
166
Lon. Math.
Soc.
Lecture
ON EUCLIDEAN RINGS OF ALGEBRAIC FUNCTIONS by University
cipal
Boulder,
1.
Introduction.
A Euclidean
with
number
ring
exist
y
a natural such
and
~ ~(~).
It
A rather
curious
great
is
that
these
in
the
are case
almost
in
of
course
in
the
never
a function
be
the
is
well
Riemann
ring
such
well
known that
be
of
in
To be
one
the
ring
finitely
generated
then
Mordell-Weil
Theorem
is
contains
an
V
that
quently,
the
however,
prove
over
k
are
It is
at
Mordell number
set
cases
in
the A:
following
If
be
least
conjecture field
*This
only
k
work
all
O (~ that
known
have
only
finitely
of
is
when
asserts
part
of
defined let
k
non-zero
this
be
an
for
or
it
is
easily
is
a principal are
with
to
few
show
that they
infinite
field
and
is
Let
k .
subset
of
U)
is
[2].
the
It
a Dedekind
When
k
is
a
numbers
every
open
domain.
numerous.
XK / k
XK / k .
rational
that
ideal
very
arise
relatively
on
seen
domains.
curve
open
of
~(6)
algebraic
example,
field
or
there
ideal
paper
an
no poles
elements
~ d 0
domains
field
be
has
domain
integral
principal
on an
constant
principal
many
fact
in
are
purpose
U
an
6 = 0
the
finite
O (Y) is
rings
See,
and
either
contexts,
let
prin-
set
U
Conse-
We w i l l ,
results.
in
was
a
O (U)
then
then
of valid
finitely
remarked
two
field
open.
and
geometric
and
shown to be algorithm.
with
and
exact
be
on the
ideal
= (~ E KI~
even
such
which
k-rational
should
not
extension
open
Theorem
need
k
be
principal
precise,
whose
O (~
In
is
more
over
domain.
the
U
K
defined
such
while
will
ring
functions
variable
for
purposes
the
all
It
algebraic
functions are to any possible
~ = ~y+6
that,
Euclidean.
Colorado
~
in
that
arithmetic
surface
fact
~
sorts
Euclidean.
known that
and
is
of
our
function
a
rings
field
valued
for
however,
all
known to
ring
all
the
fact,
of
K
for
6
frequency
of
k
MacRae*
class of rings of algebraic not Euclidean with respect
the
are
R. E . Colorado,
ABSTRACT. A large ideal domains but
together of
of
the
Euclidean
for
is
a number
field
examples
supported
satisfy
all
curves
many k-rational
by
on the
not
k
that
points
a National
167
the of
Riemann
any
proper
and
the
hypothesis genus
points
Science
at and
surface open genus
of least so
set of
Theorem two
Theorem
Foundation
for
over
K U.
K
over
A.
The
a
A applies.
Grant.
2
Theorem field
is
B:
k
Moreover,
Let
K
and w h o s e
assume
be a function genus
that
k
whose
non-trivial
automorphisms
since
Main
tion
~: ( O 6
ring
in
(0}) ~ O
such
that
2.2:
that
of
U
Riemann
on the result
k,
includes
of the
K.
over
Then,
it f o l l o w s
the f i n i t e K
that
for
a very
large
k.
of field
possesses
~(~
K
group
constant
extensions
if
surface
automorphism
for a n u m b e r
[3,
Prop.
7]
2.3:
Let
We
can
now
is not
over
number
no
k. of c a s e s
appears
to h o l d
for
of h e l p f u l
conversations
connected
with
If t h e R i e m a n n O(U)
excludes
Proposition
for
the
2.2 O(V)
is
some
finitely
since
~M
6
is
an
in
= 0
also
infinite
~ or
and
Euclidean with
if t h e r e
~ ~ 0
~(6)
results
there
from
[3].
is a funcexist
y
~ ~(~).
closed
subset
of the E u c l i d e a n
Euclidean.
O(V)
finitely
field
and
subgroup
of
first
result.
K
extension
that
K*
variable
over
such
K*
finite
a
field
= k*G
then
of K = k.
a proof. of
our
be a function
surface
Suppose
~,~
and
definition
is c a l l e d
is a m u l t i p l i c a t i v e l y
be
for
is not E u c l i d e a n
Proof:
~
for all
generated
K
following
a proof.
a proof
Let
domain
ring
k
18]
give
A:
M
for
of t h e
= ~y + 6
localized
Prop.
Theorem
of
~
is a finitely [3,
use
that
If
See
See
~(V)*
such
the
G
We make
N
then
If
that
than
exact
under
in one v a r i a b l e
of
this
An integral
~
Theorem
then
an i n v a r i a n t
extension
triviality
to P. S a m u e l
2.1:
Proposition
k.
that
whose
genus.
Results.
Definition
k.
set
t w o and field
less
any
in one v a r i a b l e
problems.
2.
and
open
of the
of large
I am i n d e b t e d these
over
be remarked
the assumption curves
most
is s t r i c t l y
for a n y p r o p e r
It s h o u l d
is at least
is a f u n c t i o n
k 0 = GF(q)
Euclidean
genus
field
K
field
contains
only finitely
for any p r o p e r
~(~ many
in one
open
is E u c l i d e a n . k-rational
Let
points.
is a l o c a l i z a t i o n generated
set
since
168
of
U V
many
the
infinite
k-rational
on the R i e m a n n be
Then O(~.
the complement
of
U
is a l s o E u c l i d e a n
Moreover of
points
surface.
an o p e n s u b s e t
O(V)
field
V
the g r o u p in the
by
of units
Riemann
3 surface
is
a finite
conditions
asserts
map o f
O(V)*
extension
of
k-rational
finite
A but for
field
is
k
Moreover, = GF(q)
no
non-trivial
not
Euclidean
algebraic that
k
k,
the
quotient
that
k
a function
genus
any
finite That
are
Galois group
is
of
of
KI .
Theorem
however,
it
But
of
the
that
O
2.3
Motzkin
the
natural
(V)/~O(V)is of
a field
Samuel
impossible
makes
since
essential
for
open
set
has
curves
applies none
use
of
defined
K
of
not
over
the
only
curves.
k0
and
the
theorem
~I
can
be
~i
over
The
latter
k
k
k'
k'(~O ~"
where
of
can
group
since
be
be
trivial,
169
k
k"
is
can is
by
Galois a subgroup hypothesis.
It
k0
be
is
clear
since
K
was
pg.
107].
K
Hence
That
field
be
is
of
in
one
described
closure
Theorem
the
over
~1 = ~,(~)~
that
algebraic
of
the
and
of
extension
this
k0
a function
[4,
k.
of
genus
certainly
the
applies
as
i, such
k"
k0
a ffnite
regarded
Theorem of
Now
Grauert-Samuel to
is
[4,
over
extensions
the
is
If
Let
points
field
O(~ K
K.
k.
possesses
that
not.
K = k Qk
Since
K
for
of
finite
points.
k-rational
points.
field
surface
algebraic
many
K~.
follows
Suppose
are
if
k-rational
and
applies
of
picked
done.
extension
finite
Hence
many
infinitely
a subfield
some
are
= koQk0k
algebraic
is
we
it
Riemann
the
Then,
constant
extensions
over
K.
k,
exact
under
variable
of
of
finitely
Grauert
ring
one
the
whose
invariant
that
on
k-rational
of
the
U
k
has
in
than
variable
an
extension
and
since
one
and
field
only
set
many
of
over
any
in
two
less
over
and
theorem
field
least
strictly
fields
and
at
applies
k0
a finite
ko
function
K
A
field
say~
group
is
that
the
defined
to
is
infinitely
field. is
a
proper
claim
two,
a curve
be
Moreover
over
the
conjecture
is
have
is
such
[3].
since
Mordell
genus
K
there
deeper
whose
of
say~
is,
first
~(V)
See
hence
This
and
respectively.
variable
K
closure
to
of
onto.
and
Let
Theorem
least
~
the
[4].
for
at
Euclidean
V.
somewhat
antomorphisms
is
by
so-called
then
to
k
in
of
the
is
B:
and
assumed
is
is
element
(V))*
= k.
is
We
proved,
K
O(V)
an
over
result
whose
Proof: be
(V)/~O
of
assume
k0
is
(V)/gO(V)
example,
Theorem
since
there
degree
0
next
Theorem
(0
points
Our
Now
that
to
and shows that
See,
set.
o£
2,
pg.
109].
k
and
the
automorphism ~r
a
= ~ .
Let,
now,
O~
be
arise.
We
the
valuation
have
ring
either
k#
of some
< O~
k-rational
or not.
point
Suppose
of
K = K'.
the
that
first
Two
case
cases
holds.
Then -
k~
is
over less
a subfield k0
by
of
equa-ls
the
that
residue
of
hypothesis.
Now
class
K
over
let
k~
field
k
~
while
be
the
= k .
the
genus
separable
However, of
k
closure
the
genus
of
over
k0
is
strictly
k#
in
k.
Since
of
Jl
k
s e
~H
and
k
the genus mula
are separably of
k~'s over
generated
T0
[I, pg. 134] applies,
implies
that the genus
points
of
K
infinitely
~H
kJ~
points
k0
-/i
of
k
over
that of
be contained
from ko-rational
many of the k-rational of
k
field
ko '
k0"
point
of
~H s
in
points
K
-k ii
and
and thus genus for-
kJ~ s"
This
and we have a contradiction.
O~.
Hence the k-rational
of
T #.
are defined
over which these k-rational
This is, however,
~11s = Tp
The relative
to the pair of fields
is at most
that
arise entirely
find a finite subfield rational
equals that of therefore,
of
Hence it is not possible
over the perfect
impossible
Now, by hypothesis,
over
k.
points
Hence we can
arise as k0-
and we have a final
contradiction.
References I.
M. Eichler,
I n t r o d u c t i o n to the Theory
Academic 2.
R. E. MacRae, J.
of
Press,
Numbers
and Functions,
New York 1966.
On unique Alg.,
of Algebraic
factorization
17(1971),
3.
P. Samuel,
About Euclidean
4.
P. Samuel,
Lectures
in certain rings of algebraic
functions,
243-261.
rings,
J.
of
on Old and New Results
Bombay, 1 9 6 6 .
170
Alg.,
19(1971),
on Algebraic
282-301.
Curves,
Tats Institue,
SUBFIELDS
OF by
This The
writer
our
results
is
an
account
(P.S.),
problems
is are
By
still
"curve"
we
coordinates
care
about
inflation
over
which
C.
We
From
C
in and
to
FUNCTION
and Pierre of Colorado
recent
work
expert
on
these
is
only
Our
work
the
an
FIELDS
Samuel
done
by
us
on
elliptic
questions, at
C
is of
projective
algebraically
resources
case
time,
nonsingular
If
curve
exclude
time
a
concerned!) the
an
known.
mean
their
are
ELLIPTIC
its
function
suspects
that
preliminary
stage,
fields. part
of
and
many
open.
have
elements
some
being
already
2 OF
R.E. MacRae* University
of
not
INDEX
algebraic
closed
exhaustion,
k
is
a
subfield
defined,
Ck
characteristic
field,
at
least
of
this
denotes
characteristic 3 will
as as
big
Points
as
far
the
set
as
which
has
also
be
excluded.
of
quite
k (char
of
needed
(we
closed
k-rational a In
(k) ~ 2)
curves do
not
transcendental
algebraically
2,
is a n e l l i p t i c c u r v e d e f i n e d o v e r a f i e l d
curve.
points
different
of
flavor.
sections
and
field
I to
Ck
V,
is non-
empty.
§I - I n t r o d u c t i o n W e fix a n o r i g i n o n points.
Any
subfield
We are interested [k(C):K]
= 2).
K
C,
so that
C
is a g r o u p for the w e l l k n o w n a d d i t i o n of
of genus
0
of
k(C)
in the s u b f i e l d s
K
of genus
is p u r e l y t r a n s c e n d e n t a l
over
0
(i.e.
and index 2 of
The following results are well known or easy
(i.i) - K
is the f i e l d o f i n v a r i a n t s o f some i n v o l u t i o n
defined
follows:
we
as
have
(1.2)
-
(u,v) on
This
T(M)
C
If
is
a
unique
point
A
C Ck
(non-exclusive"or"):
7 E AUtk(C),
such
that,
for
all
geometrically M
E C,
and
if
= -A-M
we
view
are
the
its
unique
author
There
k(C)
k.
C
as
coordinates
was,
point
in
a
plane of
at
part,
A,
cubic
curve
then
infinity)
.
supported
by
E
a
=
y
=
k(y-v) x-u
grant
171
2
from
x
3
+ ax
2
+ bx
(as
usual,
the
National
we
+ c , take
Science
for
origin
Foundatiom
2
(1.3)
- For
k(c)
any
generating
= k(s,t)
with
cubic
polynomial
(1.4)
- The
roots
2M
M
of to
of
as
of
being
a
4 tangents are
the
K
= k(t)),
the
A
solutions
to of
a
points the
root
of
cubic
the
may
write
polynomial
with
branch
from
we
quartic
polynomial
to
drawn
(i.e. squarefree
a quartic
correspond
tangents
K
at
over
curve
k
(a
infinity).
k(C)
equation
over
C.
T(M)
K,
The
= M,
or
contact i.e.
+ A = 0.
-
their
cross-ratio
it
If
takes
the
one
roots
(1.6)
(H)
- This
It
(EH)
is It
of
the
the
(E)
the
set
has
~
of
the
and
(EH)
are
group
Aut(C)
modulo
that
C
of
say
conjugate corresponding
if
then
are
as
y
The
(EH)
y
2
= x
2
= x
are
there
two is
3
3
of
involutions,
~
1
r
C
over
except
say
that
case
we
C, say
in
in
which
C
Aut(C)/T
equation
(char
~ 3):
of
of
C
(x,y)
~
(- x,iy)
+ c
,
(x,y)
~
(jx,
(i.e.
rate,
set of v a l u e s
Or
of
k.
harmonic; P(t)
are
3). automorphisms if
order
4
(H),
we
(j3
than
call
~
(resp.6),
Aut(C)/[
the
provided is
a generating
(i 2 =
if
any
M):
and
-y)
at
is
other
case
,
k
C
admits
+ bx
over
P(t)
that
3 and
cubic
or
cyclic
K;
cases:
(M~B-
is
t4,
closure
characteristic
involutions
t3,
. This
algebraic
in t h e
t2,
of
. . . r' i - r ' r - 1 the
tl,
a di-
element
of
-i)
= 1 )
i E k,
resp.
j E k),
we
k-equianharmonic)
k(C)
subfields a
1
order
independent
occur
the
defined
(resp.
subfields
r
characteristic
follows
(H)
k-harmonic
that
(In
12).
automorphisms is
C,
suitable
not
ones and
~ 3
order
T
- Conjugate We
is
only
a
1 .
we
which
does
+M)
of
of
case in
(EH)
(M ~B
i - r,
6 elements,
which
the
in
a value
class
(j3 =i)
translations
hedral
1
r, -r'
in
(case
characteristic
these
},
written has
actually
1
{_ j, _j2
translations
group
are
values
equianharmonic Cases
P(t)
isomorphism
{- i, 2, is
of
(tl,t2,t3,t4)
characterizes
§II
P
here
P(t)
the
these
(1.5)
If
t
s 2 = P(t),
is viewed
(geometrically) points
element
of
genus
E AUtk(C) this
such
0 and that
means
172
index o(K)
2,
K
= K t.
and
K I of If
T,
t
k(C) are
are the
say
3
T
Let
A,A'
be
Proposition are
the
2.2
conjugate
elements
then
This, when
simple
rank
the
is
(resp.
m
= - A - M , Tt(M)
translations
conjugacy
and
classes
of
= - A t- M .
Then:
involutions,
subfields
or
shows
= 0,
i.e.
K and
correspond
k-equianharmonic, element
that
the
of
things
AUtk(C)
conjugacy
m(A)
E 2C k
(EH)
A t-
A E 2C k
or
A t-
m(A)
E 2C k
Ck
is
-
permits
finitely
Suppose with
of
to
Kt the
computation
of
char(k)
~
3
and
2t
elements
of
order
conjugacy
classes
of
subfields
case)
k
= R ,
number
of
modulo
the
Q.E.D.
little
the
more
compli-
translations,
(2.1)
or
.
is
equivalent
A'-
m2(A)
E 2C k
number
of
conjugacy
with:
-
classes
that 2
Ck
or
is
1
a
finitely
(t =0,1,2
as
generated well
known).
is:
2 r+t
r =2s r
a
M ~ B-M),
generated:
that
(here
involution
are
relation
A'-
computations,
the
A- A t = - 2(B +A~))
or
If
the
M ~ B +M ( r e s p .
A E 2C k
r
-
is
only
A +At+2B
a generating
(EH)(here
t
T(M)
translation
k-harmonic
(H)
and
The
~
A t-
number
2.4
that
contains
the
(general
Remark
o'TO
(H)
2.3
of
such
E 2C k .
computation
group
Proposition
Then
is
some
the
group
~
we c a l l
plus
Ck
AUtk(C)
A-A'
if
C
If a
If
A - A t = 2B
When cated.
-
of
=
Ck/2C k.
fact,
means
points
iff
of
In (1)
-1
t
( 2 .i)
and
=2s
the
and
cases
(H)
connected
t =1,2)
22s+b-I
t =0,2)
1
and
(EH)
components
of
+ 2 s+l
1 + ~
(22s+t
cannot C R.
-
- i)
occur. Then
2s
We 2C
have is
t =1,2
the
unit
compo-
R nent,
and
the
Proposition index 2 sI of --
2
number
2.5 of
= Pl(t) PGL(2,k),
-
of
Let
k(C), (cf.
0(x)
K and
1.2). -
conjugacy
classes
= k(t) let
Then ax +b c x + d
'
is
and
K1 =
us
write
k(C)
K
and and --
K1 e
t.
k(t I)
are
E k*
= k(s,t) conjugate such
173
be
that
two
subfields
= k(Sl,t iff
I) there
of with exists
genus
0
and
s 2 =P(t) an
element
and e
(2.6) If that
k(t)
and
~(k(tl))
q(Sl)2
k(t I)
= k(t).
- a t + b.) = Pl~c-~-~+d
k(t),
O(Sl )2
and
d 4p at + b ) l(c---~-~-~+d ) = e2p(t).
(ct +
P(t).
quotient
Then
s2
s
differ
since
the
and
two elements e
if (2.5)
2
~
b e an a u t o m o r p h i s m
with
q ( s 1)
a,b,c,d
generate
by a square
must be a constant
Conversely,
let
at + b - ct + d
q(tl)
Since
and But,
are conjugate,
in
the
Remark
2.7.
there
of
-
exist
k(C)
Let
are
squarefree
quartic
~
be the k - i s o m o r p h i s m
holds,
and
a,b,c,d,e
let
Then
the q u a d r a t i c
isomorphic. elements
k(Sl,t
extensions
In o t h e r words, s,b,Sl,t I
of
Proposition t~e s q u a r e f r e e
be
ad-
bc
of
two
% O,
k(X)
(ct +d)4Pl(~) polynomials,
their
such that
of
k(t I)
onto
e____~s s a t i s f i e s the (ct + d) 2 may be e x t e n d e d to a n auto-
squarefree
e % O,
defined
there e x i s t s
PGL(2,k) '
of q u a r t i c
2.5 and r e m a r k
quartics ~(x)
over
+ bd ' = ac xx +
(3.1)
where P*(x,y) by
of
quartics
such
over
k
for
which
that
= e2p(x)
by
y
2
an e l l i p t i c s 2 = P(t),
2 Yl = P1 (X)
= P(X), function
field
are
k(C)
s~ = Pl(tl),
k(C)
the a c t i o n
of PGL
k-
and
4
= k(s,t) =
1)
§III - C l a s s i f i c a t i o n
of
k(C)
extension
Q.E.D.
(cX + d) 4 P (aX + b ) 1 c X+ d
(2.8)
and
st _
~ ( s I) = s /
PI(X)
C k ,
such
.
such that
P(X)
same quadratic
whence also
T h e n the e l e m e n t d e f i n e d by ~t(tl) - ab ct ++ db /2 equation s = Pl(G~(tl)) =q~(Pl(tl)). Therefore ~ q
k(C)
a d - bc f O,
k(t),
k(t)
morphism
E k,
of
2.7
k.
lead us to study
Given
we d e n o t e
such a p o l y n o m i a l
by
(cx+d)4p(@(x))
P*
is the h o m o g e n i z e d
= y4p(~)) Y
.
the same constant,
Remarks
polynomials
3.9
(artificial
The polynomial a n d is m o n i c
- The polynomial
@P
@p
the
P(x)
polynomial
(2,k)
on
and an element
defined
by
= P*(a,c)@P(x)
vitamins
added:)
does not c h a n g e
together with
(cx + d)4p(@(x))
174
if
polynomial a,b,c,d
of
P (i.e.
are m u l t i p l i e d
P
depends,
up to
4th
powers,
upon the
5
choice
of
then
a,b,c,d.
If
P
(cx+d)4p(e(x))
P*(a,c)@P(x)
give
we Later,
will
it
Proposition their
study
in
this
§ the
be
necessary
ditions
a)
k(~)
b)
there
for
our
only
the
first
3.3
- Let
P
of
necessary
acts
be
P
two
=
that
P(x)
Remark
2.2
of
the
same
P1
=
has
that
two
the
monic
been
fairly
well
conjunction
of
For
treated
quartics For
P'PI
coefficient
squares.
k).
field.
quartics
leading
~;
and
function
modulo
of
root
P(x)
elliptic
squarefree
the
rational
class
has
closure
the
shows
of
of
its
monic
9Pl) ,
ep
effect
program
algebraic
(i.e.
and
the
through
this
P1 some
PGL(2,k)
also
of
and (in
subfields
study
part
means
equivalence
purposes
roots
under is
to
it
polynomial.
conjugate
sets
equivalent
a cubic
to
which,
being,
= 0,
rise
will
P*(a,c), time
is
(a,c)
over P
by
k,
and
the
the us.
~
P1
and
to
following
be
con-
sufficient:
k(E l)
:
exists
commutes
~
with
bijection the
Galois
9t:
~ ~
group
r
~
which of
preserves
k(~)
over
the k
cross
(i.e.
0'~
ratio
and
: ~0 I
for
all
~EF) The show of
necessity
that
e'
k(~),
Then
the
whence (by
may
with
fractional
is be
c = i,
0t~
or
by
-
d = 1
a monic
led the
to list
(e.g. -
This
if
~
a
a)
and
fractional
a,b,c,d
linear
in
obtained means
that
has
4 elements.
if
c
= 0),
the
k(~).
For
by
applying
we
have
any ~
~ to
e~(t)
conclude
of
the
transformation
Normalizing we
equality
= 6(t) the
8:
E F,
the
cross
let
0~
any
coefficients
they
are
k,
with
roots
at + b t ~-~ ct +d be
coefficients for
that
ratios
the
of
8.
t E ~, a,b,c,d
in
k,
i.e.
Q.E.D.
Given thus
to
: ~e I
since
E PGL(2,k)
are
extended
transformation
condition = 0
Conversely,
coefficients
linear
@~
clear.
quartic
introduce of
the
polynomial its
type,
degrees
(i,i,2),
(2,2),
of
which the
over
tl,t2,t3,t4,
we
comprises;
irreducible
factors
of
P
(4))
necessary,
a
k(tl,t2,t3,t4)
,
in
particular
field
is
denoted
decomposition
P(t)
symbol
giving
information about here
by
175
its k t .
about
the
Galois In
nature
of
the
extension
group. the
following
list
of
types
6
we
give
the
the
cross
structure ratio
about
k(r)
values
of
r
permit
to
put
equivalent
1)
of
r
=
field
k(r)
~ l,t2,t3,t4)
the
cases
(H)
(called
here
the
"preferred
the
commutation
(i,i,i,i)
generated
but,
,
exclude
polynomials
Type
the
and
the
- Here
a
first
(EH).
In
values",
condition
have
in
by
of
same
type.
k ~ = k(r)
= k.
prop.
No
any
one
of
the
approximation,
some
cases,
PCR)
are
3.3
in
PCR.
the
certain
easier
values
the
6
handle
form.
commutation
of
results
of
to
a handy
The
six
and
Of
course,
condition
is
void.
2)
Type
(1,1,2)
and
its
their
-
Here
inverse
k t
as
is Two
PCR.
PCR's
are
equal
Type
(2,2)S
(S
for
k(r)
= k.
quadratic°
(or
We
polynomials
inverse);
take
of
then
r =
type
there
(tl,t2,t3,t3)
(1,1,2) are
are
(tl,t 2 Ek)
equivalent
iff
@t
prop.
4 bijections
(of
3.3). 3)
of
type
take
(2,2)S
r =
are
Type
(I,3)C
3-cyclic
quadratic
extension)
(tl,tl,t2,t2)
equivalent
8 bijeetions
are 4)
We
"same"
iff
and
- Here
its
their
k r
inverse
PCR's
as
are
is
quadratic
PCR.
equal
or
Two
and
polynomials
inverse;
then
there
e t
(C
for
Galois
"cyclic")-
group,
Here
and
k(r)
k t
is
= k I.
a cubic
Having
extension
chosen
of
k,
with
a
~
of
the
a generator
2 Ga!ois ing
group,
i- ~ r
mialsof
and
are
Type
(D
for
extensions,
and
is
we
the
t~o of
polynomials then
there
Type
(4)C
k(r)
is
of are (C its
"distinct" is
and
~2
their
quadratic
as
inverts sets
PCR
,
the
of
and
PCR's).
PCR's
are
r and
4
bijections
extensions)--Here
extension,
quadratic
(2,2)D
unique
~ iff
a quartic
type
for
of
(t I E k)
other
ones
Two
polyno-
equal;
be-
then
6 ~
where
them,
(tl,t2,t~,t~)
equivalent
third
(tl,t2,t3,t4) get
are
3 bijections
(2,2)D
k(r)
r =
(exchange
(1,3)C
quadratic
r =
6)
take
1 l-r
type
there 5)
we
subfield
(tl,t 2)
and
"cyclic") quadratic
of
over
equivalent
iff
a Klein
k t. We
(t3,t 4)
r t , conjugate are
with
k ~, composite
take
are k
group the
pairs
and
such
their
PCR's
of
as
PCR's
Galois in
form over
rrt =i. equal
or
two
group;
the
conjugates
that are
of
Two inverse;
~t - Here
k ~
subfield.
is
a cyclic
Having
176
extension
chosen
of
a generator
degree ~
4, of
and the
k;
2
Galois
group,
changes sueh
r
we t a K e into
that
rr'
r = (tl,t~
~); r = 1.
We g e t
Type
(4)K (K
group
two of
Two p o l y n o m i a l s
are equal or inverse; 7)
3
, t~,
t~
)
as
them,
r
and
of
type
PCR (exchange of r',
(4)C
- Here
k'
conjugate over
Then
k(r)
= k.
this value is unique. the same 8)
PCR;
Type
(1,3)~3 - Here
and
k(r)
= k'
.
the bijection 9)
Type
(4)8 -
Two polynomials of type (4)K
k'
Type ~4
Here
k'
~4"
The field
k(r)
k'
as Galois group,
ii )
J4
Type
(4)~4 - Here
and
k(r)
subgroup of bijection
is the J4" 8'
and
k(r)
in
~4
extension of
are equivalent,
in
the bijection
~4"
there are 2 bijections a group of order 2).
is the cubic cyclic extension of ~4"
k
No 8'
PCR.
k
group
correspond-
If two poly-
is unique
(the commu-
being {i}). k'
is an extension of degree 24, with
sextic extension of No PCR.
is unique
and index 2 of an elliptic
Conjugate
k
corresponding
(since
~4
~4
as Galois group,
to the invariant Klein
If two polynomials of type (4)]4
are equivalent,
the
has "no center")
fields.
function field
(1.2) is determined up to equivalence,
k(r) of
are equivalent,
which is invariant
being its center,
are equivalent,
We now return to elliptic
subfield K
(1,3)~ 3
as Galois group,
is an extension of degree 12, with the alternating
(4) ~4
in
~3
is the quadratic
ing to the invariant Klein subgroup of order 4 of
~4
iff they have
@'
of Gal(k'/k)
If two polynomials of type (4)8
(4)~4 - Here
tant of
PeR, and
is an extension of degree 8, and its Galois group is a
to the Klein-subgroup
nomials of type
as
are equivalent
If two polynomials of type
@' (the commutant of Gal(k'/k) i0)
!
of prop. 3.3 is unique.
2-Sylow subgroups of
No PCR.
with a Klein
t~ , tl~ )
is an extension of degree 6 with
No PCR.
9'
corresponding
r = (tl,t~,
then there are 4 bijections
and
Having listed as above !
the elements of the Galois group, we take
k
e' •
is a quartic extension,
{l,~,~t,q~0 t} as Galois group.
3
and
are equivalent i f f t h e i r PCR's
then there are 4 bijection
for "Klein")
~
Given a subfield k(C),
K = k(t) of genus 0
the quartie polynomial
P(t)
of
so that we can talk about the type of the
subfields have the same type.
the cross ratio is the same for all subfields
177
K,
Since the field the
8
types
which
cases
(H)
that
can
occur
and
the
(EH)
following
Classification
in
the
same
(in
which
types
may
(3.4)
- a)
field
k(r) occur
gives
(i,i,2),
c)
(1,3)C
are
not
little
(2,2)S,
arbitrary.
Excluding
information),
simultaneously
(i,i,i,i),
b)
k(C)
in
a given
the
above
the list
shows
k(C).
(4)K
(2,2)D~
(4)C,
(4)8
, (4)~ 4
d) (1,3)~ 3, (4)M 4
If
k
is a finite
Classification
Proposition "one", are
the In
(3.5) - a)
3.6
these
If
many types are excluded,
(i,i,i,i),
b)
(i,i,2),
c)
(1,3)C
the
types
are
fact,
such
one
checks
Corollar
consider to
that
(2,2)S (4)C
of
two
conjugate
7 3.7
Thus,
K
subfields
(with
involutions
and
K1 .
K,
respect
K1 to
(Look
least (cases
Classification
if
k
(H)
and
If
o_~f k ( C ) a
contain
translation),
both
and
a
the
types
of
"ones"
at
the
is
finite,
(EH)
in
cubic we
2 classes,
one of type (1,1,2),
c)
Single
b)
Single
the
TI: there
M ~ - A 1 - M ( A , A 1 E Ck ) are
points
translation
M ~
P'P1 (Pl-
of
P) +M ,
this
type
can
is
for
list
the
number
of
points
of
order
C).
the
conjugacy
classes
in
k(C)
and
excluded).
3.8:
b)
Two
that
denotes
temporarily
one of type
a)
means
equation
4 classes,
about
q
and
Q.E.D.
a)
3.9
M ~ -A-M
hypothesis
-1
- The number
at
The
T:
2P 1 = - A 1 .
T1 = ~ T q
C k.
types
the
2P = - A ,
that
2 in
Remark
and there remain only:
same.
Ck
their
-
subfields
corresponding
1 or
field,
class,
of type
~
classes, class
Using one
of
of
type
(i,i,i,i),
3 of type (2,2)S
one of type (4)C
(I,3)C remark
type
2.4,
we
(i,i,i,i),
(1,1,2)
have one
.
178
the of
following
type
(2,2)S
classification
overR:
9
IV - C o m p u t a t i o n a l Consider
method when type
The subfield
k(x)
the c o r r e s p o n d i n g
with
2
= ( x - a ) ( x - a ' ) ( x - a a)
of
k(C)
is of t y p e
s u b f i e l d is
k(t)
A straight-forward
s2 = P ( t )
A,B,C
try
are
to
with
If
distinct)
(u,v)
y - v = t ( x - u).
is a p o i n t of
Set
b =u-
shows that w e can w r i t e
Ck,
a,b'=u-
a t,
k(C) = k ( s , t )
and
the
= t 4-
elementary
b+bt
If we
(a,a',a z 6 k I
(1,1,l,1).
computation
P(t)
where
occurs.
the e l l i p t i c c u r v e w i t h e q u a t i o n
y
b#= u- a R .
(i,i,i,I)
2At 2 + 8vt
symmetric
+ b n, bb' + b b " +
decompose
P(t)
(*)
into
P(t)
+ A2 -
4B
functions
b i b n, b b ' b "
a product
of
of
b,bl,b n .
quadratics
= (t 2 + a t + 8 ) ( t 2 - at + ~ ' )
w e get the c o n d i t i o n s
8 +~'-0t
W e first s u p p o s e
v % 0.
2 = 2A
Taking
0~(8- ~')
~ = 8 + ~'
(~+2A)(~2-4A
(of course,
the c u b i c
resolvent of
= 8v,
~'
= A 2 - 4B
as m a i n u n k n o w n ,
.
w e get the e q u a t i o n
2 + 16B) - 1 6 v 2 = 0
P(t).
1 8 = ~(7+2A)
Now
is a b e t t e r u n k n o w n
s i n c e it s a t i s f i e s the e q u a t i o n
5 T h u s the v a l u e s o f
3
- A5 2 + B 6 + C = ( 6 - b ) ( 6 - b ' ) ( 5 - b
8+8'
are
S i n c e w e h a v e the v a l u e s o f equation over (if w e t a k e
k,
2(b-b'-b~'),
8 +~'
the d i s c r i m i n a n t
8+8'=2(b-b'-bH)).
is a s q u a r e in
k
and
iff
u-a
~@',
iff u - a ,
2(b~l-b-b').
~ a n d ~'
a r e the roots o f a q u a d r a t i c 4b'ba = 4(u-a')(u-a
'~)
( u - a ) ( u - a ' ) ( u - a ¢') = v2 , this d i s c r i m i n a n t
is a s q u a r e in
i n t o a p r o d u c t o f two q u a d r a t i c s
n)
to b e
and
.
2(b'-b-b
o f whi,ch h a p p e n s Since
n) = 0
k.
or u-a
179
Thus
P(t)
decomposes over
p, o r u - a ~ is a s q u a r e i n
k k,
i0
At a n y rate, tors of
P(t)
the p r o d u c t
is
( 2 _ 4~)( 2 _ 46/)
16(a t- aH) 2 (if w e t a k e
6 +~t
t o r s h a v e t h e i r roots in the same q u a d r a t i c putation
s h o w s that t h i s e x t e n s i o n is
R e m a r k 4.1 B(t)
over
In g e n e r a l , k
efficient
l-b~)),
to
-2P
s
2
w e c h o o s e a l e a d i n g co-
=eB(t)
quartic polynomial
polynomial
has a
k-rational
k(t,s),
decomposes into a linear
f a c t o r is e s s e n t i a l l y
the r e s o l v e n t
If
= t 4 + 2 t 2 ( a / + a H - 2a)
+ (a t- aH) 2
and,
e.g.,
A simple computation Furthermore,
u =a,
then
P(t)
s h o w s that the r o o t s of
it d e c o m p o s e s o v e r
( a - a t ) ( a - a H)
k
is a s q u a r e in k
P(t)
point
the t y p e o f w h i c h
f a c t o r a n d a c u b i c one, a n d t h i s c u b i c
v = 0
fac-
A f u r t h e r com-
squarefree
interpretation:
gives a subfield of
the c o r r e s p o n d i n g
thus these
k(v~).
the c u b i c r e s o l v e n t o f a q u a r t i c
The involution with respect
o f the fac-
~-2-~i)
s u c h t h a t the e l l i p t i c c u r v e
c o n t a i n s a "one";
= 2(b-b
e x t e n s i o n of
k ( ~ ,
has t h e f o l l o w i n g g e o m e t r i c
e ~ k
o f the d i s c r i m i n a n t s
of
B(t).
are
± ~ u - a t ± ~u-:-a '~ .
iff
a - at,
or
a - a H,
or
a-
a-a
z
v=O,
into q u a d r a t i c s
.
T h e r e f o r e w e h a v e the f o l l o w i n g result.
Proposition
4.2
u =a)
are
(u,v)
E 2C b)
then
bit
only
is If
P(t) A
a)
If
u-
in
k,
then
one
of
none is
type of
of
Galois
Proposition ferred
4.3
value
The
u-
a ~
P(t)
other
of
u-
a/,
.
a,u-
a t , u-
(4)K
and
u-
a ~
(resp.
is
of t y p e
(1,1,1,1)
u-
a~(resp,
of
a ~ ,
(which
for
means
that
If
in
we cross
preferred
N o t i c e that all p o s s i b l e
a H
(resp.
of
a-
a-
a t, a-
a t , a-
a H)
a ~)
is
is
a
square
a
square
case
v~0),
ink,
in
k,
.
computation
theory
the
a,
(2,2,)S u-
type
-
u-
of
straightforward of
a,
k
P(t)
then
a
squares
If
c)
-
case
chose
of C),
gives
B +~I for
r=
a'-a ~a -a
PCR's
is
ratios
the
= 2(b-
ratio
value
cross
P(t)
-
i/r°
occur
additional
in
is,
b) a n d c),
in b o t h c a s e s
a,
a ~,
(H)
suitable
plus
information:
u=a),
(~,
180
(*)
b 1-bH)(resp.v=0,
Cases for
(using
a
pre-
a#).
and
(EH)
subfields
are
not
of
k(C)
excluded .
here.
ii
Complement
4.4
subgroup curve With
Da
E. the
shows
-
The
points
of
Ck:
in fact
"at
of
Ck
such
k(x,y, x W ~ - a )
k(C) C k(E)
The inclusion origin
(u,v)
infinity"
on
E,
that
u-
a
is
is the f u n c t i o n
gives
a morphism
f
a
is
group
f:
a
square
form
a
field of an e l l i p t i c
E 4 C
and
homomorphism.
f ( E k)
= Da.
Proposition
4.2
that
(4.5)
2C k = Da n Da t
Except
possibly
correspond
for
= D a / n Daz = Dan A Da
(a,0),
(a~,0),
(aH,0),
the
to the s u b f i e l d s
of type
(2,2)S.
A computation
points
of
.
(D a U Da / U Daz) - 2C k of
the
index
(Da:2C k)
w o u l d be useful.
Example
4.6 - F i n i t e
one at least of square.
u-a,
k
u-a t , u-a"
(resp.
Then proposition
conjugacy
classes
prop.
gives
4.3 a)
are
(General
b)
(Case
(1,1,1,1),
of
case)
PCR's respectively
PCR 2
fields - Let
equal
(H)
with
one class
of
4.2
(2,2)S.
One c l a s s r,1-
i Ek, type
of r,
even in cases
Furthermore
type
2 v = (u-a)(u-at)(u-a"),
Since
let
(1,1,1,1),
(H) a n d
(EH),
a #- a r - a t -a
3 classes
must be a
the n o n - z e r o Then
of
type
(2,2)S
with
r r-1
3 classes (2,2)S
field.
o f a - a t , a - a//, ( a - a t ) ( a - a"))
shows that,
type
to
be a finite
by prep.
with
2.3).
PCR- 1,
One c l a s s
one class
of
of
type
type (2,2,)S
with
.
c) ( C a s e (i,i,I,i),
that,
4.7 -
be d e f i n e d
with
here,
Function
o v e r a field
another
(or the same)
of
into C.
C ~
J E k,
o n e class o f type
Notice Example
(EH)
The analogue
(2,2)S
w e have
with
k 0.
2.3).
O n e c l a s s of type
P C R - J.
fields -
We take for
elliptic
by prop.
(Ck:D a) = (Da:2C k) = 2.
curve Ck
Ck
is true
classes
fields as g r o u n d
As well known,
(4.8)
2
for the s u b g r o u p
k
C /
Let
the f u n c t i o n Let
~
is the direct
= Ck0
(C)y 2 = ( x - a ) ( x - a l ) ( x - a H) field
of a
181
of
be the g r o u p of k 0 - h o m o m o r p h i s m s sum
@5.
D
k =ko(Ct)
complement
(4.4):
12
(4.9)
Da = (D a A Ck0) ~
(In
fact,
mA:
C t ~
it
with
with E;
the we
the
notation
of
decompose
mA
homomorphism
In p a r t i c u l a r ,
4.4,
a
into
f:
E
a
~ C)
let us take
point
(D a Q ~ )
A
of
translation
Ek
corresponds
a
homomorphism,
and
to
ko-morphism
and
we
compose
.
k0 = ~
,
C' = C
and assume
that
C
has no c o m p l e x
q multiplications endomorphisms, whence
(thus c a s e s is a
g-module
12 " t r a n s c e n d e n t a l "
find t h e i r types,
(4.9)
the g e n e r i c
(x,y),
D~ , Da#)
point
.
(H) a n d
these
of rank 2.
conjugacy enables
as t h i r d
root of a cubic e q u a t i o n
subgroups,
of
(yq_y)2
q = 2m+l
_ z2
f
q
if w e call
besides D
point
n 8
a
the
2Ok:
its ring of
the 4 " a l g e b r a i c " •
(xq,y q)
shows that
= (yq - y ) 2 _
z =y((x-
and
Ck/2C k
~,
the
One
ones •
To
D a (nor in
(xl,Yl)
classical
16,
sees that n e i t h e r
are in
point
k 0-
= (x,y) +
computation
of
x,
gives
a)
(x q_ x)(x q +x-
a ' ) m ( x - a") m - ( x - a)
a'-a
'/);
TM) ,
the c o m p u t a t i o n
shows that
(xq-
Thus,
is a g r o u p of o r d e r
whence in
(x q- x)2(Xl-
we s e t
Then
classes
computation
is
if
so that
us to study only
(xq,y q)
but,
are excluded)
n o r the F r o b e n i u s
But an amusing in
(EH)
f q
x)2(xl-a
the F r o b e n i u s
) = z2 (,,)
endomorphism
.
of
C,
t h e r e is a
g E ~
such that
= 2g- 1, The
fact that the c h a r a c t e r i s t i c
is e q u i v a l e n t (1,1,1,1) Examples
g,
lead us to t h i n k
nor c h e c k e d
the f e e l i n g
also
This
that the
that n e i t h e r
that a c o m p u t a t i o n
E 2C k.
(Ck0) ,
the c o o r d i n a t e s
show (by p e r m u t a t i o n )
(x2,Y 2)
41Card
g
nothing
are a l g e b r a i c
new since
integers
C is o f type
k 0 = ~q
over
W e h a v e not yet c o m p u t e d to
with
r o o t s of
that
feeling
E-module (x2,Y 2)
(x2,Y 2) nor
showing, x2- a t
e.g.,
is g e n e r a t e d
that
leads us to c o n j e c t u r e
by
of the point of
(x,y) + (x2,Y 2)
and x 2- a ~
182
~
x2
and
g.
Ck corresponding
belong
a is a square
are also
1
squares,
Da. in
We have k
i.e.
would that
that the 12 " t r a n s c e n d e n t a l "
13
conjugacy occurs
classes
are of type
(4)K,
and
that
each
one
of
the
r, l-r, r-~rl
3 PCR's
4 times in t h e s e classes.
§V - I s o m o r p h i s m Remark
classes
and conjugacy
2.2 gave an
classes
a n s w e r to what m a y be c a l l e d
the c o n j u g a c y
problem:
2 two s q u a r e f r e e define
two c o n j u g a t e
morphism
problem
defined by P1
quartics
P'P1
over
subfields
(over the a l g e b r a i c
s 2 = P(t)
and
asking when these
If the types of shows that the more,
a)
5.1
and q u i t e
The elliptic
k-isomorphic morphism
iff being
b)
The
two curves P1
k
explicit
~
curves
y
C:
difficult
are i s o m o r p h i c
y
2
s
field
k(C)?
= x
3
the c u r v e s
over
to b e the
P
and
k-isomorphism
k. then
prop.
is the same as the c o n j u g a c y criterion
T h e iso-
iff the roots of
seems
a "one",
2 sI =Pl(tl)
=P(t),
is also easy:
k-isomorphic
More
3.6 problem.
Further-
can b e given.
+ b x + c,
such that
d E k*
(d2x,
2
k)
be a f i e l d of c h a r a c t e r i s t i c
exists
(x,y)
function
of
both contain
problem
curves
there
k
are
ratios.
and
- Let
closure
s~ = Pl(tl)
k-isomorphism
a well k n o w n
Proposition
P
w h e n do the e q u a t i o n s
o f the same e l l i p t i c
have the same set of c r o s s
problem,
k,
given
~ 3 ci :
and
b' = d4b
y
2
=x
and
3
+ b
'x
c'
+
c' = d6c
are
(an iso-
d3y)
and
= e(x 3 +bx+c)
y2 = e'(x 3 + b x + c)
ar___ee k - i s o m o r p h i c
iff (general (case In
fact,
isomorphism Then
it
y
2
must
have
=x
c ~ 0) c = 0)
carriesthe
with
at
form
a
square
e'/e
is
a
square
of
infinity
(x,y)
In c h a r a c t e r i s t i c
-
+ax
2
+ c(a
~ 0).
The
k-isomorphic
iff there e x i s t s
phism being
(x,y ~
is not
is
a translation
point
the
±
e'/e
~
C,
C
of
it
. may
assumed
that
to the point at i n f i n i t y
a n d a simple
(~x,~y)
be
computation
k-
the
of
C'.
gives
the
Q.E.D.
5.2
3
(H),
composing
answers. Remark
case,
(d2x,
d3y)
3, a non h a r m o n i c 2
curves
y
d E k*
such
The
= x
analogue
so handy.
183
3
c u r v e has a normal
+ ax
that
2
a' of
b)
+ c,
y
2
= d2a holds.
= x3
and
equation
+ a'x 2 + c t d=d3c
The
harmonic
(an
are isomor-
case
14
P r o b l e m 5.3 - An interesting following one:
P
particular case of the k-isomorphism problem is the
being a squarefree quartic over 2 s = eP(t)
k-isomorphic?
The condition
also necessary
if
P
In the non-harmonic k(t.) 1
"et/e
2
P)
is
k
itself: P
k,
and the square root of (2,2)D,
condition
"el/e
a square"
is of type
(4)C,
k-isomorphism means conjugacy
in general,
morphism. y
2
is also necessary
C
k = k0(x,y, x ~ - ~ )
Hence the corresponding (in fact, G E 2C k
G = A +2P since
type (2,2)S
G
Da
able point of
(4)J4
k-isomorphism.
a square"
gives,
since
has infinite order).
3.8
Ck0
have the same
).
If
P
PCR
P is
(example 4.6). for
k-iso-
ko,
and the field point
Also
G
G = (x,y)
Dat
nor in
is not in n,
Cko + 2 C k
nG=2nP,
the 3 conjugacy
corresponding
and are equivalent
to
under
G
is in Da ~ "
whence
classes of
exhaust the 3 possible values of the P'P1
if
(H), nor (EH)),
has finite order
Furthermore
In fact,
PCR's
Then the "generic"
A E Ck0
k(t ); this 1
is not necessary
(neither
intro-
.
(classification
subfield is of type (2,2)S.
Thus the monic quartics C
for
k(t.) 1
is in
in complement 4.4, but neither in
given by points of
(example 4.6).
field.
(prop.5.l,b)).
show that the sufficient
defined over a finite field
as new ground
the subgroup denoted by
"et/e
of type (1,1,1,1)
= (x- a ) ( x - a t ) ( x - aH),
(4)~4,
subfields have distinct
the condition
Take a curve
el/e
the results of §§ III-IV
then non-conjugate
and
if the intersection of the fields
ground field extension to
property holds only in types
Over a finite field
But,
= etP(t)
is a square" is sufficient in general,
case, it is also necessary
duces a "one" in the type of
of type (2,2)S,
when are the curves
is non-harmonic and if its type contains a "one"
(t.: roots of 1
intersection
s
k,
PCR
and to a suitPGL(2,k)(3, 3).
k0 At the (cheap~) P=PI"
price of a fractional
Hence we can write
s t2 = e t P ( t t) ( e , e t E k * )
.
linear transformation,
k(C) = k ( s , t ) But
et/e
= k(stt s)
with
we may th~s assume that
s 2 =eP(t)
and
cannot be a square since, otherwise,
the sub-
fields would be conjugate. The slight discrepancy
in the harmonic case deserves a further study.
due, roughly
to the fact that a harmonic
speaking,
transformations
under
PGL(2,k)
than a general quartic.
184
It is
quartic admits more self-
15
Counting being
classes prime
classes
over
to
of
6.
(general
b)
(case
finite
It
elliptic
a)
a
field,
follows
curves
case,
from y
2
c)
(case
we
( E H ) , b =0)
have
Conjugacy class except
'~
Type
q-
(H)
to
2
q~3
mod. 4
classes
if
q~ 1
mod. 6
classes
if
q~5
mod. 6
leading
the
and
Prop.
are
(EH),
the
r
5.l,b)
~
3 mod.
4
two
classes
if
q ~
1 mod.
4
q b)
(EH)
(only
Other
classes:
(EH)
(only
q - 73
q ~ 1 mod.
elements,
One if
- The (H)
of
q =7,
22
two
r, -
class
by
the of
their
classes
PCR's
corresponding possible
if
and
q-
1
is
if
if
way
in
not
an
6,
k,
and
by
quartics.
PCR's
has
to
element the
must
avoid
the
to
not
occur
here.
if
q~
1 mod.
6)
~
only
one
1
be
mul-
0,i,
q ~ 5 rood. 6. of
to
4,
be
2 classes
6,
proceed
q-
by
3
if
~ q its
such
2
own
that
rr
inverse.
if
q ~ 1 mod.
q ~
5 mod.
6.
differences.
4 classes 2(q+1)
if
Thus,
values
2 classes.
q ~ 1 mod.
does
2(q-l) 3
the
a square)
q____~5 if
q ~ 3 mod. 6)
is
q=ll.
classes
1 being
q m 5 mod.
easiest
classes;
q
k-isomorphism
q ~
1 mod.
185
6,
if
q
=
5 rood.
6.
= i. Thus
gives:
(r = - i )
Case
number
~q,
classes.
(-i
6)
PCR,
of
number
is
q ~ 1 mod.
the
such
field,
gives:
q
Here
for
characterized
conjugacy
ratio
if
classes
coefficient
of
cross
18
if
(H)
Other
if
class
-
finite
2) c l a s s e s
classes
one
(1,1,2)
(I,3)C
the
number
classes:
5.1.,
2(q-
mod. 4
type
the
if
is
the
the
q~l
given
get
Other
prop.
a
(H)
values.
l
c
that
if
q=5,
cases
(only
are
of of
Here
+
a)
be
classes
for
in
(EH)
There
Type
2
(i,i,i,i)-
whence
Type
squares
possibly by
classes
classes
modulo
tiplied
12
+ bx
5
k
0)
C
L2 Eg.,
- Let
prop.
3
b ~ 0 , c ~ O) I4
(H),
= x
5.4
4
16
Type
(2,2)S
verse;
- Here
the
other
(H)
the
ones
(only
Other
classes:
(4)C
even
in
the
same
-
if
PCR
cases
elliptic One
(EH)
(only
Other
classes:
Remark
5.5
-
mod.
6).
if
q~
q-
7
in
type
whence
r =-i,
two
Two
type
r=2)
6,
q-
(4)C
5
if
only
two
classes
own
example4
q---3
mod.
4
if
q~l
rood.
4
in-
gives:
.
5 mod.
cases
occur
its
if
q~
of
can
being Then
r=-j)
(with
Elimination
6.
and
prop.2.3
together
with
show
type
that
(1,1,2)
in
Thus:
class
if
q~3
if
q~
5 mod.
6)
Two
q-
1
q~
1 mod.
Another
with
-1
polynomials.
r= I)
classes
1 mod.
values,
menic
and
(1,1,2).
(EH),
q-2
of
r =2
q=-i
field.
(H)
r =-i, with
and
k, classes
(two
as
(H)
in
q-3 2
(with
classes
(EH)
is
give
i ~3 classes
Type
PCR
mod.
if
way
of
4,
if
q~
1 mod.
4
3
if
q~
5 mod.
6.
number
of
classes.
computing
6,
the
q-
equivalence
classes
of
monic 3
polynomials
is
to
make
PGL
(2,~q)
operate
on a suitable
set.
This
group
has
q
- q
elements. For
type
(4)C,
it
operates
on
~ q
4
q
There
2
a r e no
fixed points,
2
whence
q
orbits.
If
t E ~
2 (except
(t,t q
if
,tq,t q
n u m b e r of c l a s s e s
For elements
ments, are
type of
linearly
(~(u)
= V,
and
(u,v q)
of
(y(v) are is type
no fixed
(a,t q2)
we
-
there
are
in 1 +
(1,3)C, points, distinct
no
fixed
distinct
there
orbits,
is
in
another
operate
on
the
This
set
has
points,
(u,v)
is o n t h e o r b i t
are
tq
v % u,u q.
Fq,
whence
-
and
PGL(2,~q)
that
tq
then
of
and
t,
3
make
over
q-3 -2
'
q + 1 2
are
= u,
2
q tq
q-i 2
such
q
dependent
classes For
are
2 q since
4 - ~
q ) = - i)
1 +
is
(2,2)S, ~
and,
3
we g e t
q-2
(v,u)
except
if
are
of
(q2_
orbits.
an involution
and
set
orbit.
Thus
pairs q)(q2_
(u,v) q-
Since
~ E PGL(2,~q) in
(u,uq,v,v
the q)
same orbit. = -1.
the total
Thus
2)
of ele-
1, U +V,
such But the
UV
that (u,v) number
q-1 2
we m a k e whence except
q+l in
PGL(2,~q)
operate
orbits. case
on
The orbits
(EH)
(char.
186
~ 3).
(~q U ~) of
(a,t),
X (F 3 - Fq). q (a,t q)
and
There
17
~VI
- Curves
without
Here, over
k
its
of
k,
and
principal
on
knowledge
without
group
over a
Our
points.
is
more
k-rational
divisor
It
degree
C
over
space
real
case.
fragmentary.
of
to
homogeneous
The
points.
classes
isomorphic
over
Let
is
C
natural
O.
It
to
is
the
algebraic
J:
the
be
an
an
elliptic
introduce
elliptic
its curve,
closure
elements
of
of
curve
k.
J
defined
Jacobian
J,
defined
also
Furthermore
operate
as
i.e.
C
is
translations
C. We recall
function
1 or
sufficient admits of
a
Thus,
E
tion
of
is by
two
H + kH'
and of
of
-
conic of
F(x,y) this
=
z 2-
= 0
is
projection
conic.
vanishes
This = 0, ~E
"
over
a
of
genus
k(C)
The
an if
curve
genus
lies
denotes
C
the
that
it
is
it
existence
~
k-
in
0 and
on
the
quadratic
of
one
of
of
the
intersec-
quadrics
discriminant
index
of
2 it
degree
over
of
is
neces-
4 such
that
k.
function
3-space,
divisor
the
pencil
E
the
imbedding
curve,
field
z 2 = G(x,y),
of
2,
the a
of
have
defined
2 is
with
imbedding E
be
of
that
divisor
index
= k(x,y,z)
gives and,
0 and
writing
of
~E(C)
that
projective
linear
We
k-rational
index
necessary
use
a quartic
air).
subfields
and
the
divisors
and
existence
the
The
by
2,
0
is
D). 4,
k.
thin
genus
the
is
obtained
containing
Then
.
admit
degree
rational
further
to
~E(C)
(in
admits
C
subfield
and
cones,
to
for
implies
of
defined
of degree
equivalent
~3
admits
or
G
being
with
of
poles
cone
(or
k(x,y)
a polynom-
affine of
x
of
equations (or
y,
or
z),
"cylinder")
.
Sufficiency: the
in
of
(Remember
divisor
H'
cones
= 0.
G(x,y)
n>O
k(C)
that
~2
imbedding
F(x,y)
For
a
degree
degree
C
kH'
4 quadratic
F(x,y)
of
4 quadratic
sufficient the
4
and
H+
divisor
transcendental,
it
subfields
degree
k-rational
H
possess
linearly
purely
points;
of
n
imbeds
form
Necessity:
ial
a
L(E)
6.1
D
degree
to
k-rational
divisor
contains
Proposition
a
either
rational
k(C)
admits
such
quaternary
sary
for
quadrics
= 0
without
divisor
divisor
given
one
it
k-rational
If
the
2.
a conic
k-rational
rational
~E
of
that
a
is
that a f i e l d o f g e n u s 0
field
degree
a
rational
if from
S
S
is
induces
the a
vertex
k-morphism
of
Q.E.D.
187
the
degree
quadratic 2 of
cones %0E(C)
defined onto
over
a plane
k
18
Let fields not
us
of
only
volution C
(or,
order
now
view
genus
in
a more
index
2,
it
0 and
the
group
on
C
what
things
Jk
of
is best
amounts
to abuse
the
by
same,
language):
the
(6.2)
T6
k,
i.e.
contains
of
D
if
6
over
k
D ~
a further
deserve
For any l i n e a r
~ D t
system
6
rational
6
of
of
degree
class
(all
DIE
over
k(C)
proposition
6.3
a)
S u p p o s e AUtk(C)
iff
The
K
of
to
admit
that
AUtk(C)
involutions. divisors
T6
is
contain,
Here
of
2,
sub-
an
degree
called then
2
also
inon
6
defined
in
by
P E C)
system
D~ ~ D
6 -
k,
6
is rational
for every
over
conjugate
D~
cocycles
K6
the field of in-
. contains
6 - 6 t E 2J k (here
se___to__f c o n j u g a c y
is purely
This introduces
we denote by
only t r a n s l a t i o n s
ha__s no c o m p l e x m u l t i p l i c a t i o n s ) .
ar___e c o n j u g a t e
b)
system
such that
for e v e r y
in
d e g r e e 0).
a set
k(c)
study.
T~
J
also
iff the l i n e a r
D
of
if
k
sufficient
involution
+ P E 6
over
For
.
which
(e.g.
but
a divisor
a divisor
Then ~e have
variants
and
corresponding
is d e f i n e d
manner.
necessary
a linear
by
T6(P)
This involution
is
translations,
defined
to
algebraic
transcendental
T h e n the s u b f i e l d s
6 - 6~
classes
K6
is v i e w e d as a d i v i s o r
is e q u i p o t e n t
over
an___ddi n v o l u t i o n s
k
with
if__f 6
Jk/2Jk
contains
and
K6t
c l a s s of .
a divisor
of the
6 fo rm
P + P, c)
Le___t k'
of l i n e a r cipal
where
systems
a)
,
transcendental classes point class
containing
6i P3 E C
of
P3
6 E Ikt
P
1
+ P
i
divisors
iff
+ P3'
we
have
of
~k'
P +P
o~f
Jk
= (61-
to
188
its conjugate.
Ck, ~ ¢.
with
T h e set
2J k
Ik,
i s a prin-
P E C k,
The fact gives
i,
Ikt (Pi E C
belongs 60)
an__d P
(i.e.,
fixin~
K
purely
•
of d e g r e e
belong
k
containing
as in prop. 2.
P3 + PO ~ P1 + P2 60
over
such that
of the form
a valuation
63-
k
6 - 6 0 E ~kl)
(i = 0,1,2)
such that
quadratic
extension
from an easy c o m p u t a t i o n iff it a d m t s
of
C
space over a s u b g r o u p
we have
comes
is a point of
be a q u a d r a t i c
homogeneous
6 0 E Ikl
P
to
+ (62-80)
b). t) k'
that As t o
c),
if
the
then the u n i q u e
Ckt ; calling .
is
5
Q.E.D.
63
the
19
Example
6.4.
The
real
An elliptic
case.
curve
E
defined
C
over
c a n b e put in t h e
y
form
2
=x
3
+bx+c.
number
The
(6.5)
j'(E)
c
-
2
4 b 3 + 27o 2
is
an
absolute
j'(E)
invariant
> 0 ~
type
of
E (I) .
(1,1,2)
~ E
If
b
and
has one
e
are
real,
connected
then:
component
~ the cross
ratio
cross
is
R
I
is n o t
j'(E)
<
admits
= O,
now a
Using
type
(i,i,i,i)
~
harmonic
C
be
and
being
a monic
roots
is
must
have
case,
an
~-rational
L(D)
elliptic
divisor L(2D),
quartic
real.
one
J
there
has are
D
we
Since
the
of
see
without
type
fractional
components.
the
components
ratio
real
defined
that real
linear
=
roots.
of
C
= j'(J)
R
whence
without
with
cross has
y
ratio
the
same
transformation
2
points.
defined + P4(x)
It over
= 0,
r = (~,~,~,~) cross
ratios
case.
Then,
as
R-
P4(x) of t h e s e C,
we
~ 0
in the n o n - h a r m o n i c
subfields
real
involutions
R(x,y)
The
of
at l e a s t
classes
2,
R(C)
J
over
of g e n u s brings
0 and index
the e q u a t i o n
by p r o p . 6 . 3 , ~
2. of
C
to the
form
(6.7)
y
we
two
degree
Jacobian
(i,i,i,i),
two c o n j u g a c y
A real normal
two
or
j'(C)
and
has
curve
(6.6)
and
E
R
j'(E)
Let
0 ~
real
see
-
R(x)
-
R(u,v)
(1)It
two ,
+ (x 2 + l ) ( x 2 + k 2)
subfields
purely with
is
2
a
of
genus
0
= 0
and
index
2
+ v2 +
(k real,
~ 0,i,-i)
2:
transcendental u
Z = x~
simple
v ~
PGL(2,Q)
x
~ + x,
u
transform
of
189
(k-l)
the
2
= 0,
classical
not
modular
purely
transcendental,
invariant.
20 Thus
one
the
second
of
of
the by
the
subfields
6.8
As
seen
be
put
shows
in
that
opposite
jt
the
are
two
j'
< 0
in
Since
positive
+ bx
is
degree
also
i.
of
Since
type
j'
classes
of
> 0
carries is
such
these
two in
types
that
case.
j'
b3/c 2
Since
subfields,
(i,i,i,i)
invariant
invariant + c.
transcendental
real
fixed,
curves,
points,
prop.
thus
5.1
corresponding
to
e. still
this
case
the
]0,i[
and
have
of
2
are for
r
two
R-isomorphism
6 values
a cross-ratio values
with
purely
absolute
R-isomorphism
we
In
= x3
the of
J
given
curve
y2
by
case,
with
real
form
formed valuations
harmonic
curves a
v a l u e s of
points. are
6.4,
is
without
the
Real
there
For
2
in
-
in
classes
subfields
exist
Complement
can
conjugacy
>
of
1 ;
the
the
cross
this
curve
classes
set
with
can
be
used
to
of
r
give
opposite
ratio of
r
curves
carrying
are
real,
2 are
is
uniquely
6 values
equation
get
of
a real
<
O,
determined
(6.7)
is
kk+l]
'
and
each
one
gives
k,
real
only
by
the
two
4
values 2
of
k.
We
thus
denoted
Inverse
values
get
possible
4
by
b
4
,
a
values
for
k2
straightforward
values when
for
j'
computation
k ,
is
shows
whence
given. that
the
If
the
one
other
same
of
k
them
ones
is
are
1
Since to
two
see
curves
whether
corresponding the
R-isomorphic. gate
purely
curves If
defined
over
{i,-i
'i
\i-bJ
' -i
this
q happens
classes
of
For
is
curves
~
with
their
would
~i-~)
~
f of
carry
}
;
the
= 1 ± ~2-, given
jt = O, h a r m o n i c
w i t h real points, (1,1,2).
if
C t
subfields
necessarily
only
inverse
and
were,
transcendental
u
itself,
C
they
tion
to
j'
case,
values
equations
would,
of
and
R(C),
the
form
harmonic
= dx case.
without
w e still have
real points,
by
R-isomorphic, k2
6.4,
=
correspond
fractional
to
linear
Thus real
two
190
we
= --e (d, x have
two
conju-
points
onto
axis onto
e E R),
and
R-isomorphism
points.
B-isomorphism
o n e o f type in form
~(x)
have
transforma-
then the i m a g i n a r y or
we are
[i + ~ \i- b
set {i ,-i,i~ . . 2 ,-i~ 2 ) of branch
~(x)
and
are
and
a
it c a r r i e s
y2 = x 3 ± b x (b > 0),
T h e curves w i t h o u t
k2
k2 = ~4
with
since
> 0
of
(i,i,I,I)
(6o7),
classes
of c u r v e s
a n d one of type
correspond
to
21 2 = 2,
k- 1 they
are
to
i.e.
k = -3~
2~,
So we
have
R-isomorphico
Proposition a)
k
= 17 i
12~2--,
two inverse values;
thus
proved:
6.9 - Given a real value of the absolute
F o__r j' > 0, tw__oo R-isomorphism
invariant
j', there are:
classes o_f_freal curves,
both with real
points.
b) tw_/o
For
j' < 0 , tw__2o R-i_somorphism
R-isomorphism c)
one
Fo_r
classes
jt
= 0,
R-isomorphism Now,
have
for
jt
is
the
For
The Galois
j' > 0,
HI(~;
JC)
points ceivable
value
of
curve
which
is
as
following
and
the
j ~, those
curves
elliptic
group
¢/R
and
is
type
case
points.
described
or
in one
real
an
6.9
cyclic
or
for
the
two
0
and
in
be
associated
JC
the
order
argument
shows°
exact
sequence
that
classes cyclic
We
upon
whether
the
@
acts
trivially
in
J~
0 -9 Now,
the
group
l-coboundaries imitating that
the
contains
of
l-cooycles
is
trivial
Kummer ground all
the
theory field
E¢
of since
shows contains
"halves
and
of 0"
~
on
~
acts
all ,
~
E
JR)
= JR
we
JR)
~ HI(@
JR
is
= 0
and the
same
value
where
on
with it
HI(Q; this
curve.
jr=0,
curve
sequence,
the
for
j' < 0,
However,
is J¢)
cannot
real conmight happen
0 -9 JR-9Jc-9E
-90
is
to
191
the
exact
sequences:
; JC)
simply
= J/2.. JR"
-9 HI( ~; J¢)
deduce
trivially.
th n-- roots that
the
cohomology:
-9 HI(@;
that
course,
(~J; JC)
Jacobians,
3.
of
and
1
and
will,
points
When
exact
0 -9 J~R - 9 J c ~ -9 E ~ -9 HI(~;
Since
H
complex JC)
points,
points
group
real
real
points
HI(~;
order
consider
for
real
of
of
of
with
real
respectively.
isomorphism other
group
2 depending
(i,i,i,i),
the
with
cohomology
the
shows
curves
without
curve
is
easily
of
(1,1,2)
by
and
points..
without
is
of
real
of
situation
= 0
without
classes
curves
a given
that,
the
of
Proposition
is of
curves
tw___o_o R-isomorphism
class
a Jacobian
of
of
classes of curves with real points,
~IomE(~; Finally"
Instead
of
unity
one
say,
the
four
JR) an
of
argument
using
uses points
and
the
the fact P. 1
the
group
exactly assumption that such
JR that
of
22
2P i = 0
(remember
we
are
in
type
Homz(~,JR)
order
2 while
order
3 by
part
(b)
of
Proposition
that
HI(~
; JC)
is
of
order
2
for
- Le__t
C
be
Proposition absolute
is
(i,i,i,i).)
6.10
invariant
j'
of
l~f
a)
if
j'
b)
i~f
jl <
c)
if
j'
order
> 0 ,
4
(Klein
6.9.
real
an is
4)
Hence
each
JC
Finally
Jacobian.
group
of
HI(~
; JC)
= 0
0 ,
then
HI(~
; J¢)
is
of
= 0 ,
then
HI(~;
JC)
is
zero
on
whether
we
know
C
is
of
type
is
is
of
at
most
3 we
see
proved:
over
R and
points
order
JR
of
divide
have
defined
complex
JR/2
; JC) not
We
curve
that
HI(~ 2 does
then
depending
note
and
since
elliptic the
we
in
havin~
C
the
then:
2.
or
of
order
(1,1,2)
or
2
(i,i,i,i),
respectively. For the be
the
time
being,
of
subfields
existence useful.
k(C)
The
is a)
Can
all
be
purely
of
them
be
Can
quadratic
C
and
k(t),
as
index
which
the
2,
the
real
more
case.
A
algebraic
existence
of
criterion
than such
(6.
1
subfields
for ) would of
transcendental
over
k
without
C
admitting
c)
Given way
for
this
an
which
if
we
are
- Other
open
Even
the
in Only
part
transcendental
subfields
quartic
y2
fields
equation
of
transcendental?
curve
y = t(x 2 +2),
splitting
distinct.
explicit
plane
with
the
purely
(this
means
that
C
has
no
is
there
k).
purely the
are
a)
in
beyond
open:
over
conjugate
and
since
§VII
none
points
Non
points
0 and
problems,
seem
they
little
points?
b)
do
following
genus
assumed,
k-rational
an
of
very
for
deriving
the
an
the
to
index
elliptic
curve
for is
consider
occur:
(x 2 +x)(x 2,
curve
take
are
with
not
the
Jacobian?
curves
with
conjugate
points,
without
It
k(x)
polynomials
k-rational
curves
= @,
Then
quartic
elliptic
k
2 + 3).
but
corresponding
equations
original
content
have
of
+
seems
quadratic
k-rational to
be
possible
to
points.
questions. case of
of our
curves results
with is
rational valid
points, in
192
cases
various (H)
and
problems (EH).
A
seem systematic
open:
23
t r e a t m e n t of t h e s e c a s e s w o u l d b e u s e f u l . b)
F r o m t i m e to time,
characteristic
h a r m o n i c c a s e of c h a r a c t e r i s t i c
3 has been excluded
3, in w h i c h A U t k ( C )
(especially
can b e q u i t e large).
the It w o u l d
b e g o o d to fill the gap. c) E.G.,
The case of characteristic
i n s t e a d o f 4 b r a n c h p o i n t s of
(with wild ramification), d)
The
splitting
2
k(C)
k
over
requires quite different methods. K, w e h a v e o n l y
1 or 2 of them
so that the r e a s o n i n g s b a s e d u p o n c r o s s - r a t i o s
fields
(over
k)
associated with a given elliptic curve if
probably
is a l o c a l o r a g l o b a l
field,
of the various quartic C,
deserve a closer
their ramification over
193
polynomials,
study. k
fail.
In p a r t i c u l a r ,
could be studied.
THE GROUP OF DIVISIBILITY AND ITS APPLICATIONS
Joe L. Mott*, Department of Mathematics Florida State University, Tallahassee, Florida
32306
INTRODUCTION
Associated with any integral domain G(D).
is a partially ordered directed group
This group, the set of non-zero principal fractional ideals of
aD ~ bD If
D
K*
if and only if
aD
contains
bD,
aU(D) ~ bU(D)
D,
then
if and only if
G(D)
with
is called the group of divisibility of
denotes the multiplicative group of the quotient field of
group of units of
D
is order isomorphic to
K*/U(D),
D
and
U(D)
D.
the
where
b/a e D.
Certainly, Dedekind [9] recognized that the study of divisibility by elements of D
amounts, essentially, to the study of
and only if
D
is a
G(D).
GCD domain [16, p. 267].
Indeed,
G(D)
is lattice ordered if
The missing ingredient in Kummer's
attempted proof of Fermat's Last Theorem was that for some algebraic number fields G(D)
need not be lattice ordered.
theory of ideals. group.
To overcome this obstacle, Dedekind created the
Essentially, his idea was to embed
In the cases considered by Dedekind, the set
ideals of
D
G(D) F
into some lattice ordered
of all non-zero fractional
formed an h-group.
In this context one can trace several other developments in ring theory. F
Since
need not be a group, PrHfer [29], Krull [20], Lorenzen [21], van der Waerden
[34, 5105] and others studied other systems of ideals contained in G(D).
Krull [20, p. 665] observed that an integral domain
closed if and only if the set of v-ideals V(D)
is a complete £-group.
V(D)
D
F
and containing
is completely integrally
form a group, and, in this ease,
Prefer considered domains where the set of finitely
generated fractional ideals form a group (indeed, an h-group).
Lorenzen discovered
necessary and sufficient conditions for an abelian group to he embedded in an h-group
*This work was supported by National Science Foundation Grant GP-19406.
194
(Dieudonn6 [i0] later gave an elegant proof without using systems of ideals). [19, p. 165] proved that for a valuation ring In this case, ideal theoretic properties of properties of
G(D),
Subsequently,
D,
G(D)
Krull
is merely the value group.
are easily derived from corresponding
D
and conversely.
several mathematicians have derived, and have effectively employed,
a method involving the group of divisibility to solve certain ring theoretic problems. This method involves three general steps. in terms of the group of divisibility;
First, formulate the ring theoretic problem
second, solve the problem there; then pull back
the solution, whenever possible, to the integral domain.
Apparently, Lorenzen [21]
originally used this approach to solve a problem of Krull, and Nakayama [25] applied it to disprove conjectures of Krull [20] and Clifford [5].
Two papers by Ohm [27, 28]
greatly popularized the method, and recently Heinzer [14, 15] and Sheldon [32] employed it to negatively answer ring theoretic questions. The main tool of this method is a sequence of theorems proved by Krull, Kaplansky, Jaffard, and Ohm. group
G,
These theorems, when summarized, state that for any lattice-ordered
there is a Bezout domain
D
with
G
as its group of divisibility.
Throughout the paper, this result will be designated as the Krull-Kaplansky-JaffardOhm Theorem. Authors have generally applied the method to construct counterexamples to conjectures in ring theory.
They have stressed that the advantage of this process is
the abundance of examples of O-groups whereas examples of integral domains are not as abundant.
While not denying this aspect of the theory, I am interested in the oppo-
site end of the scale.
My primary purpose here is to show how several classical
results in ordered group theory follow naturally from a corresponding ring theory result via the group of divisibility. Krull-Kaplansky-Jaffard-Ohm
Theorem.
The principal tool in this approach is the This theorem and other results in the paper give,
among others, Lorenzen's embedding [21] of an abelian k-group in a cardinal product of totally ordered groups, an embedding of an archimedean k-group in a complete k-group [7], and Ward's [33] and Birkhoff's [2] characterization of a cardinal sum of infinite cyclic groups.
195
i.
DEFINITIONS AND NOTATIONS
The notation and terminology will essentially be the same as that of Ohm's paper [28].
In this paper, all groups are abelian; an 0-group is partially ordered, and an
R-group is lattice-ordered.
A cartesian product of 0-groups
G~
is called the
cardinal product (sometimes called the ordered direct product or a vector group) if x = (x~) ~ y = (y~) the group
G~
if and only if
will be denoted by
x k ~ y~ ~
G~.
for each
An 0-group
(usually called a subdirect sum) of the groups G
into
~k G~
such that
projection map of ~
G~,
~
G~
onto
is the subset of
Let
Z
p~¢(G) = G~
~
G~.
G~
G~
for each
~. G
The cardinal product of is a subcardinal product
if there is an 0-embedding ~,
where
p~
~
of
is the canonical
The cardinal sum of the groups
G~,
denoted by
of all elements with finite support.
denote the group of integers under the natural order and let
R
denote
the additive group of real numbers. If means
a0, al, ..., a n a0
are elements of an 0-group
ao ~ infG {al . . . . ' "na})
denote sups and infs.
If
G
If
a, b • G
A semi-valuation of a field such that for all
G
then
K
alIb
is a map
(2)
v(x + y) • sup (inf G {v(x), v(y)})
(3)
v(-l)
v
of
a $ b K*
and
and
(in Ohm's cap (A)
b $ a.
onto an (additive) abelian
= v(x) + v(y)
= o.
is called the semi-value group of
The canonical example of a semi-valuation v: K* + K*/U(D),
where
positive cone of
K*/U(D)
general,
means
cup (v)
n
x, y e K*
v(xy)
G
al, ..., a
is an R-group, let
(i)
The group
a 0 • sup (inf G {al, ..., an})
is an upper bound of the set of all lower bounds of
notation
O-group
G,
K*/U(D)
if and only if divisibility of
U(D)
v. is the natural homomorphism
is the group of units in a subring
is the set of elements
dU(D),
where
is not a directed (filtered) group; in fact,
K
is the quotient field of D.
D,
and then
D
of
K,
d • D \ {0}. K*/U(D)
K*/U(D)
and the In
is directed
is the group of
(I usually prefer to think of the canonical semi-valuation of
onto the group of divisibility of cipal fractional ideal
D
as the map
xD.) 196
v
that maps
x • K*
to the prin-
K*
If, conversely, set to
v
is a semi-valuation on
D : {x • K* Iv(x) ~ 0} u {0}
K
with semi-value group
is a subring of
K
and
K*/U(D)
G,
the
is 0-isomorphic
G. Recall, finally, that a Bezout domain is an integral domain for which finitely
generated ideals are principal.
The group of divisibility of a Bezout domain is
lattice-ordered.
2.
THEOREM 2.1.
PRELIMINARY RESULTS
(Krull-Kaplansky-Jaffard-Ohm)
group, there is a Bezout domain
D
If
G
is a lattice-ordered abelian
with group of divisibility order isomorphic to
G.
The evolution of this basic result required a period of more than thirty years. Krull [19, p. 164] made the first significant contribution, proving that for a totally ordered group
G,
there is a valuation ring
D
with
G
as its value group.
The
essentials of his proof can be described as a series of definite steps, roughly as follows. Consider the group algebra w(~ ~gg) = inf {g lag ~ 0}. w(f/h) : w(f) - w(h),
Extend
where
this case, a valuation) on
k[G]
for any field w
Define
to the quotient field
f, h • k[G]. K.
k.
Next, let
Observe that
w
w
K
on
of
k[G]
k[G]
by
by
is a semi-valuation
D = ~x • K~ I w(x) ~ 0} u {0}.
is an integral domain with group of divisibility 0-isomorphic to
Then,
(in D
G.
Jaffard, following the major steps of Krull's proof, first published the next generalization
[16, p. 263] (although Kaplansky had made the same observation some
twelve years earlier in an unpublished portion of his dissertation at Harvard University).
For an ~-group
group of divisibility.
G,
Jaffard constructed a domain
D
with
G
as its
Then Ohm [27, p. 329] astutely observed that the domain
D
in this construction is necessarily a Bezout domain. Realizations of an 0-group by totally ordered groups have received considerable attention.
The question is this:
When can a given 0-group be embedded as a subgroup
of a cardinal product of totally ordered groups?
The following theorem of Fuchs' [12]
will be useful in obtaining realizations of groups of divisibility. 197
THEOREM 2.2. overrings
V
(Fuchs)
of
D,
If an integral domain
D
then the group of divisibility
product of the totally ordered groups
The proof is straightforward;
is an intersection of valuation G(D)
of
D
is a subcardinal
G(V ).
map the principal fractional ideal
in the cardinal product of the groups
G(V ).
xD
to
(xV)
On other occasions I shall refer to
this embedding as Fuchs' embedding. Theorem 2.2 has a partial converse (Proposition 2.3).
The ideas, definitions,
and results involved in the proof of Proposition 2.3 are all contained in Ohm's paper [28].
I include these results because of their usefulness throughout the paper.
DEFINITION. C,
then
~
If
B
and
C
are O-groups and
is a V-homomorphism if for all
b 0 e sup (inf B {bl, ..., bn })'
then
o
is a homomorphism of
b0, bl, • .., b n • B
B
into
with
o(b O) ¢ sup (inf C {a(b I), ...,o (bn)}).
A
V-embedding is a one-to-one V-homomorphism. Note that a V-homomorphism
is necessarily an 0-homomorphism.
In the sequel to Theorem 2.2, I shall assume that the group of divisibility of an integral domain
D
is a subcardinal product of totally ordered groups
such a way that the map
vk = pl%v
canonical semi-valuation, G~ D
onto
and
G~).
Then
%
is a semi-valuation for each
(here
GI v
in
is the
is the projection of
is a valuation overring of
V k = {x e K* Ivy(x) ~ 0} u {0}
D = nl Vl.
The following results concerning V-homomorphisms valuation whenever
i)
Let
v
V-homomorphism
¢
be a semi-valuation on
K
with value group
B
into a partially ordered group
semi-valuation on
K
with value group
The composition of two V-homomorphisms
3)
If
C,
preserves
then infs
and B
C
(B, P).
If
B
(C, V)
then 8v
is also a
are lattice-ordered groups and
sups)
is a
is a V-homomorphism.
is a V-homomorphism if and only if and
is a semi-
(~(B), P' n B(B)).
2)
B
v~
imply that each
is a V-homomorphism.
of
into
pl
is Fuchs' embedding, and
~
G(D)
onto a sublattice of
198
C.
B 8
is a 0-homomorphism of is a Z-map (that is,
B
4) onto
The projection map
Gl
pl
of the cardinal product of directed O-groups
G
is a V-homomorphism.
5)
Suppose
ol
is an O-homomorphism of
cartesian product of all product
~l Gl
ol.
Then
if and only i f
~
ol
C
into an O-group
is a V-homomorphism of
Gl.
C
Let
a
be the
into the cardinal
is a V-homomorphism for each
I.
Proposition 2.3 then follows from the above considerations.
PROPOSITION 2.3.
If the group of divisibility
G(D)
of a domain
V-embedded as a subcardinal product of Totally ordered groups intersection of valuation rings
Vl,
where
Gl,
D
then
is D
is an
G(VI) = Gl-
An example of Jaffard The group of divisibility of a domain must necessarily be a directed group. Jaffard [18] first published an example of a directed group that is not the group of divisibility of an integral domain. Jaffard's example is The following: of two copies of the order on J
Z
J
where
(a, b) £ J
Let
J
if and only if
is the induced partial order.
as a subcardinal product of
of some domain
D,
Z @ Z.
[24, p. 38], and
J
a + b
But, then,
is lattice-ordered.
D
is even.
Furthermore,
The identity map is a V-embedding of
If, Therefore,
then, by Proposition 2.3,
one discrete valuation rings.
be the subgroup of the cardinal sum
D
J
is the group of divisibility
must be an intersection of two rank
is a Bezout domain (in fact, a
The conclusion follows at once:
J
PID) is not
the group of divisibility of an integral domain. Paul Hill showed me the following elementary proof. (2, 2)
and
(3, i)
in
J:
then for certain elements where
v
If a
J
and
Consider the two elements
is the group of divisibility of some domain b
of
D,
is the canonical semi-valuation.
(2, 2) = v(a)
and
c ~ 2
and
either case, tained in and
(3, i)
aD
d a i.
If
v(a + b) ~ v(a) or
bD,
c = 2 or
and, hence,
then
(3, i) = v(b),
Then
v(a + b) = (c, d) e sup {inf G {v(a), v(b)}} is greater than that
D,
d a 2,
and if
v(a + b) a v(b). bD ~ aD
or
aD~
(i, i) d = i
Therefore, bD.
then
(2, 0) c a 3.
(a + b)D
so In
is con-
But then, one of
must be greater than the other--a contradiction. 199
and
(2, 2)
In this context, it is appropriate to mention a result in a paper by Cohen and Kaplansky [6].
I translate their result into the language of the group of divisibility
Recall that an atom is a minimal positive element of an 0-group.
THEOREM 2.4.
(Cohen and Kaplansky)
group with positive cone that
P
P.
Suppose that
Suppose, further, that
satisfies the descending chain condition.
ibility of some domain
D,
then
contains exactly three atoms possible permutation
P
is a partially ordered
(G, P) If
is not an Z-group and
(G, P)
is the group of divis-
contains more than two atoms.
al, a2,
(i, j, k)
G
of
and
a3,
then
i, 2,
and
3.
Further, if
a i + aj = 2a k for every
Thus, Theorem 2.4 gives a simple method for constructing 0-g~oups cannot be g r o u p s
of divisibility.
P = {n • Z I n a 2
or
n = 0}
contains exactly two atoms. elements since
J
a I + a 2 ~ 2a3,
J
determines a directed partial order on Jaffard's example
J
a I = (i, i), a 2 = (2, 0)
If an integral domain expect that Fuchs' embedding is a V-homomorphism.
D
Z ~ Z.
a 3 = (0, 2).
P
on positive
Furthermore,
is an intersection of valuation rings ~
of
F,
G(D)
J
Since
V
one might
into the cardinal product of groups
denoted by
cipal fractional ideals containing
F.
V ,
tion rings
where
is the map
F + Fw,
Fv,
G(V )
F
In particular,
b 0 • (bl, ...
since
(bl,
if .
""
D
is a fractional
F ÷ Fv.
FV
is the w-ideal of means
biD , . .., bnD ,
b0D
If
D
n
200
F.
is a principal ideal.
is contained
that is,
is a Bezout domain, this means
b )D '
F
~
b0D • sup (infG(D) {biD , ..., bnD})
b 0 • (bl, ..., bn) v.
'
If
is the intersection of all prin-
= n
in every principal fractional ideal that contains
bn)D
26].
the w-operation determined by the valua-
W
Next, observe that
[13,
The v-operation is the map
is an intersection of valuation rings
'
and
d.c.c,
and that
We shall presently see, however, that this is not the case.
the v-ideal of
{V }
satisfies the
Z
AN ANALYSIS OF FUCHS' EMBEDDING
First, recall some facts about ,-operations D,
that
cannot be a group of divisibility.
3.
ideal of
(G, P)
For instance, it is easy to see that the set
can be 0-embedded in the cardinal sum
has exactly three atoms
P
.
Suppose
D = n
V ,
where
V
is a valuation ring for each
further, that Fuchs' embedding is a V-homomorphism.
boD•
sup (infG(D) {biD . . . .
, bnD}
for each
~.
In other words:
for each
~.
But then,
(b 1 . . . . Fv
, bn) v _c ( b l ,
contains
F*
b 0 • (bl, ...
b0 • n
...,
implies
~.
Then,
boV a • sup ( i n f G ( V ) {blV '
bn) v
implies
Therefore,
for each ideal
F
(b 1, . . . ,
,...,
b 0 • (bl, ...
(bl, ..., bn)V~ = (b I, ..., bn)w,
bn) w.
Suppose,
'
bnV }) b n )V
and
bn) v = (b 1, . . . ,
bn )w
since
and each *-operation [13, p. 389]; thus, the
v-operation is equivalent to the w-operation determined by the valuation rings
V a-
After verifying the reverse implications, we conclude the following theorem.
THEOREM 3 •i.
Suppose
D
_ valuation rings i_~san intersection _of
be the associated *-operation,
w: F ÷ F
into the cardinal product of the groups the v-operation is equivalent to
COROLLARY 3.2. V
If
D
is a quotient ring of
Proof.
= n
FV .
G(V )
Let
w
Then, Fuchs' embedding of
G(D)
is a V-homomorphism if and only if
w.
is an intersection o f valuation rings D,
V s.
V ,
where each
then Fuchs' embedding is a V-homomorphism.
See [13, p. 549].
Since the w-operation determined by valuation rings is "endlich arithmetische hrauchbar"
[13, p. 362], the embedding of Fuchs'
v-domain [13, p. 391].
COROLLARY 3.3.
is a V-homomorphism only if
If
D
Th___~egroup of divisibility of
D D
is V-embedded as a subcardinal is a_ v-domain.
is a v-domain, then the v-operation is equivalent to some w-opera-
tion determined by a collection of valuation rings Theorem 3.1 implies
is a
In fact, we have the following corollary.
product o f totally ordered groups if and only i f
Proof.
D
G(D)
is V-embedded in
[
{V },
where
D = n
V .
Then,
G(V ).
Remarks If
D
is a Krull domain, the only w-operation equivalent to the v-operation is
the w-operation determined by
Dp
for all minimal primes
201
P
of
D.
For if
V
is a valuation overring of then
Pv
D,
PV ~ V,
D
and
centered at some non-minimal prime ideal Pv
P
of
is the v-ideal of some finitely generated
D,
A c p
(see the proof of 36.3 in [13, p. 534]). In particular, where
Vc~
is
centered on of
G(D)
Dp
if
D : k[x, y],
for
a minimal prime
(x, y).
into
Then
the cardinal
where
P
i ~ (x, Y)v
product
4.
k
of
but
of all
is a field,
D
and
V
and
D : n
V
i s some v a l u a t i o n
i ~ (x, y)V
G(V )
then
n V,
ring
so that Fuchs' embedding
G(V)
is not a V-homomorphism.
SOME CLASSICAL RESULTS
In 1939, Lorenzen [21] proved that an k-group can be embedded in a cardinal product of totally ordered groups as a sublattice and as a subcardinal product.
Since
then, Birkhoff [2], Clifford [5], Jaffard [16, 17], Ribenboim [30, 31], and others reproved and refined this result. Theorem.
Subsequently,
THEOREM 4.1.
I shall prove other familiar results using the same approach.
(Lorenzen)
If
family of totally ordered groups of
H A Gk.
Moreover,
Proof.
Since
%(G)
G
V
G
is a lattice-ordered group, then there is a
{GI}
and an embedding
%
of
G
onto a sublattice
is a subcardinal product of the groups
is lattice-ordered,
its group of divisibility. where each
I prove it using the Krull-Kaplansky-Jaffard-Ohm
Thus,
D
G%.
there is a Bezout domain
D
with
is an intersection of valuation rings
is a quotient ring of
D.
By Corollary 3.2,
a subcardinal product of the totally ordered groups
G(V ).
G
G
as
V ,
can be V-embedded as
Since
G
is lattice-
ordered, this V-embedding is a lattice homomorphism onto a sublattice by result 3 of §2.
COROLLARY 4.2. divisibility, then
Proof. each
G%
If D
D
is an integral domain with lattice-ordered group of
is integrally closed.
By Theorem 4.1,
G(D)
is totally ordered.
valuation rings
V%
where
is V-embedded as a sublattice of
Then by Proposition 2.3,
G(V%) = G%.
In particular,
202
~% G%,
where
D
is an intersection of
D
is integrally closed.
i0
A second consequence of the Theorem 2.1 is the following expanded version of a theorem of Ward [33] and Birkhoff [2] (I have added part 3 for completeness).
THEOREM 4.3. group.
G
is a lattice-ordered abelian
i)
G
is a cardinal sum of copies of
2)
G
satisfies the descending chain condition o n positive elements.
3)
G
is the group of divisibility of a unique factorization domain.
Clearly
i)
There is a domain
d.c.c,
implies D
with
on positive elements,
non-zero non-unit of UFD
Suppose
These are equivalent.
Proof. 3).
(Ward and Birkhoff)
D
2) G
D
Z.
and
3)
implies
i).
as its group of divisibility. satisfies
a.c.c,
Then,
p
v
is a finite product of irreducible elements.
irreducible in
v(x) ~ 0
and
p
D
implies
v(p)
p
then
G
of
D
satisfies Hence, each
That
D
v(p) ~ v(x)
or
is a
is prime. D
onto
is a minimal positive element of
is prime, we need only see that if
v(y) ~ 0,
implies
Since
be the canonical semi-valuation from the quotient field of
Thus, to observe that where
2)
on principal ideals.
will follow if we observe that an irreducible element Let
We show
G.
G.
v(p) ~ v(x) + v(y)
v(p) ~ v(y).
But that is a
well-known property of minimal positive elements in a lattice-ordered group [ii, p. 71].
The next theorem apparently first appeared as an exercise in [3, p. 90, Ex. 31] (see also [13, p. 548]).
The result will be an easy corollary of Theorem 4.3 using
some well-known facts about Kronecker functions.
COROLLARY 4.4. ring
Dv
(Bourbaki)
If
D
(See [i], [20] or [13,
26].)
is a Krull domain, the Kronecker function
is a principal ideal domain.
Proof.
Since
D
is a Krull domain,
D : n
V e , where each
V
: Dp
is a
discrete rank one valuation ring and each non-zero non-unit of
D
only finitely many of the valuation rings
has finite character
on
D).
Now the Kronecker function ring
rank one discrete and
{V (X)}
V
(that is,
Dv
is equal to
has finite character on
is emheddable in the cardinal sum of
G(V (X))
203
and
{V } n Dv
G(D v)
is a non-unit in
V (X), [7].
each
V (X)
Therefore,
is
G(D v)
is lattice-ordered since
ii
Dv
is a Bezout domain.
By Theorem 4.3, Dv
is Bezout,
Hence,
G(D v) D
G(D v)
satisfies the
d.c.c,
is a cardinal sum of copies of
is, in fact, a
PID
Z
on positive elements.
and
Dv
is a
UFD.
Since
[13, p. 525].
Another accomplishment of Theorem 2.1 is the embedding of an archimedean latticeordered abelian group in a complete Z-group.
MacNeille [22] showed that the classical
method of forming Dedekind cuts can be applied to obtain an embedding of an arbitrary partially ordered set in a complete lattice. the result in general is not a group.
If this process is applied to an 0-group,
However, if the 0-group can be 0-embedded in a
complete Z-group, then the embedding can be achieved by the Dedekind cut process. This process is described in detail in Puchs' book [ii]. The embedding in a complete Z-group has been studied by several authors including Clifford [5], Conrad and McAlister [7], Krull [19], and Lorenzen [21]. proof employs the notion of v-ideal.
Recall that if
archimedean group of divisibility, then
D
D
The following
is a Bezout domain with
is completely integrally closed [13,
p. 614].
THEOREM 4.5.
If
G
is an archimedean £-group, then
G
can be embedded in a
complete Z-group.
Proof. G
Let
D
is archimedean,
Z-group.
be a Bezout domain with D
Clearly
G
as its group of divisibility.
is completely integrally closed, and
G(D) = G
can be embedded in
V(D)
Since
is a complete
V(D).
Observe that in the sense of Theorem i.I of [7],
V(D)
is the completion of
G(D). If
G
is a lattice-ordered group and
convex Z-subgroup
Mg
g e G,
g ~ 0,
such that for any convex Z-subgroup
then a value of
g
H n Mg,
A Zorn's
lemma argument shows the existence of a value for any non-zero known that p. 188].
G/M g
is totally ordered; that is '
Mg
g e H.
g • G.
is a
It is well
is a prime subgroup of
G
[7 '
I will give a proof using Theorem 2.1.
PROPOSITION 4.6. non-zero element of
Suppose that G.
If
Mg
G
is a lattice-ordered group and that
is a value of
G,
then
G/M
is a
i s totally ordered. g
204
g
12
Proof.
Without loss of generality assume
with group of divisibility v
G.
We show
is the canonical semi-valuatlon,
system since v(x) = g,
g @ H.
x % S.
Thus
Let
D \P
P
S = v-l((Mg)+)
Hence,
G
H = Mg,
S = D \ P,
and
be a Bezout domain
[233. D
is a saturated multiplicative H
D
of G/M
G
D.
If
is a saturated multiplicative
be a prime ideal of
generates a convex Z-subgroup
Let
corresponds to a prime ideal of
g
is a convex directed subgroup of
g
then
P n S = @. v(D \ P)
M
M
g > O.
If
x e D
containing
x
is such that such that
system containing
containing
M . g
S.
Hence,
Furthermore,
is the group of divisibility of the g
valuation ring
Dp
[23].
Finally, I apply Theorem 2.1 to obtain a ring theoretic proof of a well known result about realizations of an Z-group by subgroups of reals.
First, observe the
following proposition.
PROPOSITION 4,7.
If
D
.
valuation overrings of in the center of some
is a Bezout domain and if
.
D
.
{V }
.
such that each non-zero non-unit
V ,
then
is a collection of
C~
xD = na (xV~ n D)
and
x
of
D
is contained
D = n~ Ve.
The proof is similar to that of [13, p. 42] needing only the additional observation that
[zD:xD]
is finitely generated [13, p. 301].
An archimedean totally ordered group can be embedded as a subgroup of the reals. This fact led Clifford [5] to conjecture that an archimedean Z-group has a realization by subgroups of
R.
Similarly, Krull [19] conjectured that a completely integrally
closed domain is an intersection of rank one valuation rings. terexample to both conjectures in [25].
Nakayama gave a coun-
Realizations by subgroups of
R
are obtained
by applying the following corollary to Proposition 4.7.
COROLLARY 4.8.
If
D
is a Bezout domain such that each non-zero non-unit of
D
is contained in a prime ideal of height one, then the group of divisibility can be embedded as an Z-subgroup of a cardinal product o f subgroups of the reals.
Proof. P
By Proposition 4.7,
D
runs through all prime ideals of
is of height one,
Dp
is an intersection of valuation rings
Dp
D
Since
of height one.
is a rank one valuation ring, and 2O5
Apply Theorem 2.2. G(Dp)
where
is a subgroup of
P R.
13
The next theorem in its present form appeared in a paper by Conrad, Harvey, and Holland [8, p. 164], but it was originally proved for vector lattices by Nakayama [26].
I add part 3 for completeness.
THEOREM 4.9.
(Nakayama, Conrad, Harvey, Holland)
ordered abelian group. i)
G
Suppose
G
is a lattice-
The following are e~uivalent.
is 0-isomorphic to a sublattice of the cardinal product of copies of the
additive group of real numbers. 2)
For each non-zero
g • G,
there is a value
Mg
of
g
such that
G/M
i_~s g
~-isomorphic to a subgroup of the real numbers. 3)
G
is the group of divisibility of a Bezout domain
zero non-unit
Proof.
x
That
of
D
3)
such that each non-
is contained in a prime ideal of height one.
implies
Finally, assume
i) 2)
is the content of Corollary 4.8.
implies
2).
D
with
G
holds.
If
v
is the canonical semi-valuation onto
of
g
such that
g
saturated multiplicative
Suppose G,
S
of
D.
Since
x
let
is contained in the reals. system
Clearly
i)
By Theorem 2.1, there is a Bezout domain
as its group of divisibility.
G/M
D
is a non-zero non-unit of g = v(x).
By [23]
'
M
Let g
M
g
D.
be a value
corresponds to a
G/M
is totally ordered, S is g the complement of a prime ideal P of D. Further G(D S) = G/M and G/M a subg g group of the reals imply that D S is a rank one valuation ring and that P is a prime ideal of height one.
REFERENCES i.
Arnold, J. On the ideal theory of the Kronecker function ring and the domain
D(X),
Canad. J. 21, 558-563 (1969). 2.
Birkhoff, G. Lattice-ordered groups, Annals of Math. 43, 298-331 (1942).
3.
Bourbaki, N. Algebra Commutative, Chapitre 7, Herman, Paris, 1965.
4.
Brewer, J° and Mott, J. L. On integral domain of finite character, J. Reine Angew. Math. 241, 34-41 (1970).
5.
Clifford, A. H. Partially ordered abelian groups, Ann. of Math. 41, 465-473 (1940).
206
14
6.
Cohen, I. S. and Kaplansky, I. Rings with a finite number of primes, I. Trans. Amer. Math. Soc. 60 (1946), 468-477.
7.
Conrad, P. F. and McAlister, D. The completion of a lattice-ordered group, J. Australian Hath. Soc. 9, 189-208 (1969).
8.
Conrad, P., Harvey, J., and Holland, C. The Hahn embedding theorem for abelian lattice-ordered groups, Trans. Amer. Math. Soc. 108, 143-169 (1963).
9.
Dedekind, R. Ueber Zerlegungen von Zahlen durch ihre grossten gemeinsamen Teiler, Gesammelte Math. Werke, Vol. II, Chelsea, New York, 103-147 (1969).
i0.
Dieudonn~, J. Sum la Th~orie de la DivisibilitY, Bull. Soc. Math. Prance 49, 1-12 (1941).
ii. 12. 13.
Fuchs, L. Partially ordered algebraic systems, Pergamon Press, New York, 1963. The generalization of the valuation theory, Duke Math. J. 18, 19-26 (1951). Gilmer, R. W. Multiplicative
ideal theory, Queens' Papers, Lecture Notes No. 12,
Queen's University, Kingston, Ontario, 1968. 14.
Heinzer, W. Some remarks on complete integral closure, J. Australian Math. 9, 310-314 (1969).
15.
J-Noetherian integral domains with i in the stable range, Proc. Amer. Math. Soc. 19, 1369-1372 (1968).
16.
Jaffard, P. Contribution ~ la th4orie des groupes ordonn4s, J. Math. Pures Appl. 32, 203-280 (1953).
17.
Extension des groupes r@ticul6s et applications, Publ. Sci. Univ. Alger. Ser. AFI, 197-222 (1954).
18.
Un eontre-exemple
concernant les groupes de divisibilite, C. R. Acad.
Sci. Paris 243, 1264-1268 (1956). 19. 20.
Krull, W. Allgemeine Bewertungetheorie,
J. Reine Angew. Math. 167, 160-196 (1931).
Be itr[ge zur arithmetik kommutativer Integrit[tsbereiche;
I, II, Math.
Z. 41, 545-577; 665-679 (1936). 21.
Lorenzen, P. Abstrakte Begrundung der multiplicativen
Idealtheorie, Math. Z. 45,
533-553 (1939). 22.
MacNeille, H. Partially ordered sets, Trans. Amer. Math. Soc. 42, 416-460 (1937).
23.
Mott, J. L. The convex directed subgroups of a group of divisibility, submitted for publication. 2O7
15
24.
Nagata, M° Local rings, Interscience New York, 1962.
25.
Nakayama, T. On Krull's conjecture concerning completely integrally closed integrity domains, I, II, Proc. Imp. Acad. Tokyo 18, 185-187; 233-236 (1942); III, Proc. Japan Acad. 22, 249-250 (1946).
26.
_
_
Note on lattice-ordered groups, Proc. Imp. Acad. Tokyo 18, 1-4 (1942).
27.
Ohm, J. Some counterexamples related to integral closure in
D[[X]], Trans. Amer.
Math. Soc. 122, 321-333 (1966). 28.
Semi-valuation
and groups of divisibility, Canad. J. Math. 21, 576-591
( 1969 ). 29.
Prefer, H. Untersuchungen ~ber die Teilbarkeitseigenschaften
in K~rpern, J. Reine
Angew. Math. 168, 1-36 (1932). 30.
Ribenboim, P. Conjonction d'ordres dans les groupes abelian ordonn~s, An. Acad. Brasil. Ci. 29, 201-224 (1957).
31.
Un th~or~me de realisation de groupes r~ticul@s, Pacific J. Math. i0, 305-308 (1960).
32.
Sheldon, P. A counter example to a conjecture of Heinzer, submitted for publication.
33.
Ward, W. Residuated lattices, Duke Math. J. 6, 641-651 (1940).
34.
van der Waerden, B. L. Modern Algebra, Vol. II, 2nd English edition, Ungar, New York, 1950.
208
HOMOLOGICAL
DIMENSION AND EULER MAPS*
Jack Ohm, Purdue U n i v e r s i t y and Louisiana
I.
Introduction.
for
~¢
If
is a map
multiplicative
f
~¢¢
is a collection
of ~
on short
f(M)
exact sequences M
for
dimension
fractional
of
and Kaplansky's
direct certain
~
sum.
[SE]).
approach
otherwise
treatment
Roughly,
and a subset
the set of elements
O
in
of
G
is the
is that of MacRae
[M],
group of
to these what
examples.
I have followed [K].
of
~
generated
having
R-modules
free R-modules
(Compare
the paper
[M]
also Bass
[B]
are closed under
extended
finite
~,
under
to an Euler map on
~-resolution.
The
~
and
can be taken to be the set of
and the set of all finitely
respectively, are more
while
the
complicated
_~ and
in §3.
* This research was
My
I feel is a proper
~¢, both of which
it can be uniquely
give the Euler map of MacRae discussed
sum of the ranks of
the idea is to begin with a set of
for the first of the above examples
generated
free resolution,
If then an Euler map can be defined on
conditions
all finitely
if
R.
is in providing
for the exposition;
R-modules
example
is
For example,
into the m u l t i p l i c a t i v e
ideals
only claim to o r i g i n a l i t y
and Swan-Evans
which
M; here the group
Another
I shall give here a unified
of MacRae
~¢.
an Euler map
an Euler map from the set of torsion R-modules of
finite projective invertible
G
which have a finite
in a resolution
group of integers.
who has defined
setting
from
can be taken to be the alternating
the free modules additive
of R-modules,
into an abelian group
-~ is the set of R-modules then
State U n i v e r s i t y
supported by NSF Grant GP-29104A.
209
and
G
which
will be
2.
Dimension.
operation
+.
possibility avoid
Let
required
never
enter
sequences
a collection
~
n m 3, M i e ~ ,
(M n .... ,MI)
~2.
For any
(M n, • .. ,MI) (i.e.
M
the picture;
and any
can be added
For every
M ~,
~4.
(0,M',M,0)
e ~=
~5.
(Factoring
through
(M n .... ,Mi+ I, Ki, ~6. such ~7.
(0,A,B,C,0) that
(0,A',B',C,0) and
An element (Mn,...,MI) (Mi+l,Mi)
of certain
there
is
given
MI) ,
axioms:
for every
1 ~ i < n,
. . . + M, M I + M . Mi_l,
terms
,MI)
of any element
~
of ~ ) .
c ~.
= M. images exists
0) ~ ~
and
e ~
that
successive
there
and
).
i
such
and
e ~=
(0,A,B,C,0)
exists
M e ~
e such
1 ~ i < n,
e ~.
there
(0,K,M',A,0)
that
that
(0, Ki, Mi,...,MI)
(0,K,M,B,0)
e ~
there
For every K i e ~¢i such
of pullbacks).
e ~=
(0,A',M,B,0)
such
.,Mi+2, . .Mi+ .1
M'
an exact
2 ~ j ~ n-i.
(0,0,M,M,0,0)
(0,M',M,C,0)
(Existence
i
to two
~3.
that
down
In p a r t i c u l a r ,
seven
~
to
to write
to define
(Mn,...,M2,
the following
such
= (M n .
e ~
of the form
and one
instead
suppose
to an
However,
the existence
(Mi+j, .... Mi+l,Mi) j
category.
properties.
only
respect
sequence,
and then
we shall
satisfies
and
M e~ e~
(Mn,...,MI)
Thus,
e ~
i = 1 .... ,n - 2
these
with
chosen
theorems
having
of sequences
which
of an exact
we have
for our
matters.
semigroup
in an abelian
considerations,
to be something
morphisms
~i.
the notion
w o u l d be to work
the properties
exact
be an abelian
We need
extraneous
sequence
~
exists
M'
e ~¢
e 5~. 5~ and that
(0,A,M,B',0)
e ~
e ~.
of the set
is exact,
~
we shall
and shall write
Ki=
will
be called
call a
Ki
given by
im (Mi+l,Mi).
210
an exact ~5
sequence.
If
an image
for
2.1 ~i,
Consequences ~3,~4, ii)
of the axioms,
i)
(0,M,0)
¢ ~
~ M = 0,
by
and ~ 5 . For any
M',M
¢ _~f, (0,M',M'
+ M,M,0)
e ~,
by ~ i ,
~2,
and ~ 3 .
Note
that
R-modules
~¢'
of i s o m o r p h i s m
ring R) u n d e r
sum,
by the exact
~:M ÷ B, take
~':B'
is the s e m i g r o u p
(for a fixed
are s a t i s f i e d if
if
~ C,
take
appropriate
M'
sequences
= ~-I(A);
M = {(b,b')
diagrams
9
o
K
K
for ~ 6
direct
if
~ B ~ B'[~(b) and
d~7
the above
in the usual
and for ~ 7 ,
classes
sense.
of
axioms For
~
6,
~:B ÷ C,
= ~'(b')}.
The
are the following:
eT.
?
o
A'
A'
I
0- - ÷M'-_+M__÷C --+0 + + 0 + Aa + B + C + 0
Given P ¢~ any
K,K',L' there
(0,M,L'
+
0
0
an exact
will
exact,
+
sequence
be called c~
+ P,B,0)
(0,A,B,C,0),
projective
such that
exists
0-+A-÷M+ 0 +A+B i
M e Jl
relative
are exact.
that
One
+
+
0
0
A,B,C to
(0,K',L',A,0) such
_÷B'__+0 + + C + 0
c J l,
(0,A,B~C,0) and
have
if for
(0,K,P,C,0)
(0,K',M,K,0)
should
an element
are
and
in m i n d
the f o l l o w i n g
diagram:
o
9
+
0---÷
K'
....
L' 0
+
o
+
M
+
....
L'+P t
A
÷
0
B
that p r o j e c t i v e s s p l i t ,
in the sense
and
P
to
relative
+
C
÷
0
0
that
(0,A,B,P,0)
211
+ 0
P
0
Note
projective
K--
(0,A,B,P,0) implies
exact
B = A + P.
One
sees this by taking M = 0
by 2.1,
relative
K = K' = 0
and hence
to every short
then we shall
Let
O
relation
on
simply
and
L' = A
B = A + P exact
call
P
by
in the above.
~4.
sequence
If
P
(0,A,B,C,0),
Then
is projective A,B,C
e ~,
projective.
be a subsemigroup
of ~d~.
We define
an equivalence
.~" as follows:
For any
MI,M 2 e ~ ,
MI--M 2
if there exist
PI,P2
~ O
such that
M1 + P1 = M2 + P2" The equivalence P ~ O
~*
such that
M.
of
O
M + P ~ 0},
M e ~,
0,
n ~ 0,
We use
finite of
0*
an exact
{M ~ ~
Ithere exists
and one easily verifies
Res(O~)
M, do(M),
sequence
will be called
length}.
of length
If
an
to denote
n;
we define
Proposition.
and
n > 0.
Let
that
exist
P0*,PI*
~O*,
and apply
If
we define
O
M
n
for
G-resolution the
of
G-dimension
has an
G-resolution
an induction
such that
exists
are in
is exact. n > i,
be exact with
P.*I ~ 0 *
PI.* -- Pi,i
= 0,...,n
is exact.
there
P1 = PI* + Q
(K,Pn*,...,P2*,PI,P0,M,0) sequence.
of length
has an
such that
Pie
Proof.
required
(0,Pn,...,P0,M,0),
dG(0 ) = -I.
(K,Pn,...,P0,M,0)
and
{M e ~ I M
n
and such that
P0 = P0* + Q
G-resolution
(K,Pn*,... 'P0*'M'0)
Then there
Since
of the form
M # O e Res(O~),
to be the least
2.2
~5
is then
= 0*.
If Pi¢
class
factor
hypothesis.
212
Q ~ O
0.
If
such that
By J~2,
n = i,
through
this
is the
K 0 = im(P],P 0)
via
2.3
Corollary.
d~(M)
2.4
Res(~,~)
= do,(M)
if
d o , ( M ) # 0,
Proposition.
Suppose
(0'K'Pn - 1 ' "'''P0 'M'0) Pi,Pi ' g G *
Proof.
and
Proceed
(0,K',M',0) n = i:
M + P = M' suffices assume
+ P'.
P0,P0'
¢
and the s p l i t t i n g K + P0'
there
Hence
step,
factor
K 0 ~ K 0'
be an element
(0,Pn*,...,P0*,M,0).
~g
Let
having
The above
shows
via
rather For this
§2 use
only
us to a b b r e v i a t e
~2,
it
Similarly,
we may
of p u l l b a c k s that
K 0 = im(Pi,P 0)
and
n = 1
case,
K ~ K'
of
are p r o j e c t i v e ,
an
O*
than
check
reason, G*
a given all
we shall
to c o m p u t e
our n o t a t i o n
213
to
and then
and let
resolution and let
Then
that w h e n
one need only e x a m i n e
of
allows
corollary
that
lemma)
K 0 = im(P0*,M),
1 m i ~ n
remainder
M = M'.
through
by the
K i = im(Pi*,Pi_l*) ,
one.
such
on a s u m m a n d
the elements
of
and
K -- K'
Suppose
a minimal
g O
Schanuel's
2.5
do,(M),
P,P'
when
and
~a4.
(as w i t h
hypothesis.
compute
by
of p r o j e c t i v e s
by the i n d u c t i o n
projective,
= K'
by adding
= 0.
are exact with
n = 0:(0,K,M,0)
exist
~,(M)
K ~ K'
existence
K 0' = i m ( P l ' , P 0 ' ) .
M ~ 0
n.
then
are p r o j e c t i v e
from the
induction
Corollary.
~
if
It follows
= K' + P0"
For the
on
the p r o p o s i t i o n ~.
of
then
K = M -- M'
Therefore
to p r o v e
0 ~ dG(M ) ~ 1
M ~ M'
by i n d u c t i o n
M ~ M',
and
M g Res(O~/~,
(0,K',P'n-l'" . "'P0 ''M''0) If
imply
If
the elements
and
n ~ 0.
exact
Since
= Res~Y*,~.
d~,(M) =
min{ilK i ~O*}.
the e l e m e n t s
of
~*-resolution ~*-resolutions henceforth
O
are
for
M
and select
in the
dimension,
which
in turn
d(M)
~,(M)
and
for
to
Res
for
R e s ~
remainder
2.6
of
§2 that
Corollary.
then
d(M)
Proof.
Apply
be d e l e t e d
then
Proof.
If
adding
Suppose > 0,
that
(ii)
then
is in
d(M)
n > 0
implies
from
2.6 that
a summand
d(K)
are p r o j e c t i v e .
~ 0
are in Res
the h y p o t h e s i s
c Res, d(M)
Res,
and
that
P ~ O ~,
M'
and
= 1 + d(K);
then there
(0,K,P,M,0)
(0,Pn,...,PI,KI,0)
M -- M',
~ Res
can
(0,K,P,M,0)
while
if
d(M)
is = 0,
to
is exact
= n > 0, for
P
and
and
and let
M.
Factoring
is a m i n i m a l
K ~ 0, K 1 ~ 0. d(K)
= n - i.
P ~ O ~
d(M)
Suppose we may
P = K + M;
and
= 1 + d(K).
(0,Pn,...,P0,M,0) through
K 1 =im(Pi,P0),
O~-resolution
Since
M via ~ 2 ,
of p r o j e c t i v e s ,
exist
K -- K 1 now
for
by 2.4,
d(M)
assume
= 0. P,M ~ O
and hence
K I. it By •
K ~O ~
~ 0.
Let
and factor
(0,Pn,...,P0,M,0) through
is the r e q u i r e d
2.8
M,K
G~-resolution
T h e n by s p l i t t i n g and
for the
2.6.
Suppose
be a m i n i m a l
follows
M'
G
assume
2.5.
in 2.8 that
M ~ 0
such
we have
and
o~
also
~ 0.
(i)
Also,
show
d(M)
d(K)
K ~ Res
M ~ 0
2.4 and
i)
If
ii)
the elements
If
from
Lemma.
exact.
We shall
= d(M').
We shall
2.7
(= R e s ( G ~ ) .
Lemma.
If
be a m i n i m a l
K =im(Pi,P0).
Then
O~-resolution K ~ Res
and
sequence.
M c Res
and
M' ~ M,
214
then
M'
c Res.
for
M,
(0,K,P0,M,0)
Proof.
It s u f f i c e s
M + P g Res.
That
consequence
+ P,M~0)
M = 0
by 2.1,
By
and
~6,
M c Res
of ~ 2 .
(0,P,M
P' e O *
to see that
there
is exact by
such that
projectives,
P
M e Res.
2.9
Theorem
such
that
(dimension
(0,A,B,C,0)
Res,
then the third
d(B)
~ max{d(A),d(C)};
d(C)
=
Proof.
d(A)
+
Case
induction
then
the t h e o r e m
is exact.
If any two of
Assume
d(C).
If
A
is immediate.
so we may
d(C)
follows; e 6"
= 1 + d(K).
element
of
and hence
and
O
to
an
in
B,C,
hypothesis,
and
of
K + P ~ Res and
L e Res
be elements A,B,C
holds,
L e Res,
in Res.
then
of ~
are in
then
and
d(C)
that
= n > -I.
By 2.7(i~,
= 1 + d(K')
O-summand
does
not
Then
P
there
+ P,B,0)
and hence
alter
2, we may
exists
215
and
dimension
assume
L ~ ~
are exact.
B ~ Res,
there
and
2.6 and 2.8, by adding
via ~
Then
(0,K',P',A,0)
d(A)
and
and
and the
that
by
by
A = B
B = C
A # 0.
such
We p r o c e e d
C = 0
A = 0, then
e Res
Res,
(0,L,P'
are
assume
and such
is p r o j e c t i v e .
(0,K',L,K,0)
is exact.
occurs,
then that
If
also
K,K'
Since
of b e i n g
= -i,
B ~ 0.
are exact
the p r o p e r t y
this
C
Assume
theorem
(0,K,P,C,0)
A,B,C
inequality
and
d(C)
also
P,P'
is exact
Let
and if the
exist
Then by s p l i t t i n g
theorem).
C # 0, and hence
exist
+ P,0)
L = K + P; and
When
there
i.
(i):
on
By 2.7,
implies
that
(0,L,P',M,0)
is also.
is an i m m e d i a t e
M + P = 0
are exact.
implies
if and only
M + P e Res.
(0,K,P',M such
(0,L,P',M,0)
But
M + P e Res
M + P # 0.
e ~
projective
M ~ Res
Moreover,
K ~ Res
and
imply
2.1.
assume
L
P e 6,
suppose
so we may
(0,K,L,P,0)
K ~ Res.
implies
Conversely,
exists
since
for
or
on an P e 6
such that
By the i n d u c t i o n
and also
if
d(L) If
~ max{d(K'),d(K)} d(B)
> 0,
dimensions
then by
of
K',L,K
respectively, A,B,C.
If
2.7(i)
d(B)
Then exact such M
and
to
d(B)
that
and
A
there
implies
N'
By 2.4, implies
f
of
for any short f(A)f(C) sequence
J
e Res
N ~ N';
such
and hence
A e Res
by case
be a subset
into
a (multiplicative)
A simple
of
~.
as
and
B # 0.
exists Case
N e~
(i), and then
(iii):
C e Res
implies
(0,N',Q',C,0) by
2.8.
Then
is M
M i e ~,n
for
group
G
with
A,B,C
~ ~,
shows ~ 3,
An Euler map
that
such
Let a)
O i c Oi+ 1 If
be subsets
(0,A,B,C,0)
of
is exact
one has
~
such
that
and
A,C
are
in
Gi,
then
B c ~. l
b)
For any
M ~ ~i+l'
there
and
exist
216
~
is that
then for any exact
f(M1) = f ( M 2 ) f ( M 3 ) - l f ( M 4 ) . . .
3.1
is
(ii):
abelian
(0,A,B,C,0)
induction
(0,Mn,...,MI,0),
We can
(ii).
~
sequence
implies
(0,M,Q,B,0)
there
that
N ~ Res
Let
exact
= f(B).
Q' ~ G *
to
from
e Res
B e Res
by case Case
down
(i).
are exact.
N e Res
the
follows
A,B
such that
is exact.
and
carries
A e Res.
so assume
+ i.
A,B,C
case
(ii))
implies
(0,M,N,A,0)
(0,N,Q,C,0)
Euler maps.
a map
Res
(C~e
and hence by ~ 6
are
of
thus
concludes
M e Res
and
and then
than those
that
B = 0, and
= d(K')
and the t h e o r e m
B,C e Res
Q e O*
d(K)
+ i;
K',L,K
This
to e s t a b l i s h
(iii))
in
less
B e ~*
(0,A,B,C,0).
implies ~ d(L)
for
(0,N,Q,C,0)
exist
N e Res
then
= 1 + d(M);
since
exact.
d(B)
are all one
of the case
exist
and
C e Res
3.
(C~e
dispose
there
holds
2.7(i),
= 0,
It only remains
before
<
and the a s s e r t i o n
applied
C e Res
and
P,K e O i
such that
(0,K,P,M,0)
is exact;
(0,A,B,M,0)
with
projective
Theorem. group
moreover,
given any exact sequence
A,B ~ £Yi+l' this
relative
to
P
can be chosen to be
(0,A,B,M,0).
If there exists an Euler map
G,
then
f
f
of
O. i
into an abelian
extends uniquely to an Euler map of
£Yi+l into
G.
Proof. that
If
M ¢ Oi+l~£Yi,
(0,K,P,M,0)
If is exact,
P',K'
is exact.
Define
such that
(0,K,W,P',0)
K,P' e O i
implies
f(K')f(P)
since
and
W e ~i f
and
by
f(M)
(a).
P
the projectivity
P,
and by
of
(0,L,P'+P,B,0) (a), and hence
f(A)f(C)
is projective
f(P)f(K) -I =
and
L e ~
Then
= f(L). f
(0,K,P,C,0) to
K,K' ~ O i
Similarly,
with
K,P e ~i
are
(0,A,B,C,0).
such that
is multiplicative
(0,K,P,M,0)
P,K ~ O i.
By (b), there exist
relative
there exists
f(K)f(K')
of the form
= f(W) =
A,B,C e £Yi+l such that
= f(B).
are exact.
W ~
Then
f(K)f(P') Thus,
-i
(0,K',P',M,0)
are exact.
O i.
such
f(M) = f(P)f(K)
there exists
Therefore
(0,KT,P',A,0)
We have already checked that sequences
£Yi
is independent of the choice of
such that
exact and such that
P,K ~
such that
(0,K',W,P,0)
is an Euler map on
is exact,
P',P,Kt,K ~ O i
Oi
of pullbacks,
Now let us prove that for any (0,A,B,C,0)
there exist (necessarily)
are also elements of
then by the existence
f(P')f(K') -I
(b)
then by
(0,K',L,K,0)
implies f(P'+P)
By
L ~ Oi
= f(P')f(P).
on exact and
M ~ Oi+I;
so it follows that
f(A) = f(P')f(K')-l,f(B)
= f(P'+P)f(L) -I,
f(C) = f(P)f(K) -I
Putting
together, we get the
these equalities
desired result.
217
and
I0 3.2
Definition.
contain
Let
sufficient
there exist
~ c ~'
projectives
P e ~
and
contains true for any
relative
projectives
sufficient ~"
_@
Let 3
consisting
any two of
A,B,C
of
3.3
projectives of
relative
Since
an Euler map of provided
consequence
to
12:
Res(~,.~). to
~',
P
then Note
0 that
if
then the same
£Y denotes
d( )
if
is
a subsemigroup
denotes
G*-dimension.
(O,A,B,C,O)
n
Resn(O~/~)
is exact
is also;
(G, 9-)
~i
contains
and there
group
G,
into
G.
and
let
= Resi(~'Y)'i
(a) and
of the dimension
f
theorem.
an Euler map extends
it suffices
m n,
extends
This will
(b) of 3.1.
sufficient
exists
then
= UimnReSi(0,3),
-~+i~ = Resi+l(O,_~).
we verify
this
g i}.
Res(O,~
Res(£Y,_~)
that an Euler map of
A,B e J',
(O,A,B,M,O).
then the third
into an abelian Res
is exact and
are projective,
and
[ d(M)
to
to an Euler map of
with
M e J'
and let
If for some
Resn(~,j~-)
Proof.
~
= {M ¢ Res(~/o-)
Theorem.
of
such that
M 3;
if for any
E c J " c ~'
are in
Res(O~/~-) = R e s ( O ~ Resi(O~
~"
~ will be said to
(O,K,P,M,O)
relative
of projectives,
be a subset
0
~. ~
to
relative
We retain here the notation of
of
projectives
such that
to
(O,A,B,M,O)
if the elements
sufficient
of
such that
sequence
to be projective
For example, contains
relative
K e ~
if for any given exact can be chosen
be subsets
That Since
uniquely
to show
uniquely
follow
to
from 3.1
(a) is satisfied ~
f
contains
is a
sufficient
I
projectives P e ~ possibly
and
relative
to
Res(0,3),
K ~ _~ which
for the condition
satisfy that
for any
M e -~+i'
the requirements K c ~,
218
of
there exists
(b), except
and this follows
again
Ii from the dimension
theorem.
3.4
Fix a commutative
Application.
be the semigroup sum,
let
Gp
projective
of isomorphism
be the subset
modules,
torsion modules. there set
exists
2
element
sequences
the isomorphism
is called
r e R
class
if and only if there
Mn ÷ .
.÷ M .1
R.
under
to the finitely
Let direct
generated
corresponding a torsion
such that
to be the usual
is exact
Lemma.
M
identity
of R-modules
be the subset
(An R-module
5~7 of exact denotes
classes
corresponding
and let
a regular
ring with
exact
to the
R-module
rM = 0.)
Take the
sequences,
of the R-module
exist homomorphisms
M,
if
i.e.
if
(Mn ..... MI)
÷
such that
is . exact
ReSl(~p,2 )
contains
sufficient
projectives
relative
to
Res ( ~ , 2 ) .
Proof.
Suppose
M ~ Res(~p,2).
~-~
such that
regular
element
O÷K÷P÷M+O of
R
Then
there exist
is exact.
such that
Therefore
rM = O,
P ~p if
and
r
is any
then
O÷K/rP÷P/rP÷M÷O is also
exact,
an exact ~
2,
such that to
sequence
(P/rP)~e
rM = O.
R e S l ( ~ p , 2 ).
O÷A÷B+M÷O,
there exists
O÷A÷B÷M÷O.
Consider
and
a regular
We shall
Thus,
the following
with
r e R
there
such that
0
is projective
is given
a homomorphism
P
/rp B
rB =
Since
P/rP
diagram:
0 ÷A+
now we are given
A,B ~ R e s ( O p , 2 ] .
show that
suppose
Suppose
÷M÷
219
0
and hence relative P/rP ÷ M.
12 The existence from
P
being
existence exact
of a
of a
¢'
making
projective, ¢'
sequence~
above
¢
with exact
and then
making
rows
and
N
[Mi, p.20], relative
MacRae element
of
0
0
K'
K
÷
L'
+ L'~(P/rP)
0
÷
A
÷
B
By 3.3,
has
ReSl(Gp,~ )
OF
part of MacRae's
ideals
of
M
+
required
that
uniquely
paper
then by the 9-1emma
the map which
P/rP
is projective
then
invertible
[M].
A similar
ideal
assigns
of
fractional
~
ideals
Res I ( ~ F , 3 )
of
R.
Res(Op,~).
corresponding
fractional
to each
is an Euler map of
to an Euler map of
the subset
to the
is mapped
ideals,
into
and hence
into this group.
of the Euler map His main
the image under R.
commutative
0
to prove
its first Fitting
The existence
is noetherian)
diagram
÷ 0
L' ~ P/rP + B,
shown
is also mapped
I)
Now given any
0
free modules,
the group of principal
Remarks.
of
denotes
generated
ReS(OF,y )
+
into the group of invertible
if
the
Q.E.D.
this map extends
Moreover, finitely
the kernel
[M, §2]
+ P/rP
0
0÷A÷B÷M÷0.
ReSl(Op,3 )
implies
and columns:
this is the element
to
¢
results
we can find via the
the following
0
denotes
commutative
commutative.
0~K+P/rP÷M~0,
which make
0 If
triangle
rP c kernel
the diagram
0÷K'+L'+A÷0
homomorphisms
the outer
f
theorem
actually
assertion
220
f
holds
of 3.4 is only a
asserts
consists
that
(when
R
of integral
for the Euler map of the
13 first example of §I; see [KCR,p.140, Theorem 192]. and independently, hypothesis
2)
Y. Quentel,
is unnecessary
David Rush and I,
have shown that the noetherian
for MacRae's
Note that if one interprets
theorem.
the axioms
~I
-~7
for exact
sequences of isomorphism classes of R-modules by inserting arrows +
instead of
+
, then the axioms remain valid.
References [B]
H. Bass, Algebraic K-Theory, W.A. Benjamin,
Inc., New York,
1968. [KCR] [K]
I. Kaplansky, I. Kaplansky,
R. MacRae,
1966.
On an application of the Fitting invariants,
Algebra 2(1965), [Mi] B. Mitchell,
1970.
Commutative Rings, Lecture Notes, Queen Mary
College, London, [M]
Commutative Rings, Allyn & Bacon, Boston,
J. of
153-169.
Theory of Categories, Academic Press, New York, 1965
[SE] R. Swan and E.G. Evans, K-Theory of Finite Groups and Orders, Lecture Notes in Math.
149, Springer-Verlag,
221
Berlin,
1970.
CHAIN CONJECTURES AND H-DOMAINS Louis J. Ratliff~ Jr. I University of California Riverside, California 92502 Some new equivalences to the chain conjecture (the integral closure of a local domain is catenary) and to the catenary chain conjecture (the integral closure of a catenary local domain is catenary) are proved, as are some new characterizations of a local H-domain. Also, a fact which lends support to the chain conjecture is noted, and it is proved that the H-conjecture (a local H-domain is catenary) implies the catenary chain conjecture.
I.
INTRODUCTION AND TERMINOLOGY
To start with, a number of definitions are needed. be given only for an integral domain
These will
A , since our main interest in
this paper is w i t h integral domains. (i.i)
A chain of prime ideals
maximal chain of prime ideals in maximal ideal i).
ideal in P
in
A
A , and, for each such that
(f.c.c.)
A
satisfies
A
is catenary
tion fog prime ideals) A, (A/P)Q/p (1.4)
A
in
Po = (0), Pk
i = l,...,k
A is a
(that is, height
Pi/Pi_l =
k .
A
has
A .
(or, satisfies
the saturated chain condi-
in case, for each pair of prime ideals
satisfies the satisfies
is a
, there is no prime
in case each maximal chain of prime ideals in
(1.3)
(s.c.c.)
... ~ Pk
the first chain condition for prime ideals
length equal to the altitude of
in
in case
Pi-I ~ P c Pi
The length of the chain is (1.2)
A
Po c PI c
P ~ Q
f.c.c.
the second chain condition for prime ideals
in case each integral domain
B
w h i c h is integral over
A
satisfies the f.c.c. (1.5)
A
satisfies
the chain condition for prime ideals
(c.c.)
iResearch on this paper was supported in part by the National Science Foundation, Grant 28939.
222
2
in case,
for each pair of prime ideals
P c Q
in
A, (A/P)Q/p
satis-
fies the s.c.c. There are a number of relations between these last four definitions.
Many of these relations can be found in [5, Remarks 2.5-2.7]
and [7, Remarks 2.22-2.25],
and these lists are sufficient for the
needs of this paper. With the above definitions, we can now state the chain conjecture.
(1.6) CHAIN CONJECTURE: therian)
domain satisfies
This conjecture
The integral closure of a local (Noe-
the c.c.
is equivalent to:
local domain is catenary.
The integral closure of a
Some further equivalent formulations of
the conjecture were given by M. Nagata in 1956 in [3, Problems 3, 3', and 3", p. 62] (see (2.4) below). The history of the chain conjecture can be traced back at least to 1937.
It was then that W. Krull in [2, p. 755] asked if the
following condition ture, when If and
A
A c B B
is the integral closure of a Noetherian domain) holds:
are integral domains such that
is integral over
such that
A , and if
A
P c Q
is integrally closed are prime ideals in
height Q/P = 1 , then is it necessarily
(QnA)/(PnA) is "no."
(which is (formally) weaker than the chain conjec-
= 1 ?
I. Kaplansky
In the example given, however,
A
wasn't the integral cloJust prior to this,
[9] gave an example of an integrally closed quasi-local
domain which isn't catenary. domain.
height
in 1972 in [i] showed that the answer
sure of a Noetherian domain (nor a Krull domain). J. Sally
true that
B
Again,
this example wasn't a Krull
Of course, each of these examples shows that the chain con-
jecture doesn't hold for arbitrary
integrally closed domains.
Very little progress has been made on proving the chain conjecture.
In Section 2 of this paper,
(or, disproving) some new equiva-
lences to this conjecture are given in (2.4) and, using one of these,
223
3 some empirical evidence w h i c h supports in (2.7) and (2.8)
the chain conjecture
(see the comment prior to 2.8).
some characterizations in (3.1) and (3.2).
of a local H-domain
is given
In Section 3,
(Definition 2.1) are given
In Section 4, (3.2) is used to show that, of two
other conjectures w h i c h have appeared
in the literature
ture (4.1) and the catenary chain conjecture implies the catenary chain conjecture. catenary chain conjecture
(4.2)),
(the H-conjec-
the H-conjecture
A new formulation of the
is given in (4.7), and this paper is closed
with some remarks on this formulation. The undefined
terminology
in this paper is the same as that in
[4]. 2.
THE CHAIN CONJECTURE AND H-DOMAINS
To give some new equivalences
to the chain conjecture,
the fol-
lowing definition is needed. (2.1) case,
DEFINITION.
An integral domain
for each height one prime ideal
p
in
A
is an H-domain in A , depth p = altitude
A - i The condition defining an H-domain originated catenary local domains.
That is, the statement
an H-domain'
is a kind of "dual" to statement
known result
[8, Theorem 2.2 and Remark 2.6(i)]:
(2.2) domain
in the study of
'A local domain is
(3) in the following
The following statements are equivalent for a local
R :
(I)
R
is catenary.
(2)
For each prime ideal
P
in
R , height P + depth P =
altitude R . (3)
For each depth one prime ideal
P
in
R ,
height P =
altitude R - I . One reason for the name "H-domain" is that the condition defining the ring has to do w i t h height one prime ideals.
Thus, perhaps a better
name for such a ring would be an
Then an Q - d o m a i n
Hi-domain.
224
(for
4 n ~ I) could be defined analogously. the name H-domain an H-domain (2.3) notation ideal,
is it is hoped
However,
the main reason for
that every Henselian
local domain is
(see (2.4) below). REMARK.
For the remainder of this article,
is fixed:
R
denotes a local domain,
a = altitude R, F
the integral closure of In the following
is the quotient R
in
M
the following
is its maximal
field of
R , and
R'
is
F .
theorem,
the equivalence
of (3), (4),
(3'), and
(4') was stated in [3, p. 62]. (2.4)
THEOREM.
The following
statements
are equivalent:
(1)
The chain conjecture
(2)
Every local domain
R
such that
(3)
Every local domain
R
a__ssi__nn(2) is catenary.
(4)
Every local domain
R
as in (2) satisfies
holds. R'
is quasi-local
is an
H- d oma in.
(2') Every Henselian
the s.c.c.
local domain is an H-domain.
(3') Every Henselian local domain is catenary. (4') Every Henselian Proof.
local domain satisfies
(i) = (4), by [7, Remark 2.23(iv)],
(2), and (2) = (2'), by [4, (43.12)]. local domain,
if
by induction on Theorem 2.21]
R
is and
a ~ 1
p
that
a maximal
(2') = (3').
and (43.20)]). ideal in
R'
Finally, and
ideal in
b
Also,
R'
the c.c.
local domain
(4) implies
is a Henselian R , it follows
it is k n o w n
, then
R' M,
Before proceeding,
M'
if
which
the following
are needed.
225
of a
M'
is
is not in
is the integral closure
the s.c.c., by (4), and so
[7, Remark 2.23(iv)],
[6,
[4, (43.10),
(i), since,
is an element in
R[b]M,nR[b ] , hence satisfies
satisfies
R/p
(4) = (3) =
that (3') = (4') ~ (4) (since the Henselization
any other maximal of
clearly
is a prime ideal in
local domain as in (2) is a Henselian (43.11),
Since
the s.c.c.
R'
q.e.d.
results on quasi-local
H-domains
(2.5)
REMARK.
Let
(S,N)
integral over a local domain
be a quasi-local
R , and let
the following statements are equivalent
domain which is
altitude S = a > i .
Then
[7, Lemma 4.6, Proposition
4.7, and Corollary 4.10]: (i)
R
is an H-domain.
(2)
S
is an H-domain.
(3)
For all analytically
height NS[c/b] prime ideal in Moreover,
= a-I
(It is known that
Sic/b] if
S
For each
height NS[x]
b,c
NS[c/b]
in
S ,
is a depth one
[5, Lemma 4.3].) is integrally closed in its quotient field
then each of these statements (4)
independent elements
x E K
K ,
is equivalent to:
such that neither
x
nor
i/x
is in
S ,
= a-i
By the equivalence of (I) and (2) in (2.4), a new approach in investigating
the chain conjecture
is available.
Namely,
attempt to
prove : (2.6)
If, for some element
height MR[x] < a-I
, then
R'
If (2.6) can be proved, necessarily
R
x
in
F, MR[x]
is proper and
isn't quasi-local. then whenever
is an H-domain,
R'
is quasi-local,
by (i) and (3) for
R
in (2.5),
hence the chain conjecture holds, by (I) and (2) in (2.4). The following lemma is related explained following (2.7)
LEMMA.
one prime ideal in c 6 p .
Assume R
R
isn't an H-domain,
such that
independent
i_~n R
Proof.
(M,x)R[x]
let
p
be a height
d = depth p < a-i , and let
as are
y = I/x = c/b 2 , height MR[x]
d+l < height
to (2.5), as will be
its proof.
Then there exists an element
analytically and
(in spirit)
O #
b E M
such that
b,c
b2,c
and, with
x = b2/c
= height MR[y] < height
are
(M,y)R[y] =
= a .
By the proof of [7, Proposition 4.7 (last paragraph)I,
there exists an element
b E M
such that:
226
(i) cR:bR = cR:b2R
is
the
p-primary component of
pendent
in
R , as are
P = (M,y)R[y]
cR ; and,
b2,c
is a maximal
.
Let
(ii) b,c are analytically
x = b2/c
ideal in
R[y]
and
y = c/b 2
and
M
= MR[y]
and height M is a maximal Clearly
a~e depth one prime ideals
= height MR[x,y] ideal in
R[x]
= height M
and
Then, w i t h since in
A = R[x]/xR[x]
that
x .
the altitude
R'[x])
= (by
Q
Hence,
inequality
.
I ~ Rad bR . Also,
is the only maximal
ideal
altitude A = altitude
since
height Q ~ a
[5, Remark 2.11(i)]),
(since
R
it follows
height Q = a .
one prime ideal in c ~ q , then
q # p , and so Therefore
R
such that
I ~ Rad bR .
I ~ IRqNR = (b2R'NR)RqNR ~ q . b
is in the
x E qRq , hence of
Rq
i/x = c/b 2 Thus
I ~ ((b2(Rq)':C(Rq)')NR ~ qRqNR = q .
b2R'NR ~ q
If
c E q , then cR
(by (i)).
Therefore
I ~ Rad bR,
q.e.d.
there exist elements
(a) Height MA < a-I ; and, MA c N
[7, Remark 4.4(iv)] N
height
(b2(Rq)':C(Rq)')nRq ~ qRq , and this
(2.7) shows that for every local domain
such that
be a
is not in the integral
implies
domain,
q
q-primary component of
(Rq)'
as desired,
Let
b ~ q , so clearly
closure
such
is the
= (b2R':cR')AR = (b2R'nR):cR
Therefore,
Thus it remains to show that
If
onto
K
, altitude A ~ altitude R/bR = a-I .
NIx] Q/XR[x] Q = height Q - i . satisfies
=
Q = (M,x)R[x]
(where
R'[X]
is a local ring,
w h i c h contains
M
[5, Lemma 4.3]
, and assume it is known that
A = R/(xR[x]NR)
R[x]
from
(D2R':cR',X)R'[X]NR'nR
I = (b2R'NR):cR
(see
height Q = a , as will now be shown.
kernel of the natural homomo~phism
Let
Also,
Further,
xR[x]NR ~ xR'[x]NR'NR = (K,X)R'[X]NR'AR
[4, (11.13)])
Then
and height P = d+l
the proof of [7, Proposition 4.7 (third paragraph)]). MR[x]
inde-
as in (b).
and
x E F
R
w h i c h isn't an H-
such that, with
(b) There exists a maximal
height N > height MA + I
that:
A = R[x]: ideal
Also,
N
in
A
it is known
(c) There exist at most a finite number of
(2.8) below shows that if
227
R
isn't an H-domain
and
R'
satisfies
the c.c.
x E F, (a) happens because such that
(or is catenary),
there exists a maximal
M'R'M,nA = N .
ideal
Of course,
M'
in
R'
in summary,
happen because of conditions
empirical
R'
R'
is a maximal
and
ideals in
R'
There-
then (a)
Of course,
(c)
this in no way
R
isn't an H-domain,
ideal in
M' E P
let
x E F
(M,x)A , and
R']
such that
R' M,
is an H-domain,
isn't quasi-local. Assume
there exists
height M' > h .
N = M'R'M,OA h
in
Assume
If there exists
(2)
>
x E R' M,
A = R[x], h = height MA + i < height
P = [M' ; M' (I)
N
as in (b),
(2.6), but (2.7) and (2.8) together lend
PROPOSITION.
such that, w i t h
and
R'
support to the chain conjecture.
(2.8)
then
in
(c) is accounted for by the fact that
if the chain conjecture holds,
indicates how to prove
M"
N
such that
there are always only finitely many maximal
let
ideal
height M " = height M A + i , and, for each
there exists a maximal
fore,
then, for all such
Then
is a maximal
M' E ~
x
such that
o__Er I/x
ideal in
is in
A
R' M,
R' M,
such that
is an H-domain
(say, x) MA ~ N
and
and
height
.
(3) Then,
Assume every
for each maximal
N > h , there exists Moreover,
M' E P ideal M' E P
is such that N
in
A
such that
there exists a maximal
ideal
R' M,
is an H-domain.
such that
MAc
N
x E R' M,
and
N = M'R'M,NA
M"
in
R'
and height
such that
i
height
M" = h . Before proving
(2.8),
it should be noted that, since
an H-domain s there exist such is a quotient ring of
R[x]
x E F and of
(2.7). R[i/x]
Further,
N ~ (M,x)A
R[i/x]
such that
Proof.
and the maximal MR[l/x] c N*
ideals and
(I) follows from (2.5).
228
since
isn't R[x,i/x]
, it is clear that there
exists a one-to-one correspondence between the ideals that
R
N
in (3) such
N* ~ (M,i/x)R[I/x]
in
height N* > h . For (2), if neither
x
nor
i/x
is in
R' M,
, then
Corollary,
p.
thesis
(2.5).
and
diction
R'[x] , and For
and
M'
MA
A
height
height
M'B
height
M'
maximal
ideal
PnA = MA height
N > height
ideal
M"
height
, and
height
is in
integral A
R' M,
,
dependence
, NnR = M
or
(so
MA
that
MAc
let
B
~ BN,
which
.
lies over
and
by
height
N'
i/x
But
is in
R'M,.
i/x 6 N'B N,
Finally,
depth
so there
is a
P = M"R'[x]
Hence,
, for,
,
R' ~ B ~ R' M,
= N .
ideal,
with
or
so
, so
, and so
, and so
, and
such
R' M,
height
x
N
M' = N'NR' is in
in
Hence
MAc
B = R'[x]
I/x
ideal
one p r i m e
P = height
in
N , and
M'R'M,nA
such that,
such
ideal
x E R' M,
Therefore
A
inequality)
~ R'M,
Hence
R'
R ' [ x ] / M '~=
I/x
M'B ~ height MA
MA + 1 .
is a depth in
by in
the contra-
, since
or
in
x
one prime
R'[i/x]
= N' MA
, depth P = I,
(2.5),
height
M" =
MA + 1 = h , q.e.d.
3. Because interest
SOME CHARACTERIZATIONS OF H - D O M A I N S
of the e q u i v a l e n c e
to study
indicated
The fixed
H-domains.
in (4.1)
characterizations
this
ideal
Then
- 1 = (2.5)
, then
x
[i0,
~ h , by hypo-
implies
= MA
Then,
M' ~ (by the a l t i t u d e
M'R'M,~B implies
.
N' = h e i g h t R'
is a c o n t r a d i c t i o n .
MA > 0
(M'*NA
be a m a x i m a l
is a depth
N' = h e i g h t
implies
are
in
M'R'M,[X]
ideal
M' > h .
N'
height
one prime
dependence,
Therefore
be a m a x i m a l
ideal
i/x 6 M'R' M,
which
N
and
MA + 1 ~ h e i g h t
If
N ~ height
is a m a x i m a l
, so
integral
is a m a x i m a l
N > h , let
then
M'* = height
N = M'R'M,~A
, N
let
N'NA = N
if not,
by
is a d e p t h
M,* ~ height MA
Let
(3),
height
A/(M'*NA))
over
height
that
and This,
over
x 6 R' M,
of N)
20],
h ~ height
is integral say
M ' * = M'R'[x]
and
of (i) and Some
(4.5)
of an H - d o m a i n
notation
of
further
below.
(2)
in (2.4),
reasons
For
these
to study
H-domains
reasons,
some
will
be g i v e n
in this
(2.3) w i l l
continue
to be used
section.
229
it is of some
section. throughout
The m a i n it lends R
reason
support
is c a t e n a r y ,
(3.1)
for
the f o l l o w i n g
to the H - c o n j e c t u r e then
R
THEOREM.
a[xl~[x]
including
Then
is c a t e n a r y
Let
R
(4.1),
X
theorem
since
it is k n o w n
[7, T h e o r e m
that
if
4.111.
be an i n d e t e r m i n a t e ,
is an H - d o m a i n
is b e c a u s e
if and only
and let if
R
R
=
is an
H- doma i n . Proof.
Assume
height
one p r i m e
p -- 1
and
first
ideal
x = X modulo
S = R[r2x]
NS[x]
r E M
such
If
(2.5).
is the q u o t i e n t Let
are a n a l y t i c a l l y NS[x]
depth
that
p * = altitude If
rx
= a-i
over
is i n t e g r a l
over
R
is i n t e g r a l
over
R
, so
elements Thus
, then
Thus
R[x]
in
depth p
a > 1
is a non-
and w h o s e S
so
is an HAlso,
r 2 )r 2 X
[6, L e m m a = a-i
and
, hence
N = a .
ideal,
S
be a height
there
Hence
height
p
, then
R .
is a d e p t h one p r i m e
(2.5)°
let
R* /pR * = a l t i t u d e
p nR = (0)
field of
N = (MR*/p*)NS
independent
and
p = p nR # (0)
domain which
= MR[X]/(p"NR[X])
height
R
is an H - d o m a i n
is a l g e b r a i c
is a local
field
R
, as desired.
(p*nR[X])
zero e l e m e n t
domain
in
p * = p R * , hence
R/p = d e p t h p = a-i
quotient
that
4.3],
, and so
hence
R
is an
H- domain. Conversely, prime so
ideal
in
let R °
R
be an H - d o m a i n
Then
pR
is a h e i g h t
d e p t h p = (as in the first p a r t
hence
R The
is an H-domain, following
(3.2) terminate. (i)
Then R
is a h e i g h t (2)
R'
(3)
P
proof.
p
be a h e i g h t
one p r i m e
ideal
in
depth pR*
one R
in S e c t i o n
= a-I
4.
Le____tt P = R [ X ] ( M , X ) R [ X ] , w h e r e
is an H - d o m a i n ideal
and in
statements R (I) c_ R' R]
X
is an inde-
are e q u i v a l e n t : , where
R (I) = N[Rp
; p
R (I) c R'
if
.
is an H-domain. is an H-domain. It is k n o w n
,
q.e.d.
the f o l l o w i n g
one p r i m e
let
of this proof)
t h e o r e m w i l l be used
THEOREM.
and
[6, C o r o l l a r y
230
5.7(1)]
that
,
I0 and
only
p'NR
if,
= i
for each height Therefore,
dependence),
(i)
Assume If p
(2) holds
= depth p+l
x = X modulo
L = a .
domain
the q u o t i e n t (3.1)).
Let
S'
with
and
c
b
as in the p r o o f integral cS':bS' since ideal
altitude
R[x]
= S[x]
R[x]
p'NR
that = i
is k n o w n mal
R'
and
Now
[4,
Hence,
R
is a h e i g h t
of
S'
(M,x)R[x]
that
ideal
a height
x E p'
, altitude
such
that
one m a x i m a l
l-x
R[X](M,x)R[x]
isn't an H-domain.
Hence
is in all = i < a .
height
in
= a-i
,
(3) ~olds. as in the last
then
is a h e i g h t p
in
in
this
p ' N R = I , and
Then
R
such R
is an
Therefore,
implies so (3)
ideals
that
in
P
implies
(i),
q.e.d. (3.3)
COROLLARY.
Let
R
be an H-domain
23I
such
it
one maxi-
since
R'
to height
p'NR > i .
the other m a x i m a l But
(by
= depth
xR[x]
R'
Hence,
ideal
of
(11.13)],
, it suffices
ideal
d e p t h p = depth p' = depth p ' n R < a-I is a height
[4,
xS'[x]
follows
p'
one prime
and
x = c/b
Hence
height
that either
R
is the only m a x i m a l
R (I) ~ R'
Suppose
that
is an H - d o m a i n
L = a .
one prime
, where
over
(by
depth
R
is a local
Since
depth
P .
implies
(as in the p r o o f
S .
hence
in depth
to show
is integral
is an H - d o m a i n
To p r o v e
(2)
R , there
the same
x , altitude
ideal
, hence
Theorem,
Therefore,
since
then that
= pP
it remains
implies
(10.14)]),
2.11]
p'
, height
(by integral
m S'/(cS':bS')
(2)
S' - I = a-I
exists
p
S
closure
S, S'[X]/ES'[x]
[7, P r o p o s i t i o n or there
R'
one prime
over
that are
the integral
(3.1). p'
L
5.7(1)].
H-domain, with
if
such
[6, C o r o l l a r y
ideal
that
of
in
L = R[X](M,x)R[x]
and
is a l g e b r a i c
containing
(3) holds,
of the p r o o f
x
S
and
and
then let
L = P/p*
of (2.7)).
=
If
prove
in
dependence
in
, so
of
be
p = i
by the L y i n g - O v e r
~ R[x] ~ L
fields
be a h e i g h t
p NR = ~ ) ,
Since
S = R[r2x]
p
height
(since,
(p*NR[X])
p'
depth p' = depth p ' N R
and let
If
ideal
(2).
, then
= a
is an H-domain).
altitude
since
implies
p = p AR # (0)
one prime
that
R (I) c R'
II and let
R n = R[XI,...,X n]
pendent
over
R .
PNR = M, Rnp Proof. (3.1)
Then,
AQ
The proof
that
remainder
Rip
(2) If
such that prime
H-domain,
over
n .
R1
R)
Hence,
(i) If
R
in
Rn
If
such that
and
Using
n = 1 , then by
P = (M,f)R 1 , for
is integral
since
over
A = R[f]
Q = PnA = (M,f)A
, so
QR 1 = P , it follows the case for
from
n = 1 , the
q.e.d.
is an H-domain,
one maximal
is an H-domain has height
p' c M'
ideal
in
a .
and
R (I) _c R'
For,
if
height p ' N R > 1 . R
then either R'
R (I) c R'
(as in the proof
is a maximal
there exists
depth p' < a-i
But this and
ideal
a height
Since
R
[7, Proposition
AND THE CATENARY
(3.2) will be used
two conjectures
The notation
, then every maximal in
R'
one
is an
2.11]
imply
isn't an H-domain.
ON THE H-CONJECTURE
a new formulation
M'
(by (i))
such that
In this section, between
P
inde-
(i) in (3.2)).
the contradiction
4.
Then
height M' ~ a , then
ideal
on
is straightforward,
a height
R
R'
ideal
MR 1 c P , hence
is an H-domain.
(3) implies
in
(3.2).
REMARK.
or there exists
ideal
that
f 6 P .
of the proof
(3.4)
that
is by induction
is transcendental
is an H-domain
(2.5)
are alsebraically
is an H-domain.
some monic polynomial f
XI,...,X n
for each prime
it may be assumed
(hence
, where
CHAIN CONJECTURE
to show an implication
w h i c h have appeared
in the literature.
of one of these conjectures of (2.3) will continue
Then,
will be given.
to be used
throu~hout
this
section. (4.1)
H-CONJECTURE:
The converse
(4.2) satisfies
If
R
is an H-domain,
of the conjecture
CATENARY
clearly
then
R
is catenary.
holds.
CHAIN CONJECTURE:
If
R
is catenary,
then
R'
is equivalent
If
R
is catenary,
then
R'
the c.c.
This conjecture
to:
232
12 is catenary.
Some other equivalences
are given
in the following
theorem. (4.3) ments
THEOREM.
If
R
is catenary,
then the following
state-
are equivalent:
(i)
R'
(2)
Every
satisfies
the c.c.
integral
domain w h i c h
is a finite R-algebra
is
catenary. (3)
For each height one prime
ideal
p
in
R, R/p
satisfies
the s.c.c. Proof. assumed
If
that
Assume R-algebra~
a = I , then the theorem
(i) holds, and let
let
N
it is known
that
D
is integral
height N = i
in
prime
ideal
Then
depth P' = i
P' c Q , then
D
R = R/p
R-algebra. prove
~
Hence
(2) holds, , and let To prove
satisfies
an integral
domain
, for some height is catenary, and
R
A
B = AN
.
To prove
, and let
is catenary, P'nB = P
Q
and
, hence
by
P
so
be a depth
D = C(A~N ) (i).
ideal
Let
Then
P'
be a
P' = height P . in
D
such that
height Q = height QnC = a
height P = height P' = a-i
A
For this,
3.1(1)], Let
height
is a maximal
that
is catenary.
height N = a > I .
D
if
in
is a finite
[7, Propo-
, and so
B
is
(2.2).
Assume let
and
and,
that
C = R'[A]
such that
QnR = M
sition 3.1(1)]. catenary
B
ideal
domain w h i c h
or a [7, Proposition
that
B , let
over
in
be an integral
to prove
it may be assumed ideal
A
be a maximal
it suffices
one prime
so it will be
a > i .
is catenary
clearly
is easy,
and
let ~
that
p
be a height
be an integral R
satisfies
the f.c.c. A
which
it follows
are catenary)
ideal
from
p
the s.c.c,
A .
233
ideal
R ,
is a finite
For this,
R-algebra in
in
it suffices
3.11].
[7, Proposition
that every maximal
ideal
domain which
[5, Theorem
is a finite
one prime
one prime
to
there is
such that
A/p ~ =
Then
(2) implies
3.1(1)]
(since
in
~
R, A,
has height a-l.
13 Hence
(3) holds. Assume
one maximal
(3) holds, and assume, ideals.
at first,
Then it is known that
[5, Proposition 3.3], and so
R'
R
satisfies
remains to consider the case w h e n
R'
that
R'
satisfies
the s.c.c.
is a catena~y
the s.c.c. Thus it
has a height one maximal
In this case, Lemma 4.4 below shows that, for some R[x,i/x]
has no height
x E R', L =
local domain whose integral closure
R'[i/x]
has no height one maximal
ideals and the only prime ideals in
w h i c h blow up in
are the height one maximal
clearly
R'
R'[i/x]
satisfies
the c.c.,
and, as already noted, for
L
p"nR[x] 3.1(1)],
So, let , and let so
R/p
dependence,
R'[I/x]
p"
R'[i/x]
satisfies
satisfies
.
Then
satisfies
the s.c.c.,
height p = 1
the s.c.c.
satisfies
, since, by the proof of (4.4),
ideals
N 1 = (M,x)
and
N 2 = (M,l-x)
Another equivalence
Thus,
the s.c.c.; if (3) holds L , let
p' =
[7, Proposition
(by (3)), hence, by integral
the s.c.c.
R[x]/p'
R'
ideals.
be a height one prime ideal in
p = p'NR
R[x]/p'
if
ideal.
Finally,
R[x] and
L/p" =
has only two maximal
p' c N2, ~ N 1 , q.e.d.
of the catenary chain conjecture will be
given in (4.7) below. (4.4)
LEMMA.
Assume
R
i__sscatenary and
b e an integral domain w h i c h is integral over maximal
ideal
an element
N
i__nn A
x E R' R[x,I/x]
(2)
Every maximal
(3)
The prime ideals in
Proof.
ideal in
R'
height N < a , then there exists
ideal in
A[x,I/x]
R'
has height a ; and,
w h i c h blow up in
R'[I/x]
are the
height N = 1
[7,
ideals.
If such
N Let
exist,
then necessarily
B = R'[A]
Q'nA = N , and let and
If there is a
is a catenary local domain;
Proposition 3.1(1)]. such that
R
A
such that:
(I)
height one maxim~l
B
such that
a > 1 , and let
height Q = 1
, let
Q'
Q = Q'nR' [4, (10.14)].
234
be a maximal Then
Q
So, let
ideal in
is a maximal x
be an
14
element w h i c h is in a prime ideal in a height one maximal maximal
ideals in
ideal, and let
R'
such that
it may be assumed that maximal
and
local domain.
Also,
holds), and Thus,
y
Then
B[i/x]
is a unit in
and
height N 2 = a , so
R'
is integral over
N 2 = (M,I-x)R 1 , and
L = R[x,i/x]
has no height one maximal
R'[i/x] x
Clearly
has exactly two
= RIN 2
is the integral closure of
it follows from the choice of
A[x,i/x]
be an element in all other
R 1 = R[x]
N 1 = (M,x)R 1
R'[I/x]
if and only if the ideal is
x+y = u
u = 1 .
ideals, namely,
height N 1 = 1
R'
and
ideals.
L
and over
[4, (10.14)] Therefore,
is a (so (3)
A[x,I/x] that if (i) holds,
then (2) follows from [7, Proposition 3.1(1)]. To prove that
L
is catenary,
a local domain w i t h maximal R[x]
Let
and let
P'
ideal
Then
Further,
C
in
RI)
, hence
Hence,
since
C
THEOREM.
is catenary L
is
R ~ C c R1 =
L , let
(since
C
J
P = P'nR 1 ,
is the conductor
height P' = height P =
depth P' = depth P (= height N2/P ) , and
P' = height p = a-i , and so (4.5)
J = NINN 2 , and
Lp, = Rip = Cp
of the integral closure of
P = depth p .
C = R + (NInN 2) , so
be a depth one prime ideal in
p = PnC .
height p .
let
depth
[7, Theorem 3.2], height
is catenary
(2.2), q.e.d.
If the H-conjecture holds,
then the catenary
chain conjecture holds. Proof. satisfies M'
in
Let
the c.c.
R', R' M,
this holds if every in
M'
R'
R
be a catena~y it suffices
satisfies
height
and
R
height
Let
p'
p'nR > 1 , then
[7, Proposition 2.11], tion, so
the s.c.c.
a > 1
R
isn't an
p'nR = 1 , hence
is an H-domain.
Therefore
by (3.2), hence, by hypothesis,
To prove that
to prove that, for each maximal [7, Remark 2.23(iv)].
height M' = 1 , so by (4.4)
has height If
local domain.
P
ideal Clearly
it may be assumed that
be a height one prime ideal depth p' < a-i , hence, by
H-domain. R (I) ~
R'
This is a contradic-
[6,
Corollary
P = R[X](M,X)R[X ] is catenary.
235
R'
5.7(1)]
is an H-domain,
But it is known
15 [6, Theorem
2.21]
that
the s.c.c.
Hence
R'
(4.6) grally
P
satisfies
COROLLARY.
the s.c.c.,
local domain
This follows
the s.c.c,
(if and)
if and only
from
holds,
satisfies
(4.5),
only
if
R
satisfies
q.e.d.
If the H-conjecture
closed catena~y
Proof.
is catenary
then every
the s.c.c.
since a local domain
if it satisfies
inte-
the c.c.
satisfies
[7, Remark
2.23(v)],
q.e.d. (4.7)
indicates
conjecture
another
approach
to proving
the catenary
chain
w h i c h might prove more accessible.
(4.7)
THEOREM.
The catenary
if the following
condition
is such that
A
has exactly
N 2 = (M,l-x)
and
chain conjecture
holds:
If
R
if and only
is catena~x and
two maximal
NInN 2 = M
holds
ideals
is the maximal
A = R[x]
N I = (M,x)
ideal
in
and
R , then
A
is catenary. Proof. dition,
so assume
altitude assumed
the catenary
that
R
that
a > i .
R
the s.c.c.,
it suffices
is a finite
implies
and the condition
satisfies
By (4.3),
domain w h i c h
chain conjecture
is catenary
R = a = i , then
an integral B
By (4.3)
integral
holds.
hence
to prove
the con-
it may be
that if
extension
If
of
B
is
R , then
is catenary. For this,
if
B
so it may be assumed height (4.4)
one maximal
Then Let
that
ideal
it may be assumed Let
l-x
is local,
N
and
R I = R + (NInN2) Therefore,
B(ANNI)
= BN
B
is catena~y
isn't local. B , then
BN
that every maximal ideal
in
the other maximal
N I = (M,x)A
is catenary,
in
be a maximal
is in all
3.2].
B
then
ideals
RI
is integral
A
to prove over
ideal
B .
in
B
x E N Let
B(ANN2)
by a .
such that A = R[x] ideals
in
A
.
[7, Theorem
so to prove is catenary
AN1 , h e n c e i s c a t e n a r y
236
is a
Hence,
local domain
is catena~y,
that
N
3.2],
has height
are the maximal
is a catenary
by hypothesis,
it suffices
in
if
is catenary.
B , and let
N 2 = (M,I-x)A , so
Clearly,
[7, Theorem
that (since
[7, Theorem
B
16 3.2]).
Since
AN2
and
B(ANN2)
satisfy
the conditions
B , it follows by finitely many repetitions
that
B
on
R
and
is catenary,
q.e.d. Call the type of extension tion of special extensions
could prove rewarding,
might give the needed results R
is.
In the following
properties
REMARK. and let
the following
statements
is catenary whenever
If
I
be a special extension of
A/I
A , then either
Rp = Ap
R/(InR)
(if (but
an integral domain).
in
.
Then
A/I = R/(InR)
is a special extension of
There is a one-to-one p # M
R .
y E A, ~ R, A = R[y]
is an ideal in , or
catenary)
hold:
(2)
not necessarily
R
correspondence
and the prime
Moreover,
ideals
for all prime
between the prime
P ~ [NI,N2]
ideals
Q
in
in
A
given
A, QAQ = qAQ ,
q = QNR . (4)
over
in particular,
be a (not necessarily
A = R[x]
For each element
where
A
and,
a proof of some of the more evident
(R,M)
(i)
I ~ NInN 2 = M)
by
remark,
Let
local domain,
ideals
to show that
An investiga-
of special extensions will be given.
(4.8)
(3)
ring in (4.7) special.
R (5)
It follows from (2) and (3) that
is locally unramified
(see [4, pp. 144-145]). If
p # M
In fact,
if
either:
(a) qA = q~
N I) ; or, (b) that
A
q
is a prime ideal in
is a p-primary
(c) qA = q*nN 2
qANI = NIANI
similar comment holds Proof.
(if
q
ideal in c NInN2)
(if
, hence
qi
; (b) qA = q &N I
is semi-prime. q
= qRpnA
(if
q
It follows
isn't p-primary,
for
,
c N2,
in case
i ~ 2
A
in case (c).
(i) follows from
•
q
pA
R , then, with
q * c NI, ~ N2)
A = R + xR
from (i) and an easy computation, For (5), if
R , then
M = R:A , (2) follows
and (3) follows
*
~ N I , then
and
*
*
from
M = R:A . *
q nN 2 = q nM = q NR = q ~ qA ~ q NR .
237
17
It follows that either q = qA = q
(if
q
q = qA = q NN 2
~ N1NN2)
(if
q
~ NI, ~_ N 2)
or
, q.e.d.
BIBLIOGRAPHY
[z]
I. Kaplansky, Adjacent prime ideals, J. Algebra 20(197~, 94-97.
[2]
W. Kmull, Beitr~ge zur Arithmetik kommutativer Integrit~tsbereiche. III Zum Dimensionsbegriff der Idealtheorie, Math. Zeit. 42(1937), 745-766.
[3]
M. Nagata, On the chain problem of prime ideals, Nagoya Math. ~. 10(1956), 51-64.
[4]
M. Nagata, Local Rings, Interscience Tracts 13, Interscience, New York, 1962.
[5]
L. J. Ratliff, Jr., On quasi-unmixed local domains, the altitude formula, and the chain condition for prime ideals, (I), Amer. J. Math. 91(1969), 508-528.
[63
L. J. Ratliff, Jr., On quasi-unmixed local domains, the altitude formula, and the chain condition for prime ideals,
(II),
Amer. J. Math. 92(1970), 99-144o
[7]
L. J. Ratliff, Jr., Characterizations of catenary rings, Amer. J. Math. 93(1971), 1070-1108.
[8]
L. J. Ratliff, Jr., Catenary rings and the altitude formula, .Amer. J. Math., (forthcoming).
[9]
J. Sally, Failure of the saturated chain condition in an integrally closed domain, Abstract 70T-A72, Notices Amer. Math. Soc. 17(1970), 560.
[103
O. Zariski and P. Samuel, Commutative Algebra, Vol. II, Van Nostrand, New York, 1960
238
ON THE NT~]3~ OF GENERATORS OF IDEALS OF DIF~NSION ZERO Judith Sally Rutgers University
ABSTRACT. This note contains generalizations of two results of Abhyankar [!] which give a bound for the embedding dimension of certain local rings in terms of the mu!tiolicity and the dimension of the ring.
Let R be a Noetherian ring. mension of R/I is O.
An ideal I has dimension 0 if the (Krull) di-
The multiplicity of an ideal I of dimension 0 is denoted /~(I);
the length of an R-module A is denoted A(A).
If (R, M) is a local ring and I is any
ideal, v(I) denotes the number of elements in a minimal basis of I, i.e., v(I) is the 4imension of I/IM as a vector space over R/M.
When I = M, /~(M) = / ~ R )
and v(M)
is the embedding dimension of R. Theorem i. ideal.
Let (R, M) be a d-dimensional local Macaulay ring.
Then, v(i) .< d + ~ ( I )
Proof. = ~(R). d ,0.
Let I be an M-primary
We may assume I / (O).
-
~(R/I).
The proof is by induction on d.
Since ~(R)~/ X(R/I) + k(I/IM), we have that
If d = O, then/%(I)
v(I).< /~(I) - ~(R/I).
By passing to R(X), .~q(X) and IR(X) as in Nagata [3, p.18, p.71] , we may
assume that R/M is an infinite field.
By Theorem 22.3 in Nagata [3], there is a non-
zero divisor x in I but not in IM such that x is superficial for I. rin~ R/xR we have, hy induction,
v(I) = v(I/xP) + 1.
In the Macaulay
v(I/xR) ~< d - 1 +/~I/xR) - A((R/xR)/(I/x_R)).
Since x is superficial and a non-zero divisor,/~(I/xR) =/~(I).
Remark.
Assume
Since x is not in IM
Thus, v(I).~ d +/~(I) - ~(R/I).
Abhyankar's proof in [i] for the case I = M can be modified to give a proof
of Theorem I.
However, since his proof uses a fact about systems of parameters in a
239
2 local ~ c a u l a y ring which can be derived from Theorem i, a different proof is given here.
The bound obtained in Theorem i is best possible. Dles of local Macaulay rings (R, M) with v(M) = d + ~ ( M )
The following are exam-
- i, i.e., emdim(R) = d +
R(R) - I, and of an M-primary ideal I of R with v(1) = d +/K(1) - k(R/I). a field.
L e t / ~ b e any integer ~ I. 0 For d = O, take R a = K[[Xl, .... x _l]]/(x I . . . .
x~_ 1
Let K be
,X~_l)2 , where Xl, . . . .
are indeterminates, and take I to be any ideal. For d = i, take R aI = K[[x~,x~+l,
,x2~-l]], where x is an indeterminate,
snd take I = (x~,xa+l, . . ,x2~-2). For d • l ,
d 1 take R~ = R~[[tl, ... ,td_l]], where t I ....
,td_ 1
are indeter-
minates, and take I = (x~,x~÷I, ... ,x2a-2,tl , ... ,td_l).
In order to give a global version of Theorem I we need the following definition.
An ideal I of a ring R is rank unmixed if all primes minimal over I have
the same rank (i.e., height). Corollary.
Let R be a d-dimensional Macaulay ring.
Let I be a non-zero ideal of
dimension 0 and rank r.
If I is rank unmixed, then I can be generated by max(d + I,
r + ~ ( I ) - I) elements.
If R is semi-local, max(l, r +/~(I) - i) elements are enough.
Proof.
We will show that I can be !ocallv generated by r +/~(I) - i elements.
desired conclusion will then follow from the Forster - Swan Theorem [h]. Pk he the primes minimal over I P / Pl' "'" 'Pk"
The
Let PI' "'''
Since i has dimension O, IRp = Rp for all primes
By Theorem I, v(IRp.) ~ r +~(IRp.) - i, for i, ... ,k. k l l
is rank unmixed,/~(I) = i~=l~(IRp.=.). 1
Thus we have that v(IRp. )~ l
i = I, ... ,k.
240
Since I
r +/~(I) - I, for
3 Abhyankar also proves in [i] that if (R, M) is a d-dimensional local ring such that R/M is algebraically closed and A, the associated graded ring of R, is a domain, then emdim(R) & d +~(R)
- 1.
Theorem 2 below extends this result to M-pri-
mary ideals Theorem 2. closed.
Let (R, M) be a d-dimensional local ring such that R/M is algebraically
Let I be an M-primary ideal.
If the graded ring G = R/M + I/IM + I2/I2M +...
is a domain, then v(I) 4
Proof.
Assume I ~ (0).
~In/InM). p.23~]).
d +21(I)
- 1.
Let H(G, n) denote the polynomial which, for large n, gives
H(G, n) is of degree t - i, where t = dim G
(cf. Zariski and Samuel [5,
Let a/(t - I)I be the leading coefficient of H(G, n).
Since R/M is alge-
braically closed, we have as in Abhyankar [I], that a + t - 1 ~dim[I/IM
(also cf. Abhyankar [2, (12.3.5)]). t = d and that2~(I) ~ a.
: R/M] = v(I),
To prove the theorem it is enough to show that
Let B - R/I + I/I 2 + ....
B has dimension d.
B/(M/I)B, so that (~'~I)B is a prime homogeneous ideal of B.
But, since ~
some positive integer j, we have that ( ~ I ) B is the nilradical of B. dim G = dim B/(M/I)B = dim B = d. n, gives A(In/In+l).
Now G = I for
Thus, t =
Let H(B, n) denote the polynomial which, for large
H(B, n) has degree d - 1 and leading coefficient~(I)/(d
Now H(B, n) ~ H(G, n) because
- 1)I
l(In/I n+l) = l(In/MI n) + ~(MIn/I n+l) and, since both
polynomials have the same degree, we have that~t(I) ~ a .
It is interesting to note that although the bound obtained in Theorem 2 is, in general, larger than the one in Theorem i, it is best possible.
Let (R, M) be the
local, non-Macaulay domain of Nagata's Example 2 [3, 0.203-205] for the case m = O, r = 1.
(We may take K algebraically closed.)
241
Let xR' and N be the maximal ideals of
R', the integral closure of R.
Then M = xR'~ N = xN.
isfy the hypotheses of Theorem 2. v(M) =
d +~(M)
- 1 and v(I) =
Let I = x2N.
R, M and I sat-
v(M) = v(I) = 2; d = 2;/~(M) =/~(I) = I. d +~(I)
Hence,
- i.
REFERENCES 1
Abhyankar, S. S., "Local rings of high embedding dimension", Amer. J. Math 89
(1967), lO73-1077. 2
Abhyankar, S. S., Resolutions of Singularities of Embedded Al~ebraic Surfaces, Academic Press, New York, 1966.
3
Nagata, M., Local Rings, Interscience, New York, 1962.
h
Swan, N. G., "The number of generators of a module", Math. Zeit. 102 (1967), 318-322.
5
Zariski, O. and Samuel, P., Commutative Algebra, vol. II, D. Van Nostrand, New York, 1960
242
RINGS OF GLOBAL DIMENSION
TWO
by Wolmer V. Vasconcelos Department
of Mathematics,
New Brunswick,
Rutgers University
New Jersey
08903
This paper gives a change of rings theorem for homological dimensions which seems especially suited for the study of the commutative rings of global dimension two. Elsewhere we gave a description of the local rings of global dimension two that when applied to the global case yields an idea of what the spectrum of such rings look like "down" from a maximal ideal. Here an attempt is made to describe the spectrum "up" from a minimal prime ideal.
I. CHANGE OF RINGS As a general tive.
Bourbaki
elementary
proviso,
[3]
IN DIMENSION
all rings considered
will be the basic
source
TWO here will be commuta-
for terminology
and
notions.
I.Kaplansky but effective
in [12] proved
the following
result which
is a simple
dimensions
of rings and
tool in the study of homologieal
modules. THEOREM x
i.i.
a central
Let
(Change o f rings element
B = A/(x)
tive dimension
in dimension
of A which
is neither
(proj dim
for short)
With A commutative
supported
over B. If the
of E over B is finite,
= 1 + proj
by National
where
Science
243
divisor.
projecthen
dim B(E).
this result was generalized
[7,10] to the case of B = A/I
Partially
Let A be any ring and
a unit nor a nonzero
and let E be a nonzero module
proj dim A(E)
Jensen
one)
by Cohen and
I is a faithfully
Foundation
projee-
Grant GP-19995.
2 rive ideal of A. Observe Vasconoelos proved
[14]
recently
that
finitely
generated.
that projectivity
local property
for the Zariski
remark
that,
implies
such ideals
change of rings In general,
to
On the other hand~
(and thus projective
topology
in dimension
with
are according
one~
of Spec(A).
Gruson
[8]
dimension)is
In particular
it is enough to consider
a
this
the
I = (x).
if E is a module over the ring
B = A/I one always
has (2)
proj dim A ( E ) ! proj dim B(E)
This follows
for instance
from the spectral
C may be any A-module Now we will consider
be a finitely not always
generated
(cf.
[5, XVI.5]).
the dimension
where x,y
any module.
in (2) holds.
local ring of dimension
form a regular
E is a finitely
generated
In fact~
A-sequence~ B-module
p.82])
and thus we cannot have always situation
from sitting skirted
dimension
morphism.
In fact~
that its grade
ideals
overdens~.
THEOREM
1.2.
summand
(Change o_ff rings
one and let B = A/I.
holds
in (2) if
but not for
n-l,
its fini-
to Gruson
(cf.
[8.
in (2). Thus we must
>HOmA(I~A)
reasons,
I
are
be an iso-
and if A is noetherian
this
For brevity we call such
two)
generated
Let A be a commutative ideal of projective
If E is a nonzero
244
It is
I = x(x~y)
more elementary
A
in dimension
finitely
one.
of A. Both conditions
map
is at least two.
let I be an overdense
dimension
equality
such ideal is faithful
implies
ring~
according
and also prevent,for
by asking that the canonical
and
of B is
Let I
for instance,
[i, Cot 2.12])
is also n-i
in a pmoper direct
Thus,
n > 1
as the Krull dimension
projective
of (i.i).
dimension
then equality
(cf.
tistic
avoid this
two analogue
ideal of A of projective
the case that equality
if A is a regular
sequence
Ext E,C)
E ,q = Ext CE,E tX B, where
+ proj dim A(B).
B-module
with finite
3 projective
dimension
over B, then
pro~ dim A(E) Examples
of such idealstend
sion two as we shall see. manner tries
in A. Let
proj
to abound
Other examples
: Let A be a domain
of ¢. If
= 2 + proj dim B(E).
can be found
and let ~ be an
I = D(¢) be the ideal generated
= 1
is particularly
and I is overdense
easy to check
z
in the following
of A, x I C A }
domains,
zy + 1
dimen-
by the n x n-minors
from Butch's
in G.C.D.
of global
n x (n+l) matrix with en-
I -I : {x e K : field of quotients
dim A(I)
PROOF.
in domains
= A, then
theorem
[4]. This
e.g. with A = k[x,y,z]
.
We begin with some remarks.
(i) I is of finite presentation: Let
0
be a projective
resolution
is also finitely each prime Actually
>K
module
we have only to consider
Kp
the above
sequence
in P and thus
Ip = Ap
the unity" with
(cf.
for which
Kf = Afn-I
multiplicative
, where
have that for some
i.e.
denotes
E ~ (0)
fi = f
same as proj dim B(E), projectivity.
Kf
system of powers
for any B = A/I-module
is free of rank
of
At the same time
primes;
let P it
n-1 Kp = Ap .
and thus
pro~
dim (E) = -~. By a there
there exists localization
is a "partition (fl,...,fm)
= A
f = fi" While this is happening,
and
proj dim
dimension,
proj dim Bf(Ef)
to the local
245
of
of K at the
of finite projective
Ef # (0)
according
n-l.
at P. As I is faithful,
[3, Chap. II, p.138])
K is free,
to show that for
to An-l:
First we pose that if E = (0) then of Bourbaki
and we claim it
this for the minimal
(2) K may be taken to be isomorphic
theorem
>0
For that it is enough
ideal P the localized
be contained
>I
of I. Thus K is projective
generated.
be one such and localize cannot
>A n
is the
characterization
(If) = 1
since If
will
of
Af-projec-
4 tive would make it equal to Af. Assume then that we have an exact sequence (***) (3)
0
>A n-I
¢ >A n
>I
>0.
I is generated by the (n-l) x (n-l)-minors of ¢: Let D = (dl,...,d n) be the ideal generated by the minors of ~.
Using the notation in Kaplansky's dja i. Define the following map note that
is well-defined
(Zr.d.)a. = 0 i i ]
[ll,p.148,Ex.8],
~ : I
>D
as Zr.a.l i = 0
we have d.a. = l]
~(Zria i) = Zrid i. First yields
d~(Zria i ) ]
and thus Zr.d. = 0 as I is faithful. i l
=
On the other hand,
since I is overdense @ is realized by multiplication by an element d in A
and
D = dI.
Since I is torsion-free
theorem that d is a unit in A (4)
If C is a nonzero Tensor O
(***)
with
>Tor~(I,C)
By (3) above
D : I
and hence
B-module C ~ 0
0
>cn-i
~C >C n
implies that each
>0.
(n-l)-minor of ¢C Tor~(B,C)
Ext~(B,C)
>C n
is annihi-
: ker¢ C ~ 0.
~ 0 :
t¢c > c n - l - - > E x t ~ ( I , C )
it readily follows that
Now we claim that A/I-module.
>I ~ C
,C) to (***) to get
>HOmA(I,C)
from which
~ 0
to get
For any nonzero B-module C Apply HOmA(
I = D.
Tor~(B,C)
lated by D and thus by McCoy's theorem (5)
it follows by McCoy's
T : Ext~(I,A)
Ext~(I,C)
>0
: Ext~(I,A) ~ C.
is a finitely generated faithful
But this follows from the exact sequence 0
>HOmA(I,A)
>A n
>A n-I
>Ext~(I,A)
>0
and the well-known companion exact sequence 0
1 1 >EXtA(EXtA(I,A),A)
>I
.->HOmA(I,A),A)
The conclusion now follows from Gruson's result such a module T, T ~ C = 0
implies
246
C = 0.
2 1 >EXtA(EXtA(I,A),A).
[8,p.58]
that for
We are now ready to prove the theorem. Let E be a nonzero module with The first
B-
proj dim B(E) = n. Notice that by (4) pro~ dim A(E) [ 2. case to examine is then
Proj dim B(E) : 1 : Let
0
>F
>F' - - > E
.-
be a projective resolution of E as a B-module. and remark
Set
C : Ext~(E,F)
C # 0. Consider the sequences 0
>HOmA(I,F)
0
>L
obtained by applying HomB(E,
>0
>F n
>F n-I
HomA(
>L
>0
>Extl(l,F)
>0
,F) to (***) and then splicing. Applying
) to both sequences and observing
proj dim B(E) = i, we get
the diagram Ext~(E,HOmA(l,F))
>Ext~(E,F n)
>Ext~(E,L)
....>0
E~t~(E,F n-l)
with would
exact mean
row
and
column.
Notice
that
= 0
Ext~(E,Ext~(I,F))
that t
[Ext~(E,F)]n
~Ext(E,F) >[Ext~(E,r)]n-i
>0
is exact. We know however that the cokernel of this map is Extl(I,A) Q Ext~(E,F)
which by (5) is distinct from 0.
Passing now to the spectral sequence
P
we see that Ext~(E,F) ~ 0
E P2'q: 0 for and
p > 2, g > 1
but
~ 0. Thus E 2,1 2
proj dim A(E) : 3.
Proj dim B(E) : n > 1 : Write exact with F
0
>L
> F
B-projective. Then
proj dim A(L) = n+l.
>E - - ~ proj dim A(F) : 2
and by induction
Thus, by dimension shifting, proj dim A(E) =
247
6 n+2,
as desired.
REMARK.
It is clear how the procedure
and injective
analogues
of the change
2. FINITELY Let A be, unless bal dimension
two.
to be the presence results
ideals
coherence
is missing
sitting
two.
IDEALS a commutative
ring of glo-
in the study of these rings
[ 16, Prop.
provide
a clear picture ideals.
first been discovered
at least of
Examples by Jensen
where
Portions is duplicated pro~ective
and J~ndrup; to
ideals,
if and only i f Min(A),
is compact
in coherent
flat").
of local rings of global
rings
of
(This is also equivalen ~ to "the
o f A i_ss absolutely
of the structure
the subspace
of finite
global
dimension
dimension
two
with
= free.
PROPOSITION finitely
the
3.4 ] :
2.1. A is coherent
prime
seems
as in the former case the
above overdense
have
total ring o f quotients
rives
in dimension
specified,
section
the flat
we will steer away from this case and limit ourselves
PROPOSITION minimal
of rings
or lack of coherence,
of the preceding
normally
here yields
GENERATED
The great divider
the prime
quoting
otherwise
followed
2.2.
Let A be a commutative
generated
are free
coherent
ideal has finite projective
( i__nnparticular
i f A is local)
ring such that every
dimension.
If projec-
then A is a G.C.D.
domain. PROOF
(Sketch).
a domain.
[15] it follows
Let us now apply the method
each finitely "common
From Vasconcelos
generated
divisor
ideal",
and J overdense. we claim that
With
proj dim
of MaRae
that such a ring is
[13] of associating
ideal I of finite projective i.e.
an ideal
I = (a,b),
(d)
J = (e,f)
(e,f) < I. Indeed,
248
dimension
such that and
its
I = dJ
a = de, b = df
consider
the sequence
to
7 0 with
¢(x,y)
and hence
= xe-yf.
>A 2
domain
>0
xe-yf = O, x / f e
in the
for which rank one projectives
if and only if the projective
by two elements
If projectives
generated
I = P.J, with P projective
are free.
dimension
Then
of each
is at most one.
are not assumed
has that every finitely
(e,f) -I = A
by (f,e).
the rest of the argument
ideal ~enerated REMARK,
->(e,f)
K is generated
Let A be a domain
A is a G.C.D.
~
If (x,y)c K, i.e.
x eAf. Thus
We isolate LEMMA.
>K
to be free in (2.2) one still
ideal I admits
a decomposition
and J overdense.
The main use of (1.2)
for coherent
rings
of global dimension
two is the following COROLLARY
2.3.
Let I be a finitely
then A/I is an artinian I are finitely PROOF.
ring.
I__nnparticular
overdense the prime
ideal of A; ideals
above
generated.
By (1.2) the finitistic
and thus by Bass'theorem coherent
generated
projective
dimension
[2] A/I is perfect.
being a coherent
A-module,
of A/I is zero
Since A/I is also
by Chase's
result
[6] A/I is
artinian. COROLLARY
2.4.(Noetherianization)
two with Spee(A)
noetherian
and
Let A be a ring o f global Ap noetherian
dimension
for each Drime
ideal P.
Then A is noetherian. First we consider PROPOSITION
2.5.
the subspace of domains; LEMMA.
Let A be a ring of global
of maximal
ideals,
in particular
I of A (i.e.
two with Max(A),
Then A is a finite product
A is coherent.
Let A be a commutative
pure ideal
noetherian.
dimension
A/I
ring with Max(A) is A-flat)
249
noetherian.
is generated
Then each
by one idempotent.
8
PROOF.
Recall that purity for the ideal I means
ideal P the localization Let I $ 0
be pure.
subset of Max(A) is minimal. with
As
= 1
yields
For each
and
a e I
V(a)
= V(b)
if
e = ra. Notice that V(e)
(a). Pick a such that V(a)
# 0 )
then
(a,l-b)
= A. Write
A(l-e).
that
If there is an element c
ideal of A(l-e)
is properly
ra +
we claim
I = (e) @ (l-e)I observe
in every maximal V(e+c)
we may write a = ab
e is an idempotent;
= V(a); with
not contained
(l-e)I
let V(a) be the closed
and hence
(l-e)I is a pure ideal of the ring (l-e)I
or (I).
(a) = aI (checked by localization)
(e) = I. Clearly
of
0
defined by the ideal
b e I. Thus
s(l-b)
Ip
that for each prime
contained
(c exists
in V(e), which is
impossible. The proof of (2.5) now follows of A is finite. A is a domain
Assume then A indecomposable;
(cf.[17])
PROOF of (2.4): Heinzer-Ohm's generated finite.
and thus the minimal
at each l o c a l i z a t i o n primes are pure ideals.
By the preceding we may assume A to be a domain.
theorem
I = P.J
(weakly)
associated
with P projective
primes of A/I is
and J overdense.
is enough to prove this locally in the Zariski topology, Thus yields
C
>(d)/dJ
Ass(A/I)c
while a prime minimal
over
>A/I
>A/(d)
A s s ( A / ( d ) ) u Ass(A/J).
ideal is associated (d)
By
[9] it is enough to show that for each finitely
ideal I the set of
Write
: Clearly the number of idempotents
assume
P = (d).
>0
By (2.3)
to A/(d)
Since it
Ass(A/J)
is finite
if and only if it is
since A is locally noetherian. REFERENCES
[i]
M.Auslander Annals Math.
[2]
H.Bass,
and D.Buchsbaum, 68 (1958),
Finitistic
semi-primary
Codimension
and Multiplicity,
625-657.
dimension
and a h o m o l o g i c a l
rings, Trans.Amer. Math. Soc.
250
generalization
9 5 (1960),
466-488.
of
9
[3]
N.Bourbaki,
[4]
L. Burch, On ideals of finite homological dimension in local
Algebre Commutative,
rings, Proc. Camb. Phil. Soc.
Is]
H.Cartan and S.Eilenberg,
Hermann,Paris,
64 (1968),
1961-65.
941-948.
Homological Algebra,
Princeton
University Press, 1956.
[6]
S.U.Chase,
Direct product of modules, Trans.
Amer.Math. Soc. 9 7
(1960), 457-473. [7]
J.M. Cohen, A note on homological (1969),
[8]
J.Algebra
ii
483-487.
L.Gruson
and M.Raynaud,
Inventines Math.
[9]
dimension,
•
%
Crzteres
de platitude
et
°
•
•
de pro]ectzvzte,
•
13 (1971), 1-89.
W. Heinzer and J.Ohm, Locally noetherian commutative rings,
Trans.
Amer. Math. Soc. 158 (1971), 273-284.
[10]
C.U.Jensen,
Zeit.
[ll]
106 (1968),
I.Kaplansky,
[12] [13]
Some remarks on a change of rings theorem, Math.
395-401.
Commutative Rings, Allyn and Bacon,
Boston, 1970.
, Fields and Rings, University of Chicago Press, R.E.MacRae,
On an application of the Fitting invariants,
1969.
J.
Algebra 2 (1965), 153-169.
[14]
W.V.Vasconcelos, Amer. Math. Soe.
[15]
On projective modules of finite rank, Proc.
22 (1969),
, Annihilators tion, Proc. Amer. Math. Soc.
[16]
, Finiteness
430-433. of modules with a finite free resolu29 (1971), 440-442.
in projective ideals, J~Algebra
(to appear). [17]
, The local rings of global dimension two, Proc. Amer.Math. Soc.
(to appear).
251
