Partially Ordered Rings and Semi-Algebraic Geometry (London Mathematical Society Lecture Note Series 37)

an order. By 1.4.1, we must have

1.5.

•

-a G Dp, hence ^ u (-Dp) = A.

D

Remarks on the Categories (PORNN) and (PORCK) We want to extend the discussions of sections 1.3 and 1.4 to (PORNN)

and (PORCK).

Section 1.3 applies without much modification to (PORNN).

This is because the assumption assumption about

(A,*P) G (PORNN)

is really a purely algebraic

A, and does not involve *p. The only change required

concerns the souped up extension condition for f : A •> B. restrict to ideals

J C B, with

J = /T, the nil radical.

One must

Section 1.4 also applies to (PORNN), but there are simplifications. For example,

Proposition 1.5.1. (a) p

+

P2a

If

=

0

(b)

(A/P) G (PORNN), implies

If

V

a G A, then

^[a]

is an order if and only if

p, = p 2 a = 0, that is, -a ^ Dp .

(A,15) G (PORNN), then


is a maximal order if and only if

+

P U (-Dp ) = A. Proof.

The point is, condition (ii) of 1.4.1 is automatically satisfied

in (PORNN).

In fact, this condition is the defining axiom 1.1.(v) for (PORCK),

and we showed in 1.1 that (PORNN) C (PORCK).

•

Despite the redundancy of condition (ii) of 1.4.1 in (PORCK), one does not quite have 1.5.1 in (PORCK). involves more than just that

This is because the assertion

*p[a]

is an order on

D C K CP + ) = {a G A ^ x = p 2 ax ± 0, some

-a f. Drv.Q5 )

Thus

if and only if

A.

1

Specifically, define

p i G
(p, +p o a)x = 0

Ll\

(A,15[a])G (PORCK

implies

p,x = p o ax = 0.

Z

1

Z

Proposition 1.5.2. (a)

If

(A,15) G (PORCK)

(A,15 [a]) G (PORCK) (b)

If

if and only if

and

a G A, then

if and only if

+

15 [a]

is an order with

-a £ D C K C P ) .

(A,15) e (PORCK), then


is a maximal (PORCK)-order on

A

15 u (-DCKCP+)) = A.

•

We leave the proof as an exercise. (A,15) G (PORNN), then

+

Also, we mention that if

+

D C K CP ) = #P , so 1.5.2 contains 1.5.1.

The extension conditions in 1.3 also require modification in (PORCK). Suppose order

^B

(A,*p.) G (PORCK), on

B with

A -» B

fCP A ) C ^^

n 2 (X f( Pi )br)x= 0, V±eVk>

38

f:

and

bi? x e B

a ring homomorphism. (B/P J G (PORCK)

implies

Then there is an

if and only if

f (p^bfx = 0 ,

l < j < n .

This is contained in a souped up condition concerning a family of maps f : A

•+ B,

(A ,p ) G (PORCK).

Namely, there is always a smallest ideal

I C B which satisfies:

n 2 n f (p )bfx = 0 , j=l ij ij

implies

The ring

¥ = B/I

the compositions

admits an order f 7rfa:

1 < i
with

(A , $ ) -+ (B,f)

(B,f) G (PORCK)

such that all

are (PORCK)-morphisms.

The concept of contraction of order to a subring is unchanged in (PORCK)..

If

(B,
and

AC B

is a subring, then

(A/p.) G (PORCK),

where
Remarks on Integral Domains If we restrict our attention to rings without zero divisors, that is,

to integral domains, the theory of orders begins to look very much like the classical theory of formally real fields.

Proposition 1.6.1.

(A/J3) G (POR),

Let

(a) The derived set

If C A

A

an integral domain.

is an order on

Then

A.

(b) Every maximal refinement of ty contains

Dp

and

Dp

is the

intersection of the maximal refinements of *$. (c) Every maximal refinement of ty is a total order on

A.

Proof.

and

(a) Suppose

p ^ p ^

p 1 p 2 (a 1 + a 2 ) G $

and

and sums.

with

p,p 2 ^ 0.

p x ,p 2 e
Thus

Dp

Since 'P C Dp, all squares belong to

then we would have implies

e $

pqa = 0.

pa <_ 0 £ qa Again, since 1

(b) Suppose f c p . 1

of
with A

Then

Then

p ^ t a ^ ) €
is closed under products Dp.

Finally, if

p , q € f . Thus

a G Dp n (-Dp) ,

qpa £ 0 £ pqa, which

is an integral domain, a = 0. Dp C Dp 1 .

If *pf

For the second statement, suppose

is a maximal refinement a £ Dp.

By 1.4.1,

39

y> [-a] so

is an order, and can be refined to a maximal order

s

p' . Thus

-a E*p f ,

a £ *p'. Namely, if *pf

(c) This is easy from 1.4.3(b). A = f1 U (-DP1) = ^

U (^'), so
is total.

We observed in 1.2 that a field if

E

2 -aI =I (-) a a > 0.

E

•

admits an order f C E, if and only

We claim for any such order vp = Dp.

is formally real.

0 < a < b, then

is a maximal order,

0 < £D a< -, since Thus if

pa = q,

It follows from $ = Dp

Namely, if

= (b-a)/ab = ab(b-a)/(ab) 2 > 0, and

(a - Dh

p,q > 0, then

1 a = (-)q > 0. p

that any partial order ^

on a field

E

is the intersection of the total orders which refine *p. Thus any element a E E

with

a £ *p and

some refinement.

-a ^ *p can be made either positive or negative in

In particular, this applies to the weak order p

=

2

{2 a |a. G E}. Let

(A,*p) be a partially ordered integral domain and let

be the inclusion of

A

in its field of fractions.

i:

A -* E

The following is now an

easy consequence of our definitions.

Proposition 1.6.2. (a) The extended order a,b E A, belongs to (b)

i$

i*p C E

is defined.

if and only if

abp

An element

E

a/b E E , p E sp.

i*i*

(c) The functions

i*,i*

between the set of orders *p on of all orders on

E.

to total orders on

define inclusion preserving bijections A

which satisfy *p = Dp, and the set

In particular, total orders on

A

correspond bijectively

E.

•

We will prove a more general result in a later chapter on localization.

1.7.

Some Examples (1) The ring 'L of integers admits a unique order, as does its

quotient field

Q.

If A

admits an order and if every element or its

negative is a sum of squares in

A, then

weak order, which is a total order.

40

A

admits a unique order, the

(2)

Let

(A,$) G (POR)

A-valued functions on order

and let

S

S , can be ordered.

$ w , and the affine order

p g = {£:

be a set.

A S , the ring of

Then

Two natural orders are the weak S -»• A|f(s) >_ 0, all

These orders can be contracted to any subring of

s G S}.

A , and the relation

between the orders is an interesting problem in many instances.

Examples

include the ring of continuous real valued functions on a topological space X, the ring of «UX 1 ,...,X n ]

C -real functions on a C°°-manifold

M, and the ring fnl B> J.

of polynomial functions on affine n-space

More generally, let

S

be a set and let

(A $

) , s G S, be a family

of partially ordered rings indexed by the elements of J! A -*• S s € S S

of sections of the projection

ways, as can any subring.

For example, X

S.

Then the ring

can be ordered in various

might be a topological space

equipped with a sheaf of partially ordered rings.

Then the ring of global

sections of the sheaf becomes a partially ordered ring in various ways. (3)

Since the square root of a nowhere negative continuous real

function is continuous on a space affine order on

C(X) coincide.

are exactly the squares. order structure.

X, we see that the weak order and the That is, the positive functions in

Thus the algebra structure of

Any ring homomorphism from

an order, e.g., C(Y), where

Y

C(X)

C(X)

C(X)

determines the

to any ring which admits

is a topological space, is necessarily order

preserving. If

X

is reasonable, C(X)

Namely, suppose Let

^.

f(x) < 0

x E X.

for at least one point

X •> IR be a nonnegative continuous function with

p:

p E 0

f ^ '£, that is

actually has only this one order

outside some small neighborhood of

-f G Dp .

x.

This proves that the order ty on

Then C(X)

p(x) > 0 +

pf G (-£ )

and

and

admits no simple

refinements.

(4)

Let

functions on

M M.

be a

C°°-manifold,

The affine order is

C°°(M)

the ring of

C°° real

^ M = {f G C°°(M) |f (X) :> 0

all

x G M}.

Just as in the case above, *&.. admits no refinements since there always M exist

C -functions on

M

positive at a point and vanishing outside any

small neighborhood of the point.

41

On the other hand, one cannot characterize C°°(M). a

C

For example, the function function.

x +y

on

One next might guess that

^M

R?

as the squares in is not the square of

*$„ = *p , the sums of squares.

This is still not correct. Note that the zero divisors

in

C°°(M) are exactly the functions

which vanish on some open set of

M.

Suppose

f G C°°(M)

satisfies

h2f =

2 g? , h,g. G C°°(M) and h not a zero divisor. Then f G $_,, X M i=l x as is easily checked. The set of such f defines a natural order *B,,, d refining ¥> , and it turns out that ty 4 ^ w , but "2>, = 'B... W W M u M I am indebted to Paul Cohen for proofs of these results. In fact, 2 2 h f = g

one can always write x G M.

for suitable

In the other direction, there are

C

f°r

f (x) >_ 0

h,g, if

a

^

nowhere negative functions

on the real line, which vanish only at the origin, but which are not sums of squares of

(5)

C

functions.

We consider the polynomial ring

functions on affine space for all

x G R , then

f

R^.

It is true that if

n = 1

and

is a sum of squares of polynomials.

However, if

Examples are due to Hilbert.

However, if we pass to rational

functions, then every positive polynomial is a sum of squares. Hilbert problem, solved by Artin. If

f G R[X. .. . X ] in

then for suitable R[X, . . . X ] order

and

^

.

and

f (x, .. . x ) > 0 in —

h, g i G R[X X . .. X J ,

h ± 0,

(6)

42

in

h2f =

2 g 2.

If ^

is the weak

is the order defined by positive functions on

2 h f =

x = (x1 ... x ) G R ^ , then x

(x, ... x ) G R ( n ) , in

for all

is an integral domain, of course.

c *£ , . since if

borhood of

This was a

We may rewrite the result as follows.

then Artin's result becomes in our notation D£

f(x) _> 0

f(X, X 2 ...X ) >_ 0, which are not sums

n >^ 2, there are polynomials of squares.

R[X, ... X ] , ordered as real

R^

n

.

^

(

. C D^^ .

R^

,

But also,

m

2 S g.

h

Since

and if

f(x) < 0

for some

vanishes identically on some open neighh

is a polynomial, this implies

h = 0.

Return to the general construction of Example (2), (A,^) G (POR),

g S

is a set.

Suppose

BC A

is a subring.

separates points of S, by identifying all

We may as well assume

s and s

1

B

if f(s) = f(s') G A,

f e B. Suppose

Y C S

is a subset such that

f G B, f | Y = 0 implies

f = 0. We call such a Y • B-Zariski dense. the weakest topology on S

for which all sets

Y

is, in fact, dense in Z = {s g(s) = 0} are

closed, g G B. A if

B-Zariski dense

gIY — °

Y

as a f u n c t i o n

(

determines an order on B, say P Y , by Y -• A ) .

Thus ^ g

gef

Y

is the affine order on B,

which we defined earlier. Clearly, if Y, and Y_ are two such sets, then implies $ cp . Y C Y 1 l Y Y 2 l More generally, let ^ be a family of B-Zariski dense subsets of S. Suppose and

&
is a filter, that is, Y ,Y 2 G^" implies

Y- G e^f, Y D Y~ implies 5

order on

4

B, and

Y G &.

5

g G$

4

if gl

J^ C J^ are filters in <&, then 1 ^

(7) Again suppose

Then £,— = U «* x^¥<

>• 0 for some Y. G &.

Y 1 n X^ G g? ^ i

is an

Clearly, if

^^ C $ . /j /

(A,*P) G (POR),

S a set,

BC A

a subring of

functions which separates points of S. Let Z C S be a subset and let I C B be the ideal of functions in B which vanish on Z. Then the ring is,in a natural way, a subring of the ring of functions

B/I

A , and

separates points. (8) We consider as a special case of (7) , A = IR, S = R ^ , affine n-space, and B = R[X ... X ] , the polynomials in n-variables. Z CR

(n

^,

I C R[X 1 ... X ] the ideal of polynomials vanishing on Z,

then R[X 1 ... X ]/I algebraic set,

is a ring of algebraic functions on some real

specifically the Zariski closure of Z in R

However, the affine order ^ 7 L

on

Given

.

on R[X, ... X ]/I very definitely depends in

Z itself, not just the Zariski closure. (9) For example, let K be a finite simplicial complex, with vertices

v..,...,v . Embed

|K| rectilinearly in £c n ^ be sending each vertex v^

43

to the i t h unit basis vector If the

e± = (0... 0,1,0... 0), and let

Z = |K| C ]R

X.

are interpreted as coordinate functions on R , then Z is n contained in the hyperplane 2 X. = 1. The Zariski closure of Z = IK| i=l 1 is the union of all affine linear subspaces of this hyperplane through k-simplexes of |K|

|K|, k _> 0. The ideal I C IR[X1 ... X n ] vanishing on n is generated by X X. - 1 and monomials X. •...•X. , one for x x i=l x o k (e- ,...,e- )

each set of vertices of

|K|.

the

X.

of

|K| which do not span a simplex

In the affine order on IR[X1 ... X ]/I, one has become the baracentric coordinate functions on

(10) Suppose

X

0 £ X± £ 1, and |K|.

is a compact Hausdorff space, {U.}, .

is a

finite cover of X and <£. : X -»- R is a subordinate partition of unity. n That is, 2 ^ ( x ) = 1, 0 £^>i(x) £ 1, all x e X, and suppO^) C Uiil The functions ip ,.. . ,^n generate over Z (or Q or R ) a ring of real valued functions on (or Q[XX ... X n ]

X, and this ring is a quotient of Z[X, . . . X ]

or R[X ... X R ] , depending on the coefficients).

The

ideal of relations is just the ideal of relations among the <£. as functions on 0 £ X i £ 1.

X.

One has a natural partial order, on this quotient, with

One might call such rings "partially ordered rings with a

positive partition of unity". (11)

It is easy to find rings with maximal orders which are not total. A = II A^ be a direct product of rings. Then any order i is a product n ^ , where ^ C A^ are orders. This is immediate

For example, let *$ on

A

from the equation ( a ^ d j ) = (0... a. ... 0) where

1. = (0 ... 1. ... 0) = 1..

if each 'JK C A^ on

A.

44

is maximal, but

A product order II ^. np^

is clearly maximal

will essentially never be total

II - Homomorphisms and convex ideals

2.1.

Convex Ideals and Quotient Rings Fix

(A,p) G (POR).

implies

p G C.

p i G C,

i= 1,2.

are convex.

Equivalently, C

An ideal

I C A

(POR) if and only if

in

A

Thus

with I

CCA

is oonvex if

is convex if

0 <_ p £ q,

p, + p~ G C,

q6C

p. G ^, implies

Clearly, arbitrary unions and intersections of convex sets

Proposition 2.1.1.

Proof.

A subset

I

First, if

is said to be convex if it is convex as a subset.

An ideal

I C A

is the kernel of a morphism in

is convex.

I = kernel(f: (A,
f(q) = 0, then

0 £ f(p) £ f(q) = 0

in

and

0 £p

A', hence

£q

f(p) = 0.

is convex.

Conversely, suppose

I

is convex.

Consider the projection

According to 1.3, we can impose an order on n

preserving if, whenever 1 £ j £ n.

A/I

so that

IT is order

2

2

p.a. G I,

IT: A -^ A/I.

2 p. Gtyt

a. G A, then

p.a. G I,

But this is immediate from the second characterization of

convexity above.

•

The weakest order on

A/I

such that

T*CP) = {p + l|p Gf}

fr is order preserving, namely

c

A/I

will be called the induced order, or quotient order, on A/I, and will be denoted

(A/I, ^ / I ) .

Let

f: (A/P) -* (A'/p1)

convex ideal kernel

be a morphism in (POR).

(f). Then the morphism

f

Let

I C A

be the

factors as a composition

(A,?)

45

where

f,

is, ?2

is a quotient projection,

f2

is a refinement of order (that

is the identity as a ring homomorphism), and

f~

is an inclusion

of a subring with the contracted order. Notice that because of the existence of refinements of order, a map can be an order preserving isomorphism of rings, without being an isomorphism in the category (POR). on

A, then

A

In general, if

will have fewer

A

is a ring and

*£'-convex ideals than

^ C
are orders

^-convex ideals.

We now consider the operations on convex ideals of (i) intersection, (ii) forming nil radicals, (iii) sums, (iv) products, and (v) quotients.

Proposition 2.1.2.

Fix

(A/p) e (POR).

Then

(a)

An arbitrary intersection of convex ideals is convex.

(b)

If

Proof.

I C A

vT = {a e A|a n e I, some n _> 1} is convex.

is convex, then

Statement (a) is trivial.

0 < a < b 6 y T .

0 £ a11 <_ b n

Then

(b-a) (bn~l + ab n ~ 2 + • • • + a""1) .

For statement (b), suppose

for all

n, since

b n G I, then

If

b n - a11 =

a n G I, hence

a G /f.

•

It turns out that sums, products, and quotients of convex ideals need not be convex.

2.2.

Examples will be given in 2.8.

Convex Hulls Let

(A/p) G (POR),

convex ideal of

A

X C A

a subset.

which contains

I

X.

We denote by

H(X)

the smallest

Thus

convex ideal X C I

It is possible that

H(X) = A, of course.

convex hiilly or simply hull, of Here is a construction of be the ordinary ideal of

H 0 (X) = J(X)

46

A

We refer to

H(X)

as the

X.

H(X). For any subset

generated by

Y.

Define

Y C A, let

J(Y)

H n + 1 (X) = J(H n (X) U {b G A|0 < b < a, a G Hn (X)}) .

Proposition 2.2.1.

Proof. ideals of

some

m.

H(X) =

|J H (X). n >_ 0 n

Clearly, H (X) C H,(X) C ••• A, hence

Then

U H (X) n > 0 n

x G H

is an increasing sequence of

is an ideal.

(X). This proves that

m+i

Finally, for all which contains

containing

m,

H

(X)

H (X). Thus

m

Suppose

0 < x < y, " "

U H (X) n >_ 0 n

y G H (X), m

is convex.

is obviously contained in any convex ideal

n

(J

H (X)

is the smallest convex ideal

_. °

X, as claimed.

•

The following "compactness" result is useful.

Proposition 2.2.2.

Let

there is a finite subset of

Proof.

(i)

Let

be any subset and let

X, say

{ x ^ - x ^ , such that

I = \y e Aly G H({x.••-x, }), some l I I K

X C I C H(X)

and

(ii)

ideal, then necessarily Obviously, if

y 1 >y 2

I

A.

Then

y G H({x1- • -\}) •

x. e Xl. J 1

is a convex subset of

a G A, then

X p X 2 C A, e

y G H(X).

If

Then

I

is an

I = H(X).

y e I,

H(X) C H(Y), hence if easily that if

X C A

l

>

tnen

a y e I.

But also, if

X c Y c A,

H(X X ) U H(X 2 ) C H(X X U X 2 ) .

It follows

y ]_ + y 2 G- I -

•

The next result is also extremely useful.

We characterize the nil

radical of a convex hull.

Proposition 2.2.3.

v4l(Z) = {zGAJO £ z

Let

Zcf

k <_ S P i z i i=l

be any set of positive elements.

for some integer

s > 0, z ^ Z

Then

and p ^ ^ } .

Although stated for positive subsets, 2.2.3 actually characterizes the nil radical of any convex hull. X

2

2

= {x |x G X} C $.

Then

Namely, if 2

X C A

X C */H(X ), hence

AUX)

is any subset, let c Mx*)>

since

v4l(X2)

47

is convex and coincides with its own nil radical. /H(X) = 4 K X 2 ) .

so

H(X 2 ) C H(X),

But also

Thus 2.2.3 has the following corollary.

Corollary 2.2.4.

Let

X C A

be any subset.

Then

k

/H(X) = {z e Alo < z

< 2 p.xf, some s > 0 , p. e f and x. e X}. i=1

Proof of 2.2.3.

Let

i i

I = {yGA|0£y

k

2r

<_ Z p . z . ) ,

1

I = /H(Z). and (iii) imply

It suffices to prove (i) I = /T.

I D H(Z) Now,

i=i I

x

is an ideal, (ii)

Because, obviously, /H(Z) D I D Z.

and (iii) implies

We want to prove

x

I

Then (i) and (ii)

I D Ai(Z).

(ii) and (iii) are trivial, so we will prove (i).

that if y e I, say k 2 (a 2 r p i )z i . Thus i=l

0 <_ y

2r

£

2 p ^ , and

hence for any

a G A, then

It is clear

0 £ (ay) 2 r <_

i=1

ay e I.

Finally, we must show that Then for suitable

is convex,

b e A

I

is closed under sums.

and any

Let

yi> v 2

e

**

s ^> 1,

r ^> 1,

Expanding the left-hand side of (*) gives

j=0

where the

J

c.

j=l

and

d.

J

are (positive) binomial coefficients.

In particular,

C Q = 1, so transfering the negative terms to the right-hand side of (*) gives

(**)

o <

(y1+y2)4sr r •^ c . v.y-i + y o -

j-o

48

}

1

2

li But

2s 2r 2s 2r 0 0 << yy 1 << SS pp.zz . and and 00 << yy 9 5 2 p ! z ! , l 2 j=l ^ : j=i 3 J 2s 2r Replacing y, ,y~ in (**) by these larger terms

y,,y9 G I, say

p.,p! G ty, z.,z! G Z.

results in an inequality which implies

2.3.

(A;p) G (POR) throughout this section.

is a prime convex ideal, if

with

D

Maximal Convex Ideals and Prime Convex Ideals We fix

or

y, + y~ G I.

b G Q.

Equivalently,

Q ^ A

Q

and

whenever

is convex and

A/Q

A convex ideal

ab G Q, either

Q c A

a G Q

is an integral domain,

1^0. A convex ideal

whenever

Q C Q',

Q C A Q'

is a maximal convex ideal if Q ^ A

a convex ideal, either

Q' = Q

or

Q

f

and

= A.

The concept prime convex ideal just means ordinary prime ideal which is convex.

On the other hand, a maximal convex ideal may not be maximal

in the family of ordinary ideals. We first establish the existence of maximal convex ideals. Proposition 2.3.1. a convex ideal, 1 £ I

Let

(A,^) G (POR), 1 i 0

(equivalently,

I 1 A).

in

Then

A. I

Let

IC A

be

is contained in at

least one maximal convex ideal. Proof.

The usual Zorn's lemma argument applies, since the family of

all convex ideals containing

I but not containing

1

is non-empty,

partially ordered by inclusion, and satisfies the chain condition.

In particular, since that any non-zero

•

I = (0) is always a convex ideal, we conclude

(A,^) G (POR) has maximal convex ideals.

Next we characterize maximal convex ideals by the quotient ring they define.

If

and for all

(A,$) G (POR), we say a G A,

(A,?)

a ^ 0, there exists

is a semi-field if b G A

with

1 £ ab.

1 7* 0

in

Note that

this definition depends on the order ty C A, and not just the algebraic structure.

49

A

Proposition 2.3.2.

If

(A/P) G (POR), the following conditions are

equivalent. (i)

(A/P)

(ii)

is a semi-field.

Every homomorphism of

(iii)

(A,*P)

in (POR) is zero or injective.

(0) C A are the only convex ideals,

(iv)

For all

a G A,

a ± 0, we have

1 G H(a).

(v)

For all

a G A,

a ± 0, we have

1 £ pa 2 , some

Proof.

The implications

are trivial.

(i) =» (ii) => (iii) =*• (iv) and (v) => (i)

Corollary 2.2.4 gives

In general, if

Q C A

(iv) => (v).

clarified

Q

(A/Q, $/Q)

is a maximal in (POR) is

in the next section, thus 2.3.2 has the following corollary.

Let

(A/p) G (POR),

Q C A

is a maximal convex ideal if and only if

Equivalently, for all with

Q

This is more or less obvious, but in any event will be

Corollary 2.3.3. Then

•

is a convex ideal, Q / A, then

convex ideal if and only if every homomorphism of zero or injective.

pef.

a G A,

a convex ideal, Q i A. (A/Q, ^/Q)

a £ Q, there should exist

b G A

is a semi-field. and

1 £ ab + q.

q6 Q •

Partially ordered fields are semi-fields and semi-

Proposition 2.3.4.

fields are integral domains.

Proof. a i 0

The inequality

implies

a 2 ± 0.

1 £ ab

If now

implies

ax = 0,

2 2 1 <_ a b , so in any semi-field,

a ^ 0, we would get

0 <_ x 2 £ a2b x = 0,

2 for some

b, hence

x

Corollary 2.3.5.

= 0

and thus

x = 0.

Maximal convex ideals are prime.

•

We will give a second proof of 2.3.5., also based on Proposition 2.2.3. First, if convex.

Q C A Suppose

is a maximal convex ideal, then ab G Q,

b g Q.

Then

1 G H(Q,b ) .

have k

b 0 < l < 2 p.q. + pb "" "" i l X 1

50

Q = T/Q, since

2

By

v^Q

is

2.2.3, we will

for suitable

p i ? p e ty, q. G Q.

0 < a2 <

Since

Q

is convex, a

Multiplying by

a

gives

2 (p.a2)q. + pa 2 b 2 e Q. i=l

G Q,

and, since

Q = /Q,

a e Q.

Thus

Q

is

prime.

•

Finally, we relate nilpotent elements and prime convex ideals.

Proposition 2.3.6.

Let

(A,$) G (POR).

/0 =

fl

Then

Q-

Q C A prime convex ideal

Proof.

Clearly

/o c Q

if

Q

is prime.

Conversely, suppose

Then the set of convex ideals not containing any power of since

(0)

is such an ideal.

We assert such an ideal, say

let

c £ Q.

integers

b £ Q,

n,m >^ 1.

is non-empty,

Zorn's lemma applies to give a convex ideal,

maximal in this set. be G Q,

a

a g /o.

Then

a11 G H(Q,b),

By 2.2.4, there are integers

Q, is prime. a m G H(Q,c) r,s >_ 1,

Otherwise,

for some

q i G Q,

p^ G $,

i = 1,2, with 0 < a 2 r < qx + p x b 2

0 < a 2 s < q2 + p 2 c 2 .

Multiplying these equations gives

0 < a2(X+S) < q € Q ,

which is a contradiction.

Corollary 2.3.7.

Let

I C A

be a convex ideal.

n P

Proof.

Then

p.

convex prime I C p

We apply 2.3.6 to the partially ordered ring

(A/I,

and use the proposition of the next section.

51

2.4.

Relation Between Convex Ideals in Fix

(A/p) E (POR), and let

I C A

(A,?)

and

(A/I, !J)/I)

be a convex ideal.

Proposition 2.4.1. (a)

There is a natural, bijective, inclusion preserving correspondence

between ^-convex ideals J/I

of

J

of

A

which contain

I

and

p/I-convex ideals

A/I.

(b)

For any such

J D I, there is a natural isomorphism in (POR),

(A/J, $/J) 3- (A/l/j/I,

(c)

The prime convex ideals of

A

which contain

bijectively to the prime convex ideals of ideals of

A

which contain

convex ideals of

Proof.

and

J/I

Given a ^-convex

is

J C A/I

'p/I-convex. is

correspond

and the maximal convex

correspond bijectively to the maximal

A/I.

order preserving morphism Thus

I

A/I

I

p/I

J D I, the identity on

A

induces an

(A/1, *p/I) •> (A/J, !p/J), with kernel Conversely, if

convex, then

IT: A -> A/I

J = TT'^J) C A

is

J/I.

is projection, ^-convex.

The

remaining details are equally simple.

2.5.

D

Absolutely Convex Ideals Although the category (PORCK) is a full subcategory of (POR), one

sees fewer ideals as kernels of morphisms in (PORCK). of the category (PORNN), where obviously an ideal

I c A,

is the kernel of a (PORNN)-morphism if and only if "P-convex.

The same is true (A,$) E

I = /f

and

(PORNN),

I

is

In this section we will investigate kernels of morphisms in

(PORCK) . Let convex if

(A;p) E (POR), 0 ^ SL ^_ b

absolutely convex if

52

and

I C A

an ideal.

bx E I

(p+q)x E I,

implies

We say that a x e I.

p,q E*p, implies

I

is absolutely

Equivalently, px E I

and

I

is

qx E I.

Proposition 2.5.1. (a) Absolutely convex ideals are convex. (b) Arbitrary intersections of absolutely convex ideals are absolutely convex. (c) Any convex ideal

I with

I = /f

is absolutely convex.

In

particular, convex prime and maximal ideals are absolutely convex. (d)

If y c $ »

is a refinement of order and

convex, then (e)

If

I

is £'-absolutely

is ^-absolutely convex.

(A,?) -> (A 1,^ 1 )

f:

I c A

absolutely convex, then

is a morphism in (POR) and I = f" (If) c A

I ! C A1

is

is absolutely convex.

Proof. (a)

Let

x = 1, the unit in

A.

(b) Obvious. (c)

If

0 ± a <_ b

hence

and

(ax) 2 e I.

bx e I, then Since

I = /f,

0 £ ax 2 ± bx 2 , hence

ax 2 e I,

axel.

(d), (e) Obvious.

Any subset

X C A

•

is contained in a smallest absolutely convex

ideal, namely the intersection of all absolutely convex ideals containing X.

We denote this smallest absolutely convex ideal

hull of

AH(X), the absolute

X.

It is easy to give a construction of AH(X)

as a countable union

of an increasing chain of ideals of A, just as was done in 2.2 for H(X).

Namely, define

AHQ(X) =

J(X)

AH n + 1 (X) = J(AHn (X) U {ax|0 < a < b, bx e AH

where

J(Y)

is the ordinary ideal generated by

Y.

Proposition 2.5.2. (a) AH(X) = U A H (X). n

53

(b)

If y G AH(X), then there exists a finite subset of {XJ-'-XJ^}, with

(c)

X,

y E AH({x 1 ---x k }).

/AH(X) = M(X).

(d) AH(X)AH(Y) C AH(XY)

where

XY = {xy|x e X, y G Y}.

Proof: (a), (b) Very similar to 2.2,1 and 2,2.2 (c) Clearly

H(X) C AH(X).

2.5.1(c), hence

Also, M(X)

AH(X) C Af(X).

(d) We prove by induction that

is absolutely convex by

Thus,

AHn (X)AH m (Y) C AH n + m (XY).

n = m = 0, this is clear.

If

bx G AH

If

(X), 0 £ a £ b, and

y £ AH (Y), we must show m

(ax)y £ AH (XY), since n+m

typical new generator of

AH n (X).

ax

is a

But, by induction, (bx)y =

b(xy) e A H n + m _ 1 ( X Y ) and 0 £ a £ b , hence

a(xy) e A H n + m ( X Y ) .

•

Proposition 2.5.2(d) allows us to construct graded rings naturally (PORCK).

in have AH(I

Namely, if

(A,$) G (PORCK)

2

3

A 3 AH(I) 3 AH(I ) 3 AH(I ) = > • • • .

and

I CA

is any ideal, we

By 2.5.2(d), AH(In )AH(I m ) C

) . We thus have a graded ring associated to

I,

GA = 0 AH(I n )/AH(l n+1 ) . n>0 This construction does not seem natural in

(POR) because the analogue

of 2.5.2(d) for hulls, rather than absolute hulls, is not true.

Our next result is that absolutely convex ideals are closed under the quotient operation, (I : J) = {b E AjbJ C I}.

Proposition 2.5.3. subset,

Let

(A,$) E (POR),

(I : Y) = {b E A|bY C I}. Then

54

(I : Y) = (I: AH(Y)).

an ideal, Y C A

any

I is absolutely convex if and only if

(I :Y) is absolutely convex for all subsets convex, then

IC A

YCA.

Moreover, if

I

is absolutely

Proof. if

Since

(I: {l}) = I, one direction is trivial.

I is absolutely convex,

0 <_ a <_ b

b(xy) G I for all y G Y. Hence Thus

a x G (I : Y ) , which proves

and bx € (I : Y ) , then

if

(I : Y)

is absolutely convex.

b*AH (Y) C I. It suffices to prove 0 <_ a <_ c,

bJ(Y) C I. Assume inductively

b'AH _(Y) C I, and, in particular,

ex G AH n (Y), it suffices to prove

c(bx) G I, hence

(bx)y =

a(xy) = (ax)y G I for all y G Y.

For the final statement, if bY c I, then that

Conversely,

b(ax) G I. But-, b(cx) =

a(bx) = b(ax) G I.

•

In 2.8 we will give examples of convex ideals which are not absolutely convex.

We will also give examples of sums and products of absolutely convex

ideals which are not absolutely convex. We now establish the relation between absolutely convex ideals and the category (PORCK).

The defining axiom 1.1.(v) for (PORCK) is exactly the

condition (0) absolutely convex.

That is, (p1 + p 2 )x = 0 ,

p^ G ^

implies

p^x = P2X = 0. In fact, we have

Proposition 2.5.4. (i) (ii) (iii) (iv) (v) (vi)

If

(A,^) G (POR), the following are equivalent.

(A/P) e (PORCK). (0) C A

is absolutely convex,

(0 : a) is absolutely convex for all a G A. (0 : a) is convex for all a G A. (0 : X) is absolutely convex for all subsets (0 : X)

Proof.

is convex for all subsets

We have just observed that

X c A.

(i) ^ (ii). Proposition 2.5.3

gives (ii) =* (iii) and (iii) =* (iv) is trivial. if

0 £ a <_ b

hence

and bx = 0, then

X C A.

(0 : x)

Now, (iv) =* (ii) because

convex implies

a G (0 : x ) ,

ax = 0. The equivalences (iii)** (v) and (iv) <=> (vi) follow from (0 : X) =

D (0 : a) . a G X

Remark: The proposition is true with

(0) replaced by

•

I, for any

55

ideal

I c A, if statement (i) is replaced by

(A/I, ^/I) G (PORCK).

The

proof is the same. As a consequence of the results above, we state Corollary 2.5.5. (a) I C A

Let

(A,$) G (PORCK)

[respectively

(A,q3) e (PORNN) ].

An ideal

is the kernel of a (PORCK)-morphism [respectively (PORNN)-morphism]

if and only if (b)

If

I

is absolutely convex [respectively

(A/P) G (PORCK)

and

Ic A

I = /F].

is absolutely convex [respectively

(A,$) G (PORNN),

I = / F ] , then there is a natural bijection between absolutely

convex ideals

of

J

A

A/I [respectively ideals

containing J

of

A,

I

and absolutely convex ideals

J = / j , containing

I

J/I

and ideals

of

J/I

of A/I, /J7T = J/I]. (c) The inclusion functors (PORNN) •* (PORCK) •* (POR) have adjoints (POR) -> (PORCK) •> (PORNN), defined by assigning to partially ordered quotients

(A,?) G (POR)

(A/AH(0), ?/AH(0)) G (PORCK)

and

the

(A//o, ^//o) G

(PORNN). Proof.

2.6.

Whatever remains to be proved will be left as an exercise.

•

Semi-Noetherian Rings It should not be surprising that various finiteness conditions on partially

ordered rings lead to structure theorems and results which cannot possibly be proved in general.

Since a partially ordered ring consists of a ring

A,

together with an order '£ C A, natural finiteness conditions can involve either the '^-convex ideal structure or the order 'p itself, as an extension of the weak order *pw c A.

We are ultimately interested in partially ordered

structures on finitely generated extensions of real closed fields. Chapters 7-8.)

(See

Such rings are, of course, Noetherian in the classical

sense, hence no finiteness assumptions are necessary on chains of convex ideals.

However, even for this restricted class of rings, many classical

results fail to generalize to the partially ordered context.

For example,

one cannot always decompose convex ideals in such a ring as intersections

56

of primary convex ideals. the order.

One needs additional finiteness conditions on

When one examines the classical proofs of primary decomposition

and other results which fail to generalize the point which causes the difficulty often involves the fact that in commutative algebra the multiples of a single element always form an ideal, while in partially ordered algebra the smallest convex ideal containing a given element may be quite large. On the other hand, chain conditions on convex ideals have certain interesting consequences, regardless if the underlying ring in the classical sense or if the order

'p

A

is Noetherian

satisfies extra conditions.

Some-

times these results are proved most easily in the classical context of Noetherian rings as applications of primary decomposition, rather than by exploiting the chain conditions directly.

Thus it actually gives one some

added insight into these classical results to investigate chain conditions in the partially ordered context, where certain techniques of proof are unavailable. It is the purpose of this section to initiate this study.

We work

entirely in the category (PORCK) because of our frequent use of the quotient construction

(I : X)

for ideals

I C A

and subsets

X C A.

However, 2.6.1 through 2.6.4 have obvious (POR) versions as well.

Definition 2.6.1.

A ring

(A,'£) e (PORCK)

is semi-floetherian if any

of the three equivalent conditions below hold: (i)

The absolutely convex ideals of

(A,'p)

satisfy the ascending

chain condition. (ii)

Any non-empty collection of absolutely convex ideals of

(A,}5)

contains a maximal element with respect to inclusion. (iii)

Every absolutely convex ideal I = AH(x1,...,x, )

I

of

(A,'.J5) may be written

for some choice of finitely many elements

x r . . . , x k e I.

As simple applications of the definition, we state some standard results.

Proposition 2.6.2.

If

(A,'.j>) e (PORCK)

is semi-Noetherian,

I C A

57

A 1 = A/I, then

is absolutely convex and any refinement

Proof.

'£' of

(A','-P!)

p/I.

This is immediate from 2.5.5 (b).

Let us call an absolutely convex ideal I.

is semi-Noetherian for

absolutely convex, implies

Proposition 2.6.3.

If

I = I.,

•

I

or

irreducible if

1 = 1 ^

i

I = I~«

(A,'£) G (PORCK)

is semi-Noetherian, then

every absolutely convex ideal is a finite intersection of irreducible absolutely convex ideals.

Proof.

An absolutely convex ideal maximal among those not so

expressible leads to an immediate contradiction.

Proposition 2.6.4. I = /F

(A/p) e (PORCK)

is semi-Noetherian and

is a radical convex ideal, then there is a unique expression

I = P.n...n p 1 K if

If

i ^ j.

where the

P. j

Moreover, the

containing Proof.

P.

are prime convex ideals and

are precisely the minimal prime ideals

In fact, one only needs the ascending chain condition for The proof is just as in the classical case.

Uniqueness is clear since if a prime ideal

P

P

contains some

D P! D P .

P^

for some

also shows that the taining

P. £ P. 1 J

I, which are necessarily convex.

radical convex ideals.

then

•

Thus if

i,j , hence

P.

P

contains

P^.-np^,

P ^ - . - n p ^ = p|n...np^, j = 1

and

Pj = P!.

then

The argument

which occur are exactly the minimal primes con-

I, which are thus necessarily convex.

(This result will be

generalized in section 3.9.) Existence of the stated decomposition is established as follows.

Each

radical convex ideal is a finite intersection of irreducible radical convex ideals, by the chain condition on radical convex ideals. is an irreducible radical convex ideal, then ab € I, let

a,b £ I.

{P^,}

containing

58

Let

{P }

is prime.

I = /F

Otherwise, let

be all the prime convex ideals containing

containing

a

and let

be those

Pa

b.

I = n P by 2.3.7. a oi

Then

I

But if

{PaiJ

be

those

I,

Pa

On the other hand, {P }= {P ,}U{P „} a a a

since

I = I1 n I"

ab G I, hence 1

a e I ,

I1 = n p , , I" = n P „ .

where

b e I", this contradicts the irreducibility of

Since

I.

•

Directly from 2.6.3, absolutely convex ideals in a semi-Noetherian ring can be understood in terms of irreducible absolutely convex ideals. However, irreducible absolutely convex ideals need not be primary, even if the ring is Noetherian.

Despite the lack of primary decomposition,

the associated primes of an absolutely convex ideal are accessible in any semi-Noetherian ring.

Proposition 2.6.5. I C A

Suppose

(A/p) G (PORCK)

an absolutely convex ideal.

is semi-Noetherian,

Then the following sets of primes

coincide: (i)

{P|P

prime and

P = (I : x ) , some

(ii)

{P|P

prime and

(I : x)

(iii)

{P|P

prime and

P = /I : x, some

(iv)

{P|P

a minimal prime containing

is P-primary, some

Passing to

x G A}

x G A} (I : x ) , some

Moreover, this set of associated -primes of

Proof.

x e A}

I

x G A}.

is non-empty and finite.

(A/I, 'p/I), we may assume

I = (0). Certainly,

Set (i) c Set (ii) c Set (iii) c Set (iv). Moreover, among all the absolutely (0 : y ) ,

convex ideals (0 : z) = P

is prime.

y i 0, choose a maximal one, say For suppose

abz = 0 ,

hence by maximality, (0 : z) = (0 : bz) Next, suppose ideals if

(0 : y)

c £ P,

P G Set (iv), say

with

(0 : z) = (0 : cz) C P.

bz ^ 0.

(0 : z) = P.

abz = 0 ,

Namely, if

This proves

Then

In fact, if

bz ^ 0, then

(0 : z) C (o : b z ) ,

(0 : x ) .

(0 : z)

Thus

(0 : bz) C P,

c £ P, hence

(0 : z)

Among the

be maximal.

bz t 0, then

We now claim

Then

Thus Set (i) i 0.

minimal over

Otherwise, bcz = 0, some

= (0 : z ) , contradicting

az = 0.

P

az = 0.

(0 : x) C (0 : y) C P, let

(0 : z) = (0 : bz).

hence

and

bz ^ 0.

(0 : z).

b G (0 : cz)

is prime, hence

(0 : z) = (0 : bz), hence

Set (iv) c Set (i).

Finally, we must prove there are only finitely many associated primes. Suppose the set of associated primes is there are infinitely many.

Let

{P } = {(0 : x )}

AH({xrv}) = AH(x

••• x

).

and suppose If

x,, / xrt ,

59

then

P

2.5.3.

^ P

K P = ( O : x ) D H ( O : x ) =

, but

Thus

Pa D Pa

f

i

, some

i.

3,

0

3^33

many

P Y

all

j , then

infinite chain

P

C Po

Proposition 2.6.6. I,J C A

PD D Po P ^ Pi

properly contain some C P

I.

Then

P.DP . P t %

some

PQ . Po

Suppose

J c P., some

Let

P.,

Conversely, if

xJ C I,

ideal of the form prime of

x £ I, then

(I : J) ^ I.

In

consists of all elements

x.J c I, hence

(I : J) t I.

j

J C (I : x)

is contained in some maximal

I.

•

If

(A/p) e (PORCK)

I.

P

Proof. Let

is semi-Noetherian and

containing

I

I, and

Suppose

(I : x) £ P, where

xy G I,

y <£ P.

Then

(I : x)

(I : z)

I,

{P.}, are

/F = n p^.

(I : y) c P

of 2.6.5, a maximal ideal among the

I c A

contains an associated

In particular, the minimal associated primes of

exactly the minimal primes containing

I.

1 <_ j <^ k, be the associated

(0).

J

is absolutely convex, then any prime

of

A

•

(I : y ) , which, by the proof of 2.6.5, is an associated

Proposition 2.6.7.

prime of

So infinitely

is semi-Noetherian,

j , if and only if

J C p. = (I : x . ) , then j

i, by the same argument.

(A,'])) e (PORCK)

belonging to some associated prime of

If

AH({xo}) = AH(x o •••x o ) . P Pi Pn

Iterating this process leads to an

particular, the set of zero divisors in

Proof.

Let

ou 1 , call it

c ..., which is impossible.

absolutely convex ideals.

primes of

by Proposition

Choose an index among the

a , such that for infinitely many If

K n p

since

with

be prime, hence will be an associated prime of

is some associated prime

I

I C p.

By the proof

(I : y) C (I : z) c P

will

contained in

•

P.

Although irreducible absolutely convex ideals of a semi-Noetherian ring are not necessarily primary, we can prove that an irreducible ideal has a unique maximal associated prime. ideal

I C A

and prime ideal

P,

I C p C A, define

I(P) = {x G A | xs G I, some

60

In fact, for any absolutely convex

s £ P}.

(The ideal where

I(P)

A, p .

is the kernel of the composite homomorphism

is the localization of

considered in the next chapter.) properties of the ideals

A

at

P.

A**A.p^A.p./IA. . ,

This point of view will be

The following proposition lists some (A,'£) e (PORCK),

I(P). We assume

I C A

absolutely convex.

Proposition 2.6.8. (a)

I(P) =

(b)

I c i(P) c P.

(c)

If

P

n

(I : s ) , hence

primes of

is absolutely convex.

is a minimal prime containing

in fact, I(P) Moreover, if

I(P)

(A,'p)

I, then

I(P)

is P-primary;

is the smallest P-primary ideal containing

is semi-Noetherian and

{P.}

I.

is the set of associated

I, then

(d)

Px C P 2

if and only if

(e)

I = n I(P.)-

I(P 2) c I(P 1 ).

In particular, if

I

is irreducible, it has a

unique maximal associated prime. (f)

The associated primes of of

Proof.

I

contained in

I(P.)

are exactly the associated primes

P..

(a) and (b) are trivial.

The result (c) is actually true in

arbitrary commutative rings, for arbitrary ideals

I.

This point will be

discussed again in the next chapter, so we postpone the proof of (c). If

?1 C P 2 , then certainly

I(P 2) C I(P X ), but y £ P2

implies

x G I(P)

Px £ P 2 .

x^ €

J

CP 2 )

Let

c

is equivalent to

by the proof of 2.6.5.

c

But

(I : x) £ P.

(I : x) Thus

I^)*

yGP1-P2,

!(?]_)•

Also, we deduce (e) since if But any proper ideal

I(P 2)

tnis

is

(I(P.) : x i ) = P ^

Then

x^Gl,

ridiculous, since clearly

x G n I(P.), then

(I : x) £ P., all

is contained in an associated prime of x e I.

If

j. I,

The second statement of (e) now

trivial unless there is a unique maximal

above show

P1=(I:x1).

This proves (d).

follows from (d), since the intersection formula

Finally, we verify (f).

Conversely, suppose

I = n I(P.)

is non-

P..

(I : x.) = P. c P., then arguments just as

Also, if

(I(P-) : x) = Q.

is prime, then

61

Q. C P..

We must show

Q. = AH(y, •••yv.). K

Then

Q.

xy.s. E I

•••

3

verify

s- £ P.. 3

Let

It is now easy to

3

ideals.

•

Note that 2.6.8(c), (d), (e) imply that if

associated primes, then

I

I

has no embedded

is an irredundant intersection of convex primary

A standard argument shows such a representation is unique.

Convex Ideals and Intersections of Orders Let

A

be a ring, '-P-j/K

is an order on for

for some

3 3

I.

(I : xs •••s,) = Q., using 2.5.3. IK J

Remark.

2.7.

is also an associated prime of

$v

A.

? 2 > and

orders on

!p-2

C S, as in Example (6) of 1.7.

n $

S, with values

That is, for

j = 1,2,

s € S.}.

is clearly the affine order corresponding to the subset

s e S.

functions

'£ 12 = ^

'#., '$„ could be affine orders defined by suitable

? 1 2 = {f e A | f(s) >. 0, all

Points

Then

could be a ring of functions on a set

•j). = {f 6 A | f(s) >_ 0, all

Then

A.

$u.

in an ordered ring, and S.^ S

two

We are interested in relations between the convex ideals

For example, A

subsets

c A

define

f G A

with

S.US^S,

s e SjUS^.

'p.-convex ideals, j = 1,2, namely, the ideal of f(s) = 0.

Thus points of

S1US2

define

£12

convex ideals. More generally, for any

A

and

•J^-convex or ^-convex, certainly j = 1,2, then

I. n I

I

'.p, and

'^ , if

is '^12-convex.

is '^12-convex.

I c A If

I.

is either is '^.-convex,

Both these remarks are equally obvious

for absolutely convex ideals. If

I C A

is

'^12

below that we can write convex ideal, j = 1,2.

convex and

I = /F

I = l^ n I , where

I. = /T.

is a .p.-radical

For non-radical ideals, even primary ideals, such

a decomposition is impossible in general.

62

is radical, then we will show

We will illustrate these phenomena

with examples in 2.8. The main positive result in this direction is the following very useful fact, discovered by Andrew Klapper.

Proposition 2.7.1 (Klapper). ideal, p 1 2 = ^

Proof.

n

Then

'P 2 «

Suppose

P

p

is

Suppose either

was neither

P C A

is a prime 'pi:?-convex

'^-convex or

'p.-convex nor i

0 £ x £ y (rel 'pj), 0 £ u £ v (rel 'P2), with

'^-convex.

'p0-convex. £

y,v e p,

x,u

0 < x 2 u 2 < y 2 u 2 < y 2 u 2 + x 2v 2

(rel $.)

0 £ x 2 u 2 £ x 2v 2 £ y 2 u 2 + x 2 v 2

(rel 'P ) .

£ P.

Choose

Then

and

Thus

0 <_ x 2 u 2 <_ y 2 u 2 + x 2 u 2

2 2 x u € P, contradicting

Corollary 2.7.2.

(rel ' P 1 2 ) .

y 2 u 2 + x 2 v 2 e p, hence

But

x,u £ P.

•

Any maximal

p --convex ideal is either a maximal

'p,-convex ideal or a maximal '^-convex ideal.

Proof.

Maximal convex ideals are prime.

Corollary 2.7.3. I = 1^. n i

where

Proof.

P a

I.

'p12-convex ideal

is a radical

I.

which are

Let

I

can be written

'p.-convex ideal, j = 1,2.

I = n P , the intersection taken over all

ideals containing those

Any radical

•

'pl9-convex prime

L

= n p ,, L = n p where the P , are 1 a 2 a a 'p.. -convex and the P „ are those P which are 1 a" a

f

£2 -convex.

•

Corollary 2.7.4.

If ^ . . . . ^ C A

•p n ••• Op,-convex prime ideal, then IK 1 £ j £ k.

If

P

is a maximal

^

P

are orders and is

P C A

'p.-convex for some J

n ... rvp-convex ideal, then

is a j,

P

is a

maximal 'p.-convex ideal, some j. If I is a radical 'Pj n ••• n P k - convex ideal, then I = L n - . - n i , where the I. are radical 'p.-convex ideals. IK j j

Proof.

Each statement follows by induction from the results above.

63

•

Proposition 2.7.1 has a very useful application to the following problem.

Given

can the order

(A,'.p) G (POR),

Q C A

'£ be refined, keeping

seems difficult.

a convex ideal, to what extent Q

convex?

In general, this problem

However, for prime ideals in integral domains we have the

following.

Proposition 2.7.5. be a subring and let exist total orders

Proof. p[-x]

Q C A

Suppose

be a

x G K,

f = p1 = p2

or

that

is either

with

K

a field.

'pnA-coiwex prime ideal. 'p, with

x,-x £'£.

Q

still

(q2 + q x )f = q 2 P x + q ^ .

f = (q 2 P 1 + q 1 p 2 )/(q 2 + q x ) •

Let

A C K

Then there

$ n A-convex.

Then from 1.6, both

Moreover, ']) = 'p [x] n^[-x]

K.

pi,q;. e p, then

QC A

(K, '£) G (POR)

"jT c K, refining

are orders on

P2-q2x,

Let

$ [x]

since if

Either

and

f = p , +q..x =

qx - q 2 = 0

and

We now apply 2.7.1 to deduce ('£ [-x] n A) -convex.

Cp[x])n A-convex or

Zorn's

lemma then completes the proof.

Corollary 2.7.6. Q C A

be a prime

refining

'p, with

Proof.

Let

d

(A,'p) G (POR),

^-convex ideal. Q

still

'^-convex if and only if

D'£

Q

is

Let 'p c A.

D'^-convex.

were studied in 1.6.

is the intersection of all total orders on

'£, so necessity of the condition is clear. D']5 C A

an integral domain.

Then there exist total orders

Properties of derived orders

particular, D'£

A

A

In

refining

Conversely, the derived orders

are exactly the orders contracted from orders on the field of

fractions of

A, hence sufficiency follows from 2.7.5.

•

The next result generalizes 2.7.1, but the first step in the proof is the argument used in 2.7.1.

Proposition 2.7.7. on

A.

Let

QC A

is not

64

A

be a primary

absolutely convex ideal]. P

Let

'.p^-convex, Q

Let

be a ring, 'j^, £ 2 > $ 1 2 = 'j^ n f2

orders

'.p, 2-convex ideal [resp. primary P = v^

be the (prime) radical.

i-s '^-.-convex [resp.

1.-

Then if

"£, absolutely convex].

Proof. Since

P

Suppose

0 <x
(rel

P 2-convex, choose

is not

'£•.)»

Y

we conclude

x

x u

Q-

We need to show

0 <_ u £ v (rel

Raising to a large even power, we may assume The argument for 2.7.1 shows

E

G Q

v E Q

and since

:p 2 ), v E P, and

Q

x E Q. u £ P.

0 < u,v (real -Pi)-

is primary, u

£ P,

€ Q.

We now observe

p = x 2 + 2ux + v 2 = (x + u ) 2 + (v2 - u 2 ) e •£

since the first term is obviously in in

'^2.

f^

and the second term obviously

Similarly,

q = (y-x) 2 + 2u(y-x) + v 2 = (y-x+u) 2 + (v2 - u 2 ) 6 $

Now

p + q = x + (y-x) + 2uy + 2v .

and since

Q

Again, since

is Q

'£

Since

-convex, we deduce

is primary, we conclude

x ,y,v E Q, we have

p,q E Q.

Thus

p + q E Q,

p - x - v = 2ux E Q.

x E Q.

The proposition for absolutely convex ideals is a slight extension of the argument. must show (rel 2

xz E Q.

'£12)>

2

Assume

x u z E Q

as

Choose

in the

and

If and

u,v

'.p,) and

as above.

proof of 2.7.1, and

2

x z E Q.

as is easily checked. finally

0 <_ x _< y (rel

if

Thus

xz E Q

as desired.

Q C A

is a primary

p,q E -p12 pz, qz E Q,

yz € Q, some

We have 2

2

2

0 £ x u 2

(y u + x v )z E Q.

We

£ y u + x v Hence

are also as above, then

(p+q)zEQ

2

2 pz - x z - v z = 2uxz E Q, and •

'^12-convex ideal and

v^ = P

'^2~convex, it seems difficult in general to relate

'•P2-convex ideals.

z E A.

In 2.8 we give an example of such a Q = Q-, n Q2> where

be written as an intersection

We give one final generalization of 2.7.1.

Q.

Q Q

is

is both to

'^

%

£, and

which cannot p.-convex.

Again, the proof is very

similar.

Proposition 2.7.8. ideals with H i

Let

A

= (0), '£ C A

be a ring, = A/I

I

C A

a finite collection of

orders, and set

$ = A n n $

with

65

respect to the obvious inclusion

A -* IT A .

'£-convex if and only if for some

a,

Proof: to be

We certainly know some

and

I C Q .

C A ), or '

Multiplying by a suitable square in all Ig

with

Ig £ Q.

Q/I

is

is

'$ -convex.

If for all these, Q/I

failed

(Note for those

y

with

I

x

Q

is

x

x a

£ Q. x

IQ - Q, we may assume

2 )y 2 a a

x ,y

G I

for

(rel * ) . Y

(A,'£) .

Since

$ = A n n $ ,

The right-hand side belongs to

'£-convex, we have a contradiction.

part of the proposition.

2.8.

e Q,

£ Q, both sides vanish.)

we have the same inequality in so if

y a

J

Then, just as in the proof of 2.7.1, we get for all y

0 i n X 2 < Z ( n a' ctfa1 a a

Q,

This proves the "only if"

The "if" part is trivial.

•

Some Examples (1)

The ring

and

Z, and

(2)

Partially order the ring

on the line. Then p

CQ

Q C A

'£ -convex, we get inequalities

0 < x 2 < y 2 (rel $ r a — a — Ja

(0)

I

Then a prime ideal

Let

(f) C £[T]

TL has a unique order.

The only convex ideals are

TL is a semi-field.

Z[T]

as a ring of real valued functions

f(T) be an irreducible polynomial which has real roots. is a convex prime ideal, since, if

must vanish at all roots of

f, hence

0 £ p £ q

and

q E (f),

p G (f).

In fact, (f) is a maximal convex ideal.

The only ideals of

properly containing (f) will also contain some integer If such an ideal is convex, it must contain

1.

Thus

m E Z,

Z[T]

m ^ 0.

Z[T]/(f), with the

quotient order, provides another example of a semi-field which is not a field. (3) If

C

Let

(A/P) G (POR).

is convex, then

In fact, 0 <_ a £ b convex, then

66

C2

C2

If

CCA

is a subset, let

is convex, since

implies

2 2 0 £ a x

is absolutely convex.

0 <_ a 5 b

2 2 £ b x , all

C^= {a€ A|a 2 e C}. implies

0 £ a

x e A, hence if

C

<_ b . is

Now suppose

IC A

is a convex ideal.

an absolutely convex ideal. a G A.

Also, if

yvy2

ordered ring

I*

is an ideal, hence

2

ay E I 2 , all

Certainly, if y G I , then (yx + y 2 ) 2 + ^

G 1%, then

(y, + y 2 ) G I 2 , because

hence

Then

I

( 0 ) 2 = {a G A|a

is convex.

= 0}

- y 2 ) 2 = 2y 2 + 2y 2 G I,

For example, in any partially

is an absolutely convex ideal.

We can say slightly more about I 2 . If y-^^ e *2> t n e n (vi + Y^) " 2 2 y^ - y 2 = 2 v i y 2 G *• *f either I is absolutely convex or 2 G A is invertible, then

(4)

We have

Let

y^y?

G

implies

(A,?) e (POR).

Define

K = {c G A | S C = 0, some

1 £ s}.

T

and

s G A,

K

We regard the convex ideals

are convex ideals, as is easily T

and

(A/P) G (PORCK), necessarily

(5) Order Z[X]

Suppose

f(X), g(X) G Z(X).

X

same time that

Letting

f(X)

is convex.

is divisible by

T = K = (0).

0 2

X .

shows that Thus

On the other hand, (2X,X )

Consider the quotient ring expressible as

as somewhat pathological.

0 £ f(X) <_ 2nX+X 2 g(X), some

approach

2X G (2X,X 2 ), but

0 £ 1 £ 2,

K

as a ring of real valued functions on the line.

'Consider the ideal (2X,X 2 ).

since

*•

n ^ 0}

Note that if

(2X,X )

e

y ^

T = {b e AJnb = 0, some integer

T C K, and both

checked.

*'2

n + eX, n G 2,

n = 0

n G Z,

and at the 2

f(X) G (2X,X )

and

is not absolutely convex

X £ (2X,X 2 ).

Z[X]/(2X,X ) . Elements are uniquely e = 0 or 1.

Since

2 2 2 2 _ 2 (n + eX) = n +2neX + e X = n ,

the quotient ring Z[X]/(2X,X ) has a unique non-zero convex ideal, consisting of the two elements and

{0,X}.

We have

{0,X} = T = K

(as defined in Example (4))

{0,X> = (0)^ (as defined in Example (3)). (6) We continue with the example in (5), (2X,X2) C Z[X]. 2

(2X,X ) order.

2

3

4

= (4X ,2X ,X ) C z[X]

is not convex since

0 £ X

2

£ 4X 2

Thus, products of convex ideals need not be convex.

H((2X,X2) ) = (X 2 ), which is, in fact, absolutely convex.

The ideal in any

We have

Namely, if

67

0 1 P 0 0 £q(X) at the origin. or

f

q(X)f(X) e ( X 2 ) , then

and Either

q

In any case, pf

second order at the origin, hence

q

and

f

vanish,

will also vanish to

pf e (X ) .

Z[x]/(2X,X 2 ), consider the ideal

In the quotient ring 2X = 0,

vanishes to second order

vanishes to second order, both

vanishes to second order.

Since

qf

2 <= ((0): (X)), but

((0): (X)).

1 £ ((0): (X)). Thus quotients of

convex ideals need not be convex.

(7)

Order

£[X,Y]

as a ring of real valued functions on the plane.

A typical non-convex ideal is the principal ideal Certainly

X 2 ,Y 2 e H(X 2 + Y 2 ) .

XY G H(x 2 + y 2 ) .

hence

(X + Y ) .

0 £ 2X 2 + 2Y 2 + XY <_ 4X 2 + 4Y 2 ,

But also

In fact, H ( X 2 + Y 2 ) = AH(X 2 + Y 2 ) = (X 2 ,Y 2 ,XY), the

absolutely convex ideal of functions vanishing to second order at the origin. The argument in Example (6) shows that Z[X,Y].

convex ideals of

But

2 2 (X ),(Y )

are also absolutely

H((X 2 ) + (Y2)) = (X 2 ,Y 2 ,XY), hence sums of

absolutely convex ideals need not be convex.

(8)

1R[X,Y,Z]/(Z2-X-Y)

Order the ring

on the piece of the surface that is, 0 £ x,y,z. morphism

z

= x+y

Evaluation at

R[X,Y,Z]/(Z2-X-Y) -> R ,

as a ring of real valued functions

lying in the first octant of

(0,0,0)

BS

\

defines an order preserving

with kernel

(X,Y,Z).

Thus

(X,Y,Z)

is

a maximal convex ideal. (X,Y,Z) 2 = (X 2 ,Y 2 ,Z 2 ,XY,XZ,YZ).

Consider the square and

0 <_ X, 0 £ Y, we see that

(X,Y,Z) 2

is not convex.

Since

Z2 = X + Y

In fact, H((X,Y,Z)2)

(X,Y,Z ) . This example shows that products of absolutely convex ideals need not be convex. The ideal

(X,Y,Z )

on our piece of surface example, if

c:

If we regard

is, in a real sense, exactly the ideal of functions S

[0,e] •*• S

is any smooth curve,

R[X,Y,Z]/(Z2-X-Y) 2

on the entire surface

68

which vanish to second order at the origin. c(0) = (0,0,0), then

as a ring of real valued functions (3}

z = x + y in Ii?lv J, then the functions

X,Y are

For

no longer positive.

There are curves on the surface through the origin

along which

Y

X

and

c(t) = (t,-t,0).

do not vanish to second order at

In fact, with this order

(X,Y,Z)

2

(0,0,0), say,

is convex.

This

discussion illustrates how the convex ideal structure of a ring of real valued functions on a space can reflect some of the geometry of the space.

(9)

The zero set

Z

topologically a circle.

Order the ring

ring of functions on this zero set. the ideal class group group

K Q (A).

1 = y 2 + x 2 (x 2 +l)

of the equation

C(A)

2

2

in

IR^2'

2

R[X,Y]/(1-Y -X (X +1)) = A

Now,

A

is

as a

is aDedekind domain, hence

is isomorphic to the (reduced) Grothendieck

This example has been studied by Evans [27],in connection

with the problem of relating

K Q (A)

Evans showed that the ideal

(Y-1,X)

contains elements not of order

to the real K-theory of the zero set C A

is not principal, hence

Thus, the natural map

2.

Z.

C(A)

K Q (A) •*• KO(Z)

is not injective. However, suppose in defining

C(A), we replace ideals by convex ideals

and principal ideals by ideals of the form = H(Y-l), which is "principal". X 2 G H(Y-l).

hence

Y2 -l = X 2 (X 2 +1)

Then

H((Y-1),X) 2)

Namely, (Y-1)(Y+1) = Y 2 -l E X 2 (X 2 +1) € A,

(Y-1,X) 2 C H(Y-l).

Thus

shows

H(a), a G A.

Y2 -l G (Y-1,X) 2 .

(Y-1)((Y+1)-(Y-1)) = 2(Y-1) G (Y-l,x) 2 .

Conversely, the same equality

Also

(Y-l) 2 G (Y-1,X) 2 , hence

We conclude

H(Y-l) = H((Y-1,X) 2 ),

as claimed. Actually, we prefer a (PORCK) version of

C(A), that is, absolutely

convex ideals should be used in the definition. K-theory is not really so relevant.

The correct (PORCK) analogue of

should really correspond "to a "real divisor 1

turn should be

H (Z, 2/2). The group

through projective modules. zation

Ag, where

zeros on

Z.

S C A

Evans

1

The relation with real

class group" of

KO(Z)

C(A)

Z, which in

should still be approached

method was to replace

A

by a locali-

is the multiplicative set of functions with no

We will see that this localization occurs quite naturally

in the category (PORCK).

(10)

Let

K

be a finite simplicial complex, with vertices

and associated barycentric coordinate functions

X ,...,Xn-

Let

v ,...,vn A (K)

69

be the partially ordered ring of real valued functions on over Z where

by the functions I(K)

K

generated

XQ,...,X . Thus,

A^(K) = Z[X Q . . . X n ]/I(K), n by polynomials 2 X. - 1 and X- •...•Xi

is the ideal generated

1=0

for all

i0 < ... < i r

such that

(v^ ,...,v^ ) do not span an r-simplex of or In particular, if K is an n-simplex A n , then

(See 1.7, Example (9).)

Thus

A~(K)

is an integral domain if and only if

A (K) we have the order relations

K

is a simplex.

K.

In

0 £ X i <_ 1.

We have the following assertions as exercises. A^(K) •> Z = ^ ( A 0 )

(i) The non-zero (POR)-morphisms exactly to the vertices

v^

0

A •> K

of

K.

(ii) The (POR)-automorphism group of % ( A r ) group

S

which acts on

A

exactly to the inclusions of r-simplices in If

K, L

is exactly the symmetry

by permuting vertices.

(iii) The surjective (POR)-morphisms

(iv)

correspond

A ^ K ) •* ^(A 1 *)

correspond

K.

are two finite complexes,

We thus obtain a faithful embedding of the category of finite simplicial complexes in the dual category of (POR).

(11)

In this example we study the ring of polynomials

rational numbers.

It is convenient to think of

say on the real line. think of

Q[X]

Q[X]

Q[X], Q

the

as a ring of functions,

For algebraic purposes it is slightly preferable to

as a ring of functions on

Q, the real algebraic numbers.

Before beginning the discussion, we indicate two directions of generalization.

First, Q

real closure dense in

F.

could be replaced by any totally ordered field The discussion below will apply to

F, in the interval topology.

examples, no element of

70

F

F, with

F[X], provided

(This is a tricky point.

F

is

In all

will be infinitesimally small relative to

F.

That is, all intervals around F.

0

in

F

will contain non-zero elements of

However, there are examples where some intervals of

elements of

F.)

F

contain no

Secondly, the number of variables could be increased.

This would require much more work.

In fact, the study of polynomial rings

in several variables over an ordered field is, in some sense, the final goal of the whole theory.

Polynomials in one variable is essentially

zero-dimensional algebraic geometry, which is trivial from a geometric point of view. All ideals in convex ideals rings

Q[X]

We will (i) characterize the

(f(X)), and (ii) characterize certain orders on the quotient

Q[X]/(f(X)),

be given.

are principal.

f i 0.

Since this is only an example, proofs will not

We fix the weak order ty C Q[X]

is known that

throughout this example.

It

$ w , the sums of squares, coincides with the affine order,

the polynomials which assume no negative values on the real line.

Assertion 1.

An ideal

irreducible factor of Assertion 2.

Suppose h.

(f(X)) C Q[X]

f(X) has real roots.

All convex ideals

(f(X))

is convex if and only if every

is convex.

(f(X)) C Q[X]

Factor

f, say

are absolutely convex.

k i f = h. 1

k r h^ , where the

are distinct irreducible polynomials, with real roots.

real roots of

h.

by

a-••

We denote the

By the Chinese Remainder Theorem, we have an

isomorphism of rings

i=1

(hj1)

From 1.7, Example (11), we know that any order on a direct product of rings is a direct product of orders. h-

Let

Thus we are reduced to studying

k i Q[X]/(h. ) ,

irreducible. g = g Q + g 1 h i + ••• + g ^ ,

If

g(X) e Q[X], write

j,

0 <_ j £ m, be least such that

Assertion 3. only if

j > k.

We have (whence

where

degree (g.)< degree

g. i 0.

g e ? w /(h i i ) C Q[X]/(h i i ), the weak order, if and g = 0) or

j < ki

is even and

g i (a i -) > 0

for all

71

real roots

a.,

of

h..

Assertion 3 can be interpreted geometrically. zero set of

h^.

Regard

neighborhoods of to

Z..

k.-th order on

Q[X]

Z. = {a--}

be the

as a ring of germs of functions, defined on

The ideal

Z..

Let

k i (h. )

is the ideal of germs which vanish g £ (h.1)

Assertion 3 says that a function

is

k

i Q[X]/(h. )

positive in the weak order on is locally non-negative near

if and only if the germ of

g

Z..

The geometric interpretation of the weak order suggests other orders on

k i Q[X]/(h i ) .

denote by

Each root

ou . , aT. .

1>S C Q[X]/(h i i ).

ou . has two sides on the real line, which we

Any non-empty subset

A function

g £ (hu1)

S C {a.. , aT.}

belongs to

$g

is locally non-negative on those sides of those roots the subset

k i The orders P s C Q[X]/(hi )

Q[X]. The maximal orders on

by single elements of h^. y>s

if the germ of a..

g

picked out by

S c {at. , aT.}.

Assertion 4. on

defines an order

k i Q[X]/(h. )

are all quotients of orders are precisely the orders defined

{a-. , aT.}, that is, by one side of a single root of

The maximal orders on

k i Q[X]/(h. )

are thus total orders.

All the orders

are (PORCK)-orders.

(12)

If we refine the order on

convex ideal changes. algebraic numbers.

For example, let

Y

is

*PY

in Example (11), the notion of

Y C Q

be any infinite subset of

(Infinite is equivalent to Zariski dense.)

1.7, Example (6), define Let

Q[X]

<$Y C Q[x]

by

f(X) e $ y

be the usual topological closure of

Y.

if

f(y) >_ 0, all

Then an ideal

convex if and only if every irreducible factor of

one real root in (13)

As in

h(X)

y € Y.

(h(X)) c Q[X] has at least

Y.

In several examples above we have studied polynomial rings with

orders defined in terms of values of functions on subsets of affine space. There is a more intrinsic, algebraic approach to these same orders. Let with C] I

(A/P) e (POR)

{l a }

= (0). Then the natural map

define a refinement of

72

and let

s

£

s

be a collection of ^-convex ideals, A -»• IT A/I

by ]5 = A Q ^1 Q)/I ) . a a

is injective, and we can

We are in the situation of

1.7, Example (2), where

A

is now interpreted as a "ring of functions" on

the index set {a}, with value at a The order ^ C A (i) All I

in the partially ordered ring

(A/I ,ty/I) .

can also be characterized as follows,

are ^-convex

(ii) ?/I a =F/I a CA/I a (iii) f

is the union of all orders on A

satisfying (i) and (ii).

As a specific example, let A = H[X,«*'X ] , with the weak order ^ . Each

y €= IR> ' defines an evaluation homomorphism

a maximal *$ -convex ideal.

(In fact, the I , y e Ec ^, are exactly the

maximal ^ -convex ideals of A.) Certainly, quotients

on R

(n)

f] f % I = (0), and the

y € R are all isomorphic to R, which has a unique order. The

A/I

order ty C A

A -*• IR, with kernel I

is thus the affine order, of polynomials nowhere negative

. (2X,X 2 )CZ[x]

(14) Reconsider the convex but not absolutely convex ideal of Example (5). The associated primes are is not convex.

(X) and

(2,X).

However, (2,X)

This illustrates an advantage of the category (PORCK), where

associated primes are always convex. (15) Consider R[X,Y]

and the family of orders p n = '£W[X,Y, 1-Y, Y n -X],

We have P n c $ n + 1 since

n^l.

(X2, XY)

The ideal 2

is !p -absolutely convex for all n. To see this, 2

write

(X , XY) = (X) n (X , XY, Y m ) .

and

2

m

(X , XY, Y )

Y n - X = Y n + 1 - X + Yn (l-Y) . Let p = U ^ .

It is relatively easy to check that

are $ -absolutely convex if m > n.

Thus

(X)

2

(X , XY) is

'^-absolutely convex, since a contradiction of this assertion would involve only finitely many elements, hence would already be a contradiction in some :p . The associated primes of observed that

(X , XY)

(X2, XY)

are

prime

(X,Y).

We have just

has an absolutely convex primary decomposition for

each of the orders '$> . However, (X , XY) for

(X) and

has no convex primary decomposition

$> since, first, the primary component corresponding to the minimal (X) is necessarily

(X) itself in any decomposition, and secondly,

73

any

'£-convex ideal with radical

0 < X < Y

m

(rel

(X,Y)

Y m , hence

contains some

'£) . 2 (X , XY)

The ideal

is actually irreducible among

'^-convex ideals. If

but

(X^, XY) i I

since

X, since

then clearly

X £ I- n l o . 1 z

^

/ i ^ = (X),

= (X). We must then have

It follows that

(X) C / T ^ ,

/I~ must be a finite intersection 2

(X, Y - c i ) , corresponding to points on the Y-axis.

of maximal convex ideals

(X2, XY) = Ij n i 2

But now Proposition 2.6.8(c), (e) would imply that

has a

primary convex decomposition, which we know is impossible by the above discussion.

(16)

(Andrew Klapper)

'$. C A, neither

j = 1,2, prime £ ..-convex nor

We give an example of a ring

"p.-convex ideals

'p2-convex.

Note

P., such that

A, two orders I = Px n P 2

is

I = /F , so Proposition 2.7.1

does not extend to radical ideals. A = R[X] x IR[Y].

Let Then

'J^ n $ 2 = ^

P 2 = (X) x H [ Y ] . (but not P,

n

?2

^

= iyX-1] x ^ ,

x -j^, the weak order on Then

P2

is

Let

similarly, P x n P 2

Geometrically, the ring

A

n P2

E(E-l) = 0,

is

$ -convex But

would imply

'^2-convex.

can be rewritten as a quotient of a

polynomial ring in four indeterminates, 1R[X,Y,E,F]. by the relations

Px

'^,-convex).

(X,Y) ^ p

is not

x ^[Y-l].

?x = H[X] x (Y),

.p -convex (but not

'^-.-convex since otherwise

(1,Y) e P 1 n p 2 .

A.

£2 = ^

?x n P 2 = (x) x (Y). Clearly

'^-convex) and is not

Let

F(F-l) = 0,

Namely, one divides

E + F = 1,

XF = YE = 0.

The

maximal convex ideals for the weak order then correspond to the points on the two lines in F = 1,

X = 0.

IR

defined by

The variable

parametrizes the second. P x = (E,X,Y),

(17) ideals convex,

Q X ,Q 2 Q2

F=0,

Y = 0

and

E = 0,

parametrizes the first line, while

In this notation

'^ = '^[X-E],

Y

£ 2 = £ w [Y-F],

P 2 = (F,X,Y).

We give an example of two orders

'^,-convex nor

74

X

E = 1,

with is

^

= v ^ = (X,Y)

such that

^-absolutely convex, but

.p2-convex.

Since

Q

'p., '.p2 on

Q

=

Qi

Qx n

is primary with

IR[X,Y], primary is

Q2

is

'^-absolutely ne

itner

T/Q = (X,Y), we see

that Proposition 2.7.1 does not extend to primary ideals. •p1 = ? W [ X , Y , Y 2 - X ] ,

We let 2

Q 2 = (X ,Y). convex. if

* 2 = '^[X,Y,X 2 -Y].

It is not difficult to check that

In fact, Q x

is

Let

Q1

is

^-absolutely convex, where

0 £ x,y,y -x}. Similarly,

•^ = {£ e H[X,Y] |f (x,y) ^ 0

is

Q2

2

),

'^-absolutely

^

= {f eH[X,Y] |f (x,y) >_ 0

'p'-absolutely convex, where

0 £ x,y,x 2 -y}.

if

Q^CX^

Q 1 n Q 2 = (X 2,Y2,XY)

But

2 is not

'PT-convex since

0 < X < Y

(rel

'$,) and is not

'po-convex since

2

0 £ Y £ X (rel -P2). (18) a

The example in (17) still leaves open the hope that if

'^ n ' ^

convex primary ideal in some ring

'^-convex and

'^-convex, then

where

£.-convex.

Q.

is

Q

A

with

P = v^Q

Q

is

both

Q = Q-^ n Q 2

might always be written

However, we now give an example where such a

decomposition is impossible. Let Clearly

A=H[X,Y], Q

is not

£ -convex since In fact, let

^

= $ W [Y,X-Y],

'^-convex since

0 £ Y £ X (rel

Q

is

Suppose

Then

'P(S)-convex.

(In fact, Q

0 £ f £ g

as functions on

(3f/3y)(0,0) = 0.

and

is

v = i + j

S

and

vectors along the

x

v

S, and both

w

lie in

and

y

and

Q

is not

*^(S) = and we will

g e (X,Y ) = Q.

Equiva-

f(0,0) = 0

and

£(0,0) = 0. f

and

g

g

i, j

axes, respectively. f

and

'£.. n $ -convex.

We must show

w = -i + j , where

and

Q=(X,Y2).

'^(S)-absolutely convex.)

(8g/9y)(O,O) = 0.

It is obvious that

Let

'^ n $ 2 c -p(S)

We consider directional derivatives of

and

is

S = {(x,y) e IR2 | 0 £ y, y 2 £ x 2 } , and let

lently, g(0,0) = 0

the vectors

'p^

'P 2 ). However, Q

0 £ Y £ -X (rel

{f e H[X,Y] |f (x,y) >_ 0, (x,y) e S}. show

? 2 = ? W [Y,-X-Y].

with respect to

are the usual unit Since

vanish at

0 £ f £ g

on

S,

(0,0), we must have

0 <^ (0,0) £ ^ | (0,0) 3v 3v and 0 <^ (0,0) £ ^ | (0,0) . 8w 9w But then (suppressing evaluation at (0,0))

75

- 9+

3x

+

3y — 3x

3y

3x

and Q <

3f _ -3f

which implies by adding that The ideal R[X,Y].

<

-3g

t

3g

=

-3g

2 -r— (0,0) = 0, as desired. oy

Q = (X,Y2) C R[X,Y]

is irreducible among all ideals of

Thus there is certainly no decomposition

is '£.-convex.

76

3f

Q = Q1

n

Q2

where Q.

Ill • Localization

3.1.

Partial Orders on Localized Rings Let

A

be a ring and let

T c A

be a multiplicative set, that is,

a subset closed under products and containing the unit denote the ring obtained by "inverting elements of

T".

1 £ A.

We let

A

That is,

A T = {a/t|a e A, t e T } / ~

where

a^tj ~

a

2/

t

if

2

there

is

t e T

with

t ( a 1 t 2 - a 2 t 1 ) = 0.

Sums

and products are defined by

tit 2

There is a canonical ring homomorphism such that

i T (t)

is invertible for all

i™:

t ^ T

A -*• A™,

i T(a) = a/1,

and, in fact, (A T ,i T )

is

the universal morphism in the category of rings with this property. We ask if partial order

A ^T

has a partial order, say on

AT

such that, first, i T :

preserving and, secondly, the triple

Proposition 3.1.1.

(A/P) G (POR), is there a natural

Define

^T

c

(A^ ^ T ) ,

AT

(A/£) -> (AT,?PT) iT

is order

is universal in some sense?

by

!PT = {a/t|ats2 e p , some

s e T}.

Then

(a) P T (b)

iT:

is an order on

A™.

(A,P) -* (A-., ^ )

is order preserving.

77

(c)

Given a morphism

(A,$) -* (A'/p1)

f:

is invertible in A

1

for all

in (POR) such that f(t)

t e T, then there is a unique

(A T /£ T ) -* (A 1 ,? 1 ), such that

morphism in (POR), g:

(A,?) - ^ > (AT,?T)

( commutes. (d) !PT = (i^) ^ , the weakest order on A~ such that

i T is order

preserving.

Proof.

We first show that our definition

a/t E $ T

does not depend

on the choice of representative, a/t, of the element of A,p. at't" = a'tt"

then

for some

t" E T.

We thus have, for any

If a/t ~ a'/f s E T,

Cat)(t')2 (t") 2 s2 = (a l t')t 2 (t") 2 s 2 .

ats 2 E ? , then

Thus, if (a)

a ' t ' ( s t t " ) 2 € $ , hence

a f /t' E ?

r

The equations in A

(alt2 + V l

5 1

^

s

l S2

=

show that if a ^ t j , a 2 t 2 E ? T , then Also, since

(a/t)

2

2

= a /t

2

2

Vl

S

l *2

a^t^

and a t

2

2

+

¥ 2 S2 * !

&2/t2, (a^tj) (a 2 /t 2 ) E ? 2

= a t -l

2

r

E $, we see that P T

contains all squares. Finally, suppose

a/t E ^

T

and -a/t E !PT.

ats 2 s 2 £ 0 £ ats 2 s 2 , hence

Then

^T C N 1

Say, ats 2 £ 0 £ ats 2 .

at(s s + ) 2 = 0 and a/t = O G i ^ .

Thus

is an or< er

^*

(b) is trivial. (c)

We already know that there is a unique such morphism

in the category of rings. check that

78

g

Namely, g(a/t) = f(a)f(t)~

is order preserving.

1

1

E A .

g: A™ -* A' We need to

Lemma 3.1.2. x,y

invertible.

(A 1 ,? 1 ) G (POR), and let

Let

Hence

A ! , with

in

0 ± y"1 ± x" 1.

Then

a/t G ^L,, then for some

The lemma implies our result, since if a t s 2e p .

0 < x f y

f (a)f (t)f (s) 2 G $'

s G T,

f (a)f (t)f (s)2(f (t)"1£ (s)" 1 ) 2 =

and

g(a/t) The lemma itself is easy, since

3.2.

y~

= y(y~ )

y

- x~

Sufficiency of Positive Multiplicative Sets In general, if

T c A

other than elements of

T

is a multiplicative set, there are many elements which are invertible in

T = (a G A|ab G T, some

then if

and

T

is exactly the set of elements of

ab G T, then

then

1

b/ab = a.' G A T

axs = ts G T

for some

A .

b G A> ,

A

invertible in

and, conversely, if

s G T.

In fact, if

Note that

T

A^.

First,

(a/1) (x/t) = 1/1 G A T , is a multiplicative

set.

Proposition 3.2.1. TCA

as above.

If

T

f

Let

(A/p) G (POR),

T C A

a multiplicative set

C A

is any multiplicative set with

T C T' C f,

there is a natural isomorphism in the category (POR)

defined by i T ^ r ( a / t ) = a/t.

Proof. clear that

It is routine that i T T , CPT) c ? T i -

morphism in (POR).

course.)

is a ring isomorphism.

It is not quite so clear that

However, since elements of

the universal property of -* (A T ,^ T )

iT T,

(AT,^T,)

is a morphism in (POR).

T

1

It is also

i~ T ,

is a

are invertible in

in (POR) implies that

i^ T ,:

AT,

(A^,^,

(This could also be proved directly, of E

79

Turning the argument around, we see that there are multiplicative sets smaller than

T with the same localization in (POR).

Proposition 3.2.2.

Let T C A be a multiplicative set

(A/P) e (POR).

Define T + = {t t G T n T 2 = {t2|t e T}. T c T+ c T

Then

are multiplicative sets and the natural maps

(A i 2, ? T 2 ) ~* (AT+, ? T +) ~* (AT, ^ T ) i li 11 are isomorphisms in (POR).

Proof.

The first assertion is trivial and the second follows from

Proposition 3.2.1 and the observation that

T C T + C T C (T 5 ).

D

Thus, when localizing in (POR), we can get by with positive multiplicative sets.

3.3.

Refinements of an Order Induced by Certain Localizations Suppose

i T : A -> A T

is injective, (A,!p) G (POR).

Then there is a

natural refinement of *£, namely, the contraction of *PT to A, that is, ^ c i*(iT)^p = f T n A = {a e A|as 2 G f, some

It is easy to see that *$„ n A = *p + n A = !B 2 IT I 3.2.2 of the preceding section. then

T

+

If A

= ^ , the strictly positive elements of A.

This is exactly the derived set Dv£

80

A, say from Proposition

is an integral domain and T = A - {0},

^T+ H A = {aGAlpaG'P

domains.

n

s G T} .

of $

some

In this case

p G "P+ } .

discussed in 1.6 for integral

More generally, if

(A,^) G (POR)

is arbitrary and

plicative set of non-zero divisors of We conclude that any order where

T C A

iT:

coincides with

$

is the multi-

A -• k^, is injective.

has a natural refinement

is the set of non-zero divisors.

der-ived order of

*p, = ^

We will call

n A,

^,

the

(hopefully remembering that the derived set

D^

*p , only in special cases).

Proposition 3.3.1.

Proof.

^

A, then

T

If

If

T C A

(A,$) € (POR), then

CPd)

= 1^.

is the set of non-zero divisors, i^:

A -• A T

the

canonical map, one has (as a special case of a more general principle) that (iT) i*(iT) $ = (iT) p C A T . i

*

l

i

*

i *

Remark.

Now intersect with

Note that if

A

A. Q.

of

D

is an integral domain

the following "simpler" definition of an order fractions

A.

i

A

(A,*P) G (POR), then

^/-Q^ °n the field of

does not work:

The condition doesn't always say the same thing for the equal elements and

as/bs

2.4.

of

A, . unless the order

Convex Ideals in The morphism

enables to

(A,P)

iT:

and.

^

is sufficiently strong, say

$ = ^

(A^ ? T )

(A,p) -> (AT, !f T ),

T C A

a multiplicative set,

us to define correspondences between ideals of

I C A, assign the ideal

a/b

I T = IA T C A^, and to

A

and

A™.

Namely,

J C A T , assign the ideal

i^(J)-C A. If *P

(A/P) G (POR)

convex, since

i^

and

J c AT

is

$T

is a (POR)-morphism.

convex, then

i^.1^) C A

is

The other correspondence also

preserves convex ideals. Proposition 3.4.1.

If

I c A

is ^-convex, then

IT C AT

Moreover, there is a natural isomorphism in (POR), induced by

is iT:

PT A

81

where

IT:

(A/p) -* (A/I, y>/I)

Proof.

One knows from commutative algebra that

a ring isomorphism

(A/I) ,„,. 25, A^/IA

with forming quotients." to be

is the quotient projection.

(Namely,

.

iT:

A -»• A T

induces

That is, "localization commutes

Acp/IA^

has the right "universal property"

(A/I) (T>jO Consider the commutative diagram

A/I

Following the arrows clockwise from be extended, first to order

$

that =

A, we see that the order

$/I C A/I, then to

c (A/I)^..

can also be extended going counterclockwise from

*PT C A,j, can be extended under is

IT

flVl).^

^ C A

A.

can

Thus the

It follows

TF, which can only happen if kernel (IT)

1$T- convex.

The equality of orders weakest extensions of

Remark.

CP/I)^,-^ = ? T / I T

follows since both are the

$ C A, under either i • TT or

The result that

1^

is

i ° iT»

*£„,-convex could also be proved by

direct computation.

Proposition 3.4.2. (a) spondence If

J

If

J C AT

J ^ i

is an ideal, then

(J)

is injective from ideals of

is a primary ideal of (b)

If

Q C A

disjoint from

Q «• QA T T

A ^ then

i^. (J)

is a primary ideal and

a primary ideal, v ^ H T = 0, correspondence

J = i~ (J)AT.

Thus the corre-

AT

to ideals of

is a primary ideal of

Q n T = 0, then

/

v Q A T = /QA T , and

AT>

A.

QAq, C A ?

1

Q = i~ (QA T ).

is

Thus the

is a bijection between all primary ideals of

and all primary ideals of

A.

A

The correspondence commutes

with the nil-radical operation and preserves inclusions, hence also provides a bijective correspondence between prime and maximal ideals of

82

A^. and

prime and maximal ideals of I C A, then

A, disjoint from

T.

(If

I n T t 0, any ideal

IA ? = A r )

Proof.

This is a result in commutative algebra and doesn't really

involve (POR).

Of course, we get a result in (POR) as a corollary by

replacing the word "ideal" in the proposition by "convex ideal."

This

is justified by Proposition 3.4.1 above.

3.5.

•

Concave Multiplicative Sets (A/p) e (POR).

Let p 6 S

implies

is convex.

q e S.

A subset

S C A

equivalently, S

is ooncave if

is concave if and only if

Y C A, we denote by

S(Y)

Thus, S(Y) = f l S a

containing

Y.

S S Q (Y)

Given any subset

the smallest concave multiplicative set containing

where the

S

a

run over all concave multiplicative sets a

We call

We construct

Then

A- S

Arbitrary unions and intersections of concave sets are concave.

We are interested in concave multiplicative sets.

Y.

0 £ p £ q,

S(Y)

S(Y)

the concave shadow (or, simply, shadow) of

explicitly as follows.

Y.

Let

o°°

is the smallest multiplicative set containing

is the empty product.)

Y.

(1 G S Q (Y)

Let

S n + 1 (Y) = S Q (S n (Y) U { q | 0 < p < q, p € S J Y ) } ) .

S m (Y)

Then each

is a multiplicative set, S m (Y) C S

is the concave shadow of

Proposition 3.5.1. Let in

(A/P) -> (A1 ,$•)

f: A

and

morphism

t E T, then g:

(

A

S(Ty

(Y)

and

S(Y) = U S n ( Y )

Y.

Let

T C A

be a multiplicative set

(A/£) e (POR).

be a morphism in (POR) such that wherever f(s)

is invertible in A?

^s(T)^ "* (

^')

such that

A1. the

0£t£s

Then there is a unique dia

gram

83

commutes.

Proof.

First we prove a Lemma.

(S(T)) 2 C S(T 2 ) C S(T + ) C S(T).

Lemma 3.5.2.

Proof.

(

m (T))

in

(S Q (T)) 2 = T 2 = S Q ( T 2 ) .

S Q (T) = T, so

and s

Only the first inclusion is not obvious.

2

2

C Sm(T ).

S m (T)

(Sm+1(T))

S

Any element of

2

C Sm+1(T ).

Thus

(S(T))

2

S(T) = U S (T) n n

Suppose, inductively,

m+1(T)

is a product of elements either

or larger than positive elements of 2

But

S m (T).

It follows that

2

C S(T ).

D

Returning to the proposition, from 3.2 we have canonical isomorphisms in (POR) (we suppress the symbols for the orders)

A

We now replace suitable

g:

( A S m' ^SfTp

S(T 2 ) ^ A S(T + ) ^ A S(T) '

by

^AS(T+V ^S(T+ )^

(AS(T+^> ^SfT+'P "*" (A* $'^ '

But

Siven

a n d look for a a

positive multiplicative

+

T ,

set

S(T + ) = {s|o £ t £ s, t € T + } ,

as one verifies easily by showing the set on the right to be a concave multiplicative set.

Thus, the hypothesis of the proposition clearly gives

( A s ( T + ) , ? S ( T + ) ) •+ (A',P f ), as desired.

g:

3.6.

The Shadow of Let

1

(A/p) G (POR).

set, we study

As an important example of a concave multiplicative

S(l) = {b € A|l £ b } , the shadow of

First, an element

84

D

y G A

is invertible in

1.

Acr .

if and only if

1 <_ xy

for some

x G A; that is, if and only if

1 G H(y).

In a commutative

ring, any element not contained in some ordinary maximal ideal is a unit. If

(A,*p) G (POR), an element

ideal if and only if making all such

y

y e A

1 G H(y).

is not contained in any maximal convex

Thus, passing to

Acri.

has the effect of

units.

Secondly, the natural map

icri>.:

A -• A c r i .

oil)

need not be injective.

Z>{i.)

In fact, kernel (i c r i .) = {a G Alab = 0, some

If

1 < b, then —

0 < a —

implies

a

= 0.

kernel(ig.,.), which one sees by

2

(a.^ + a 2 )

pathological.

G

a1,a2

2 expanding

a G kernel(i c r ,O ^HlJ

< a b , so —

Consequently, 2a1a2 = 0 for any

b, 1 < b} .

and

^

- a2) .

It is the ideal

K

We regard this kernel as somewhat

of

2.8, Example (4).

We now give an alternate characterization of semi-fieIds. Proposition 3.6.1. if and only if

igriy

such a case, Ag,..

Proof. A

Let A

(A/p) G (POR).

"*"Asfl1

is

in

Then

3ective

is the field of fractions

Semi-fields were defined in 2.3.

is an integral domain, so

paragraph above.

i,.,...:

and

A. .

If

i s a fielc

^sfl") of

^#

In

A.

(A,$)

is a semi-field,

Moreover, directly from the definition of semi-field and A

are invertible in

Thus, A S ( 1 ) = A ( o ) , a field.

Conversely, if

igQ-j = A ^ A g , .

then all non-zero elements of

A

is injective and

are invertible in

paragraph above, this says exactly that

We point out that any convex proper S(l).

is a semi-field

A -*• A« <-•.-. is injective by the second

the first paragraph above, all non-zero elements of AS(1).

(A,$)

(A/p)

If

gflV

S

Q ^ is a field, Bv tne

fi

is a semi-field.

ideal of

Thus, at least the primary convex ideals of

bijectively with primary convex ideals of

A

A

A

•

is disjoint from

(A,sp)

correspond

A

( g(iV ^ s f l ^ "

(A,?) G (POR), we define the Jacobson radical of

(A,?), R(A/p),

to be the intersection of all maximal convex ideals

85

n Q = maximal convex ideal

We have the following analog of a classical result.

Proposition 3.6.2.

Let

IC A

be a convex ideal.

Then the following

are equivalent. (i) (ii) (iii)

I C R(A,?) 1 + (a/b) is invertible in For all

Proof.

A

for all

a e I, b e S(l), there is an

a e I, b e S(l).

x e A

(ii) and (iii) are clearly equivalent.

with

1 £ x(b+a).

We will deduce the

equivalence of (i) and (ii) from the classical result in commutative algebra.

Namely, one first checks that

R(A,P) = R(Ag. ., ?

S

n

Q O

A

and

using the correspondence between maximal convex ideals of (A

S(l)'»sa)>-

ThuS

ICR

(A'?)

Secondly, one checks that in (POR).

^

and only if

(A s ( 1 ) , ? g ( 1 ) ) =

IAg ( 1 ) C R(Ag ( 1 ) , ? s ( 1 ) ) . (CAs ( 1 ) ) s ( 1 ) .<$ S (l)> s ( 1 ) )

In particular, those elements of Ag,_. not contained in any

maximal convex ideal are already invertible in Now, if

IA c r .

then the elements

1+ (a/b),

Conversely, if all

a/b e IA c r i ., Thus

1+ (a/b)

= b, for suitable

1 = b/b = q/b + pa 2 /b is impossible since

and

are not in any maximal convex

(i) =» (ii). are invertible, a e I, b e S(l), but

a £ Q, for some maximal ^-convex ideal 2.2.4, 1 < q + p a

A«,-v.

is contained in all maximal convex ideals of A c

ideal, hence are invertible.

86

(A/P) and

Q C A, then

q e Q,

1- (pa2/b) = q/b

pelf.

1 e H(Q,a). Computing in

is invertible in

q/b e Q A « Q N , a proper convex ideal.

By Corollary Acri. ,

A g . ^ , which

Thus (ii) =*• (i) .

•

3.7.

Localization at a Prime Convex Ideal Another important concave multiplicative set S

a prime convex ideal

Q, that is, S = A - Q.

localized partially ordered ring unique maximal ideal

Q A ^

More generally, if = n (A - Q )

A - U Q

We denote by

( A ^ ,ty( Q O the

(A. ~,ty.~ ) . A.^. is a local ring, whose

is in fact convex.

{Q }

is a set of convex prime ideals, then

is a concave multiplicative set.

ideals of the corresponding localization Q

is the complement of

The maximal convex

i^ffn \}> ^(fr\ T.0

are tne

A,r Q .,., (at least assuming there are no non-trivial inclusions If

A

Q

is an integral domain, then for all multiplicative sets

A^, is a subring of A.-., the field of fractions of A.

C Q).

T C A,

The following is

also an analog of a standard result in commutative algebra.

If

Proposition 3.7.1.

(A/j>) e (POR), A

an integral domain, then, in

convex ideal

Proof. if

(A s(1) , P S Q ) )

x/b G A.Q.

x/b = x Q /b 0 , (since

x

0/b0

c

(A(Q)> ^(Q))

belongs to all

b Q ^ Q.

Then

A,Q. C A

S(1) C A

'Q-

Conversely,

, we can write for each

1 E H({b Q}).

We may as well assume

Q,

b

e

= x Q b 0 /b 0 ), hence we can write

1

± *, PQ- bQ1=1

for suitable

pn e p vi

X

l

X

l

{Q.} , a finite subset of the maximal x 1
convex ideals.

which shows that

and

x/b G A gflV

Finally, suppose convex ideals

since

Q.

Then

1 £ b,

0 £ x/b

0 £ xbb

1 e H({b Q }), hence for suitable

in pQ

in

(km)> ^(Q)^

for a11 maximal

(A/p), for suitable

b~ £ Q.

Again,

G ?,

87

1 <

0 £ xbb 1

But now

3.8.

in

2 p b n = b1 ' i=l i i

0£xb(b')2

(A,^), hence

and

0 <_ x/b

in

(Acri^,1

Localization in (PORCD The results in this chapter extend to the category (PORCK), as we verify

in this section.

In fact, in (PORCK) certain desired results become true,

that are not true in (POR).

Proposition 3.8.1.

Proof. i' b i

is a

Suppose

a multiplicative set.

(a1/b]L + a 2 /b 2 ) (x/b) = 0 G A T ,

2 2 a b s i i i^^ b i s i^

by

assume

w e ma

y

as wel1

b^ = b 2 , by a similar trick. t G T

with

a,xt = a o xt = 0. 1

a^ G
and

Then

•

1 1

(A,$) G (POR),

an absolutely convex ideal.

implies

(A-,, ?„) G (PORCK).

1

1 1

Let

Replacing

Also, we may

Now, (A,!})) G (PORCK)

Thus, (a./b.)(x/b) = 0 e L

Z.

I C A

assume

a i /b i G S P T .

The hypothesis then implies that there

(a1 + a2)xt = 0 G A.

Proposition 3.8.2.

of

(A,$) G (PORCK), T C A

(A T ,? T ) G (PORCK).

Then

a

Let

T C A

IA T

a multiplicative set,

is an absolutely convex ideal

AT.

Proof.

Actually, we don't have to prove this, since we know that

IA T

is convex and that in (POR) there is an isomorphism

By Proposition 3.8.1, the left-hand side is in (PORCK), hence

so is the

right-hand side.

•

Proposition 3.8.3. set

S(l) = {b G A11 < b}

Let

(A,$) G (PORCK).

consists of non-zero divisors.

is always injective in (PORCK) and the order refinenent

Proof.

Then the concave multiplicative

p

bx = 0,

0 < 1 < b, then

L

n

• A -» A

admits a canonical (PORCK)

? s ( 1 ) = P s ( 1 ) n A.

If

Thus

l»x = x = 0.

Proposition 3.8.4. set and

I C A

Let

(A/p) G (PORCK),

an absolutely convex ideal.

S C S(l)

Then

a multiplicative

I = IAg n A.

Thus, there

are natural bijective correspondences between the absolutely convex ideals of

(A,?),

(A, ? s n A)

Proof. _________

and

(Ag, $ s ) .

In general, IA^C n A = {a e Alsa G I, some i

is absolutely convex, s a £ I,

s

0 _: 1 __ > "then

s G S}.

la = a £ I.

But if

I

This proves

IA S = I. The second statement follows, since by 3.8.2 convex, hence

I = IA g H A

Lemma 3.8.5.

by some

Let

2

q G Q,

q1 E I.

Then

ab G I,

a , we assume

I

is

ty

absolutely

absolutely convex.

I C A

•

absolutely convex with

/f = Q

is a primary ideal.

a £ /f = Q.

a G $.

p E p.

^s n A

(A/J3) G (POR),

Let

a maximal convex ideal.

Proof.

is

IAC

We must prove

b G I.

Replacing

a

Now, 1 G H(Q,a), hence by 2.2.3, 0 £ 1 £ q +pa,

Raising to some large power, we get

But now, (q1 +p ! a)b G I

and since

I

0 <_ 1 < q f * p f a ,

is absolutely convex,

1 • b = b G I.

•

We can apply Lemma 3.8.5 and Proposition 3.4.2 to define symbolic powers in (POR).

Let

(A,Q. , ^ ( - Q O

G

(A,!p) G (POR), (POR)

and

absolutely convex ideal

Definition 3.8.6.

QA^

Q C A

is a maximal convex ideal.

AH(Q n A. J

Q ^

a prime convex ideal.

c A

= AH(Q nA

^-absolutely convex, primary ideal, with

3.9.

and certainly

) n A. yQ

Then, Q ^

Then We have the

AH(QnA

C A

) = QA,

is a

= Q.

Applications of Localization, I - Some Properties of Convex Prime Ideals In this section we first give an alternate proof of the fundamental

result that maximal convex ideals are prime (Corollary 2.3.5).

The argument

is due to Andrew Klapper and is more elementary than the proof given in Chapter II, where we appealed to the characterization of the radical of a convex hull (Proposition 2.2.3).

We then use localization to give a

89

.

quick proof that the nil radical of a convex ideal is an intersection of convex prime ideals (Corollary 2.3.7).

Finally, we use localization to

study minimal primes. We begin by pointing out that the existence of maximal convex ideals in any

(A,'-p)

Next, if

I C A

(A/1, '$/!) ideals of Q C A

is a trivial Zorn's lemma argument (Proposition 2.3.1). is convex, then the construction of the residue ring

is elementary, as is the correspondence between A/I

and

'p-convex ideals of

is maximal convex, then

A

(A/Q, 'p/Q)

containing

p/1 -convex

I.

Thus, if

has no nonzero proper convex

ideals.

Proposition 3.9.1. "^-convex ideal, then

Proof. (Klapper)

(A,1^) e (POR)

If

A/Q

Q C A

is an integral domain.

(A,'£)

domain.

First notice that if

is a maximal

Thus

By the paragraph above, we

that if

Q

0 f p e '|5, then

is prime.

are reduced to showing

has no nonzero proper convex ideals, then

(0 : p) = (0). Namely, if also

and

A

(0 : p) C A

0 <_ x <_ y, then

is an integral is convex, hence

0 £ xp £ yp, so if

yp = 0,

xp = 0. Now, suppose

principal ideals P 1 ,p 2 ,q 1 ,q 2 ^ f P

ab = 0, (a), (b)

with

a,b / 0.

a

and

b

are not units.

The

cannot be convex, hence we can find elements

p x + p 2 G (a), q x + q 2 G (b) , but

Multiplying, (p, +p 2 )(q, + q 2 ) (0 : p) = 0,

Then

=

0* hence

p.q- = 0.

p i £ (a), q.. £ (b) .

But this contradicts

p G 'J>.

•

Our next observation is that if set, then the construction of

(A/J5) G (POR),

(A^/p^) E (POR)

S C A

a multiplicative

in 3.1 is elementary, as are

the results in 3.4 concerning the correspondence between prime

'£c-convex o

ideals of

S, and the

Ag

and prime

f-convex ideals of

A

disjoint from

permutability of localization and residue ring constructions. just proved, we know there exist prime

'pg-convex ideals of

enables us to shorten certain arguments. Proposition 3.9.2.

Let

A g , and this

For example, we can reprove 2.^.1.

(A,'£) G (POR),

I C A

a convex ideal.

/T = n p , the intersection taken over all prime convex ideals

90

By the result

P

Then

D I.

Proof.

Passing to

(A/I, *-p/I), we may assume

trivial step is showing that if ideal

P,

f £ P.

ideal of {f }

n,

f

£ 0

all

Consider the localization

A-^ corresponds to a prime

I = (0). The non-

n, then for some prime convex (A,,, 'P~).

'^-convex ideal of

Any prime A

'p--convex

disjoint from

as desired.

•

The use of localization also aids in the study of minimal convex primes.

If

(A,^) e (POR)

of multiplicative sets S(l), the shadow of any such prime

S

S C A

I

is a convex ideal, consider the family

with

S n I = 0.

1, are such sets.

is contained in a maximal

'$„ -convex ideal containing

P , with Thus

and

I C P Q C A- SQ

S Q C A - PQ

and

Since our original

S .

Consider

gives a prime

IA g

(A - P Q ) = 0.

could have been

have proved the following

(POR)

and

Zorn's lemma easily implies that

(such primes do exist since

in

S

For example, {1}

(Ag , '£s ) . Any 'p-convex ideal A g /IA g

By maximality of A-P

t 0).

SQ,

for any prime

S Q = A - PQ.

P 3 I, we

extension of a familiar result in com-

mutative algebra.

Proposition 3.9.3.

If

(A,'.p) e (POR)

ideal, then the minimal prime ideals of 'p-convex.

3.10.

Any prime ideal containing

A I

and

I C A

containing

is a I

£-convex

are always

contains a minimal such prime.

Applications of Localization, II - Zero Divisors Our next application of localization will be to zero divisors.

The

results are most satisfactory if attention is restricted to the category (PORCK). If

A

is any ring, let

and let

N

denote the non-zero divisors.

algebra is that

D

D

denote the set of zero divisors of

A

A standard result in commutative

is a union of prime ideals.

More precisely,

D

is the

union of those prime ideals which are minimal over ideals of the form (0 : a ) , a e A.

Since

(0:1) = (0), the minimal primes of

The proof of these assertions is as follows. union of the ideals

A

are contained in

Certainly

(0 : a ) , so if we prove that whenever

P

D

is the

is a prime

91

D.

minimal over

(0 : a ) , then

P C D, we are finished.

maximal multiplicative set disjoint from x £ P-D

(0 : a).

But If

S = A- P

is a

P £ D, choose

and observe that

(J S x is a multiplicative set strictly n>0 S, but disjoint from (0 : a ) . This contradiction proves P C D.

larger than

Using the results that if and that if

(A,'£) e (POR), any minimal prime is convex

(A,'p) G (PORCK), all ideals

minimal primes over

(0: a)

are convex, hence so are

(0 : a ) , we see that we have proved the following.

Proposition 3.10.1. (a) then

If

(A,'p) G (POR)

and

P

is a minimal convex prime ideal,

P C D. (b)

If

(A,'p) e (PORCK), then

ideals minimal over ideals

D

(0 : a ) ,

is the union of all convex prime

a £ A.

•

For the sake of argument, we give two alternate proofs that in (PORCK),

D

is a union of convex prime ideals.

These arguments are

analogous to standard arguments for commutative rings. First, if fact, if

0 <_ a £ b

then so is such that let

I

(A,'£) G (PORCK), then the subset

a.

and

be = 0, then

ac = 0, so if

For example, (0)

be a maximal such ideal.

Otherwise, let

ab G I, y f. D.

0 £ y 2 n £ q" + br"

is such an ideal.

Then

a,b £ I.

I = /F.

Choose

By 2.2.4, we can write

for suitable

n _> 1,

convex set, so

x £ D

xy £ D

and hence either

We claim

0 £ x

q f ,q" G I, But

is a zero divisor,

(A/p) G (PORCK).

2n

I

and

£ q' + ar f ,

r',rM G A. and

or

y £ D. D

I C A

is prime.

x g D

I C D

The second proof also exploits the fact that

D

Then

is a

is convex if

This is, of course, equivalent to the statement that

is concave.

sense that if

b

In

By Zorn's lemma,

x G AH(I,a),

0 £ ( x y ) 2 n £ q = (q1 + ar 1 )(q"+ br") G I.

N = A- D

is convex.

Now consider the family of absolutely convex ideals

I C D.

y G AH(I,b),

D C A

N

is also a saturated multiplicative set, in the

st e N, then

s G N

and

t G N.

We digress a moment to establish a lemma of independent interest. This lemma clarifies in some abstract sense how suitable localizations of partially ordered rings will have a "good" maximal ideal spectrum.

92

Lemma 3.10.2. set.

Let

(A,£) e (POR),

Then any maximal ideal in

Proof, prime ideal

Q = PA~

for some ordinary

P C A, maximal among those disjoint from q G P, hence

q £ P, the ideal

a' e A, then

s

P

and

P+ (q)

= p" + a"q

Remark.

Q

S.

Let

0 < q < p G P.

are convex.

intersects

for some

2 2? 0 £ s = p" + a"q £ p M + a"p a contradiction.

A~

a concave multiplicative

is '.pg-convex.

If Q C A,, is maximal, we know

We will show If

Ag

S C A

p" e P

S.

s = p 1 + a'q,

If

and

p' € P,

1

a" = (a ) G A.

But

? and since

S

is concave, p" + a"p

e P n S, •

As a corollary of 3.10.2 we deduce that all semi-units in

are actually units, since the semi-units belong to no maximal convex

ideal, hence, by 3.10.2, to no ordinary maximal ideal.

This result could

also be verified by direct calculation. We now complete our second alternate proof that if D

is a union of convex prime ideals.

By the above discussion, this

follows immediately from the next lemma, by taking

Lemma 3.10.3.

If

(A,'p) e (POR), then

multiplicative set if and only if

Proof.

A- S

Ag.

S c A

S = N = A- D.

is a saturated* concave,

is a union of convex prime ideals.

The if statement is trivial.

and concave, consider

(A,'p) e (PORCK),

Conversely, if

S

is saturated

By the Remark above, the set of non-units in

Ag

is the union of the maximal convex ideals, hence the set of elements

of

A

which are non-units in Ag

ideals which are disjoint from

S.

is the union of those prime convex But the assumption that

simply says that the set of elements of A exactly 3.11.

S

is saturated

which are non-units in

is

Ag

A- S.

•

Applications of Localization, III - Minimal Primes, Isolated Sets of Primes, and Associated Invariants. Suppose

A

is a ring, P C A

a minimal prime ideal.

result in commutative algebra is that associated to

P

A standard

there is a minimum

93

P-primary ideal, Q = kernel (A -• A . p . ) . decomposition, then

P

If the ideal

(0)

has a primary

is necessarily an associated prime of

(0)

and

Q

is necessarily the associated primary component. If

(A,']5) e (POR), then any minimal prime

Q = kernel (A -*• A , p . ) .

P

is convex, as is

This remark establishes part (c) of Proposition 2.6.8,

which was postponed in Chapter II. In the interest of completeness, we prove the classical result. if

Q

1

is P-primary and

s £ P.

Thus

1

x G Q , and

xys = 0, some we must show

x E Q = kernel (A -• A . p O , then

f

s £ P.

Since also

P C I/Q.

set disjoint from

Q C Q .

Let

Secondly, if

(0). But if

Then

yep

and

is a multiplicative set disjoint from If

A

is a ring and

xy G Q,

sy £ P, we conclude

S = A - P.

S

(0) = n Q.

If

P

Js

s

component

Q

associated to

A

and

P

Finally,

is a maximal multiplicative

y11 £ Q

all

U Sy11 n>0

n, then

is a finite intersection of primary

We then have

is a minimal prime of

u £ P, then

x e Q.

S,

(0) = n Q^ A g c A .

this intersection reduces to the intersection of those Q.AC = A c .

for some

(0), contradiction.

ideals, then for any multiplicative set

since otherwise

xs = 0

First

Q.AC

with

kernel (A -*• A c ) = s

Of course, Q . n s = 0,

0

Q--

J

Q.ns = 0

S = A - P, then only the primary

can be disjoint from

S.

We state these results in our partially ordered context in the following

Proposition 3.11.1.

If

(A/P) G (POR), [resp. (PORCK)], I c A

convex ideal [resp. absolutely convex ideal] and prime over

I, then

P

is a minimal convex

Q = kernel (A -• A, p ./IA, .) is the smallest P-primary

convex [resp. absoltely convex] ideal containing

I.

Moreover, if

a decomposition as a finite intersection of primary ideals, then primary component belonging to

Remark.

Q

has is the

P.

•

If

(A,'p) e (POR)

and

PC A

convex prime ideal, consider the family of convex ideals This family is closed under the operations

(and also, of course, I, n I ) .

94

I

We can apply 3.11.1 to construct convex primary ideals associated

to certain sums and products.

/f = P.

is a

Given any convex

I

is a fixed such that

H(I 1 + l~) I

with

and

H(I,I 2 ),

/T = P, Proposition

3.11.1 gives the minimal P-primary convex ideal containing kernel (A -• A, J I A , J take

I = H(I, + I 2 )

, which we abbreviate

or

H(I,I 2 ), if

I(P). For example, we could

/T7 = P.

an analogue in (POR) of the symbolic powers of kernel (A -• A. p ./H(P n )A. p O .

I, specifically,

This construction includes P, by taking

P^ ^ =

As an exercise in the use of the explicit

construction of a hull, following:

H(X) = (j H.(X), given in 2.2, we state the , . , . , . i>0 1 I = H ( P ( n V m ) ) , then I(P) = P ( n + m ) .

If

There is a completely analogous construction for absolutely convex ideals, using absolute hulls, A H ( I 1 + I 2 ) ,

AH(I 1 I 2 ), etc.

As final applications of localization, we discuss the semi-Noetherian case.

If

(A,'£) E

(PORCK)

any multiplicative set

is semi-Noetherian, then so is

Suppose

is a multiplicative set, I C A Then the associated primes of P C A

Proof.

(A,'£) E (PORCK)

I = (0). If

I Ac

in

Ac

are exactly those primes

I

disjoint from

P = (0 : x) C A

is prime and disjoint from

PA g = (0 : x)A g = (0 : x/1) C A g .

prime of

in

Ag.

P = AH(y1...y,)> with

Conversely, suppose

(A/p)

P

exactly the ideal

PA C

S.

I, then the ideal {x E A|xy E I

some

S, then it is

PA g

is an associated

PA g = (0 : x/s) C A g . s. E S.

is an associated prime of

is semi-Noetherian, I C A

associated prime of

Thus

xy.s. = 0 E A, for suitable

(0 : xs-...s,) = P C A, hence

If

S.

(A g/IAg , '-Pg/IAg) = ((A/I) g , CP/I) S ), we may assume

easy to check (0)

is semi-Noetherian, S C A

an absolutely convex ideal disjoint from

is an associated prime of

Since

for

S c A, by 3.4.2(a).

Proposition 3.11.2.

where

(Ac, $ c )

Let

Then (0)

in

absolutely convex, P C A

A.

an

I (P) = kernel (A -* A-.p./IA,p.) y £ P}

considered in 2.6.

•

is

According

to 2.6.8(e), I = n I(P.), the intersection taken over the finitely many associated primes of

I.

by 3.11.2 and 2.6.8,

IA g = n IAgCPJVg), the intersection taken over those

associated primes and

Pi

If now

of

I

S C A

disjoint from

I(P-) = kernel (A -• A C /I(P.)A C ). 1

O

is a multiplicative set, we have

1

S.

Easily, IAgCPJVg) = I(P i )A g

Thus we have proved the following.

O

95

Proposition 3.11.3.

If

absolutely convex, S C A

(A/£) E (PORCK)

is semi-Noetherian, I C A

a multiplicative set, then

kernel (A -• A^/IAg) =

n I(P.), the intersection taken over those associated primes disjoint from

Remark. of

I

of

of

I

S.

•

As a special case of 3.11.3, let

and choose

I

P.

f. E A

not contained in

with P.

only associated primes of

f. E P. - P

Let I

f = ITf.

disjoint from

P

be an associated prime

for each associated prime n

and let S

S = {f }

_.

P.

Then the

are those contained in

P.

(J (I : f*) • Since (I : f) c (I : f 2 ) C . . n>0 is an increasing chain of absolutely convex ideals, we must have I(P) = I(P) = kernel (A -> A /IA f ) =

Thus

1

(I : f n ) , for suitably large

n.

In the classical Noetherian case the ideals the intersection of the primary components of of associated primes contained in of an -isolated set

E

P.

I

I(P)

belonging to the set

is a union of

Ep.

of associated primes, meaning

E

contains all asso-

The most general such

Thus for any isolated set of associated primes

the intersection of all primary components of E

Ep

This is a special case of the notion

ciated primes smaller than any one of its elements. E

are interpreted as

I

E,

belonging to primes of

is invariant and is characterized as

Q I(P)« In our semi-Noetherian PGE case we do not have primary decomposition in general, but the results above and in 2.6 show how many of the classical results involving isolated sets of associated primes do extend.

3.12.

Operators on the Set of Orders on a Ring Let

to

(A,£) £ (POR).

'$ a refined order

potent operators

We are interested in operators

Aj>.

( A $ = Ap)

A

which assign

In particular, functorial operators and idemare natural objects of study.

The motivation is ultimately this:

We seek to interpret partially

ordered rings as "rings of functions" in some generalized sense. formulas involving a "function"

96

f £ A

ought to imply

f

Certain

positive.

Thus

if

f

is positive and

a set, then

f

so should be '.p W

(A,'.p)

is a ring of suitably valued functions on

should be positive.

f.

But if the order

If 'p

p

and

(l + p)f

are positive,

is too small, say a weak order

or a finitely generated refinement

'p [g-,*#*gv], such function theoretic W 1 K

results will not follow by simple algebraic manipulation. natural algebraic extensions of

Thus we seek

'p, which more closely capture the behavior

of functions. We present four such operators here.

The first three are related to

localization. The operator

^.

If

(A/P) e (POR), (PORCK), or (PORNN), and

NCA

is the multiplicative set of non-zero divisors , then we have seen that iN:

(A,'p) -* (A N /P N)

'£d = A n !pN p

of

'p.

is injective and induces a natural refinement Specifically, f G '£

not a zero divisor.

The operator shadow of

$ .

(A,'p)

Let

p,q G 'p,

originally.

The operator

'£, is idempotent,

(A,'£) G (PORCK) or (PORNN).

Then

S(l), the

1, is a multiplicative set, contained in the non-zero divisors. iaf,^'b(.lj

We thus have an injection

fs

pf = q, for some

(A/p ) e (POR), (PORCK), or (PORNN),

We have

whichever category held

if

t h e r e f i n e m e n t of

'£

g i v e n by

(A,'£) -• (A c r i ., f c n j , and we denote by oil) oil)

Anf

Thus

f e f

if

( l + p ) f = q,

e

P,q ' P We have

(A,$ ) G (PORCK) or (PORNN), accordingly.

the absolutely convex ideals of verified that

operator

'pg

Given

'£ S

(A,'£)

and

(A/p )

is idempotent, 'P = CP ) • S S c

is a functorial operator in

In 3.8 we saw that

are identical.

Certainly

fc c 'p,. S Q

AN)

f = q/p,

p,q E'p,

From the point of view of functions, this gives information about p.

On the other hand, if

gives a globally defined formula for the function 'p

The

(PORCK) or (PORNN).

f E '£,, we can write formally (or in

least off the zero set of

We also

f.

f G 'pg, then

f, at f = -j^-

Thus functions in

are "non-negative" in rather a strong way. Suppose

with

A

is a ring, and

n p^ = (0). Then

A

Pi C A

is a finite collection of primes

has no nilpotent elements and

A

has only

97

finitely many minimal primes, say Moreover,

Hp

= (0)

and

Up

P , which are included among the

P..

is exactly the set of zero divisors of

A.

Proposition 3.12.1. In the situation of the paragraph above, assume either an order

'£ C A

with all

Pi

or assume orders

'.p. C A^

A -> II A^.

$ d = A H nCP a ) •

Then

Proof:

and set

convex, and set

'•]) = A n n tp^.

There are two points here.

then 'J5

p/P

Pi

than

and

£ C A

will be strong enough to agree with 'p, C IICP )

not a zero divisor,

suppose given

(A/P ,'£/P ) •

f £ A

a

G

P , since

h f = q (mod P ) , where

and since E h

a

tp C A H n 1^i. £ PQ p

A .

Still

h

pf = q,

p £ P .

£ P

p,q *='$,

Conversely,

and q (mod P )£ '£

Adding these equations gives

for any minimal

(A^'-P^,

n p - p , we may assume ijta 1 A, hz f = q £'£, since a a

n p. - p , and we may assume equality in i a and

Secondly, if

on

is clear since an equation

i^a

(0) = H p

'£

can

A n nCP ) •

can be reduced modulo

with

$,

is defined using all the

Multiplying each such equation by a square in h

(A,'p), then

will generally be a much weaker order than

The inclusion p

Pa

Consider the inclusion

First given

be computed by passing to the integral domains there are more

(A^'^) = (A/P^'p/P

P o , we conclude P

(Eh^)f = E q a G 'p

f £$,. a

•

The geometric significance of 3.12.1 will come out in Chapter V M , when we relate derived orders with positivity conditions on certain "non-degenerate" subsets of real varieties.

The non-minimal primes

correspond to proper subvarieties (hence lower dimensional or "degenerate") of the irreducible components of a variety.

The operator

$ .

Let

(A/p) G (PORNN)

canonical injection where the Consider those elements p,q G'.p. order

Such

f

{P }

and let

A -> n A/P

be the

are the minimal convex primes of

f G A, such that for some

A.

n

(f + p ) f = q,

n > 0,

form a set closed under products and contained in the

A H ITCP/P ) .

It is not clear under what conditions this set will

be closed under sums. In any event, define

n^^ >^ 0,

98

p.,q i G'p}.

Then

'£

!p

= {£ G A|f = E f ^

is an order and

2n. (fi + Pi) f i

:p C rp .

=

If

^i*

A

£or

some

has only

finitely many minimal primes U p

P , then the zero divisors

This can be used to show

The operator

$

Rewrite

is functorial on

(f

+p)f = q

positivity condition on the shadow of

'J>d = A H IICP/Pa)

f

as

f = q/(f

n

+ p) .

off the zero set of

(f n + p)f = q £ |

i(f) E *-P «(£"»•

are exactly

In this case

f.

In fact, if

S(f)

e

(A/-P) -+ ( cff) »'

can be seen to be equivalent to A^-*

ought to be a ring

f.

Q> hence each summand

these summands are in on

k, all

'^.

is

A

It is not difficult to see that any 'p-convex prime ideal Q C A k 2m +Vi)£± = <\i> P ^ i e ?• T h e n '^-convex. Suppose X f± € Q, {f± k k 2nd S ( n (£. + p O J f x J j=l i=i 1

'$ c'$

This provides a strong

i = ig/-£<»:

This is reasonable since

of functions off the zeros of

A

(A/p) e (PORNN).

f, we have the localization

and the condition

C A.

of

is

k 2n^ II (f. +p.)f- G Q, since 3 i=l x

It follows that some

f. G Q, hence by induction

f i e Q.

The operator ) G (PORNN)

y .

As a last example of an operator, consider those

so that the Jacobson radical vanishes,

R(A,?) =

f|

Q = (0) .

Q = maximal convex ideal Then n p : max. Q ^

(A,?) -

n (A/Q, ' max. Q

is injective and we define

B = A n ( n 'P/Q) . max. Q That is, '.J5

consists of elements of

A

positive at each maximal ideal.

The analogy with functions is clear, and, in fact, in a later chapter we will see that for finitely generated integral domains over real closed fields

A = R[x,'--x ] , with finitely generated extensions '$ = *P,t[g, • • -gv] in w i K

of the weak order on

A, the orders

'^

and

'£

coincide* and consist of

exactly those functions nowhere negative on the semi-algebraic set defined

99

by the relations In fact,

'P

f^x^'-x ) = 0 G A

= *J5m = {£|(f

2n

and inequalities

+ p)f = q, some

p,q G 'p,

g.(x ---x ) >_0.

n _> 0}.

This is

essentially a theorem of Stengle [22], and generalizes in several directions the work of Artin on representation of positive functions [4 ] . It is easy to check that all and that

*p m

is an idempotent operator

Finally, suppose relating the operators

(a) CP»

'^-maximal convex ideals are

n

T)

CPm) m

=

f • m

'p!, "p" are two orders on *p , 'p , p ,

'P -convex

A.

We state some formulas

above and intersections.

s

(b) CPf n ' P M ) p

('£' n ' P M ) d

The inclusions (a), (b) are trivial, as is If

f e T

p f ,p M

d

n f » ,

let

p'f = q',

not zero divisors.

Then

p"f = q",

pSq'Gf,

c

'Pd n '^ d •

p",q" € J>»,

(pfp") 2f = p'q'Cp") 2 = p"q M (P') 2 G 'P' n '^

Thus f'd n •pjj c fpi n -p") d . It is not quite as easy to formulate a result for the operator since even if the Jacobson radicals of clear when the Jacobson radical of

1

'p

'£' n $"

and

tp" vanish, it is not

vanishes.

However, in the

algebro-geometrical situations to be studied later, our rings will have the property that for each maximal convex ideal a fixed real closed field

'p ,

(A/p)

Q,- A/Q

R, therefore with a unique order.

is

In the

reduced case, (A/p) G (PORNN), the Jacobson radicals will vanish and (A,'P )

identifies with a ring of R-valued functions (ordered by func-

tional values) on the set of maximal convex ideals of such orders

'£','£" C A

the maximal convex ideals of

exactly the union of the maximal convex ideals of and clearly

100

$'

n

'P") m = '£'m n '£"• m

'P'

(A,p). p

1

n 'p"

and

For two will be

'p!f by 2.7.2,

IV • Some categorical notions

4.1.

Fibre Products Proposition 4.1.1.

The categories (POR), (PORCK), (PORNN) admit

arbitrary fibre products.

Proof.

The construction is identical in all three categories.

We

first show that direct products exist. Let fa:

(Aa, p a ) G (POR).

(C,^c) -* (Aa, $ a )

a unique (POR)-morphism diagram commutes for all

Consider

(A,?) = (II A a , II? a ).

If

is a family of morphisms in (POR), then there is f:

(C,^c) -• (A/p)

such that the following *

a:

(A,

Thus, (II A a , n f a )

is the direct product of the family

More generally, suppose

ga:

morphisms over the fixed base (Aa, $ a )

over

(B, *PB)

contraction of the order

(Aa, P a ) -> (B, *Pg) (

(B, ]5fi) .

is the subring II^ .

(Aa, $ a ) . is a family of

Then the fibre product of the n A a c IlAa, together with the

B That is,

II

(A , y )

is the ring

together with the order

n
4,2.

Fibre Sums Proposition 4.2.1.

The categories (PORNN), (PORCK) and (POR) admit

finite fibre sums.

Proof.

We first make the construction in (POR).

The fibre sums in

(PORNN) and (PORCK) require a slight modification of the construction. Let

(C/Pc) •+ (A,$A)

We begin with the ring

and

(C/Pc) -* (B/PR)

A ® B. C

We would like to impose an order on

such that the natural ring homomorhisms order preserving.

be morphisms in (POR).

A -> A ® B C

and

B -* A ® B C

A ® B C are

In general, however, this cannot be done.

Instead, we construct the smallest ideal

R C A ® B C

which satisfies the

following condition: k 2 .2 (Pi ® q±)xl G R, i= 1

Pi

G '?A , q. G P R , x. G A ® B C

4.2.2. implies

(p. ® q-)*- G R,

1 < j £ k.

This is exactly the generalized extension condition of 1.3 for the pair of maps

A-»A<8>B, C

is then the ring

B -• A ® B.

The fibre sum

(A/BJ

c

A

® (B,^D) B (C,?c)

(A ® B)/R, together with the order C

k

? = {^2

? (p. (8) q.)x*) |p. G ? A , q. G ^

x. G A ® B}

1— 1

where

L

IT: A ® B -• (A ® B)/R C C We verify that

is the projection.

(A ® B/R, !p) C

Suppose given (POR)-morphisms

f:

has the desired universal property. (A,!PA) -»> (D,?D)

such that the diagram below commutes.

102

in (POR)

and

g:

(B,?fi) -* (D,

Then there is a unique ring homomorphism' f ® g: A ® B-*D such that f is A-^A® B-*D and g c

c

k

? B -* A ® B -> D. Let I = kernel (f ® g). Suppose 2 (p. ® q.)x7 G I, x C i=l * p. G ( P A , q i e ? B , Xj_ G A ® B. Since f ® g(( P j ® q.)x2.) = f (p..)g(q..) (f ® g(x. ) ) 2 e ^ D is

1 < j < k, we deduce

(p. ® q.)x. e I, 1 < j < k. Thus,

~ "~ 3 3 3 — — factors A ® B -» (A ® B)/R -» D. Clearly, (A ® B)/R •* D C C C hence

((A ® B)/R, $ ) = ( A ® ? . ) A

c

®

R C I, and f ® g

is order preserving,

(B;f R ).

(c;pc)

In the category (PORNN), the ideal

R c A ® B must satisfy R = v^R, C There is a smallest such R, and the rest of the proof is

as well as 4.2.2. unchanged.

In the category (PORCK), the condition 4.2.2 is replaced by

R, p i e f /

q.

1=1

G p B > x. ,y e A ® B,

1

1

1

,

,

4.2.3. implies

(p. ® q.)x?ye R,

Again, there is a smallest such

1 < j < k.

R, and the rest of the proof is the

same.

•

4.3.

Direct and Inverse Limits Let

I

be a directed, partially ordered set, with relation

{(A., ? - ) } - F T

be a family of partially ordered rings.

Proposition 4.3.1. k < Jj —

in

<_. Let

I, such that '

the direct limit

Suppose given morphisms & g..

j3

= Id

lim (A.,?.) X 1

g, .:

(\>?v) "*" (A->?-)>

and 5 g,5.g , =5 g ., if m < k < Jj . kj mk mj — —

Then

exists in all categories (POR), (PORCK),

(PORNN).

~T* Proof.

The construction is the same in all categories.

Begin with the

disjoint union, k G I,

V A.. Identify a. G A. and a. G A. if there is a x 1 3 i G I x 1 i,j < k, with g.,(a.) = g.v(a.) G A v . The resulting ring is the —

IK

ring theoretic direct limit

1

J K J

Impose an order ty on A in

A

of all the $. C A..

K

lim A. = A. by taking as positive elements the images

It is easy to see that

(A/P) = lim (A i ,P i ),

in the sense of universal property.

•

103

Proposition 4.3.2.

Suppose given morphisms

f., :

(A.,p.) -> (A, , P V ) , J J K K f. . = Id and f. f.. = f. , if m < k <J i. Then 33 km jk jm' — — JK

k < j in I, such that — the inverse limit

lim ( A ^ . ) exists in all categories (POR), (PORCK) and

(PORNN). Proof.

Again the construction is the same in all three categories.

The answer is the usual ring theoretic inverse limit

jim

i C O A .

together with the contracted order

•

4.4.

Some Examples (1)

Let C = Z[X], ordered as a ring of functions on the line. Let

A = B = Z, with the unique order. by

ct(X) = 1, 3(X) = -1. Thus

at the points

Define

a: Z[X] -* Z

and 3: 2[X] -»- Z

a, 3 are simply evaluation homomorphisms

+ 1, - 1, respectively.

In the category of commutative rings with unit, 'L ® Z = Z / 2 , but in Z[X] (POR), Z ® Z = (0). Geometrically, this is preferable, since the fibre Z[X] sum should be a ring of functions on the intersection of the two subspaces {+ 1} , { - 1} , which is empty. (2)

Let

(A,$) G (POR),

(A, (A/I, ?/I) and

(A/I, $/I) ®

I,J C A

convex ideals.

(A/P) -* (A/J, ^ / J ) .

We have projections

Then

(A/J,

(A,?) There are obvious (PORCK) and (PORNN) analogs. Note that in this example the fibre sum is (0) convex ideal of (3)

Let

a convex ideal. 104

A

which contains both

(A,$A )G (POR), Then

I

and

(A,^) -* (B/PR)

if there is no maximal

J. a (POR)-morphism and

I C A

(A/I,

Again, there are obvious (PORCK) and (PORNN) analogs. (4)

If

(A,SPA), (B,$fi) G (POR), then their direct sum in (POR) is the

fibre sum over the unique ring maps sum

Z -• A, "£ -*• B.

We denote the direct

(A,PA) ® (B,? B ). As an example, let A = Q[X]/(X 2 ),

B = Q[Y]/(Y 2 ), ordered as rings of

germs of functions at 0, module those vanishing to second order at 0. In the category of rings, A ® B = Q[X,Y]/(X 2 ,Y 2 ). 2

2

2

Moreover, an element

However, in (POR),

H(X ,Y ) = (X ,Y2,XY)

A ® B s Q[X,Y]/(X ,Y ,XY), since

2

2

by 2.8, Example 7.

a + b X + cY will be strictly positive if and only if

a > 0 G Q. So, the direct sum is the ring of germs of functions at modulo those which vanish to second order at (5) We continue with

A = Q[X]/(X 2 ),

(0,0) e ^

(0,0).

B = Q[Y]/(Y 2 ), but with refined

orders induced by functions nowhere negative for all sufficiently small positive real arguments. As a ring, the direct sum in (POR) will still be Q[X,Y]/(X 2 ,Y 2 ,XY).

However, now an element

a + b X + cY will be positive

if and only if a > 0 or a = 0 and b,c >_ 0. by ordering (6)

Q[X,Y]

This is the order induced

as a ring of functions on the first quadrant in R

Let (A/p) G (POR),

Ic A

(2) .

a convex ideal. There are various

inverse systems associated to I. One has the (POR)-inverse system (A/H(I n ), $ / H ( I n ) ) n > 1 , and the (PORCK) inverse system If

(A/AH(I n ), s £/AH(I n )) n > r

I is a prime convex ideal, we have the symbolic powers

and we can form

[k/I^n\ $/I

)

n > 1

I

of 3.8.6,

- E a ch of these inverse systems will

have a partially ordered inverse limit. (7) and

Let (A,?) G (POR),

QC A

S(a) is the shadow of a, then

Thus there is a natural (POR)-morphism

a prime convex ideal. S(a) c A - Q, since

If a £ Q

A - Q is concave. A

(A g ( a ) , ? s ( a ) ) "> ( (Q)> *(Q))"

We will see in the next chapter that the rings

(A cr .,tyQf. ) ,

a £ Q, form a natural directed system, and that there is an induced isomorphism in (POR),

^(A

s ( a )

, ? s ( a ) ) =V(A ( Q ) . ? ( Q ) ) -

105

\

V • The prime convex ideal spectrum

5.1.

The Zariski Topology Defined Let

(A/p) e (POR).

ideals of A

We denote by Spec(A,$)

the set of prime convex

and by Spec (A/£) the subset of maximal convex ideals.

We define a topology on Spec(A,p), called the Zccrishi topology, in the usual way.

The closed sets are

Z(I) = {Q|Q€E Spec(A,$), I C Q}

where

IC A

is a convex ideal.

In fact, if X C A

is any subset, let

Z(X) = {Q|Q e SpecCA,^), X C Q} .

Proposition 5.1.1. (a)

Z(X) = Z(H(X)) = Z(/H(X))

(b) 0 Z ( J ) = z ( 2 ! ) a a a a

for an

for any subset

y collection of ideals

I C A. a

for an

(c)

Z(I ) U Z(I 2 ) = Z(I 0 I 2 )

(d)

Spec (A,?) = Z((0)) and 0 = Z(A).

(e)

If X,Y C A

Proof.

X C A.

X Pair

are subsets, Z(X) c Z(Y)

o£

ideals

I X ,I 2 C A.

if and only if Y C /H(X).

The proofs are routine, in view of the Corollary 2.3.7, which

asserts vE(xT

H

Q•

D

QGZ(X) As a basis for the open sets of Spec(A,*J3) we have the distinguished open sets

106

D(a), a £ A, defined as follows:

D(a) = {Q|Q e Spec (A,?), a g Q}.

That the

{D(a)}

.

do form a basis follows from the formulas

D(a)f| D(b) = D(ab)

Spec (A,?) - Z(X) = U D(a) . a G X

We give

5.2.

Specm(A,![)) C Spec(A,$)

Some Topological Properties Proposition 5.2.1.

Q

the subspace topology.

A point

is a maximal convex ideal.

Spec(A/P)

TQ.

is

Q e Spec(A,*p)

Thus

Spec(A,^)

is closed if and only if is not

That is, given two points

T 1 , in general.

P , Q G Spec(A,p), at least

one is not contained in the closure of the other.

Proof.

For any

V C Spec(A,^), the Zariski closure of

V = Z( H

V

is given by

Q) •

Q G V This proves the first statement, and also the last since Q e {?}

implies

P C Q

Proposition 5.2.2. In particular,

Proof. all the

ai

and

P e {Q}

Q C P.

•

The basic open sets s

D(l) = Spec(A, £)

and

D(a) C Spec(A/£)

are compact.

is compact.

Suppose

D(a) C U D(a^). Then any convex prime ideal containing i must also contain a. By Corollary 2.3.7 , a G >/H({ai>). By

Proposition 2.2.2, there is a finite subset a G i/H({a1,...,av}).

Thus

D(a) C

U

{a.,...,a,}

of

{a.}

with

D(a.), as desired.

•

5.3. Irreducible Closed Sets in Recall that a closed subset of a topological space is irreducible if it is not the union of two proper closed subsets.

Proposition 5.3.1. there is a unique

Q G V

If

V C Spec(A,!p)

with

is closed and irreducible, then

V = {Q}.

107

Proof.

Uniqueness of

preceding section.

Q

(Specifically, Spec(A/£)

We prove existence of is a convex ideal.

Q

I

{b}.

as follows.

is Let

It suffices to prove that

Suppose, then, ab e I, u

is a consequence of Proposition 5.2.1 of the

Any prime

P

a £ I,

b £ I.

containing

I

T .) V = Z(I), where

I

is prime.

Consider the sets

contains

I = /f

a

or

b.

I U {a},

Thus

Z(I) = Z(I U {a}) U Z(I u {b}) .

On the other hand, since containing

I

but not

I = /F, a.

Thus

a £ I

implies there is a prime

Z(I u {a}) C Z ( I ) , and similarly,

Z(I u {b}) C z(I), which contradicts the irreducibility of

Corollary 5.3.2.

Spec(A,£)

/(0)

V.

•

is irreducible if and only if

a unique minimal convex prime ideal irreducible if and only if

P'

P.

A

Equivalently, Spec(A,$)

has is

is prime.

Consider the family of convex ideals

• I C A

such that

I = /F.

Note

that this family of ideals satisfies the ascending chain condition (every increasing chain the closed sets of decreasing chain

I, c I

C ...

Spec(A,'p) Z, D Z« ^ ...

is eventually constant) if and only if satisfy the descending chain condition (every is eventually constant.)

thesis it follows that every closed subset of of irreducible closed sets.

Spec(A,tp)

Under this hypois a finite union

Translating this in terms of ideals of

A

we

have

Corollary 5.3.3.

Suppose the family of convex ideals of

A

which

coincide with their nil radical satisfies the ascending chain condition. Let

J C A

P C A

be any proper convex ideal.

containing

Then among the convex prime ideals

J, there are finitely many minimal such

P, say

P..,...,P,,

and

/r = n p, •

108

a

5.4. Spec (A ,'ft) as a Functor If

f:

(A,£) •+ (A1 ,^1 )

convex prime ideal, then f

induces a function

is a morphism in (POR) and Q ! c A1 (Q!) C A

Q = f

f*:

X C

is a convex prime ideal.

Spec(A',£') -* Spec(A,£).

continuous in the Zariski topologies since

is a Thus

Clearly, f* is

(f*)~ Z(X) = Z(f(X)), any

A.

As special cases, we have morphisms I C A

is a convex ideal, and

iT:

IT: (A,$) -* (A/I, $/I), where

(A,'])) -• (A^/p-p), where

T C A

is a

multiplicative set. In the case of a projection, homeomorphism from

TT*: Spec (A/I,ty/I)-* Spec (A ,^)

Spec(A/I, $/I)

to the closed subset

Z(I) of

is a Spec(A/p)

This follows from 2.4. In the case of a localization, homeomorphism from

Spec(AT,'£T)

This follows from 3.4. If T set (that is, if T

i*: SpecCA™,'^) -* Spec(A,'p)

is a

to the subspace

Q D(t) of Spec(A,'p). t€T is finitely generated as a multiplicative

is the smallest multiplicative set containing, say,

t p . . . , ^ ) , then k

(1 D(t) = 0 D(t.), t6T

i=l

and this intersection is an open set in Spec(A,'£). We also point out that because the convex prime ideals of A from

T

coincide with those disjoint from the shadow

the natural map

5.5.

A

(AT,:pT) -» ( opry^S(Tp

disjoint

S(T), it follows that

induces a homeomorphism

Disconnectedness of Spec(A,'ft) In the category (POR), there is not quite as close a relation between

disconnectedness of Spec(A,'£) product

and decomposition of A

as a nontrivial

A 1 x A 2 , as there is in the category of commutative rings. We do

have the following semi-analogue. Proposition 5.5.1.

Spec(A,'^)

is disconnected if and only if there

109

2 Q

i>e2

are elements 1 2 HCe^,

n

'^

C A

* with

e

1 £ e, + e 2 ,

=

Under the hypothesis, it is clear that

Z(e 2 ) = 0, and

Z^)

t 0,

°>

e

i £

e

i »

Z(I i ) t 0,

i = 1,2, then

1 G H(I 1 + I 2)

By 2.2.4, we can write

Spec(A/p) = Z{e^) U Z(e 2 ),

i = 1,2.

Conversely, if Spec(A,'p) = Z(I X ) U Z(I 2 ),

elements.

i'e2

i = 1,2.

Proof. Z(e x )

G

Z(I 1 )

and

n

Z(I 2 ) = 0, and

I1 n I2

1 £ ej + e 2 ,

consists of nilpotent

e^ G I i n p .

(ej e p n = 0,

If

we get 2n-1 2n-l n n < (ej + ej) = a^e') + a 2 (ej) , 1 = lZn

with

a i G!p,

hence

i = 1,2.

1 ^ H(ei),

If

e i = a i ( e p n G!p.

Set

i = 1,2.

Also, e ^

1 £ e1 + e 2

1 £ H(ep,

and

A -> Ag.,.

and an order

$

S

Q^

C

Recall Acfli

Spec(A,'^) = S p e c ( A s ( 1 ) / p s ( 1 ) ) .

Proposition 5.5.2.

1'A2C

A

in

(A,^) G (PORCK), then

e . , e . e f C A,

Ag,.y

Spec(A;p) for

suitable

is disconnected subrings

i

In 5.5.1 we proved that if

are elements Now,

A

S(1)'

Proof.

If

( A S Q ) » ^ S ( 1 ) ^ ~ ( A i^i) x (A2'^2^

if and only if A

e| e I.,

(A/p) G (PORCK), we can push this result somewhat further.

that by 3.8.4 we have an injection with

= 0,

Since

u

Spec(A,$)

1 £ e, + e 2 = u,

is invertible and

is disconnected, there

e^ £ e^ ,

e.e2 = 0,

1 ^ H(e i ).

1 = e,/u + e2/u = ej + e'.

Consider the natural projection

The kernel since gives

110

of

TT is

AH(eJ) n AH(e^).

e!e* = 0, we have

But if

e!x' = elx1 = 0.

(0: e!) = (0: AH(e!)).

Thus

x

1

xf G AH(eJ) n AH(e^), then,

This follows from 2.5.3, which

= (ej + e^)x! = 0, and

IT is injective.

If

b ! ,c' £ A S . 1 . , then

7T(b'e^ + c'ep

Thus,

cr = c f e ^ + c l e ^ , and hence

b 1 = b'e^ + b f e^,

= (b',c») e Ag (1) /AH(eJ) x Ag (1) /AH(e») .

IT is surjective. b',c' e f p s ( 1 ) , then

Next, if Thus,

TT~

b'e' + c' e[ e ? s ( 1 ) , since

is order preserving and

Finally, 1 £ H(e i ) C A

TT is an isomorphism in (PORCK).

implies

1 £ AH(e i ) c A, since

Then, also, 1 £ AH(ej) C A g ^ . , since A n AH(e i )A s . 1 . = AH(e i ).

e!G

Ai^e^

AH(e£) = AH(ei)Ag

Thus the factors

= /AH(e7)

and

Ag^./AHCep

are nonzero.

D

5.6. The Structure Sheaf, I - A First Approximation on Basic Open Sets We will eventually impose on rings.

The classical idea is this:

a E A, we know "a hence

Spec(A/p)

a ^ 0 e A/Q.

is never zero".

on a basic open set That is, if

Q e D(a).

set

{al}i>0-

a

Specifically, let If

D(a) C Spec(A,*P),

Q G D ( a ) , then

So we localize and invert the element

But in our partially ordered ring if

a sheaf of partially ordered

A

S(a)

over D(a).

we know more than just

a £ Q

be the shadow of the multiplicative

b G S(a), Q e D(a), then

is positive, since then

a

a £ Q,

b £ Q.

<_ b, some

S(a) = {b|0 £ a

harder in the general case, but since

The proof is easy if

D(a) = D(a )

and

n}.

It is not much

(A cr 2^>$Cr * O D(^a j

(A cr \>'#Qr O >

we mav

as w

^ll assume

a

is positive, (see 3.2 for details).

Thus it is natural to begin by trying to invert all elements of over the basic open set

D(a). That is, to the basic open set

we assign the localized (POR) this assigns

(Ag^^g^^)

Lemma 5.6.1.

If

(A c . .,'£c, O £>laJ ^laJ to Spec (A,?).

D C a ^ = D(a 2 ),

a

^

b^a j

-,,a2

G

Since

S(a)

D(a) C Spec(A,'p),

Spec(A,'^) = D(l),

A, then there is a canonical

isomorphism in (POR),

Proof.

DCa^^) = D(a 2 )

primes containing

a«

implies that the primes containing

coincide.

Thus

y/H(a,) = Ai(a^).

Since

a^^ a

and the a

i> o

e

P»

we have by 2.2.3

111

0 £ a^1 <_ pa 2 0 < a r 2 < Hqa — 2 — I

for suitable integers

r.^,^ > 0

since now all elements of

S(a 9 )

and

p,q e:p.

The proposition follows

are invertible in

In fact, this argument really shows that if

" = *D(a2);D(ai):

defined by so that

and vice versa.

r

DCa^) C D(a 2 ), then

0 < a"^1 < pa 2 , and there is a canonical morphism in — * —

a, G ,/H(ao), hence l z (POR),

A

(A

r


p b > (Pa?)

^.a i

S(a 2 )'*S(a 2 ) ) -* ( A S(a 1 )'*S(a l ) ) - *"

r

e Ag,

1#

. , where

a € A, b >^ a^

These morphisms


,

,

,

and

is

pa2 > a ^ ,

are our

candidates for the presheaf restriction maps corresponding to the inclusions D(a,) C D(a 2 ).

of open sets

Lemma 5.6.2.

If

D(a Q ) C

D(a 1 ) C D(a 2 ), then the diagram below

commutes

where

Proof.

This is easy using the universal property of the localizations

(A cr ^y^cr 0 o^a^j b^a.j

If

Q c A

(rather than the explicit formulae for

is a convex prime ideal,

the concave multiplicative set There is thus a natural map

112

A - Q

a ^ Q, then

contains


Q G D(a)

•

and

a, hence also S(a).

D

Lemma 5.6.3.

The induced map

is an isomorphism in (POR).

•

We leave the proof to the reader, with only the reminder that one must not forget to check that the inverse ring isomorphism is order preserving. Thus, from Lemma 5.6.3 we expect the stalk of our structure sheaf at the point

Q G Spec(A,]5)

to be the usual local ring

A. Q.

with the natural

order „( Q r

5.7. The Structure Sheaf, IT - The Sheaf Axioms for Basic Open Sets We formulate the "first sheaf axiom" for basic open sets in the following.

Lemma 5. 7.1 (a) . Suppose

x/b e A

and

Lemma 5.7.1(b). is injective.

(A,$) G (POR), *>

^^ a S(aa])

From 5. 7.1 (a), the map

$cr on

. C Acr .

l

Secondly, D(a.) C D(a)

the element Since

x/b

means that

and integers

r. > 0.

.) n A C , ..

That is,

= x

Moreover, since

p >^ 1,

U l
D(a.)» 1

a^ G v/H(a), hence

0 £ a^1 £ p^

We can replace all the

p.

by

in these inequalities.

A c r a >. , we assume •H J

*D(a),D(a.) ( x / b )

D(a) =

K

p = 1 + P j + ••• + p k G

Then

is compact, hence there is a finite with

a , say, a,,...,av

p. Gtp

a.

( A g ( a ) f ? s ( a ) ) - H ( A g ^ ,» s(gfl) )

. = (npc,

fcr

G •£.

is the contraction of the natural product

First, D(a)

a

for some

all

a

IlA c r v a S(a a )

Proof 5.7.1.(a). subset of the

D(a) = U D(a ) , a,

(x/b) = 0 e A

We claim moreover that

the natural order order

Let

b > a s , some integer ~~

P S / b P S ' Cbps > a s p s > a ^ S , hence a s p s >_ a s , hence

bp s >^ a s

s > 0.

Then

bp s e S(a.).) and we can write

113

x/b = xp s /bp s e A s ( a ) .

To assert that

means that there exists a

*D(a)^

c. > a. 1 , with

r

0 < a

integer s

x/b = xp /bp

s

But now,

= 0 G Acr . o ^aj

Proof 5.7.l(b). above.

i -

a

for suitable

X X

xps c = 0, and since

q. G •£, and some *

c G S(a), we have proved

as desired,

The proof is a slight modification of the argument

xp S /bp s >_ 0

i1

with

in all

Ag,

.,

1 £ i £ k, then there exist elements

in

A.

If

xbp2s(cp2^0

xbp S (c') ^ 0

0 £ q! , we see that But

q.c. = c

aGv / H({a 1 1 ,.. . ,a. k } ) , Ik

implies

We continue with the same notation.

If C

J

D(a) = U D(a.) x i=l

k t. k < 2 q.a. < 2 X x ~ i=l ~ i=l

r > 0.

11

F

l—i

hence

(xp s/bps ) = 0 e A ^ ^

xp s c. = 0 G A.

l—i

We now use the fact that

^

in

0 < a r ' <_ S q l q . 1 ^ xbp S (c f )

A, hence also

xp s /bp s >_ 0

c' G S(a), so we conclude

in

2 q ! ( c ! ) 2 = c'

>_ 0

A g . ..

D

The "second sheaf axiom" for basic open sets would say this: D(a) = U D(a ) ot o^

and suppose

x / b G Acr . ot ot, o (_a I

fx /b ) = ip

if> (Note

Ot p

such that

Ot

DCa

, for all

Then there is an element

ct, 8

x/b G A c r . o (_B-J

p

} (x/b)

Suppose

have the property that

(x /b ) G A

D(a a o ) = D(a ) n D(aD).)

in A.

= x a /b a G A s ( a ^

for all a.

This assertion does not seem to be true in general.

However, the failure

of the second sheaf axiom does not really affect the way one defines the structure sheaf.

What it does affect is the evaluation of the global sections.

We will return to this point in the next section. What we can show is that in the category (PORCK), the second sheaf axiom for basic open sets is true.

Lemma 5.7.2. a G iB, and let x /b

a a S AS(a )

Let

(A/p) G (PORCK), D(a) G Spec(A,'f)

D(a) = U D(a ) a a have the

Specifically

be a cover,

114

Suppose the elements

Pr°P«rtX * hat *D(a ))D(a a . ) a

W),D(aaJ(VVeAS(aaq) (X p

G •£.

a

a p

a

01 p

a basic open set,

f o r a11 a

C

W

=

a p

> 6 - Then there exists

x/b6Ag(a) with » D ( a ) > D ( a a ) W W = * A 6 A S ( a a )

f

° r a 1 1 "•

Proof.

The proof is essentially the usual proof for commutative rings. k First, D(a) is compact, so D(a) = U D(a.), for some finite subset of the

a .

If

x/b G X

and

x/b

restricts to

x./b.,

^laJ

CX

1

have the same images in all

A~,

1 < i < k, then

x /b

,

1

..

But

01 Ot

D(a)= U

D(aa.),

and we already know from the "first sheaf axiom", Lemma 5.7.1, that an element in

A

. *-acr

x/b

Thus

is uniquely determined by its restrictions to the

r

must restrict to

x /b , all

a.

Acr

.. ctai

This reduces our problem to

the case of a finite cover. From the hypothesis, there are elements such that

x i - b . - c i . - x . * b i - c i . = 0, all r

0 £ ( a ^ . ) 1 ^ £ c^.

c^. G A,

i,j. Since

(A,$) G (PORCK),

r

x^b. ( a . a . ) - x . b i ( a . a . ) = 0. Replacing x i /b- by x ^ . / b ^ T , we may assume x.b. = x.b.. W em a y also assume that 0 < as . < b., l < i < k. 1331 — 1 — 1 — — Since the Pi G £

D(a i )

cover

D(a), a G /H{a ••-a,} , and we can find positives

such that

t 0

In

A , . b(^aj

1 a•

we have

1 = b/b =

k Pi 2 (-=--)b. .

^_^

D

1

Consider the element

\c

If

This proves that

vn. . n .

.(x/b) = x./b. G A c r

.,

x/b = V

1 < j < k, as desired.

5.8. The Structure Sheaf, IH - Definition Let

X = Spec(A,'p)

with the Zariski topology.

€L, of partially ordered rings on If

U C x

X.

is open, consider the product

jects naturally to

U C X. A section

s:

II ( A r m ,'£r m ) , which pro-

QGU W j U ->• n A QGU

x

n/bn

G

ArnV

and elements

We will define a sheaf

w

Q G U , is continuous if there exists a cover x /b G A c r ., such that for all ot ot o^a 1

WJ

say

s(Q) =

U = U D(a )

Q G D(a ) ,
115

D

is the natural map.

Definition 5.8.1.

Define

U,^) = {s: U ^

(T(U,^X), ^(U,#x)) e (POR) by

II A rn . |s = continuous section}

^

Qeu x

x

Lemma 5.8.2. U y:

CQ}

Qeu

Let V C u C X be open sets. There is a canonical map

(r(U,^x),^(U,^)) -> (r(V,^x),'j3(V,^x)) induced by the obvious projection

s: U -• n A.o. , s(Q) = xo/bo G A r o v ^ ^ ^J Q€U ^ J

Proof. Suppose

is continuous

»

say with respect to a cover U = U D(a ), and elements x /b e Acr .. a a a a ^*-aa^ Cover V, V = U D(a'). Since V C U, we have V = U D(a a'). Then the P a, 3 a ^ 3 sL : V -• n A. . is defined locally by the elements v W QGV

section

^D(a ) , D ( a a ' ) ( W (X

d

*^U V ^

= s

eA

S(aaL)'

ThuS

S

I V ^ continuous and we define

CX p

p

lv G ^W,0y)>

^y y

Proposition 5.8.3. The ^x),!p(V,^))

is obviously order preserving.

(r(U, (^) ,'jJ (U,^)) and ^

^

y:

form a sheaf on X.

Proof. The proof is routine. The first sheaf axiom is a consequence of Lemma 5.7.1.

The second sheaf axiom is a consequence of the definitions. D

Remark 1. The significance of the failure of the second sheaf axiom for basic open sets is that we cannot easily describe global sections T(X,^X),

X = Spec(A,^), or more generally

r(D(a),^x),

D(a) C Spec(A,'p)

a basic open set. Lemma 5.7.1 implies that

116

(AS(-a) ^ s ( a ) ) "* (r(D(a) ,0£ $ (D(a) ,^))

is

injective and that However, F(D(a),^Y ) A

'p g, . = A g , . n p ( D ( a ) , ^ ) may be bigger than

is the contracted order.

A c r .. b ^aj

In any case, one shows in

the usual way that there is a natural sheaf isomorphism

Remark 2.

It is perhaps good that

r"(X,/?Y)

is larger than

A

The natural map

(A,'£) -*- (Ac,,>. ,'.pcr,O oil) o^lj

In such case, T ( X , ^ )

Ac ^

is not always injective in (POR) .

may contain information about

(A,'p)

lost in the

localization In the category (PORCK), the second sheaf axiom for basic open sets, Lemma 5.7.2, does allow computation of the global sections of the structure sheaf.

We have

Proposition 5.8.4.

Let

(A/p) e (PORCK),

sections of the structure sheaf

X = Spec(A,'p).

€L over the basic open set

Then the D(a) C x

A

is the ring

F(D(a),^ x) = (A g . •x/^Sr ^ '

is the ring

T ( X , ^ ) = (Ag ( 1 ) ^ s ( 1 ) ) •

In

Particular»

tne

global sections D

The Zariski topology on the prime convex ideal spectrum of a partially ordered ring has the following property not shared by unordered rings. A finite union of basic open sets is still a basic open set. The proof is easy since D(f) = D(f 2 ) and D(IIf?).

U D(f?) = D(E f?), which is symmetrical with

H D(f?) =

If, further, our ring is Noetherian (more generally, if the radical

convex ideals satisfy the ascending chain condition),then every Zariski open set is a basic open set.

117

VI • Polynomials

6.1.

Polynomials as Functions Let

over

A

A.

be a ring, A | X ... X ]

Let

A ^

f e A [ X 1 ... X n ] If

(A/p)

A X , from for all

X

x e X,

1

If

to

(A, ]))

denote affine n-space over

defines a function

f:

A ^

A.

-» A

n-variables

Each polynomial by evaluation.

is a partially ordered ring, then so is the ring of functions A, for any set

X.

f (x) e '$. Note

belongs to

(PORCK)

injective?

Namely, define

A C AX

or

A very natural question is: -* A A

the polynomial ring in

'£x c A X

by

(the constants) and

(PORNN), then so does

f e$

^x

x

if

extends '£.

X

(A , £ x ) .

when is the ring homomorphism

A[X.^ ... X n ]

The following at least gives an easy sufficient condition.

Proposition 6.1.1.

If the ring

A

is

z (n) A f X ^ . .X^]-• A

Z-torsion free, then

is injective.

Proof.

We first establish the result for

a Q + a x X + • • • + adX

is a polynomial such that

in particular, f(0) = a

=0.

Next, let

equations in the coefficients

a.,...,a.

2 ad = 0

118

o

+---+da, = d

Suppose

f(k) = 0

k = l,2,...,d.

a. + a_ 1 2

da.. + d a 1 2

n = 1.

f(X) =

for all We obtain

k € Z C A. d

Since

2 d\ x, • • • x, \ x X 2

det

Xx

2

2

••• Xx d 2

x) n

, xx \ n

xx

2

n

n (x - x ) , l
... x x d n

we deduce that for some non-zero integer

/ i,

m,

ma. = 0 ,

1 j< i <_ d.

Namely,

i, •••, i \ 2

2, 2 , ..-, 2 d

m = det

\ d , d 2 , -.., d d /

Thus

a i = 0, If

1 < i < d , and hence

f(X) = 0.

n > 1, we use induction.

A[x1...xn_1][xn]

C

Corollary 6.1.2.

If

A

is

A

Z-torsion free and

B c A

is any subring,

(n) then the homomorphism

...XJ^A"

A[X, . . . X ] -•• A

is injective.

In particular

is injective.

Corollary 6.1.3.

If

(A/£) e (PORCK), then

A

is

Z-torsion free,

A (n)

hence

A[X X . .. X^-* A

is injective.

From Corollary 6.1.3 we see that if

D (A,'p) G (PORCK), then there is a

natural (PORCK) order on A[X, ... X ] , namely the contraction of the order A (n) y , . on A to A[Xj ... X n ] . We denote this contracted order by the same symbol

(A[XX ... X j , $

Next suppose

(R)

).

(A,£A) C (B,'PB), both in (PORCK).

Suppose as an

119

algebra

B

is finitely generated over

A[T.. ... T ] where

be the polynomial ring in

projection

n-indeterminates, so that

Let

B=A[T,...T ]/I

A[T.. ... T ]

is order preserving, where

extending

p..

p

is the

In particular, I C A[T, ... T ]

'^-absolutely convex.

Proof.

A typical element of

G A[T X ... T n ] .

Remark.

The image in

p B

is is

Ep.g.(T)

In words, the proposition shows that if

pure polynomial extension of

A

(B,p R ).

p

D

(A,'PA) C (B,'pg)

(B,pR)

In the case

a totally ordered integral domain over

That is, first the order

p. G p ,

is obtained from a

on

A

a totally ordered field,

A, this procedure can be reversed.

A[T, ... T ]

can be refined to a total order,

This assertion will be established in Chapter VIII.

We conclude this section with an example.

Consider the ring Z[T],

ordered as a ring of real valued functions on the real line. is a convex ideal. expressions

Let

n + eT,

A

the order on

A.

6.2.

.

A = Z[T]/(2T,T ) .

e = 0 or 1,

vanishes identically on functions on

is

by first dividing by a convex ideal and

then, possibly, refining the order.

inducing

where

Ep.g^(y) e 'PR.

a finitely generated algebraic extension, then

B

... y ] , y. G B.

(A,p A ), (B,pR) G (PORCK) , then the

If

( A p ^ . . . T n ] ,'p) -* (B,'PB)

weakest order on

gi

B = A[y

I = {f e A p ^ ... T j | f iyy . . y j = 0 G B}.

Proposition 6.1.4.

is

A, say

A^

n G Z.

Elements of

A

The polynomial

, so we cannot regard

On the other hand, A[X]

Then

(2T,T )

have unique f(X) = TX+ TX

A[X]

G A[X]

as a ring of A-valued

does admit an order, extending

Adjoining Roots Let

(A,'p) G (POR),

a G A,

We ask if there is an order on

A -• A[T]/(T2-a)

the obvious inclusion.

A[T]/(T - a ) , extending

'p. The answer is

similar to well-known results for (totally) ordered fields, and extends the results of 1.4.

120

Recall from Chapter I the notation

Dp = {aGA|pa G'p

some A.

pG'j) }. We established that - a £ D'£ implies ']>[a] is an order on

We now refine this result. Proposition 6.2.1. (a)

If -a £ Dp, then

B = A[T]/(T2-a)

can be ordered extending

p,

and if B can be ordered extending '£, then '£Ja] is an order on A. (b)

If either

A

has no nilpotent elements or a E A

divisor, then the existence of an order on B to '$ [a] (c)

is not a zero

extending ty is equivalent

being an order on A. If

(A,'£) e (PORNN), '£[a] is an order on A and a 2 zero divisor, then B = A[T]/(T -a) has no nilpotent elements. (d) If A is an integral domain, then either 2 A[T]/(T + a) admits an order extending $.

is not a

A[T]/(T -a) or

Proof. (a) The second statement is obvious since if B can be ordered extending '£, then contains

a = T

is positive in B, hence the contraction of the order to A

£[a]. For the first statement, note that elements of the ring

have unique representatives

b + cT,

b,c e A.

B

If B can't be ordered extending

£, then we have a nontrivial relation

k

i=l

with

2 X

X

X

p i e 'j5 C A. Thus, in A, k k 0 = ( 2 Pi b^) + ( 2 p ^ J a . i=l i=l

(**)

We will prove

k 2 2 p.c. ^ 0 i=l X x

in A.

2 2 Assuming the contrary, then p.c. = 0 and p.b. = 0 , 1 £ j <_ k. Also k 0 = 2 2 p . b . c , b y (*) . On t h e o t h e r h a n d , 0 < p . ( b . + c . ) 2 = p . ( b ? + c ? ) + 2 p b . c . • _ -i

i l l

so the terms

j

2p.b-c.

are all positive.

j

J

J

j

J

j

j

Since their sum vanishes, 2p.b.c. = 0,

121

J

1 £ j £ k. (b)

This contradicts the assumption that (*) is a nontrivial relation. If

'£[a]

is an order and if there is a relation (*) in

then from (**), p.b. = 0, hence

p.b. = 0, if

2

2

Also, p.c.a = 0, hence

p. (b. + c.T)

A

2

A[T]/(T 2 -a),

has no nilpotent elements.

2

= p. (b. -f-c.a) + 2p.b-cT = 0

and relation

(*) is trivial. If we assume that

is an order on A and a is not a zero divisor, 2 2 2 then a relation (*) in B = A[T]/(T -a) implies p.b. = 0 and p.c. = 0, 1 £ j £ H- The argument now proceeds just as in the proof of (a). (c) b

( b + c T ) 2 = b 2 + c 2 a+2bcT = 0 G B, then the assumptions imply

If

= c a = 0, hence (d)

Hence if

'p[a]

If

A

b = ca = 0

and

c = 0.

is an integral domain, then - a £ Dp

a ^ 0, either

admits total refinements, either

or

DjJCA

a £ DJ5.

^ [a]

or

is an order on

A.

Alternatively, since 'p

£[-a]

is an order on

A.

These

two assertions have already been discussed in 1.6. Thus (a) or (b) implies (d). •

Remark.

It does not seem clear whether there is a reasonable

analogue of parts (a) and (b) of the proposition. (A,'£) e (PORCK),

One would hope that if

(*]5+), then one can adjoin a square root of

-a £ D

That is, there ought to exist a

(PORCK)

order on

However, imitating the proof of (a) in the

(PORCK)

a.

A[T]/(T - a ) , extending 'p.

(PORCK)

case does not lead to a

simple relation like (**), but rather to a matrix equation

(***)

k

where

r=

2 i=1

2 2 p. (b. + c.a) i

i

The usefulness of (***)

Proposition 6.2.2. f(T) e A[T] orders on

Proof.

122

i

k

and

s =

Z i=1

2p.b-c., I

I

i

p. e '£, I

ii

is not apparent.

Let

(A,'£) 6 (POR),

A

an integral domain.

an irreducible polynomial of odd degree.

A[T]/(f(T))

p. ,c. ,x,y e A.

Let

Then there exist

extending 'p.

This is a familiar result in the Artin-Schreier theory of

ordered fields. K

First, we assume

be the field of fractions of

we proceed to show

'p A.

K[T]/(f(T))

is a total order on Then

K

A, and we let

is also totally ordered, and

can be ordered, extending the order on

K.

If not, there would be a relation

- 1 = Zg?(T)+h(T)f(T) e K[T]

where degree ( g ^ then

g.^

cannot all be constant, because

h = 0, and we would contradict the fact that

degree than of

< degree (f). The

(Igi(T)) _< 2n-2, and is even, hence

n-2.

h(T)

6.3.

is ordered.

Now,

has odd degree, no greater

By induction we have a contradiction since some irreducible factor

also has odd degree.

•

A Universal Bound on the Roots of Polynomials Let

where

h(T)

K

©

A

be a ring.

For simplicity in what follows, we assume

CCA,

is the field of rational numbers.

Proposition 6.3.1.

f(T) = T 2 n + a T 2 "" 1 + ••• + a

Let

monic polynomial of even degree.

be a

Then there are universal polynomials

3 + (u 1 ... u 2 n ) , 3~(u]L ... u 2 n ) e
€ A[T]

and polynomials

h!(T),

such that

T-3+(a

... a ? j + n

2 ht(T) 2 = £(T) i=l

and

k

3 (a ... a _ ) - T + l zn

Corollary 6.3.2.

If

2 h.(T) 2 = f(T) . i

I i=1

(A/p) e (POR) ,

Q C A

is a monic polynomial of even degree, and if

and

f (a) < 0,

f (T) = T 2 n +

a^2*'1*

a e A, then


Proof.

Substitute

a

for

T

in the polynomial equalities of the

proposition.

•

123

t^T

Corollary 6.3.3.

Same assumptions as above.

~ + ••• + b 2 n - l

a monic

a e A,

be

Let

g(T) = T

polynomial of odd degree.

If

+

g(a) = 0,

then

Moreover, if

a e '•]> and

g(a) <^ 0, then

0 < a < 3 + (b 1 ,...,b 2n _ 1 ,0)

Proof.

Apply Corollary 6.3.2 to

Proof of Proposition 6.3.1.

f(T) = Tg(T).

•

The proof is by induction on

n.

If

n = 1,

f(T) = T 2 + a 1 T + a 2

Thus

and

Assume the proposition for polynomials of degree

2i < 2n.

Let

X = T + (a 1 /2n), and write

f(T) = g(X) = X 2 n + b 2 X 2 n - 2 + b 3 X 2 n - 3 + . . . + b 2 n

where the

b.

are polynomials in the

ai

with rational coefficients.

~±, . n fb2-l^vn-2 hjCX) = X + (^—2—J X

so that

124

Let

h- (X)

= X

+ b2X

and Y 2n ~ 2 +^ bbKXXY 2 n " 3 +^ fu b b

= XX

r^2-l-) 2 W 2 n-4 +Abb, XXY 2n-5

l 4 " ll~2~J J Jx

3

5

Now apply induction to the right-hand side to find

...b J )

Then, since

+

Z h±(X)2 j==2 2 J

2n

3 , h.

such that

= g(X) - h*(X) 2 .

X = T + (a../2n) , we set

hJ(T * 2i) and check the desired formulas.

Remark.

The assumption

ordered ring zation.

If

A

we have

ma > 0

•

J}C A

is not too important.

Z C A, and one can include

implies

a>0,

m G Z ,

Q

In any partially

by a simple locali-

a G A , even this is unnecessary.

For, one can go through the proof above, "clearing denominators". event, one obtains bounds in of

6.4.

f(T), where

m

A

for some integral multiple

ma

In any

of a root

a

depends only on the degree of f(T).

A "Going-Up Theorem" for Semi-Integral Extensions We apply the result of 6.3 to prove an analogue in

theorem in commutative algebra. Then each prime ideal of

A

if the convex prime ideals of B.

A C B,

is the intersection with

If we now assume an order

ideals of

Specifically, let

'pfi C B, extending A

(POR)

A

B

of a well-known integral over

A.

of a prime ideal of B.

'£A C A, it is natural to ask

are intersections of

A

with convex prime

In this generality, the answer is surely no, because one could

have a very weak order on

A

and a strong order on

B.

Thus it is philosophi-

cally sound to make some additional assumption on the orders

'.p. and '£R.

In categories of partially ordered rings, there is perhaps a more natural notion than integral extension, at least for certain purposes.

If

(A,'^A) C (B,'^g

125

we say that

B

is semi-integral over

B.

A

if for all

2n

1

b e B

there is a monic,

+ a 1 T " + • • • + a. , a. e A, with i zn 1 Clearly, integral implies semi-integral.

even degree polynomial in

2n

f(T) = T

We first characterize semi-integral elements in several ways. simplicity, assume

C

For

J}C A.

Proposition 6.4.1. of

f(b) < 0 —

Let

(A/p.) c (C,'Pr) . Then the following subsets

coincide. (i)

{b|a' < b < a"

some

a', a" e A}

(ii)

{b|-a < b < a

some

a e ^A}

some

a e PA>

(iii) (iv) (v)

2

{b|0£b £a 2

{b|f(b ) <_ 0

some monic polynomial

(b|f(b) <_ 0

some monic, even degree

Moreover, if subring of

B C C

f(T) e A[T]} f(T) e A[T]}

is the subset so described, then

B

is a convex

C.

Proof. (i) C (ii)

we have

a' < b < a", then

0 <_ ( a " - % ) 2 , so

-a < b < a

(ii) C (iii)

with

a n < (a") 2 + l. M

2

f

Thus, if

2

a = 1+ (a ) + ( a ) .

if - a < b < a, then

a 2 - b 2 = (a-b) (b+a) _> 0, hence

b 2 < a2.

(iii) C (iv)

(iv) C (v)

If

f(T) = T - a

Obvious,

(v) C (i)

Finally, B

b 2 £ a, let

This is exactly Corollary 6.3.2.

is a subring because (i) is obviously closed under sums and

(iii) is closed under products and negatives.

Also, (iii) (or (ii) or (i))

is clearly convex.

Remark.

The extremely simple characterizations (i), (ii), (iii) of

semi-integral elements make the original definition (v) seem unnatural. However, geometrically, semi-integral extensions are quite analogous to

126

•

the integral extensions of algebraic geometry, as we will see in later chapters. Also, in some situations, it is exactly the characterizations (iv) or (v) which are easiest to verify.

We now turn to our going-up theorem.

The proposition below is surely

capable of some improvement and clarification.

We will present a different

approach for integral domains in the next chapter.

Proposition 6.4.2. (a)

Let

Assume

(A,'£A) c (B,'£B),

Then for any convex prime

c

Q B

A. semi-integral over

A,

fA = A n ^ g .

P C A, there exists a convex prime

Q C B

with

A n Q = P. (b)

Suppose

prime of and

B

}>A/Po

clear.)

C P PQ.

above

coincide on

c A

are convex primes, and

If

'^./P, then

Q C B

P C A If

prime and

A/P Q .

(Only

C B C£ B /Q 0 )

is a convex n

(A/PQ)

$A/PQ = C£ R ^A/P o ) C C^B/QO)

Q, C B

above

P.^

with

is

Q Q C Q1«

is a maximal convex prime, P = Q n A, and

C-Pg/Q)

n

A/P =

is a maximal convex prime.

A C B

Q C B

Q

Assume further that the orders

Then there exists a prime

(c)

(d)

P

is an integral ring extension, P C A

a convex prime above

P, then

Q

a maximal convex

is maximal convex.

Proof. (a)

Let

S = A- P, a multiplicative set.

is a semi-integral extension and

Then

"pR n A^ = '.pA . s S S

(Ac, 'pA ) C (B c , £ R ) b b S S

Thus we may assume

P C A

is a maximal convex ideal. We now claim

/H(PB) c B

is a proper convex ideal.

In fact, yfa(PB)n A = P.

denote the set {y | y G P C A } . To see this, let (for purposes of this proof) P Then vft(W) = {z|0 < z 2 n < Z p ^ ? , j>± e ^ R , y± E P}. Since (B,$

is semi-integral over

(A,'PA), l e t 0 < p . < a . , A

in

£ A = '^B H A .

If

—

1 —

z e A, then convexity of

Finally, with our assumption that it is only necessary to choose

Q C B

P C A

a . G A.

Then

0 < z

P

n

—

1 1

implies

< Ea.y. —

11

z e P.

is a maximal convex prime ideal,

any prime convex ideal containing

127

/H(PB).

(A/PQ, '^A/po)c (B/Qo>'£{

(b) Apply (a) to the semi-integral extension (c) Apply (b) to deduce a contradiction if

P C P C A,

P,

convex

prime. (d) A/P C B/Q field. and

If

is an integral extension and

S = S(l) C A/P

(B/Q) s }> '^R/Q)

is the shadow of

is integral over is a

(A/P) g .

Thus

(A/P, P A /P)

1, then (B/Q) g

is a semi-

(A/P) g

is a field

is a field, hence

semi-field (see Proposition 3.6.1).

D

We conclude with a counter-example to a "going-down" property for semiintegral extensions. Let

Let

R be a real closed field

A = R[X,Y],

B = R[X,Y,Z].

3

S = {(x,y,z) G R 10 <_ x,y,z, z <_ 1, xz <_ y ) . We partially order

a ring of functions on

S, and contract this order to

is certainly a semi-integral extension since

prime ideals

(Y) = P Q C p^ = (X,Y) C A

and

QQ

by a trivial argument.

) C

Then

Q x = (X,Y) C B. QQ C Q1

0 £ XZ £ Y (rel 'pg)

Then

with

did exist, P Q = (Y) C Q Q C Q 1 = (X,Y) But

as

0 £ Z £ 1 (rel '£„). Consider

Q x H A = P 1 ? but there is no 'J>B convex prime Namely, if such a

A.

B

implies

Q Q H A = PQ. gives

(Y)

Q Q = (Y),

is not

'po- convex. The primes

Q1 C B

Geometrically, P the origin and

128

Q,

and

^0>^i

c A

are

convex for '£g, $

corresponds to the (positive) X-axis, P, corresponds to the (positive) Z-axis.

respectively. corresponds to

A potential

Qo C Q

lying over

PQ

would correspond to a 2-dimensional surface in

S

containing some of the Z-axis and intersecting the XY-plane in some of the X-axis.

The only possible such surface is the XZ-plane which has only a

1-dimensional intersection with

S.

129

VII • Ordered fields

7.1.

Basic Results In the first few sections of this chapter, we recall the basic results

of Artin-Schreier on ordered fields.

The last few sections develop the

notion of signed place, which seems very natural in real algebraic geometry. A signed place of a field is a place with values in a totally ordered field and for which we distinguish fashion.

+ °° and

- °° in an arithmetically coherent

In the next chapter, we apply these ideas to establish the basic

results of affine semi-algebraic geometry.

Our organization follows expo-

sitions of Lang [13], [66], up to a point.

In particular, Lang emphasized

the usefulness of real places in real algebraic geometry. To begin, then, recall that a field of squares in

E.

partial orders on

E

is real if

-1

is not a sum

From Chapter I, this is equivalent to the existence of E.

If

ACE

is any ring which has

E

as field of

fractions, then partial orders on E coincide bijectively with derived partial orders on

A

CP ='£<»)•

order refinements.

In fact, if

are refinements of

p.

A real field R

is real.

R

Any partial order

'|J on

a, - a £ p, then both

E

'p[a]

admits total and

is real closed if no proper algebraic extension of

A simple Zorn's lemma argument shows that any real

in a real closed

p[-a]

R, algebraic over

E.

E

embeds

A more precise understanding of

this situation stems from the following, which we established in Chapter VI.

Lemma 7.1.1.

Let

E

be a totally ordered field.

Let

f(x) e E[X]

be an irreducible polynomial, either of odd degree or a quadratic with

a - 4b > 0.

Let

a

be a root of

f.

Then

(hence totally) ordered, extending the order on

130

E(cQ E.

X + aX+b

can be partially D

Proposition 7.1.2.

The following conditions are equivalent on a field

(i)

R

real closed

(ii)

R

real, every element or its negative is a square in

polynomial of odd degree has a root in (iii)

R

not algebraically closed, but

R:

R, and every

R.

R[/^F]

algebraically closed. D

We omit the proof, which can be found in many elementary texts [64], [65]> [66], [67]. The main points are that Lemma 7.1.1 provides the necessity of the conditions in (ii), and Galois theory and the solvability of 2-groups gives the implication (ii) =*• (iii). The following corollaries of 7.1.1 and 7.1.2 are routine, the first two by Zorn's lemma, the third directly from 7.1.2 (ii).

Corollary 7.1.3.

Let

E

be a totally ordered field.

algebraically extended to a real closed field the original order on

E.

Then

E

can be

E, whose order contracts to

That is, each positive element of

E

is a square

in I".

•

Corollary 7.1.4. istic

0.

Let

be an algebraically closed field of character-

Then there are real closed fields

Corollary 7.1.5. Then

K

Let

E

be a real field

If, the algebraic closure of

Remark 7.1.6.

Thus the

R

in

with

E C R,

R

of

R

K = R[/Tj.

E.

D

K = C, the complex

uncountably many pairwise non-isomorphio real IR/with

C

isomorphic (algebraically) to

R

of 7.1.4 is highly non-unique.

Remark 7.1.7.

•

real closed.

R, is a real closure of

Artin and Schreier showed that if

numbers, then there are closed subfields

E

R C K

Among other results of Artin-Schreier theory is the

fantastic characterization of real closed fields as the only fields of finite codimension under their algebraic closure.

We will not use this

result.

131

T

7.2. Function Theoretic Properties of Polynomials If

E

is a totally ordered field, any polynomial

be interpreted as a function

f:

E -*• E.

f(T) G E[T]

can

We will use the usual symbols

(a,b), [a,b], [a,b), (a,b], for open, closed, and half open intervals in We also allow

a = - °° ,

b = + ° ° , with the obvious meaning.

Similarly, it

is clear what me mean by

f

on an interval and by

positive or non-negative on an interval.

a € E , we define

f

|a| € E

E.

(strictly) increasing or (strictly) decreasing

by

|a| = ± a,

|a| >_ 0.

If

The following is proved

in general just as it is for real numbers.

Lemma 7.2.1(a). ordered field.

If

f (T) = a Q T n + a^ 1 1 " 1 + • • • + a R G E[T],

Let x,

b E E,

E

a totally

|x| £ |b|, then

|f(x)| < 2 |a. Mbl""1 . i=0 (b) If a € E

is a root of

f, then n

|a| <_ I |a. |/1a. | . i=0

then

Corollary 7.2.2.

Polynomials are "continuous".

Corollary 7.2.3.

If

f

at

p GE

T = 0

•

is a monic polynomial of odd degree,

is negative on some interval

some interval

If

f(T) £ E[T]

D

x

(- «», a)

and

f

is positive on

(b,«>).

•

is positive, the polynomial

and positive if

T > 1 + p.

T -p

is certainly negative

From this fact, 7.1.2 and 7.2.3 we

obtain the "if" direction in the following characterization of real closed fields.

Proposition 7.2.4.

A totally ordered field

and only if for every polynomial

f(T) E R[T]

f(a) < 0 < f(b), there is a root

c E R

132

of

R

and f

is real closed if a,b G R

between

a

with and

b.

Proof:

The algebraic closure of a real closed

R

is

R[/T].

We

thus can factor

where the

r. E R

are the real roots and the

pairs of non-real roots. b

It is then clear that if

must lie on opposite sides of a real root

Remark: that if

a. _+ $. *^T

r.

are the conjugate

f(a) < 0 < f(b), a

of odd multiplicity

and

m..

•

The existence of this simple factorization also easily implies

f(T) G R[T]

f(T) = gj(T) + g*(T)

is nowhere negative on for suitable

g][ ,g 2

R,

R

real closed, then

e R[T].

Next we consider the geometric behavior of a polynomial function near a point

x = a, over any totally ordered field

f(x)-f(a) =

E.

f(x)

We can write

.,(a) (x-a) 1 .

I

Using the estimates of Lemma 7.2.1, one can verify the following.

Proposition 7.2.5.

Let

f(T) G E[T],

E

m ^ 1

be smallest such that

f ^ C a ) ^ 0,

a £ E.

in

such that:

E

(i) If

m

even, f

(a) > 0, then

(a - £, a + e ) , decreasing on (ii) If

m

odd, f

(a - e, a + e ) , negative on Moreover, on these positive when

f

a, the derivative

We obtain statements for the two cases when f(T)

by

-f(T).

f

(a,a+e).

is increasing on

and positive on

is increasing, negative when

e > 0

is non-negative on

and increasing on

f(x) - f(a)

(a-e,a)

e-intervals around

Then there is

f(x) - f(a)

(a-e,a)

^(a) > 0, then

totally ordered, and let

ff

(a,a+e). of

f

is

is decreasing.

fW(a) < 0

O

by replacing

We draw the four possibilities for local behavior below.

Combining this local behavior of polynomials with the characterization 7.2.4 of real closed fields, we can deduce certain global results about the behavior of polynomials over real closed fields.

133

i,

f ( m ) (a) > 0

m

fW(a) > 0

odd,

f• > 0

f' > 0 /

x= a

< 0

f1 > 0,

m

even,

vf' < 0

f ( m ) (a) < 0

m

odd,

f^(a)

< 0

Figure 7.2.6

Proposition 7.2.7. a < b

in

If

is real closed, f(T) G R [ T ] ,

R

R, then the derivative

More generally, for any

of

f

has a root

f(T) G R[T], there exists

*ir^

Proof:

f'

c,

c G (a,b)

f(a) = f(b) = 0, a < c < b. with

- f(b) -f(a)

The second statement (Mean Value Theorem) follows from the first

(Rolle's Theorem) in the usual way.

To prove Rolle's theorem, one first

assumes

a

f(T)

has no roots between

root closer to

a

if necessary.

and

b, by replacing

By 7.2.3, f

b

by another

does not change sign on

(a,b).

Then a short case-by-case consideration of the local possibilities for near

a

f'(a+e)

and and

b, drawn in Fig. 7.2.6, implies that for suitable f' (b - e)

Corollary 7.2.8.

134

have opposite sign.

Let

R

f

e > 0,

Now 7.2.4 completes the proof. •

be real closed, f(T) e R[T].

Then

f

is

monotonic on any interval [a,b],

closed interval the endpoints

a,b

(a,b) f

not containing roots of

f'.

On any

assumes maximum and minimum values, either at

or at roots of

f'

(a,b).

in

•

7.3. Sturm's Theorem Let

E

be a totally ordered field, f(T) G E[T]

polynomial.

A priori, the number, or even the existence, of roots of

in a real closure of

E

E

any real closed field extending

Proposition 7.3.1. E.

Let

for counting the real roots of

E.

uniqueness of the real closure of

real closures of

f

might depend on the choice of real closure.

theorem provides an algorithm in

over

an irreducible

E

Sturm's f(T)

in

We first show how this implies the

E.

be a totally ordered field,

R,,R~

Then there exists a unique homomorphism

two

R.^ -> R 2

E.

Proof: gives

Uniqueness follows from existence, since, first, symmetry

R 2 •> R,

over

over

E, since any

E.

But

R, •> R,

R,

admits no non-trivial endomorphisms

maps squares to squares, hence preserves order,

but also induces a permutation of the roots in

R,

of any polynomial over

E, which are, of course, finite in number. Sturm's theorem implies that if f(T) G E[T], then E(a) -> R 2 a

o

< a

l

<

over

''' <

a

k

f

has roots in

E.

Let

^"^

a

a £ R-,, with minimal polynomial

R2-

a Q ---a^ G R^

-+i " a i

=

2 "Y'+i"

Thus we obtain algebraic maps be all the roots of The

tneorem of tne

f, and let

primitive

element implies find

E(a ••• a,, y, • • • y v ) = E(3), and by the above, we can O K 1 K E(3) -> R~. This gives by restriction E(a ••• a,) -*- R o , keeping Z O K Z

the roots of

f

in order.

by symmetry, since other hand, if extension

f

There is clearly a unique such map over

has no more roots in

0 < y G E(a • •• ou )

E(3,y) -*• R2> hence

Zorn's lemma now gives

and

R2

y = y

E(a Q ••• a^) -»- R 2

than G R

in

R,.

E,

On the

then we get an

is order preserving.

R, ->• R ?.

•

135

We turn now to Sturm's algorithm. £(T) £ E[T], or more generally, any f 1 = f»

and define

deg(f i _ 1 ).

f^

i >_ 2

f

by

(f,ff) = 1 , we get a sequence

Since

f^, let

(f Q(x) ^

sequence g.-f. - f.

fi ,

of all the

of

f,

[a,a,), (a , b ] . and

x',x"

E [T] , d e g C ^ ) <

£ Q ,f 1 ,...,£ k

a non-zero constant. f..

of Let

For each

x G [a,b],

From the defining equations

f.

of

fj, in the interval

fQ = f,

f^

1

=

is a non-zero constant, we see that if

i >_ 1, then R

e

be the number of sign changes in the

(x) , ... ,f, (x)}.

In some real closure

x' < x"

5(x)

and the fact that

is a root of

and on

f,

be two points which are not roots of any

not a root of any

a

(f,f!) = 1, let

with

f±_2 = g±_1^i_1 - f ±

polynomials o£ decreasing degree with a,b €= E

Beginning with an irreducible

(a)f.+1 (a) < 0. ct-^ < • • • < a n

E, let (a,b).

Then

6(x)

is constant on 6(x f )

We ask the relation between

are separated by a single

f = f Q , the sequences defining

f

6(x )

a..

and

be all the roots

If M

6(x )

a.

(ou,a-

),

6(x M ), if

and

is not a root

may differ at

{... f.^Cx'^f.Cx'^f^Cx') ...} and {... f^Cx'O.fiCx'^f^Cx") ...}. But even if

f^

changes sign at its root

a., so that

f^(x')

and

f i(x")

have opposite sign, one checks that the number of sign changes, reading across the two sequences, is invariant, 6(x') = 6(x"). is a root of

f = f Q , then the irreducibility of

sign across f

a.

but

f

does not.

Comparing

M

On the other hand, if f

implies xl

{fo(x' ),f1(x ),---}, we see, in fact, 6(x ) - 6(x") = 1.

{+>+t •••}•)

changes

x

{f Q ( )>?iC ')»•••}

f

7.2.6 shows the two sign sequences are

f

with

(The local picture

{+,-,•••}, {-,-,•••} or

{-,+,•••},

We conclude then, with the notation as above:

Proposition 7.3.2.

The number of roots of

hence is independent of the real closure

Since we can first compute all the all roots of all

fi

by

R

of

f

in

(a,b)

E

chosen.

f., and then choose

is

6(a) - 6(b), D

(a,b)

We state a consequence of 7.3.2.

Proposition 7.3.3.

containing

7.2.1(b), we can count all the real roots of

by this algorithm.

136

a.

For each

n >^ 1, there is a finite number of

f

systems of finitely many inequalities g.ji

e

sucn tnat

Qt^i'•'"tjj]* n

given any totally ordered field

n

f(T) = T + a 1 T " + • • • + a n G E[T] roots), then systems

Si

f

E

with

E

(or, more generally, f(T)

has real roots over

holds in

S. = {g..(t-...t ) > 0} , with

E

and irreducible with no multiple

if and only if at least one of the

t^ = a^, that is, for some

i, all

g..(a ...a ) > 0.

•

Remark 7.3.4. set of systems

The Sturm algorithm itself would give a rather inefficient

S^.

Of course, 7.3.3 could be improved to a statement con-

cerning the precise number of real roots in a given interval

(a,b), again

using nothing but the Sturm algorithm. Despite the totally elementary character of the proof of Sturm's theorem, (which is basically no harder than the statement itself), it seems reasonable to believe that added insight into 7.3.3 could be derived from Galois theory. The coefficients

a.^ of

f(T) G E[T]

are the elementary symmetric functions

of all the roots in

R[/-T], and somehow the signs in

combinations of the

a.^ should determine not only the real roots and their

E

of various rational

location, but also the location of the real and imaginary parts of the complex roots and other information about real subfields of the splitting field of

f.

The determination of whether a quadratic has real roots is trivial: T +aT+b

has real roots if

r

d = a - 4b > 0.

d = -(4p + 27q ) , we have one or three real roots.

d < 0

or

d > 0

T + p T + q, with

accordingly as the cubic has

It turns out that a quartic

two real roots if the discriminat p < 0

For the cubic

2

d < 0

T + p T + qT+ r

and four real roots if

has

d > 0,

8pr- 2p 3 - 9q 2 > 0.

and

7.4. Dedekind Cuts; Archimedean and Non-Archimedean Extensions Let

E

such that if and if D

be a totally ordered field. b G D,

a < b, then

D = E, we write

is rational if

D

D = + <» .

a G D.

By a out we mean a subset If

D = 0, we write

D = - °° ,

All other cuts are called finite.

has a maximal element or

E-D

D C E

A cut

has a minimal element

137

b G E.

We write

D «* b

two cuts rational at f

D » D

E C F

a e F

D

is rational at

b, one contains

if the two cuts

Let Each

if

D, D

1

b

the other does not.

D

of

E

D E(a)

of

E

algebraic if

a €= E, the real closure of

Proposition 7.4.1. ordered fields

E.

We write

differ by at most a single element of F

E.

extending

E.

by

D E (a) = {a e E|a < a

(i) If

Note there are exactly

be totally ordered fields, the order on

defines a cut

We call a cut

b.

in

F} .

D = +_ °° or if

D »

D

E (°0

for some

All other cuts are called transcendental.

The following are equivalent for an extension of

E C F:

a G F

(ii) D E (a)

and

Dg(a)

is rational, then

is finite for all

(iii) For all

e > 0

Proof: (i) =* (ii)

in

F

a € E.

a 6 F.

there exists

Suppose some

e > 0

in

D £ (3) = +_ °° •

E

Then

with

e > e.

3 £ E, but

D E (l/3) » 0. (ii) =* (iii) elements of

E.

Suppose

Then

(iii) =* (i)

e > 0

in

F

was smaller than all positive

D (1/e) = + °° . D E (a) » a G E, a £ E.

Suppose

is positive, but is smaller than all positives of conclusion holds for

If

a < a, then

E.

If

I/a- a

a < a, the same

I/a-a.

D

Extensions satisfying the conditions of 7.4.1 will be called Archimedean.

Lemma 7.4.2. (a) If over (b) If

E C F

and

F

is algebraic over

E, then

F

is Archimedean

E. E C F

is Archimedean and

F C K

is Archimedean, then

E C K

is Archimedean.

Proof:

138

The first results follows from the

bounds on algebraic quantities

given in 7.2.1.

The second result is clear from any of the conditions in 7.4.1.

Remark 7.4.3.

© C IR, the real numbers over the rationals, contains all

Archimedean extensions of

Q

and

IR admits-no proper Archimedean extension.

IR is unique with this latter property. properly containing Dedekind cuts of Q

E

E

is a totally ordered field

IR, then one cannot make sense out of arithmetic of

E.

For example, the smallest cut

cannot be distinguished from

sequences in

If

2D, whatever

2D

D

of

means.

E

containing

Also, Cauchy

are not generally interesting, for example, if

E

has

uncountable transcendence degree over IR.

Remark 7.4.4. over

Although the real closure of

E, it does not follow that

E

is order dense in

certain topological reasoning precarious. of

real closure of

IR(t).

F.

E.

This fact makes

R(t)[X]

which are strictly

IR(t), yet which have roots and change sign over the (See 8.2 for specific examples.)

We now point out how an extension valuation ring in

E, is Archimedean

For example, given any ordering

IR(t), there are irreducible polynomials in

positive as functions on

E, say

E C F

of ordered fields defines a

This construction and its uses goes back to Krull [12].

We define A E = {a e F|DE(a)

If

3 £ A £ , then

is finite} .

D E (3) = +_ °° , hence

D E(l/3) » 0, so

A£

is a valuation

ring.

Proposition 7.4.5. maximal ideal is

residue field over

A£

is a totally ordered valuation ring in

Q c = {a £ F|Dc(a) » 0}, and E E

A £ = A £ /Q E

Qc E

F.

is a convex ideal.

is totally ordered, contains

E

The

The

and is Archimedean

E.

D

This result is checked routinely, so we omit the details. Note that given over

E.

If

E C F, we have

E C K C F

and

K

A£ = F

if and only if

is Archimedean over

E, then

F

is Archimedean

A n = kv. E

Lemma implies that any

K C F, Archimedean over

Zorn's

K

E, is contained in a subfield

139

•

maximal with this property.

We say

maximal Archimedean over itself in F

implies

E

in

F.

F

If

E

In particular, E

and

A£

F

if

E

is

Archimedean closed in

F.

is Archimedean closed in

is algebraic over

Ap/Qp = A p

closed in

is Archimedean closed in

algebraically closed in

Proposition 7.4.6. field

E

E.

F, then the residue

Conversely, if

is algebraic over

E, then

E

E

is algebraically

is Archimedean closed

F.

Proof: over

E.

Suppose

Then

t £ Ap

ECE[t]

is Archimedean over

and the image

and

t

in

Ap

Q E n E[t] = (0), hence

is transcendental E(t) C A £ .

Thus

E(t)

E, contradiction.

Conversely, if

t €= F - E

Q £ fl E(t) = (0), hence

A£

and

E(t)

is Archimedean over

E, then

contains the transcendental extension

E(t).

D

7.5. Orders on Simple Field Extensions We will classify all total orders on a simple field extension E(x), extending a fixed total order on If E

and

x

is algebraic over

E(x)

E.

Let

E(x)

in

f(t)

not separated by any element of

E

E(x)

However, let

E(x) if

E(x 1 )

and

in

and

(see 7.4.4).

be two roots of

E(x 2 )

coincide in

in

R.

E,

Dg(x).

R, we don't distinguish the orders on x

to

x.^

x 2 , respectively.

and

R

over

However, E

E, by the proof of

But this is impossible, by the same proof.

has a real closure, which contains

Thus orderings on

140

R

Of course, if the two

We now turn to orders on the transcendental extension

way.

f(t). This is

are distinct, any order isomorphism between them over

uniqueness of real closure.

E(t)

correspond

Thus, there may exist several

f (t)

would extend to a non-identical endomorphism of

ordering on

E(x)

may have two roots (or more) in

*pX2

E(x 2 )

The orderings on

R, which define the same cut of

induced by the two maps taking E(x 1 )

R.

E.

f(t) € E[t], then

R, that is, to the real roots of

not completely obvious, because

embeddings of

be the real closure of

E, with minimal polynomial

have the same real closure

to embeddings of

fields

R

E(t)

E(t). Any

R(t), in a natural

are seen to correspond bijectively to orderings

on

R(t). Recall total orderings on

Proposition 7.5.1.

R(t)

A total order

correspond to orderings of R[t].

p

on

R[t], extending

D = D R (t)

closed, is uniquely determined by the cut

of

R.

R,

R

real

The possibilities

are: Case (i)

D

infinite.

If

D = oo , ']5 = ' ^ = {f G R[t] |f((b,~)) > 0, some

If

D = - oo, p = $_m=

Case (ii)

D

rational, say

p = -p

If

Then:

a £ D

^= ^a Case (iii)

=

D

•p = p

{£ G R[t] | f ((- oo,b)) > 0, some

D *» a G R.

in

aGD

^f G R^ lf ^ a " e » a^ > °> some

transcendental.

Moreover, R(t)

b G R} .

(that is, a < t

e > 0

in

in

R(t)), then

R} .

R(t)), then

e>

°

in R

-^ *

Then

= {f G R[t] | there exists

Proof:

If

= {f G R[t] |f ((a,a + e))> 0, some

(that is, t < a

bGR}

is Archimedean over

R

a,b G R, a < D < b, such that

if and only if

First one checks the orders

^m, '$ „, '-P

D

+,

f((a,b)) > 0} .

is transcendental

'-P

, '^D

are, in

fact, total orders on R(t). Secondly, the case with

D

infinite is dealt with by replacing

t

by

1/t,

D(l/t) » 0, hence rational. Thirdly, suppose

D » a, as in case (ii), with

which we may assume monic, factor

f(t)

where

a. ± /-T 3-

(Recall

r.

are the real roots and R(/TF)

if and only if

over

is the algebraic closure of r. < a 1 —

in

R, if and only if

R.

a < t.

Given

f(t) G R[tJ,

Thus

are the non-real roots. R.)

Obviously, (t-r.) G £

(t - r.) G '$ 1

.

Thus

'p = '£

3-,+

3.,+

141

as asserted.

If

if and only if

t < a

in

'£, then

(t-r.) G'.]}

a,—

We sketch a second proof that

'$ = '$

if

+

a < t,

D = D R (t) » a.

f(t) = f(a) + f ^ (a)(t-a) + ••• + (f^(a)/n!)(t-a) n =

Begin by writing

(f^(a)/m!)(t-a) m (l + terms in (t-a)), for some In the abstract order

'^ on

small relative to

Thus the 'p-sign of

f^

R.

R[t], t - a

m > 0, with

f^

is positive, but infinitesimally f(t)

'£ = '$

, as asserted.

The case

is simply the sign in

t < a

We point out that the replacement of

R

t

is analogous.

by

1/t, converting behavior

of functions "at +_ «>" into behavior "to the right or left of formulated in terms of coefficients. n

a Q t + • • • + a n , then then

f 6'|5 T

if

n

f G ^

is even, a

> 0

odd,

D = D R (t)

and

closer to

D

f

R(t), then the polynomial some real field containing between

a

f

(a,b)

D(t) = - oo ,

One simply

R. and

By Sturm's algorithm, f

(t-ri) G $

r i G D, if and only if

A R C R(t)

in the case

R.

so we may assume

t G AR.

f.)

If

D

f(t)

QR C AR

is a non-trivial, convex prime ideal of D = D D (t) » a.

had

'£

on

and

b

Thus

in

'p = '$~.

transcendental can be

(t-r^ G $

Obviously,

D

-

be the valuation ring associated to

Conversely, assume If

a

(Simply

would therefore have

II (t-r.)((t-a.)2+ 3?). x i,j 3 ^

is infinite or rational, then

Archimedean over

(a,b).

b, contradicting our choices.

f(t) =

Finally, let

f G R[t], choose

would have a real root between

based on the factorization

if and only if

For each

in the abstract order

'p = '$

142

If

r

has constant sign on

A simpler proof that

a G R, hence

< 0.

than all the roots of

the opposite sign as its values on

D = D R (t)

R.

O

is transcendental.

choose

b

a

in

t .

such that the function

R

n

aQ > 0

f(t) =

n

a < D < b

a root in

or

0" is easily

D(t) = + oo , and

O

Next, suppose

a

Namely, if

if and only if

-oo

divides through by

some

(a) ^ 0.

(a). By the discussion of local behavior of functions of 7.2, we

deduce

If

r. < a,

(t-r.) G'p 1

of

if and only if

A R ± R(t)

and

A R f R(t). Either

R(t) t

or

is the maximal ideal, then

R C R(t). is non1/t G A R , Q R fl R[t]

R[t]. Necessarily, then, Q R n R[t] = ((t-a]), D

Corollary 7.5.2. Suppose

§!>•••»§!<

e

R

Let t

t ]

R(t)

are

be partially ordered, R

finitely many elements positive in the order.

Then there is an interval of values

Proof:

real closed.

(a,b)

in

R

with

gi((a,b)) > 0, 1 £ i £ k.

Extend the partial order to a total order and consider the

classification of total orders directly.

Corollary 7.5.3. Suppose on

E,

E

E

is order dense in

f(t) G E[t].

Proof:

Then

be totally ordered with real closure R

f(t)

and

f(t) < 0.

f(t)

f

is non-negative as a function

2 2 h f = Zg.

By 7.5.2, f(t)

(a,b) C R, hence

R.

is a sum of squares in E(t).

If there is no equation

be ordered with interval

Let

•

in

Eft], then

Eft]

can

is strictly negative on some

is negative at some point of

E, which

contradicts our assumption.

Remark. f(t)

•

It is known that if

f(t)

is actually a sum of squares in

is a sum of squares in

E(t), then

Eft]. See [ 4 ] , [ 5 ] , [ 6 ] for

detailed discussions of this point.

Corollary 7.5.4. (a) Let

E

be totally ordered with real closure

is ordered, non-Archimedean over residue field

A = A£/QE

R, and suppose

E, with valuation ring

A p C E(t).

is then a simple algebraic extension of

E(t) The

E,

A = E(a). (b) If in

t 6 A E , let

E(t). Then

Eft]

HEft],

g(t)€E[t]

with degree

g i (t) < degree

and only if

g(a) > 0

Proof:

f(t)

A = E[t]/(f(t)), and the ordering on

corresponds to a choice of a root of (c) If

Eft]

(f(t)) = Q

in

and

irreducible, A

f(t) > 0

induced by that of

f(t), a £ R.

g(t) = f(t)m (g Q(t) +g x (t)f(t) + ••• + g k (t)f k (t)),

f(t), g Q (t) i 0, then

g(t) > 0

in

E(t)

if

R.

Assertion (b) implies (a) since either

is a principal ideal domain, the localized ring

t

or

1/t £ A £ .

Eft].^^..

Since

is the

143

valuation ring

Ag, and

Q E = (f (t))E[t] , f , _ . .

This proves (b). Finally,

(c) is more or less obvious.

•

7.6. Total Orders and Signed Places Just as valuation rings in fields correspond to places, we will show that totally ordered valuation rings with convex maximal ideal correspond to a special type of place, in which we distinguish place on a field

F

+ °° and

immediately yields a total order of

totally ordered valuation ring in

- °° .

A signed

F, in fact, a

F, with convex maximal ideal, and con-

versely, such a valuation ring yields an equivalence class of signed places. Let to

A

A

be a totally ordered field.

We adjoin symbols

- ©o

and extend the operations of addition, subtraction, multiplication,

and division between

°°, - <» and elements

then justify the notation

Addition

a E A

- °°.)

00+00=00 (_ 00) + (_ 00) = _ 00 a

+ 00 = 00

a + (- 00) = _ 00

Subtraction

00 - (- 00) = 0 0 r_ 00} - 00 = - 00

a

- 00 = _ 00

a - ( - 00) = ° o

Multiplication

°° • °° = (- °°)• (- °°) = °° 00 • (_ 00) = _ 00

ifi

a > 0

I - 00 ifi

a < 0

r

a • (-00)

144

oo and

oo 00

r-00 = j u 00

if

a > 0

if

a < 0

as follows.

(These rules

Division

a/°o = a/-00= 0

o°/a = (l/a)« 00

0* ( ± ° ° ) , ± °°/0,

the four possibilities

a ^ 0

if

a / 0

(l/a)-(- 00)

- 00/a =

The symbols

if

00 + (_ 00),

00 - 00 ,

± 00/+ 00 are undefined.

(- 00) - (- 00)

and all of

Of course, we want the usual

commutative and associative rules to hold, in all defined expressions. that we do not distinguish the signs of both

0. If

a/- °°, a ^ 0, but call

These elements can be distinguished by "inverting". K

function and

a/°° and

is a field by a signed place, with the values in A, we mean a

p:

K->A, ± °°, such that

p(x+y) = p(x)+p(y),

p(l) = 1, whenever the terms are all defined.

guarantees

p ^ 0

and

the ring

A

Let

p:

= {x G K|p(x) G A}

The last condition simply

K -• A, ± 00 be a signed place on is a valuation ring.

Moreover, K

ordered if the strictly positive elements are defined to be p(l/x) = °° , or

p(x) > 0

the maximal ideal

Proof: p:

Q

in

C A

A}.

If the order on

products

K

x, -x G p . xy

and

becomes a place on

K.

It remains to show that

x + y, for

The possibilities for

Then

is totally

'£ = {xEK|p(x)= °°,

is restricted to

First, if we simply suppress the distinction between K -* A, 00

K.

A ,

is convex.

Thus

Secondly it is clear that for all non-zero one of

p(xy) = p(x)p(y),

p ^ °° .

Proposition 7.6.1.

- °° ,

Note

A

is a valuation ring.

x G K,

'£

+ °° and

x

G '.p+, and exactly

is closed under sums and

x,y G ']> .

p(x), p(y)

are covered by (i) p(x),p(y) > 0,

(ii) p(x) = 00 , p (y) > 0, (iii) p(i) = °° , p(y) > 0, (iv) p(x) = °o , p(y) = 00 , (v) p(-) = 00 , p(i) = 00 , (vi) p(x) = 00 , p(I) =00. it is trivial to check directly that evaluate either p(xy) ^ 0.

If

p(xy)

or

p(xy) < 0

This contradiction shows

p(l/xy) or

xy G'|) . directly.

In case (vi), one cannot By symmetry, we may assume

p(xy) = - «> , then

p(xy) > 0

or

i n all cases except (vi)

p(x) = p(xy y) = p(xy)p(-) =

p(xy) = °° .

The six cases for sums

145

are all trivial except (v). But

x + y = xy(— + — ) , x y by product cases, and case (iv) for sums. Finally, Q homomorphism

C A

A

is convex as the kernel of an order preserving

-*• A.

A signed place

•

p:

K -• A, ± °° determines a total order on

given a valuation ring ideal

Specifically, if

in a totally ordered field

1/x £ Q, then we set

and

p(x) = - °° if

and

p(xy) = p(x)p(y)

with in

A

Q C A, one recovers a signed place

that since

A

x

is negative.

This shows that if

Verification of

ap^

Here

valuation ring of

and

p^:

p i (K)

p..

valuation rings in

K -• A^, ± °° ,

Remark 7.6.2.

g:

K

P-^A

A

is all of

K.

maximal ideal, one sets becomes the "addition" in

of

K,

K

G Q.

Thus

) C A ^ where A

C K is the

and totally ordered A signed place

p

of

K

K.

correspond to a third concept, that of

A valuation is a function

If

A C K

T = K*/A*, T

and

A - (0) = {x G K|v(x) :> 0}

A* = A - Q. 1 E K*

and

v:

K* -* V, where

K* = K - (0), with

is a valuation ring with

The natural projection

valuation, v(x) = v(-x).

146

0 <_ 1 £ 1 + (y/x)

In commutative algebra, equivalence classes of places on

and valuation rings in

y/x E A.

K

with convex maximal ideal.

v(x+y) >_ min(v(x) ,v(y)).

if

1

p-^ (K), ± ° ° ^ p 2 ( K ) , ± °° such

a totally ordered (additive) abelian group and

x <_ y

x + y = x(l + (y/x))

We then see that there is a bijective correspondence

a (Krull) valuation of

and

Note

i = 1,2, are equivalent if

means the subfield

is trivial if its valuation ring

K

K

p(x+y) = p(x) +p(y)

l/(x+y) = (1/x) (1 + (y/x))"

between equivalence classes of signed places of

a field

is positive in

p(x) = °° , p(y) = °° , then

there is an order preserving isomorphism p2 =

x

A = A/Q.

This is perhaps the trickiest verification.

We say signed places

that

K -* A, ± °° where

p(x) = °° if

Conversely,

consists of a routine case-by-case discussion.

1 + (y/x) £ Q

p(x+y) = °° .

p:

K.

K, with convex maximal

is a valuation ring, we may always write

y/x E A.

A, hence

so this case is covered

"0" 6 F .

v: K* -*• V

Q CA

We order

the

K* T

by

is then a valuation

Q - (0) = {x € K|v(x) > 0}.

is

v(xy) =v(x)+v(y)

The multiplication in

becomes

T

For any

Now, if

K

is totally ordered and

A C K

maximal ideal, then the sign homomorphism a(x) = + 1 to

v:

if

x

K* -> T

the

x.

hence

v(x-) > 0.

x^ positive and

a(x i ) = + 1

and

v(Ex i ) > 0, then

x

is negative, is related

v(x i ) >_ 0,

If

a(x i ) = + 1, and

x- € Q.

v(x i ) > 0

In fact, we can state if

for any

But then

1 <_ E(x./x )

in

x i G K. Q, then

Namely, if v(l/x ) >^ 0,

A, which contradicts

Q.

cr(-l) = - 1 ,

K, and a sign homomorphism

cr(xy) = a(x)a(y)

one recovers a total order on is a valuation such that then the valuation ring such a pair

if

x , did not belong to

Conversely, given a field with

a(x) = - 1

Ex. £ Q, then

v(E(x./x )) > 0.

K* -»• {± 1}, defined by

Translating, this condition simply says if

of minimum value, say

convexity of

a:

in the following manner.

v(Ex-) > 0, then x- £ A,

is positive and

is a valuation ring with convex

(v,a)

the elements

K

and

by

$

cf(x-) = + 1 A

of

v

a(x+y) = 1

+

a(x) = a(y) = 1,

and

v(Ex-) > 0

implies

has convex maximal ideal K, since

not distinguished by

Although the residue fields

K* -* {± 1}

= {x £ K* |o*(x) = + 1}.

a split valuation on

+ x, - x

if

a:

A/Q = A

a

Q.

If

v:

K* •> r

v(x.) > 0, We call

serves to distinguish

v.

and the domain

K

of such valuations

are certainly restricted (they are totally ordered), no general restrictions are implied on the value group

T

of a split valuation.

We have seen in 7.4 that if subfield, then pE:

place and

E1

is

E

Archimedean over pp

Let, then, p:

A = A £ /Q E . E.

p| p

is a

A E C K, hence a signed

Moreover, p E = p £ l if

ECE1 C K p

We may assume

Consider the family of subfields is trivial.

For example, Q C K

p

E C K is such a

E.

Proposition 7.6.3. (a) A

K

is

Zorn's lemma implies that there exist maximal such subfields

and we fix such an

on

E C K.

K •> A, ± °° be a signed place.

such that the restriction

E C K

We show now that any signed place

for some subfield

surjective, that is, p(A ) = A.

subfield.

is totally ordered and

defines an (ordered) valuation ring

K -+ A, ± °° where

is equivalent to

K

is algebraic over its subfield

p(E) C A.

147

E C K,

(b) The place (c) E

p

is equivalent to

is Archimedean closed in

pE<

K.

Proof: (a) Let

t £ A - E, so

p(t) £ A.

homomorphism by the choice of (b) We prove

= A£.

A

the induced order on p(E) C A.

If

E, hence

p:

p(t)

The signed place

x £ A , then

p(x) £ A

for some

for some

of

PE = p

p:

K -> A, ± °° orders

is algebraic over

e E E. Thus, x €= A £ .

e *= E, hence

and

!

E , hence

1

E

in

A

and

x £ A£, xE A .

C K were Archimedean over

E = E', by the maximality

E.

•

is a signed place on to the order on

K

K

Suppose

A

and

L ->• K, ± °° is a signed place on

q:

induced by

p.

is totally ordered, p:

Then

L, which induces the same order on

po q: L

as

a signed place on

K

and surjective, then L.

(If

f

q:

q (a) G A

CK

L

qf:

K.

•

L -*• K, ± °° is q1

If

is non-trivial

L -> A, ± °° will never be a signed place

"changes sign" and

q' (b) = <» , then

Nonetheless, if we suppress the distinction between is a place on

L, relative

L •> K, ± oo .

relative to some new order on r = poq':

K -* A, ± °°

L -> A, ± °° is a signed place

The proof is routine. On the other hand, suppose

on

K

p(E) by (a), hence

Conversely, if

p(-e) < p(x) < p(e)

would be trivial on

Proposition 7.6.4.

on

is a non-trivial

is algebraic over p(E).

(c) This is a consequence of (b) since if E, then

E[t] -> A

E C K of course coincides with the induced order on

p(-e) < p(x) < p(e) -e < x < e

Then

+ °° and

and we will see in the next section that

L

r(ab) t r(a)r(b).)

- °° , po q1 : L-*-A,°° can be ordered

so as to convert any real valued place into a signed place.

7.7. Existence of Signed Places We have seen in previous sections how signed places of a field closely with total orders on

K

together with a subfield

E C K.

K

In this section

we show how, beginning with less data, signed places can be constructed.

148

correspond

The

first main result, which is due to Krull [12] (and in a different form to R. Baer, according to Krull) is that if of

K

and

A

refined to a signed place, p : (K,:p)

K •> A, °° is an ordinary place

is a real field, then given any total ordering of f

if

p:

K •*- A, ± °° .

is a partially ordered field, A C K

a subring, P C A

p:

on

(A ,Qp)

(that is, A C A , Q

and maximal ideal of to refine

'p

ideals of tf w

n

A

A

that is,

A = A/Q

Let

p:

Suppose are

A

is a real field.

£ w n A - convex, where

Suppose

Ex. = 0,

x i /x Q ^ A.

1 + 2 p(x./x ) 1 i^O originally. Thus

Then

in

say

x. € K. 2

Since

Q.

G Q.

Finally, let

If some Thus

'£

Let

K

with valuation

is real and all prime

is the weak order on

x

K

(so

have minimum value for

= 1 + 2 (x^x ) i^O

2

=0

in

p,

A, hence

0 =

x. = 0

be a prime ideal of

P, that is, p A, p . = P.

first part of the proof, A p

A. p . C K

Let

H A - convex in

A.

P

f

n A

= 0 €= A

is a contradiction.

A standard result of valuation

is a valuation ring with maximal The

pair

A p , with residue field

is a real field.

is

A.

A p = A. p ./P.

with valuation ring

part of the proof that

x^ €= Q, it follows that

( 1 / X Q ) 2( E X ? ) = E(xi/xQ)2 =

But, again, 1 + Ep(x./xQ)

P C Q

K

If we show all

l/xQ 6 A , hence

valuation ring and maximal ideal in

$

Then

K

x. £ Q, then the term of minimum value,

theory is that the localized ring

of the place of

will be Archimedean over the

is real, we must have had all

A

is real. 2 Ex. £ Q, x. €= K.

x Q , is not in

ideal

can be arranged

of the weak order on A ) ,

E(x./x 0 )

A.

'j$ n A - convex.

1 + 2 (x^/x ) i^O

K

K

Next suppose is

P

is the valuation ring

K -»• A, °° be a place of a field

1

Q

'$ n A-convex

K -* A, ± °° with center

p ) . Moreover, the induced order on

= ("B ) , the derived order w ^

Proof:

a

A/P.

Lemma 7.7.1. (A,Q).

where

and the residue field

fraction field of

ring

H A = P

can be

The second main result is that

prime ideal, then there exist signed places A

A, p

Since

A

(A/P, Q/P) A.

is a

Thus, by the

is the residue field

(A. p .,P), we deduce from the second -convex in

A r D . , hence is also

(In other terms, we are using the specialization chain

K -> A , °°+ A, °° corresponding to the pair of valuation rings

A Ck,^

149

in

K.

But see the Remark following the proof of the next proposition.)

Proposition 7.7.2. (Krull) with valuation ring

(A,Q).

p H A - convex ideal, and

Let

Let

']) C K

$" = 'p/Q

be any total order refinement of p! 3 $

refinements

on

K

Moreover, all prime ideals

Proof:

Let

Q~

pf:

and by

A = A/Q

such that P CA

a:

Let

K,

Q

a

p"' D f

Then there exist total order 'p1 n A - convex

is

p'/Q=Tf-

and

p' n A - convex.

1/x G K

K* -> {± 1}, then refine

p:

with

x G Q.

K -> A, °° to a

'JT1 C A, by p 1 (x) = p(x), x E A

K •> A, ± °° , relative to The order

f

CK

will be the order induced

p'. A*

denote

x = yz, with

product

A - Q, the units in

z E A*.

x-j^-.-'x, ,

A subset

x^ G s,

possibly

a:

then there is a relation xi

define

occur.)

Define

a(y) G {± i}

are fovoed on us if orders on

p

x,y G K*

are associates

'p -independent if no finite

'p - independent subsets

SCR*

although

S H A* = 0.

S •> {± 1}

a(z) £ {± 1}

be an arbitrary function. x. G S , p £ j ) ,

to be the sign of

z

If

y G K* - S,

z G A*.

(Possibly

(A, p"1)

in

and

cr(y)a(x1) • •-a(xk) = a(z). The point is, these choices is to refine

p

and if we are to have

p'/Q = 'p!

as

A.

We first check that yx| .. • x^ = p'z an associate of choice of

S.

1

with

a:

Thus

K* •* {± 1}

x! G S,

pp'/y

and it follows that

p. !

verify that

p'

Verification of

Gp,

is well defined. z

1

G A * , then

This can occur only if f

pz = p z , cr(y)

p

1

f

z = (p'/p)z ,

If also x-^ • -\*[- • - ^

{x^} = {x!}

(p'/p) G A * , hence

pf:

a(z)=a(z'),

K •> A, ± °° and we only need

is a signed place, relative to the total order f

p ' O y ) = p (x)p'(y)

is

by our

is independent of the choices made.

At this point we have a function

150

is

v x ^ — x ^ = pz, with

by 1

We say

x^ ^ x., is an associate of an element of 'p.

S = 0, and certainly

In any event, let

A.

S C K*

By Zorn's lemma, there exist maximal

no

K, with

the induced order.

A. Q

are

be a place of a field

be a partial order on

p" on

p 1 (y) = a(y) • °°, y E Q" 1 .

Let if

c

K -> A, «

denote the subset of elements

We will define a function signed place

p:

•

'p' on

is quite simple and verification of

A.

p'(x+y) = p 1 (x) +p f (y) Finally, let

is also easy if use is made of the identity

P CA

signed place

p : K -> A, ± °° to

The positive elements in in

A. p . C K.

A p = A, p ./P.

Restriction of the

induces a signed place

A p -* A, ± °° .

are clearly the images of the positive elements P c A, p >.

|5f n A. p . -convex.

is

•

The last paragraph in the proof can be established more directly.

In fact, if

(A,Q)

f = '£d

Q

and

is a valuation ring, '£ C A

-^-convex, then any ideal

Namely, suppose is

Ap

A, p . C K

But this just says

Remark:

Q

be a prime ideal,

1

x+y = x ( l + — ) .

0 <^ x <_ y,

:p-convex,

y/x £ Q.

yz £ I. Thus

Since

is any partial order with

I CA

is '^-absolutely convex.

£ = .p^,

x/y £ A, hence

0 <_ 1 <_ y/x, and since

x = y(x/y)

and

xz = y(x/y)z £ I.

Example:

Let

E

be a totally ordered field, We obtain the f-adic place

irreducible polynomial. A = E[x]/(f(x)). If

'£ = p

If

f(x)

has real roots over

is the weak order on

of a single element each ordering of

K = E(x), f(x) £ E[x]

{f(x)}.

K, a maximal

p:

K •> A, °° , where

E, we obtain orderings of

A.

-p-independent set will consist

Thus there are exactly two orderings on

A, determined by whether

an

f(x) £ K

K

over

is made positive or negative.

We turn now to our second main existence theorem, the construction of signed places on a field

Lemma 7.7.3. P C A

a prime

'£' D'$

on

K

Proof:

Let

K

with preassigned center on a subring

(K, £)

be a partially ordered field, A C K

'£ n A - convex ideal. such that

P

A C K.

is

a subring

Then there exist total order refinements

tp1 H A - c o n v e x .

We use the basic result of

A.

Klapper, Proposition 2.7.1, that

if a prime ideal is convex for an intersection of two orders, it is convex for one of them separately. p[x]

and

since if y £•£.

'p[-x]

x £ K,

are (derived) orders on !

!

y = p + qx = p -q x, Thus our prime ideal

^P[-x] n A-convex.

If

x, -x £ p , then both simple refinements K.

Moreover,

P>p'>q,q' ^'P> then P CA

is either

p[x] H^[-x] = '$

(q+q')y = pq' + p f q, hence

'.p[x] O A - convex or

Zorn's lemma now implies 7.7.3.

•

151

Proposition 7.7.4. subring, order

PCA

a

Let

(K,'£)

be a totally ordered field,

'$ n A - convex prime ideal

1

Q , such that

Af = A'/Q',

We may as well assume

by passing to /HPA[x]

or

assuming

1

A C A ,

(A,p. , PA, .) /HPA[l/x]

x > 0.

m

A

a^jb.

e

111 1

P,

m <_ n

since

Thus

is 1


"

'$ n A - convex.

+ ...+b]Ji)

we must have

1 -b

b! = b./(l-b Q ).

a > 0, or the term n

1 < a +a,x+ ••• +a x n . — o 1 n

But now

n

A

bQ < 1

and

Q

f

in the total order

where

A, = A[x]

P, . Then

A[x] If

Archimedean over defined by

x.

x

A. If

D

or

n.

(A',Q')

in

K

A[x]

because if

A[l/x].

Zorn's lemma now implies the

with the properties

Q' H A = P

A1 = A'/Q 1

Archimedean over

If

x

is transcendental, consider the cut of

is a transcendental cut of A.

If

D

is an algebraic cut of

k^ = A[l/x]

D = °° .

rather than

b! = b . / ( l - b Q ) .

x = x

A[x]

is is

A,D = D»(x),

A, then from 7.5 we conclude

were improper, we would have

x m £ bjx"1"1 + ••• + b ^ ,

to

P, C A.,

A, certainly the field

would not be a semi-field unless

/HPA[l/x]

(A,P)

Now, say, A,/P, = A[x], where

is algebraic over

then we return and work with

152

P, C A, = A[xJ, say, with

is a totally ordered semi-field since

is Archimedean over

from 7.5,

or

and

is least possible,

Return to the inductive step, where we passed from

maximal convex.

A(x)

n

A

$

£ - convex.

A = A/P.

modulo

Then

1 < a + a,x + • • • + a 1 x n " 1 + a (b'x n~ 1+. • .+b'x nJs~ m ), — o 1 n-1 n^ n m

It remains to show that we may construct

(A,,P,)

+

could simply be omitted from r J

is a local ring, P 1 H A = P.

existence of a valuation ring

1 <_ b

least possible.

Now we simply choose any maximal convex ideal Since

and

Finally, since

a x n

P,

x £ K, we claim either

is a positive unit in

which is a contradiction of minimality of

P C P1#

n A, and residue field

First, we lose no generality by

and

We must have

in

with

(A, '£A) .

1 £ a + a, x + ••• + a x

x (l - b Q ) £ bj^x " + ••• + b m . P

Given

is a proper ideal.

with

A' C K

is a local ring with maximal ideal

if necessary.

Then suppose

b 1 /x + .. .+ b m / x m

x

P = Q

1

-p1 = p H A'/Q', Archimedean ordered over

Proof:

a

A = A. p ./PA, . , with induced

'|5. = tp n A. p ./PA. . . Then there exists a valuation ring

p - convex maximal ideal

ra

A C K

A, then again But if

D = °° ,

A[x]. This is justified

1 <^ b

+ b,/x + ••• + b /x

But then from 6.3, x

would be

bounded by an element of

A, so we could not have

D.(x) = °° . This completes

the proof of 7.7.4.

•

Combining 7.7.3 and 7.7A, field, and

P C A

is a

K

refines

(K,£)

K •> A , ± °° such that the induced total

p:

'£, p

is finite on

in knowing if first the order

A

A

of

'$ n A/P

with center A/P.

on

For example, if

(A,P)

A/P

ring over a totally ordered ground field

A/P.

pf = q + r g

in

so must be A/P.

f.

A'

is

K

constructed lifting

is already a valuation ring and maximal We will see in the

A = R[X1«-.X ]

next chapter that this is also the case if

the order on

and

can be refined to a total

K, Proposition 7.7.2 says this is the case.

field of rational functions.

P

One might be interested

order in an arbitrary way, then a total order on

ideal in

K, then

f

Archimedean over the fraction field

this choice.

is a partially ordered

•£ n A - convex prime ideal of a subring of

we can find a signed place order on

we see that if

R

is a polynomial

K = ( ]/'* x n )

R, and

X

is

tne

However, in general, one cannot first assign

A typical difficulty is that one could have an equation A, with

p>q>r £ P.

Thus, if

However, this relation between

g f

is made positive in and

g

A,

disappears in

The geometry of this problem is interesting, and will be clarified

by results in the next chapter. Our proof of 7.7.4 is directly modeled after the standard proof of Chevalley of the place extension theorem in algebra. f

additional conclusion that the residue field over

A = A. ./PA,p..

x

can be constructed algebraic

A

Now, in our context, this cannot always be done.

difference is, a simple extension unless

is algebraic over

A[x]

of a field

A, whereas

A[x]

Specific examples arise as follows.

of

K

will be trivial.

Then

K

in two indeterminates over

Q,

P = (x,y).

x

A. A = Q, the rationals,

'£ the total order on

is Archimedean over

Secondly, suppose

The

can never be a field

First, suppose

P = (0), K = C O ) , IT = 3.14159... GIR, with IR.

A

can be a semi-field if

defines a transcendental cut of the ordered field

by the embedding in

That proof has the

A = Q[x,y]

induced

Q, hence any signed place is a polynomial ring

The extension field

contain enough elements so that even the weak order on

K

K

forces

K

could p(x/y) = T\ £ IR,

153

where

p:

even if

K -*IR, ± °° is any signed place with center K

is smaller, say

P

on

A.

Thirdly,

K = ©(x,y), our method of proof of the signed

place extension theorem involves first choosing a total order on (x,y) C Q[x,y]

is convex.

so that

If this choice is made "carelessly", it is possible

p(x/y) = TT G IR, for the resulting signed place on

that again

K

In a positive direction, we can at least say this.

K.

Suppose

A = R[x,---x ]

is a finitely generated integral domain over an ordered ground field K = R(x 1 *-*x n )

is the fraction field of

to a partial order

field

A

1

(So

P

tr.deg. R (A/P), since otherwise

p

would necessarily be the trivial place.

will be algebraic over

P C A

tr.deg. R(A 1) =

with center

must have

A, the fraction field of

A/P.

actually be a finite algebraic extension of R, of transcendence degree over

R

Moreover, Af

[68 , Vol. II, Chapter VI, §14, Theorem 31], the residue field

over

is

Then from classical valuation theory, the residue K

A'

p

is a convex prime relative

on

Thus

of any place

P C A

tr.deg.D(A/P) = tr.deg.n(A) - 1. K K

'p C K, with

a minimal non-zero prime.)

A,

R,

will

A, that is, a function field

one less than that of

K.

We next give some applications of the signed place extension theorem to the notion of semi-integral elements introduced in 6.4. Proposition 7.7.5. a subring.

Let

(K/-P)

be a partially ordered field, A C K

Then the following subrings of

K

coincide

(i) The ring of elements semi-integral over

A.

(ii) The ring of elements bounded by elements of refinements of

]5 on

A

in all total order

K.

(iii) The ring of elements finite in all signed places of

K

finite on

and refining the order '£. (iv) The intersection of all valuation rings of

K

containing

A

with

'^-convex maximal ideal.

Proof:

(i) C (ii)

bounded by elements of

since semi-integral elements over A

in the order '$.

(ii) C (iii) since a signed place induces an order.

154

A

are already

A,

(iii) C (iv) since if an element was not in some such valuation ring containing

A, it would be infinite in the associated place and in any signed

place refinement. (iv) C (i) since if an element 2

x G K

z

then the ideal

Ai(l/x )k[l/x ] C A[l/x ]

be an equation

n

1 <_ a,/x

is not semi-integral over

2

is proper.

+ • • • + a /x , or

x

n

Otherwise, there would

<_ a,x n "

Thus, there exist by 7.7.3 and 7.7.4 valuation rings A[l/x 2 ]CA',

with

l/x2GQ«.

x 2 £ A1

Thus

hence

A,

+ • • • + an,

(A',Q')

in

K,

Q

a. € A.

1

xfEA1.

convex, D

The following corollary was proved more generally in 6.4.

Corollary 7.7.6. '•P = '^

a

Let

derived order,

(B,'J>) be a partially ordered integral domain,

A C B

over

(A, '.p H A ) .

exist

'$ - convex prime ideals

Proof: defined by

a subring, such that

Then for each

Let

K

Q C B

with

(Since

'.p = 'Pd

on

where

B C A , where

Q

C A

P

the "order

B, the notation is justified.) p

of

K, with center

is the valuation ring of

p.

Let

P C A.

Q = B H Q ,

is the maximal ideal.

a local ring, with closure of

elements

B, f C K

D

P

Proposition 7.7.7.

(A',Qf)

A

P C A, there

Q H A = P.

denote the field of fractions of

'-p C B.

is semi-integral

f H A - convex prime ideal

Using 7.7.3, 7.7.4, construct a signed place By 7.7.5

(B,'.p)

A

in

with

Q1

x

Let

(K/p)

be a partially ordered field,

'|5 - convex maximal ideal K

P C A.

Then the semi-integral

is the intersection of all valuation rings in

£-convex, A C A 1 , and

P = Q 1 n A.

which satisfy a semi-integral equation

A C K

x

n

K ,.

Moreover, the + b,x

+ ••• + b-

l with

bi £ P

are exactly the elements in the intersection of the above

maximal ideals Proof: 2

Q' C A 1 .

Suppose

x G K

2

2

, l/x )A[l/x ] C A[l/x ] with since

bQ G P , A

< 0

zn —

a i e A, hence

is a local ring.

is not semi-integral over is proper.

Otherwise

A.

Then

1 <_ b Q + a^x2

x 2 n <_ a|x 2 n " 2 + • • • + a R , where

+ • • • + a n /x 2 n ,

a! = a./(l-bQ) € A,

(Strictly speaking, we should first refine the

155

order '$ to insure that f

(A',Q )

valuation ring

1-b with

is positive.) 2

1

The last statement follows since if and

p:

(P, 1/x2) C Q « .

A[l/x ] C A ,

does not belong to all valuation rings over x

Thus, we can construct a

A n

with maximal ideal over

+ b ^ n ~ 1 + ••• +^2n

K •> A, ± °° is a signed place finite on

p(x) £ A

since

x

Conversely, if

is semi-integral over

x

This shows

A

and zero on

A, hence

p(x

b

-°>

) <_ 0

i

x P.

G

P

P, then and

p(x) = 0.

satisfies no such equation, then one argues as above that

2

2

/HPA[l/x ] C A[l/x ]

is proper.

Q1 H A = P

constructed with

(A f,Q !)

Thus a valuation ring 1/x2 e A 1 , hence

and

can be

x f. Q 1 .

D

Our final results in this section will concern chains of prime ideals in an integral domain containing of

P C P- C ••• C P C A.

In particular, if

K

is a field

A, we are interested in specialization chains of signed places

K, K -»• A

with

pi

o , ± °° -*• A ,1 , ± oo -»••..-»- A n , ± ° °

finite on

the composition ideal of

and

P^

the center of

K -* A. , ± °° . If

p i , then

VV

A

(A-,Q.)

A n C A n l C ••• C A Q

and

p^

on

A.

Here

p^

denotes

is the valuation ring and maximal Q Q C Q 1 C •. . C Q^.

In fact,

VcQ.r

The following general result provides some added insight into the close relationship between valuation rings and totally ordered integral domains. Proposition 7.7.8. Let I,J C A

(A,p) be a totally ordered ring.

are two convex ideals, either

I C J

or

Then, if

J C I. All convex ideals

are absolutely convex.

Proof:

Suppose

is convex, no element elements of

yGJ

x E I-J.

We may assume

can be greater than

x.

Thus

J, in particular for all positive elements.

y G I, hence

positive, hence

Next let

x > 0. y < x

Since

I

Since

J

for all is convex,

J C I.

Also, if

156

I <j- J, say

I

is convex, 0 < x < y 0 < xz < yz

hence

and

yz G I, then we may assume

xz G I and

I

(K/p) be a partially ordered field,

z

is absolutely convex.

A C K a subring and

•

p

i>

° -

i

n>

-

^

n A

order refinement of form a chain, say

convex

"

P r i m e ideals of

f, keeping all

Pi

A.

When is there a total

convex?

P Q C P 1 C • • • C P^.

Necessarily, the

Also, all

Pi

P./Pi C A / P ^

must

i < j , are

convex for a total order refinement of But now if

'P/P-. Thus P./P. must be CP/P-) -convex. •*J -*i d Pv/P. 3 P./P- are CP/P-) -convex ideals of

i < j < k, then

K

A/P i

and

l

3

l

i ^

P^Pj = (P k /P i )/(P j /P i ) C ( A / P ^ / C P ^ ) = A/P j

is convex for a

total order refinement of

C£/P.) /(P./P.) C A / P . . If we abuse notation 1 d 3 1 J slightly and write this last order as CP/P-) /p-> t n e n w © deduce P v/P. C A/P. 1 K d 3 J J is necessarily (CP/P-) /P-) -convex. Continuing, we find the following 1 d J d

necessary conditions for the existence of a total order refinement of which keeps all

(*)

If

P. C A

i < j , then —

(K,'J>)

convex:

P./P. CA/P. ji

is

i

Proposition 7.7.9.

( • • • ( f f / P J . / P , ) -1• - / P .1) °

d

d

d

- convex.

In the notation above, the conditions (*) are

necessary and sufficient for the existence of a total order refinement $' D'P

on

K

Proof: orders on

such that all

Let K.

P.CA

x E K, with

x, - x £ ' p .

As in the proof of 7.7.3,

generally that if

if

'£, '£', tp

that conditions (*) hold for

p ! HA-convex.

are

Let

'P'=1>[x],

']>' 0 $ " = f.

are orders on

A

with

'£ M =1>[-x]

as

Now we will prove f

'P n •£" = $

'$, then conditions (*) hold for

and such

'£' or for '£".

The proposition then follows from Zorn's lemma. We distinguish two cases, case (I) if case (II) if

P

is both

conditions (*) hold for with prime ideals

'£' and

P

'^"-convex.

CA

'£'. In case (II), we will pass to the ring

P i /P Q = P±, and orders

to the chain of prime ideals $'

or

'£" on

number of prime ideals on

A = A/P Q .

"^"-convex, and

In case (I) we will prove

f' = C P 7 P O ) , , and

Then we will establish conditions (*) for the order

for either

is not

P^

A

P.,

1 £ i <_n.

tp"1 n^»»

A/P Q = A

f" = CPff/Po) • on

A

relative

The assertion that (*) holds

then follows in case (II) by induction on the (We do not assert

'JF1 n ff" = (p/P )

as orders

If this last assertion were true, case (II) would be easy.)

157

Consider first case (I). Fix an equation u £ PQ.

First, we check that

P.j/P0

are all

makes sense because 2.7.1 implies if

P

r

'£'

convex.

Suppose

Pj/P Q

0 < u < v (rel 'p") , v G P , CP'/P0) -convex.

is not

J5

Note this

convex, then

P o

T

o was not

M

Cp'/P0) -convex.

is

There would then be

an equation 0 £ ax +z 1 £ ay + z 2

ae-jj 1 ,

with z k G P Q ,

a £ PQ,

(rel -p')

y^p.,

x

£ P^

Combined with

0 £ u £ v

(rel -p"), we obtain as in the proof of 2.7.1

0 £ u 2 ( a x + z x ) 2 £ u 2 (ay+z 2 ) 2 + v 2 ( a x + z x ) 2

But

(rel $)

v G P Q , z^ G P Q , so this equation simplifies to

0 £ u2a2x2 £ u2a2y2 + w

w G PQ.

with

in the ring x G p^

But

(rel 'P)

u,a ^ P Q , and we deduce therefore

A/P Q .

But

P^/PQ

0 <_ x 2 <_ y

p

is assumed

C^/ o) -convex and

(rel Cp/PQ) )

y G p^

hence

which is our contradiction.

Next, we verify the "second level" conditions of (*), namely that if then

P./P-

is

(Cp'/P ) /P-) -convex.

part of the proof above.

x £ P./P .

This makes sense because of the first

Assuming our desired conclusion false, we would obtain

0 < ax+

an inequality in

A/P Q , with

ay

a £ Pi,

a G'p'/P Q , z^ G P i /P Q ,

From this we obtain an inequality in

0 £ b(ax+z 1 ) + zj £ b(ay+z2 ) + z^

where

b f- P ,

b G«p',

z£ G p

0 £u £v

A

P

j/p0'

A,

(rel $•)

(rel 'p")

and its image in

A/P Q .)

Now

to obtain as in the proof of 2.7.1

0 £ u 2 b 2 (ax+z 1 ) 2 £ u 2 b 2 (ay+z 2 ) 2 + w

158

y ^

(By slight abuse of notation, we are

using the same symbol for an element of we combine this with

i <_ j ,

(rel 'P)

with

w £ P .

Since

u,b f- P , we have

0 <_ a 2 x 2 <_ a 2 y 2 + w'

w1 £

where

in the ring

P

i/p0

c A

/ p 0 > since z^ €

A/P.. i

P

(relCP/PQ) )

j/Po-

From this follows

0 < x 2 < y2

(rel(Cp/P ) /P.) ) ° d x d

But

and

y6p./P. ji

P./P. j i

is assumed

(more precisely, (*) easily implies this), hence original choice of

(Cp/P ) /P.),-convex ° d x d

x G P./P^, contradicting our

x.

The general technique involved here should now be clear. is not

((•••(Cp'/P ) /Pi) "-/P ) /P.) °d d d d

inequality in

A/P.

convex, m < i, we unravel a suitable

and get a (complicated) inequality

is combined with the inequality

0 <_ u <^ v

obtain a complicated inequality

(rel p)

again to get an inequality

(rel $") , in

A.

(rel '£')

v £ PQ,

Finally,

P./pi

given by (*).

m

d

in

A.

u £ PQ

This

to

this is unraveled

rel ((• • • (('p/P ) /P.) **•/?) /P.)

°d-Ld the assumed convexity of

P./p-

Assuming

1

which contradicts

d

We will leave the actual details

of this induction to the reader. We next consider case II ,

is both '.p1 and ^"-convex. We

when P

need to establish conditions (*) for the order relative to the chain of prime ideals that each P i /

p o

is

Cp7P0)

p

-j/

p 0

n

CP7PQ)

C ... C

p

p

n/ 0-

CP"/P0)

on A/PQ,

We f i r s t argue

n C-P"/Po) - convex, or, what is the same by

2.7.1, that each P ^ P Q is either

Cp7PQ) -convex or

(^M/Po) -convex.

Assuming otherwise, we would have inequalities 0 _ < a x + z 1 £ a y + z 2 ( r e l '£') 0 <^ bu + w : £ bv + w2 ( r e l

where w ^ z ^ P Q , y,v e p ^

x ,u

^ Pi,

a^1])1,

a £ PQ ,

b G f ,

b £ PQ.

From this (ax+z 1 )2 (bu+w 1 )2 <_ (bu+w x )2 (ay+z 2) 2 + (ax+z x) 2 (bv+w 2 ) 2

or,

(rel $

simplifying,

159

0 <_ a 2 b 2 x 2 u 2 <_ a 2 b 2 (u 2 y 2 + x 2 v 2 ) + t (rel

f o r some

t Gp . o

Then

a,b £ P ' o

0 £ x V <_ u y in t h e r i n g

z

i mrp l i e s

+x V

A/P , which i m p l i e s s i n c e

convex, t h a t e i t h e r

x G p.

or

~~

J

n

(CCP'/P )

VT,

y , v £ P.

and P . / P

is

Cp/P )

u G p . , contradiction.

The second l e v e l v e r i f i c a t i o n i < j , P./P . is 1

(rel

i n c a s e ( I I ) c o n s i s t s o f showing t h a t f o r

CP"/P ) ) / P1 - ) - c o n v e x .

° d

°

d

d

Assuming o t h e r w i s e , we

g e t in A/PQ 0 < ax+ z. < ay + z o —

where

z K

G P., 1

i—

y G p . , J

(relCB'/P )

z

x ^ P . , J

n

O f j

a ^ P . , 1

C^"/P ) )

a G (-^'/P

o&

) O (p"/P) . O J O(J

This

inequality becomes two inequalities in A

where

z^,z£ G P Q ,

+ z» < b ' ( a y + z 2 ) + z^

( r e l '])')

+ zy < b"(ay+z 2 ) + z'^

(rel ^")

b ' , b " £ PQ,

b 1 G^p», b " Gtp».

Then we g e t

(rel

in A < (b') 2 (b") 2 (ax+z 1 ) 4 < 2(b') 2 (b") 2 (ax+z 1 ) 2 (ay+z 2 ) 2 + t

with

t Gp

This g i v e s i n A/PQ

0 — < a4x4 — < 2a4x2y2 + u

where

u Gp

s i n c e z, G P^.

F i n a l l y , in A/P.

0 < x 4 <_ 2x 2 y 2

since

a ^ P..

imply

x G p.

But y G P.

(relflVP o )^ )

(rel(CP/PQ)d/Pi)d)

and convexity conditions (*) for

P./P.

which is our contradiction.

Again, we leave the rest of the details of case (II) to the reader. The general step is to show that

P./P. is ((• • • ((CP'/PJ J

160

1

° d

n CT/PJ °

1/PJ ...PJ /P,) d

i

d

m

d

d

convex for

m < i.

P ) /PJ

Assuming otherwise ultimately contradicts the assumed /PJ

•*• d

d

'"/?

m

•*• d

) /P-) d

1

- c o n v e x i t y of

P./P..

d

3

So e n d s o u r p r o o f

1

of 7 . 7 . 9 .

D

Remark:

The geometric significance of 7.7.9 for function fields over a

fixed real closed ground field P C R[X^...X ]

R

is roughly as follows.

Convex prime ideals

will have d-dimensional zero set in affine space

d = tr.deg.R(R[X1-..X ]/P). local dimension less than

However, P

d.

R n , where

may also have "degenerate" zeros, of

Passing to derived orders has the effect of

eliminating these degenerate zeros from the maximal convex ideal spectrum of R[X,,,,X ]/P.

Similarly, total orders on

R[X,...X ]/P

can be interpreted

as orders obtained by evaluating functions on (infinitesimally) small open sets in the non-degenerate part of the zeros of

P.

in 7.7.9 that assure a chain of convex prime ideals

Thus the conditions P.

can be kept convex

in a total order amount to adherence conditions on the non-degenerate zeros of the P_

All this will be clarified by the results in the next chapter.

Proposition 7.7.10. a subring

P

C p

Suppose

C ••• C P

the conditions (*) of 7.7.9. places of

(K,'p)

a chain of

Q^ H A = P^.

± °° -*• A 1

Proof:

±<x>->... -> A n

K -> A ^ ± °° denoted by

Moreover, each residue field

over the field of fractions of

First, choose a total order on

Archimedean over

(A^Qp

P]/P Q

c A

/PQ

such that

A^ = A^/Q^

A/P .

pQ:

K, keeping all

A C Ai

is Archimedean

P.

We now have the convex prime

and with

A^

convex.

K •> A Q , ± °° , with center

and we apply 7.7.4 again to get a suitable signed place with center

i °°,

A/P^

Now apply 7.7.4 to get a signed place A

p n A-convex prime ideals satisfying

K,

with valuation rings of

and

A C K

Then there is a specialization chain of signed

K -* A o

and

is a partially ordered field,

Archimedean over

PQ C A

P^/P Q CA/P

CA

A Q •* h^, ± °° , A/?^.

Continuing

by induction gives the desired specialization chain.

•

161

VIII - Affine semi-algebraic sets

8.1.

Introduction and Notation Throughout this chapter we fix a real closed ground field

E C R

n

of affine n-space over

R

is a semi-algebvaio set if

the smallest Boolean ring of subsets of

R

U(g) = {x G Rn|g(x) > 0>,

R. E

A subset belongs to

which contains the sets

g G RtX^'-Xj .

We also define sets

W(g) = {x G R n|g(x) > 0} Z(g) = {x G R n|g(x) = 0} .

If

g. G R[X •••X ]

is a finite collection of polynomials, define

u{ gi } = nu(g.) w{ gi} = n w( gi ) z{g.} = nz(g.) . Any semi-algebraic set sections of

(*)

Zf s

E C R

and

can be expressed as a finite union of inter-

U's

E = u (Z{fik> n u { g j k } ) ,

for finite collections of polynomials is of course highly non-unique. properties of the sets tations (*) .

162

E

f-v> S-v*

Such a representation of

We are primarily interested in "geometric"

themselves, not in the particular represen-

E

Finite unions of the Their complements in

R

n

U{g^}

will be called open semi-algebraic sets.

will be called closed semi-algebraic sets.

typical closed semi-algebraic set

S = U si ,

S

A

can be represented

where

(**)

f..} n w { g . k } .

Of course, equalities and

f = 0

could be avoided altogether by writing

f > 0

f £ 0, but this is psychologically less natural.

Lemma 8.1.1.

If x G U{g i >, then there is

g G RfX^.X^

such that

x 6u(g) Cu{g.}.

Proof:

The estimates of 7.2 can be used to find a ball around the

point

x = (x.....x ) G R n , contained in

U{g.}.

where

B(x,e) = {y| lly-xll < e} ,

= £(y.-x.) .

given by

2

Hy-xll

That is, B(x,e) C u { g . } , A suitable

Rn.

topology on

•

U(g)

form a base for a

But we now emphasize that semi-algebraic geometry is

precisely not concerned with this strong topology. ring of semi-algebraic subsets of

R

Instead, only the Boolean

is relevant.

If

E CR

is semi-

algebraic, then we also get a natural Boolean ring of subsets of by intersecting

closure

is then

g x > £ (y) = e - E ( 7 j - X j ) .

Of course, Lemma 8.1.1 states that the sets

If

g

2

E, F ¥~

of

E

with other semi-algebraic sets.

are semi-algebraic and F

in

E

F° = {x e E | exists

F"E = {x € E | for all

Of course, F°

E, simply

and

F

in the topology on

E

F C E, define the interior

F°

and

as follows.

e > 0

such that

y G E, lly-xll < e

e > 0, there exists

yEF

implies

such that

yEF}

lly-xll < e}.

are the ordinary toplogical interior and closure of with base the sets

U(g) H E .

F

But, for our purposes,

the point is that it follows easily from the Tarski-Seidenberg theorem, to be

163

discussed below, that we will simply write

F° F°

and and

F

are also semi-algebraic.

(If

E = Rn,

F.)

We now make the following confusing (but crucial for understanding the differences between algebra and topology) definition. F C E F

A semi-algebraic subset

will be called an open, semi-algebraic subset of

will be called a closed, semi-algebraic subset of

commas should not be ignored, at least for now.

E

of

U{g.} n E

E.

if

if

F = Fp

F = F .

and The

It is easy to check that

the complements of the open, semi-algebraic subsets of semi-algebraic subsets of

E

E

are the closed,

It is also easy to check that finite unions

are open, semi-algebraic and (hence) finite unions of

Z{f.} <^w{g.} H E

are closed, semialgebraic subsets of

E.

In other words,

open semi-algebraic subsets are open, semi-algebraic subsets.

In fact, the

converse is also true, which we now state as

Unproved Proposition 8.1.2. open semi-algebraic subsets of

Open, semi-algebraic subsets of

E

are

E.

D

Thus, the comma can be ignored when talking about open or closed semialgebraic subsets. difficult to prove.

However3 this foundational point seems to be rather In particular, it does not follow directly from the

Tarski-Seidenberg principle, as does the fact that

F

and

F

are indeed

semi-algebraic. The paragraphs above are a bit of a joke, written in order to emphasize the subtlety of

8.1.2.

We will develop in this chapter an extensive theory

of semi-algebraic sets without using 8.1.2.

This will require some care

with words, or, at least, with commas.

It would be nice to have a simple

proof of 8.1.2 right at the beginning.

On the other hand, all the results

we will prove in order to circumvent 8.1.2 are results we would want anyway. The Tarski-Seidenberg theorem, [56], [57], [62], is sometimes regarded as a theorem in logic.

It asserts the existence of a decision procedure

for deciding the truth of any elementary sentence, built up from finitely many inequalities

P.(x....x ) > 0, where the

P.

are polynomials over a real

closed field, using basic logical connectives "and", "or", "not", and quantifiers "exists

164

x.", "for all

x.".

The decision procedure amounts to checking

whether or not certain polynomial inequalities involving the coefficients of the

Pi

hold.

For example, the sentence "there exists

2

x G R

such that

2

ax + bx + c = 0"

is equivalent to

"b - 4ac >_ 0".

Also, Sturm's theorem in 7.3

is a special case of such a decision procedure. The proof of the Tarski-Seidenberg theorem is not difficult. Cohen's proof [62] in an appendix.

We will give

From the algebraists point of view, what

is involved is just an argument making use of (1) induction on degrees of polynomials and number of variables and (2) explicit calculations in polynomial rings involving partial derivatives and division algorithms.

In other words,

elimination theory. The applications of the theorem are rather striking. sharp distinction between two types of application.

We will make a

The first type, which

is almost a reformulation of the theorem itself, says that any set defined in terms of semi-algebraic sets by an elementary sentence is still a semialgebraic set.

For example, this includes closures and interiors mentioned

above, images of semi-algebraic sets under polynomial mappings and other frequently used constructions.

R n -> H m ,

(However, the theorem gives little

insight into the question of whether a set is, say, open.)

It is amusing that

even the simplest special cases of this type (say the projection to zeros in

R

R

of the

of a single polynomial, or the closure of the set where a

single polynomial is strictly positive) are no easier to analyze than the whole theorem.

Thus the Tarski-Seidenberg theorem is a very efficient tool.

We emphasize that this type of application actually provides a proof that the asserted set is semi-algebraic, simultaneously for all real closed fields, in fact, an elementary proof.

The reason is, any single elementary sentence

is just a special case of the theorem.

The more subtle application is this.

Given an elementary sentence, suppose it can be checked in one real closed field where it makes sense and

is true.

For example, it might be checked for

the classical real numbers by transcendental methods (use of completeness, possibility of integration, etc.)

Then the sentence must be true for any

real closed field to which it applies, because there exists a decision procedure which is independent of the field in which it is carried out. Thus, one might say that this method of application amounts to proving by non-

165

elementary methods in one special case that an elementary proof of the statement in general does exist.

It certainly might be very tedious, if not

physically impossible, to actually work out this elementary proof. In this book we absolutely and unequivocally refuse to give proofs of this second type. fields.

Every result is proved uniformly for all real closed ground

Our philosophical objection to transcendental proofs is that they may

logically prove a result but they do not explain it, except for the special case of real numbers.

Also, one of our central themes is that the real numbers are

totally irrelevant in algebraic topology, so it would not do to rely on them at some point in our chain of reasoning.

Finally, there is already a respect-

able tradition in this century of finding non-transcendental proofs of purely algebraic results concerning algebraically closed fields.

We think real closed

fields deserve (at least) equal time and effort. We do not at all mean that only elementary proofs are acceptable. we use Dedekind cuts, total orders, and signed places repeatedly.

In fact,

The point is,

in the form we use these concepts they apply uniformly to all real closed fields. One advantage to developing such techniques is precisely that one is not tied down to "elementary sentences".

There is rich non-elementary theory to be

studied for arbitrary real closed fields, and even if a statement turns out to be equivalent

to

an elementary statement, it may be unnatural to dwell

on this fact, and even worse to be forced to depend upon it. There are obviously aesthetic questions involved in this discussion.

We

admit that many of our proofs are long and could be replaced.by the single phrase "Tarski-Seidenberg and true for real numbers". effort is worthwhile.

However, we feel the

In fact, we do not use the Tarski-Seidenberg theorem at

all, until 8.7. We now change tacks somewhat and indicate a more invariant approach to semi-algebraic geometry. R-algebra of finite type.

Suppose

(A,p) £ (PORNN), and assume

Define a set

X = X(A,'£)

A

is an

by

X(A/p) = H6m ( p 0 R ) ((A,tp),(R^ w )).

(Of course, 'C

166

is the only order on

R.)

We have the canonical adjoint

homomorphism from the elements

A

to the ring

R

of

R-valued functions on

f £ 'p define functions nowhere negative on U{g i >, W{g.}, Z{g.} C X

the subsets

X.

X, and If

g. E A,

are defined in the obvious way. Interiors

and closures of subsets of X are also defined just as above, essentially using the topology X with base the collection of sets U(g), g^A. (The U(g), g e A, are invariants of (A,'£) and replace the e-balls in the earlier definition.) If we choose a specific presentation I CRpC^.-.X ]

A = R[X,...X ]/I, where

is some ideal (necessarily radical and

n

gets identified with a subset of the zeros of

I, Z(I) C R .

X = X(A/p) = {x G Z(I) |g(x) ^ 0 all

Obviously, X

is a closed subset of

Rn

$ -convex), then

X

Specifically,

g € tp}.

in the topological sense. y

Now, the problem is, the homomorphism In fact, X

could be empty.

Definition.

RHJ-algebra (real, Hilbert-Jacobson) vif

is an R-algebra of finite type P C A , and

f £P

and

need not be injective.

Thus, we make the following definition.

(A,'£) is an

ideal

A -• R

A

(A,'£) £ (PORNN), and for each prime 'p-convex

g £ P, there exists

x E X

with

f(x) = 0

for all

g(x) t 0.

Corollary 8.1.3.

If

(a) the homomorphism (b) the set

X

(A,'£) is an A -* R

RHJ-algebra, then

is injective

is identified with the maximal convex ideal spectrum of

(A,?)

(c) for any subset

J C A, we have

vtt(J/p) = {£ e A|f(x) = 0 all

where

Z(J) C X

is the set of zeros of

xC-Z(J)},

J.

Proof: (a) The intersection of all '^-convex prime ideals of (A,'p) G (PORNN).

167

A

is

(0), since

(b) Each all

e

f '£.

x £ X

corresponds to a surjection

Thus kernel (p) C A

the definition of

contain

/H(J,$) J.

f €= A

is already maximal, then

"jj-convex maximal ideals containing

which vanish on all zeros of

J

in

P.

Thus

P

is /H(J,$)

X.

D

RHJ-algebra is attractive, there is still a

Without some control on the order

p C A, the resulting subsets

of affine space (relative to a presentation

can be rather chaotic. an

P

P

x £ X.

Although the notion of

X = X(A,p)

Conversely,

Again, the definition and (b) says each p-convex prime

is the set of

p(£) >^ 0,

is the intersection of the prime '^-convex ideals which

the intersection of all

problem.

with

RHJ-algebra states that every '^-convex prime ideal

corresponds to some (c)

A -*• R

is a maximal, '^-convex ideal.

is contained in such a maximal ideal. Thus if P

p:

A = R[XX ...X n ]/I)

The control needed in order to guarantee that such

X(A,'£) is semi-algebraic (necessarily closed, semi-algebraic) roughly

amounts to finiteness conditions on orders naturally associated to •£. Our study of semi-algebraic geometry will essentially amount, then, to identifying a large class of

RHJ-algebras

(A/p) with

X(A,'p) semi-algebraic and,

conversely, given a closed, semi-algebraic set RHJ-algebras easy if

E

(A,'p) with

E = X(A,'£).

E C R n , construct natural

This last turns out to be fairly

is closed semi-algebraic, but not so easy if

E

is only closed,

semi-algebraic.

8.2. Some Properties of RHJ-Algebras In this section we will construct many

RHJ-algebras, assuming the

following basic real Nullstellensatz, which will be proved in 8.4.

Proposition 8.2.1. is an

The polynomial ring with the weak order

(R[X^.. .Xn] ,'p

RHJ-algebra.

Of course, X(R[X,...X ] ,'£ ) = R n . l n w

•

Once we have this one

RHJ-algebra,

there are natural constructions of others. We indicate several such constructions in the propositions below.

168

Proposition 8.2.2. 'P(X) = { f G A|f (x) :> 0

If

(A/-P)

all

is an

RHJ-algebra

x G X}, then for any order 1

$ O p ' C ^ ( X ) , we have that

(A,? )

In particular, the radical

is an

X = X(A/P), and f

CA

with

RHJ-algebra and

X(A/p») = X(A/P).

'p-convex ideals and the radical

'.p1-convex ideals

coincide.

Proof:

Any

'£'-convex ideal is also 'p-convex.

obviously define

'-P(X)-convex ideals, hence also

Since the points

x ۥ X

f

'p -convex ideals, the

proposition follows easily.

Corollary 8.2.3.

•

If

(A/p)

and

(A/PJ m are RHJ-algebras. X(A,!Pm), and $m

Proof: P

=

'-P(x)

is an RHJ-algebra, then

(A/p ) , (A,'P ) ,

We have X = X(A,tp) = X(A/PJ s = X(A,:pj p =

The orders

and

it

'£ , 'p , :p were defined in 3.12. r s rp rm eas Y t 0 check that for an RHJ-algebra

is

By definition, J ' (A/p), we have

• Proposition 8.2.4.

If

(A,'p.)

are

RHJ-algebras, 1 £ i £ k, then

(A, H'p.) is an RHJ-algebra and X(A, rvp.) =iUX(A,!p.). l I l Proof:

The results of 2.1 show that any

'.p.-convex for some

j.

The result follows easily.

Proposition 8.2.5.

Proof: P C A

Any prime

is a prime

(A,'p )

Proof:

is an

Write

(A,'p)

is an

RHJ-algebra and

'^/I-convex ideal of

If

A

•

RHJ-algebra and

'.p-convex ideal containing

Corollary 8.2.6. then

If

(A/I, (]5/I) is an

'^-convex ideal, then

H *p. -convex prime ideal is

A/I

I CA

is a radical,

X(A/I, -p/I) =

is of the form

P/I, where

I.

D

is a reduced, real R-algebra of finite type,

RHJ-algebra.

A = R f X ^ . .X n ]/I, where

I CRj^.-.X^

is a radical,

169

'P -convex ideal.

(This is exactly what the hypotheses on

A

mean.)

Now

apply 8.2.1 and 8.2.5.

•

Proposition 8.2.7. such that

A

is an

is finitely generated over n

(A/£ A ), where

over

(B,£B)

If

£ A = A '-Pg, then

X(A,'£A) = image (X(B,$B))

R

and

(A,'£A)

A C B

(B,'£ )

is an

under the natural map

induced by the inclusion

Proof:

RHJ-algebra,

a subring

is semi-integral

RHJ-algebra and

Spec(B,'^B) -• Spec(A/pA)

A C B.

This is immediate from the going up theorem for semi-integral

extensions proved in 6.4.

Proposition 8.2.8. let

f CA

Let

A

(0) = O p . ,

Then

Proof:

(For example, the

(A.,'p.)

Next, suppose g £ Q^.

P i C Q, and g(x) t 0. Thus

A

is an

(A,'£)

Then

g £ Q.

Then

(Ai,']5i)

Qi Find

A

contains some

is an

is an

x G X(A,'£) x

Proposition 8.2.9. Pi C A

given orders

Let

A

by 3.11.)

A -*• R P-.)

under the inclusion

(A^p^) = is an

RHJ-algebra

contains some

This implies

Q^ C A.

is a

^-convex prime

so that

x

P. .

X(A,'p) =

'.{K-convex

is a zero of

Q

and

g(x) t 0.

The converse is equally routine, using A

contains some

P^.

•

be a reduced R-algebra of finite type,

and define an .order

(0) = n p . , '|5 C A

by

Suppose

'|5 = A f l l f - ,

A -• n A. . Then X(A/p) = U X(A. ,'p.) and (A,tp) l l l i RHJ-algebra if and only if each (A^^K) is an RHJ-algebra.

170

(In

Q C A , with

Q i = Q / P ^ and

be a finite collection of primes with !p. C A . = A/P.

Let

Moreover, (A/P)

is a zero of

RHJ-algebra.

the fact that any prime ideal of

could be the minimal primes

RHJ-algebra and

corresponds to a

x e X(Ai>^i),

P.

'p-convex

RHJ-algebra.

The kernel of any homomorphism

fact, any prime ideal of

prime

be any finite collection of

X(A/P) = U X(A.;j5.).

if and only if each

an

P. C A

A, which are convex for any order on

(A/Pi, '-P/Pp.

let

be a reduced, real R-algebra of finite type,

be an order, and let

primes with of

•

is

Proof: for some

By Proposition 2.7.8, if

i,

P. C Q

and

Q/P.

is

Q C A

is any

$.-convex.

'^-convex prime, then

All parts of the proposition

above follow routinely from this fact.

Proposition 8.2.10. and

'£ = '£ [g.] w j

is an

A

is a reduced, real R-algebra of finite type

is a finite refinement of the weak order on

A, then

(A/p)

RHJ-algebra.

Proof: refinement

Let

P. C A

'p [g.] = 'p. I w j

Of course, some •}K.

If

•

Now, A^

g.

be the minimal primes

will also be an order on

may be

0

in

A.

B

.

is a domain, unless

g

1

§m+1

and go on to

steps, we have a domain

B

= A.,

where

*-Pw(B) C B

is the weak order.

Applying 8.2.7, (Ai,tpi)

(A,p)

RHJ-algebra.

Remark.

g

. = B [/g

. ] , m > 0. B .

The first

, £'J> [!•••••§ ] C A . .

instead.

B = lim B , integral over

RHJ-algebra. is an

B

g. ^ 0, consider the

is already a square in

Bm[>/gm+2]

R, and real, say be Proposition 6.2.1.

g. = g. (mod P . ) . j j i

and these can be omitted in studying

time this happens, an easy computation shows we can skip

The finite

A., where l

is an integral domain, and assuming all

finite sequence of integral extensions Each

A. = A/P..

Thus,

After finitely many A., of finite type over

Moreover, f. = 'p [g.] = A. n •£ (B), Applying 8.2.6, (B,$w(B)) is an RHJ-algebra.

is an

Finally, by 8.2.8, •

The proof of the basic Nullstellensatz 8.2.1, to be given in

8.4, will, in fact, yield directly that any order of type

finite real integral domain over R gives an

'£ [g-] w 3

RHJ-algebra.

be used instead of 8.2.7 in the proof of 8.2.10.

on a

This could then

On the other hand, both

proofs use the device of adjoining square roots. If

(A/P)

is an

X(A,'p) C X(A,? w ). Now, if

RHJ-algebra, then so is

In fact, X(A/£) = {x € X(A,? w ) |g(x) > 0, all

A = R[X1...X ]/J, where

^-convex ideal, then n

Z{f.} C R . 1

then

If

(A/p w ), by 8.2.6, and

X(A,'^w)

£ = ^ [g.] W

J = (f/) C R p ^ . J ]

g

is a radical,

is identified with the real algebraic set

is a finite refinement of the weak order on

J

X(A,'£) = Z{f.} Hw{g.}. 1 3

More generally, if

!p = H •£K

where each

171

A,

$

= 'P w [g- k ]

is a finite refinement of the weak order, 1 £ k £ m, then by

8.2.4, X(A,p) = U z{f i > n W{g. k >.

Thus orders of this type give rise to

closed semi-algebraic sets. We now want to prove that any closed semi-algebraic set

(**)

s = us.,

can be represented as A

a quotient of

RfX^.-X^. S.

Trivially, I(S) = n I(S i ), S.,

is an order on Let

P.

A ^ , since the

We now have

k(S/) = R[X X . . .X J/ICS^ g.v

g-k

A i a = R[Xr..Xn]/Pia.

I(S.)

Then

f±a

in

P.

A(S.)-

= » w [g i k l

P i a C P.0

implies

and thus we have an inclusion

P.

CP.»

A(S)

occur among the

hold, even

are the minimal primes of

RHJ-algebra and

Proof:

we have

I(S.), we know

If

$ = A(S) n

n

^

C A ( S ) , then

(A(S),^)

is

X(A(S),'^5) = S C R n .

X(A(S),tp) =

X(A(S),'^) C S.

I(S.)

is a zero of

as an

A. -ideal, since to ik ia

U

*(\a>Via)'

Conversely, since

P. , some

a.

g., (x) > 0. -

If

Since clearly

x

Of course, the order

is an

XCA^,^)

I(Si) = C\ P ^ , any zero

x G S., then

Thus

(A(S),'P)

is clearly

x

c S

'^ -convex

•

y

p C A(S)

constructed above depends on the

i>

of

S. C U X(A. ,'£. ) , and l a ia la

S C X(A(S)/P).

172

= P.^,

1

Propositions 8.2.9 and 8.2.10 guarantee that

RHJ-algebra, with

P.

i i j.

Proposition 8.2.11. an

is an order

RfX^^.. .X n]/I(Si ) = A(S^).

10L

that

IK

R|X . . .X ] , so that

are convex for any order on

It is possible that some relations i, the

$ T[g.,] W

may be zero in

Note that all the minimal primes of

but since for fixed

S., hence 1

Of course, some

P.a/I(S^)

S.

is a ring of functions

are non-negative on

I (S) = H I(S.) = flflP.

A(S) -> II A. . P. .

is a radical, f -convex ideal

denote the minimal primes over Let

denote the ideal of

IK

A(S.)«

I(S i ) = n PiQj. on

I(S)

(A,'p), with

is a ring of R-valued functions on

1

1J

RHJ-algebra

I(S) CR[X 1 ...X n ]

Then

f.. £ I(S.), and the

1

for a suitable

Let

A(S) = R[X,...X ]/I(S)

since

on

s i = z{f..} nw{g ik }

S = X(A,'.p)

functions which vanish on

S C R ,

representation (**)

of

{f GA(S)|f(x) > 0

all

RHJ-algebra. S C Rn.

S, and is a rather weak order. xGS}.

We refer to

'p(S) =

Then by 8.2.2, (A(S)/P(S))

(A(S) ,'£(S))

The fact that it is an

Let

is also an

as the affine coordinate ring of

RHJ-algebra is geometrically very satisfying.

However, a useful question is, how does one go about deciding algebraically, in terms of a representation (**) to

'P(S)?

f G'-P(X')

for some

n >_ 0,

this implies

f G A(S)

(A',p'), with

RHJ-algebras

if and only if there is an equation

p,q G ^ 1 .

belongs

X' = X(A',p'), (f 2 n + p)f = q,

(The "if" part is trivial.)

'£' = $(X f ) = '£' , where

defined in 3.12. f

S, whether a given

One answer is provided by a theorem of Stengle, which asserts

that for a certain class of we have

of

'£' and

In particular,

'£' are the operators on orders

In fact, it also shows that in this case the collection of

satisfying such a formula is closed under sums, thus simplifying the con-

struction of

p1

in these cases.

The most general class of orders satisfying

Stengle's theorem is obscure, but it includes at least all is a reduced R-algebra of finite type, and

f CA

(A/p), where

A

is a finite refinement of

the weak order. In particular, reconsider the closed semi-algebraic set by (**)

above.

R[X r ..X n ]/P i £ x ,

We constructed an inclusion P.a

prime.

S = X(A(S),A(S) H n 'Picp-

A(S) -• II k^a>

We had the orders

^

S

described

where

A^a =

= 1>w[gikl C A.^

and

As a consequence of Stengle's theorem, we can

now state

Corollary 8.2.12. $(S) = A(S) n nCP- ) , where .

ia

CPia) = {f e

AicJ(f2n+

p

p)f=q, some n > 0 , p , q 6 y .

D

This result characterizes algebraically the collection of polynomials f G R[X 1 ...X n ]

non-negative on the closed semi-algebraic set

S C Rn .

We

will prove Stengle's theorem in 8.5. Note that recovering

S

from the

(A. ,'p. )

an algebraic set from its irreducible components. setting, where orders

f^a

is analogous to recovering In this semi-algebraic

are carried along with the integral domains

173

A i a , or with the prime ideals

P.^ C R[X^...X ] , we can have the same

occurring several times with different orders

'£ [g-k]

corresponds to several patches on the variety

Z(P.^)

W

^g'k^

n

Z

( P icP J

P

ia5 P j0 J ° r

a11

contained in

e< uivalentl

l

belonging to

S.

Remark.

with

P a t c h e s on

The orders

$ [g.,] W

on

A.

IK

so we can find

V w [ g i k | g i k £ Pia] •

x G X(\a$ia)

hence the patches

^ia*

z p

( j^)

= R[X,...X ]/P.

XCc

Let

x

n

g =

also

g(x) ^ 0.

)

contain points

are strict, g i k (x) > 0, for all tJie c o n

d

ition

g^k >. °

is

g ik -

Then

g £

P^,

ik ^ P ia

with

W{g.^} H Z ( f

considered above

10c

II g

ik

of the form

nw( gl )

can be replaced by

e

This

We can even have proper inclusions

y» Z(?j^ $ Z ( P i c P '

— — — —

g i k (x) :> 0

A*/y"

P^a

S.

s=

g

on

It follows that x

where the inequalities

gik ^ P ^ .

superfluous on

z

g i k 0 0 > 0,

Of course, if

P

( £a)-

We have indicated above how certain operations on orders defining RHJ-algebras yield other

RHJ-algebras.

Certain other operators, which one

might hope do the same, turn out to be more subtle. are

RHJ-algebras

A

is

( /P[g])>

(A,'p)

such that some simple refinement of the order '^,

not an RHJ-algebra.

i: A -* A[t]

where

indeterminate over

We will show that there

Also, (A[t],iJ>)

is the inclusion of

A

need not be an RHJ-algebra,

in the polynomial ring in one

A.

The basic example, which is a good source of counterexamples, is (A,p) = (R[x,y], ? w [x] n ? w [ y ] ) . Suppose

h(x,y)

Thus, X(A,?) = {(x,y) G R 2 |x > 0

is nonnegative on

X(A,'^).

Since

h

or

y > 0}.

cannot change sign

across either the positive x-axis or the positive y-axis, it follows that

174

h(x,y)

is divisible by an even power of

for some

r < 0,

{(x,0)|x < r},

h

x

-x,-y

y.

It then follows that

is nonnegative on a neighborhood in

{(0,y)|y < r}

on the negative

A somewhat surprising consequence is that is an order.

and

This follows since for any

x-

and

R

of the two rays

y-axes, respectively.

$[-x,-y] = CPw[x] n •£ [y])[-x,-y] Si >•••>&]< *= P> the elements

are simultaneously positive on a non-empty open set in

sets in

R

R .

are Zariski-dense, relative to the ring of polynomials

g.,

(Open R[X,...X ] ,

because of the finite Taylor expansion of a polynomial about any point. holds

for any real closed

This

R, since the estimates of 7.2 show that the

formal algebraic partial derivatives of a polynomial agree with the "limit" definition of partial derivatives.) It is unclear whether or not

(A,'.]S')

2

X(A,'^') = {(x,y) G R |x < 0 < y}. be an

RHJ-algebra.

cannot be an

is an

RHJ-algebra.

It is then obvious that

(A/£) = (R[x,y], ?

RHJ-algebra.

W

M

n

above concerning functions

and

g

vanishes on

cannot

(A[t], i*£)

The point is, one can find an algebraic surface {(x,y,t)|x,y < 0 < t}

X(A[t],iJJ) = {(x,y,t)|x >^ 0

ijp

(A,1])11)

'£w[y])> we argue that

x-y plane to precisely the open third quadrant

is

In any case,

RHJ-algebras.

in three space in the octant

I(S) C A [ t ]

'£" = y [-y] = P[-x,-y].

In either case, we conclude simple refinements of

RHJ-algebras need not be With this same

p f = $[-x],

Let

y >_ 0}.

h(x,y)

convex. S, then

or

which projects on the

{(x,y)|x,y < 0}.

Now

But the discussion in the paragraph

non-negative on

The reason is, if f

S

X(A,'p)

implies the ideal

0 £ f(x,y,t) <_ g(x,y,t) (rel

must also vanish on at least a non-empty open

175

subset of

S, above a region near the (sufficiently) negative

x- or

y-axis.

In general, open subsets of varieties need not be Zariski-dense relative to the coordinate ring of functions

A(S), because of "degenerate zeros".

in this case, such an open subset of the hypersurface hence

S

But

will be Zariski-dense,

f e I(S). The ring

(A[t],i^)

is also easily identified with the fibre sum over

(A,?) ® (R[t],'£w ) . Thus direct sums of R in general.

In case

(A,'£ [g.])

RHJ-algebras need not be

and

(B,'£ [h.])

as finite refinements of weak orders, then

If

S

and

A(S) ® A(T) R

T

V

h

j

] )

RHJ-algebras,

RHJ-algebras obtained

(A,'p [g.]) ® (B,'p [h.]) =

w 1 R w j w i l l b e a n R H J - a l g e b r a w i t h X ( A & B , '£ [ g . ® 1 , R

(A & W B , 'p [ g . ® 1 , 1 ® h . ] ) R 1 J X(A, ? w [ g i ] ) x X(B,

are

R,

Wl ® 1

h.]) ^

'

are closed semi-algebraic sets, so is

is a ring of R-valued functions on

S x T.

S x T, and Thus

A(S) & A(T) R

is

a reduced, real

R-algebra, in fact, A(S) ® A(T) = A ( S x T ) . If A(S) and R A(T) are integral domains, so is A(S) &> A(T). (This is most easily proved R geometrically, using the irreducibility of the Zariski closures S and T of

S

and

T.)

Since a homomorphism

A(S) & A(T) •> R is just a pair of R homomorphisms A(S) •> R, A(T) -*• R, it is clear that X(A(S)® A(T) ,$ (S) ®fP(T)) = R R S x T. However, in general even £(S) ® '£ (T) will be too weak an order and R will have convex prime ideals with no zeros in

S x T, as in the above example.

Contracted orders are also hard to work with, in general. if the order

p = 'p [xy - 1] C R[x,y]

the weak order.

is contracted to

R[x], one obtains

Geometrically, it is hard to reconcile

X(R[x,y] ,']$) C R ,

which lies in the half plane X(R[x],'|5 ) .

x > 0

with the entire x-axis, which is

Presumably, a contraction of an

an RHJ-algebra.

For example,

RHJ-algebra need not even be

On the other hand, 8.2.7 states that contractions are very

well behaved for semi-integral extensions. Suppose and an order where

Rn.

176

is any subset.

Then we can define

p(Y) = {f e A(Y)|f(y) ^ 0

Y = Z(I(Y))

to see that in

Y C Rn

is the Zariski closure of

X(A(Y) ,'p(Y)) = 7, where

Namely, any

all

f

Y

non-negative on

y e Y}. Y

in

A(Y) = R[X][.. .Xn]/I(Y)

Of course, I (Y) R .

= I (Y),

Also, it is easy

is the topologieal closure of Y

is also non-negative on

Y.

Y

If

Y

is semi-algebraic, then so is

Y

by Tarski-Seidenberg. n

Y C R ,

we will prove that for all closed, semi-algebraic RHJ-algebra.

semi-integral extensions A C B

f e A,

i*:

with

(B,p)

an

RHJ-algebra with

X(B,p) -* X(A, 'p n A ) .

Zariski-dense in f

closure

i*(X(B/P) C X(A/P w ).

Y = X(A(Y) ,p(Y)) -• X(A, A H']5(Y)).

is non-negative on

Pa C B

A -> B

Thus

X = X(B,'|5), then

is injective, no function

That is, i*(X(B,'£)) f 6 'j5 H A

is

if and

A

-• B a

A a = A/i" (P a ). We have

and a diagram of inclusions

i • IIB a

is the contraction of an order on n B , namely, n '£(X(B /-P ))>

QJ

f C B, but

(X

'^

OC

f

= $/P . (X

(We could also contract IItp ,

CC

OL

could be definitely weaker than

the Noether Normalization Lemma, we can obtain of some pure polynomial extension

A , and then contracting

n(A

the only point at which we might lose pure polynomial extensions

B

A [X,...X, ]

AHf by first contracting the \i\), \

of

A .

= (VV> to

Thus we can study t0

A.

form

ViY^

for some subset

that

(A , Aa n>]3(Y1))

Perhaps surprisingly,

In fact, by induction we could

RHJ-algebra order on

Y x C X(A ,']$w) x R.

was an

VXr"\]'

RHJ-algebras by contracting is in the

A -• A [X,...X, ] . K^ a <x l

a finite, real domain, and with an

By

as an integral extension

X

H 1 ^ (X ))

'JHXCB^/p )).)

attack this extension by attacking the case of one variable A

is

X(A, '£ H A ) = i*(X(B,p)).

be the minimal primes, B a = B/P a,

X(B,'P) = U X(B ,']) ) , where

then to

f CB

i*(X(B/j$)), or, equivalently, on the topological

I f CB

i:

i*(X(B/|3)) C X(A/£ ) .

IIA a

and get

is an

RHJ-algebra, under

'£ = '^(X), where

Since

inclusions of finite integral domains

since

an

X(A,'p ) . Moreover, it is obvious that

only if

The order

(A(Y),'^(Y))

A C A(Y), inducing

f ^ 0, can vanish on

Let

Y

is any extension of reduced, finite R-algebras, and

an order with consider

(A(Y),'^(Y))

So far, we have established this only (a) for closed semi-algebraic

sets and (b) for images of sets

If

Ultimately,

RHJ-algebra, with

A

-•A [X,], with A [ X ^ , of the

We would then like to conclude X(A , A^ H']5(Y1)) = T T Q ^ T ,

177

where

IT: X(A^, ^^)

R -• X(A ,^ w)

x

is projection.

However, it is not clear

how generally this holds, and even in the semi-algebraic case, we need the Tarski-Seidenberg theorem to prove it. We conclude this section by pointing out that we have established an invariant criterion that

(B,'p) G (POR)

is the affine coordinate ring of some

closed semi-algebraic set in affine space over R-algebra of finite type. Pi C B B

i

= B

with

/ p i»

R.

Namely, B

must be a reduced

Also, there must be finitely many p-convex primes

(0) = O P i , and finite refinements of the weak orders

with

^

= B n

nC-Pi) > under the inclusion

B^IIB^

^

C Bi,

(Strictly

speaking, we must still prove the basic real Nullstellensatz, 8.2.1, and 'P(X(B. /£•)) = CP-) • 1 1 p these tasks in the next three sections.)

We proceed now to

establish Stengle's theorem that

8.3. Real Curves Let

K = R(x,...x )

closed ground field

be a real field, finitely generated over our real

R, with tr. deg. D(K) = 1. K

fractions of the integral domain P C R[X 1 -..X n ] zeros in

is a prime

Thus

K

is the field of

A = R[x-,...x ] = R[X,...X ]/P, where

'^-convex ideal.

We will prove that

n

R , that is, there exist homomorphisms over

R,

R C K, with

the valuation ring of elements finite relative to place of

K,

K

A C A R , where R.

must be Archimedean closed in

K

R.

(Or, since

for any order on

can be AR C K

K

over

tr. degR(K) = 1 , K

is

The associated signed

p R , is then R-valued since a non-trivial place of

lowers the transcendence degree over

has

p: A -• R.

Actually, we will prove something stronger, namely that totally ordered non-Arehimedean over

P

R R

non-Archimedean over

R.

Thus the residue field Thus we obtain

pR:

A n /Q n is algebraic over R by results of 7.6.) K K R. K -• R, ± °° and, by restriction, p: A -• A R

Proposition 8.3.1.

Let

K = R(x1-..x )

one variable over a real closed field Let

x £ K - R, so that

the contracted order

178

K

R.

Let

is algebraic over

'$ n R(x).

be a real function field in ']) C K

be any total order.

R(x), and let

'£ C R(x)

Then (using the notation of 7.5):

denote

Case (i) a >b

If '£ = '$m [resp. 'p J , we can find

[resp. a < b ] , both orders

f

and "J5 a,—

a,+ orders on K with Case (ii)

b G R

on R(x) extend to total

A C AR.

If "jT = $c

such that for all

+

[resp. $c

] , c G R, we can find

[resp.

a G (c-e,c)], both orders

a G (c,c+e)

!P on R(x) extend to total orders on K with a, ~ Case (iii)

If f = £ D where

such that for all a G ( a 1 , a " ) , both orders total orders on K with

c = 0, by replacing

'p

v j>J .i

be the minimal polynomials for f^ . (x) G R[x].

a 1 , a" G R,

R(x), say

Suppose

y

and

m .l«y ij

i j

x.,

1 <^ j _< n, with

g.(x),

K, the polynomial

(«(x),j3(x)) c R(x)

The last

aG

(c, c + e)

[respectively

G i (a(a)), Gi(3(a))

have

e > 0

has real roots over

{G Q (a(x)),.. .Gk(a(x)) = h(x)}

f-signs in

in case (ii) or

R(x). We now appeal [resp.

a' < D < a"

in

(c - e, c ) , depending on whether

a G (a 1 ,a")], then the R-signs of all

agree with the

We also make our choices so that

G(y)

is an interval containing (at least)

The terms in the sequences

case (iii)] so that if 'B c,-

G Q (y),... ,G k (y).

h(x) ^ 0 G R(x).

to the results of 7.5 to choose

terms

K = R(x,y). Let

m.-l J-»i ,J

I J

{Go(j3(x)),...,Gk(j3(x)) = h(x)}

or

on R(x) extend to

••• + g m (x)

+

G Q (y) = G(y), G ^ y ) = ^

by our assumption on

one of these roots.

•IT = *B c,+

, '$

a f < D < a",

Consider the Sturm sequence for the irreducible polynomial

term is a constant

and

K

y G K so that

m. j j J J

(R(x), f).

A C AD.

x by 1/x.

0 = G(y) = g o ( x ) y m + g 1 ( x ) y m - 1

Now,

and

A C AD. K

In cases (ii) and (iii), choose

over

'J5

Case (i) occurs only if D = D R (x) = ± °° . We can deduce case (i)

from case (ii) with

G(y)

e > 0 in R

D = D R ( x ) , the cut of R defined by the

order 'p C K , is transcendental, then we can find

Proof:

such that for all

-f-signs

a(a) < 3(a)

in

of the R

G i (a(x)), G i (3(x)).

and so as to insure that

179

a

is not a root of any of the numerators or denominators of the rational

functions R(x).

a(x), 3(x) , or the coefficients of the polynomials

G.(y)

over

Finally, and this is the most important restriction, we insure that the

conditions on a guarantee that coefficients

a

is not a root of any of the leading

f n .(x) of the minimal polynomials

F.

for the elements

x.,

1 £ j <_n. With all this accomplished, we now start over with either order '$ a, + or '£

on

R(x), and consider the Sturm algorithm for real roots of

in the interval

(a(x), 3(x)) G R ( x ) .

place defined by our new order 'p a,+ pCG^cxCx))) = p(Gi(a(a)))

and

If p:

R(x) •* R, ± »

or 'p a,-

on

G(y)

is the signed

R(x), then, by our choices,

p(G/(3(x))) = pCG^GCa))) G R

since

p(x-a) = 0.

The signs of terms which occur in our new Sturm algorithm thus coincide exactly $*-Sturm algorithm for G(y).

with the signs which occurred in the original

Thus G(y) has real roots over R(x) in (a(x), 3(x)) with respect to either order •£ or '£ , and this order on R(x) extends to K. a, + a, — Finally, since 0 = £ .(x)xJ +••• +f (x) G K, f± . (x) G R[x] and ^(a ) t °> fu, n j

i1:

Remark.

follows trivially that

x.JKG A D C K, hence

Because of our careful choice of

additional conclusions can be drawn.

in the proof above,

Note that the signed place

associated to our order '$ or 'ft on a, + a, ~* sequence

a G R

A = R[x1J...x Jn CA DK. D .

p: R(x) + R, ± °°

R(x) has the property that the

p(GQ(y)),...,p(GR(y)) = p(h(x)) = h(a) t 0, is actually a Sturm

sequence over

R

for the polynomial

G(a,y) = Eg.(a)ym " 1 .

This is not

completely obvious because construction of a Sturm sequence involves division, so our choice of

a

coefficients of the the polynomial

avoiding roots of numerators or denominators of the G.(y)

G(a,y)

over

is essential here. R

has no multiple roots.

little further care we could arrange that roots in

R

as

G(y)

or *jT, and that the G(y).

has over

In any event, since

G(a,y)

In fact, with a

has the same number of

R(x) with either of the orders '£

At these points

180

in

R

, '$

R-roots of

G(a,y) are "close" to the R(x)-roots of 2 (a,b) G R on the curve G(x,y) = 0, we have

(d/dy)G(a,b) ^ 0. Thus we have a special type of simple point: G = 0

h(a) £ 0,

crosses the line

x = a

transversally.

the curve

,

Examples.

Consider the curves below:

xy - 1 = 0

y -x = 0

(b)

(a)

y 2 -x 3 +x 2 = 0

(d)

(c)

(e)

(f)

y 6 -2xy 3 +x 2 -x 3 = 0

(g)

In each case we have an irreducible ring

G(x,y) = 0, an affine coordinate

A = R[x,y]/(G(x,y)), and a function field

K = R(x,y).

We study the

181

behavior near

x = 0, to illustrate certain aspects of the proof above.

Example (a) illustrates that p:

R(x,y) •> R, ± «

curve.

If

with

(R(x), '£n

(R(x), fQ^J

)

R(x,y) may admit a signed place

p(x) = 0, but with no points is extended to

(0,b) G R

R(x,y), then

on the

p(y) = °° and if

is extended, p(y)

Example (b) has the origin as a simple point, but only the order (R(x), :j3

)

extends to

cross the line

x = 0

R(x,y)

since

> 0.

transversally, but does cross

The curve does not x = a

transversally

a G (0,e).

for

3 2 In example (c) , G(x,y) = y -y -x

specializations a>0,

x = a, small

y - y - a

has three real roots over

(R(x), !pQ

a.

If

a < 0,

y -y

has one real root, and if

-a

has three real roots,

a = 0,

y

-y

has a double

root and a simple root. Example (d) has a zero at the origin, but D R (x) « 0.

total order with

Since

R(x,y)

does not support a

G(0,0) = 0, we get a homomorphism

p: A = R[x,y]/(G(x,y)) •> R, hence (x,y) C A is f -convex. However, (x,y) C A is not ('Cw ) -convex. If it were, the signed place extension d theorem of 7.7 would imply the existence of a total order on

K = R(x,y),

with signed place

Note specifically

that

2

pR:

K -»• R, ± °° extending

2

x (x-l) = y , hence

(x,y)

is not

x-1 G ('£ ) w d

Cpw) -convex.

or

p:

A •*- R.

1 < x rel ($ ) . w d

The conclusion

relative to any total orders on

x-1 £ ($w)

K = R(x,y)

we must have

Clearly, then

also shows that 1 <^ x, which

agrees with the picture. In both examples (e) and (f), R(x,y)

supports total orders at the

origin, but the origin is a singular point. extends to

R(x,y)

does not extend. in two ways to

in two ways (with

In example (e) , (R(x), 'pQ + )

y > 0

In example (f), both orders

or

y < 0 ) , but

'pQ + , 'pQ _

on

(R(x), 'pQ _) R(x)

extend

R(x,y).

In example (g), (R(x), 'pQ + )

extends to

R(x,y)

this case, it turns out that there is no element of over

in two ways. R(x)

In

between the two

real roots of

G(x,y)

y = f(x)/g(x)

lies between the two infinitesimal pieces of the curve

182

)

(R(x), 'pn ) . This behavior is reflected by the u, +

but only one real root over

if

x = y

(R(x), 'pQ + ) .

(Geometrically, no graph of

G

(x,y) = 0

in

at

(0,0).)

(R(x), 'pQ + )

The Sturm algorithm for any interval

will yield either

2

roots or

(a(x), 3(x)), depending on whether or not

R(x)

roots for

0 G (a(x) , 3(x)).

when we specialize to small positive values This example shows how

0

(a(x), 3(x)) G(x,y)

in

Of course,

x = a, these two roots separate.

is not order dense in its real closure (see 7.4.4).

The proof of 8.3.1 leads to a rather nice picture of the real closure of the field

R(x)

with any order

Also, all the orders

$

a,i

$.

If

look the same.

The points in the real closure of polynomials

G(y)

over

x

Thus we look at only

(R(x), 'Pn

)

by

1/x.

pn and u, +

are roots of irreducible

R(x). We can think of these roots as infinitesimal

"connected" pieces of plane curves x-side.

f = 'p+oo, we replace

G(x,y) = 0, near the y-axis on the positive

One point is greater than another if its graph is above that of the

other, infinitesimally near the y-axis.

A point is finite relative to

its graph meets the y-axis, and is infinite relative to asymptotic to the y-axis. relative to

R

if

R if its graph is

The elements which are infinitesimally small

are those graphs through the origin.

The addition and multi-

plication corresponds to addition and multiplication of functions on The case of the transcendental order First, the real closure of

R

(R(x), $ D )

'$„ c R(x)

(0,e) C R.

is slightly simpler.

is actually Archimedean over

R.

We visualize points as infinitesimal pieces of graphs of plane curves

183

I Cut D I

I "crossing" the cut at

x » D.

graph can be extended to cuts of

R.)

closure of

(The "values" of the function defined by the

x = D.

In the case of

These "values" are necessarily transcendental

'£n, R

is actually order dense in the real

(R(x), £ D ) .

The notion of "connected" piece of a curve, together with rigorous formulations and proofs of the statements in the preceding two paragraphs, would require some effort to develop. intuitive discussions only.

After

Thus, we mean these paragraphs to be

all the machinery of this Chapter VIII

is developed, for semi-algebraic sets of all dimensions, it would still be a respectable project to investigate curves carefully.

This study would

include a detailed analysis of singular points, especially branches at a singular point, and a classification of orders on any function field in one variable, extending the results of 7.5 (which is the special case

R(x)).

8.4. Signed Places on Function Fields In this section we establish the existence of nice signed places on real function fields, in order to study prime convex ideals for various orders on integral domains of finite type over P C R[X 1 ---X n ]

be a

R.

A.

Thus,

K

K = R(x 1 -..x n ), the fraction field

is a real function field.

Proposition 8.4.1. (Lang) with

Let

'^-convex prime ideal in the polynomial ring, let

A = R[x 1 -.-x n ] = R[X 1 -..X n]/P, and let of

We begin with the weak order.

If

K = R(x 1 -..x n)

is a real function field

tr.deg.D(K) = r, then there exists a sequence of signed places over R

184

R,

K - > K 1 , ± ° ° ^ K 2 , ± oo -» * •. -*K r ,± oo, with each tr.deg.D(K.) = r - i . K 1

(Thus, K = R.) r

ring of the composite where

p^:

A = R[x,...x ] .

hence

IC

Moreover, if

a function field with A. C K l

is the valuation

K -*• K.^, ± oo , then we can arrange that

In particular

p | : A -*• R

(a-^-.a ) = (p ( x ^ ,.. . ,p (x )) £ R

n

A CA.,

is a homomorphism,

is a zero of the prime ideal

P C R[Xr..Xn].

Proof:

The proof is by induction on

follows from 8.3.1. of

K

over

K,

F CT K1

Thus

r > 1, assume

R, so that

R(x^...x ) . of

If

K

the real closure of

r = 1, the proposition

{x,...x }

1

p:

K

K

and let

L

C F

Let

p 1 = p| :

denote the real closure Kf = F(x r...x n ) C r.

Let

tr.deg.D(K,) = r-1

with

F, such that

K -• F, ± «> . since

Remark.

J

J

—

However, when we

K

K

2

~*K

± ©o is dropped.

l'±O°

i s a t

least

K

2

-• K^

using

place if the distinction K

, ± oo relative to this "new" order on

Continuing, we work backwards until all

reordered so that all

a

K _^

Using Krull's Theorem 7.7.2, we then reorder

and recover a signed place .

r-1.

K, -• K^, ± °° , a new order

K -* K.^, ± °° -••••-• R, ± oo , then we order

K

—

•

ultimately arrive at

between

is a

The proposition now follows

may need to be chosen, as in the proof of 8.3.1.

Now

p1

1^.

In continuing the proof to obtain

, -* R, ± °° .

is

of transcendence degree

(See [68], Vol. II, Chapter VI, §14, Theorem 31.) by induction applied to

The image of

±

R

p(x.) G F

F.

p,(x.) = x., 1 < j < r-1.

K l

K, is actually a function field over

K,

r

R(xr . . x r l ) .

-* F, ± oo t over

1

on

is a transcendence base

is a function field in one variable over the real closed field

finite, r <_ i <_ n.

Thus

If

is algebraic over the pure transcendental extension

Choose any total order on

Using 8.3.1 we find

subfield

r.

K., including

K = K , are

K^ -• K^ +1 , ± °° are signed places.

Our proposition has many consequences.

The first is the following "weak"

weak Nullstellensatz.

Corollary 8.4.2. ideals

The maximal

(X^ - a^,...,X - a )

'.J3 -convex ideals of

corresponding to points of

R[X ...X ]

are the

n

R .

185

Proof:

In fact, any

f -convex ideal w

P

'•]) -convex ideal

and if

there is a chain of prime

I

is contained in a prime

r = tr.deg.n(R[X., .. .X ]/P), 8.4.1 shows that K i n '^-convex ideals

(Xx - a 1 ,...,X n - a n ) C R [ X 1 . . . X n ] .

P = PQ C p

C . .. C p^ =

Namely, P^^ = kernel (pi| :

A -» IO) , in

the notation of 8.4.1.

Proposition 8.4.3. transcendence degree

r

Let

K = RCxj.-.x )

over

R.

Suppose

be a real function field of

finitely many elements positive in some total order on signed places over

R,

p:

g. i- 0, are

gi>«-«>g^ ^ K, K.

Then there exist

K ~* R, ± °° , with rank(p) = r,

p(x.),p(g.) E R,

p( g j ) > 0.

and

Proof:

Apply 8.4.1 to the real function field

E = R(x.,y.,z.) l

j

3

where

y]-zy Yfr-i. Remark.

n

An alternate proof of 8.4.3 could be given as follows.

as in 8.3.1, first any order extending one of the orders

!|5 C K 'p

+

If

r = 1,

is chosen, the replaced by an order

C R(x^) C K, by a careful choice of

a E R.

This choice can be made exactly as in the proof of 8.3.1, so as to preserve the signs of the relative to 8.4.1

R.

g.

and keep the

g.

finite, but not infinitesimally small

This technique is now combined with the inductive proof of

to give 8.4.3. Using either method of proof of 8.4.3, the conclusion can be strengthened

to include an approximation statement. defines a signed place The new order on

K

pR:

Precisely, the original order on

K -* A, ± °° , where

A

is Archimedean over

corresponds to an R-valued signed place

The conclusion of 8.4.3 can then be extended to say that the approximate the

p R (g-) E A, ± °° as closely as desired.

this just means

p(g.)

p:

K

p R (g.) = ± °° ,

can be made large.)

The next corollary constitutes a "weak" strong Nullstellensatz.

Corollary 8.4.4. g ^ P.

186

Let

P C R^.-.X ]

Then there exist zeros of

P,

be a prime n

a E R , with

'p^-convex ideal,

g(a) ^ 0.

R.

K -• R, ± °° .

p(gO ^

(If

K

Proof:

g / 0 £

R ( x r . . x n ) = K, the fraction field of

Choose a total order on

K

R [ X r . -XJ/P.

and apply 8.4.3 to the single element

g

or

-g,

whichever is positive.

Remark.

•

Corollary 8.4.4 is exactly the basic Nullstellensatz stated

previously as Proposition 8.2.1. established.

For example, if

n

{a e R |h(a) = 0 all of

a €= z(I)}.

all

Thus all the consequences of 8.2.1 are now

I CR[x i ...X n ]

h E I}, then

is any ideal, Z(I) =

*ti(I/Pw) = {f € R [ x r . . X j |f (a) = 0

If we combine this result with the characterization 2.2.4

vfa(I,pw), we conclude that the ideal of functions which vanish on

consists of exactly those

f

such that

Pi,q. E Rp^.-.X ] , h. E I.

Z(I)

S

f + Z p . = Z q . h . , for some

s _> 1,

This result, or similar versions, is known as

the Dubois-Risler Nullstellensatz in the recent literature, [14], [16], [17]. We point out that our proposition 8.4.3 as well as Corollary 8.4.4 date back to the early work of Artin, at least in essentially equivalent forms.

The

other ingredient of the Dubois-Risler Nullstellensatz is then the characterization of the radical of a hull for general partially ordered rings, given in 2.2.4. In fact, little extra work is now required to characterize those functions f E RfXj.-.X ]

which vanish on a basic closed semi-algebraic set of the form

Z(I) H w{g.} C R n . V = $„[g-] w j

is

First, assume

an order on

is an

RHJ-algebra,

f E A

which vanish on

Z(I) H W{g.}

is a radical ideal so that

and assume

J C I.

D Z(I) Hw{g,}

f

s >_ 1

...g

g X

l

X

= r

E(k2 J
l

and polynomials

allow the "empty product"

f E /H(I,^) C A.

vanishes on

1

q

g 3

...g 1

J

h,k,y,z, with

of the

g.

) 2 q

J

y GI

l'"\

and

z E J.

I(W{g^».

Certainly,

'^[gj]

(We

here.)

Secondly, it is not hard to see that the smallest such ideal exactly

(A/P)

and the functions

are exactly the functions

f E R[X....X ] , we conclude

Then

if and only if there is a formula

f2s+Zh2 1 l*-' 1 r

for some

A = R[X,...X ]/J in

X(A,$) = Z(J) Hw{g,}

In terms of polynomials Z(I) n w { g . }

J C R[X1 ...X R]

is an order on

J

is

R [ X r . .X n]/I(W{g.}).

187

If

J

is any radical ideal with this property and

vanishes on if

X ( R [ X r . .Xj/J, '^w[g. ]) , hence

J = I(W{g.}), then

h G I(W{g.}), then

h = 0 eRp^.J^/J.

Z(I) n W{g.} = Z(I + J) n W{g.}, and

The two paragraphs above reduce the determination of to the determination of all

I(W{g.}).

j , then, of course, I(W{g.}) = 0.

Even if

n

R , in a given case we may know an ideal

R[X,...X ]/J.

J C I

W{g.} with

Moreover,

J C I + J. I(Z(I) nw{g.})

x G Rn

If there exists

h

with

g.(x) > 0,

has no interior in '$ [g.] w j

an order on

It would then be unnecessary to actually determine

I(W{g.}).

Next we state a weaker form of 8.4.3 which is as useful for most purposes. Corollary 8.4.5 •P [gi'**gv]

a

(Artin)

A = R[x....x ]

be a real integral domain

finitely generated refinement of the weak order on

Then there is a homomorphism

Proof:

Let

Extend

p:

'^w[g^]

A -• R

with

p(g.) > 0 ,

to a total order on

A

g. ^ 0.

1 < i < k.

and apply 8.4.3.

In particular, this result applies to the polynomial ring In fact, we deduce the following, also due to Artin.

A,

•

R[X,...X ] .

(Part (b) is Hilbert's

1 7 t h problem.)

Corollary 8.4.6. (a)

g. G R[X 1 ...X n ]

If

is an order on

R[X 1...X n ]

are nonzero polynomials, 1 <_ i •< k, then a G Rn

if and only if there exists

with

'^[g^] g i (a) > 0,

1 < i
The polynomials

^, J

Proof:

188

some

nowhere negative as functions on

h,h. 6 R [ X .L . . . Xn J , h ji 0 } . J

The "only if" statement in (a) is 8.4.5,

true because the set nonempty.

f G R[X^...X ]

The "if" statement is

U{g.} = {a|g.(a) > 0, 1 <_ i <_ k}

is Zariski dense if

Similarly, a function

f(a) < 0 around

and

a.

f G ('|5 ) w d

h f = Eh. , then

Conversely, if

h

is clearly nowhere negative since if

would vanish identically in some ball

f £ (:pw) , then

'^w[-f]

is an order on

R[Xr..X

and (a) applies.

Q

Remark 8.4.7.

The "if" statement of 8.4.6(a) is not true for arbitrary

finite real domains p:

A -• R

point

with

A = R[X ...X ]/P.

p(g i ) > 0

That is, the existence of a homomorphism

does not imply

a = (pCXp,...,p(X n ))

$ [g^]

is an order on

could be a "degenerate" zero of

Example (d) of 8.3.

The argument of 8.4.6(a) shows that if

is Zariski dense in

X(A,'£ ) , then W

Z(P) C R n , the zeros of * w [gil

c A

is an order

is an order on that

h

M O G A ,

A, since

Remark 8.4.8.

A is a domain.

as in

U{gi> n X(A/}5w) Here

X(A,'B) = W

Conversely, if or

^[g^-h]

In either case, 8.4.5 guarantees U{g i> n X(A,'£w).

Thus

U{gi>

is

C Q> ) C CP ) C cp ) = (B ) =?(R n W WS Wp W m W(J In the next section we will prove a

Consider the orders R[X^...X n ].

theorem of Stengle which implies distinct orders here, f

w

C (tp ) w s

V

('£ ) = '£(R n). Thus, there are only three w P C'^(R n ). These inclusions are definitely

For example, Hilbert [ 1 ] gave an example of a strictly positive f(X,,X2), which is not a sum of squares.

polynomial in two variables, proof that

(£)

= *p(Rn)

W P polynomial belongs to

hand, by definition, term of

^[g^h]

The

X(A,'p ) .

on the polynomial ring

proper.

then either

does not vanish identically on

Zariski dense in

is an order.

U{gt> = {b G R n |g ± (b) > 0}.

P, and

and

'^1T[g.] C A W 1

P

A.

f

(p ) . w s Q3 ) w s

Thus

'B / (? ) , if n = 2. On the other w w 9 9 = {f|(l + Eh.)f = Z g . } . Immediately, the homogeneous 1 3

of lowest degree must be a sum of squares.

Hilbert's example

f(X,,X2)

But now if, say,

is made homogeneous, F(X ,X.,X 2 ), so that

f(X 1 ,X 2 ) = F(1,X 1 ,X 2 ), then hence cannot belong to

Our

will show, in fact, that any strictly positive

F(XQ,X;L,X2)

Cp ) . w

s

Thus

cannot be a sum of squares and

Cp ) w

f f-P(Rn)

if

n = 3.

s

The last result in this section is a signed place perturbation theorem for function fields.

The statement and proof are similar to 8.4.1, but

instead of trying to make elements finite relative to a discrete rank

r

189

signed place, we assume certain elements infinitesimally small relative to one signed place, then try to keep them that way relative to a discrete rank r

signed place.

This result will be the basis of our study of derived orders

in 8.6 and dimension theory in 8.10.

It provides a tool for studying the

geometry of a semi-algebraic set near a point, or more generally near a semi-algebraic subset.

Proposition 8.4.9. R

with

p':

Let

tr.deg.R(A) = r.

A = R[x,...x ]

Let

K

be the fraction field of

K ""* A, ± °° is a signed place over

p'(x^) = 0.

(Necessarily

p

1

R, with

signed place

Proof: p1

hence

p:

If

A

A

K -• R, ± °° with

and suppose

Archimedian over

= p R , relative to the order on

This follows easily from results of 7.6.) r

be a finite integral domain over

and

induced by p 1 .

K

Then there exists a discrete rank

p(x^) = 0.

r = 1, there is nothing to prove since we assume

is already discrete rank

R

1

and R-valued.

If

p'Cx^) = 0,

r >_ 2, we prove the

proposition by induction. By the Noether Normalization Lemma we can find of the

x., say

p'(u.) = 0.

u,,...,u , such that

A

r

linear combinations

is integral over

Thus we may as well assume that

A

R ^ . - . u ].

Certainly

is already integral over

R[xr..xr]. Let x

.,

f.(x

.) = Ea..(x, ...x )x

j > 0, with coefficients in

polynomials, hence applying polynomial

pf

. = 0

be the minimal polynomial for

R[x,...x ] .

we get that

Za i -(0...0)T 1 £ R[T].

Choose

The

0 £ R

0 < e G R

than the absolute value of all other roots

{a,}

f.(x

.)

is a root of the nontriviat such that

of the

L

be the algebraic extension of

j > 0, and replace on

K

defined by

namely, take

p

hence replacing rank

r

place of

190

1

A p'

extends to R C L.

K

L

by

p:

L.

is less

/e 2 - x z + . ,

y. = / e 2 - x 2 + . - e.

In fact, the place

p

1

The order

extends to

The residue field is Archimedian over

does not change our hypotheses.

K = L.

L,

R,

Also, any discrete

L -• R, ± °° restricts to a discrete rank

Thus we may assume

over

IJ

obtained by adjoining

R[x^,y.], where

= pn,

signed place K.

by

K

e

Ea..(0...OjT1

K

Let

are monic

r

signed

R.

With this assumption, given any signed place with

p(x i ) = 0 ,

1 <_ i £ r, then necessarily also

The reason is that interval

Let

0 G R

(-£, e)

Let

F

F C r

is the only root of the

p(x + .) e (-e, e)

and

be the real closure of

y

x

x.

by

-x.

F(x )

Dp(x r) = D

of

F(x ,y) of

F.

in the

in

K. p1.

R(x 1 -..x r 1 )

over

of

K.

Write

F(x ,y)

is a

F, totally ordered as a subfield x^, 1 <_ i <_ r, are infinitesimally

We may assume the

x^

positive, be replacing

induces an order on the simple transcendental

This order is completely determined by the cut

F, as in 7.5.1.

regarded as a subset of D

2

if necessary.

The order or extension

R.

x +. < e

R(x,...x ) . Then

With respect to this total order, the

small relative to the subfield

Ea^^. (0- . .OjT1 2

R

j > 0.

relative to the order defined by

algebraic over

function field in one variable r.

K

K -*• R, ± °° over p(x + .) = 0,

since now

be the real closure of the subfield

K = R(x,...x ,y), with

of

p:

F.

We have

Now, D

0 < D, that is, 0 E D

if

D

may be algebraic or transcendental.

is algebraic, we will not yet perturb the order on

F(x ,y).

If

D

transcendental, we appeal to Proposition 8.3.1 to choose an element 0 < 3 < D, so that if

F(x ) r

is reordered over

as in 7.5.1, then this new order extends to we still have

xr

is

F

infinitesimally small relative to

Since R C F.

is

3 6F,

with either order

F(x ,y).

If

!p , p,+

0 < 3 < D> Of course, the

x., i < r, are also still infinitesimally small relative to R since F has not changed. We now consider the signed place (possibly) new order. F-valued.

We have forced

on

F(x ,y), with respect to our

pp

to be non-trivial, thus

In fact, we have forced the restriction

to be non-trivial.

pp:

(Specifically, p p (g(x 1 ...x r )) = 0

is the minimal polynomial for trivial on

pp

R(x,...x

) , hence

3 ^ F

over

pp:

R[x,...x

pp

is

R(Xj. . .xr,y) = K + F, ± °°

where i]«)

gCxj...x r_1,3) = 0 But

p

is still

K -» F, ± °o is a discrete rank 1 signed

place. The image degree

r-1

p p (K) = K' C F over

R.

is thus a function field of transcendence

In fact, the proof of this result in, say, [68, vol.11,

Chap. VI, §14, Theorem 31] shows more. closure of

A = R[xi ]

in

K.

Then

Let 1

K

A' = R[x.,z.]

denote the integral

is the field of fractions of

A'/P',

191

P1 C A 1

where

is some minimal prime ideal. 1

by the inclusion integral over

K

A.

C F, the

z.

With the order on

are finite relative to f

Thus we can write

K'

induced

R, since f

z.

is

x| = x.(mod P 1 )

A /P' = R[x!,z!] C K , where

and

z! = z. - p (z.)(mod P'). By induction, applied to the signed place

pR:

K1 -> A 1 , ±°o,

signed place

A1

K1 -*• R, ± °° , with

q:

We now compose altering

pp

Archimedean over

pp

and

q.

r-1

q(xp = 0.

K -* K', ± °° -* R, ± °° , and reorder

K

K!

induced by

pp

We end up with our desired rank

Remark.

Here is an extension of

r

is not the same as that induced signed place

p:

K -> R, ± °°.

In the notation of 8.4.9, suppose given finitely

many elements

g. ± 0, which are positive in the order on

g. G A ,

p1 :

K -*- A, ± °° .

statement that the discrete rank

r

g.

are also positive in the order induced by the

signed place

p:

is the composition of the first gj

K

Then we may add to the conclusion the

K -* 1^, ± «>->... -> K p r ^Cg^) t °» where

and moreover, we may arrange that

If

•

8.4.9 which is useful in studying

geometry near a point.

induced by

by

on infinte elements, as in 7.7.2, to take into account the

fact that the order on by

q,

R, there is a discrete rank

r-1

g.

P r j_ •

±°°->R, ± ° ° , K

~*K r y

±

°°

maps in the chain.

£ ( x r . . x n ) C A = R [ x r . . x n ] , then already

are various ways to keep the

^

p( g j ) t 0.

There

positive relative to our new order on

K.

For example, using the technique of the proof of 8.2.10, one can adjoin certain P'(g-)

/gT =e

to

K, then replace

-^R>

0 < c.

there is only a single

so that

g

{x^.-.x^,}.

R[x,...x , Jg7 - £ . ] , where

This technique works whether or not

Thus we can concentrate on the case

just try to arrange that

formula for

by

(See also the proof of Proposition 8.6.5 in a

later section for this technique.) g. €= (xj.-.x^.

A

p g.

,(g-) £ 0.

Setting

g = Ilg., we may assume

Now, we just add the dummy generator

g

to our

A = R[x,...x ,g]. Then we might as well rename the generators becomes

x^, the first element of the transcendence base

If one inspects the proof above, one finds

so that at the crucial inductive step constructing we obviously have

192

g. G (x-^.-.x^, and

p F (g) = g / 0.

pp:

g = x^ & R(x 1-. .x r _ 1 ) F(x ,y) ~* F, ± °° ,

Our extension of 8.4.9 is thus proved by

induction.

(Note that, again, if

r = 1

there is nothing to be proved.)

8.5. Characterization of Non-Negative Functions In this section we prove the theorem of Stengle [22], characterizing the functions nowhere negative on a closed semi-algebraic set. Begin with a reduced, real R-algebra

A

of finite type, and a finite

refinement of the weak order ty = p [g.]. w (Stengle).

Proposition 8.5.1 X(A,'p)

j

An element

f E A

if and only if there is an equation

(f

n

is nowhere negative on

+ p)f = q, some

n :> 0,

p.q eVProof,

The "if" statement is trivial.

We prove the "only if" statement A 1 = A[t]

by applying the Nullstellensatze of 8.2 and 8.4 to the ring polynomials in one indeterminate over Let on

A

f

= f [g.] C A'.

We know

extends to an order on

simply a homomorphism follows that

f

x:

!

X(A ,^ )

A1

A -* R

of

A. '.|5! is an order on

by 6.1.

A homomorphism

together with a value

is identified with

Af

since any order xf:

A1 -> R

x'(t) E R.

X(A/p) x R.

is

It

(This we also dis-

cussed in 8.2.) Suppose

f E A

is nonnegative on

t2+ f

on all zeros of

f E /H(t z + f/p1) C A 1 .

in

X(A',^').

X(A,'p).

Then

f €ACA'

vanishes

Thus, by the Nullstellensatz,

From 2.2, we obtain an equation

P(x,t) = Q(x,t)(t2 + f(x))

for some

n > 1,

functions on

P(x,t) E:p'.

We are regarding elements of

G(x,t) E A' can be uniquely written

G (x,t ) + tG,(x,t ) . Applying this decomposition to P(x,t), Q(x,t)

by

F(x,t2) = 0 E A' a E A.

as

X(A,'p), X(A',^'), respectively.

Now any element

any

A,A'

2

P Q (x,t ), QQ(x,t ) , respectively. implies

F(x,t) = 0 E A1

Thus we obtain a relation in

and thus

G(x,t) =

(*), we can replace But any identity F(x,a) = 0 E A,

A,

193

f 2n (x) + P Q(x, -£(x)) = 0.

(**)

Now, we investigate

P Q (x, -f(x))

more closely.

Since

P(x,t) ^ ^ ' ,

we can find a formula

P(x,t) = EPj(x,t) gj (x),

Specifically, the

gT j

gj £'j5CA.

are finite products of the

g., where 3

'p = '£ [g.] C A. w j

Thus

P Q (x,t 2 ) = (Pj )0 (x,t 2 ) + t 2 P^ 1 (x,t 2 ))g J (x)

by writing

PT(x,t) = P T

by

) +tP

-,(x,t ) , expanding, and using uniqueness 2

of the decomposition t

ft(x,t

P(x,t) = P Q (x,t ) + tP,(x,t ) .

Replacing

t

Thus (**)

f n (x) + p(x) = q(x)f(x)

gives an equation

We obtain 8.5.1 by multiplying this last by

CoroHary 8.5.2. '-P = '-P [g • ] w j

a

and

f)g(x)) j

p(x),q(x) G'^.

f(x).

•

be a reduced R-algebra of finite type,

Suppose

g GA

A,

vanishes on

f £ A

a function

Z(f) H X(A,f) , so 0 £ g2n <_ qf (rel

Then, there is an inequality

q£'|J.

Proof. = A[t]

A

X(A,']>).

g G v^H({f}, $) C A.

for some

A

Let

with

finite refinement of the weak order on

nowhere negative on

1

t

- f(x), we find

PQ ((x, -f(x)) = (Z P x f)g(x)) -f(x)(E f(x)(E Pj(# 1x Pj() 0 (x, -f)gj(x))

that

by

This is a consequence of the proof of 8.5.1. again, and

'£' = ? w [g k l

c

1

A .

Then

We introduce

g e >^M({t^ + f}, f1) C A 1 .

Going through the proof of 8.5.1 8.5.] yields an equation

g

+ p = qf,

p,q

which is exactly our assertion.

Remark. the hypothesis

Corollary 8.5.2 should be compared with 2.2.3. f £'|5

of 2.2.3 by the weaker hypothesis

f

We have replaced non-negative on

X(A,']>), and have established the same conclusion in both 8.5.2 and 2.2.3.

194

Corollary 8.5.3. '£

=

a

'P [§•] w 3

function

Let

A

be a reduced R-algebra of finite type,

finite refinement of the weak order on X(A/p).

strictly positive on

1+p = (l+q)f, some

Proof. setting

a

p,q ^='p.

But now

p'

(l+q')f = p f ,

is also strictly positive on f

8.5.2 again gives

f €= A

Then there is an equation

Corollary 8.5.2 gives equations

g = 1.

A, and

1+p" = q"p ,

p",q" £'p.

p! ,q' e 'P X(A,'.p).

by

Applying

1

Now check that

(1+q )(l+q")f =

1 +p' +p".

D

Note that the equation e

CA

*s(i)

Let

1+p = (l+q)f

A

be a reduced, real R-algebra of finite type, fp = 'p [g.] w j A.

is an RHJ-algebra and that the functions coincide with

the orders

consists of all functions

(Y - X 3 )

(p/I)

A

I

C A/I

n >_ 1, is

p,q G ])}.

A = R[X,Y],

wil1

2

f G A

I = (Y - X ) , then

R 2.

congruent to

(Consider

3f/3X

f £ (*p /I)

then

(l+p)f = q, some

t

f £ A

P

er

is nowhere X

modulo

at the origin.)

('P /I)

is easily seen to be non-

C (p/I)

by Stengle's theorem.

is a reduced real R-algebra of finite type and

the ring of functions $ = 'Pw[t5]>

X

P

r0

= CP/I)p> since, first, 'P/I ° P / I , which gives

2(1) C x, which gives

with

be a

3

C A/I

On the other

of the functions

a finite refinement of the weak order, then the order all

By 8.5.1, fly I)

Z(I) C X.

Z(I)

f (X,Y)

Suppose

!p -convex by 8.2.2, and

In general, ']) /I C CP/I) p

X.

(A,'.p)

X = X(A,'p)

1

are defined.

Z(I) C R , but no polynomial

Cp/I)

nonnegative on

non-negative on

Cp /I) , and, secondly, any

negative on If

Then

can be non-negative on all of

We do have c

f G A/I

For example, if

negative on

f £ A

consists of the restrictions to

non-negative on all of inclusion.

CP /I)

a

We have now established that

+p)f = q, some

^-convex ideal.

Cp/I) , 'p / I ,

'p /I

n

•£ = {f G A|(f

is a radical

hand,

f = 1+p/l+q

s(i)-

finite refinement of the weak order on

I CA

can be written

A

'$ = 'p [g.]

is

'P C A, consisting of

p,q G'p, is not a geometric invariant of

on the set

X(A,'p).

For example, if

A = R[t] ,

^^s"

On the other hand, if

f

is strictly positive on

X(A,'p), then 8.5.3

195

guarantees that

f G'j5 .

contraction to

A

Also, recall that

of the order

'pcfl. C A

characterized as the ring of functions on all

f G A

with no zeros on

X.

C f n

.

Now, A c f 1 . can be invariantly

X = X(A,'p)

f

pG'P,

is perhaps a more natural geometric invariant of '£ by

£ ( X ) , then

$(X)cn-\

(A(X) c r i ., $ ( X ) c r i O

than the ring A

X

( )CM^ X.

is

Ag^.

A.

If

invariantly

Recall from Chapter V

Stengle's theorem has a nice interpretation

in terms of the structure sheaf associated to X

The ring

is the ring of global section of the structure

sheaf associated to (A(X), '^(X)).

nowhere negative on

X, then

g G A.

X c

characterized as the functions nowhere negative on that

obtained by inverting

is nowhere zero on

and by 2.2.4, 1+p = fg, some

we replace the order

is defined as the

This is immediate from 8.5.3, or directly

from the Nullstellensatz, since if 1 G H({f}, p)

'p C A

if and only if

(A/p).

Namely, f G A

is

f G ' P g ^ C Ag, f . . That is, f

be "positive" in the ring of sections over the basic open set

should

D(f) C X.

Stengle's theorem also generalizes Artin's result 8.4.6(bJ, in the case A = R[X X .. .Xn] , p = £ w . nowhere negative on Eh^, some

h^}.

Namely, it is trivial from 8.5.1 that if

n

R , then

$ ww)) . f G ($

f

is

2

($ w : f) = {h G R[X 1.. .XR] | h 2 f =

Let

It would perhaps be interesting to characterize the zeros

Z(pw : f) = Z(/H(Cpw: f) , ?J) C Rn.

By 2.2.3, if ZCPw : f) = 0 , then

and conversely.

By Stengle's theorem

Z('^ ' f) C Z(f).

s t include all zeros

x

Z($ w : f)

that

mu

of

of lowest degree in the Taylor expansion of

f f

f G Cp w ) g ,

It is also easy to see

such that the homogeneous term about

x

is not a sum of squares.

8.6. Derived Orders Let

A

be a reduced, real R-algebra of finite type, 'J5 = '-P.Jg-] w j

finite refinement of the weak order have identified

X

X = X(A,'^3).

some

n >_ 1, p,q ۥ]>}.

Also, if orders

196

I C A

In previous sections we

with the maximal convex ideals of

that the functions nowhere negative on

X

(A,'jJ) and established

coincide with

']3 = {f G A| (f n +p)f = q ,

In this section we study the derived order

is a radical

a

'^^

of 3,12.

p-convex ideal, we investigate briefly the

(1> / I ) d and, when defined, the orders

'^/I

and

C^d/I) •

All of

these orders are geometric invariants of the rings of functions X(A,p),

p,q G p

and

p

CP/I)^-

*4>,/1, when defined.

zation of

']3^ consists of those

not a zero divisor.

include, of course, the orders

on

Our study of

An equation

pf = q,

nowhere negative on where

f

with

pf = q,

'£,, ^ = ^^[g.]

(Hp.), = nCPO

Note that since

! d

we will also have described

U(-f) C x

f GA

will

Also, we will then have a characteri-

Q

f

A/I

X(A/I, 3 V I ) , respectively.

Recall that the derived order some

A,

by 3.12,

! d

'£, for a finite intersection of orders of type

p,q £'.p,

p

X = X(A,'|5).

is negative.

not a zero divisor, does not imply

The function

p

could vanish on the set

We need to distinguish between "degenerate"

and "non-degenerate" points of our semi-algebraic sets. In general, let

(B/p)

is degenerate if for some and

be any RHJ-algebra, f,h G B,

y e U(f).

h(y) = 0, all

h

f

points of

X,.

X

by

with

(We will write

X = X(B,'^).

We say

not a zero divisor, we have h(U(f)) = 0.)

definition of degenerate point depends only on particular order

X = X(B/P).

X

and

x G X x G U(f)

Note that the

B, not on the

We denote the set of non-degenerate

The subscript refers to "derived" or "dense" for

reasons which will appear shortly. Given any point as follows. U C U' I

=

For each open

implies U

I(U)

h i (U(f i )) = 0. hence

I

U(f) C X If I in

x £ X

we can associate an ideal of degeneracy U,

x £ U, let

f

I(U ) C l ( U ) .

Since

I(U) = {h G B|h(U) = 0}.

Now, if

U(f) C U{f.}

^,...,11,, with

is chosen by 8.1.1, then

Choosing an affine embedding

as a small open ball in

R n , intersected with

is an integral domain, it is clear that

For general

the minimal primes.

B,

x G Xd

if and only if

This is equivalent to

I

Recall that the

order

CP.

We claim that

I x

I A = ( 0 ) C B/P . J

x £ X,

h.^ e I(U(f)),

x

if and only if is Zariski dense

I x C U p . , where

P-

if and only if

J

From 8 . 2 , Z ( P . ) = X ( B / P , , tp/p ) J J J

x ^ U ^ ) ,

X.

C P . ( some

argument in commutative algebra. '$ C B.

Clearly

X C R n , we can interpret

= (0), which just says that any open neighborhood of X.

C B

is Noetherian, the ideal

is finitely generated, say by

= I(U(f)).

B

B

I

P. C B

are

j , by a well-known

are convex for any x e Z(P.)

and

J

and

X(B/P) = U X ( B / P . , p / P . ) i

197

The "if" part of the claim is easy, since if some a neighborhood

U(f) of

x G X, then

U(f) H Z ( P . ) , contradicting 3

on

I C p., x J U(f)

f G p., 3

since

f( x) > 0.

hf = 0 G B.

h G B/P.

Thus

x G Z(P.)-

vanished on

is non-zero, but vanishes

I = (0) C B / P . . x 3

Then any element

h £ P.

Conversely,

h G n ^

P. - p. i 3

Finally, if

suppose vanishes on

hj £ P.

vanished

on a neighborhood

U(f) n Z(P-), choose h 9 G O p. - p and consider 3 I ± j . i 3 We have h £ P., h(Z(P.,)) = 0 , i / h, hence h(U(f)) = 0,

h = h1h2. since

U(f) = U (Z(P.) n u ( f ) ) . x i

prove

the following.

X(B,1>) , =

Proposition 8.6.1.

Proposition 8.6.2, (0) = n p

with where

U X(B/P.,?/P.) . d P i minimal

Suppose

and suppose

B -•IIB.

The arguments in these last two paragraphs

p. C B

is a finite collection of primes

'^ C B i = B/P^ are orders.

is the natural inclusion, then

1

Proof: U X(B., p . ) . ^ l l

x G X(B, '^) , a

If

x G X(B., '^.) , then 3 3 d

but

x £ X(B., 15.), let

X(B,'.|>) , = U d Pj minimal

U(-f) .

if and only if

f G B,

f(x) < 0,

Thus

U m

3 d

I C P., some minimal x 3

and, again, any

The proof that

just like the last step of the proof of 8.6.1, with the the

3

If

P.. 3

I C P., x 3

f (mod P.) G -jj. C B., then

X(B., '^.) i i

x G X(B., f.) . J j

X(B., ']>.) .

X(B, %]>) =

I C P., just as in the proof of 8.6.1. ^ J

xGU(-f)CX(B,f)-X(B.J.)C 3 3 vanishes on

If '£ = B nl!']).,

The starting point is 8.2.9, which asserts that We know

D

hG

n P. -P. . ^ 1 3

x G X(B., '^.) 3 3d

is

X(B., 'p.) replacing

Z(P i ).

•

Proposition 8.6.1 seems more natural than 8.6.2, since it applies to any RHJ-algebra.

But 8.6.2 is better since it applies more directly to the

situation of 8.2.11, where we showed how any closed semi-algebraic set is of the form

X(B, •]>) .

We next show how non-degenerate points behave with respect to intersections of orders.

Proposition 8.6.3. i £ i £ k, and suppose

198

Suppose

P C B

(B, 'JK)

are RHJ-algebras,

is a '^-convex prime ideal.

']$ = n-j^,

Let ]3. ,

denote

those

p.

'p..)

Then

1

so that

P

is

X(B/P, $/P)

Proof:

a

'£., convex.

(Thus if

P = (0), we get all the

= U X(B/P, ?.,/P) . i1 -1 d

First note that by 2.7 we know some

'p.,

exist.

If

x G X(B/P, .p. ,/P) , but some f G B, f £ P, vanished on a neighborhood U 1 d of x in X(B/P, $/P), then f also vanishes on U nx(B/P, f.,/P). This contradiction shows

U X(B/P, *.p.,/P) C X(B/P, 'P/P) . 1 d d i1

Conversely, if

Let P

f. , £ P is not

vanish on a neighborhood of

'£. ,,-convex, where the

follows that f.,, 4 P

P

x,

x £ X(B/P, p.,/P) 1 d

vanish identically on

Since

'.p.,, are the other orders among the H(P, 'p.M) C B.

if.

all

U(h. ,) C\ X(B/P, 'p.,/P).

is properly contained in the hull

From 8.1.1, choose f

x G X(B/P, 'p/P) , assume d

'p., it

Thus let

Z(P) n X(B, '£.„) = X(B/P, p/P) n X(B, 'p.,,).

x G U(h) G n U(h i f )

vanishes on the neighborhood

U(h)

and let of

x

f = II f±1- II f i M £ P.

in

X(B/P, '.p/P)

Then

since

u(h) = u [u(h) n x(B, 'p.,) n z(P)] U U [u(h) n X(B, £ in ) n z(P)] . i1

x

Remark 8.6.4.

•

i"

The study of derived orders of the type we are interested

in has already been reduced to the case of integral domains in Proposition 3.12.1. Thus, if

B

is a ring,

P. C B

a finite collection of primes with

and we have either

(1) an order

'p. C B. = B/P.

$ = B nil p., then

i

l

l

'P, = B n n('p.) d x

and

f C B l

with all

P.

convex, or

'P, = B O IlCp/P.) a

i d

(0) = n P^ (2) orders

in case (1) and

in case (2).

d

The next result is the central result of this section, and gives the relation between non-degenerate points and derived orders in a crucial special case.

Proposition 8.6.5. domain over on

A,

R.

g i t 0.

Moreover,

Let

Let

']3 = 'Pw[g-]

Then

(A, £ d )

p , = { f G A| f (x) > 0 a

A

•

—

be a real, finitely generated integral be a finite refinement of the weak order

is an RHJ-algebra, with all

x G X(A, '£),}.

coincides with the following subsets of

a

X(A, 'Pd) = X(A, '£) d .

Finally, X(A, 'p), also a

X = X(A, 'P w ), (the irreducible real

algebraic variety associated to A ) : 199

(i) {x G X |exists total orders p D p

(ii) {x G X|for all

g e A

with

on

A

g(x) > 0,

with

!p[g]

x

'^-convex}

is an order on A}.

Before giving the proof, we discuss some applications. the Proposition gives us a class of RHJ-algebras not at all obvious that

Corollary 8.6.6. 'P = V [g-1 Vd

a

If

B

for which it is

is a closed semi-algebraic set.

is any reduced real R-algebra of finite type,

finite refinement of the weak order, X = X(B/p), then

= {f G B|f(x) _> 0

Proof:

X(A/pj)

(A/p^)

Let

all

x G X d >.

P. C B

be the minimal primes.

and by Remark 8 . 6 . 4 , ' p , = B n i I C P / P . ) . a 3d

But

By 8.6.1, X d =UX(B/P., 'P/P.) ,

' p / P . = 'P [Wg . | 1g . £1 P . ] , and by J J

8.6.5, f G CJ5/P.) if and only if f is non-negative on J d Thus f G *.p if and only if f is non-negative on X,.

Next, let with

S

S i = Z{f ik >

in n-variables. primes above

n

W{g ik > C R , as in 8.1, where

Then

= R

I(S.)«

i a ia We also have the order p.

P.

FrOm

S 2Al

-

S = X(A(S), 'p), where

among the

we have

>

the

occur.

are polynomials are the minimal

= '$ [g., ^ P. ] W

IK.

delusion

p = A(S) nil']). .

On each ring

A

P

on

XCt

A(S) -• n A i a

and

There may be repetition

P. , but a given prime occurs at most once among the

We can collect similar terms, and denote by

S = U S^ ,

ia

ICt

[ x i " 'XJ/Pia'

we have

•

f i k , g.fc

I(S) =Hi(s.) = f l O p . , where

1

ia

X(B/P., '-P/P.) . J J d

be any closed semi-algebraic set, and represent n

1

A

First note that

P. , fixed i.

the various primes which

= [R[X,...X ]/P , put the order

'P

= O'p. , the

intersection taken over those indices with A(S) -• IlA

and

'P = A(S) H n -p .

ot

P. = P . We still have inclusion ia a Denote by P o the minimal primes of A(S).

ot

Necessarily, they all occur among the

Moreover,

p

P . a

$o = H ' p . o , so from 3.12, CPJ |i

ip

From Remark 8.6.4, "p, = A(S) nn(']5o)d a P

= n C-P-o) • 1

p d

refinement of the weak order on the integral domain

negative on

X(A O , 'P1 • o) . P P d

iK

200

ik

Q]• p

p. o 1

is a finite

P

A o , namely, 'p. o = P

'P [g-vlg-i ^

P

w

Now

Each

d

P

8.6.5 describes

CP-Q)

ip ^

!p

as the functions in

Finally, from 8.6.2,

a

S, = X(A(S), 'P) , = d

A

p

nowhere

U X(A O , *pj . From 8.6.3 (with Pg minimal P P d U Pg minimal

P = (0)), we conclude

U X(A O , p. ) , and moreover, pd , C A(S) P ^ = Pg P XP d

of functions

S, a =

consists precisely

f nowhere negative on S,. We summarize these arguments in

the following.

Corollary 8.6.7.

Let S be a closed semi-algebraic set, 'p C A(S) a

specific order of the form considered in 8.2.11 with

S = X(A(S), 'p). Then

for any order p' with p C p ' C'P(S), we have that

(A(S), 'Pp

is an

RHJ-algebra and S d = X(A(S), p ^ ) .

Proof: 1

p ,

The argument above gives the result for p' = p. For any such

p d C ^ Cp(S)d-

on

But if f eip(S) d , trivially

f is nowhere negative

S d . Thus ? ( S ) d Ctp(S d ) =1> d .

We remark that if '£" C $

•

is a weaker order with

S = X(A(S), '£"), then

it is not clear when '£" = 'p(Sd ). In fact, we do not assert that if (A,'p) is any RHJ-algebra, then necessarily

(A,'p,)

is also an RHJ-algebra. Our

arguments definitely use properties of orders of specific types. Next let (A/p) be an RHJ-algebra with '$ = '^^[g^ of the weak order.

a finite refinement

Let I C A be a 'p-convex, radical ideal

X = X(A,'p).

Corollary 8.6.8 (a) I is '^j-convex if and only if each minimal prime I

is p d -convex if and only if P^^ = I(Z(Pi) n x d )

P. C A

over

if and only if

I = I(Z(I) H x d ) . (b) If I is £ d -convex, then X(A/I, *Pd/I) = Z(I) to

Z(I) ^ Xj

n

Xd -

Moreover, ^ d / I C A/I

of functions

X(A/I, Cp/I)d) = X(A/I, 'P/I) d-

functions

is an RHJ-algebra with

consists of the restrictions

f ^ A nowhere negative on X-..

(c) For any radical 'p-convex with

(A/I, $ d /I)

I C A,

(A/I, CP/I)d)

Moreover, CP/I) d

is an RHJ-algebra,

consists of the

f ^ A/I nowhere negative on X(A/I, $/I)j.

(d) Op ) = 'pd C A, hence for any 'p-convex radical

I C A,

CP/i)d = CPp/i)d = C^/i) p ) d . 201

Proof:

(a), (b), (c) are routine restatements of various results.

first statement in (d) holds, since by 8.5.1 and 8.6.5 CPd)

have

Tne

= '-Pd-

last

P

art o f

d

( ) holds because CP/I) d

finite refinement of the weak order, so p p /I C CP/I) , so we have

c

Q}/I) d C CP p /I) d

=

and we

always

is still a

Also, by 8.5.1,

(CP/I)p) •

•

c

(1*d/I)

A/I, when

I C A

'$£-convex. We state the following without proof.

(A/I, CPd/I) )

Proposition 8.6.9. (Z(I) H X,) .

Moreover, f £ (;£,/I)

is an RHJ-algebra, with

if and only if

f

I = P C A

too hard. 8.6.5.

a prime

*£d-convex ideal.

X(A/I, CPd/I) ) =

is non-negative on

Using techniques above, one can reduce to the case and

Pd>

£/I C A/I

(CP/I) ) •

An interesting order not covered by 8.6.8 is is

C:

£

The

A

an integral domain

The last two assertions are not

But the RHJ-property seems to be a fairly strenuous extension of

We will provide the necessary ingredient for this extension in 8.10.

We now return to the proof of Proposition 8.6.5.

We follow the notation

in the statement of that proposition.

Proof of 8.6.5: we must have

f

First, if

non-negative on

neighborhood of a point of Thus, points of

X(A, '$),

are exactly the points of

-p on

A.

h(U(f)) = 0.

If Then

x

p G £,

Otherwise, h

h ^ O , then

would vanish on a

X(A,'$),, contradicting the definition of are

$d-convex. x

h f £ 0

on

By 7.7.3, the '^-convex points

f ,h G A ,

X(A, $ ) .

h j- 0,

By 8.4.3, f

'£, that is, -f

We have now proved

X(A, *#)J-

is convex for some total order refinement

x £ X(A, ']5)d, choose 2

h 2 f = p,

say

X(A,*.J3) ,.

for which

in any total refinement of Pd -convex.

f G •£

e>

^d-

Since

f (x) > 0

and

is therefore negative -f(x) < 0,

x

is not

X(A,v£d) = X(A,'J))d> and have established the

characterization (i) of this set. Next, suppose find a signed place Here, K

f £#d. p:

Then

'£[-f]

is

K -• R, ± °° , with

is the fraction field of

A.

an order on

By 8.4.3, we

p(A) C R , p(g i ) > 0,

But then

which is convex for the total order refinement of

202

A.

p

gives a point '£ defined by

p(-f) > 0. x £ X(A,'p) p.

Thus

x

is

'^-convex and

f(x) < 0.

This proves that

x e x ( A / P ) d } , since

X(A/P) d = X(A,$ d ).

Third, if

and

x ^ X(A,'£)d

orders for which we see that whenever

x

£[g]

'^[^1

is convex.

A

In particular, U{g^,g}

g

is positive in all total

Since some such order exists, refining •£, A.

is an order

x £ X(A,'|5)d, we can find

an order.

g(x) > 0, then

is an order on

c

p d = {f G A|f(x) _> 0, all

Conversely, note that by Remark 8.4.7,

U{f i> C X(A,'.pw)

h,g e A,

h ? 0,

is Zariski dense.

g(x) > 0, and

is not Zariski dense in

X(A/Pw)

This establishes characterization (ii) of

Finally, we come to the hard part of the theorem. P C A with

is a

open, hence

of

X(A,'^)

is closed.

is not

X(A,']3)... We must show that if x G Z(P) n X(A,'£)d, X(A,'£)

is obviously

X(A,'^) ,, which will imply by Tarski-Seidenberg that

contains functions in

A

(This is not clear yet since the definition as quantified variables.

finitely many distinct ideals of degeneracy we could conclude that

I

C A

If we knew only x G X(A,1^),

occur, for

X(A/.p) , is semi-algebraic without the new characterization.)

But even granting all this, we would still only know that semi-algebraic.

*])[g]

In a later section we will give another

is a semi-algebraic subset.

X(A,p),

hence

Note that the set of degenerate points of

characterization of X(A,'.J3)j

h(X(A/P) n U(g)) = 0.

'^-convex prime ideal, g £ P, then there exists

g(x) ? 0.

Now if

X(A,'p)d

was closed,

If unproved Proposition 8.1.2 were available, then we would

have a proof that

is an

(A>Pd)

RHJ-algebra by the constructions in 8.2.

However, we must make do without 8.1.2 and give a direct proof of this part of 8.6.5. Assuming '£, with a domain

P

P C A

is

']5 convex, we can find a total order

still convex, by 7.7.3. B, integral over

By adjoining

/gT

to

A, with a total order extending

'^' on

A, refining

A, we can construct '£' on

A.

(We

use the method described in the proof of 8.2.10 and adjoin no more of the than we need to insure that the weak order on

B

contracts to

By the going up theorem 6.4.2, we can find a convex prime order on

B, lying over

any total order on Let that on

K B.

B

P C A.

The reason for passing to

will refine

'P [g.]

be the fraction field of Write

on

'-P = '-P [g«]

Q C B B

/gT on A.)

for this total

is because now

A.

B, with the total order induced by

B = R[x,...x ] , and suppose

x. = x. (mod Q ) , 1 £ i <_ s,

203

gives a transcendence base for of

K

and let

field of

E

B/Q

B/Q

over

be the real closure of R(x.....xn), with

is

x

R.

Let

RCx^.-x ) , ...,x

transcendental extension

R(x ...x ) . Consider

F C r, with convex ideal

Q C B

was an order on

A.

K C r be a real closure in

T. The fraction

algebraic over the pure B = R[x,...x ] C E(x +-,---x ) =

for our total order '£'. (Originally, '])'

We extended it to

B, K

and

T, and we continue to call

the extension '£'.) By the signed place existence theorem 7.7.4, we can find a signed place 1

p : F -• A, ± °° , inducing the order 'p1, finite on and with p1

A

Archimedean over

R(x1-..x ) . Since

is also trivial on the algebraic extension

p'(E). over

Since E.

A

is Archimedean over

Since the

polynomials

f.

It follows that 1 _< i £ n.

x +.

f

p (x s+ .) G E.

R [ x r . . x n ] , then

E

with

Rfx^-.x ]

p (x. - £.) = 0

f.(xs+.) G Q.

and with

p'(xi) = ^ £ E, all

and also, of course, if

Q = (h.)

1

Let

F = E(x +-,---x )

r = tr.deg c(F).

J

over

(Thus

E

is the fraction

r+s = tr.deg (B).)

c

By

K

the signed place perturbation theorem 8.4.9, there exists a discrete rank signed place over

E, p:

p(x i - Z,/) = 0, and Let

Q1 C Q

rings

F -* E, ± «>

p(h.) = 0.

C ... C Q

(inducing a new order on

Certainly, p

is finite on

E

I X S +1* ' < x n J

in

are all convex for the total order defined by Q 1 C Q 2 C ... C Q r .

Also, Q C Q r

function field over

E

with

since

F

'

assoc

p.

Let

p(h^) = 0.

tr.deg^Fp = r - i .

iated

204

F ) , with

E[x s + 1 -..x n ].

tr.degR(B/Q^) = s+r-i.

t0

P-

The

Qj. C *i

Q i = Q- n B, so that Each

B±/q± = F±

is a

This follows by repeated

application of [68 , vol. II, Chapter VI, §14, Theorem 31]. below we deduce that

r

denote the chain of maximal ideals in the valuation

B ^ B ^ - ' O ^ D

degrees.)

c

p'(h.) = 0.

E[x. -£.,h.]. 1

RCx^.-x ) ,

R(x.....x ) , j > 0, there are

We can then write

Certainly, the function field field of

is trivial on

R(x1-..x ) , it is also Archimedean

f

Thus we have

p

Q C B

E, and we identify

are algebraic over

with coefficients in

B with center 1

From the diagram

(The integers indicate transcendence

R[x r ..x n ]/ Q i

R[x r ..x s ]

In particular, tr.degR(B/Q r) = tr.degR(B/Q) = s, and since Q = Qr.

Note this implies that if g £ Q,

We now restrict the signed place K = R(x 1 ...x n ) r

p(g) ± 0.

p: F -* E, ± °° to K C F . (Recall

is the fraction field of B.) We then have a discrete rank

signed place

p: K -* E, ± °° over the ground field

B = R[x 1 -..x n ], with

kernel(p|B) = Q C B, hence with

Again, the image field tr.degR(L) = s. rank

g G B, then

Q C Q^, we have

If g G B,

s signed place

We then compose

L = p(K) C E

p

R C K, finite on tr.degR(p(B)) = s.

is a function field over

R, with

g £ Q, then by 8.4.3 we can find a discrete

q: L -* R, ± °° , finite on B/Q C L, with

and q, K -* L, ± °° -• R, ± °° , and change

q(g) ^ 0 € R.

p

elements as in 7.7.2 to get a new order on K and a discrete rank q op:

place

on infinite r+s

signed

K -* R, ± °° .

We have thus constructed a strictly increasing chain of convex prime ideals

Q x C Q 2 C • • • C Q r = Q C Q r + 1 C • .. C Q ^ C B, with

g £ Q

. We now contract to our original ring

The (new) total order on B of

B. The ideals

increasing chain A/P

+

= R.

P^ = Q. H A

A C B of Proposition 8.6.5.

'£ = *Pw[gi]

are then all

on A by construction

'pd=convex and form a strictly

P 1 C P 2 C ... C P r = P C P r + 1 C .•• C P r + g C A, with

If g € A , g £ P, then

g £ Q + , hence with

still refines

B/Q r + s = R,

g £ P + . Thus, P +

g £ B,

g £ Q, and we may assume

corresponds to a point

x £ Z(P) n X(A,^^

g(x) i 0, completing the proof of 8.6.5.

D

For later reference we state a corollary of this proof of 8.6,5.

205

Proposition 8.6.10.

Let

A

be a real finite integral domain, '$ = '$ [g.] C A

a finite refinement of the weak order, P C A r = tr.degR(A) - tr.deg R (A/P), and let

K

a '£,-convex prime ideal.

Let

denote the fraction field of

A.

Then there is a sequence of discrete, rank 1 signed places P

P

l

P

2

I* ~ inducing a total order on where

p

order refinement of

r>

*• K

is the composition

chain of prime ideals

r

refining p:

(0) C p

-

»

'£, all finite on

K ~* K , ± °° .

C • • • C p

A, with

P = kernel(p| ) ,

In particular, we obtain a

= P C A , all convex for a total

$.

D

8.7. A Preliminary Inverse Function Theorem In order to make sense of differential topology over an arbitrary real closed field, it is imperative to investigate purely algebraic versions of the inverse function theorem.

The result proved in this section is a rather

special case of a better algebraic inverse function theorem, but is strong enough to provide a good picture of a real algebraic variety near an algebraic simple point.

This application will be given in the next section.

This, in

turn, will be used to stratify arbitrary closed, semi-algebraic sets and prove that any such is the maximal convex ideal spectrum of an RHJ-algebra.

Also

crucial for this discussion will be the work on derived orders in the previous section.

(We refer again to 8.1 for the distinction between closed, semi-

algebraic sets and closed semi-algebraic sets.) Before stating the main theorem of this section, we digress a bit in order to put in perspective a consequence of the Tarski-Seidenberg theorem which we seem to require at this point. is a polynomial and n

The result is that if n

B = {(x....x^ € R |a. £ x^ <_ b^}

is a closed bounded

(or a closed ball), then the set of values {t e R|t= f (x),

rectangle in

R

some

is a semi-algebraic subset of the line

x £ B}

fCx^.-x^

R.

This is trivially a consequence of the Tarski-Seidenberg theorem (see the Appendix), but it does not seem any easier to prove directly than the

206

Tarski-Seidenberg theorem itself. Let F in

F = {(x 1...xn,y) e R is the graph of Rn+ .

f

We can reformulate the result as follows.

x R = R

over

The values of

via the map

n

f

n+1

1 (xr . .xj £ B , y =

f(x 1 ...x n )}.

Thus,

B, and is obviously a closed semi-algebraic set on

B

are obtained by projecting

F

onto

R,

R n x R -> R.

y:

As discussed in 8.1, the Tarski-Seidenberg theorem easily can be used to prove certain sets are semi-algebraic, but it does not give much information on whether sets are closed or open, without further work.

On the other hand,

our going-up theorem for semi-integral extensions provides a nice tool for concluding that certain sets are closed. 8.2.7, where we showed that if

A C B

This was formalized in Proposition

is a semi-integral extension of finitely p C B, then image (X(B,'p)) =

generated R-algebras relative to an RHJ-order X(A,£ H A ) , under the projection is always closed. image(X(B,'P))

If

X(B,'£)

X(B,$W) -+X(A/£ W ).

Of course, X(A/P H A )

is semi-algebraic, Tarski-Seidenberg implies

is also semi-algebraic.

As an example, we can deduce that a polynomial

f(x,...x )

maximum value on any bounded closed semi-algebraic set the graph of

f

over

S, say

F C R

closed semi-algebraic set and the Moreover, (A(F),$(F)) We project

F

n

x R = R

n+

RHJ-algebra

is semi-integral over

.

Then

n

S C R . F

(A(F)/£(F))

assumes a Consider

is a bounded has

R, hence also over

X(A(F),^(F)) = R[y] C A ( F ) .

onto the y-axis and get a bounded closed, (comma) semi-algebraic

subset of the line.

But in dimension one, the distinction between closed,

semi-algebraic and closed semi-algebraic obviously is unnecessary, both notions simply corresponding to finitely many closed intervals (including single points and closed rays).

Thus, bounded, closed, semi-algebraic

implies a maximum element in dimension one. Now, as another way of applying the Taski-Seidenberg theorem, one could draw the same conclusion about maximum values directly from the fact that it is true in the case of the real numbers.

But this is a "transcendental proof",

whereas we have just given a "purely algebraic proof". We now state an inverse function theorem.

Proposition 8.7.1.

Suppose

Y ...Y

£R[X,...X ] .

We regard the

207

Rn

as functions on R

n

n

-• R .

Assume

and we regard the n-tuple

Y(0) = 0

B(0,e) C R n

ball

(b) Given

as a ma

e > 0

of radius

Then:

such that the map

Y

restricted to the closed

e at the origin is injective.

e > 0, there exists

6 > 0

B(0,<5) Cy(B(0,e)).

such that

(c) From (a), (b), near the origin the coordinate functions functions of the

Y..

—

i

X^

are

These functions are in fact algebraic functions in

the sense that for suitable polynomials 1

P

and assume that the derivative matrix

is non-singular. (a) There exists

Y = (Y1--«Yn)

f^, we have

f i ( Y 1 ..« Y »X^)

E

°>

n

— (d) Let

A(x) = ((3Y./3X.)(x)), the derivative of

Y

Then sufficiently near the origin the inverse function tiable with derivative

A

, in the sense that if

Proof:

x e Rn.

X = X(y)

is differen-

y = Yfx ). oo

X(y) - X ( y 0 ) - A " 1 ( x 0 ) ( y - y 0 ) H X ~ X0 H

lim

y -• V

at

=

The point to be made at the outset is that once we have the maximum

value property of polynomials on closed bounded semi-algebraic sets (which we have just established purely algebraically) one can write out word for word one o f the standard proofs of the inverse function theorem for real numbers entirely in elementary algebraic terms. Part ( a ) , the local injectivity of Y necessary to find

e > 0 and a constant

is very easy since it is simply c > 0

such that for all x,x f £ B(O,e),

we have

Hx-x'll <_ c||Y(x) -Y(x')|| .

By a linear change of coordinates, we m a y assume that is the identity matrix, that i s , Y

208

i

= X

i

+

Cterms

of d e

g r e e 1 2)«

((3Y./3X-)(0)) = A(0)

Now using the estimates of 7.2 we can, in f a c t , find x,x

!

e BCO,^)

e-^ > 0

so t h a t for

we have

\ llY(x) - Y(x')ll <_ l l x - x f l l <_ 2||Y(x) -

(Note that the first inequality will guarantee that the inverse function X = X(y)

is continuous near

0.)

We next prove (b), the local surjectivity of llxll < e < e, , then

if that

A(x)

P(x)=

llxll = e 2 , then the function

X

e o

2

If 2

= liY(x) - yll

£~ < £i

so small

Ilyll < •—• , consider

on

B(0,e o ).

If

Z

P(x) >_ (e 2 /4) 2 > Ilyll2 = P(0). Thus

llY(x)ll > -| , hence P(x)

If

2 (Y. (x) - y.) x 1 i=l

Directly from (*),

Let us now assume x G B(0,£ 2 ).

is non-singular for all

the function

B(0,e 2 )-

-|- < ||Y(x)ll.

Y.

does not assume a minimum value on the boundary of B(0,e2)

is an

interior point at which

P(x Q )

is

minimum, then

0 = (3P/3Xj)(xo) = E2(Y i (x 0 ) - y ^ ( O Y . / S X ^ (xQ))

for all

1 < j
singular, we deduce Since R(Y 1 -..Y n )

That is, A ( X O ) ( Y ( X Q ) - y Q ) = 0.

Since

A(x Q )

is non-

Y(x ) = y , as desired.

A(0) = ((8Y./3X.)(0))

is nonsingular, it is easy to see that

has transcendence degree

is an algebraic extension.

n

over

R.

Thus

R(Y1...Yji) C R(X][...X )

This proves (c).

Finally, to prove (d), apply the linear isomorphism

A(x )

to the

limit statement of (d), transforming (d) to

lim y

Ao(X(y) -X(yo)) - (y-y o l |fy - yo|| y

o

which does hold by continuity of

=

±

v -> v y y o X(y)

A 0 (x-x 0 ) - (Y(x) - Y ( x 0 ) ) f x - x o | = || x - x o || I y-yo || "

and differentiability of

Y(x).

We discuss further the differentiability of the inverse function

X ( y y lim — — £-0

+£,...y) - X.(y —

D

X = X(y).

Once we have (d), it is easy to see that the partial derivatives

(3X./3Y.)(y ) = J X °

'

y ) —

209

exist and, in fact, that the matrix yQ = Y(x Q ).

where

((3X./3Y.)(yQ))

Thus, Cramer's rule expresses the

coincides with (8X./3Y.)(y )

det(A(x ) ) .

Continuing this discussion, it can be shown that the infinitely differentiable functions of x Q = X(y Q )

computable in terms of

X. - x . , where

x

y.

X. - x.

= (y,...y ) .

Y^-y^

(il) ,..*..(y

Alternatively, the total differential

dY

as

The constant

((3Y i /8X.)(X Q ))(X-x Q ).

as power series in the

CY1-y1)V...C Y n- yn) « will then be

dY:

The higher derivatives are formally

= (x.....x ) , y

i

function

are actually

Y. - y. = Y.(X,...X )-Y.(x.....x )

terms vanish and the linear terms will be solve for the

X.(y)

as follows.

First, rewrite the polynomials polynomials in

as

(8Y i /8X.)(x Q ),

specific rational algebraic functions of the polynomials with denominator

A(x ) ~ ,

Then formally

The coefficient of

, O^/BY 1 ) (yp) , I = Ci r ..!„).

can be regarded as a polynomial

R n x R n -*• R n x R n , linear on the second factor at each point of

the first factor, and, moreover, dY has non-singular differential at (0,0). 2 The second derivatives (3 Y./9X.8X,) then occur as part of the first derivative i 3 K of

dY.

Applying the general discussion of first derivatives to

the inverse function of of the

Y

dY

shows that

is twice differentable, and leads to a computation

(3 2 X./8Y-3Y V ). 1

J

K

Here is perhaps a more algebraic approach to the derivatives The derivation to a derivation

Z/dY±: D±:

R[Y1...Yji] ->R(Y 1 ...Y n ) CR(X 1 ...X n )

R [ X r . .XR] -» R(Xj .. .X n) .

a o (Y)X m + a 1 (Y)X m " 1 + ... +a m (Y) = 0

(3X./3Y.).

extends uniquely

Specifically, if

is the minimal polynomial for

f.(Y,X.) = X.

over

R[Y,...Y ] , then we must have 0 = D i (f j (Y,X j ))

= ( O a o / 8 Y . ) X m + . . . +(3a m /3Y i )) + ( O f ./3X.) (Y,X.))Di(Xj) ,

and this last equation can be solved for

D.(X-).

Using the inverse function theorem 8.7.1, we can easily establish the following implicit function theorems.

Proposition 8.7.2.

210

Suppose

Yj ,... ,Y R e R[XX .. . X j ,

k < n,

Y^O) = 0 ,

and suppose the vectors

d Y ^ O ) = ( ( B Y ^ B x p (0),.. ., ( S Y ^ B X ^ (0)) ,

are linearly independent.

Reordering the variables if necessary, assume that

{dY.(O), 1 < i < k; dX.(O), k < j < n}

i

—

—

3

0 G R k , the equations

Then

F

Define

Y^x) = c ^

is locally

F: 1-1

span

—

uniquely define the coordinates

Proof:

1 < i < k,

1 < i < k,

x-^.-.x^

Rn

Rn.

by

Then, sufficiently near

c± G

x-(x 1 ...x k ,x k + 1 ...x n )

as functions of

x

ic+]/--xn

F(x) = ^(x),...,Y k (x),X R + 1 (x),...,X n (x)).

and onto by 8.7.1.

The result then follows easily.

n-k

Surfaces

Remark.

i i i k-

Surfaces

Y..(x) = c±,

D

Proposition 8.7.1 and the discussion of derivatives above also

gives the tangent plane of the surfaces (8Yi/8X.)(O) = 0 ,

Y i (x) = c^.

j > k, then the tangent plane to

at the origin is the coordinate plane general, the tangent plane of

x =

(^

Y^x) = 0

For example, if Y i (x) = 0 ,

••• = x ^ = 0 } = R

1 <_ i <_ k, x {0}.

In

at the origin is the annihilator

with respect to the usual scalar product on

Rn

of the k-vectors

(dY^)(0),

1 <_ i £ k, dY± = (3Yi/8X1,.. • ^ / a ^ ) .

Proposition 8.7.3.

Suppose

Yl.. .YR G R[X X . . . X j , k > n,

and suppose the vectors

dY.(O),

1 £ i £ k

variables if necessary, assume that

span

R n.

like the standard inclusion, mapping ||x||

Reordering the

dY^^ (0),. .. ,dY (0)

after smooth algebraic change of coordinates the map to

Y ^ O ) = (0) ,

Y:

span

Rn .

R n ~> R

Then looks

(x 1 ...x n , 0...0), for

sufficiently small.

211

Proof: (0 ... 0,x

Define

G ( x r . .x n ,x n + 1 .. .xk) = (Yj (x1.. .xn) ,... ,\(*v

... x v ) , so that K

n+1

G:

R k -• R k

+

• -\))

is locally 1-1 and onto near

0

by 8.7.1.

D

8.8.

Algebraic Simple Points, Dimension, Codimension, and Rank We continue our study of the polynomial ring

real closed field

R.

n-vector of functions. of

P

to be

oodimension of df. G A .

If

f e R[X], we have P C R[X]

P

A = R[X]/P.

df = (9f/9X 1 ,...,3f/3Xn ), an

Thus

Suppose

dim(P)

It is routine to check that P, using the formula

is an invariant of the

P = (f 1 ...f s ).

to be the maximum number of

of generators of

space of

R

codim(P)

is independent of the choice

d(hf) = f dh + h d f = hdf G A n

derivations

D:

A -* K, where

K

P.

Such a

D

is determined by

precisely the linear conditions words, we have

codim(P)

codim(P) < codim(Q).

212

K

of the

is the fraction field of D:

f € p.

R[X] -* K

A.

which vanish

DX = (DXX,...,DXn) G K n , subject to

0 = Df = df-DX, if

linear conditions on

We will use the fact that if

if

dim(P) + codim(P) = n.

can also be characterized as the dimension over

This coincides with the space of R-derivations on

Define the

A-linearly independent vectors

A standard result in commutative algebra is that In fact, dim(P)

over a

is a prime ideal, define the dimension

dim(P) = tr.degR(R[X]/P).

finite integral domain

n

If

R[X] = RfX^.-X^]

P C Q, then

f = f(X) G P.

DX, so

In other

n = dim(P) +codim(P).

dim(Q) < dim(P), hence

It is also clear that (Sf^/SX.), where r x r

at

x

P = (f ...f ) .

is the rank over

Let

submatrix of maximal rank over Z(P) C R n

Let

codim(P)

to be

denote the zeros of

V E R[X]

be the determinant of some

A.

V £ P, that is, V £ 0 G A.

P.

Thus If

x E Z(P), define the rank of

K ok .

Suppose

g i ^ 0 E A,

A = R[X]/P,

and

X

affine embedding of

Then there exist zeros

with

rank (P) = codim(P) x

x E Z(P)

with D

X(A,*p ) . This definition refers to a particular

X(A,$ ) . However, from commutative algebra, algebraic

local ring associated to the point

x

and let

is a simple point if and only if

If we refine the order on

A, say to

m

C A

figures below illustrate what might occur.

Let

A

be the

be the maximal ideal.

dimD(m /m ) = dim (A) = tr.deg D (A). K X X K $ = '£ [g^] , we are not especially

interested in all the algebraic simple points belonging to

'£

is an order on

will be called

simple points can be invariantly characterized as follows.

=

A oA •

In particular, rank x(P) = codim(P).

x € X(A,'l> ) w

algebraic simple points of

x

—

P C R [ X ] is a prime and '^ ='^w[gi]

1 < i < k.

V(x) ^ 0.

The points

Then

P

From the Nullstellensatz 8.4.3 we obtain the following.

Proposition 8.8.1.

g.(x) > 0

of the matrix

rank (P) = rankD (-5—" (x)). Obviously, rank (P) < rankA(—^-) = X

codim(P).

A

In both cases

X(A,'£).

The

X(A,'-p), where

'P [g] > contains degenerate points which are algebraic simple points of g >. 0

Z(P)

Figure (a)

Figure (b)

X(A,$ ) . We will define the algebraic simple points of

X(A,p)

to be the

non-degenerate algebraic simple points, that is, simple points of belonging to

X(A,'£,).

X(A,£ )

Propositions 8.8.1 and 8.6.5 show that these simple

213

points are, in fact, dense in X(A,£ d ). and

g.(x) > 0, then

'£[g-]

Namely, if x G X(A,£ d) = X(A,£) d

is an order on A and 8.8.1 guarantees simple

points exist in the neighborhood

x G U{g-} C X(A,'£),.

Now, however, we should check that an algebraic simple point of X(A,'p ) is automatically non-degenerate.

This requires a somewhat careful argument.

In fact, we will base the proof on the inverse function theorem.

The following

is the central result of this section.

x G R n , with

Let g,f ,... ,f r G R[X] ,

Proposition 8.8.2.

f^x) = 0

df i (x) linearly independent, g(x) > 0, and assume the neighborhood x

is sufficiently small.

In fact, P Moreover,

Then the ideal

I(Z{fi>

n

U(g)) = P

is the unique minimal prime of the ideal P

is convex for any order

is prime.

(^...f ) , with

£ = p [g.] C R[X], with

at

U(g) of x

vanishing of the

f., 1 < i < r. l — —

A = R[X]/P

in the variety

Z{f^}

g. are positive in the order

for

•£ [g-]. This proves the last part of the proposition, w 3

If A

Proof:

A = R[X1...X ]/P.

that

P

df.(x)

zeros.

f^ near

'•P

'P [g-]

214

x and P

a

If A

R and

f 1 ...f r G P,

Rank considerations show

The zeros of P near

Thus every neighborhood of x

Corollary 8.8.4. =

By assumption, there exists

linearly independent.

we are in the situation of 8.8.2.

is convex

x G x^.

is necessarily a minimal prime of (^...f ) , and since

the zeros of the

P

is a real finite integral domain over

is a simple point, then

r = codim(P), with

U(g) C u { g . } . l

$(Z{f.} fiu(g)) C A, hence

x G X = X(A,'J)w)

Write

is a ring of

defined by the

If g-(x) > 0, then choose 3

The

Corollary 8.8.3.

In practice,

can be thought of as small balls centered

x. The first part of the proposition says that

functions on any neighborhood of x

x G Z(P).

g.(x) > 0.

Before indicating the proof of 8.8.2, we give applications. the "small" neighborhoods

U(g) of

x G Z(P),

x thus coincide with

is exactly the ideal vanishing on these is Zariski dense in X, as desired. •

is a real finite integral domain over R,

finite refinement of the weak order on A, g. ^ 0, then the

non-degenerate set simple points

Proof:

X(A,'£),

x £ X(A,£ )

is exactly the closure of the set of algebraic with

g i (x) > 0.

We know from 8.8.1 and 8.6.5 that any neighborhood of a non-

degenerate point contains such simple points.

Conversely, 8.8.2 guarantees that

a function which vanishes on a neighborhood of such a simple point is already 0

in

A.

•

Corollary 8.8.5. X(A,£) d

With the same assumptions as in 8.8.4, the set

is a closed, semi-algebraic set.

Proof: —————

If

A = R[X]/P,

P = (f.), then the simple points of I

are the points in the open, semi-algebraic set

X(A,$ ) w

{x|rank((8f./8X.) (x)) =codim P}.

(In fact, by looking at the possible subdeterminants, this is even an open semi-algebraic set.)

U{g.}, then take closure.

We intersect with

Tarski-

Seidenberg guarantees that the closure of a semi-algebraic set is semialgebraic.

•

Proof of 8.8.2. result„

Denote by

First, part of the proposition is a purely algebraic

Ax

the local ring obtained by localizing

and dividing by the ideal generated by the

fi#

R[X]

The assumption

at

x

df^Cx)

independent implies that the graded ring associated to the maximal ideal m

C A

is a polynomial ring, in particular, a domain.

a domain, or equivalently, (f1...f ) i r this says precisely that of

(f^.,^)

in

x

Therefore,

is a prime ideal in

R[X] . x

A

In turn,

is a zero of precisely one minimal prime

R[X], Geometrically, Z(P)

a Zariski open neighborhood of

x £ Rn .

and

Z(f io ..f r )

is

P

coincide in

For details of this argument, see

[63, Chapter 11], or texts on algebraic geometry. Now we must look more closely at small semi-algebraic neighborhoods of We must show that if

h

vanishes on

Throughout the argument we may assume assume of

P

df,(y)...df (y) and

Z{fi> H u ( g ) , U(g)

g(x) > 0, then

as small as desired.

h e P.

Thus, we

are independent, y £ U(g), and we assume the zeros

(^...f^) coincide in U(g) o

Suppose otherwise,

that is, P C l(Z{fi>

n

U(g)) = H p

where the

215

P.

x.

are convex prime ideals, strictly containing

P.

say

and by 8.8.1 we can find

x G Z(P ) . Then

y G Z(P Q ) n u ( g ) independent.

U

Z(P j ), such that

Since

we know that near Thus, f

codim(PQ) > codim(P)

G P

Z{£.} n u ( g ) = U (Z(P.) nu(g)) i j J y

the zeros of

df,(y),...,df (y). 1

f

o o

G

P

° *

=

(f,...f )

I

z

f

^ ^ i^

nu

f

y £ Z(P.), J

and

j t 0,

y, yet

df (y) 1

(Alternatively, if we choose by

f fJ

without changing

( g ) ) » which certainly vanishes on

€= n

f

yet still has differential independent of

p.,

j^o J

df Q (y).

Z{f i )

P .

is

° f

G P ,

coincide with the zeros of

Z{f.}, near

j,

are linearly

r

f Q (y) = 1» then we can replace f

x G Z(P-)> some

df Q (y) ,df: (y),. .. ,dfp(y)

vanishes identically on

independent of

We know

near

Now, y,

From this,

df,(y),...,df (y).)

we will derive our contradiction, by appeal to the implicit function theorem 8.7.2. By translation and linear change of coordinates, we may assume

and

f i = x i + (terms of degree •> 2) G R[X],

theorem states that near graph of a map y

i

= y

0

R n " r •* R r

i^ x r+l* ''Xn-^

are

the surface

f^ = ••• = f

The implicit function =0

• • -\h

where

algebraic functions of the last

n-r

variables.

Moreover, the tangent plane of the surface is the plane equivalently, (8yi/8x.) (0), function with have

df Q (0)

l < i < r ,

independent of

(8f /dx.)(0) £ 0, some o j

shows that

fQ

fQ

x, = ... = x

r+1 <_ j <_ n.

If

df i (0) = dx i (0),

j > r+l, say —

calculation of derivatives of t,0...0)

is given by the

R n = R r x R n ~ r , { y r . 'YT>\+1

in

smootn

1 < i < r.

y = 0 G R

j = r+1.

along the curve

= 0, or

fQ G R [ X ]

is a

1 < i < r, we must

But then a simple (y^(t,0...0),...,yr(t,0...0),

cannot vanish identically on the surface

• The last part of Proposition 8.8.2 states roughly that a prime convex if it has enough zeros. prime, •£ = 'p [g.] C R[X] P

is

'^-convex if

P

Precisely, one can say that if

has algebraic simple zeros

x G R

with

is

P C R[X]

a finite refinement of the weak order, n

P

is

g. £ P, then g i (x) > 0.

We will now establish another criterion for convexity with this same flavor. The result can be found in [21].

It could be derived from the implicit function

theorem and existence of simple zeros, but we will follow the proof of which avoids the implicit function theorem.

216

[21],

Proposition 8.8.6. domain over

R,

Let

U(g) C R

n

A = R[x,...x ] = R[X,...X ]/P a non-empty open set,

be a transcendence base over

R

R[x 1...x d] -»A.

defined by '£ [g] C A

and let

Then

P

is convex for

Proof:

First assume

y6R ,

z£z

of

P.

1

and

are

R[x 1 -..x d ] -•A -•A/P i ,

P

is

Choose

A.

Let

R[xr..xd].

d+1 < j < n . such that 2m=1

$ [g] C A

11

f(T) = I !

is

Suppose not, that is,

f = Hf i # f = 0.

is an order, let

Then

in the kernel

Then

f

vanishes on

This contradiction

t G A

R(x 1 ...x d ) C K, where

a. (x, .. .x-JT

We can write

1

a m ( y r . . y d ) + 0,

be a primitive K

is the fraction

be the minimal polynomial for

t

1

Xj

= 2?;J ( b ^ (xy . . x ^ / c ^ ( x r . .x d ))t G K, y = Cy1--->rn)

By the Nullstellensatz we can find c ij (y 1 ...y d ) + 0, and

i-ai(y1.. . y ^ t 1 " 1 t 0.

Thus

t(y 1...y d ) G R

G

Z(P) H U(g)

3f / 3t (yx . . . y d )

=

is a simple root of the

f (yx,... ,yd,T) 6 R [ T ] ,

polynomial Suppose

(z-^... zd )

will have a root f(z-...z,,T) id

x

is very near

near

T = t(y,...y ) . in

•

Tnen

f

(z^ • • • z d > T )

If

d+1 < j < n, define — — z = (z.....z ) £ R .

z G U(g) C R .

We are finished

z G Z ( P ) C R n.

the kernel of

f(xr..xd,T).

R

by the rational formula above and consider

Watching our estimates carefully, we may assume if we prove

in

(y^'-'y^)

tCy^.-y ) , by a simple estimate argument showing

changes sign near

z. = x. (z, .. .z ,,T)

Now,

Now, P

£ i GR[ X;L ...x d ]

f i i 0, and let

element for the field extension

over

contains a disc

'£w[g] -convex after all.

Conversely, if

field of

(equivalently,

W

TT(Z(P) flU(g)), which contains a disc, hence shows

CR[X]

'B[g]-convex primes.

1

tr.degR(R[x]/Pi) < d.

^[g]

contains a disc.

P = I(Z(P) n u ( g ) ) .

P CI(Z(P) nU(g)) = H P . , where the

x-^.-.x^ E A

e > 0 G R.

TT(Z(P) HU(g))

\[g]-convex if and only if

Let

be the projection

TT(Z(P) nij(g)) d

B(y,e) = {z G R | llz-yll <_ e}, some

P CPi

TT: Z(P) -• R

is an order), if and only if d

g £ P.

be a finite integral

Define

f(z 1 ...zd,x) = 0

R[x,...xd,T] -* R[x,...xd,t] = B C A p:

implies

R[x1...x d,T] "• R p

factors through

by

p C x ^ = z±, p:

a little further care with estimates will insure

B -* R.

is precisely p(T) = x.

Since

Then

c^. (y^. . .yd) f- 0,

c..(z,...Zj) j- 0

and thus

217

p

extends to

p:

A "• R,

p(x 1 ...x n ) = (z1 ...z ji ).

Thus

z^Z(P)

as

desired.

D

8.9. Stratification of Semi-Algebraic Sets In this section we give procedures for stratifying semi-algebraic sets, that is to say, for breaking arbitrary semi-algebraic sets up into "simple" pieces.

We should emphasize that we deal only with the most elementary

aspects of stratification questions.

For example, we do not consider the

Whitney-Thorn regularity conditions, nor questions of equisingularity. Actually, our primary goal in this section is to show that if

F C Rn

any closed, semi-algebraic set, I(F) = {f e R [ X r . . X j | f ( x ) = 0, all A(F) = R[X r ..X n ]/I(F), fl(F) = (g e A(F)|g(x) >_ 0, all (A(F),!P(F))

is x e F},

x e F } , then

is an RHJ-algebra (with, of course, X(A(F) ,'£(F)) = F.)

Strati-

fication considerations arise naturally in the proof of this result. E C Rn

Let

be any semi-algebraic set, and represent

E. = Z { f . . } n u { g I

IJ

Then the

I = U ^ P^

},f..,g GR[X ij iK l

and

n

X]. n

I(E) = 1(1") = n 1(1..)

are the minimal primes of

g i k £ P^ . hiQ e

iK

I(E^).

Let

I C R

n

E

as

E = U E.,

be the closure of

E.

Write I (Ei) = l(E/) = H p ^ , where Then

f. . E P. , all

j,a, and

This last holds since if, say, g i o E P^ , we could choose

Pi3-Pia-

Then

g i Q h i o e I(E i ), but

gio

is strictly positive on

pjFCX

E^, hence Let not have

h^o£

I(E i ), contradiction.

A. = R[X. ...X J1/P. , 'B. L 1 n / ia y i a ia

X(A. , tp. ) C E*., in general, because the XOc

10t

zeros far away from

1

ia' ^ia1 *

Proof:

Then

S

S.

= X(A. , 'JK ) , the non-degenerate points of

Moreover

> I^i^ J I^i " s i a ) •

We use 8.8.4, which characterizes

set of algebraic simple zeros

218

could have degenerate

However, we do have the following.

Let

iaC f i*

g.v IK.

U{g ik >.

Proposition 8.9.1. X(A

=r $ L 6[g.,1 C A. . Now, we definitely do 3 w ikJ la

x

of

P. , with lOt

S.

In

as the closure of the

g., (x) > 0. IK

Call this last

set

V. a , so

thus clearly Let

S. a = V. a.

V i a C E. = l{£^}

gi

n U{g ik >. Thus

= n gik.

Then

g^A^ x

Via = (

algebraic simple zeros

A

E

minimal primes of the inclusion

n u

to be

A

^i* ia^

with

vanishes on We have

and

S.a = V i a C E.. denote the set of

A. = 2 det(A ) 2 , where ia p p

x £ Z(P i )

We are now in Fat City.

{A }

Z(P. a) C Z{f..}

r x

where

g^OO

Via

> °>

as

Moreover,

is the set of above.

1". - V^ , hence on

I(E) = 1(1) =

r

r = codim(P. ) . la

/ O G A i a , that is, g i -A i a £ P i a .

ia^ia)

gj_#A^

it is obvious that

dvnens'ion of

Let

(3h./3X,), and let j K

it is clear that

gj_'A^ae15

Since

IL - S^ .

H p. . i,a i a

•

Define the

max dim(P. ) . This is the maximum dimension of the

I (E), which all must occur among the

A(E) = A (I) "• II A i a i,a

S = U s.ia C U E .I = L

with

f.. €= P.^ we have

P i a = (h.) C R [ x i . . . X n ] .

submatrices of Let

Since

and the order

Then,* by J 8 . 6 . 5 ,> each

X(A ia ,(!p ia ) d ) = S. a

and

P. .

A(E) H n($ i ) .

v(A.i a » ,('£• w10LJ)£>)

<#±a)^ = V(S±a).

We consider Let

i s an RHJ-algebra, & >

By 8.2.9,(A(E),A(E)

is an RHJ-algebra, with X(A(f),A(l) H (tp. ) ) = S.

Obviously,

nnflj.^)

A(E) nn(tp. ) =

icx ^

the functions in Consider

A(E)

nowhere negative on

E - S.

By Tarski-Seidenberg

Clearly, F - S = U I. - U S. i 1 i,a l a But

I(f.) C i(f. - U s . ) . I :£ i a ia

must properly contain some < dim(E.), hence

ia ^

S. E -S

is a semi-algebraic set.

C U ( Ix. - U Si.a ) , thus i a

dim(¥- S) < max dim (IT. -US. ) 1 ~ a ia

In fact, by 8.9.1, each minimal prime of P. . Thus, we have strict inequality

dim(E - S) < max dim(E.) = dim(E).

X(A/p) d C R n , where

dim(E. - U S . )

We can now repeat the

whole process above, beginning with the semi-algebraic set many steps we succeed in writing

1(1". - U S . ) I a la

E - S.

After finitely

E C

A = R[X1...X ]/P,

P

prime, and where

$ C A

is a

finite refinement of the weak order. As consequences, we have proved the following two results.

Proposition 8.9.2. (A(F),$(F))

F C Rn

be a closed, semi-algebraic set.

Then

is an RHJ-algebra.

Proposition 8.9.3. type.

Let

In order that

Let

•

(A,$) €= (PORNN), with

(A,«p) — (A(F),$(F))

A and R-algebra of finite

for some closed, semi-algebraic set

F, it is necessary and sufficient that there exist (1) finitely many primes

219

P. C A %

C A

with

i

(0) = n p .

= A p

/ -[»

and (2) finite refinements of the weak order

such that

$ = A H Il(^i) , under the inclusion

A -^IIA^ D

Proposition 8.9.3 should be compared with the results of 8.2, especially 8.2.11, 8.2.12, and the last paragraph of 8.2. '^

Working with derived orders

instead of ty , we can now deal with all closed, semi-algebraic sets, not

just the closed semi-algebraic sets. If we combine the Tarski-Seidenberg theorem, the going-up theorem for semi-integral extensions, and 8.9.2 we can prove the following.

Proposition 8.9.4. inducing set.

cp:

Then

R

n

m

-> R .

cp(S) C R

Proof:

m

Let

cp*:

Suppose

R[Y.....Ym] -•R[X1...X ]

S C R

The image

A(cp(S)) ->A(S), where

cp(S)

is a closed, bounded, semi-algebraic

is semi-algebraic by Tarski-Seidenberg and The homomorphism cp* induces an inclusion

A(S) = R [ X r . .Xn]/I(S)

Moreover, ^(cp(S)) = A(cp(S)) H ^ ( S ) .

Since

and

A(S)

is certainly semi-integral over A(cp(S)).

RHJ-algebra, hence by 8.2.7, so is

Remark.

A(cp(S)) = R ^ . . .Yj/IfcpCS)) .

is semi-integral over By 8.9.2, (A(S),$(S))

(A(cp(S)) ,$(cp(S))) , and

Thus, cp(S) C R m

X(A(cp(S)),'^(cp(S))).

be a homomorphism,

is a closed, bounded, semi-algebraic set.

bounded by simple estimate arguments.

A(S)

n

R, is an

cp(S) =

is closed.

D

As an immediate corollary of 8.9.4, we get that any polynomial

function on a closed, bounded, semi-algebraic set assumes maximum values.

This

generalizes the discussion at the beginning of 8.7, where we proved this result for bounded, closed semi-algebraic sets. In many ways the non-degenerate sets integral domain over

R,

$ = $ [g-] C A

X(A,-.p)d, where

A

is a finite

a finite refinement of the weak order,

w j are more natural "building blocks" for semi-algebraic sets than the Although

X(A,'p),

X(A,*£).

is not a manifold, as we have seen it is the closure of a

d-manifold, where

Specifically, it is the closure of the

set

x

V

d = tr.deg D (A). K of algebraic simple points

of

X(A,'B )

w implicit function theorem of 8.7 guarantees that that is, locally like

220

R .

Thus, X(A,$),

with

g• (x) > 0, and our

j V

is an algebraic d-manifold,

is a sort of closed "d-manifold with

boundary and singularities". contractions to fractions

If

K

E

of

A

Secondly, these orders

'£, C A

are exactly the

of finite refinements of the weak order on the field of

A.

Thus we can view

X(A,$,)

as an "affine model" for

is a semi-algebraic set, the procedure above for obtaining

E~ = UX(Ai,>|5i) , where

Pi C A ( E )

are primes,

A i = A(E)/P i , and

^

C A^^

a finite refinement of the weak order, definitely depends on a specific presentation of

E = UE-,

E. = Z{f..} n u { g . v } .

1

1

1J

It is also not really a

lK

stratification of E, since the pieces

X(A.,$.) can overlap, and, in fact, 1 -1- d

can overlap on more than their "boundaries", as in Figure (a). Even if they overlap of

only on boundaries, we may end up with a very unnatural decomposition

E, as in Figure (b).

it Figure (b)

We can rectify these problems somewhat by replacing the orders

(•£.) 1

by finite intersections of such orders.

Specifically, let

{P }

C A. d

1

index the

distinct primes which occur among the P.. We assume our decomposition E = U X(A.,'£.) is irredundant, in the sense that no term can be omitted. 1 -1- d On

A

a

= A(E)/P , we impose the order a

n p p

'£• = '£ . i a

Then we have

1

(•£ ) = fl(p.) 1 d d RHJ-algebras

and

X(A ,$a ) = u X(A.,$.) . i : L ^ d p = p d i a

(A , ('£ ) )

are geometric invariants of

= (A(E),A(E) if

a / 3.

A(E)

The primes

E,

E = U X(A ,$ ) ,

dim

) , and

The proof is not hard.

certainly minimal primes of primes of

Now, we claim that the

P

of maximal dimension are

A(E), hence are characterized as the minimal

of maximal dimension.

which is a geometric invariant.

Now

Moreover, X(A ,'£ ) = (E n X(A $ )) , i Y d "Y w d 1" - U X(A $ ) has strictly lower

221

dimension than invariant.

E.

Thus, by induction, the whole decomposition of

E

is

This discussion indicates that very nice basic building blocks

for semi-algebraic sets are affine models of orders on function fields where

'£

is a finite intersection of finite refinements of the weak order.

We can also use the ideals of degeneracy stratifications.

By definition, I

=

lim

= I(E n U )

x G E" C Rn.

if

U

Write

minimal primes of

C A(E),

I

x G E, to study

I(E fiU), where

smaller and smaller open neighborhoods of I

'£ C K,

x.

U

parametrizes

Of course, the limit stabilizes

is small enough, say a small ball centered at

I x = I (I H U ) = H P

where the

P i x C A(E)

are

I .

Proposition 8.9.5. (a) over

E". Each (b)

the

Only finitely many distinct prime ideals

£

a

If

C A

a

P. = I , ix y

for suitable

E = U X(A ,$ )

y

near

P.

P ot

The subset

is an irredundant representation of

P

= kernel (A(E) -*• A )

X(A ,•£ )

C E

is a minimal prime of

Proof:

which occur are exactly the prime, y G E . x G E

such

I . x

P. .

x

in

In fact, we can find

P.

is

y G I" O U

H z(P. )

y G U,

XX

such that

E H n

X

entirely of simple zeros of

P. .

such that

Then

I Pix,

y f. Z(P. ) , ~J X

= Z(P. ) O IL XX

X

The finiteness of the set of all

nl)-convex,

XX

0

j- P. , and an open set

I(U

If there will exist algebraic simple

XX 1X

I

is characterized as those

For example, since each

we know that arbitrarily near

P.

E", where

The proof consists of reviewing the various results of 8.6, 8.7,

8.8, and this section.

zeros of

varies

x.

prime ideals which occur as ideals of degeneracy,

that

x

are finite intersections of finite refinements of the weak

order, then the primes

(c)

occur, as

consists

X

= P. . x G E, follows from (b) and (c).

These two statements can be proved readily with all the machinery at hand. For example, suppose Consider the

X(Ag,«£g)

E H U , and since

222

I

= P

is prime, and assume

which intersect

U H X(Ag,$g)

E H U.

P = I(E H U ) , y G U. Since

is Zariski dense in

P

vanishes on

X(Ag,$ w ), we have

P C p

But also, some

X(A,# ) i

of

P

in

will contain a whole neighborhood of simple zeros

T d

E H U, so

P

C p , hence

P

= P.

We leave the rest of the details

of (b) and (c) to the reader.

•

The ideals of degeneracy can be used to define local notions of rank and dimension.

Namely, if

I

= H p. , define

d i m ( E ) = max dim(P. ) A

J.A

rankx(E) = rank x (I x ) .

(Rankx (I) = rank{dh.(x)}, where J fications of

E

I = (h.)> x G Z(I).) J

based on rank and dimension.

{x £ E|dimx(E) >^ k} an open subset of

is a closed subset of

E.

E

We can define strati-

Note that for any integer

k,

and

is

{x £ E|rankx(E) _> ^}

We do not quite want to begin a stratification with

all points of maximal rank, since this set will have (possibly rather singular) boundary points due to inequalities defining

E.

to be the "interior points of maximal rank".

These are obtained as follows.

Take the irreducible components take only those points

y

X(A ,'£ )

Cf

However, we can define

of least dimension.

E.

We will have

but will terminate.

Then

which are simple points, lying on a unique

X(A ,'£ ) , and for which an entire neighborhood of to

E

I(E-E^ -*) 3 I(E)»

so the

y

in

Z(P )

belongs

process can be iterated,

Using the implicit function theorem of 8.7,

E^ *

is

a manifold.

In the figure above, E^ ^

consists of the circle minus the vertex of

the triangle.

Then at the next stage

the triangle.

Next, E ^

triangle and finally

E

E ^

we get the interior points of

consists of the interiors of the edges of the is the set of vertices.

223

(x 2 +y2 )(y2-x-l) = 0

y 2 -x 3 + x 2

Figure (a)

Figure (b)

We point out that by defining rank locally, our stratification by rank is not the same as that of Whitney [44] for algebraic varieties

Z(I) C R n.

In our stratification, the origin has rank 2 in both varieties above, hence is the first stratum. origin has rank

0

On the other hand, in Whitney's stratification, the

in Figure (a) and the curve has rank 1, whereas in

Figure (b), the origin has rank 2 and the curve rank 1. 2 the polynomials

x(y

(In Figure (b),

2 - x - 1)

and

y(y - x - 1)

vanish on the variety,

but have independent differentials at the origin.) We can also find a manifold stratification of

E

by dimension,

of E E D E. Q . 3 E.j. D ••• . We begin with the pieces X(A ,'£ ) of greatest dimension, and take for the "interior" simple points, lying on a unique

X(A ,$ ) .

Then

dim(E-E, 0 .)< dim(E), and the

stratification continues inductively.

8.10. Krull Dimension Let

A

be an integral domain, •]} C A

an order.

By a weak %-oha-in of

prime ideals we mean a strictly increasing sequence of '^-convex primes (0) C P 1 C ... C p^ C A.

The length of the chain is

r.

We define the

weak Krull d-imens'ion, dim (A,$) , to be the maximum length of such a chain. In complete generality, this notion is probably uninteresting. we make the drastically simplifying assumption that

224

A

Even if

is finitely generated

over a ground field example, if then

R[T]

R[T]

R, the order

•£

is ordered with

can make things complicated.

T

For

infinitely large relative to

is a semi-field, with weak dimension zero.

R,

This pathology is

caused by lack of finiteness conditions on the order. Suppose, then, that we begin with a finite real domain R

real closed, and an order

the weak order. Thus

If

Q Cp

*jp = *J3 [g.] w 3

primes of length

any '^-convex prime, we can apply 8.4.3 to '^-convex primes above

Q,

Q C p

=£

l

R,

tr.degR(A/P) < tr.deg R (A/Q).

But from 8.4.3 we have chains of ^-convex

dim (A,'£) = tr.deg(A).

r, thus

over

obtained by finitely extending

are prime ideals, then

dim (A,'£) <^ r = tr.deg(A).

A

Q C A

is

(A/Q, :J3/Q) and obtain chains of

C ••• C p

=£

In fact, if

=£

C A, where

s = tr.deg(A/Q).

s

By analogy with classical Krull dimension theory, at least two other questions come to mind. •^-convex primes below

First, given such a Q,

(0) C Q

Q C A, can one find a chain of

C • • • C Q

= Q?

More generally, given

Q C p C A , both $-convex, can one find a chain of $-convex primes between and

P,

Q = P Q £ P x J •**

c

P t = P, where

Q

t = tr.deg(A/Q)-tr.deg(A/P)?

The answer turns out to be "yes", but before discussing this further, we want to argue that there are other, perhaps more natural, notions of dimension.

Again, we begin with

(A,'p) , '£ = $ [g.] w j

the weak order on the finite domain

A.

Define a strong 'p-chai-n of prime

ideals to be a strictly increasing chain some total order refinement

'p ^ $ , all

a finite refinement of

(0) C p P-

are

C ••• C p «p-convex.

9

such that for

Note that a weak

•jj-chain corresponds to a sequence of (non-injective) epimorphisms of domains (A,p) •+ (A/P1, tp/Pp -*...-> (A/Pr, tp/Pr).

From Proposition 7.7.9, a strong

#-chain requires much more restrictive convexity hypotheses on the primes

P^,

and by 7.7.10, a strong «j)-chain corresponds to a sequence of signed places K -• Kj^, ± «>-• ... "* K r , ± °° over where

K

dim (A,$)

is the fraction field of

R, finite on A.

P i = kernel (A -• K i ) ,

We define the strong Krull dimension

to be the maximum length of a strong $-chain.

actually produces strong '^-chains in tr.deg(A).

A, with

Of course, 8.4.3

A, hence we still have

dims(A,'.p) =

But what is more important, we have available the signed place

perturbation theorem 8.4.9 and its consequence Proposition 8.6.10 which gives more delicate information immediately.

225

Proposition 8.10.1.

Let

Q C A

there exists a strong 'p-chain

be a

(0) C P

'^,-convex prime

C • • • C P

= Q C • • • C P

'=f=-

' =F

=p

•$ = $ [g-]. Then

'=fz

^

C A, where

^

s = tr.deg(A) - tr.deg(A/Q) = codim(Q).

Proof:

The chain below

Q

K -» K-j^, ± oo-> ... -^ K g , ± «>.

comes from 8.6.10, which gives signed places

Then

Kg

is still a function field, to which

we apply 8.4.3, and extend the strong $-chain above Actually, a little more is required if some that the total order one finally obtains on one constructs an

integral extension

A^B

necessary in order that the weak order on Q

lifts to a convex prime of

field of

K

B

Q.

g. G Q

in order to conclude

actually extends $.

Specifically,

by adjoining as many

fgT

contracts to

The prime

$ C A.

as

B, and we apply 8.6.10 and 8.4.3 to the fraction

B, then restrict to

K.

This argument was actually used in the proof

of 8.6.10.

•

The reason strong dimension is a more natural concept than weak dimension is that strong dimension is a looal concept, giving geometric information about a semi-algebraic set infinitesimally near a point. prime x

of 8.10.1 is a point

Q

is a non-degenerate point

then any !pd-convex prime of

x

in

P.

in

x E X(A,$),. is

Then

P.

'^,-convexity of

Moreover, if

p[g]-convex.

X(A,'p)j, the prime ideals

have lots of zeros. of

P

x E X(A/p).

Specifically, suppose the

g(x) > 0,

says gGA,

Thus in any neighborhood

U

of the strong *p-chain of 8.10.1

Specifically, P. = I(Z(P.) n U ) , we can find simple zeros

U, and so on.

The strong chain of primes going down from

correspond to a chain of subvarieties going up from the point l,2,...,r.

Q

Q

x, of dimension

In other words, i/nfinitesimally near* x, we can move about on the

semi-algebraic set

X(A,'£)

with

r-degrees of freedom.

By way of contrast, we reconsider weak dimension.

The result which

allows the "desired" conclusion is the following.

Unproved Proposition 8.10.2. Then there exists a fpj-convex prime tr.deg(A/Q) = 1.

226

Suppose

Q C A

Q 1 , with

is '^-convex, $ = •£ [g.].

Q1 C Q

and

tr.deg(A/Q') -

Corollary 8.10.3. in A.

Suppose

Pi

C ... C p . _ C A

(0) C p^ C •. • C P^ C A

Then there exists a ^-convex refinement

of

r = tr.deg(A).

maximal length

Proof:

is any $ -convex chain

It suffices to insert an appropriate chain between any two

•^-convex primes we may assume

P C Q.

Passing to

(A,$) , where

A = A/P,

= tp/P = $ w [g.]» j Q1 C Q

P = (0). Apply 8.10.2 to find a $,-convex prime

dimension one greater than that of

Q.

of

Then apply 8.6.10 to go down from

Q'.

•

A proof of Proposition 8.10.2 can be extracted from [21]. Here is a rough outline of the geometry involved. R

By extending the ground field from

to an appropriate transcendental extension, we may assume that

dimensional, that is, a point. of points.

Now, we know the variety

X(A,$)

For example, if tr.deg(A) = r, near a simple point

Q

is zero

has lots X = X(A,'p)

looks like affine r-space smoothly embedded in some higher dimensional affine space point

Q.

Rn.

We consider sections of

X

by

n-r+1-planes through the

The implicit function theorem guarantees that many of these sections

will be 1-dimensional semi-algebraic sets.

(Simply take the

general position with respect to the tangent r-plane of The difficulty is, the point Y, and

X

n-r+1

plane in

at a simple point.)

Q may be an isolated point of such a section

Y may not be algebraically irreducible.

However, the method of [21]*

is essentially to argue that generically these sections irreducible 1-dimensional sets through

Q, although

Y

are algebraically

Q may indeed be a

degenerate point. Such a set corresponds to the desired prime

Q1 C Q .

227

This

Qf

£,-convex since, by construction, Qf has sufficiently many zeros in

is

V Note the proof of this dimension theorem 8.10.2 uses completely different concepts than those required for the study of strong dimension.

This is because

8.10.2 really is not a local geometric result at all, but a global property of semi-algebraic sets. not be described by

Near a degenerate point

Q, the variety

X(A,$)

will

r-independent parameters; although globally we can pass

a curve, then a surface containing the curve, and so on, up to the r-fold itself, through the point

Q.

Also, we point out that the commutative algebra

analogue of the dimension theorem 8.10.2 can be proved using integral extensions and various going-up and going-down theorems for prime ideals.

In our real

setting this method seems to break down because of the special hypotheses needed in 6.4, especially in Proposition 6.4.2(b). The analog of 8.10.3 for strong chains does not follow routinely from 8.10.1. A

and

The reason is that given Q/P

a

P CQ CA

C£j/P) -convex prime of

strong ^-chains between

P

and

Q

with

P

a ^.-convex prime of

A/P, we cannot immediately construct

by passing to

(A/P, ($j/P) ) . The

finiteness condition on the order is lost by this process.

What is needed

instead is a more vigorous version of the signed place perturbation theorem 8.4.9, which was the basis of 8.10.1.

Proposition 8.10.4. over

R, and

:p' C A

small relative to

ideals of where that on

A.

if

induced by

r

Suppose

(0) = Pi

and

and all

K = R(x-,-..x ) P^.

r

x.

are infinitesimally are

P r i m e $'-convex

signed place

p:

is the fraction field of

are'^-convex, where

$ CA

K -*R, ± °° ,

A, such

is the total order

p.

The proof is by induction on

is quite small.

Let

we may also assume that the

r, there being nothing to prove

r. = tr.deg(A/P^.), so that

As in the proof of 8.4.9, we may assume

228

is a finite integral domain

C P " S '** S p # o ^ 1 i =£ =£ x s

Then there exists a discrete rank

p(x.) = 0

Proof:

A = R[x,...x ]

is a total order such that all

R.

r = tr.deg(A)

A

Suppose

A

r=r

is integral over

> r, > ••• > r .

R[x,...x ] , and

{x, = xv(mod Pi.), 1 < k < r.}, give a transcendence

base for If

A/P^., all r

o

- r, = 1, then the place l

the total order rank

1

A

£'

on

K

A

1

C K

= R[x.,y.], with

and the subfield

x.

and

y.

P^ • of

Af

fraction field of a quotient of

the convexity of the

Write

R[x,...x

i j K i r i , K

F

CF

relative to

R, the

A

A'

to

q:

+1]-..[xn],

Also, K1 is the

A'/P- )> hence by induction, K ! -* R, ± °° , preserving

and for

generate the prime

f. v (x v ) 3 > K- "•

is integral over

x, K

over

Eb ik (O...O)T 1

x, K

over

P- Ac j j

in

AQ . b j

R[x,...x ] . l r

Let

f. ,, with J>^

•£' C K.

Let

r

denote

F. C r be the real closure of R(x,...x ) , J i rj C ••• ^F-i- Since all x. are infinitesimally small x.

are certainly finite relative to

R(x1...x

) . In j p! = p f x .. (This Kv. i. . .x r j 3 i

A

of the place

p!(x i )

R(x 1 ...x p ) .

r

Thus elements of It follows that the

R(x 1 ...x r .), in the appropriate residue field,

with some

xa. . G F.. f. ,

is a root of the non-trivial polynomials

for

then gives

K, and let

are all algebraic over

k > r

K

r - r., > 2. I — o k > r., let

be a monic polynomial equation for

to our integral dependence relation for

if for

of

is the multiplicative set of non-zero elements

is the center in

and we can identify

Pj

Thus we may assume

are infinitesimally small relative to

p'-tx^)

R, we

$'-convex primes

follows easily from Proposition 7.7.4 and results in 7.6.) P^.

tr.deg(K') =r-l.

K, we can write

by 7.7.8.

1

fact, P^. C A 3

with

in

R[x1...x ] . We have our total order

a real closure of

is a discrete

Composition and reordering

{f. ,} J K ' k>r-

0 = Ea. ., (x ...x )f. ,

so that

j > 2.

S. C A

k > r = r , then o

coefficients in

A

R

represent the minimal polynomial for

] , then j

1

If

If

) C K

over

signed place

A/Pi. = R [ x r - . x r ][x r

R[x1...x, .] .

K1

(in fact, of

K -• K ! , ± °° -*• R, ± °° .

p:

f. , = f. (xv) £ P| x j,K i,k K j

of

r-1

P! ,

K -* A, ± °° defined by

infinitesimally small relative to

A', and they necessarily form a chain in

we can find a discrete rank

if necessary.

R(x,...x

is the integral closure of

can lift all the '£' -convex primes

the desired

x,...x

p', .: R(,x1...xriJ

place, with image a function field

Moreover, if 1

j , by simply rearranging

in

Applying the place K,

p!

k > r, we get that

Eai.-^(C1.. • C r )T 1

over

F..

Also,

)x11

we let

0

0 = Eb.v (x, .. .x be an integral dependence relation IK i r K R[x.,...x ] , then 0 is also a root of the non-trivial polynomials 1 r

over

R.

229

Let R

0 < e £ R

and let

over

0 < e. € F. 3 3

F..

to a field

where

L

by adjoining the elements

1 <^ j <_s,

1 £ i <_ n, and

e. < x...

We now extend

/ e 2 - x2., and

£• ., 3 >i

K

/e? - f 2 , , 3 3 >K

K

r < k £ n.

over

Ea. ., (£,...£ )T 1 i-JK i r

be less than all other roots of the

x 1 £ F., we may also assume

Since

Eb., (0. . .OJT1

be less than all other roots of the

(Note these elements belong to

B = [x.,x.-£. ., / e 2 - x 2 - e, 1 1 j,1 K small relative /e? - f? , J1. Since the f. , and x. - s£ . . are infinitesimally 7 3 3>k j,k I j,i 2 2 to F., hence also relative to R, and since e. < x. , all these generators

T

of

so

L

B

is real.)

We replace

A

by the ring

are infinitesimally small relative to

The places

p!

are still defined on

L

R. since

R^.^x

J

center of

p!

on

B

is a prime

all our hypotheses hold for

B.

restricts to a discrete rank

r

K = L.

) C L.

l

The advantage in this:

Q^.

which contracts to

Since any discrete rank place of

The

3 P^. C A. r

place

Thus p:

K, we may as well assume

if we now construct any total order

L -• R, ± °°

A = B, '£ C K

such that the {x-...x } are infinitesimally small relative to R and the {f. , |r. < k < r} are infinitesimally small relative to R(xn...x ) , then J »^ J 3 necessarily all

x. 1

are infinitesimally small relative to

are infinitesimally small relative to generate

Pi.'Ag,, where

we have degree

E C r F, C E. 1

over

recall, f

S. = R[x,...x r ] . Thus if

be the real closure of

E, which can be written

R[x x .. • x r _ 1 ] ,

(x ) £ P^

x .

pn f

.

If

D

E

so that

D £ (x )

is the place

E(x ,y) r

or

Since ...x ) E(f

of transcendence

, z ) , where, i ,r

1

on

K.

p

D = DEOr)

x of

over E

is non-trivial, hence

We are then finished by induction, We propose to reorder

E(x ,y)

is algebraic and so that all our hypotheses on

K C E(x ,y) are preserved in the reordering. It is better to work with the transcendental element

230

r - r 1 >_ 2,

represents a minimal polynomial for

as in the second paragraph of this proof. over

E(x ,x

is algebraic, the place

is discrete rank

f. , K 3 >

$ , we still have that the

x i = xi(mod P i ) . We consider the cut

defined by

p.

R(x 1 ...x r p .

Consider the function field

= £-

and all

R C x , . . ^ ^ ) . As we observed, {f. ,} 1 X 3 J » k k >r-

pDr . on K relative to this (new) order K^X]^. . .X r ) center of p. on A is Pj_.c A. J J Let

R

(x r) G E(f 1 ^,z)

= E(x , y ) , since

f,

R(x 1 ...x r ) C E.

Thus let

sign of

f,

is infinitesimally small relative to the (large) field D 1 = D £ (F 1

if necessary.

so that if

E(fn

)

E(f1

R(x,...x

We may assume

0 < D', by changing

We appeal to 8.3.1 to choose

is reordered over

new order extends to relative to

).

r

E

with

, z ) . Of course, 3

D c (f 1

3 £ E,

) = 3,

0 < 3 < D',

then this

is still infinitesimally small

).

We now consider our hypotheses on the elements

x.,

i <_ r

and

f. ,,

1

r. < k <_r. xr

Since the order on

and the

f.^,

1 < j < s.

Pp (x^) = C-, •

nas not

i

c^(£, •!•••£•,

ficients of

(x )

l,r of

pp

on

But

f ^

i)

= f ^ r ( x r ) = E c ^ . . . x ^ x * , say, -JPp (Xp) 1 * where

0 = Ec.(C, i---Ci

changed if

that is,not all f..

is unchanged, we need only worry about

p f , we get

and if we apply the place G F

E

J ,K

i < r.

This polynomial is non-trivial,

vanish, because, by definition, the coef-

are non-zero modulo

P-

r

C A

and

P^

i1

A.

We see that when we made our finite extension from

K

to

graphs back, we should have added two more elements, /6 - x r y,6 G F 1 ,

Cx

-xx < y < q

r

contains only the one root

r

< 6 < Cx

t>1

of

conditions indicated, the elements small relative to

R.)

r

G

F

Ec i (C 1 1"-^>1 /6 - x

and

E(f;.

to

Moreover, if one writes out the polynomials

i«

and

^T

1

= °-

in

Fj

(With the

particular, x r

,z) = E(x r ,y), we have

is still infinitesimally small relative f.

= f.

i < r, have not moved at all, and

xr

(x ) , 1 < j < s,

r

j,r

one sees that they are still infinitesimally small relative to

on

(y,6)

are infinitesimally

j,r

our reordering.

/x r - y, where

But there is no problem doing this retroactively, and

Pp (xr)

r

r

/x ^~y

=

^i

In

L, four para-

+ x x , and the interval

the conclusion is, even with our new order or

R.

is the center

ij

—

F., since

has not moved relative to

x^

F^^ 3 F., by

This establishes our original hypotheses for our new order

E(x r ,y), now with

D £ (x r )

algebraic, and hence completes the proof

of 8.10.4.

•

We can now prove the strong chain analogue of Proposition 8.10.3. Proposition 8.10.5. £,,[g-] w J

is

a

Suppose

A

is a finite integral domain over

finite refinement of the weak order.

R

and

Then any strong '£-chain

231

of prime ideals

P^

£ * * * £ ^i

(0) C P 1 C .. . C P^ C A

Proof:

g..

where

of maximal length

r = tr.deg(A) .

The proof consists of arguments used previously.

adjoining suitable ho

can be refined to a strong (£-chain

^ A

/gT

to

First, by

A, we are reduced to the case where there are

Secondly, by extending the ground field, we are reduced to the case P.1: C A s

is zero dimensional, that is, A/P^

= R.

Both these reductions

s

are just as in the proof of 8.6.10.

We then apply Proposition 8.10.4 to get

a maximal strong $-chain refinement of the P^

below j

discrete rank

fraction field of

pf:

signed place

r - rg

A, r

P- . x s

K -* K 1 , ± °° , where

= tr.deg(A/P^ ) , and

Kf

This gives a

K

is the

is a finite algebraic

extension of the fraction field of rank r signed place «B-chain above P, . s Remark:

If

A

p":

1

K

A/P- . Finally, 8.4.3 gives a discrete x s -* R, ± °° , which completes the strong D

is a finite integral domain, $ = $ 7[gi] c A, and

a prime $J-convex ideal, then the RHJ-property of rather easily from 8.10.5. if

g E A,

follows

In fact, in 8.10.5 one can add the conclusion that

g £= P^ , then the refinement chain can be constructed with

Of course, P x

(A/P, (pj/P) )

P C A

is zero dimensional, hence is a point

is highly non-degenerate.

g jE P .

x E X(A,$), and moreover,

This argument would complete the proof of Proposition

8.6.9 which we omitted previously.

8.11. Orders on Function Fields Let Write

K

be a real function field over the real closed ground field

K = R(x,...x ) , A = R[x,...x ] .

X = X(A,$ ) C R n w

Thus

is an algebraic variety and

algebraic variety of non-degenerate points of

where

£ W (K)

is an order on

is the weak order on A

if and only if

All non-empty open sets

232

K.

A

is an integral domain,

X, = X(A,($ ) ) a w jj X.

Note that

We know that if

is the (semi)($ ) = A n $ w d

g i ^ 0, then

U{g i> H X d ^ 0, where

U{g.} n x,

R.

are Zariski dense in

(K) ,

'^[g^]

U{gi> = {x E x|gi(x) > 0}. X^.

We will find it convenient to forget the degenerate points of and work in

X^.

Thus, the notation

U{g^}

now means

Let

V

Vlg^} ^ ^

(In

Xd,

W{g i }-U{g i >

% [g^]

V{g^} =

has no interior.) V{g.}.

be the collection of non-empty sets of the form if and only if

altogether,

{x £ Xd|g.(x) > 0}.

We will also find it convenient to work with the larger open sets U{gi> = W{g i> C X d .

X

is an order on

A.

The

Vfg^}

Thus

are regular

o_

open sets, that is, V{g.} = V{g.}, since, in fact, V{g.} = U{g.}. Seidenberg, the

V{g.}

are open, semi-algebraic sets.

in any semi-algebraic set) the the

U{g.}.

That is, each

finite union of

Vs.

V{gi>

V

is a finite union of

U's

much the same statement as Unproved Proposition 8.1.2. fact, although there are strong indications that the

By a prefilter

V

£ &.

U

is a

In fact, it is pretty We will not use this V{g.}

provide a more

we mean a non-empty subset of ~//, closed under

A prefilter is a filter if

V G ^ ,

V C V ' G ^

implies

A filter is an ultrafilter if it is not properly contained in any

other filter in

X

Note that

This is the property of the U{g.}

and each

U{g.}.

^^- ^

finite intersections.

In affine space (hence

form a base for the same "topology" as

This is certainly not obvious.

natural base than the

By Tarski-

V{g i> C v ( h )

V{g i >

if and only if

hCVCg^) _> 0.

which makes them more convenient than the

for the purposes of this section.

The set of filters in

V

is partially ordered by inclusion, an arbitrary

intersection of filters is a filter (every filter contains a union of a chain of filters is a filter. an ultrafilter.

Each prefilter

^ C ^

X^ = V(l)), and

Thus every filter is contained in

is contained in a smallest filter

d Let if

$ C A

g. E •£.

only if

«^('p)

is a prefilter since

Define ^tp) C f tp [g.]

by

V{g.} e^"($)

is an order on

A

if and

V{g.} £ ^ , which implies the desired finite intersection property

for ^ p ) . If Let

be any partial order.

&CV

^

C $ 2 , then

an order on

2

be an arbitrary prefilter in

if there is V Cv(g).

p

V £&

Since the sets A.

with V

g(V) :> 0.

V.

Define

<$(&) C A

by

Equivalently, this says

are Zariski dense, it is clear that

In fact, '$(0) is a derived order, since if

f g = h

233

f ± 0, and

h(V) :> 0, then

g must be non-negative on

would vanish on the Zariski dense set ^

c

Also

C ^ , then yi&fi 'p(^>)-

V.

Otherwise, f

U{g.} ^ U ( - g ) , where

$ W

=

V = V{g.}.

'£(«£§)» where ^

If

is the filter

generated by &". We will study the compositions p ( ^ W ) and

&^&(${&)')

and ^^(0)).

First, $

are completely obvious from the definitions.

Proposition 8.11.1. (a)

If

«^C /' is any prefilter, then ^ ( ^ ) = ^

(b)

If <3* is an ultrafilter, then '$Q?)

c

A

C X

is a total order. Also,

if $ C A is a total order, then &($) C "#" is an ultrafilter. (c)

is any order, then pG^Op)) = P d -

If 1) C A

Proof: (a) Since $££) = '£G^), we may as well assume know

3FC£T(:p@r)).

Conversely, if

g.(V.) ^ 0 , V. e ^ T Thus

Then

^ is a filter.

g^^ € $ 0 9 , so that

V. CV(g.)

and since &

We

V{gi > £&($(&)),

let

is a filter, V(g.) e ^

V{g i ) G J T and (b) Suppose f = #(£}

admits a proper refinement, $ C'£[g].

and since ,F is an ultrafilter, 3~= g'&ig]).

Then But now

, contradiction. Secondly, assume

'^ C A

properly contained in a filter

is a total order and suppose (

S. Let

? C^C^QJ)) Ctp(®, we have fl = ?G?).

&- &($') is

V = V{g.} G ^ - ^ T . Since But

g± e ? ( ^ ) , hence

V{ g i >

contradiction. (c) Finally, ^ c'p(^(^)) and $(^f(^)) ).

Also, 'p, = fl !p

total order refinements of $. c

is a derived order, so

where the intersection is taken over the Write $ n

Then ^ p ) C ^ , fl(^CP)) ' P ( ^ )

cn

= $C^T)

where

<2" = &{$ ) .

' ^ C ^ ) = ?d.

•

We have thus established a natural, bijective, refinement preserving correspondence between orders on the function field as derived orders on

234

A) and filters in the family

K

(which are the same

V of subsets of

X,.

This result makes it clear how an arbitrary order on a function field

K

is related to the particular orders defined by evaluating polynomial functions X C Rn

on Zariski dense subsets of any affine model

for

K.

Total orders

correspond to evaluating functions on infinitesimally small open subsets, that is, germs of open sets. Of course, the totality of all orders on of

X.

K

is a birational invariant

However, with respect to the particular model

X, we can distinguish

orders obtained by evaluating functions on bounded sets and orders obtained at infinity. writing element

A = Rfx^.-.x ] , we say a filter V{g^}

unbounded. If

With reference to the embedding

^"C V n

which is a bounded subset of

If &

X C R , which amounts to

R .

is hounded if it contains an Otherwise, we say &

has no bounded refinement, we say

V{g t } C Ty we have the closure

is a filter, by the center of &

&

V{g i> = U{gi>

is

is at infinity. in

Xd.

If

^C

V

we mean

n v{g.} ex, x d .}e^ If

^ = ^ ' C ^ w [ g i ] ) , then

center of

&

C(0) = V{ g i >.

could be empty, even if &

In particular, C { M £ W ) ) = X d . is bounded.

The

In the case of filters

at infinity, homogeneous coordinates and hemispherical models of semi-algebraic sets makes these notions quite analogous to the centers of places in algebraic geometry over algebraically closed fields.

This is especially true for total

orders, because of the intimate relations between total orders and real places. In the real case, we have the added geometry provided by partial orders on function fields.

Proposition 8.11.2. &CV

of subsets of

order refinement £P =^*CP»), then

If $ C K

X d , then

'p! D $

with

C ^ " ) = x.

is an order with associated filter

x € C(^j x

if and only if there is a total

a 'p1 -convex maximal ideal of

In general, if

&

A.

If

is an ultrafilter, C(^)

consists of one point or is empty, and is always empty if

&

is at infinity.

235

Proof:

The main point here is that if

g £ A,

necessarily positive in any total order on ideal.

K

for which

$ [-g] C A

in

X H W(-g)

of

X H W(-g).

if

is an order and

g(x) > 0

x G V(g), then

and even if

g(x) = 0 ,

The proposition is then proved as follows. family of elements

g £ G A, all

e > 0 G R, with

This is easy using the technique of 8.1.1. #[§£> all

e

]

is

an

exists with

x

'£[g ] .

f

£ -convex.

Otherwise, x GV(-g.)> some

Then

j , and

g.

If

x

g

is

is a convex

x

For

is clearly not

is a degenerate point

X E Cf^O, we choose a

g (x) > 0,

The assumption

order, clearly with center

any total order refinement of

•£' 3'£

x

This assertion can be deduced from either 8.6.5 or 8.11.1.

example, if

'p'

x E V(g), then

{x}.

n v ( g ) = {x}. x G Cffi

implies

We can then take for

Conversely, if a total refinement x G V { g . } = O V*(gi)

all

would be negative in

$'.

g-,,...,g^

The other statements of the proposition are also easily deduced from the remark in the first paragraph of the proof.

Remark.

•

If our ground field is R, the real numbers, then a compactness

argument implies that the center of a bounded filter is never empty.

Remark.

If the order

$ C A

is finitely constructed using the operations

of finite extension, finite intersection, and the operators of 3.12, then

C(^('^)) C X,

will be semi-algebraic.

£ , !p , # , $ ,

In general, however,

this will not be expected.

Remark. • *

If $ C K = R C x , . . ^ ) in

the associated signed place, with center

{(a,...a )}

A

if and only if

is a total order and Archimedean over

pD: K

K -» A, ± °°

R, then &ty)

has

P R (^ i ) = a^ G R C A.

It is pretty easy to see intuitively (but not necessarily easy to prove) what all orders are like on function fields in one variable.

236

An affine model,

with degenerate points excised, will look like a smooth curve, with finitely many singular points. A branch at

x

At each point some even number of branches comes in.

is a connected component (see 8.13) of

a very small open neighborhood of

x.

U - {x}

where

U

is

The total orders correspond precisely

to the branches at all affine points, together with a finite number of branches at infinity.

The individual sets

V{gi>

are finite unions of

open intervals on the curve, with disjoint closures. In 7.7 we studied places if

A

is real and

p:

K •> A, °° of real fields

K.

In particular,

A

is given any total order, then by 7.7.2

p

can be refined

to a signed place, p:

K •*• A, ± °° , thus inducing an order on

K

compatible

with the order on

To conclude this section, we study a finite variant

A.

of this problem of lifting orders in the case of function fields. Suppose A = A/P, and '3 C A P

A

is a real finite integral domain, P C A

$ CA

such that

will be

a total order.

TTCP) = $"> where

a prime ideal,

We ask when there exists a total order IT: A -• A

'^-convex, hence necessarily

P

is the projection. is

($ ) -convex. w d

In particular,

This condition

will be included in our proposition below. Let

X = X(A,'£w),

X = X(AyB ) , X . C X W

X^ C X

the non-degenerate points, and similarly

the non-degenerate points.

We first prove a lemma

Q

of independent interest.

237

Proposition 8.11.3. X = X(A,£ w ), in

X

E C X

Suppose

A

is a real finite integral domain

a semi-algebraic subset.

Then

(that is, I (E) = (0) C A) if and only if

E

is Zariski dense

E H ^

has non-empty

interior.

Proof: dense. some

The "if" statement is clear since open sets in

Write E^^

E = U E^

Ei = Zff^} n u { g i R} .

is Zariski dense.

Thus, f i . = 0 e A

I(X-X,) j- (0), we must have

Proposition 8.11.4.

^{g.^} n X,

If

P C A

then there exists a total order all finite sets in

{f } C $,

and

is Zariski dense,

E_L = U(g i k ) H X.

Since

non-empty as claimed.

is prime, f C A

$ C A

E

are Zariski

with

f. £ 0, the set

= A/P

IT (•£) = $

•

a total order,

if and only if for

U{f\} H x"d n x d

has interior

X,. d

Proof:

First a comment on notation.

Also, we write

U{f;[} C x,

F = {f G A | ? ^'F, ? 7* 0}. is

Then if

X,

U{? i> C x , if Suppose

$

We have

X CX

f± e A,

Tr(fi) = 7 ^

= $ W [F]

Obviously, fr(t|J1) = $ C A . *

CP-i) -convex. A d

TT(^) = '^5, since

is already a total order.

order and

f C A P

is

exists, certainly

IT: A ->A. Now, let

is an order and

P CA

From 7.7.3, we can choose a

£ ^ ^

total order

P

A

total order refinement '|5 C A

such that

c

from

is 'J3-convex.

Then also

Conversely, if our desired

F O p , hence

•£, = '£W[F] C A

is an

(#..) -convex.

We now prove the hypotheses of the proposition are sufficient. prove

$.. C A

is an order and

order, then some

^ [£^]

P

is

If

'J3, is not an

is not an order, for finitely many

is equivalent to

U{fi> H X d = 0, so

U{f-} n X,.) i (X

'p, l

If

($,) -convex.

is an order, but

U{? i ) n ^ P

n X d = 0.

is not

We must

f^ C- F. (Note

This

u^}

H x"d

('p..) -convex, then again Id

there is a finite set of (A, e p j f i l ) ^

f. G F with P not ('£ [f-]) -convex. By 8.6.5, 1 W 1 (J is an RHJ-algebra, with X(A, ( ^ [ ^ 1 ) ^ = U{f i> H ^ . (Here,

the bar denotes closure.)

But if

U{fi> H X d n f ^

is exactly the ideal of functions which vanishes on C U{f i } H X d H X = U{fi> n X d n Z(P). convex.

238

has interior in U{f.} n X, H X,

But in this case, P

would be

X*d, then

P

Finally, we prove the hypotheses are necessary.

Assuming

P

is

(•£ {f.}) W

convex, we know that

P = I(U{fi> H X d O Z(P)).

Zariski dense in

Since

X.

f = IT£.

That is, U{f i ) H X

vanishes on

f £ P, we must have

8.11.3 to

l

U{f i ) H x ^ f l x

A, we conclude that

Zariski dense in

U{fi> H x d H x"

1

a

X.

i

d

Applying

has interior in

x"d, as

desired.

•

Corollary 8.11.5. if

X

In the situation above, if

X = X(A,'J>w) = X d , that is,

has no degenerate points, then any total order

total order on

Proof:

'£ on

A

lifts to a

A.

The point is if

X = X,, then a

I

Gf,

I

f 0, then

U{?i> n X, + 0.

If

U{?.} n x . O X = U{f\} n l , I a d ! d

For example, if

D

A = R[X,...X ] , the polynomial ring, then 8.11.5 applies.

However, the conclusion admits a trivial proof in this case. the fraction field of

A = A/P.

give the polynomial ring small relative to C A[X,...X ] . of any

A

is

U{f.} H X, - U{f.} H X,

l

and

H X

A.

Then

A[X.....X ]

tp

A

induces a total order on

any total order with

denote A

(X. - C O

We then restrict this order to the subring

We see this order lifts

f G R[X][...Xn]

Let

and we infinitesimally

R[X,...X ]

'£, by writing out the Taylor series

in powers of the

(X± - q ) . Here

q

= X^mod P) G A.

The geometry of 8.11.4 is roughly illustrated by the picture below, where

we have orders

X = X,

for simplicity.

*& C A, we can think of

subsets of

X,.

Using the ultrafilter interpretation of total 'jf as

picking out infinitesimal ly small open

If these small sets do not contain enough points in

there can be no ultrafilter of sets in

Xd

giving an order on

A

Xd,

lifting $".

239

In the figure, orders

*p

"centered" in the right half of

X,

will lift,

while orders centered in the left half of

X will not. d We will state a slight generalization of 8.11.4.

Proposition 8.11.6. Suppose $

[g-1 *- A 3

w

c

A.

is a real finite integral domain,

a finite refinement of the weak order and

'•Pw[g-1-convex ideal. $ w [g-] j

A

Suppose

£ C A = A/P

Then there exists a total order

= t, i f and only if for all finite sets n

f^g.} nx(A,!D w[g.]) d X(A,-B w [g j ]) d

Proof:

P C A

a prime

is a total order refining !p C A

with

{?.} C $",

and

g.l

f\ ^ 0, the set

has interior in

First, 8.11.3 generalizes routinely to sets

Then the proof of 8.11.6 is exactly like that of 8.11.4.

X = X(A,£ [g.]). w 3 The reason one

must work with the closure of may belong to

P, that is,

U{f.,g.} H X(A;J [g-]) is that some g. 1 w J 3 d 3 g. = 0. If, in fact, all g . f- P, then one

can just require that

U{f.,g.} H X ( A ^ [g.]) meets X(A,'£ [g.]) in 1 W W 3 3 d 3 d a s e t with non-empty i n t e r i o r . For e x a m p l e , i f g . f- P and X(A,$ [ g - ] ) = w 3 3 d X(A,$ w [g.]), this always holds, giving a generalization of 8.11.5. D

8.12.

Discussion of Total Orders on Suppose given a total order

$

R(x,y) on

R(x,y), the rational function field

in two variables over the real closed field

R.

Let us assume that

'£

is

2 centered at the origin in pD: R

where

R(x,y)

R , that is to say, P

R

M

= P R (y) = 0 G R C A,

R(x,y) -* A, ± °° is the signed place associated to our order on

and the subfield

R CR(x,y).

The results of the preceding section show that the order describable by infinitesimal behavior of functions On the other hand, the order pR:

R(x,y) -* A, ± °° .

f £ R[x,y]

is necessarily near the origin.

is also described by the signed place

In this section we will reconcile these two descriptions

by comparing invariants of

240

$

*£

pR

(the residue field

A, value group

r, rank

of

Pn, etc.) with the particular geometric behavior of functions

origin which determines whether

f

is positive or negative (rel

f

near the

'£). We do

not attempt to state a theorem, and our discussion is meant to be enlightening, not rigorous.

Our discussion of signed places on

R(x,y)

is a watered down

version of the discussion of valuations on function fields of two variables given by Zariski in his work on resolution of singularities of surfaces. There are two possibilities for the residue field p •

R(x,y) -• A. ± oo #

Namely, A

Archimedean ordered over A ^ R. over

R, and

cut of in

There is then

R

R

A

R, or

converging to

T.

may be very large.

converge to have

is isomorphic to

0.)

R(t). Let

and let

(a!,aV)

This means

is either smaller than some {i}

A

R.

t = f(x,y)/g(x,y) G R(x,y), with

t

of the signed place

is either a function field in one variable,

is algebraic over

defined by

A

Assume first that R(t)

T = D R (t)

Archimedean

be the transcendental

be a nested family of intervals

a! < x < aV

and every element of

a!^ or larger than some

aV.

R

(The index set

Also,it does not follow that the differences

aV - a|

$ C R(x,y), we

In any event, with respect to the order

a! < t < a'.'. Consider the neighborhood in

R

defined by

U i j £ = {(x,y)|0 < x 2 + y 2 < e , a!g2(x,y) < f(x,y)g(x,y) < aV g 2 (x,y)}.

We decompose these

U.

into their "connected components".

are non-empty and have the origin in their closure. and

g(x,y)

vanish at the origin.)

These components

(In particular, f(x,y)

Moreover, the behavior (number and

approximate location) of these components stablizes for sufficiently small and

a!,a'.'

close enough to

T.

The order

one of the component families of the in the next section, 8.13.) if

h(x,y)

U.

$ .

is then determined by selecting (Connected components are studied

Specifically, h(x,y) G R[x,y]

is positive in

$

assumes only positive values on the selected component, infinitesi-

mal ly close to the origin.

We interpret this selection of a component of

as selecting a branch at the origin of the curve larger field

e

R ( T ) . If

A = R(x,u),

u

f(x,y) = Tg(x,y)

algebraic over

U.

over the

R ( T ) , this choice is

just the choice of one side of a real root of the minimal polynomial for

241

u.

As a concrete example, let

R

be the real algebraic numbers

component of a

Given any polynomial h

h(x,y), we can find numbers

has no zeros in the intersection of the cone

disc

0 < x

2

+y

2

< e.

of this region, hence In case h(x,y)

ry < x < sy

U.

e > 0

so that

and the puctured

Then by requiring, say, 0 < y, we single out a component h(x,y)

will have constant sign in this component.

A ^ R, the value group

is the order to which

r

h(x,y)

branch of the transcendental curve first non-vanishing derivative of Note that if

r < TT < s,

t = x/y,

is always

Z.

The value in

Z

of

vanishes at the origin along our chosen

f(x,y) = Tg(x,y). h(x,y)

That is, we find the

on the branch at the origin.

R = ]R , the real numbers,this example cannot exist since

]R

admits no Archimedean extensions. We now turn to the R-valued signed places are four possibilities for the value group

Case I.

r

non-Archimedean.

(m,n) < (m',n')

Case II.

if

r

Case IH(a).

m < m

1

T

242

There

r = Z 1

m = m ,

x

Z, ordered lexicographically,

n < n1.

That is, F = Z.

Archimedean, non-discrete, and containing two incomWe can then assume

an irrational real number.

F

R(x,y) •> R,± °° .

r.

Archimedean and discrete.

mensurable elements.

Case HI(b).

or if

Then

pn: K

F

F = {n+mx|n,m £ z ) , where

is ordered as an additive subgroup of

Archimedean, non-discrete, but a subgroup of

T

is

IR .

Q, the

rational numbers. the F

p.

The prime integers fall into two classes, p i ? q., where

occur with arbitrarily high powers in denominators of elements of

and where the powers of

say by

b..

q.

occuring in such denominators are bounded,

Then

a 1

= p

•••

r Pi

3, 3_ q / '•' *j . 0 < a. < oo, o < 3. < b. } .

We now analyze orders on

R(x,y)

which yield signed places

p:

R(x,y)-*R,± °°

with these value groups.

Case I.

Here we have a rank 2 valuation.

defines an algebraic cut of the subfield there is an algebraic function f.(x) G R[x]

with

irreducible in

f(x,y)

R(x) C R(x,y).

y,

In other words,

f(x,y) = fQ (x) +f 1 (x)y+--« +f m (x)y m ,

infinitesimally small relative to

R(x), and

R[x,y].

The curve

f(x,y) = 0

discussion in 8.11 or 8.3.

V

One of the variables, say

/

thus has branches through the origin, as in our The order

$ C R(x,y)

picks out not only a real

\ \ \ \ \ \

root of in

R

2

f(x,y) over

R(x) (which is a branch of the curve

f(x,y) = 0

at the origin, as indicated by solid lines in the figures), but also

a side of this real root in the real closure of

R(x) (as indicated by the

normal arrows to our branches in the figures). With these choices made, the £-sign of a function determined as follows.

If

g(0,0) > 0, then

g

g(x,y) G R[x,y]

is positive.

If

is

g(0,0) = 0,

243

but

g(x,y)

does not vanish on our branch

x

g( >y))> then to the origin.

g

(that is, f(x,y)

has the same sign as its values on Finally, if

small normal curve to the values of

C

g

g

vanishes on

does not divide

C, infinitesimally close

C, then

g

will not vanish on a

C, near the origin, and the sign of

g

is the sign of

on such a normal curve, on our preferred side of

C.

(In

fact, this last "test" actually covers all cases.) The value group

V

is

determined by first writing is, h h

x

z

0 =

2 a..x y , i+j=l l j

h

f ) . The invariant

vanishes at the origin, along

r = Z,

The value

g = f h, where

is not divisible by

Case II.

Z-

A = R.

(m,n)

assigned to

g

does not vanish on

C

n

is (that

is then the order to which

C.

In this case, there is an "analytic curve"

a.. £ R, with non-trivial real "branches" at the origin, 1J

and the $-sign of a function

g

selected branch of the curve.

is the sign of the values of

g

on a

Now, what does this mean in light of the

fact that convergence may not even be sensible, and even if it is sensible, a. .x1y-) = h(x,y)

2

may not converge?

The answer is provided by looking

13

i+j=l

a t t h e honest curves

0 = h r ( x , y ) , where

hr(x,y) =

r I

i

The point is the infinitesimal behavior of branches of origin will stabilize.

precisely because

g

h(x,y)

h (x,y) = 0

at the

g(x,y)

on these branches, near the origin.

is positive or negative will be stable.

It is

is not a polynomial, that our branch for

0 = h

differs slightly from that for the branches chosen for all

r.

the (stable) order to which

g

branches of the curves

r -• °° .

•*

We can coherently select branches in the limit, as

r -> °° , and measure the sign of The decision of whether

i

a^.x y J , as

i+j=l

0 = h .

0 = h , and no The value in

g(x,y) ^ 0 r = Z

can vanish on

assigned to

g

is

vanishes at the origin on the selected (In other words, the degree of the first

non-zero term in some power series.) As a concrete example, consider

2 n!x . Given g(x,y), n=l we look at the behavior of g(x,y) infinitesimally close to the origin, along the honest curves y = 2 n!x . As r gets larger, we must look nearer and n=l

244

h(x,y) = y -

nearer the origin, but "convergence" of

Case III (a). one example here.

h(x,y)

is irrelevant.

Y - {m+nT|m,n £ Z, T irrational}, Intuitively, the '£-sign of

A = R.

g(x,y) £ R[x,y]

yielding this value group is computed by restricting dental curve", for example, r./s-

converging to

T

y = x .

We give only

g(x,y)

for an order to a "transcen-

What one really does is find rationals

and then restrict

g(x,y)

to the curves

y

= x

,

or more precisely, to coherently chosen branches of this family of curves.

Case m ( b ) .

F Cj},

A = R.

Again, we give only one simple example.

We might have a series representation (mi/m) < (m2/n2) < ••• ^ F.

y = x *

2

* +x

2

+ ..., where

This representation is purely formal.

finitely truncated formulas define

The

honest algebraic curves with "stable"

infinitesimal behavior near the origin.

We test

g(x,y)

by restricting to

suitably selected branches. It is clear that the "general" total order on

R(x,y) is complicated.

This is consistent with our set theoretical characterization in the preceding section in terms of ultrafilters of certain open subsets. algebraist should only tolerate the orders of type I valuations on fields of functions in

r

variables).

In general, the

(discrete, rank

r

These are more in

line with our philosophy of finite algebraic computability.

The other types

of orders are perhaps interesting to analysts. For example, consider a first order differential equation Q(x,y)dy = 0,

P , Q G R[x,y],

Q(0,0) t 0.

P(x,y)dx +

We ask what interpretation can

245

be made, for an arbitrary real closed field say through (0,0)?

R, of the "solution curve",

Although we do not expect an honest curve, we do have a

procedure for deciding if a polynomial

f(x,y)

is infinitesimally positive

or negative at the origin, along our "phantom" curve.

One approach is to

00

just take the formal power series solution

order, as in Case II above.

y =

2 a.x i=l X

and construct an

However, it is much better to just use the

differential equation itself to decide if a polynomial ought to be positive or negative along this curve, say in the positive x-direction, near the origin. Specifically, first look at (df/dt)(0), where

know

is the evaluation of

f

f(0,0).

y = y(t)

If

f(0,0) = 0, then we want to

is the "phantom" curve and

on the curve

t > 0.

f = f(t,y(t))

This doesn't make sense, of

course, but the result of applying the chain rule does make sense,

Sf <«) - If (o.o) - If (0,0)>$$. since the differential equation says computation of order.

If

(df/dt)(0)

dy/dt = - P(t,y)/Q(t,y).

is positive, then

algebraic solution

is positive in our total

(d 2 f/dt 2 )(0), and so on until we

(df/dt)(0) = 0, we compute

finally reach a decision.

f

If this

Of course, if the differential equation has an

f(x,y) = 0, we are in Case I above, rather than Case II.

This sort of interpretation of differential equations seems quite reasonable and worth further study.

Equations of higher order and behavior

near singular points are topics to be investigated.

Of course, one wants to

make sense not just of the germ of a solution curve at an initial point, but also the continuation of the curve. one wants to associate an order

c

$ x

values of

xQ

in some interval.

o

That is, given an initial point,

R[x,y], with

D R (x) = x , for all K o

This collection of orders will play the

role of the solution curve of the differential equation through the initial point.

246

8.13.

Brief Discussion of Structure Sheaves The material in this section is partly in the form of an outline of a

discussion to be worked out in detail elsewhere.

On the other hand, the

ideas are quite fundamental for our program of algebraizing topology. Let

S

be a closed, semi-algebraic set, identified with the maximal (A,'£) = (A(S) ,»B (S)).

convex ideal spectrum of an affine coordinate ring

Recall from 8.9.3 that this situation is intrinsically characterized as follows. A

is a reduced R-algebra of finite type.

primes

P. C A

$• = 'P [g-J I

w

with on

A -* ITAi.

(0) = Hp., and finite refinements of the weak order

A- = A/P.

ij

The set

There are finitely many '^-convex

such that

*J3 = A HII0&.) , under the inclusion •*• d

1

1

S

is the union of the sets

S i = X(Ai,('Bi) ) , each of

which is the closure of the set of algebraic simple points variety

X(A.,# )

with

g.-(x) > 0.

non-empty strong open set We will refer to such

x

of the real

In the semi-algebraic sets

U{h.} = {y £ S-|h.(y) > 0}

S., every

is Zariski dense.

as -irreducible components of

S. = X(A.('J$.) )

S.

In Chapter V, we constructed rather generally a structure sheaf relative to the Zariski topology on

Spec(A,$).

The global sections turned out to be

the ring

(A CM >., ']) en J

than

Because of the Nullstellensatz, this amounts to inverting all

1.

functions

f £ A

obtained from

with no zeros on

S.

A

by inverting all elements greater

Also because of the Nullstellensatz

we can simplify the discussion by restricting this sheaf to the maximal convex ideals

X(A,'£) = S.

Zariski open set elements of on

A

D(f) C S

x £ S.

f £ A, the ring of sections over the basic

is

(Acr,~ , '£C/-.pO> obtained by inverting all

with no zeros in

D(f). Thus elements of are

are functions nowhere

D(f). The stalks of this sheaf are the local rings

(Ax>'£x)>

Elements of

neighborhoods of

x

Ax

^cr^

those functions in

Acr4~ Ag .f .

D(f). The elements of

negative on

If

may be regarded as germs of functions on Zariski open

and such a germ belongs to

where negative on a Zariski open neighborhood of On the semi-algebraic set

S

£x

if and only if it is no-

x.

we also have the "strong topology", that

is to say, the collection of open, semi-algebraic subsets

U C S.

We would

like to study "sheaves" for this strong topology, but classical sheaf theory

247

is intimately tied to the infinite procedures of point set topology, so we should proceed with some caution.

Perhaps the most natural finiteness con-

dition to impose is that we seek sheaves for the Grothendieok topology on the set of open subsets {IK -* U}

are allowed.

U Cs

in which only finite covering families

When discussing open, semi-algebraic sets and

finite open coverings, it is obviously very convenient to assume Unproved Proposition 8.1.2. of any sets

U C S

This assures us that, essentially, finite open covers

just amount to writing

U{f.} C S.

U

as a finite union of basic open

On the other hand, all the propositions proved in this

section are independent of Unproved Proposition 8.1.2. We have in mind three sheaves of rings, in fact, associated to any semi-algebraic set

E C Rn.

We call these sheaves the rational structure

sheaf, the semi-algebraic structure sheaf, and the smooth structure sheaf. Each structure sheaf corresponds to a category of morphisms between semialgebraic sets,

although we do not study these morphisms here.

These

three categories can be interpreted, within real algebra, as delineating the three subjects, algebraic geometry, algebraic topology, and differential topology. (I)

Let

E C Rn

be a semi-algebraic set, A(E) = A(!) = R f X ^ . ,Xn]/I(E)

the affine coordinate ring of embedding of

E

E.

Throughout this section the particular

in affine space can be suppressed.

invariantly as some dense subset

We can think of

E

n

E = UE.,

E. = Z{f..} U { g . v } C X (A (!) ,'£ (!)) ,

1

of the maximal ideal space of an RHJ-algebra

1

1J

(A,'£)

of a certain type.

IK

None-

theless, it will sometimes be convenient to refer to the distance between points in embedding. form.

Rn,

||x - y||. Thus we do not go out of our way to avoid an affine

The reader can reformulate for himself all statements in invariant

For example, if we say a subset

S C E

then this can be reformulated by saying that all

h £ A

are bounded as functions on

is closed and bounded in S C X(A,'£)

R ,

is closed and that

S.

The rational structure sheaf is very similar to the structure sheaf for the Zariski topology.

We first define it, then prove a few propositions

which enable us to compute rings of sections, in some sense. is (relatively) open, we define

248

I(U) = {f

E

If

A = A(E)|f(U) = 0}.

U C E Then

A/I(U)

is a ring of functions on

U

and we define

A(U)

to be the

localization of this ring obtained by inverting all functions with no zeros on

U.

We define an order

negative on

U.

•£ (U) C A ( U ) , consisting of all functions nowhere

We obviously have a presheaf of partially ordered rings

for our Grothendieck topology and the rational structure sheaf of the sheaf associated

E

is

to this presheaf.

In order to make computations, let us first consider the case where E = S = X(A,*0), where

A

is a finite integral domain over

R, and

is a derived order of a finite refinement of the weak order. $ = 0#,T[g.]) w 1 d Any non-empty basic open set U = U{h.} C s is Zariski dense, hence A(U) = A.

Proposition 8.13.1. order

$' = (:p[h-]), J d

an RHJ-algebra.

C

A

Let

coincides with

Secondly,

D(h) Cx(A,$»), where

U = U{h.} C S , as above.

U CX(A,'|5)

h=nh..

Then, first, the

('£w [g-,h.]) , hence ! J d

Proof:

C

C&Jg^V )d

m

i

j

(CVgi] ) d [ h j ]

w

j

I

since

h. > 0, J ~

II h. ^ 0 J

means all

}

= d

m _> 0}.

*[hj]V

BUt

is nowhere negative on

J d by Proposition 8.6.5.

tp [g.,h.]

It is completely obvious that

D(h) CX(A,'£ ! ),

£ f, rel •£' , some

also, i t is easy to argue that any f ^ 0B[h.]) U(g.,h.}, hence belongs to

which

Finally, (A(U),fl(U)) = (Ag ( h ) ^ s f h ) ^

S(h) = {f <= A|0 £ h

Certainly

is

coincides with the Zariski open set

is the ring of sections in the Zariski structure sheaf over obtained by inverting

(A,$')

d

U{h.} CX(A,$) h- > 0. J

coincides with

D(h) C X C A

The last statement follows from

the Nullstellensatz.

•

In the case under consideration here, a function

f G A(U)

of the

presheaf is completely determined by its restriction to any non-empty smaller open set

V C u.

components".

In particular, this presheaf has no respect for "topological

The effect of the sheafification is exactly to allow independent

rational functions over the components of each More precisely, let us say

U

is connected if

union of two open semi-algebraic subsets. a finite cover of

U

U. U

is not the disjoint

Equivalently, if

by open sets, then for any two indices

{U. -• U} i,j

is

there is

249

a chain

i = i , i1,...,i

= j

Suppose we have functions assume all

g. > 0, and I —

A(U), since

Zg._ > 0

on

such that

U.

f-/g- ^ A(U.)« g. > 0 I

on

U.. I

U = UU..

n U.

Since Then

Suppose

£ 0,

f-/g- = f-g-/g-> we may f/g = If./£g1 1

( f ^ ) \u

i,j.

Then if

A, since

U. n u.

(f/g)I,,

then

U. n u . f- 0, we have is Zariski dense.

= f-/g-> all

sheaf defined by the

belongs to

= C^/g^lu nu ' j i j in the integral domain

n u

i all

k = l,2,...,n.

f-g. = f-g-

It follows that if

U

is connected,

i, hence the section of the rational structure

f-/g- ^E A(U-)

already comes from the presheaf element

f/g e A(U). On the other hand, we will prove below that any semi-algebraic set is a finite union of connected semi-algebraic subsets.

Thus, if an open

is not connected, a section of the rational structure sheaf over of a section over each connected component of

U.

U C E

U

consists

Thus we have proved the

following.

Proposition 8.13.2.

If

U C S

is (relatively) open, S = X(A,$)

ducible and non-degenerate as above, and U

U = U u^

is the decomposition of

into disjoint, (relatively) open connected components, then the ring of

sections of the rational structure sheaf over

If

E

S. = X(Ai, Bi)

and each

A^

is a domain with

finite refinement of the weak order. A -• II A i I(U) =

U is

is any semi-algebraic set, we can write

<

A

irre-

and

$(E) = A n n ^ i -

n p. U ns. £ 0 J 3

= A/I(U)

since if

If

If

p^

A

O

E = U s^, where the derived order of a

A = A(E), we have the inclusion

U C E

is (relatively) open, then

U n s. £ 0, then 3

with inclusion

II(A(U.) /p(U.)) •

I(U n S.) = P.. J J

Let

"* II A/P., the product taken over those

j

with U n s. £ 0. Proposition 8.13.3.

If

U = U{h.} n E

the presheaf sections (A(U) ,$(U)) the localization of such that (That is, $

250

(Ao>'£ )

are obtained as follows.

$

C A oo

is the order

consists of the functions nowhere negative on

E, then

(A(U),'B(U))

obtained by inverting all functions

Z{f} n u { h . } n f C f - E , where I

is a basic open set in

U.)

is

f £ AQ A

nn(D.[h.]) . 031^

In particular,

if

E = E

is closed,

then

(A(U) ,$ (U))

is the ring of sections of the

Zariski structure sheaf associated to the RHJ-algebra Zariski basic open set

Proof: each

D(h) C X ( A Q , ^ o ) , where

h = II h^.

This is essentially the same as 8.13.1.

tp.[h^]

c

A-

(A ,£ ) , over the

Note that from 8.6,

is an order giving an RHJ-algebra, precisely if

V{\} n S. t 0.

D

In general, in the reducible case, it is not true that if

U C E

connected, then every section of the rational structure sheaf over comes from the ring

A(U).

of the connected plane set U 2 , U,

is

U

already

In the figure below, the affine coordinate ring

E

is

and the rational functions

A = R[X,Y]. y^AQjp,

But

E

is covered by

0 €= A(U.),

together to give a globally defined rational function on

U, ,

j = 2,3, fit

E.

A precise

description of the global sections of the rational structure sheaf seems somewhat complicated.

Remark.

As a general rule, when sheafifying a presheaf with respect

to anon-topological sort of Grothendieck topology, the description of sections in terms of elements of stalks at each point which locally fit together is not appropriate.

However, it turns out that the rational structure sheaf defined

above actually coincides (on semi-algebraic open sets) with the classical topological sheafification of the presheaf defined by the

(A(U) ,'£

251

The stalks are clearly the rings the localization of Here

I

=

X

I(U)

lim

A/I

I(U)

(A ,'£ ) , where, if

defined by inverting functions

is the ideal of degeneracy at

x

x^E, g

A

with

g(x) f 0.

(which coincides with

~G%

if

U

is sufficiently small).

a function at

x E E.

s E A y y

U

s

£ A , for all

with elements X

for all

y G u . x

'£ C A x x

A

is a germ of

are the germs which

x.

Suppose given an element has a neighborhood

Thus an element of

The positive elements

are locally non-negative near

f /g

x £ U, such that each

E A(U ) X

X

which restrict to

X

Now, U

not compact, so we cannot assert that finitely many of the

cover

However, let

f

f /g

l^gl'

f

V

= D(g ) C E

a

§ree

on a

those irreducible components of

E

with

U is U. g

i 0.

Moreover, if two functions

u^ n u*2, then they will

U(S. - Z(g 1 ) - Z(g2 ))

where the

S. n \J^ n U 2 ± 0.

is possible to replace the data defined by the fx-/gx-

V .

semi-algebraic open set

also agree on the much larger set

functions

U

denote the Zariski open set defined by

is defined on the much large set

2^ g 2

x

Such data exactly describes a section over

of the classical sheaf associated to the presheaf on question.

Thus

is

f /g

€E U

are

In this way, it

by finitely many

defined on larger neighborhoods of the form

This says exactly that the sheafification of our presheaf

S.

UHvXi

(A(U),$(U))

HS^. with

respect to the classical topology and the finite Grothendieck topology actually coincide.

II.

We next define the sheaf of semi-algebraic functions on a semi-

algebraic set

E C Rn.

Again, by "sheaf" we mean sheaf for the Noetherian-

Grothendieck topology of finite, open, semi-algebraic covers of open subsets of

E.

We first define semi-algebraic functions and establish many basic

properties. Let

f:

E -* R

consider the graph of

Definition.

be a function. f,

Then

E x R

is semi-algebraic, and we can

F C E x R.

The function

f:

E -*• R

is a continuous (semi-algebraic)

function if the graph

F C E x R

semi-algebraic subset

S C E, which is closed and bounded in

252

is a semi-algebraic set, and if for each R n , the graph

of

f

over

R" X R =

R

S,

n+1

F n (S x R ) , is a closed, bounded, semi-algebraic set in

.

Note the definition involves both local and global properties of the graph of

f.

with graphs (f^f^: <

If we are given two such functions F, C E x R

""l ^ F l^ n ^2 ^ F 2 ^ ' S C E, then R

n7T

E •> R

and

F 2 C E x R, consider the graph

f2 :

F, x F 2

E + R x R, that is, {(x.fjCx) ,f2(x)) |x G E } C E X R X R.

^1(x,y1,y2) = (x,yx)

n

and

f •

F,

and

wnicn

TT2(x,y1,y2) = (x,y 2 ), then obviously

i s semi-algebraic.

x

is certainly bounded. 9 i s Fz

If

S

E -*- R, of

If

Fj x F 2 =

is closed, bounded, and

Also, F, x F 9

is

z

is closed in

x R x R , which is directly seen from the formula F

2 ^

2

n

^

S X R

^'

since

sum and

F, x F o = IT" (F, H ( S X R ) ) 1 l l S l Product are polynomial maps R x R -> R,

we conclude from this paragraph and 8.9.4 the following result, by projecting F, X F - J C E X R X R - ^ E X R ,

Proposition 8.13.4.

usin

g

sum or

product

R x R -> R.

Sums and products of continuous semi-algebraic

functions are continuous semi-algebraic functions.

•

As another immediate corollary of 8.9.4, we get a generalization of the remark following 8.9.4.

Proposition 8.13.5.

A continuous semi-algebraic function assumes a

maximum value on any closed, bounded, semi-algebraic set.

•

We have chosen to define continuous functions without the usual and

S's.

The

e-6

e's

definition turns out to be equivalent, although this

is not completely trivial.

Since the

e-6

definition is more convenient

for certain arguments, we will now establish this equivalence.

First, we

need a lemma which guarantees that the local part of our definition of continuous function really carries some information.

The following

proposition is a sort of "curve selection lemma".

Proposition 8.13.6. x £ E

Let

E

a point in the closure of

be a non-discrete semi-algebraic set, E.

Then there exist closed, bounded,

253

one-dimensional, non-degenerate semi-algebraic sets

C = C,, with

x e C C E U {x}.

Proof:

First, we may assume

finite union of such sets.

E = Z{f.} HU{g.}

Then if

E = US.

into irreducible components, we have

since any

A

is a domain, and

the weak order.

$

In fact, the

$, that is, we may assume

is a

is the decomposition of

E

E = U(S. H E ) , and therefore we have,

say, x £ S , H E . This allows us to assume, in fact, that where

E

E = U{g.}CS =X(A,

the derived order of a finite refinement of g.

will actually be subsumed into the order

•$ = ($ [g-]) C A. w J d

Also, we may as well assume

Now we are in a position to apply the signed place perturbation theorem, 8.4.9 (or its extension in the Remark following the proof of 8.4.9) to the domain refining

A = R[x 1> ..x ] .

We know that there is a total order

0 £ R , is 'B'-convex.

order and find a strong chain of 'B-convex ideals (x,...x_) C A, where in

tr.degD(A/P.) = r - i. K l

8.4.9 allows us to assume The set of zeros of

g. £ P r P

semi-algebraic set, x = 0 a discrete subset of

C • .. C P

=

is thus a one-dimensional

is a non-degenerate point, and the . ) . Also, Z(P

Nonetheless, we obtain our "curve"

secting

Z(P

degenerate points of

0 C p

1.

points.

.) C s

We may thus perturb this

Moreover, the Remark following

, x € Z(P _.) C s

Z(P

A,

(x, .. .x^) C A, in

'B = (ftw [g-]) , such that the maximal ideal J d

corresponding to the point

'£' on

g.

vanish on

.)

may contain some degenerate

C

of the proposition by inter-

with a closed ball sufficiently small to exclude all Z(P

.)

and all zeros of the

g., except

x = 0.

•

Next, we need to know that a semi-algebraic function satisfies a "polynomial equation" with the coordinate functions.

Proposition 8.13.7.

If

E C Rn

a function such that the graph of is a polynomial

cp:

E -* R

cp, F C E x R, is semi-algebraic, then there

P(x;....x ,y) such that

x = ( x r . . x n ) € E.

254

is a semi-algebraic set and

P(x,...x , cp(x;....xn)) = 0

for all

Proof:

Express

E = U Z{£. -} H u { g . , } . If i 1] IK Z{£..} U { g . , } , then at least one f. . occurs. ij IK 13 n

(x,t),

a whole interval of values be the graph of a function.

Proposition 8.13.8. function.

Then

graph of

f,

0 < e G R

f

Otherwise

cp(x), and thus

We can now set

P = FI ( 2 f. . ) . i j 1J

E

be a semi-algebraic set,

would contain

F

would not •

f:

E -* R

a

is a continuous semi-algebraic function if and only if the

F C E x R, is a semi-algebraic set and if for all

there exists

Proof: bounded set.

0 < 6 = 6(x,e) G R, such that

First, we assume the

e-6

property.

x G E,

|jx-y|| < 6,

Suppose

yGE

is closed, bounded. since

S

If

is closed.

t = f(x), and

F

a closed,

We must show that

t / f(x), the

near

(x,t)

e-6

FH(SxR) x G S,

property easily implies

and this gives a contradiction.

(x,t) G F H ( S x R ) , as desired.

We will prove that

F H (sxR)

The result is obvious if

is a domain

is bounded by induction on

dim(S) = 0, since then

Also, we may assume

A

S

(x,t) G F H ( S x R ) C R " X R, then certainly

Now, if

there are no points of

points.

S C E,

(We will frequently drop the adjective semi-algebraic, it being

understood that all sets must be semi-algebraic.)

where

F

|f(x) - f(y)| < e.

implies

Thus

(x,cp(x))G

near

Let

t

x € E,

•£ C A

S

f

is a finite set of

is "irreducible", that is, S = X(A,'B) , d

a finite refinement of the weak order.

is justified because from 8.9 any closed ducible pieces and if

S

dim(S).

is unbounded on

S

This

is a finite union of such irreS, then

f

would be unbounded on

one of these pieces. Suppose where

f

satisfies the polynomial equation m

P(x,t) = p Q (x)t + ••• + p m ( t ) .

We may assume

since otherwise we could just drop this term from is a closed, bounded subset of lower dimension. for all

xf G s 1 .

Consider

0 = P(x,...x , f(x,...x ) ) , P Q (x) / 0 G A P(x,t).

Thus

= A(S), Z(p (x)) = S'C S

We may thus assume

U = {x G S|f(x) < b}.

Then

S - U

f(x') < b G R ,

is a closed, 2

bounded, semi-algebraic set, disjoint from all zeros of assumes a minimal, non-zero value on of

P(x,t), for

S - U.

x G S - U , henc * a bound on

P o (x).

Thus

(pQ(x))

This puts a bound on all zeros f(x), for

x G S-U.

As , f

255

is

also bounded by definition on

U, we have

Next, we go back and assume Suppose for some y = y(6) E E EQ EQ

x G E,

with

f

0 < e £ R

|x-y| < 6

= {y E E| |f(x) - f ( y ) | >^ e}.

but

f

bounded on

S.

continuous and prove the and any

e-6

0 < 6 £ R, there is a point

|f(x) - f (y) | _> e.

Since the graph

F

of

Then f

x E EQ, where is semi-algebraic,

is indeed semi-algebraic, as the image under projection

a semi-algebraic subset of

F.

f

is continuous, F H ( S x R ) = F Q

In general, if

FQ C R n + 1

E x R -* E

of

By Proposition 8.13.6, choose a closed,

bounded, one-dimensional, non-degenerate Since

property.

S

with

x E S C E Q U { x } C E.

is closed, bounded.

is a closed, bounded set, and

closed, the squared distance function

d'.

F Q -• R, defined by

B C Rn+1

is

dg(y) =

min ||y-b|| , obviously has the e-6 property. (For fixed y, |y-b| is b E B large for b ^ B far from y and does assume a minimum on any closed, bounded subset of

F

B, hence definitely assumes an absolute minimum on

as above, x E E

the t-axis above

our fixed point, and

x = (x^-.x^.

B.)

We take

B = {(x,t) |f(x)-t| >_ e}, part of

In this case, d g

is easily computed

explicitly:

fz(x!-x,) 2 if (x'...xM ,t (X....XJ if

256

(X-...X-) = (X....X1 .

In particular, it is clear that the graph of and, as already observed, has the

e-6

d R , above

property.

Thus

continuous by the first part of the present proof. a minimum value on establishes the

is semi-algebraic,

2 dfi:

F Q •> R

is

It must therefore assume

F , which it obviously does not do.

e-6

F ,

This contradiction

property for our original continuous function

f.

•

We have some immediate corollaries of 8.13.8.

Corollary 8.13.9. and

U C R

If

f:

E -* R

is a continuous semi-algebraic function

is an open, semi-algebraic set, then

open, semi-algebraic in

E.

Corollary 8.13.10.

If

z €= E

is (relatively) D

E

is a connected semi-algebraic set, f:

a continuous semi-algebraic function, then there exists

f" (U) C E

with

x,y G E, and

E -*> R

f(x) < t < f(y) *= R,

f(z) = t.

•

The following result will be useful in the last part of this section.

Proposition 8.13.11.

Suppose

semi-algebraic subset such that semi-algebraic function.

E

E Cf ,

fQ

Suppose

fQ:

E Q -* R

f:

and which has the

e-6

E "* R.

a

is a continuous

Then there is at most one extension of

continuous, semi-algebraic function which extends

is a semi-algebraic set, E Q C E

fQ

to a

Moreover, any function

property on

E

f

is, in fact, a semi-

algebraic function.

Proof:

The uniqueness of

continuous functions.

f

is obvious from the

Also, given an extension

property, it is clear that the graph of in

E x R

of the graph of

fQ,

f,

f

of

e-6 f

property of with the

F C E X R, is just the closure

F Q C E Q X R C E X R C E" X R.

We made use above of the distance function to a closed set say

dR:

R n "* R,

d R (y) =

e-6

min |y-b|. b £ B always continuous, semi-algebraic.

D

B C Rn,

We now want to prove that

dg

257

is

Proposition 8.13.12. :

R

n

-*• R

Proof:

If

B C Rn

is closed, semi-algebraic, then

is continuous, semi-algebraic.

The

e-6

property of

dn

is immediate from the triangle inequality

D

for distances.

Thus we only need to prove that the graph of

is semi-algebraic.

dg

Rn x R

in

To see this, begin with the subset

D = { ( x , b , ||x-b||)|x G R n, b G B} C Rn x B x R . D

is obviously semi-algebraic, hence by Tarski-Seidenberg, so is the image

of

D

TT: R n x B x R -• R n x R.

under projection

Now, ir(D) C R n x [0,co),

and the subset

D1 = { ( x , t ) | x G R n , 0 < t e R ,

is also semi-algebraic.

This again follows from Tarski-Seidenberg since

is defined by an elementary sentence. now be described as

[ 0 , t ] n ^(D) = 0}

1

(F

dg :

Note that the graph of

!

n

- D ) U (B x {0}) C R

x R.

Thus

dg

D'

R n -• R

can

is a semi-

algebraic function.

•

As a corollary of this argument and 8.13.8, we can prove that continuous semi-algebraic functions are uniformly continuous on closed, bounded sets.

Corollary 8.13.13. and

f:

S -* R

there is a

Proof:

If

S C Rn

is a closed, bounded semi-algebraic set,

a continuous semi-algebraic function, then for all

0 < 6 G R

such that if

For each

x,y G S

x G S, the set

S

—————

and

= S

X,£

0 < e £ R,

||x-y| < 6, then

= {y G si If(x)-f(y)I > e} is X

I

—

a closed, bounded semi-algebraic set, by Corollary 8.13.9 (possibly but the set of such

x G S

with

S

obviously any modulus of continuity points

x.

= 0 6

so is

258

Then

BQ

It is not hard to prove that

AS = {(x,x)|x G s} C s x S, say using the

= 0,

e-value at these

Anyway, the rest of this proof goes through even if some

B = B Q C s x s.

S

is open and can be discarded, since

will work for our

B Q = {(x,y)|x G S, y G Sx> C S x S.

Let

|f(x)-f(y)| < e.

e-6

S

=0.)

is semi-algebraic, hence S H B = 0, where

property of

f.

Thus, the

distance function 6'

on

AS.

dR:

AS -• R

assumes a minimum, strictly positive value

We are here computing distances in

no greater than the distance from if

0 < 6 < 6', then

for our given

e.

x

to

S

R n x Rn.

in

Obviously, 6 !

R , for any

<5 is a uniform modulus of continuity for

(If

B = 0, we define

is (relatively) open in

E, we define

C°(U)

with

f -Li fin = fi.

^

=

U = Uu.

Remark:

E.

If

U C E

'p°(U) C C°(U)

the

It is obvious that this presheaf is a

is a finite open cover of

(U)

S •

U, and

^*ln fin ' then there is a unique

Moreover, f

on

to be the ring (by 8.13.4) of

continuous R-valued semi-algebraic functions on subset of nowhere negative functions.

f

Thus,

d g = + °° .)

We now define the sheaf of semi-algebraic functions on

sheaf, that is, if

x £ S.

is

if and only if all

U

and

f £ C (U)

f i G C°(U i) with

£i ^ ^ ( U p .

Semi-algebraic functions are not infinitely collatable in

general.

Graph of

f

U = (-1, +1)

In the figure, f

is a function on an open interval which is locally but not

globally semi-algebraic.

Remark:

Very little work is required to extend the notion of continuous

R-valued semi-algebraic functions to vector valued functions where

E C Rn.

First, the graph

one can give either the

e-6

F C E x Rm

f. :

E -* R,

f = (^•••fm)

must be semi-algebraic.

Then

S C E.

With either definition,

continuous exactly if all the coordinate functions

1 <^ i <^ m, are continuous.

equivalent, by 8.13.8.

E -* R m ,

definition or the closed, bounded property of

the graph over any closed, bounded subset one proves

f:

Thus, the two definitions are again

It is also easy to prove that compositions

E ~> E' -*• E"

259

of continuous, semi-algebraic functions are continuous, semi-algebraic, where E C Rn,

E1 C Rn ? ,

if and only if

E" C R n", and that

go f G C°(E), for all

f:

E -+E 1

is semi-algebraic continuous,

g G C°(E').

We end this subsection with a proof of the fact that any semi-algebraic set E

E

has only finitely many components.

is connected if

open and closed in

E E.

Recall that a semi-algebraic set

has no proper semi-algebraic subsets which are both We want to prove that any

E

is a finite union of

connected semi-algebraic subsets which are both open and closed in

E.

These

subsets are obviously maximal connected subsets, which can then be called the connected components of

E.

Although we have notions of open sets, closed sets, connected sets, and so on, we must be careful not to assume facts from general topology which use infinite techniques inappropriate in our case.

For example, we cannot

routinely make use of the union of all connected semi-algebraic subsets containing a point

x G E

(which is how connected components are constructed

"topologically") because there is no reason a priori that this union is semialgebraic. If

E

E 1 ,...,E m> then we

is any union of finitely many connected sets

can consider maximal connected subsets of the form

are routinely shown to be both open and closed in components of

E.

sets, we can write

E^ U ... U E• . x x l k

These

E, hence are the connected

If an E exists which is not such a finite union of connected E = Uj U uj

are open and closed in

E.

where

l^ n uj = 0

Then, say, U-,

splits

producing an infinite strictly decreasing chain and closed subsets of

E.

The sets

and both

l^

and

U|

U-, = U 2 ^ Ui, and so on, U^ ^ U 2

V i = LL - U. + ,

c

U- ^ •••

of open

then give an infinite

collection of pairwise disjoint, open and closed, semi-algebraic subsets of

E.

We refer to this process as a splitting process. Clearly, we need only consider E of the form Z{fi> n U { g . } C R n , where 1 _< j _< k, say. Then E is the image of the real variety V = Z{f.,y.g. I 3 j - 1} C R n x R k = R n + k , under projection and we adjoin new variables since if

260

V. C E,

IT: R n + k -+ R n , where

Xi***^-

Tnus

it:

f^g. e

R^.-.xJ,

suffices to study varieties,

i > 1, are disjoint, open and closed subsets, then so are

IT

(V^)

c

assume

V.

V

If some

V

is not a finite union of connected sets, we may

is irreducible and of least dimension with this property.

that any O-dimensional semi-algebraic set is finite, so Let

VQ C v

singular set. where

E^V

be the algebraic simple points and

Then

is connected.

closed subsets.

Each

We split

E.V

V = l^ U UJ

is either in

continues with, say, Uj = U*2 U U£.

U,

Those

Now, the conclusion we want is that

VQ

dim(V) > 1.

EV = V - V Q

dim(EV) < dim(V), so we can write

Note, if

U|

in

into disjoint, open and

or

Ui.

Z^V C ^

This splitting process

are either in

V^

U1

less than

contains all

as above.

l£J, then

of the

splitting process.

U£.

V\ ,

i >_ 1, which

U1 C V Q

is one such set.

U*2.

E^V

If the

U^ = U 2 U U', then we

If not, we must start over and go back and split

V^ = U^.

and some in k

or

R .

UJ, but at least we have are in

U2

actually contains infinitely many

splitting process continues infinitely, starting with get all our

the

EV = Z^V U ... U E k V,

pairwise disjoint (relatively) open, semi-algebraic subsets are actually dosed

Note

Another case to consider is if some

Then we haven't even found

in whichever of

l^, U'

EiV

V 1 , but there are

begins the infinite

It follows, then, that sooner or later the splitting

process gives infinitely many disjoint, relatively open in

V\ C v , which are semi-algebraic, pairwise Rn.

V , and closed in

We are now in a position to use Whitney's proof [44] of the finiteness of the number of components. V

and a point

cuts the manifold

x

G V,

V..

and

xQ.

Let

g:

V -• R

transversal ly at x

be the function

x .

V^

connected sets. arranged that

In fact, we can assume that

g(x) = ||y-x| 2 . V

is a closed, semi-algebraic set in

value at, say, x^ G v^.

Then

x^ G V , hence

V

of lower dimension.

V

If

I(V) = (f±)

= {x1 G v|rank(df^(x),dg(x)) = r}. R n,

g| v

assumes a minimum

is not a finite union of

On the other hand, by our choice of x Q f- V , hence

||y- x || centered at

lies in the tangent plane to V..

codim(V) = r, we consider the subvariety

Since each

We bhoose a point

so that the sphere of radius

the normal vector to this sphere at at

dim(V-) = dim(V ) > 0, since, in fact,

is homogeneous, in the sense that it is a manifold.

y GR y

We know

y

and

is a proper subvariety of

x Q , we have V, necessarily

This contradiction proves our desired result.

261

Proposition 8.13.14.

Every semi-algebraic set is a finite union of pair-

wise disjoint, connected, open and closed, semi-algebraic subsets.

Remark: of

E

A further result is that if

which contains

x

is the set of

semi-algebraic "path" p:

•

x G E, then the connected component

y £ E

[0,1] -> E with

such that there is a continuous,

p(0) = x, p(l) = y, where

[0,1] C R

is the unit interval. We will not prove this result here, although the results of this chapter are sufficient for constructing a proof.

Essentially, one

must look closely at global stratification and local geometry near a point. Note that if our ground field is the field of real numbers, it is not obvious from the considerations leading to 8.13.14 that a semi-algebraically connected set is topologically connected.

However, since

[0,1] C K

is both

topologically and semi-algebraically connected, the fact that components are path components in general, does imply this fact about real numbers.

IE.

Finally we discuss briefly smooth semi-algebraic functions.

simplicity, we restrict our attention to open subsets

n

U C R.

For

We begin by

giving a little more structure to the graph of a continuous semi-algebraic function

f:

U -• R.

Proposition 8.13.15. and suppose

P(x,...x ,y)

P(x,f(x)) = 0 ,

x £ U.

(b)

If

P^

f:

U -* R

is continuous, semi-algebraic

is a polynomial of least degree in

Let

irreducible factors in (a) The

Suppose

P =

p # #p 1 --- r

be the

are distinct, that is, P

P± = P/Pi

and

P

into

is square free.

U, and

P^CxjfCx)) = 0

P.

vanishing on the graph of

IL

are

if x *= U^.

(P^) C RfX-^.-.X ,Y] are convex prime ideals.

(d) dimension (U - U U\) < dimension(U) = n.

262

factorization of

U\ = {x e uli^U.f(x)) * 0}, then the

(c) The principal ideals

If some

such that

RfX^-.X ,Y], then:

non-empty and disjoint open subsets of

Proof:

y

divided

In particular,

UCUJj..

P, we would find a polynomial of lower degree

f by dividing by

P^.

Similarly, if some

u\ = 0,

then

Pi

and

P.

would vanish on the graph of

have lots of zeros in

R

(P.) .

that the algebraic simple zeros of IJL. of

u\ f

are obviously disjoint, over

U-.

follows now from

8.8.6, since the

Moreover, the other results of 8.8 imply P^

are dense on the graph of

f

over

In particular, no polynomial can vanish on any open subset of the graph f

over

IL

Suppose graph of

f

unless

P^

U - U U. over

V.

divides it.

contained an open set Then

But all

P-

vanish on

dense in

P. U

Let

F C R

denote the

is convex and must have

n, which is therefore principal, say

(Q).

F, and applying the arguments of the paragraph above

Q, we would deduce that all

since the

V.

I(F) CR[X,...X ,Y]

some associated prime of dimension

to

The

obviously must vanish on the graph of

The convexity of the ideals (P.)

f.

P. G (Q). As this is clearly impossible

are distinct irreducible polynomials, we conclude

Uu.

is

and (d) follows.

•

We now want to define the subring entiable semi-algebraic functions

f:

C 1 (U) C C°(U) U -• R.

of continuously differ-

We first assume

f G C°(U),

then assume the limits f (x, . .. ,x.+e,.. .xn) - f (x, .. . x j (3f/3x.)(x) = 1

exist in

R

the

property.

£-6

x G u

for all

lim e - 0

x G u

and define functions

(3f/8x i ):

U -* R

We can apply the chain rule to our relation

with

0 = P(x,f(x)),

of lowest degree and this gives

0 = 3P(x,f(x))/3xi

= (3P/3Xi)(x,£(x)) + ((3P/3y)(x,f(x))((3f/3xi)(x))

Since

(3P/3y)(x,f(x) ^ 0

where

P = P^-.P

is dense in

U.

U, and because no

as in 8.13.14, we deduce that The function

since it can be written 8.13.11,

on

3f/3x i:

U -• R

Sf/Sx^

Pi

divides

3P/3y

V = { x G u | (3P/3y) (x,f (x)) i 0}

is obviously semi-algebraic over

(-(3P/3xi)/(3P/3y))(x,f(x)) also belongs to

if

x G V.

Thus by

C°(U).

263

V,

We refer to such functions The set of such

C (U) C C°(U)

U -* R

as

is a subring.

r

by requiring that r,

f

C -semi-algebraic functions. By iterating the procedure,

f G C r (U) C C ^ ^ U ) C ••• C C°(U),

C -semi-algebraic functions

we can define

including

f:

have continuous partial derivatives of order up to and

1 < r < °° .

hence we have a sheaf of

Clearly,

Cr-functions are finitely collatable,

Cr-functions, defined on (variable) open subsets

U C Rn. There is no difficulty whatever now in extending the inverse function theorem, Proposition 8.7.1, to the case of a (Rn,0) -* (Rn,0) injectivity of

C^-map

with non-singular derivative Y

Y = (Y....Y ) :

( O Y - ^ X . ) (0)) .

The local

is proved by standard estimate arguments using the

hypothesis of differentiability.

The local surjectivity of

Y

is proved just

as in the earlier proof of 8.7.1, using the minimum value property of continuous semi-algebraic functions on closed, bounded sets. In fact, we will sketch another proof, similar to a standard proof in the classical case of real numbers.

Beginning with any

x 1 ...x n

y, ...y

near

0, and

C -functions

(3y./9x.)(O) = id, one shows that

of the

y.. ,x_,... ,x

C -coordinate system

x., with

is a

C

y(0) = 0

and

coordinate system.

This argument uses "completeness" in the form of the intermediate value property on intervals for continuous functions.

Since an interval is

connected (in our sense), we have the intermediate value property from 8.13.10 at our disposal.

The inverse function theorem is then proved by

iterating this substitution procedure. Note that even if originally the functions and the the

y"2»»«»»y

will be

y\

x^

are the standard coordinate

are polynomials, at the second step of this proof,

will not generally be polynomials in

C -semi-algebraic functions of

y, ,X2»**«>x , but

y,,x~,...,x . Thus, this proof

would not have been feasible in the special case dealt with in 8.7. The

C°° theory is quite different from the

example, if

U C R

is open, r < °° , then

Specific examples are easy. if

x _> 0

x £ 0.

264

and

Then

f(x) = 0 f-g = 0.

if

Let

Cr

C r (U)

f,g G C ((-1,+1))

x £ 0,

g(x) = 0

if

theory, 1 £ r < °° .

For

is not an integral domain. be defined by x >_ 0

and

f(x) = x

g(x) = x 4

if

If U

is connected, then

if U = U U.

U

is an integral domain.

is the decomposition of

U

r

components, then obviously on

C (U)

(In general,

into disjoint open connected

r

C (U) = IT C (U.)»

0 £ r £ ° ° .) The C°°-functions

are known as Nash functions and have been widely studied in the case

of ordinary real numbers, [30] through [42]. following.

We use the notation n

algebraic functions at

x £ R ,

Proposition 8.13.16.

Cx

for the ring of germs of

C°° =

lim

P(x,f(x)) s o ,

(b) The function C (U) "* C

U

a connected, open,

f £ C (U). Then:

(a) There is an irreducible polynomial such that

C°° semi-

C°°(U).

0 ^ U ^ Rn,

Suppose

semi-algebraic set, and suppose

The basic result here is the

P(X 1...X n ,Y) G R ^ . - . X ^ Y ]

x G U.

f

is determined by its germ

[f] E C~, that is,

is injective.

(c) The germ

{f} =

2 K.—U

[f] £ C^

2 2JX

is determined by its formal power series

(i ) !

(i ) !

-L

Tl

• ~"iC

3

i 1

that is, C* "*-R[[X . ,.Xn]] (d) The power series

in (°)X1

'"

X

R

n

tE x i•' t X n ] ] >

C/A._ • • • O-A. n

is injective. {f}

is a formal solution of the equation

P ( x r . . X n , {£}) = 0. (e) Given an irreducible polynomial series

{f}

PCX^.-X ,Y) and a formal power

P(X 1 ...X n , {£}) = 0, then there is a germ

such that

[f] G C~

with underlying power series {£}.

Sketch of Proof:

Let

P(x,f(x)) = 0 ,

P = P^-.P

8.13.15 into distinct irreducible factors, UL C u P/P.

is non-zero.

factored as in

the open subset where

To prove (a), we must prove that no

x E U

belongs to

U. n U.,

i ^ j , because it then follows from connectedness that there is

only one

U..

If

We may as well prove

0 G u" , then we can find (xQ,f(x Q)) ^ 0.

Near

O ^ U . HU., x G u,

i^j.

arbitrarily near

(x Q ,f(x o )), the graph of

f

0

such that

coincides with

265

the zeros of

P^

The non-vanishing of

compute all partial derivatives the coefficients of

(3P1/8y) (xQ,f ( X Q ) )

(8 f/8x )(x Q ),

I = (i,...ijj, in terms of

P.^, by simply iterating the chain rule and using the

identity

0 = P^XjfCx)), near

variable

x, then

xQ.

For example, if there is only one

0

= fr tvf<xo» + IT cvf<xo» f
0

• i & (v f < x o» + 2 £ & (xo'f(xo» f

and so on.

In fact, these expressions are just the coefficients (up to constant

factor) of powers of

(x - x Q )

obtained by formally computing 00

where

allows us to

f(x)

is replaced by the power series

P^(x,f(x)),

no

k

(l/k!)f^ ; (x )(x-x ) .

2

k=0

The

°

multivariable analog is also true, of course. Now, {f},

we want to assert that the formal power series of

f

at the origin.

as in parts (c), (d) of the proposition, is a formal solution of

P,(X, ...X , {f}) = 0. finite expression by continuity of

But each coefficient in this formal expansion is a

which, by continuity of all P,

9 f/8x , and also of course

and its derivatives, can be evaluated as

1

lim

x

the corresponding coefficients of the formal computation about

of

0 -*°

x

^U,.

But we know these coefficients vanish, hence so do the coefficients at the origin.

This proves (d).

The polynomial ring and power series ring are integral domains R[X 1 -..X n ] C R[[x l ...X n ]]. over

Thus, if some power series

f

is algebraic

R[X 1 -..X n ], there is a unique irreducible polynomial (up to factor in

R[X . . . X J ) ,

P e R [ X r . . X n ] [ Y ] , with

P(X 1 ...X n , f ) s 0.

conclude from the paragraphs above that we cannot have

We therefore

0 *= U. n u.

This proves (a). But also, we conclude that if the power series then the irreducible polynomial This means that the function

P(X 1-..X n,Y)

i^j.

{f} = 0,

is just the polynomial

Y.

f = 0, and (b) and (c) are proved.

Part (e) is quite interesting.

266

if

However, we prefer to defer the proof

until we make a more detailed study of power series rings in a subsequent chapter.

•

Remark:

The

C

structure sheaf is analogous to the rational structure

sheaf, in the sense that a function

f:

U -* R

function in each germ, that is, [f] £ C , all C (U).

In other words,

C

which is a

C -semi-algebraic

x G U, is in fact already in

functions are infinitely collatable.

267

Appendix

THE TARSKI-SEIDENBERG THEOREM

In this appendix we give Paul Cohen's proof [62] of the Tarski-Seidenberg theorem [56], [57], which we state in the following geometric form.

Proposition A.I.

Let

E C Rn

the projection onto the last

(n-1)

be a semi-algebraic set, coordinates.

Then

ir: R n -> R11"1

TTE C R

is semi-

algebraic.

Actually, the proof provides an algorithm for writing down semi-algebraic set, in terms of the polynomials used to define

TTE as a E.

However,

we will leave to the reader the details of keeping track of this algorithm through the various steps of the proof. Proposition A.I can be reformulated in logical language as "elimination of quantifiers".

Specifically, by a polynomial relation

we mean a finite sentence built up from basic relations where

p

is a polynomial with coefficients in

tives "and", "or", "not". such that the sentence algebraic set in the type 3 G

or

R

p(x, ...x ,t.....t ) > 0,

R, using the logical connec-

Thus, the set of points

A(x.....x ,t,...t )

(x_...xn,t,...t ) £ R n + m

is true, is just a general semi-

- B y an elementary sentence, we mean a sentence of

(Q.^) • • • (Qnx n)A(x,.. .xn ,t, .. .t ) , where the

V, and

A

A(x . . .x jt^. . .t )

is a polynomial relation.

R m |(Q 1 x 1 )•••(Q n x n )A(x 1 ...x n ,t 1 ...t m )}.

Qi

Consider the set

are quantifiers, ^C t i--- t m )

Another form of the Tarski-Seidenberg

theorem is that this set is semi-algebraic.

Proposition A. 2.

268

There is a polynomial relation

B(tj...t )

(i-n

an algorithm for producing

B), such that

BC^.. .tm) •(Q 1x 1 ) • • • ( Q ^ ) •

A(x 1 ...x n) t 1 ...tJ. Since

(Vxn)A

is equivalent to

~(3x n )(~A), it is clear by induction

that Proposition A.2 needs to be proved only for is what we actually prove below. m

R |(3x 1 )A(x 1 ,t 1 -..t m )} set

n = 1

In this case, the set

is just the projection to

{(Xpt-L..-tm) E R m +

\k(xlitl.. -t m )}.

R

m

and

Q^ = 3.

This

{(t,...t ) £ of the semi-algebraic

We see therefore that Propositions

A.I and A.2 are equivalent. As an example of the sort of sets described by elementary sentences, we mention the topological closure of a semi-algebraic set E = { ( t r . . t m ) eR m |(Vx o )(3x 1 )-..(3x m )(x o = 0 or m o i

E C Rm,

^...xJGE

and

. y w < *o}In order to prove Proposition A.2, we need to introduce temporarily a new notion.

Consider functions

values in

Note that the set of polynomials in one variable of degree

<_ d

R.

is an affine space over

R

f

with domain

of dimension

(a . . . a ^ ) , the coefficients of a polynomial

a

subset

X C R

and

d+1, with coordinates

q = a x + ••• +a^.

Thus, it

makes sense to talk about semi-algebraic subsets of the space of polynomials of degree

<_ d.

Definition A.3. all

D

The function

f:

and all semi-algebraic subsets

Rm|q(f(x)) G E}

X -»- R

is psenri-algebraic if for

E C R, the set

+

{(q,x) G R

x

is semi-algebraic.

Note that the definition guarantees that the domain of a psemi-algebraic function is a semi-algebraic set. to whether

t > 0, t = 0, or

If

t < 0.

rather simple, it is easy to see that and

X = +1,0,-1

algebraic.

the sets

t £ R, let

sign(t) = +1,0,-1

according

Since semi-algebraic subsets of f

R

is psemi-algebraic if for all

{(q,x) e R d + 1 x Rm|sign q(f(x)) = X}

are d

are semi-

It is also clear that polynomials are psemi-algebraic.

We will

need to know that differences of psemi-algebraic functions are psemi-algebraic. However, it is not much harder to prove the following.

Lemma A.4.

If

Si»'##>Sic

are

psemi-algebraic functions of

m

variables

269

and

f

is a polynomial in

k

x

psemi-algebraic function of

Proof:

ar

+

Now

q(f ( y ^ . .yk)) = Q(y 1 ...y k ) =

Q^ = Qi( v]/--y k p

only on the coefficients of

f.

xR m |sign q(f(g (x),...

{(q,x) £ R

ficients are polynomials in the coefficients

it follows that the set

is a

x = ( i--- m )-

e semi-algebraic.

* *' + Q e , where the

f (g-^Cx), ... ,gk(x))

x

We must show that the sets

•••»g k 00)) = A} Q o yf

variables, then

are

polynomials whose coef-

(a^.a^)

Therefore, since

of

g k (x)

q, depending

is psemi-algebraic,

{(q,y1-•.yk_1,x)|sign Q ( y r • - y k _ r g k(x) ) = A}

is

semi-algebraic, hence can be expressed as a Boolean combination of sets of the form

sign F(a Q . . .a^y.^.. .y^-^x) = u, where

similar reasoning, the set

x

x

k i( )>Sk( ))

=

is

^

Continuing this type of argument, we

{(q,x)|sign QCg-^Cx),.. . ,gk(x)) = A}

is semi-

algebraic, as desired.

£j_(b) < £ 2 (b) < • • •, defined.

+

*•• + b n

be a polynomial with real roots

b = (b Q ...b n ).

Of course, the

£^

are not everywhere

The main result needed to prove Proposition A.2 is the following.

Proposition A.5.

Proof: n = 1.

D

p(x) = b Q x n +b^x 1 '

Let

The

^(b)

are psemi-algebraic functions of

We argue by induction on

Since the derivative

(nbQ,(n-l)b1,...,b

p'(x)

The data

has degree

_< n-1, with coefficients

^), we have by induction that the roots of

p'(x), say

sign(bQ),...,sign(bn),sign pCnjCb)),...,sign p(nk(b))

the number of roots of the intervals

of

p(x)

as well as locating each

and on

This is so because C~°°>Tli)> (iiv*00)*

an(

p(x)

p(x)

270

^(b)

in one

^ because the

at

sign(b-)

determine the

J

± °°. {(q,b) G R d + 1 x R n + 1 1 sign qC^Cb)) = \}

We must show that the sets semi-algebraic.

determines

is monotonic between any two roots

IK behavior of

b = (b Q ...b n ).

(- °°,ri-.(b)), (n- (b) ,n-+ 1 (b)) , (nk(b),°°), or among the points

n-^Cb) ,.. . ,n k(b) . p'(x)

£. (b)

b.

n, the Proposition being trivial if

n-j^b) < n 2 (b) < ••• < Tik(v<j (b), are psemi-algebraic functions of

of

By

{(q,y.....yk 2,x)|sign Q(y,.-.yk 2»

semialgebraic.

eventually conclude that

is a polynomial.

{(q,y1-..yk_2,x)|sign F(a Q ...a d ,y 1 ...y k _ 2 ,g k _ 1 (x),x)

is semi-algebraic, and it follows that g

F

We can divide

q

by

p

and write

are

q(x) = s(x)p(x) +r(x),

where

degree(r) < degree(p).

Then

q(^(b)) = r ^ C b ) ) , hence

{(q,b)|sign qCqCb)) = A} = {(q,b)|sign r(q(b)) = X}. ficients of

r

are polynomials in the coefficients

b =

(b0---bn)

of

r(x)

of

q

and

p.

b

psemi-algebraic functions of

= 0, b, ^ 0, and so on.

(a,b).

signCr^Cb) -Yj(a,b)), sign p(y^(a,b)). a psemi-algebraic function of

E $ = {(a,b) e R sign r(^i(b))

is constant on

(a,b).

x R n+1 |(a,b)

(Strictly speaking, there

contains $

$

E$.

sign(b.), sign p(r|.(b)),

By Lemma A.4, r ^ O ) - Yj(a,b)

is

Therefore, any one of the finitely

$}.

Therefore, the set

is determined by

which interval

^ ( b ) , or whether

We claim that

{(q,b)|sign r(^i(b))=

$, we observe first that

(- °° ,ru (b)), (n- (b) >r\., (b)), or 1 3 j+1

^ ( b ) = n-(b), some

Y-(a,b)

of

r(x)

j.

lie in this interval.

^(b),

or

whether

a b

^ ( b ) = Y-( > )>

(- 00 ,Y 1 ), (Yj>Yj+ 1)>

is constant on the intervals

K

some

3-

Thirdly,

p(Y-(a,b))

(-°° ,Y-i(a>b)), (Y- (a,b) >Y- + 1 C a > b )) > o r

determine the interval

(ru (b),°°)

Secondly, we know from"

is monotonic on this interval, hence the values of sign

which contains

bQ ^ 0

E,, hence is semi-algebraic.

sign rC^Cb))

which of the roots

p(x)

r(x), namely if

realizes sign data

is a finite union of some of the

we know from

y. < • • ' < Y m

$ , for this sign data determines a semi-algebraic set

d+1

To see that

and

This causes no problem.)

Consider the sign data consisting of

many possibilities,

a = (a o>..a,)

Thus, by induction, the roots

are cases to l?e distinguished in the formula for or if

Moreover, the coef-

But

(Ym>°°~) > arid

(Ym(a,b),°°) s

ign

our

r

(x)

claim is

established.

D

Proof of Proposition A.2: Seidenberg theorem. polynomials

It clearly suffices to show that given finitely many

^Cx-j^.-.x^

sign f.(x,...x ) = A.} in

Xj

signs \^3 the set

and

is semi-algebraic.

(x 2 ...x n ).

f^, say

^•(x2-•-xn)»

Consider the sign data

^ i -, say

£^ <_ ^

are

orderings, which partitions

R ~

f-

as polynomials

(x 2 ...x n ).

The real

psemi-algebraic functions of

sign(^i- - ^ . , . , ) , which orders the

£ * * * £ £t* n

{(x2 ...xn) G R n" | (aXj)

Regard the

with coefficients which are polynomials in

roots of the

roots

We can now complete the proof of the Tarski-

There are finitely many possible

into disjoint, semi-algebraic subsets.

271

If

(x 2...x n)

is fixed, the

C - 0 0 , ^ ) , ( £ . , £ . , ) , (£,«). sign data

sign ^ ( £ ^ - 1 ) , sign f j ^ — — 2 sign

is obvious that the truth of

Hence

have constant sign on the intervals

(In fact, these signs are computed by the

one could use the sign data

the data

f^

sign f ^ - " * ^ ) ,

3+

)* sign

f

i^t+1^'

(3 f ^ B x ^ (£.), f

x

x

k > 0. See 7.2.5.)

(3x^) sign i( ;j/-' n)

sign f^.),

sign f ^

{(x 2 -. .xR) | (3x1)sign f ^ x ^ . - x ^ = Xi>

Alternatively,

=

\

It

depends only on

,£ j+ 1 ) , sign

f^,-).

is semi-algebraic.

•

We can now use elimination of quantifiers to eliminate the notion of psemi-algebraic functions.

Proposition A.6.

A function

if and only if the graph of

Proof: function set.

f

f,

f:

X -> R,

X C R m , is psemi-algebraic

F C R m x R, i s a semi-algebraic set.

A very special case of the definition of a psemi-algebraic shows that the set

But this is Conversely, if

polynomial relation {(q>x)I(3t)(A(x,t) Proposition A.2.

272

This observation is due to Efroymson [36].

{(t,x)|f(x) -t = 0}

is a semi-algebraic

F. F

is semi-algebraic, let

A(x,t). and

Then

F = {(x,t)|A(x,t)}

for some

{(q,x)|sign q(f(x)) = X} =

sign q(t) = X]

which is semi-algebraic by •

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VUT. Basic Algebra Texts 63. M.F. Atiyah and I.G. Macdonald, Introduction to Commutative Algebra, Addison-Wesley, 1969. 64. N. Jacobson, Lectures in Abstract Algebra, Vols. 1-3, Van Nostrand, 1951, 1953, 1964. 65. N. Jacobson, Basic Algebra I, W.H. Freeman, 19 74. 66. S. Lang, Algebra, Addison-Wesley, 1965. 67. B.L. van der Waerden, Modern Algebra, Vols. 1,2, Frederick Ungar Publishing Co., 19 53. 68. 0. Zariski and P. Samuel, Commutative Algebra, Vols. 1,2, Van Nostrand, 1958, 1960.

277

Notation

(Introduction not included)

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278

Index

(Introduction not included)

absolute hull, 53, 54 affine coordinate ring, 173, 201, 219 affine order, 41 Archimedean closed subfield, 140 Archimedean extension, 138, 139 Artin, E., 42, 100, 122, 130, 131, 187 188, 196 associated primes, 59, 60, 61, 73, 95, 96 bounded filter, 235 center of a filter, 235 closed semi-algebraic set, 163, 172, 201 closed, semi-algebraic set, 164, 215, 219, 220, 258 closure of a set, 163 codimension, 212, 213 Cohen, P., 42, 165, 268 concave multiplicative set, 83, 93 connected semi-algebraic set, 249, 257, 262 continuous sections of a sheaf, 115, 116, 117 continuous semi-algebraic function, 252, 253, 255, 256, 257 contraction of an order, 35 convex set, 33 convex hull, 46, 47 Dedekind cuts, 137, 138 degenerate points, 197 derived order, 39, 42, 64, 81, 97, 98, 188, 199, 200 derived set, 36, 37, 38, 39 dimension, 212, 219 direct limit, 103, 105 Dubois, D., 187 Efroymson, G., 216, 227, 272 elementary sentence, 268 extension of an order, 35, 40, 120, 121, 122, 130 Fat City, 219 fibre product, 101 fibre sum, 102, 104, 105 filter, 43, 233 filter at infinity, 235 formally real field, 34, 149

Hilbert, D., 189 Hilbert 17th problem, 42, 188 ideal absolutely convex, 52, 53, 54, 55, 88, 89, 156 convex, 45, 46, 52, 81 maximal convex, 49, 50, 63, 87, 93 167, 185 minimal prime, 58, 60, 91, 92, 108 primary convex, 57, 64, 74, 75, 82, 89, 94 prime convex, 49, 51, 63, 66, 74, 90, 93, 127, 186 Jacobson radical, 85, 86, 99 Krull dimension, 224, 225 Krull, F., 139, 149, 150 Krull valuation, 146 Lang, S., 130, 184 localized order, 77, 79, 80, 81, 82, 111, 112, 113 maximal order, 33, 37, 38, 39, 44 morphism, 32 Nash functions, 265 nil radical, 37, 46, 47, 48, 51, 90, 167 Noetherian-Grothendieck topology, 248, 252 non-degenerate points, 197, 198, 199, 201, 202, 215, 218 open semi-algebraic set, 163, 164 open, semi-algebraic set, 164 order, 32 partially ordered ring, 32 partition of unity, 44 polynomial relation, 268 (POR), 32, 34 (PORCK), 33, 34, 55, 56, 88, 110, 114, 117, 119 (PORNN), 33, 34, 56 (PORPP), 33 prefilter, 233, 234 product order, 44

279

psemi-algebraic function, 269, 270 272 quotient order, 45, 52, 56, 58, 81, 201, 202 rank, 213 real closed field, 130, 131, 132, 134 135 refinement of an order, 33, 80, 81 96, 97, 150, 151, 157, 189 RHJ algebra, 167, 168, 169, 170, 171, 199, 200, 201, 202, 219 Risler, J. J., 187 Schreier, 0., 122, 130, 131 semi-algebraic set, 162, 268 semi-field, 49, 50, 85 semi-integral extension, 126, 127 154, 155, 170 semi-Noetherian ring, 57, 58, 95, 96 shadow of a set, 83, 84 shadow of 1, 84, 88 sheaf of partially ordered rings, 41, 112, 249 signed place, 145, 148, 161, 184, 186 190, 206, 228, 242 simple point, 213, 215 simple refinement, 36, 37 simplicial complex, 43, 69 split valuation, 147

280

Stengle, G., 100, 173, 189, 193, 195, 196 strong 'j^chain, 225, 226, 228, 231 strong topology, 247 structure sheaf rational, 248, 250 semi-algebraic, 248, 252, 259 smooth, 248, 262, 265 Zariski, 112, 115, 116, 249 Sturm algorithm, 135, 136 symbolic powers, 89, 95 Tarski-Seidenberg theorem, 163, 164, 165, 203, 206, 207, 215, 233, 258, 268, 271 total order, 37, 64, 141, 143, 145, 156, 178, 234, 238 ultrafilter, 233, 234 valuation, 146, 242 valuation ring, 139, 149, 152 weak order, 33, 168 weak $-chain, 224, 227 Whitney, H., 261 Zariski dense set, 43, 238 Zariski, 0., 244 Zariski topology, 106, 107, 108, 109, 110 zero divisors, 42, 60, 91, 92