Introduction to Finite Fields and their Applications

ds Introduction to finite fiel and their applications RU D O LF L1 DL , /iubart, AWilraliu Unirersil)' (If Tasmania

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ER SI TY PR ES S CA M BR ID G E UN IV Cambridge NeH' Rochelle Lo nd on Ne w York Melhourne !>:vdn<'j

Published hy the Press Syndicate of the Universit)' of Cambridge The Pitt Ruilding, Trumpington Street, Cambridge CB2 I RP 32 East 57th Street, New York, NY 10022, USA 10 Stamford Road, Oakleigh, \r1elbourne 3166, Au~tralia (L; Camhridge

First

pub1i~hcd

Univer~ity

Press 1996

19R6

Printed in Great Britain at the

Univer~it)'

Press, Camhridge

British I.ibrary Cataloguing in Puhlication Dala Lidl, Rudolf Introduction to finite fields and their applications. 1. Finite ricld~ (Algebra)

I. Title

512'.3

lI. l\iiederreiter, I larald QA247.3

Library of C01tKr£>ss Cara/oRillg in Publication Data Lidl, Rudolf. Introduction to finite fields and their applications. Bihliography: p. lndudes index. 1. Finite fields (Algebra) 1. Niederrciter, Harald, 1944 Il. Title. QA247.3.L54 19R5 512'.3 85-9704 ISBN 0-521-30706-6

Contents

vii

Preface Chapter I

AI~ebraic

Foundations

I Groups

2 Rings and Fields 3 Polynomials 4 Field Extensions Exercises Chapter 2 Structure of Finite Fields I Characterization of Finite Fields 2 Roots of Irreducible Polynomials 3 Traces, Norms, and Bases 4 Roots of Cnity and Cyclotomic Polynomials 5 Representation of Elements of Finite Fields 6 Wedderburn's Theorem Exercises Chapter 3 Polynomials over Finite Fields I Order of Polynomials and Primitive Polynomials 2 Irreducible Polynomials

I

2

" 18

30 37 43

44 47 50 59 62 65 69 74

75 R2

Contents

3 Construction of Irreducible Polynomials 4 Linearized Polynomials 5 Binomials and Trinomials Exercises Chapter 4 I 2 3

Factorization of Polynomials Factorization over Small Finite Fields Factorization over Large Finite Fields Calculation of Roots of Polynomials Exercises

87 98 115 122 129 130 J 39 150 J 59

Chapter 5 Exponential Sums I Characters 2 Gaussian Sums Exercises

162 163 168 181

Chapter 6 I 2 3 4 5 6 7

Linear Recurring Sequences Feedback Shift Registers. Periodicity Properties Impulse Response Sequences. Characteristic Polynomial Generating Functions The Minimal Polynomial Families of Linear Recurring Sequences Characterization of Linear Recurring Sequences Distribution Properties of Linear Recurring Sequences Exercises

185 186 193 202 210 215 228 235 245

Chapter 7 I 2 3 4

Theoretical Applications of Finite Fields

251 252 262 271 281 294

Chapter 8 I 2 3

Algebraic Coding Theory Linear Codes Cyclic Codes Goppa Codes

Finite Geometries Combinatories Linear Modular Systems Pseudorandom Sequences Exercises

Exercises Chapter 9 Cryptology I Background

299 300 311 325 3:12

338 339

v

Contents

2 Stream Ciphers 3 Discrete Logarithms 4 Further Cryptosystcms Exercises Chapter 10 Tables I Computation in Finite Fields 2 Taoles of Irreducible Polynomials

342 346 360 363 367 367 377

Biblio2raphy

392

List of Symbols

397

Index

401

To Pamela and Gcrlindc

Preface

This book is designed as a textbook edition of our monograph Finite Fields which appeared in 1983 as Volume 20 ofthe Encyclopedia ofMathematics and Its Application.,. Several changes have been made in order to tailor the book to the needs of the student. The historical and bibliographical notes at the end of each chapter and the long bibliography have been omitted as they arc mainly of interest to researchers. The reader who desires this type of information may consult the original edition. There are also changes in the text proper, with the present book having an even stronger emphasis on applications. The increasingly important role of finite fields in cryptology is reneeted by a new chapter on this topic. There is now a separate chapter on algebraic coding tbeory containing material from tbe original edition togetber with a new section on Goppa codes. New material on pseudorandom sequences bas also been added. On the other hand, topics in tbe original edition that are mainly of theoretical interest have been omitted. Thus, a large part of the material on exponential sums and tbe ebapters on equations over finite fields and on permutation polynomials cannot be found in the present volume. The tbeory offinite fields is a branch of modern algebra that has come to the fore in the last 50 years because of its diverse applications in eombinatories, coding theory, cryptology, and the mathematical study of switching circuits, among others. The origins of the subject reach back into the 17th and 18th centuries, with such eminent mathematicians as Pierre de Fermat (1601-1665), Leonhard Euler(1707 1783), Joseph-Louis Lagrange (1736-1813), and Adrien-Marie Legendre (1752 1833) contributing to the structure theory of special finite fields --namely, the so-called finite prime fields. The eeneral theorv of finite fields mav be said to begin with the work of

Preface

viii

Carl Friedrich Gauss (1777-1855) and Evariste Galois (1811 1832). hut it only became of interest for applied mathematicians in recent decades with the emergence of discrete muthematic~ as a serious discipline. In this hook wc have aimed at presenting hoth the dassical and the applications-oriented aspects of the suhject. Thus, in addition to what has to bc eonsidcrcd the essential core of the theory, the reader will find results and techniques that arc ofimportancc mainly heeause of their usc in applications. fleeause of the vastness of the suhjeet, limitations had to be imposed on the choice of material. In trying to make the book as self-contained as possible, we have refrained from discussing results or methods that helong properly to algebraic geometry or to the theory ofalgebruic fum.:tion fields. Applications are described to the extent to which this can he done without too much digression. The only noteworthy prerequisite for the book is a haekground in linear algehra, on the level of a first course on this topic. 1\ rudimentary knowledge of analysis is needed in a few passages. Prior exposure to ahstraet algebra is cenainly hdpful, although all the necessary information is summarized in Chapter I. Chapter 2 is basic for the rest of the hook as it contains the general structure theory of finite ficids as well as the discussion of concepts that arc used throughout the book. Chapter 3 on the theory of polynomials and Chapter 4 on faetori/ation algorithms for polynomials arc dosely linked and should hest be studied together. Chapter 5 on exponential sums uses only the dementary structure theory of finite fields. Chapter (> on linear recurring sequences depends mostly on Chapters 2 and 3. Chapters 7. 8, and 9 are devoted to applications and draw on various material in the previous chapters. Chapter 10 supplements parts of Chapters 2, 3, and 9. Each chapter starts with a hrief description of its contents, hence it should not he necessary to give a synopsis of the hook here. In order to enhance the attractiveness of this hook as a text hook, we have inserted worked-out examples at appropriate points in the text and inciuded lists of exercises for Chapters I -9. These exercises range from routine problems to alternative proofs of key theorems, but contain also material going beyond what is covered in the text. With regard to cros~-rcferences, we have numbered all items in the main text consecutively by chapters, regardless of whether they are definitions. theorems, examples, and so on. Thus, "Definition 2.41" refers to item 41 in Chapter 2 (which happens to be a definition) and "Remark 6.23" refers to item 23 in Chapter 6 (which happens to he a remark). In the same vein, "Exercise 5.21" refers to the list of exercises in Chapter 5. We gratefully acknowledge the help of Mrs. Melanic Barton and Mrs. Retty Golding who typed the manuscript with great care and efficiency. R. LIDL

H.

r-;tEllI'RRI'.ITI.R

Chapter I

Algebraic Foundations

Thi' introductory chapter contains a survey of some basic algebraic concept' that will be employed throughout the hook. Elementary algebra uses the operations of arithmetic ,uch a, addition and multiplication, hut replaces particular numbers hy symbol, and thereby ohtains formulas that, by suhstitution, provide solutions to specific numerical problems. In modern algebra the level of abstraction is raised further: instead of dealing with the familiar operations on real numhers, one treats general operations

-processes of t:omhining two or more clements to yield another element·- in general sets. The aim is to study the common properties of all systems consisting of sets on which are defined a fixed number of operations interrelated in some definite way-for instance, sets with two binary

operations behaving like + and· for the real numbers. Only the most fundamental definitions and properties of algehraic systems- that is. of sets together with one or more operations on the

set will be introduced, and the theory will be discussed only to the extent needed for our ,pecial purposes in the study of finite fields later on. We state some standard results without proof. With regard to sets we adopt the naive standpoint. We use the following sets of numbers: the set I'\J of natural numbers, the set 7L of integers, the set Q of rati,mal numhers, the ,et IR of real numbers, and the set C of complex numhers.

2

I.

Algehraic Foundation.,

GROUPS

In the set of all integers the two operations addition and multiplication arc well known. We can generalize the concept of operation to arbitrary sets. Let S be a set and let S X S denote the set of all ordered pairs (s, I) with s E S, IE S. Then a mapping from S X S into S will be called a (billa~v) operalioll on S. Under this definition we require that the image of (5, t) E S X S must be inS; this is the closure property of an operation. By an alf{ehraic structure or algehraic system we mean a set S together with one or more operations on S. In elementary arithmetic we are provided with two operations.

addition and multiplication, that have associativity as one of their most important properties. Of the various possible algebraic systems having a single associative operation, the type known as a group has been by far the most extcnsively studied and developed. The theory of groups is one of the oldest parts of abstract algcbra as well as onc particularly rich in applications.

1.1.

Definition. A group is a set G together with a binary operation

0

on

G sueb that the following three properties hold;

1.

* is associative; that is. for any a, h, c E

(j,

ao(boc)~ (aob)oc.

2.

3.

There is an identity (or unity) elemelll e in G such that for all aEG, For each a E

a*e=e*a=a. G, there cxists an inverse element a- J E G such that

If the group also satisfies 4. Foralla.hEG, a*h=b*a,

then the group is called abelian (or commutative). It is easily shown tbat the identity element e and lhe inverse clement a J of a given element a E G are uniquely determined by the properties above. Furthermore, (a 0 b) J ~ b- J 0 a J for all a. bEG. For simplicity, we shall frequently use the notation of ordinary multiplication to designate the operation in the group. wriling simply ah instead of a 0 h. But it must be emphasized that by doing so we do not assume that the operation actually is ordinary multiplication. Sometimes it is also convenient to write a + h instead of a 0 hand - a instead of a J. bUI this additive notation is usually reserved for abelian groups.

I.

(jroup~

3

The associative law guarantees that expressions such as a 1Q 2' •• an with aj E G, I ~ j ~ n, are unambiguous, since no matter how we insert parenthcses, the expression will always represcnt the same clement of G. To indicate the n-fold composite of an element a E G with itself, where n E I'll, we shall write an=aa"'a

(nfactorsa)

if using multiplicative notation, and we call an the nth power of a. If using additive notation for the operation' on G, we write na=a+a+'" +a

(nsummandsa).

Following customary notation, we have the following rules:

Multiplicative Notation

Additive Notation

a-n=(a- I )" a lla m = an t m (a")m=a"m

(-n)a=n(-a) na + ma = (n + m)a m(na)~(mn)a

For n = 0 E Z, one adopts the convention aO ~ e in the multiplicative notation and Oa = 0 in the additive notation, where the last "zero" represents the identity element of G. 1.2.

Examples (i)

Let G be the set of integers with the operation of addition. The ordinary sum of two integers is a unique integer and the associativity is a familiar fact. The identity element is 0 (zero), and the inverse of an integer a is the integer - a. We denote this group by Z. (ii) The set consisting of a single element e, with the operation' defined bye' e ~ e, forms a group. (iii) Let G be the set of remainders of all the integers on division by 6-that is, G ~ CO, 1,2,3,4, 5}-and let a • b he the remainder on division by 6 of the ordinary sum of a and b. The existence of an identity element and of inverses is again obvious. In this case, it requires some computation to establish the associativity of '. This group can be readily generalized by replacing the 0 integer 6 hy any positive integer n. These examples lead to an interesting class of groups in which every element is a power of some fixed clement of the group. If the group operation is written as addition, we refer to "multiple" instead of "power" of an element. 1.3. Definition. A multiplicative group G is said to be cyclic if there is an clement a E G such that for any bE (j there is some integer j with b = a i .

Algchrail.: roundation:;

4

Such an dement a is called a gel1eralOr of the cyclic group, and we write G = (a). It follows at once from the definition that every cyclic group is commutative. We also note that a cyclic group may very well have more than one dement that is a generator of the group. For instance, in the additive group il. both I and - I arc generators. With regard to the "additive" group of remainders of the integers on division hy 11, the generalil.ation of Example 1.2(iii). we find that the type of operation used there leads to an equivalence relation on the set of integers. In general, a subset R of S X S is called an equivalel1ce relaliol1 on a set S if it has the following three properties:

(a) (', s) E R for all s E S (reflexiGitv). (b) U (s, I) E R, then (c, s) E R (symmelry). (c) U(S,I), (c.u)E R, then (s.u)E= R (cral1siliviZv). The most ohvious example of an equivalence relation is that of equality. It is an important fact that an equivalence relation R on a set S induces a partition of S -that is, a representation of S as the union of nonempty, mutually disjoint subsets of S. If we collect all clements of 5 equivalent to a fixed S E S. we obtain the equiwlel1ce class of s. denoted by [sl~ (I E S:(S.I) E= R}.

The collection of all distinct equivalence classes forms then the desired partition of S. We note that [s J = [I J precisely if (s. r) E= R. Example 1.2(iii) suggests the following concept.

1.4. Definition. For arbitrary integers a. h and a positive integer 11, we say that a is cOl1gruent to h modulo n, and write a'" hmod 11. if the differcnce a - h is a multiple of n -that is, if a ~ h + kn for some integer k. It is easily verified that I'congruence modulo Jl" is an equivalence relation on the set il. of integers. The relation is ohviously rel1exive and symmetric. The transitivity also follows easily: if a ~ h + kl1 and h = c + In for some integers k and I. then a = c +(k 0 1)11, so that a'" hmod 11 and b '" c mod n together imply a '" C mod 11. Consider now the equivalence classes into which the rclation of congruence modulo n partitions the sct il.. These will bc tic scts

[0]

= {

[i]=(

-211, - n.D.I1,2n .... }. -2n+I,-n~I,I.n-I,211,I .... },

[11 - 1] = { .... - n - I. . I. 11 - I. 211 - I. 3n - I, ... }. We may definc on the set ([D].[I]..... [I1-IJ) of equivalence classes a binary

5

operation (which we shall again write as ordinary addition) by

+, although it is eenainly not

[al+[b]~[a th],

( 1.1)

where a and h are any clements of the respective sets [aj and [b] and the sum a I h on the right is the ordinary sum of a and b. In order 10 show that we have actually defined an operation-that is, that this operation is wcll defined- we must verify that the image element of the pair ([aj,[h]) is uniquely determined by raj and [b] alone and does not depend in any way on the representatives a and h. We leave thi1:i proof as an exercise. Associativity of the operation in (1.1) follows from the aS1-.ociativity of ordinary addition. The identity clement is [0] and the inverse of [a] is [- oj. Thus the elements of the set ([OJ,ll]..... [n· Ij) form a group. 1.5. Uefinition, The group formed by the set ([OJ,[lj.... ,[n -I]) of equivalence clas.ses modulo n with the operation (1.l) is called the group of illlexers modulo n and denoted hy 1'.".

71" is actually a cyclic group with the equivalence class [I J as a generator, and it is a group of order n according to the following definition. 1.6, Definition, A group is called finite (resp. illfinile) if it contains finitely (resp. infinitely) many elements. The number of elements in a finite group is called its order. We shall write: CI for the order of the finite group C. Th~rc is a convenient way of presenting a finite group. i\ table displaying the group operation, nowadays referred to as a Cayler IOhle, is constructed hy indexing the rows and the columns of the tahle hy the group clements. The element appearing in the row indexed hy a and the column indexed hy h is then taken to he ah.

1.7.

Example,

The Cayley tahle for the group 1'., is:

~[OJ

[Il

[0]

[II [21

[0]

[II [Il

[21 [2]

[3]

[3] [4] [5] [3] [4] [5] [4] [5] [0] [5] [0] [ I]

[3] [4] [3 J [3] [41 [5] [0] [4] [4] [5] [0] [Il [5] [5] [0] [ I ] [2] [21 . [2]

[I] [21

[2]

[3] [3] [4]

u

A group (j ~ontains certain suhsets that form groups in their own rig,ht under the "peration of G. for instance, the subset ([OJ, [2j, [4j} of 1'., is easil:v :,een to have thi:-. property.

Alg,chra:c

roundation~

1.8. Uefinition. II subset fI of the group G is a subgroup of G if 1I is itself a group with re:-.pect to the operation of G. Subgroups of G other than the frivinlsuhgroup.\ {e} and G ihclf are called nontrivial suhgroups of G. One verifie' at once that for any fixed a in a group G. the set of all powe", of a is a subgroup of G. 1.9. Definition. The subgroup of G consisting of all powers of the clement a of G is called the subgroup generated hy a and is denoted by (a). This subgroup is necessarily cyclic. If (a) is finite. then its order is called the order of the clement a. Otherwise. 11 is called an dement of infillite order. Thus. a is of finite order k if k is the least positive integer such that e. Any other integer m with am = e is then a multiple of k. If S is a nonempty subset of a group G. then the suhgroup /I of G consisting of all finite products of powers of clements of S is called the subgroup genCfaled hy S. denoted by /1- (5). If (S) ~ G. we say that S generate., G. or that G is generated hv S. for a positive element n of the additive group 71. of integers. the subgroup (n) is elo~ely associated with the notion of congruence modulo n, since a '" hmod n if and only if a - b'" (II). Thu, the subgroup (n) defines an equivalence relation on 71... This situation can be generalized as follows.

(lA

.

1./0. Theorem. If H is a subgroup of G. thell the re/otioll R II on G defilled hy (a. h) E R II if and only if a ~ "h for sO/lle h r 1I. is WI equimlellee relaTion.

The proof is immediate. The equivalence relation R If i' called left congruence modulo II. I.ike any equivalence rehnion, it induces a partition of G into nonempty. mutually disjoint suhsets. These subsets ( - equivalence classes) are called the left coselS of G modulo /I and they arc denoted hy all

~

{ah: h

C

H}

(or (/ -'- H·· (a - h: h'" H) if G is written additively). where a is a fixed clement of G. Similarly, there i:-. a decompositilm of G in\{) right coset5 modulo /I, which have the form I/a .. (ha: h E /I). If G is abelian. then the di:-.tinction between left and right cosclS modulo II is unnecessary. 1.11. Example. Let G ~ 71." and let 1I be the subgroup {IO].13j.[6].[9]}. Then the distinct (left) coset; of G modulo 1I arc given by:

[0]1 H·· ([01.[3J.[6].[9]). [i]+ /I ([il.[4].[7j.[IO]). [2J-II" ([2].[S].{KJ.lJ I]). 1./2. Theorem. If /I is a fillile suhgroup of (;. then ,,"err (I"ft or risht) ('oset of G moduln H has the same number of dements as H.

1.

(jroup~

7

1.13. Definition. If the subgroup I/ of G only yields finitely many distinct left eosets of (; modulo I/. then the number of sucb cosets is called the index of fl in G. Since the kft eosets of G modulo I/ form a partition of G, Theorem 1.12 implies the following important result. 1.14. Theorem. The order of afinire group G is equal TO rhe producr of rhe order of any subgroup I/ and rhe index of H in G. In parricular, rhe order of H dieides rhe order of G and rhe order of any e1emenr a E G divides rhe order of G.

The subgroups and the orders of elements arc easy to describe for

cyclic groups. We summarize the rdcvant facts in the suhsequent theorem. 1.15.

Theorem

Every suhgroup of a ()'Clic group is ,:rdic. /n a finire cvdic group (a) of order m. rhe elemenr a k generares a suhgroup of order m/ged(k. m), where ged(k, m) denores rhe greatesT common dil:isor of k and m. (iii) If d is a posiriee dieisor of rhe order m of a finire ,ydic group (a). rhen (a) conrain; one and on!r one suhgroup of index d. For any posirive didsor f of m. (a) conrains precisely one subgroup of order f. (iv) Ler f he a posiriee dieisor of rhe order-of a finire cvclic group (a). Then (a) conrains ( elemenrs of order f. Here 9(/) is Euler's function and indicaTes the number of integers n with l.:s;;; n .:s; ; f rhar are relarively prime TO f. (v) A finire cyclic group (a) of order m conrains ( m] generarors-rhar is. e1emenrs a' such rhar (a') = (a). The generarors are rhe powers a' wirh gcd( r, m) = 1. (i)

(ii)

n

Proof (i) Let I/ be a subgroup of the cyclic group (a) with 1/ (e). If a" E H. then a "E I/: hence I/ contains at least one power of a with a positive exponent. Let d be the least positive exponent such that a d ,= H. and let a' E H. Dividing s by d gives s = qd + r, 0", r < d. and q. r E 71.. Thus a'(a-d)q = a' E H. which contradicts the minimality of d, unless r = O. Therefore the exponents of all powers of a that belong to Hare divisible by d. and so I/ = (ad). (ii) Put d = gcd( k, m). The order of (a k ) is the least positive integer n such that a'" = e. The laner identity holds if and only if m divides kn, or equivalently. if and only if mid divides n. The least positive n with this property is n = mid. (iii) If d is given, then (a J ) is a subgroup of order mid. and so of index d. because of (ii). If (a') is another subgroup of index d, then its

=

order is m / d, and so d ~ ged( k, m) by (ii). In particular, d divides k, so that a' E (ad) and (a') is a suhgroup of (ad). But since hoth groups have the same order, they are identical. The second part follows immediately because the subgroups of order I arc precisely the subgroups of index m / f. (iv) Let I(a) I ~ m and m ~ df. By (iii, an clement a' is of order I if and only if ged(k, m) = d. Hence, the numher of clements of order lis equal to the numher of integers k with I,;; k,;; m and ged(k, m) = d. We may write k ~ dh with I,;; h,;; I, the condition ged(k, m) ~ d heing now equivalent to ged(h,j) ~ I. The numher of these h is equal to $(/). (v) The generators of (a) are precisely the elements of order m, so 0 that the first part is implied by (iv). The second purt follows from (ii). When comparing the structures of two groups, mappings hetween the groups that preserve the operations play an important role. 1.16, Definition. A mapping/: (i ~ II of the group (i into the group 1/ is culled a homomorphism of G into 1/ if I preserves the operation of G. That b. if * and· arc the operations of G and 1/, respectively. then I preserves the operation of G if for all a.hEG we huve l(a*h)~/(a)·/(h). If. in adctition, I is onto 1/, then I is called an epimorphism (or homomorphism "onto") and 1/ is a homomorphic image of G. A homomorphism of G into G is called an endomorphism. If I is a one-to-one homomorphism of G onto 1/. then/is called an isomorphism und we say that G und 1/ arc isomorphic. An isomorphism of G onto (j is called an tlUlomorphism.

Consider. for instance, the mapping I of the additive group 1L of the integers onto the group 1L" of the integers modulo n, defined hy I(a I ~ ra]. Then I(a+h) ~ ra+hl~laJ+lhl~/(a)I I(h)

fora,hElL,

and I is a homomorphism. If I: G -. II is a homomorphism and e is the identity clement in G, then ee ~ e implies I( e )J( e) ~ I( e), so that I( e) ~ e'. the identity clement in II. fromaa-'=eweget/(a ')~(/(a))-' forallac(i. The automorphisms of a group G are often of particular interest, partly heeause they themselves form a group with respect to the usual composition of mappings, as can he easily verified. Important examples of automorphisms are the iflller llUlomorphisms. For fixed t1 c- (j, define Ju hy fo(h) ~ aba I for beG. Then la is an automorphi.sm of G of the indicated type, and ",e get all inner automorphisms of G by letting a run through all clements of G. The elements hand aba I arc said to he ('(JIIjugat", and for a nonempty subset S of G the set aSa I ~ (asa -, : .\ C S) is called a conjugale of S. Thus, the conjugates of S arc just the images of S under the ,arious inner automorphisms of G.

9

(jf()L1P~

1.17. Definition. The kernel of the homomorphismf: G G into the group /I is the set kerf~

~

II of the group

(a'=G:f(a) . e'},

where e' is the identity clement in JI. 1,18. Fxampl<'. For the homomorphism f: 7L ~ 7L" given hy f(a) ~ [aJ, ker f consists of all a F 7L with la I ~ [OJ. Since this condition holds exactly for all multiples a of n. we have kcr f ~ (n), the suhgroup of 7L generated hy n. It is easily checked that kerf is always a subgroup of G. Moreover, kerf has a special property: whenever a (" G and bE: kerf, then aha I EO kerf. This leads to the following concept. 1.19. Definition. The subgroup II of the group G is called a normal subgroup of G if aha I E II for all a (" (i and all h'= H. Every ~uhgroup of an ahelian group is normal ~ince we then have 1/1 eli = h. We shall state some alternative characterizations of the property of normality of a subgroup.

aha

1

= aa"' 1.20, (i)

(ii)

Theorem The subgroup H of G is normal if and onlv if /I is equal II! ils conjugates, or equi1;a/em/y. if and only if JI is im;ariant under al! t/1(' inner QUlOmorp/1isms of (i. Ihe subgroup /I of Ii is normal if and only if Ihe lefl COSe1 all is equal (() the rig/1t Cosel Jla for rrery a C
One important feature of a normal suhgroup is tbe fact tbat tbc set of its (left) cosets can be endowed with a group structure.

1.21. 7heorem. If II is a normal subgroup of G, Ihen Ihe sel of (/ef1) 10sels of G modulo /I forms a group wilh respeCI II! Ihe operalion (aH)( bH) (ab)lI. 1.22. Definition. for a normal suhgroup /I of G. the group formed by the (left) easets of Ii modulo H under the operation in Theorem 1.21 is called the faclor group (or qU01ienl group) of Ii modulo H and denoted by G/I/. If (i/II is finite. then its order is equal to the index of H in G. Thus, by Theorem 1.14. we get for a finite grclup G.

. II

I( '/ I ~

IGI iHi

b.lch nnrmal suhgroup of a group (; uetermines in a natural way a hpIll0mDrphbm of G anu vice v~rsa.

10

Al~chraic Foundation~

/.23.

Theorem (Homomorphism Theorem). LeI f; G ~ f(G) = G , be a homomorphism of a group G omo a group 0 I' Theil kerf is a lIormal subgroup of G. alld Ihe group G I is isomorphic 10 Ihe faclOr group GIker f. COllcersely. if 11 is allY 1I0rmai subgroup of G. Ihen Ihe mappillg 1/-; 0 ~ GI H defilled by I/-(a) ~ aH for a EGis a homomorphism of G 01110 G/H wilh kerl/-~H.

We shall now derive a rdation known as the dass equacion for a finite group. which will be needed in Chapter 2. Section 6.

1.24. Definition. Let S be a nonempty suhset of a group G. The 1I0rmalizer of S in G is the set N( S) ~ {a '= 0; aSa- 1 ~ S}. /.25. Theorem. For allY 1I0llemply subsel S of Ihe group G. N( S) is a suhgroup of G and [here is (l one-to-one correspondence berween the left cosels of 0 modulo N( S) alld Ihe dislillcl cOlljuKales aSa I of S.

Proof We have e F N(S), and if a, b E N(S), then a-I and ab arc also in N(S), so that N(S) is a subgroup of 0. Now aSa l~bSb '=S=a-lhSb 'a~(a 'h)S(a 'h)

1

=a 'bEN(S)=hEaN(S).

Thus, conjugates of S arc equal if and only if they arc defined by elements in the same left coset of G modulo N(S}, and so the second part of the theorem is shown. 0 If we collect all clements conjugate to a fixed element a, we obtain a set called the conjU?;aly class of a. For certain elements the corresponding conjugacy class has only one member, and this will happen precisely for the clements of the center of the group.

1.26. Definition. For any group G. the cemer of (; is defined as the set C -, (c Eo 0; ac~· ca for all a E G}. It is straightforward to check that the center C is a normal subgroup of G. Clearly, 0 is abelian if and only if C ~ G. A counting argument leads to the following result.

/.27, Theorem (Class Equation). center C. 71zen

LeI 0 be a finile group wilh

,

IGI '-ICI~

L: ""

, -I

where each n, is ~ 2 Gild a diri.wr of IGI. In jact, "l' n 2 ..... n k are the numhers of elements of the distinct conjuy,acy c1a.'ises in G containillf? more than nne member.

II

Pruo/ Since the rclati()n "(1 is conjugate to /I'" is
2.

RII\GS ANn FIELDS

In mo:-.t of the numhcr systems used in ckmcntary arithmctit: there .He two distinct binary operations: addition and multiplication. Exampks are provided by the integers, the rational numbers. ami the real nUlllher..,. We now define a type of algehraic structure known as a ring that sllJres some of the basic properties of these number systems.

1.28.

Definition.

A ring (R, ,-,.) i, a set R, together with two hinary

operations. denoted by I- and·. such that:

1. R i~ an ahelian group \kith respect to 2. . is assoeiativc--that is, (a·h)·,- a·(h·c) for all a,h,n" R. 3. The distriburice laws hold: that i,,,,. for all a. h. (' E-.:.: R we have a·(h.c c) ~ a·b + a'c and (Ii· c)·a h'a I c·a. We ,..,hall use R as a designation for the ring (R. - .. ) and stress th'll the operations t- and . arc not necessarily the ordinary operations with number:-:.. In following convention. we use 0 (called the :('(0 e!enW1l!) to d~n{)te the identity ckment of the ahelian group H with respect to addition. and the additive inverse of a is denoted hy u: abo. a + (- h) i... ahhrc\'i~ ated h y a-b. Instead of (l·b we will u,..,ually write uh. As a consequence of the ddinition of a ring one ohtains the g~n~ral property (J() On 0 for all a cR. This. in turn. implies (- alb ~ a( - Ii) ali for all u. Ii (" R. The most natural example of a ring is perhaps the ring of ordinary integers. If we examine the properties of this ring.. we realil.e that it has properties not enjoyed by rings in general. '( hus. ring.., (,:an he fun her classified according to the following definitions.

1.29.

Definition (i)

!\ ring b called a ring with identity if the ring has a multiplicative identity that b, if there b an clement c ~uch that (1(' "- ('(J ~

(ii)

a fpr all a

E

R.

!\ ring i~ c<.lllcd omllnututhe if . is commutdtive:.

(iii) (iv) (v)

A ring is called an illlexral domain if it is a commutative ring with identity f -r 0 in which ah ~ 0 implies a ~ 0 or h ~ O. II ring is called a dirision ril1g (or skew field) if the non7.ero dements of R form a group under·. A commutative division ring is called a field.

Since our study is devoted tD fields, we cmphasii'.c again the

dcfini~

tion of this concept. In the first place. a field is a set Fan "hich tWI' binary operations. called addition and multiplication. are defined and which con-

tains two distinguished clements 0 and

e

with O"f:"

c.

Furthampre. F is an

abelian group with respect to addition having 0 as the identity ekment. and the ekments of F that are -r 0 form an abelian group with respect to

multiplication having e as the idt..·ntity

c1~ment.

The two operations of

addition and multiplication arc linked by the distributive law II( h -+ e) ~ ah + ae. The second distributive law (b·\ da·· hll -,- ea follows automatically from the commutativity of multiplication. The element 0 is called the ~ero element and t' is called thc multiplicative identity elemem or simply the idelllil}'. Later on. the identity will usually be denoted by I. The property appearing in Definition 1.29(iii)-namely. that lib·· 0 implies a ~ 0 or h = 0 is expressed by saying that tbcre arc 110 ~ero divisors. In particular, a field has no 7.ero divisors. for if ah -= 0 and II =1= O. then multiplication hy a \ yields h ~ a' \0 ~ O. In order to give an indication of the generality of the concept of ring.

we present some examples. 1.30.

Examples (i) (ii)

Let R be any abclian group with group operation -. Define ah ~ 0 for all a. b EO: R: then R is a ring. The integers form an integral domain. but not a field.

(iii) The l:\'cn integers form a commutative ring without identity. (iv)

The functions from the real numhe", into the real numhers

form a commUlative ring with identity ufl.der the definitions for f • g and fl( given by (j T g )(x) - f( x) • g(x) and (fg)( x) " f(x)l(x) for x c-Ihl. (v) The set of all 2 x 2 matrices with real numhers as entries forms a noncommutativc ring 'with identity with respect to matrix addition and multiplication. L1 We ha\'e :-,een uhove that a field is. in particular, an integral domain. The converse is not true in general (sec Example 1.30(ii)). hut it will hold if

the structures contain only finitely many elements. 1.31.

Theorem.

F."ery finite integral domam is

II

field.

['roo! I.et the clements of the finite integral domain R be a l • Q 2 · ... ,a'1. For a fixed nonzero element a CR. c()Osider the products "" "', nil Thp,p Mf' di"tincl. for if aa" = all . then a( ll. - ll") = o. and

13

2. Rings and ridd:-.

since G '=t= 0 we must have at - a, := O. or a/ = a i . Thus each clement of R is of the form Ga i , in particular. e = aa l for some i with I ~ i ~ n. where e is the identity of R. Since R is commutative, we have also a/a ,-= e, and so a/ is the multiplicative inverse of a. Thus the nonzero clements of R form a commutative group, and R is a field. U

1.32. Definition. A subset 5 of a ring R is called a 5uhrinf!, of R provided S is closed under - and . and form~ a ring under these operations. 1.33. Definition. A subset J of a ring R is called an ideal provided J is a >uhring of R and for all a Eo J and r F R we have ar F J and ra r J.

1.34.

Examplc, (i)

Lei R he the field 4)
int~gers i~ a

(ii) (iii)

Oefinition. Let R he a commutative ring. An ideal J of R i~ ~:lid to be principal if then~ is an a C R such that J . (£1). In this case. J i.;, also called the principal ideal generaTed hv a. 1.35.

Since ideals arL' normal suhgroups of the additive g.rl1Up llf a rin~. it immediatc1) that an ideal J pf the ring. R defines a rartitil)1l lIt' R into disjoint cnsets. culh.:d residue cla;;s('~ modul0 J. The re~iduc L'1a~s of thl' dement lJ of R modulo J will he denoted by laJ a + J. ~inl.'L' it L'l)n~i:-;ts l)f all clements of R that arc of thL' form lJ + (" for :'l~lme (" L J. I:klllcnh a. h E:= R arc called congrllem modulo J. written lJ·- hlllod.l. if they are ill the samL' residue class modulo J. or e4uivalelltly. if (1 h \ J (~l)mparL' Vvith Definition 1.4). One can verify that a = bmod.l implit:s IJ • I' -::-: h ' rJlll)d J. ar -= hrmod J. and ra == rhmod J for any r E:= Rand na =. nh mod J f~\r any 11 E:= Z. If. in additipn. r ==.\ mod J. then IJ + r - h I smod J and arhSInodJ. It is shown hy a straightf0f\I,:an.l arg.ument that the set of re~idu~ L'lasse:-. of a ring R tnl)dulo an ideal J forms a ring with r~spcL't III the opcrutions f()llow~

(u' J)T(h I (u I J)( h

l-

J)~

J)

~

(u+/o)+.I.

( 1.2)

u/o + .I.

(U)

1.36. Definition. ("he ring of resiJuL' da:-.~e ... of th~ ring. R nwdulo the ideal J undL'r the opt:ration~ (1.2) and (1.3) is called the residue da.\.\' ring (or

14

Algehraic l;oundations

1.37. Example (The residue class ring Z/(I1)). As in the case of groups (compare with Definition 1.5). we denote the coset or residue class of the integer a modulo the positive integer n by [a], as well as by a +(n), where (nl is the principal ideal generated by n. The elements of Z/(n) are

°

I I ~ 0 • (n). [I] ~ I ~ ( n ) .... . [n - 1]

.~ n - 1+ (n ).

D

• 1.38. Theorem. Z /( p), th" ring of residue classes of the integers modulo the principal ideal generated hy a prime p, is a field. Proof Ry Theorem 1.31 it suffices to show that Z/( p) is an integral domain. 'low [1] is an identity of Zj(p), and [al[bl=[abl~[OI if and only if ail '. kp for some integer k. But since p is prime, p divides ab if and onlv if p divides at least one of the factors. Therefore, either raj ~ [OJ or [hi ~ [01. s<) that Z/( p) contains no 7ero divisors. D

1.39. Example. Let p"~ 3. Then Z/( p) consists of the elements [0], [I], and [21. The operations in this field can be described by operation tables tbat are similar to Cayley tables for finite groups (see Example 1.7);

:J [0-,--]----,[~1]

[2]

. I [0]

[I]

[2]

[11 12] [0]

[2] [01

~I[0]

[0] [I]

[0] [2]

[2]

[I]

[0]" [0] [I] 1[1] [2] 12]

[1].[0] [211 [0]

[11

D

The residue e1ass fields Z/( p) are our first examples of finite field, that is, of fields that contain only finitely many elements. The general theory of sueh fields will be developed later on. The reader is cautioned not to assume that in the formation of residue e1ass rings all the properties of the original ring will be preserved in all eases. For example, the lack of 7ero divisors is not always preserved, as may be seen by considering the ring Z/(n), where n is a composite integer. There is an obvious extension from groups to rings of the definition of a homomorphism. 1\ mapping 'P: R --> S from a ring R into a ring S is called a homomorphism if for any a, h EO R we have 'f ( a + b ) ~ 'P ( a ) + 'P ( h)

and

'f (ab ) = 'P (a) 'P (h).

Thus a homomorphism 'f: R --> S preserves both operations + and . of R and induces a homomorphism of the additive group of R into the additive group of S. The set ker'f - {a

E

R ; 'f ( a ) ~ 0 (= S}

is called the kernel of 'f. Other concepts, such as that of an isomorphism, are analogous to those in Definition 1.16. The homomorphism theorem for rings, similar to Theorem 1.23 for groups, runs as follows. 1.40.

Iheorem (Homomorphism Theorem for Rings). .'

01

..

l. __ ..

:_

,. •.

;.J __ "

n~

If 'P is a D

~.~.I

C' ;"

2. Ring.:. and

Fi~'ld~

;-:>/(C' U

.'

>.:-

\ ':.

l~

.~'.

k

i,omorphic 10 Ihe faclor ring /ker
r,

r,

/<1.41. Definition. For a prime p. let Fp be tbe set (O.I. ...• p - I) of integers and lot
Examples (i)

Consider Z/(5), isomorphic to 11',. {a, 1,2,3,4}, witb the isomorphism given by: [OJ--. 0. [I] --.1, [2j--. 2, [3j--. 3. [4]" 4. The tables for the two operations + and . for elements in 11'5 are as follows:

. ,. I ° I 2 3 4 ° I 2 3 4 I 2 3 4 ° ° ° ° 0 0 ° 0 I 2 3 4 ° I ° I 2 3 4 I 21 () 2 4 1 3 2 2 3 4 ° I 3 3 4 0 I 2 ° 3 I 4 2 4 0 I 2 3 0 4 3 2 I 4 An even simpkr and more important example is the finite field II' 2' Tbe clements of tbis field of order two arc 0 and I. and the operation tables have the following form:

+

~I

(ii)

°

+

0 I I °iOIO,OO I ; I I I 0 I In this context. the clements ° and I arc called binarv elemell!>. ~

°

0

16

If h is any nonzero dement of the ring 7L of integers. then- the additive order of b is infinite; that is. nh _. 0 implies II O. lloweva. in the ring lL/( p). p prime. the additive order of every nonl.ef\) ekment b is p: that is. pb ~ O. and p is the kast positive intcger for which this holds. It is of interest to formali7c this property. ::..0

If R is an arhitrary ring and thcre exists a positive such that nr = 0 for every r E:::: R. then the least ~uch p{)sitivc integer n is called the charaClcrislic of Rand R i, said to have (positive) characteristic n. If no such positive integl'r 11 exists. R is said to have characteristic O. 1.43.

Definition.

integer

11

1.44. Theorem. A ring R '1=- {O} of posilire cJll1f(lCleristk hoeing WI identizv and no ;ero divisors mUll have prime clwracfuisfir.

Proof Since R contain~ nonzero clements. R has chara~teristic 2. If n were not prime. we could write n = km with k. m E lL. I < k. m < Ii. Then 0 ~ nc ~ (km)e ~ (ke)(me). and this implies that either ke ~ 0 or mc ~ 0 since R has no zero divisors. It follows that either kr ~ (ke)r .~ 0 for all r E R or mr - (me)r- 0 for all r E: R. in contradiction to the

Ii ~

definition of the '.:haractcrislic n. 1.45.

Corollary.

_J

A finil£" field has prim£" charaumsliL

Prvvf Ily Theorem 1.44 it sufficcs to show that a finite field F has a positive charactcristic. Consider the multiples c.2e.3e.... of thc identity. Since F c.:ontains only finitely many distinct elements. there exist integers k and m with 1 ~ k < m such that k" m£". or (m - k)c ~ O. and so F has a positive characteristic. ..J The finitc field lL/( p) (or, equivakntly. IF p) obviously has characteristic p. wherea, the ring lL of integers and thc field Q of rational number,

have characteristic O. We note that in a ring R of charal'tcristic 2 we have 2a a -'- a ~ O. hence a = - Q for all Q E R. !I. useful property of commutative rings of prime characteristic is thl.: following. 1.46. p.

Theorem.

I.e! R be

£l

cvmmUlatire ring vf prime charaClt'fi'ilic

1Ileli

(a + h)r·· ~ v p " + h P· and (a - b)"" ~ aP' for

Q.

h E: R and

Proof

Ii

hr

"N.

We usc the facl that .. ·(p i+l) l'P)i ., -p(p-I) - I T:':. -'i- -

'= II mod

p

for all i E lL with 0 < i < p. which follow, from I;') heing an intcgcr and the observation that the factor p in thc numerator cannot bc cancellcd. Then hy

2. Ring:'. and held:.

17

the binomial theorem (see Exercise 1.8),

(a+b)P~aP+(~)ap

'b·· ... \

(p~l)ahP-'+hP~ap+bP.

and induction on n completes the proof of the first identity. By what we have shown, we get aP'~ «a- h) 1 h)P" ~ (a-h)P' +h P'.

o

and the second identity follows.

Next we will show for the case of commutative rings with identity which ideals give rise to factor rings that are integral domains or fields. ror

this we need some definitions from ring theory. I.ct R bl.: a commutative ring with identity. An element u E:: R is called a diui",r of heR if there exists e Eo R such that ae ~ h. A unil of R is a-J-livisQr_91J.h.e.id..c.ntity; two dements a, b E R are said to hi_~Q5,~)~f there is a unit t: of R :"luch that a _. hL An element (' E R is called a prime dement if it is no u~i.~_~.~_d i~ .i.t ha~.~..>n}.x~~h~-~.~i~;~"~~f B.,~.l:!d_ the associatcs of (' as divis(;r'CAnideal p; R of the ring .K.is.qllc.dLPJ:i!.ntJdeaJ. if for a. hER we have ah E' P only if either a r: P or h E' P. An ideal M < R of R is called a maximal ideal of R if for any ideal J of R the property Me J implie, J R or J ~ M. Furthermore. R is said to be a prillcipal ideal domain if R is an integral domain and if every ideal J of R is principal-that is. if there is a generating clement a for J such that J (a) ~ (ra: r E R).

1.47.

Theorem.

l.et R he

II

commuTative ring with idenTity. Then:

An ideal M of R i\ a maximal ideal if and only if RIM is a field. lin ideal P of R is a prime ideal if and onlv if RIP is all illlegral domain. Fr;ery maximal idelll of R is u prime ideal. If R i' a principal ideal domaill, Ihell RI(e) is afield if alld ollly if (' i'i a prime element of R.

(i)

(ii)

(iii) tiv)

!)roo;' (i)

I.et M he a maximal ideal of R. Then for a if M, a

Eo

R, the set

J ~ (ar . m: r r R. m EM) is an ideal of R properly containing M, and therefore J . R. In particular. ar -'- m I for some sliitahle r cR. m E= AI. where I denotes the multiplicative iden-

tity element of R. In other words. if a - M' 0 1 M is an clement of RIM different from the zero clement in RIM. then it posscsses a multiplicative inverse, hecau:-.c (ll + M)( r + M) = ar ~ M ~ (I m) \ M ~ I + M. Therefore. RIM is a field. Conversely. let RIM be a field and let J ~ M. J .. M, be an ideal of R. Then for a c. J. a
11<

Algebraic

roundation~

plicative invc"e, so that (a + M )(r + M) = I + M for some r E R. This implies ar + In = 1 for some In EM. Sincc J is an ideal, we have IE J and therefore (I) ~ R c J, hcneeJ = R. Thus M is a maximal ideal of R. (ii) Let P he a prime ideal of R; then RIP is a commutative ring with identity 1+ P "" 0+ P. Let (a + P)(b + P) = 0+ P, hence ah E P. Since P is a prime ideal, either a E P or bE P; that is, either a + P = 0+ P or h + P = 0+ P. Thus, RIP has no zero divisors and is therefore an integral domain. The converse follows immediately by reversing the steps of this proof. (iii) This follows from (i) and (ii) since every field is an integral domain. (iv) Let e E R. If c is a unit, then (c) = R and the ring RI(c) consists only of one element and is no field. If c is neither a unit nor a prime element, then c has a divisor a E R that is neither a unit nor an associate of c. We nOie that a 0, for if a = 0, then e = 0 and a would be an associate of c. We can write c = ah with hER. Next we claim that a'f- (c). For otherwise a = cd = ahd for some d E R, or a( I - hd) = O. Since a "" O. this would imply bd = I, so that d would be a unit, which contradicts the fact that a is not an associate of c. It follows that (c) c (a) ~ R, where all containments arc proper, and so RI(e) cannot bc·a field because of (i). Finally, we are left with the case where c is a prime clement. Then (c) R since e is no unit. Furthermore, if J;;) (c) is an ideal of R, then J = (a) for some a E R since R is a principal ideal domain. It follows that c E (a), and so a is a divisor of c. Consequently, a is either a unit or an associate of c, so that either J = R or J = (c). This shows that (c) is a maximal 0 ideal of R. lIence R/(c) is a field hy (i).

=

=

As an application of this theorcm, let us consider the case R = Z. We note that Z is a principal idcal domain since thc additive subgroups of Z arc already generated by a single clement because of Theorem I, 15(i). A primc numher p fits the definition of a prime clement. and so Theorcm 1,47(iv) yields another proof of the known result that Z/( p) is a field. Consequently, (p) is a maximal ideal and a prime ideal of Z. ·For a composite integer n. the ideal (n) is not a prime idcal of Z, and so Z/(n) is not even an integral domain. Othcr applications will follow in the next section when we consider residue class rings of polynomial rings over fields.

3.

POLYNOMIALS

In elementary algebra one regards a polynomial as an expression of the -+Y I- ... -+- 11 y'l Thp 11 '", ;Irt~ CBlkd (:oefficicnts and are usuallv

f ..... rm "

1/

3. Polynomiab

14

real or complex numbers; x is viewed a!:l a variable: that is, substituting an arbitrary number a for x, a well-defined numher Go + ala + ... + Gna n is ohtained. The arithmetic of polynomials is governed by familiar rules. The concept of polynomial and th~ associated operations can be generali7ed to a formal algehraic setting in a straightforward manner. Let R be an arbitrary ring. A polynomial over R is an expression of the form

f(x)

"

La/x':..:.. an + u1x + ... + GnX",

=

1=0

where n is a nonnegative integer, the coefficients a , . 0.:::;; i .: : ; n, are clements of R. and x is a symhol not belonging to R. called an indeterminate over R. Whenever it is clear which indeterminate is meant, we can use f as a designation for the polynomial fix). We adopt the convention that a term a,x i with a, ~ 0 need not he written down. In particular, the polynomial fix) above may then also he given in the equivalent form f(x) ~ a o + a,x + ... + anx n +Ox n ' I + ... +Ox n I h, where h is any positive integer. When comparing two polynomialsf(x) and g(x) over R, it is therefore possible to assume that they hoth involve the same powers of x. The polynomials n

f( x) ~

L

n

a,x'

and

g( x) ~

/,... 0

L

h,x i

1=0

over R are considered equal if and only if the sum of fix) and g(x) hy

0i =

hi for 0

~

i ~ 1l. We define

n

f(x)+g(x)~

L

(a,+h,)x i •

,- 0

To define the product of two polynomials over R, let n

f(x) ~

L

m

a,x'

and

g(X) ~

i=U

L

hjx i

J-O

and set n+ m

f(x)g(x) ~

L

1<=0

c,x',

where c, ~

L

a,h)"

ilJ="1< 00;;.':;;; n.O:;;; ,.,;;. m

It is easily seen that with these operations the set of polynomials over R forms a ring. 1.48. Definition. The ring formed by the polynomials over R with the above operations is called the polynomial ring over R and denoted hy R[x]. The 7ero element of R[x] is the polynomial all of whose coefficients are O. This polynomi~1 is called the zero polynomial and denoted hy O. It should always be clear from the context whether 0 stands for the zero clement of R or the zero polynomial.

2U

Algebraic f(lundations

1.49. Definition. Letf(x) = [:_oa,x' he a polynomial over R that is not the zero polynomial. so that we can suppose a" '" O. Then a" is callcd the leadinr, coefficiellt of f(x) and 0 0 the cOllstant term, while n is called the der,ree of f(x), in symbols n = deg(/(x)) = deg(/). By convention, we set deg(O) = - x. Polynomials of degree", 0 are called constant polyllomials. If R has the identity I and if the leading coefficient of f(x) is I, then f(x) is called a monic polyllomial. By computing the leading coefficient of the sum and the product of two polynomials. one finds the following result. 1.50.

Theorem.

Let f. r,

E

R[x]. Then

deg(f + g) '" max(deg(f), deg( g )). deg(fg) '" deg(f) + deg( g).

If R i, all integral domain, we have dcg(fr, ) = deg( f} + deg( r, ).

( 1.4)

If onc identifies constant polynomials with elements of R, then R can he vicwed as a suhring of R[x]. Certain properties of R are inherited by R[x]. The essential step in the proof of part (iii) of the suhsequent theorem depends on (1.4).

1.51. (i)

(ii) (iii)

Theorem.

Let R be a ring. Then:

R[x] is commutati"e if and only if R is commutative. R[x] is a rinr, with identity if and on(v if Rho" an identity. R[x] is all intewal domain if alld only if R is an integral domain.

In the following chapters we will deal almost exclusively with polynomials over fields. Let F denote a field (not necessarily finite). The concept of divisibility, when specialized to the ring F[x], leads to the following. The polynomial g E FIx] divide" the polynomial f E FIx] if there exists a polynomial h E FIx] such thatf = r,h. We also say that g is a divisor of f, or that f is a multiple of r" or that f is divi,ible by r,. The units of Ff x] are the divisors of the constant polynomial I, which are precisely all nonzero constant polynomials. As for the ring of integers, there is a division with remainder in polynomial rings over fields.

1.52, Theorem (Division Algorithm). Let r, '" 0 be a polynomial in Flx]. Then for any f E FIx] there exist polynomials q, r E F[x] such that f = qr, + r.

Example. Consider f(x) = 2x' + x' +4x +3 E !',lx], r,(x) = 3x' + !',Ix]. We compute the polynomials q, r E !',Ix] withf = qr, + r by using

1.53. IE

where deg( r ) < deg( r, ).

21

3. Poiynomiab

long division: 4x'+ 2x'+2x+ 3x' +

II

2x'+x' -2x' -4

+4x+3 X"

x4 + x3

-x'

-2x'

x' +3x'+4x

- x'

-2x 3x'+2x+ 3 -3x' - I 2x-r2

Thus q(x) ~ 4x' +2x' +2x + I. r(x) ~ 2x +2, and obviously deg( r) < C deg(g). The fact that F[x] permits a division algorithm implies by a standard argument that every ideal of F[x] is principal. 1.54. Theorem. F[x] is a principal ideal domain. In fact, for every ideal J ~ (0) of FIx] there exists a uniquely determined monic polynomial g E FIx] with J = (g). Proof F[x] is an integral domain by Theorem 1.51(iii). Suppose J ~ (0) is an ideal of F[x]. Let h(x) be a nonzero polynomial of least degree contained in J, let b be the leading coefficient of h (x), and set g( x) ~ b 'h(x). Then g E J and g is monic. If f E J is arbitrary, the division algorithm yields q.rEF[x] withf~qg+r and deg(r)<deg(g)~deg(h). SineeJ is an ideal. we getf - qg ~ r E J, and by the definition of h we must have r = O. Therefore, f is a multiple of g, and so J = (g). If g, E F[x] is another monic polynomial with J=(g,). then g=c,g, and g, ~c2g with c,.c,EF[x]. This implies g~c,c,g. hence c,c,~I. and c, and c, are constant polynomials. Since both g and g, are monic. it follows that g = g,. and the uniqueness of g is established. 0 1.55. Theorem. Let f" . . ..fn be polynomials in F [x] not all of which are O. Then there exists a uniquely determined monic polynomial dE F[x] with the following properties: (i) d divides each 1;. I" j" n: (ii) any polynomial c E F[x] dividing each 1;, I" j" n. divides d. Moreover. d can be expressed in the form d~h,j,+'" +bnfn

Proof

withb" .... bnEF[x].

(1.5)

The set J consisting of all polynomials of the form c,j,

+ ... + cnfn with c" ... ,cn E F[x] is easily seen to be an ideal of F[x].

Since not all 1; arc 0, we have J

~

(0). and Theorem 1.54 implies that J

~ (d)

22

1\1).!,t'hra:c rOUnOalll.ln<;

for some monic polynomial d'- Flx I. Propeny (i) and the representation (1.5) follow immediately from the construction of d. Propeny (ii) follows from (1.5). If d l is another monic polynomial in Flx] satisfying (i) and (ii). then these properties imply that d and d, are divisible hy each other. aneLso (d) ~ (d l ). An application of the uniqucne" part of Theorem 1.54 yields d-d l • U The monic polynomial d appearing in the theorem ahove is called the greUl"t common divisor of 11.· .../.,. in symhols d ~ gcd(j, .. ...1,). If gcd(j, .... ./,) ~ I. then the polynomials II ... · ·1" arc said to he relUlively prime. Thcy are called pairwise relativel,. prime if gcd(j;. I,) ~ I for I", i < j ~

n.

The greatest common divisor of two polynomials I. g E FIx I can he computed by the Eudidean algorithm. Supposc. without 105' of generality. that g ~ 0 and that g docs not divide f. Then we repeatcdly use the division algorithm in the following manner:

I

~

q,g -'- r,

0", deg(r,) < deg(g)

g

=

Q'!Jl +'!

0", deg(r,) < deg(r,)

rl

= q~r2

+

'1

0", deg( r,) < deg(r,)

0", deg(rJ < deg(r, I)

's

1= qj' 1',_

Here q, ..... q", and r, ..... r, arc polynomials in Fix]. Since deg(g) is finite, the procedure must SlOp aftcr finitely many stcps. If thc last nonzero remainder r, has leading coefficient b, then gcd(j. g) - h 'r,. In ordcr to find gcd(jl .... ./,) for n > 2 and nonzero polynomials I" one first computes gcd(j,./,), then gcd(gcd(jl./' )./3)' and so on, hy the Euelidean algorithm. 1.56.

Example.

The Euclidean algorithm applied to

l(x)~2x6+x'+x'+2<=0'3[X].

g(X)~X4 +X'+2XE0'3[X]

yields:

2x' +x 3 + x',.2 ~ (2x' -'-1)(x 4 + x' +2x)+ x-2 x 4 ,. x" -t 2x ~ (x' + x' + 2x -'- 1)( x .. 2) + I x+2~(x+2)1.

Therefore gcd(j, g)

= I andl and g are relatively prime.

0

A counterpart to the notion of greatest common divisor is that of least common multiple. Let 1, ... .,1, be nonzero polynomials in F[xl. Then one shows (see Exercise 1.25) that there exists a uniquely determined monic

J. Polynomials

2J

polynomial m, Flxl with the following properties: (i) m is a multiple of each f i , I ~.i" n; (ii) any polynomial hE F[xl that is a multiple of each fj, I .:S.; j.:S.; n. is 3 multiple of m. The polynomial m i~ t:alh:d th~ Icast common mulliple of fl .... ,j" and denoted bv m - lcm(/I'"·. Jo)' For two nonzero polynomials J, g E FI x] we have

a

'f~

-

lcm(f, K )ged(f, g),

(1.6)

wher~

a is the kading coeffkient of fK. This relation t:onveniently reduc~s the calculation of lcm(/. g) to that of gcd( J, K). Ther<' is no direct analog of (1.6) for three or morc polynomials. In this c(Jsc. one uses the identity Icm(/,.···./o)c Icm(1em(/,'" .. ./o _I). fo) to compute the least common multiple. The prime clements of the ring FIx] arc usually called irreducible polynomials. To emphasizc this important concept. wc givc the definition again for the present contcxt. 1.57. nefinition, A polynomial p E FIx] is said t,) he irredueihle ocer I' (or irredueib!::.!.!!-..tJ x]. orJ'c!,!,eIn-n,,jfiTp"ii'us posit;v"~~d p he wifhh,CCi-'[x] i~s that either h or e is a e'2E'~~~lial.

Briefly stated. a polynomial of positivc degrec is irreducihle ()ver F if it allows only trivial factorintions. A polynomial in Fix J of positive degree that is not irreducihle over I' is called redueihle vrer F. The rcducihilitv or irreducibility of a given polynomial depends heavily on the field under consideration. For instance. the polynomial x' 2, O[x] is irredueihle over the field Q of rational numbers. hut x' ,·2,' (x' /2 )(x ,Ii) is reducihle over the field of real numbers, Irreducihle polynomials are of fundamental importance f,)r the structure of the ring FIx] since the polynomials in Fix] can be written as products of irrcducibk polynomials in an e~sential1) unique manner. for the proof we need the following result.

1.5/1. I. emma. If an irredueihle polynvmial p ill F[x] dil;ides a produel f, ... f rn of polp/omiols ill Fix]. Ihen iJl I('(iSI olle of liz.. faoors fj is dieisihle hy p. Proof Since p divides J, ... fM' we get the identitv (/, - (p)) . .. (/rn ~ (p) ~ 0 - (p) in the factor ring Flxll( p). 'Jaw Flxl/( p) is a field by Theorem 1.47(iv). and so /, - (p) ~ 0 - ( p) fe'r Sl)me j; that is. p divides fj. C

f

1.59. Theorem (Unique factorization in I"[.\j). E F[x] of posilive degree ran he I>'rillell ill the form

f

~

ap; ... p;',

AllY polynomial

(1.7)

where a E F, PI ....•PJ.. are distinct monic Irreducihle polynomials in F[xl. and el, .... e" are posiTive imegers. Moreocer. this factorization is unique lIparr

from Ihe order in whieh Ihe faclOrs occur.

24

Alg".:bra:(,: Foundation:.

Proof The fact that any nonconstant f E F[ x] can be represented in the form (1.7) is shown by induction on "the degree of f. The case deg( f) = I is trivial since any polynomial in F[ x] of degree I is irreducible over F. Now suppose the desired factorization is established for all noneonstant polynomials in F[x] of degree < II. If deg(f) = 1/ and f is irreducible over F, then we are done since we can write f = a( a 'f), wbere a is tbe leading coefficient of f and a If is a monic irreducible polynomial in FIx]. Otberwise,f allows a factorization f = gh with I ~ deg( g) < n, I ~ deg( h) < n, and g, h E F[x]. By the induction hypotbesis, g and h can be factored in the form (1.7). and so f can be factored in this form. To prove uniqueness, suppose f has two factorizations of the form (1.7). say

( 1.8)

By comparing leading coefficients. we get a = h. Furthermore, the irreducible polynomial PI in FIx] divides the right-hand side of (1.8), and so Lemma 1.58 sbows that PI divides ql for some j. I ~ j ~ r. Rut q, is also irreducible in F[x]. so that we must have ql = CPI with a constant polynomial c. Since ql and p, are both monic, it follows that qj = P ,. Thus we can cancel PI against ql in (1.8) and continue in the same manner with the remaining identity. After finitely many steps of this type. we ohtain that the two factorizations are identical apart from the order of the factors. 0 We shall refer to (1.7) as the canol/ira/facTOrizatiol/ of the polynomial in FIx]. If F= Q. there is a method due to Kronecker for finding the canonical factorization of a polynomial in finitely many steps. This method is hriefly described in Exercise 1.30. For polynomials over finite fields, factorization algorithms will be discussed in Chapter 4. A central question about polynomials in F[x] is to decide wbether a given polynomial is irreducihle or reducihle over F. For our purposes, irreducible polynomials over IFp are of particular interest. To determine all monic irreducible polynomials over IF, of fixed degree n, one may first compute all monic reducihle polynomials over IF, of degree n and then eliminate them from the set of monic polynomials in IFp!x] of degree 1/. If P or n is large, this method is not feasible, and we will develop more powerful methods in Chapter 3, Sections 2 and 3.

f

1.60, Example. Find all irreducible polynomials over IF, of degree 4 (note that a nonzero polynomial in 1F,[x] is automatically monic). There are 2 4 .~ 16 polynomials in 1F,lx] of degree 4. Such a polynomial is reducible over IF, if and only if it has a divisor of degree I or 2. Therefore, we compute all products (a o - a,x + a,x' + x')(b\) • x) and (a o I- a,x + x' I(b o -r b,x + x') with ai' bl E IF, and obtain all reducible polynomials over IF, of degree 4. Comparison with tbe 16 polynomials of degree 4 leaves

25

us with the irredueihle polynomialsf,(x) ~ x' f,(x) ~ x 4 -.- X" + X" + X + I in 1F,[x].

+x

~ 1. f,(x) ~ x 4

+ x' r I, -J

Since the irreducible polynomials over a field F are exactly the prime clements of F[x], the following result. one part of which was already used in Lemma 1.58. is an immediate consequence of Theorems 1.47(iv) and 1.54. 1,61, Theorem, For f E FIx], the residue class rillg F[xl/(j) is a field if alld only if f is irreducihle o"er F.

As a preparation for the next section, we shall take a closer look at the structure of the residue class ring F[xl/(/), where f is an arbitrary nonzero polynomial in F[x]. We recall that as a residue class ring Flxl/(/) consists of residue classes g ~(/) (also denoted hy [gj) with g E F[x], where the operations arc defined as in (1.2) and (1.3). Two residue classes g + (j) and h + (j) are identical ptecisely if g " h mod f -that is, precisely if g - h is divisihle hy f. This is equivalent to the requirement that g and h leave the same remainder after division hy f. Each residue class g + (f) contains a unique representative r E FlxJ with deg(r) < deg(j), which is simply the remainder in the division of g hy f. The process of passing from g to r is called reduction mod f. The uniqueness of r follows from the ob,ervation that if r, E g 'i-(j) with deg(r,) < deg(/), then r - r, is divisihle by f and deg( r - r,) < deg( I). which is only possible if r ~ r,. The distinct residue classes comprising F[xl/(j) can now he descrihed explicitly; namely. they are exactly the residue c1assesr + (/). where r runs through all polynomials in F[ x] wi th deg( r ) < deg( fl. Thus. ifF o. IF, and deg( /) ~ n ? O. then the numher of elements of IF ,[x]l( j) is equal to the numher of polynomial, in 1F,[x] of degree < 11, which is p".

1.62.

[xamples (i)

(ii)

Let I(x) ~ x E 1F,[xJ. The p" ~ 2' polynomials in 1F,[xJ of degree < I determine all residue classes comprising 1F,[xl/(x). Thus. n=, [x III x) consists of the residue classes [0] and [I] and is isomorphic 10 iF 1I.et Ir.,) ~ X' -I- X - I E iF ,[xJ. Then IF ,[xl/(/) has the p" ~ 2' clements [OJ, [I], [x], [x -IJ. The operation tahles for this residue class ring are obtained hy performing the required

operations with the pnlynnmials determining tht.: residue classes and by carrying out reduction mod I if necessary: t

[0] [ I] [x] Ix+1]

[0] [0] [ I] [x] [x

+IJ

[IJ

[x J

[IJ [oj [x+ IJ 1x ]

I x]

[x + I] ----[x + I]

[x - Ii [0 J [I J

[x]

[I] [OJ

26

Algcorruc hlUnJ:J.lions

~O]

I [OJ

[OJ [1] [xJ

[0] [OJ [[oj

'I

[x-'-I]

[xl

[X' I]

[OJ [I] [xJ [x-I]

[OJ [x] [x -,-IJ [I]

[01

I [a]

[I I

[2J

1°1 TiOlli] [21 [i]

[21

lui

[,...:.1_.:...[' _.' _I_['_1_2]_

[.'J

[,+i]

[,'1[

Ix·2[

[,-,I ['I

I']

[dl]

12" ' I[

I~·'· 21 12 ,]

['J Id2J

[21 ['I

[2.,]

1"lJ

[2,,21

[X'2]1

[2"lJ

[2., J

12.,

I

'I

I2I-2J

12, 12J [II]

[2'1 [ IJ

\211 i] 12J

1'1 I'] 1'1

12 1

1°1

IIIJ

Ii]

1"'1

[,.21

I ' . 2]

I 'I [, , I[

[I]

12]

1'1

[Ui1[O]T0]

[0]

[0]

[21

IxJ pxJ

10J

I"

ii]

II[

\21 [,J I x + I] [x 121 [2x]

P,-i], [h ,

[x]

12'1 12x·11- [2,12] ---. 2 \2,] 1 '.'J 12,.2] [2, . II [2" 12 1 [I., I

I

I I[

[x ·1] [IJ

By inspecting these tahles. or from the irreducihility of lover IF, and Theorem 1.61. it follows thatIF,lxl!(f) is a field. This is our first example of a finite field fl'r which the numher of clements is not a prime. Let I(x) ~ x' + 2 EO: IF ,[x]. Then IF ,[x I!(f) consists of the p" ~ 3' residue classes [OJ. [I]. [2J. [xl.[x , l]. Ix + 21.lh].[2x f I]. [2x +2]. The operation tables for IF,[xl/(f) arc again produced hy performing polynomial operations and using reduction mod I whenever necessary. Since IF.,[ x 1/( j) is a commutative ring. we only have to compute the entries ()n and ahove the main diagonal.

(iii)

;

[Ij

211

[x·lJ [,,21 121 1 '-- --[01 IU] 1°1

'--

[II

I,·IJ 12 "-21

1,-1] [2,. II [';'] 12.,,1[ [2.,'21101 I ' ' 2J

12.,J [.,]

[2., . lJ 111 1

12 ) , I J

1,1

I ' , 2] I ' -i I

[2,,21 I ' . 21

[0] [Y., . I[

[I]

\2'

I

'J

I ) + '·1

[2, .

21

°

I I

\2' . 2]

I ' . I] 12 )

-

21

[ , . 'I I I I ' I '] 1°]

°

[7.x

·21

Note that IF,[xl/(f) is nol a field (and not even an integral domain). This is in accordance with Theorem 1.61 since x' + 2 = (x + I)(x +2) is reducible over IF,. [1 If F is again an arbitrary field and I( x) F F\ x]. then replacement of the indeterminate x in I( x) by a fixed element of F yields a well-defined

27

3. Polynomial:-.

clement of F. In detail, if I(x) ~ a o '" alx + ... + anx n E F[x] and bE F, then replacing x hy h wc gct I(b) ~ a o + alh + ... + anb n E F. In any polynomial identity in F[x] we can substitute a fixed b E F for x and obtain a valid identity in F (prillciple 01 subSlilutioll). 1.63. Definition, An element bE F is called a rool (or a zero) of the polynomial IE F[x] if I(h) ~ O. An important connection between roots and divisibility is given by the followi ng theorem. 1.64. Theorem. All elemem hE F is a rool F[x] il alld ollly il x - h divides I(x).

01 Ihe

polyllomial f

E

Proof We use the division algorithm (sec Theorem 1.52) to write q(x)( x - b)+ c with q E F[x] and c E F. Substituting b for x, we get I(b)~c, hence/(x)~q(x)(x-b)+/(b). The theorem follows now from 0 this identity.

I(x)

~

1.65. Definition. Let b E F be a root of the polynomial I E F[x]. If k is a positive integer ,uch that/(x) is divisible by (x - b)', but not by (x - b)' I I, then k is called the mulliplicily of b. If k ~ I, then b is called a simple rool (or a simple zero) of I, and if k ;;, 2, then b is called a mulliple rool (or a multiple zero) of f. 1.66, Theorem, I.et I E F[xt wilh degl = II ~ O. II bl,. .. ,bm E F are dislillct roots 01 I wilh multiplicities kl,. .. ,k m , respectively, Ihen (xb, )"'" (x - bm )'" divides/(x). Consequemly, k l + ... + k m ~ n, andl can hace at most n distinct roots in F.

Proof We note that each polynomial x - bi' I ~ j ~ m, is irreducible over F, and so (x - hi)', occurs as a factor in thc canonical factorization of f. Altogether, tbe factor (x - b l )" .. • (x - bm )'· appears in the canonical factorization of I and is thus a divisor of f. By comparing degrees, we get k I .......... + k m .:s;;; n, and m : : ; k 1 + ... + k m ~ n shows the last state· ment. 0 1.67. Definition. If I(x) ~ au + alx + a,x' + ... + anx n E F[x], then tbe derivative I' of lis defined by!,~ !,(x) ~ a, + 2a,x + ... + nanx n- I E F[x]. ;'

1.68. Theorem. The demem b E F is a multiple rool and only il it is a root 01 hOlh I and 1'.

011 E

F[ x]

if

There is a relation between the nonexistence of roots and irreducibility. If I is an irreducible polynomial in F[x] of degree;;, 2, then Theorem 1.64 shows tbat I bas no root in F. The converse holds for polynomials of degree 2 or 3, but not necessarily for polynomials of higher degree.

2X

Algebraic Foundations

1.69. Theorem. The polynomial! E Flx J o! degree 2 or 3 is irredacihle in F[x] i! and only i!! has no rool ill F. Proof The necessity of the condition was already noted. Conversely. if! has no root in F and were reducible in FlxJ. we could write !~gh with g.hEF[x] and I';;deg(g),;;deg(h). But deg(g)+deg(h)~ deg(/),;;3. hence deg(g)~I; that is. g(x)~ax+h with a.bEF. a"'O. Then - ba" 1 is a root of g. and so a root of! in F. a contradiction. [] 1.70. Example. Because of Theorem 1.69. the irreducible polynomials in 1F,[x] of degree 2 or 3 can be obtained by eliminating the polynomials with roots in IF, from the set of all polynomials in IF,lxJ of degree 2 or 3. The only irrcducible polynomial in 1F,[x] of degree 2 is !(x) = x' + x + 1. and the irreducible polynomials in 1F,[x] of degree 3 are!,(x)~x3 +x + I and !,(x)=x-'+x'-t 1. 0 In elementary analysis there is a well-known method for constructing a polynomial with real coefficients which assumes certain assigned values for given values of the indeterminate. The same method carries over to any field.

1.71. Theorem (Lagrange Interpolation Formula). for n" 0. leI u(), ... ,a n be n -l.. I distinct demems of F. and let ho, ...• bn be n + I arbitrary elemems o! 1-". Then Ihe~e exisls exacllv Olle po(vnomial! E F[x] o! degree ,;; n such Ihal !( a, ) ~ b,jor i = O-.~.~ -: This polynomial is given hy n

!(x)~

L: I·

n

b,

0

n (a,-a

k )"

'(x-a,).

k - 0 k .,

One can also consider polynomials in several indeterminates. Let R denote a commutative ring with identity and let x, ..... x n be symbols that will serve as indeterminates. We form the polynomial ring R[xd. then the polynomial ring R[x,.x,J= R[x,Ux,J. and so on. until we arrive at R[x, •...• xnJ= R[x, ..... xn_lJ[x n]. The clements of R[x, ..... xnJ arc then expressions of the form

f:.:.. /(x 1····.x ll ) = La, ... ,.. X;:··· x~~ with coefficients Q/ .. I C R, where the summation is extended over finitely many II-tuples (i ".'".in ) of nonnegative integers and the convention ~I (1 ~ j ~ n) is observed. Such an expression is called a po~vnomial in Xl" .• , X n over R. Two polynomials f. g E R[x, ..... x n] are equal if and only if all corresponding coefficients arc equal. It is tacitly assumed that the indeterminates x l ••••• x l1 commute with each other. so that. for instance, the exprcs51ions x 1x}.X 3X 4 and X 4 X\X 3 X 2 arc identified.

x?

1.72.

Oefinition.

Letl'= Rlx, •...• x"l be given by (tv

....

\_"""

.... 1 • • • • yin

}

29

P01~-nomiab

It' all"" "I,,:.J. O. then all" ",,,X~I •.. x~' is called a Term of f and i j + ... + in is the degree of the term. for! '" 0 one defines the de!!,ree of!, denoted by deg(f), to he the maximum of the degrees of the terms of f. for ! ~ 0 one sets deg(/) ~ - oc. If ! ~ 0 or if all terms oll.fulVe the same degree, then! is called ho,,:ogel!!_o.-us. -- - -,

Any! E Rlx" ... ,x"l can be written as a finite sum of homogeneous polynomials. The degrees of polynomials in R[x, .... ,x"l satisfy again the inequalities in Theorem 1.50, and if R is an integral domain, then (1.4) is valid and Rlx" ... ,x"J is an integral domain. If F is a field, then the polynomials in Ff x " ... ,x,,] of positive degree can again be factored uniquely into a constant factor and a product of "monic" prime clements (using a suitahle definition of "monic"), hut for n;;, 2 there is no analog of the division algorithm (in the ease of commuting indeterminates) and F[x,,. .. ,x,,l is not a principal ideal domain. An important special class of polynomials in n indeterminates is that of symmetric polynomials.

1.73. Definition. A polynomial! EO R[x,,. .. ,x"J is called symmetric if !(x, ,. . .,x; ) ~ !(x,,. .. ,x,,) for any permutation i" .... i" of the integers L .. ~.n. ~ 1.74. Example.

Let: be an indeterminate over R[x,,. .. ,x,,l, and let

g( z ) = (z - x, )( z - x,)··· (z - x,,). Then

!!,(:)~z"-a,:" '+a,:" '-I ... +(-l)"a"

with

L

a,~a,(x" .. .,x,,)= 1,,;;

II

< ."" <

x,,···x;. I~ ~

(k~1.2, .. .,n).

n

Thus: 01 = Xl I Xl

+ ...

a;? -'- XIX:!"";" X1X~

I XI/'

+ '" + XIX n

t- X 2 X 3 "";" .••

+ X:!X n "";'"

... ...... XfI·" IXn'

As g remains unaltered under any permutation of the x" all the a, are symmetric polynomials: they are also homogeneous. The polynomial a, = a,(x" .. .,x")ER[x,, ... ,x"l is called the kth elementary symmetric polynomial in the indeterminates xj ..... x n over R. The adj~ctive "elementary" is used heeause of the so-called" fundamental theorem on symmetric polynomials," which states that for any symmetric polynomial! E R[x".,.,x"l there exists a uniquely determined polynomial h EO R[x" .. .,x"l such that !(x,,. .. ,x,,) ~ h(a" .. .,a,,). ::J

1,75.

Theorem (NeWlOn's formula).

~""""_'_:n

~~/ •. ~"~.:,.,f,,

;..

'V

..-w'n ..

I.CT a" .. .,"" he the elemen1) n."n '0' .. c::: 71 lInn =."

Algehraic Foundations

)1)

s,(x, .... ,xn)~x;~···+x~ERlx,,.. .,xnjfork"'l. Then the formula

s,

holds for k;, I. where m

= min(k, n).

1.76. Theorem (Waring's Formula). Theorem 1.75, we have

With /he same I/olatiol/ as il/

for k;, I, where the summaliol/ is extended over alll/-lUples (i" ... ,i,) of nonnegative integers with i J + 2i 2 + ... + /lin = k. The coefficient of (J~la~~' •• a> is always an imeger.

4.

FIELD EXTENSIONS

.. Let F be a field. A subset K of F that is itself a field under the operations of F will be called a .,uhfield of F. In this context. F is called an ex/ension (field) of K. If K ~ F, we say that K is a proper subfield of F. If K is a subfield of the finite field IF p' p prime, then K must contain the elements 0 and I, and so all other elements of IF p by the elosure of K under addition. It follows that IF p contains no proper subfields. We arc thus led to the following concept. 1.77. field.

Definition,

A field containing no proper subfields is called a prime

By the above argument, any finite field of order p, p prime, is a prime field. Another example of a prime field is the field (I of rational numhers.

The intersection of any nonempty collection of subfields of a given field F is again a subfield of F. If we form the intersection of all subfields of F. we obtain tbe prime suhfield of F. It is obviously a prime field. 1.78, Theorem. The prime suhfield of a field F is isomorphic to either IF l' or Q. according as [he characterislic of F is a prime p or O.

1.79. Definition. Let K be a subfield of the field F and M any subset of F. Then the field K( M) is defined as the intersection of all subfields of F containing both K and M and is called the extension (field) of K obtained by adjoining the clements in M. For finite M = (0" ... ,On} we write K (M) ~ K(O" ... ,On)' If M consists of a single clement E F, then L ~ K(O) is said to be a simple ex/elISion of K and is called a defil/ing element of I. over K.

°

°

4. Field Extcn!'.ions

31

Obviously, K( M) is the smallest subfield of F containing both K and M. We define now an important type of extension. 1.80. Definition. Let K be a subfield of F and 0 E F. If 0 satisfies a nontrivial polynomial equation with coefficients in K, that is, if anon + ... + 0 10 + 0 0 ~ 0 with a, E K not all being 0, then 0 is said to be algebraic over K. An extension L of K is called algebraic over K (or an algehraic extension of K) if every element of L is algebraic over K. Suppose 0 E F is algebraic over K, and consider the set J = K[x): f(O) ~ 0). It is easily checked that J is an ideal of K[x]. and we have J". (0) since 0 is algebraic over K. It follows then from Theorem 1.54 that there exists a uniquely determined monic polynomial g E K[x] such that J is equal to the principal ideal (g). It is important to note that Ir is irreducible in K[x]. For, in the first place, g is of positive degree since it has' the root 0; and if g ~ h,h, in K[x] with I", deg(h;l < deg(g) (i ~ 1,2), then ~ g(O) ~ h,(O)h,(O) implies that either hi or h, is in J and so divisible by

(j

E

o

g, which is impossible.

1.81. Definition. If 0 E F is algebraic over K, then the uniquely de· termined monic polynomial I( E K [x] generating the ideal J ~ (j E K[x): f( 0) ~ 0) of K I x) is called the minimal polynomial (or defining polynomial, or irreducible polynomial) of 0 over K. By the degree of 0 over K we mean the degree of g. 1.82. Theorem. If 0 E F is algehraic over K. rhen its minimal po(vnomial g ocer K has the following proper/ies:

g is irreducible in K[x). For fE K[x] wehavef(O)~O ifandonlv ifgdividesf. (iii) I( is lhe monic polynomial in K[x] of leasl degree hacing 0 as a (i) (ii)

rool.

Proof Property (i) was already noted and (ii) follows from the definition of g. As to (iii). it suffices to note that any monic polynomial in K [x 1having 0 as a rool must be a multiple of g. and so it is either equal to g or its degree is larger than that of g. D

We note that both the minimal polynomial and the degree of an algebraic element 0 depend on the field K over which it is considered. so that one must be careful not to speak of the minimal polynomial or the degree of 0 without specifying K, unless the latter is amply clear from the context. If L is an extension field of K, then I. may be viewed as a vector space over K. For the elements of L ( ~ .. vectors") form, first of all, an ahelian group under addition. Moreover. each "vector" a E t. can be multiplied hy a .. scalar" r E K so that ra is again in L (here ra is simply the

32

Algebrail.:

foundation~

product of the field elements r and a of L) and the laws for multiplication by scalars arc satisfied: r( a -'- {3) = ra + r{3, (r +,)a = ra + sa, (rs)a = r(sa), and la = a, where r, s E K and a, {3 E L. " 1.83. Definition. Let L bc an extension field of K. If L, considered as a vector space over K, is finite-dimensional. then L is called a finiTe e:.aension of K. The dimension of the vector space I. over K is then called the degree of I. over K. in symbols l J.: K]. 1.84. Theorem. If J. is a finile eXlension of K and M i, a finile eXTension of /., [hen M is a finite eXlension of K wilh

[M:K]=[M:L][L:K]. Proof PUlrM:Ll=m,II.:Kl~n.andlet{a,.... ,a",}beabasis of M over I. and ({3, .... ,{3,) a basis of I. over K. Then every aC' M is a linear combination a = YIO':I; ••. - YmCt m with y, E:: I. for I.:;:;: i:::;; »1, and writing each Yi in terms of the basb clements 13, we get m

n,

!

m

11

1J

I: Y,a, = I: II:, r,A In, = I: I: r, ,Il,a,

a~ I

I

,.- I \ J - ,

1-1 / -

I

I

with coefficients rlt E K. If we can show that the: n1n clements f3J D.,. I ~ i ~ m. I ~ j s:;; n, arc linearly independent over K, then we are done. So suppose we bave m

,

I: I: I

witb coefficients s'J

E

s,,{3,a,

=0

-= I / -!

K. Then

and from the linear independence of the a, over I. we infer

L s,/l, -- 0

for I ~ i ~

In.

1·1

But since the {I, arc linearly independent over K. Wl.: concluul.: that all

SII

Q 1.85.

Iheurem,

are 0

LGcry finile cXlensioll of K is algehraic IiDer K.

Proof Let l. be a finite extension of K and put 1/.: K J = m. For 0("" L. the m + I clement> I, 0, .. .,0"' must then be linearly dependent over K, and so we get a relation a o I il 1(J •..• + ilfYI()m =-= 0 with a i C K not all being O. Thi,just says that 0 is algebraic over K. CJ

4. Field

r..xtension~

33

For the study of the structure of a simple extension K(O) of K obtained hy adjoining an algebraic element. let F he an extension of K and let 0 E F he algebraic over K. It turns out that K(O) is a finite (and therefore an algehraic) extension of K. 1.86. Theorem 1RI 0 E F he algebraic o[ degree n ocer K and leI g he Ihe minimal polvnomial o[ 0 over K. Then: (i)

(ii) (iii)

K(O) is i.wmorphic 10 K[x]j(g). [K(O):K]~nand{I,O, .... On ') is a basis o[K(O) over K. t'cer} a E K (0) is algebraic neer K and its degree over K is a dicisor o[ n.

Proof (i) Consider the mapping T: K[xJ ~ K(O), defined hy T(j) for [E K[x], which is easily seen to he a ring homomorphism. We have kerT ~ {j E K[x]: [(0) = O}~ (g) by the definition of the minimal polynomial. Let S be the image of T: that is, S is the set of polynomial expre"ions in 0 with coefficients in K. Then the homomorphism theorem for rings (see Theorem lAO) yields that S is isomorphic to K [x]j( g). But K[x]j(g) is a field by Theorems 1.61 and 1.82(i), and so S is a field. Since K >; S >; K(O) and 0 E S, it follows from the definition of K(O) that S ~ K(O), and (i) is thus shown. (ii) Since S = K(O), any given a E K(O) can he written in the form a ~ [(0) for some [E K[x]. By the division algorithm, [~qg -i- r with q,rEK[x] and deg(r)<deg(g)~n. Then a~[(O)=q(O)g(O)Tr(O)~ r( 0), and so a is a linear combination of 1,0, .... 0' 'with coefficients in K. On the other hand, if a o + a,O + ... + an ,0'-' ~ 0 for certain a, E K, then the polynomial h(x) ~ a o + a,x + ... + an ,Xn 'E K[x] has 0 as a root and is thus a multiple of g by Theorem 1.82(ii). Since deg(h) < n ~ deg(g), this is only possible if h = O-that is, if all a, ~ O. Therefore, the elements 1,0.... ,0'-' arc linearly independent over K and (ii) follows. (iii) K(O) is a finite extension of K by (ii), and so a E K(O) is algehraic over K hy Theorem 1.85. Furthermore, K(a) is a subfield of K(O). If d is the degree of a over K. then (ii) and Theorem 1.84 imply that n=[K(O):K]~[K(O):K(a)][K(a):K]=[K(O):K(a)]d.hence d di= [(0)

vides n.

LJ

The clements of the simple algehraic extension K(O) of K are therefore polynomial expressions in O. Any element of K(O) can be uniquely represented in the form a o + 0 1 + ... + an Ion 1 with a/ E K for 0 ~ i ~ n-1. It should he pointed Out that Theorem 1.86 operates under the asoumption that hoth K and 0 are emhedded in a larger field F. This is necessary in order that algebraic expressions involving make sense. We

°

°

now want to construct a simpk algehraic extension ah oeo

that is, without

34

Algebraic roundations

reference to a previously given larger fidd. The clue to this is contained in part (i) of Theorem I.R6. 1.87. Theorem. l.et f E K[x] be irreducible ""er the field K. Then Ihere exisls a .,imple algebraic extension of K wirh a roOi of f as a defining element. Proof Consider the residue class ring 1. = K[xl/(/), which is a field hy Theorem 1.61. The clements of L are the residue dasses [h] = h + (/) with hE K [x 1. For any a E K we can form the residue class ra] determined by the constant polynomial a, and if a, h E K arc distinct, then [aj'" [h]' since f has positive degree. The mapping a""""" [a] gives an isomorphism from K onto a subfield K' of L, so that K' may be identified with K. In other words, we can view 1. as an extension of K. For every h(x) = au + a,x + ... + amx m E K[x] we have [h]= [aD + a,x + ... + amx m]= [au]+[aJ![x]+ ... +[am][x]m = aD + a,[x]+ ... + am[x]m hy the rules for operating with residue classes and the identification [a,] = a,. Thus, every clement of L can he written as a polynomial expression in [x] with coefficients in K. Since any field containing both K and [x] must contain these polynomial expressions, 1. is a simple extension of K obtained hy adjoining [x]. If f(x)=bu+b,x+'" +hnx n. then f([x])=bo+b,lx] + ... + bn[x)" = [bo + h,x + ... + hnx n ] = [f] = [0], so thatfx] is a root of f and L is a simple algehraie extension of K. 0

1.88. Example. As an example of the formal process of root adjunction in Theorem 1.87, consider the prime field IF, and the polynomial f(x) = x' + x + 2 E 1F,[x], which is irreducible over IF,. Let 6 be a "root" of f; that is, 6 is the residue class x + (/) in I. = 1F.,[x 1/(/). The other root of fin L is then 26 +2. since f(26 +2) = (26 +2)' +(20 +2)+2 = 6' + 6 +2 = O. By Theorem 1.86(ii), or by the known structure of a residue class field. the simple algebraic extension I. = 1F,(6) consists of the nine dements 0,1.2.6,6+1.6+2,26,26+1,26+2. The operation tables for 1. can he constructed as in Example 1.62. 0 We observe that in the above example we may adjoin either the root 6 or the root 26 + 2 of f and we would still obtain the same field. This situation is covered hy the following result, which is easily established. 1.89. Theorem. Ler a and f3 he fWO rools of the polynomial f E K l x] Ihat is irreducible over K. Then K(a) and K(f3) are isomorphic under an isomorphism mapping a 10 f3 and keeping the elemellls of K fixed.

We are now asking for an extension field to which all roots of a given polynomial helong. 1.90. Definition. Let f E Klx] he of positive degree and F an extension field of K. Then f is said to splil in F if f can be written as a product of

4 field

LXIl·m.ion~

35

linear factors in Flx] that

that is, if there exist elements "" "" ... ,". E F such

/(x)~a(x- "I)(X

- ",) ... (x-".),

where a is the leading cocfficicnt of f. The field F is a splilling /ield of / over Kif / splits in F and if, morcovcr, F= K("I' "" ... ,,,.). It is clear that a splitting field F of / over K is in the following sense the smallest field containing all the roots of /: no proper suhfield of F that is an extcnsion of K contains all thc roots of f. By repeatcdly applying the process used in Theorem 1.87, onc obtains the first part of the subsequent result. Thc second part is an extcnsion of Thcorem 1.89. 1.91. Theorem (Existcncc and Uniqueness of Splitting Field). 1/ K is a /ield and / any polynomial 0/ posilive degree in K rx], lhen lhere exisls a splilling /ield 0/ / over K. Any IWD splilling /ields 0/ / oeer K are isomorphic

under an isomorphism which keeps lhe elemenrs into each other.

0/ K fixed and maps roOlS all

Sincc isomorphic ficlds may he identified, we can speak of lhe splitting field of / over K. It is obtained from K hy adjoining finitely many algehraic elements over K, and therefore one can show on the basis of Theorems 1.84 and 1.86(ii) that the splitting field of / over K is a finite extension of K. As an illustration of the usefulness of splitting ficlds, we consider the question of deciding whether a given polynomial has a multiple root (compare with Definition 1.65).

1.92. Definition. Let / E K [x J be a polynomial of degree n;, 2 and suppose that/(x)~au(x-"I)"'(x-".) with ", ..... ". in the splitting field of / over K. Then the discriminanr D(/) of / is defined by D(f) ~

a6·'

n (", - "y. 1 .... 1

< , .... n

It is obvious from the definition of D(/) that/has a multiple root if and only if D(/) = O. Although D(/) is defined in terms of elements of an extension of K. it is actually an clement of K itself. For smalln this can he seen hy direct calculation. For instance. if n ~ 2 and /(x) ~ ax' + bx + c = a(x-"I)(x-a,). then D(/)~a'(al-a,)'~a'«a,+a,)'-4a,a,)= a'(b'a-' -4ca- I ), hence

D(ax' + hx + c) ~ h' -4ac. a well-known expression from the theory of quadratic equations. If n ~ 3 and /(x)~ak'+bx'-rcx+d~a(x-a,)(x-a,)(x-- a,). then D(/)= a 4("1 - "')'("1- ",)'(a 2 - a,)". and a more involved computation yields

D(ax'

-r

hx' ~ ex

f

d) ~ h'(" -4b'd -4ac' - 27a'd' + 18ahcd. (1.9)

Algehraic

roundation~

In the general case, consider first the polynomial s E K[x" ... ,x"] given by 2n' S ( X1'''·'X n ) --a O

n (

X/-X;

1 ,;;

I <.

)2 .

j ., ; : n

Then s is a symmetric polynomial, and by a result in Example 1.74 it can he written as a polynomial expression in the elementary symmetric polynomials 0, .... ,on -that is, s = h( 0, .... ,on) for some h E K [x " .. .,x n]. If f(x) = auxn+a,x n '+ ... +an=aO(x-a,)"'(x-a n ), then the definition of the elementary symmetric polynomials (see again Example 1.74) implies that o,(a" ... ,an)~(-I)kakao'EK for I.;k.;n. Thus, D( j) ~s( a" ... ,an) ~ h (0, (a" ... ,a") ... .,0n (a p

•••

,a"»

~ h ( - a,a o ', .. ., ( - I) n ana o ') E K.

Since D(f) E K, it should be possihle to calculate D(/) without having to pass to an extension field of K. This can he done via the notion of resultant. We note first that if a polynomial f E K [x 1 is given in the form f(x) = aox n + a,x n-' + ... + an and we accept the possibility that a o ~ 0, then n need not be the degree of f. We speak of n as the formal degree of f; it is always greater than or equal to deg(f). 1.93. Definition. Let f(x)= aux n + a,x n '+ ... + an E K[x] and g(x) = box m

+ h,x m '+ ... + bm E K[x] be two polynomials of formal degree

n resp. m with n, mEN. Then the resultant R(f, g) of the two polynomials is defined by the determinant ao

a,

0

00

an a,

0 an

0

~

:I 1

mrows

0 hu

R(j,g)~

0 I 1

of order

JJI

"T"

0 h, bo

ao

a, bm

h,

0 hm

:

0

0

hu

b,

a ' " 0

0

1--

hmi

n.

If deg(/) ~ II (i.e., if a" = 0) and fix) ~ ao(x - a,)'" (x - an) in the splitting field of f over K, then R(f, g) is also given hy the formula n

R(j,g)~aong(a,).

(1.10)

1=1

In this case, we obviously have R(f, g) ~ 0 if and only if f and g have a common root. which is the same as saying that f and g have a common divisor in K rx 1of positive degree.

37

Theorem 1.68 suggests a connection between the discriminant D(j) and the resultant R( I.et E K rx 1 with deg( /) = tl ;;, 2 and leading coefficient au' Then we have, in fael. the identity

t, /,).

t

D(f)~(-I)"("

II!2 a ,,'R(j,j'),

(1.11)

where /' is viewed as a polynomial of formal degree tl - I. The last remark is needed since we may have deg(j') < tl -- I and even /' .~ 0 in case K has prime characteristic. At any rate, the identity (1.11) shows that we can obtain D(/) by calculating a determinant of order 2tl- I with entries in K.

~:XERCISF:S

1.1.

Prove that the identity element of a group is uniqucly determined. For a multiplicative group G, prove that a non empty suhset 11 of G is a suhgroup of G if and only if II, b c H implies ab 'Eo N. If H is finite, then the condition can be replaced hy: II, b c H implies lib Ell. 1.3. Let II be an element of finite order k in the multiplicative group G. Show that for til EO Z we have am - e if and only if k divides m. 1.4. For mEN, Euler's function ¢( til) is defined to be the numher of integers k with I,,; k ,,; til and gcd( k, m) = I. Show the following pr()pcrtie~ for 111, II, S F N and a prime p: , I ) (a) ¢(p')=p,II-. : \ p (b) ¢(mtl)=¢(m)¢(tl)ifged(m,n)~I: , I \ ' I' Ie) ¢lm)=til!1 - / ... 11--- \' where m=p; "'p; is the , PI I \ Pr J 1.2.

prime factor dccompo~ition of m.

1.5. 1.6.

1.7.

Caleulate <1>(490) and ¢(68). Usc the class equation to sliow the following: if the order of a finite group is a prime power p~. p prime, S ~ I, then the order of it~ center is divisihle hy p. Prove that in a ring R we have (- a)( - hI -- IIh for all II, "EO R.

I.g,

Prove th(jt in a commutative ring R the formula

I.Y. 1.10.

holds for all II," F R and II c N. (Binomial Theorem) Let p he a prime number in Z. for all integers II not divisible by p, show that p divides a f I - I. (Fermat's Little Theorem) Prove that for any prime p we have (p -I)' -- -- I mod p, (Wilson's Theorem)

1.11. 1.12.

J. D.

1.14. 1.15.

Prove: if p is a prime. we have ( p

Provt:: if m 1 , ... ,n1 A arc positive integers that arc pairwise relatively prime-that is. ged(m,. r>lJ! ~ I for I,;; i < j,;; k -then for any integers a l' ...• a" the system of congruences y == a I mod mI ' i = 1,2, .... k. has a simultaneous solution,. that is uniquely determined modulo m 111 1 '" m". (Chinese Remainder Theorem) Solve the system of congruences 5x:c. 20mod6. 6x '" 6mod5. 4x '" 5 mod 77. For a commutative ring R of prime characteristic p. show that (11

"" + ... -a,) p.:....:af . . . ...

p

,

L ;

1.18.

p~ +(1\

for all a, ..... a, E Rand n E' I'll. Deduce from Exercise 1.11 that in a commutative ring R of prime characteristic p we have

(a-h)" : 1.17.

I ) '"" (- I)J mod p for 0 ,;; j ,;;

p- I.jE7L. A conjecture of Fermat stated that for all n ;, 0 the integer 2'" + I is a prime. Euler found to the contrary that 641 divides 2" t I. Confirm this by using congruences.

(

1.16.

7

alh" , ,

foralla.hC" R.

II

Let F be a field and f0 Fix]. Prove that (K(j(x)):KE F[x]) is equal to FIx] if and only if deg(j) - I. Show that p'(x)- xq'(x) ~ xr'(x) for p. q. r E D;l[ x] implies p ~ q ~

r ~ O. 1.19. 1.20.

1.21. 1.22. 1.23. 1.24.

Show that if f. K Eo Fix]. then the principal ideal (j) is contained in the principal ideal (K) if and only if K dividesf. Prove: iff. K E Fix] are relatively prime and not both constant. then tbere exist a. b E F[x] such that af + hK ~ I and deg(a) < deg(g). deg( b) < deg(j). Let f, .....f" E Fix] with gcd(j, !.,) ~ d. so that !, ~ dl(, with K, E FIx] for 1,;; i,;; n. Prove that g, K" arc relatively prime. Prove that ged(j, ... ..f,) ~ ged(ged(j, J, ,). f,) for II ;, 3. Prove: if f.l(. h E FIx]. f divides I(h. and ged(j.1( ).- I. then f divides h. Use the Euclidean algorithm to compute ged(j. g) for the polynomials f and I( with coefficients in the indicated field F: (a)

F~Q. f(X)~X7+2x'+2x'-x+2. l(x)~x'-2X'-X41

x 2 +2x.-+ 3 (h) (c)

(d)

F=1F2,f(X)~x7~1.I(x)~xs+xl-jx

tl

F~IF,,f(x)~x'+x+l.l(x)~x6I x'+x 4 +1 F~1F3' f(X)~X8-i-2xs~X3 ex'+I. g(X)~2X6+X5+2xl

+2x'

+2

.19

Exercises

1.25.

1.26. 1.27.

1.2B. 1.29.

Let 1, .... ./" be nonzero polynomials in F[x]. By considering the intersection (/,)(1 ... n(/,,) of principal ideals. prove the existence and uniqueness of the monic polynomial mE F[x] with the properties attributed to the least common multiple of 1, .... ./" Prove (1.6). If I, ...../" to FIx] are nonzero polynomials that arc pairwise relatively prime. show that Icm(/, ... .. I,,) ~ a "/, ... I". where a is the leading coefficient of I, ... I". Prove that lem( I,.··· .f~) ~ Iem(lem( f, .... ./" ,). /,,) for n '" 3. Let 1, .... ./" to F[x] he nonzero polynomials. Write the canonical factorization of each j" I ~ i ~ II, in the form

where u/

E::

F, the product is extended over all monic irreducihle

polynomials pin Flx]. the e,( p) arc nonnegative integers. and for each i we have e,( p) > 0 for only finitely many p. For each p set m( p) ~ min(e,( p)..... e,,( pi) and M( p) - max(e,( p)..... e,,( pl). Prove that {)-nU'IP' gc d(11'··· 'In (J •

IemU, ... ../,,) ~ 11 pMII". 1.30.

Kronecker's method for finding divisors of degree.;; s of a noneanstant polynomial IE Q[x] proceeds as follows: (I) By multiplying I by a constant. we can assume I E Z[x]. (2) Choose distinct elements ao, ...• a,E Z that are not roots of I and determine all divisors of I(a,) for each i.O';; i.;; s. (3) For each (s + I)-tuple (h" ..... h,) with h, dividing I(a,) for 0.;; i.;; s. determine the polynomial g E Q[x] with deg(g).;; s and g(a,)=h, for O.;;i.;;s (for instance. hy the Lagrange interpolation formula). (4) Decide which of these polynomials g in (3) are divisors of f. If deg(j) = n ?> I and s is taken to be the greatest integer <; n/2. then I is irreducible in Oix1 in case the method only yields constant polynomials as divisors. Otherwise. Kronecker's method yields a nontrivial factorization. By applying the method again to the faclors and repeating the process. one eventually gets the canonical faetori/,ation of f. Use this procedure to find the canonical factorization of

I(x)" :x'-ix s f2x 4 -x' t5x'- \'x-IEQ[xl· 1.31.

Construct

the addition

and

multiplication

tahle for IF ,Ix 1/

(x' - x' + x). Determine whether or not this ring is a field. 1.32. Letlx+lj be the residue class of x+1 in IF,lx1/(x 4 +1). Find the residue classes comprising the principal ideal ([x + III in IF,lxJl ( .. 4

I

1\

Algchmic f-'oundalit)n . .

4{\

Let I' he a field and a. h. g r FI x 1 with g ~ O. Prove that the congruence af ~ h mod g has a solution 1 '" FI x] if and only if gcd( a. g) divides h. 1.34. Solve thc congruence (x' I IJ./h) = I mod(x' + I) in f "lxJ, if possible. 135. Solve (x' I x" I x' - I)!(.<)=(x' T I)mod(x' .,,1) in r,rxl if

1.33.

possihle.

1.36.

Prove that R[xl!(x 4

T

X"

+ x'" I) eannot he a field. no malter what

the commutative ring R with identity is.

1.37.

Prove: given a field F. nonzero polynomiabli"",f, c. Flxlthat are gl g" E }-·[xl. then the simultaneous c(mgruenccs h ::== XI mod!,. i = 1.2 ,k. have a unique solution hE FIx 1 modulo 1 . 1, ... 1,. (Chinese Remainder Theorem for Ff x 1> Evaluate 1(3) for 1( x) - X 214 -:1.<".' + 2x 4 " I 2 c IF ,I x]. Let p h~ a prime and a v'" .. il" int~gcrs with p not dividing a l1 • Show that all + illY' ... I a,tY n - Omoo p has at moM n different :o,olutions y modulo p. If P > 2 is a prime, ~how that there ar~ exactly two dements il F U- p sueh that a) - 1. Show: if fElix] and 1(0) '" 1( I) '" I mIld 2. then 1 has no ro,'ts in 1. Let p he a prime and 1 E l[xl. Show: l(a)" Omod p holds for all a c 1 if and only if l(x)'" (XC x)g(x) I ph(x) with 1'. h E 1lxl· Let p be a prime integer and (' an ekment of the field F. Show that x P - c is irreducihle over F if and only if x" - c has no root in F. Show that for a polynomial I r Fix] of positive degree the following puirwis~ rdatively prime. and arbitrary polynomials

1.3H. LN.

lAO.

1.41. 1.42. 1.43. 1.44.

condition~ are cquivi;llcnt:

1.45.

(a) I is irredueihle over F: (h) the principal ideal (/) of F[x] is a maximal ideal: (e) the principal ideal (/) of Fix] is a prime ideal. Show the following properties of the derivative for polynomials in FIx]: (a) (h) (c)

(/,-'" (f~)'~

+ I",)'~ I{+ ... -

'" Ll,"'I, ,/,'/,.,"-/,,,.

(/""/,,,)'~ I

1.46.

1.47. 1.48.

1.49.

I~:

j'g'" II": =- I

For I c FIx) and I' of characteristic 0, provc that j' ~ 0 if and only if 1 is a constant polynomial. If I' has prime characteristic p, prove thatj'=O if and only if/(x)~g(xP) forsomcgE Flxl. Prove Theorem 1.68. Provc that thc nonzcro polynomial I E F[ x] has a multipk root (in some extension field of F) if and only if I and j' arc not relativcly prime. Usc the criterion in the previous exercise to determine whether the

41

following polynomials have a multiple root: f(x)~x'-5x'+6x'i4x-8EOQ[x]

(a)

(h) f(x)~x' ex ' 1 x' 1 xJ-IEf,lx] 1.50, The nth derivativef'"' of f C Fix] is defined recursively a, follows: 4 f'"'·= 1./'''' ~
(fg)''''~

L" I""

1,51,

1.52.

Let F he a field and k a positive integer such that k < p in ca,e Fha., prime characteristic p, Prove: bE F is a root of f c= FI x J of multiplicity k ifand only iff"'(h)~O forO"i"k -·1 andf'k'(h)~O. Show that the Lagrange interpolation formula can also be written in the form

~

f(xl=

1...J I

1.53,

1.55. 1.56,

. L57,

1.58,

L59. 1.60. 1.61.

o

1.62.

()

n

b( '(a») ,g(x) I

g

I

X -

withg(x)~

a

I

n (x-a,).

k=U

Determine a polynomial fEIF,[xJ with f(0)~f(1)~f(4)~1 and ~ f(3) ~ 3, Determine a polynomial f EO Qlx] of degree" 3 such that f( 1) = - I. flO) ~ 3, f( I) = 3, and f(2) = 5, Express s)(xl.X2.x.~.X4)=xf+x~-1-x~+X~ElF3rXI,X2.X1,X4J in terms of the elementary symmetric polynomials ° 1_ 02' °3 _ °4 , Prove that a subset K of a field F is a subficld if and only if the following conditions are satisfied: (a) K contains at least two elements: (b) ifa,bE K, then a-hE K: (c) ifa,hEK and h~O, then ab 1 EK, Prove that an extension /, of the field K is a finite extension if and only if /. can be ohtained from K hy adjoining finitely many algebraic clement> over K. Prove: if 0 is algebraic over /, and L is an algehraie extension of K, then 0 is algehraie over K. Thus show that if F i, an algehraie extension of /., then F is an algehraie extension of K. Prove: if the degree [L: KJ is a prime, then the only field, F with KeF r;; L are F = K and F = L. Construct the operation tahles for the field I. ~ ~J(Ol in Example 1.88. Show that f(x)~x4-x+IEIF,lxJ is irreducible over IF,. Then construct the operation tahles for the simple extension IF.,(O), where is a root of 1. Calculate the discriminant D(f) and decide whether or not f ha> a multiple root: (a) f(x) = 2x 3 -3x' -x + I E Q[xJ

f(2) 1.54,

(~)f'" "g''',

0

42

Alg,chra;c

Foundat~ons

(0) I(x)' 2x'l-k;+X2+2x-2c~,,[x] 1.63. 1.64.

1.65. 1.66. 1.67.

Deducc (1.9) from (1.11). Prove that I. g E K [x] have a common root (in "orne extension field of K) if and only if I and g have a common divisor in K [x] of positive degree. Determine the common roots of the polynomials x 7 - 2x' - k' + 2 and x' - 3x 4 - Xi 3 in O[x]. Prove: if I and g arc as in Definition 1.93. then R(/. g) =

(- l)m'R(g. fl. Let I. g E K[x] be of positive degree and suppose that I(x) ~ Go(x-a,)"·(x-a,,). Go*O. and g(x)~h(](x-fJ')"'(x-fJm)' bo * O. in the splitting field of Ig over K. Prove that m

n

J

1.6~.

1.70.

I"'"

1j - I

where nand m are also taken as the formal degrees of I and g. respectively. Calculate the resultant R(/. g) of the two given polynomials I and g (with the formal degree equal to the degree) and decide whether or not I and g have a common root: (a) I(X)-X'+X+l.g(x)~2x'· x'· 2FIF,[x]

(0) 1.69.

1

n (a;-fJjl. 111

R(f.g)= (_l)m"h~n/(fJ)~a;;'h;;1l

I(X)~X4+X'll.g(x)~X4_X2+X-ICIF,[x]

For 1'= Klx, ..... x,]. n" 2. an II-tuple (a, ..... a,) of clements n. od<1llging to some extension L of K may be called a zero of I if l(a, ..... a,)~O. :-.low let /'g'= Klx, ..... x,,] with x, actually appearing in I and g. Then I and g can be regarded as polynomials fix,,) and )(x,) in Klx, ..... x, ,Ux,,] of positive degree. Their resultant with respect to x, (with formal degree ~ degree) is R(/.)() ~ R,.
zeros of the polynomials/(x.y)~x(y'-x)'·v' and g(x.y)y4 + r'; - x' in O[x. y].

Chapter 2

Structure of Finite Fields

This chaptl.:f i~ of central importanl:l.: :-.incc it contains various fundamentul properties of finite fields and a dc:-.cription of mcthoos for L'on~tructing

finite fields.

The field of integers modulo a prime numher

i~.

of course. the most

familiar example of a finite field. hut many of its properties extend to arhitrary finite fields. The eharaeteriLation of finite field, bee Section I)

shows that every finite field is of prime-power order and that. conversely. for every prime power there exists a finite field whose numha ()f elements is exactly that prime power. Furthermore. finite fields with the same numher of clements are isomorphic and may therefore he identified. The next two sections provide information on roots of irrcducihlc polylwmiab. leading to

an interpretation of finite fields as splitting fields of irredueihle polynomi-

als. and on traces. norms, and

has~s

rdative to fidd extensions.

Section 4 treats roots of unity from the viewpoint of general field theory. which will he needed occasionally in Section 6 as well as in Chapter 5. Section 5 presents different ways of representing the elements of a finite field. In Section 6 we give two proofs of the famous theorem of Wedderburn

acc(xding to which every finite

divi~ion

ring

i~

a field.

Many discussions in this chapter will he foll,)wed up. continued. and partly generalized in later chapters.

Strudurl' of hnill' rkld:-

:j.4

/1.

C1IARACTERIZA1'101'1 OF FINITE FIELDS

In the previous chaptcr we have already encountered a basic c1a~~ of finite field., that i,. of fields with finitely many elements. For every prime p the residue class ring Z /( p) forms a finite field with p clements (see Theorem U~). which may be identified with the Galois field I'p of order p (sec Uefinition 1.41). The fields I' p play an important role in general field theory since every field of charactcri~tic p must contain an i~omorphic copy of IF p hy Theorem 1.7X and can thus he thought of as an extension of I'p. This ohservation. tngether with the fact that every finite field has prime charaucristil: (sec Corollary 1.45), is fundam~ntal for the classification of finitc fields. We first e~tahlish a simple neces~ary condition on the numher of e1cmenb of a finite field.

2.1. I. emma. I.e[ F he u finite field containing elemems. nlell F hw qnl ell'mems, where m .;..·1 f': K 1.

U

SlIlifield K with q

Prvoj. F i~ a vcctor ~pace over K. and since F is finite, it is finite-dimensional as a vector space over K. If [F: K J ~ m. then f' has a hafo.is over K consisting of In element~, say hi' b'!. .... ,hm • Thus every element of F can hc uniquely represented in the form alh j + a 2 b.,. - ...• a...,hm • where at. £1'!., .... a m 0" K. Sincc each £1 , can have q values, F has exactly qm clements. D 2.2. Theorem. I.e[ F he u Jini[e field. Then F has p" ('I('mellls. where the prinze p is the charu(,(t'fistic vf F £1lld 11 is the degree of F Ol::er irs prime suhfield.

Pmo! Sinl:c F i~ finite, its charal:tcri~tic is a prime p according to Corollary I.4S. Therefore the prime suhfield K of F is isomorphic to I'p by Theorem 1.7X and thus contain" p ekments. The r~st follow~ from Lemma

2. I.

"

Starting from the prim~ fields I!-". W~ can construct other finite fields h) thl,.' proce~:-; l)f root adjunction desnibed in Chapter I. Section 4. If f ( 1',,[ x I is an irredueihle polvnomial over ~ p of degree n. then hy adjoining a f()()t of f to 1f fJ we get a finite fidd with p" elem~nts. llowever. at this stage it i.. . nl)t dl.:ar whether for every positive intege.:r II there.: exists an irrcducihlc polyll0mial in I' p lx1 of degree n. In order to estahlish that for every prime p and e.:vcry n E N then.~ i:-- a finite field with pfl elemenb, we use an approach suggested hy the following results.

2.3. slillsjies a" th~

I. emma.

If F is u filli[(' field with q elemems. [hen everv a

F

F

= li.

PrO(~f The.: identity £11{ '- a i~ trivial for a = O. On the other hand, mml'e.:ro elements of F form J group of order q - I under multiplication.

4.'

Thu:'l a
E:

F with a

=1=

O. and mUltiplication hy a yic:lds

thL'

Il

2.4. Lemma. If F is a !lIli(e jle/d \1I"ii/l (/ c1eml!III.\ and K i.\ a suhj/('/d of F. rhen rhe polynomial x'· x in KI.,jfatiors in Fix] as

.,'1- x

~

n (x -

'I l:.

and F is a splilling field of x q

'.

II)

I"

x vrer K.

Proof The polynomial .>.:" - X of Jcgrt::e q has at !n\l..,t q wots in F. By Lemma 2.3 "e know q ,ueh rool> namely. all the elements of F. Thu, the given polynomial splits in F in the indicated manner, and it can no! :-:.plit in any smaller field. c:: W...:- ar~ now ahle to prove the main t.:haracteriLution thcon:m for finite fields. the leading idea being contained in Lemma 2.4.

2.5. Jheorem (Existence and Lniquenc>s of finite helds). For every prime p and erery positive inlef.{er n there exists a finite field WiTh p" f~ elemen/s. Any jinife field Ifilh q = p" elemenl.\ is isomorphic 10 Ilze s1;jliil;,~ ,field of x q • x orer IF p' Proof (F.xiwncfl For q . p" consider x a x in "pix I. ani let I-" he its splitting field over IF p. This polynomial has £I distinct roots in F since I in IFplx] and so can h<.lve no common ront its derivative is £IX" 1 - I with x" -.\ (compare with Theorem L6K). Let S ~ {a C F: a 4 - a ~ Oi. Then S is a subfield of F since: (i) S contain, () and I; (ii) a. I> r S implie., by Theorem 1.46 that (a- I»q ~ a' 1>' ~ a- b. and so a- I> c S: (iii) for a. b Eo S and I> ~ 0 we have (al> ,)q a'l> 'I = ab '. and so Ill> 'c S. Hut. on the other hand, x l( - x must split in S since S conwin.. . all its roots. Thus F = S. and since S has q elements. F i, a finite fidd with q elements. (Uniqueness) Let F be a finite field with q p" elemenl>. Then F has characteristic p by Theorem 2,2 and so contain.;,lF p a~ a ...uhfield. It follows from Lemma 2.4 that F is a splitting field of \"" -.' o,er ~ p' Thu, the desired result is a consequence of the uniquene~." (up !l) isomorphisms) of splitting fields, which was noted in Theorem 1.91. .1

The uniquene.:ss part of Theorem 2.5 provides the ju~tification for speaking of rhe finite field (or rh" Galois field) with q element;. M of I!I" finite field (or rhe Galois field) of order q. We shall denote thi, field hy If". where.: it i~ of course understood that q is a power of the prime characterisl it' p of Fa' The notation (iF(q) is abo used by many authors. q

=

2.6. Theorem (Subfield Criterion). Ler} q I", Ihe fill ire field wah pI/ elen/ems. Then el;ery suhjhdd of IF" has order p'''. where m is a fO'ii{in'

divi'ior of 11. Convene/)'. if m is a fosiffl:e diuSOf of 11. {hen (here is eXl1l'fl)' one wl>field of F" wirh p'" elemellls.

Proof. It is dear that a subfidd K of IF q has order pm for some positive integer m :;;; II. Lemma 2.1 shows that q ,- pll must be a pow~r of pn,. and so In is neces~arily a divisor of n. Conversely. if In b a positive divisor of fl, then pm - 1 divides pi' - I.

and ~() Xf'~1 1 -I dividct' x p "' 1 I in iF-vrx]. Consequently. x P "'.- x divides x P " - x "x" - x in IF{,lx J. Thu~. every root of ),./,." - x is a root of Xii - x and so hehmgs to ~ q' It follows that '-" must contain as a subficld a splitting field of x p " x over IF p' and as wc have seen in the proof of Theorem 2.5. such a splitting field has order p"'. If there were two distinct suhfields of order pm in IF q' they would together contain more than pm fools of X p'~ - X in IF
IF

21()

/~

1F2~

D-i"

!F l ,·

1F , .-

IF). I

lF1s

IXIX "~I/ IF,

By Thcon.:m 2.6. the containment relations are equivalent to divisibility r.: relations among the positive divi~()rs of 30. For a finite field IF q we denote by' IF; the multiplicative group of nnnzero elements of iF,," The following result enunt.:iates a useful property of thi, group. 1.8. Theorem. J-I" e"ery finile field IF q lhe muliiplieali"e group IF; of nonzero e/emems of IF q is (yc!ic. Proof We may a"ume q;;, 3. Let h ~ P;'P2"" p;';- be the prime factor decompo,it;on of the order h ~ q - 1 of the group IF;. For every i. I ~i~m. the polynomial X hIP . I ha, at most hlp, roots in IF q • Since h/ PI < II. it follows that there are nonzero clements in IF,! that are not ronb of this polynomial. Let £1 1 he such an element and set b l = a;/P>. We have b(- I. hence the order of b,. is a divisor of p;' and i~ therefore of the form P; with () ~ s, ~ r,. On the other hand.

h('

= a:I!Jl:

J

I,

and ~o the order of h, i~ 1';-. We: claim that the element h = hlh., ... bm has order II. Suppose. on the contrary. that the order of h is a proper divi~or of II

2.

Root~

47

of lrn:dudblc Polynomials

and is therefore a divisor of at kast one of the m integers hip" I <;; i <;; m, say of hlp\, Then we have 1 = hh/p,

= h7/P'b~/P'

...

h::/

P1 •

:>low if 2 <;; i <;; m, then P~' divides hip" and hence b;/p, ~ I. Therefore b;/P, ~ I. This implies that the order of h, must divide hlp" which is impossible since the order of b, is P,'. Thus, IF; is a cyclic group with [j generator b.

2.9. Definition. elemen! of IF q'

A generator of the cyclic group IF; is called a primiliee

It follows from Theorem 1.15(v) that IF q contains ¢(q - I) primitive clements, where ¢ is Euler's function. The existence of primitive elements can he used to show a result that implies. in particular, that every finite field can be thought of as a simple algebraic extension of its prime suhfield. 2./0.. Theorem. LeI IF q he a finile field and IF, a finile extension field. Then IF, is a simple all;ebraic extension oflF q and eeery primitive element of IF, can sen;e as a defining element of IF r over IF q'

Proof Let 1 be a primitive element of IF,. We clearly have IF q(n <;; IF,. On the other hand, IFq(n contains 0 and all powers of I, and so all elements of IF,. Therefore IF, ~ IF q(n. I 0 2.11. Corollary. For every finite field IF q and every posiliee in!eger n lhere exiS1S an irreducihle polynomial in IFqlx] of dewee n.

Proof Let IF, he the extension field of IF q of order q", so that llF,; IF q] ~ n. By Theorem 2.10 we have IF, = IF.
2.

ROOTS OF IRREDUCIBLE POLYNOMIALS

In this section we collect some information about the set of roots of an irreducible polynomial over a finite field. 2.12. Lemma. Let f E IF q[x] be an irreducible polynomial ocer a finile field IF q and let a he a root of f in an extension field of IF q' Then for a polynomial h E IF qlA] we have h( a) = a if and only iff divides h.

Proof Let a be the leading coefficient of f and set g(x) ~ a-'f(x). Then g is a monic irreducible polynomial in IFq[x] with g(a) ~ 0 and so it is the minimal polynomial of a over IF q in the sense of Definition 1.81. The rest follows from Theorem 1.82(ii). 0

Structure (If rin:tt' T"ickb

degree

2.13. I.emma. I.el [ C',IFql X j be all irreducible polvnomial orer IF, o[ lhell [(xl dirides x q - x i[ alld only i[ m divides n.

Ill,

Proof Suppose [(xl divid"" x,' - x. Let a be a root of [in tbe splitting field of [over IF q. Then a q" ~ a. so that a E IF q•. It follows that IFq(a) is a subficld of IF q... But since IlFq(a):lFqj~m and [lFq":lFql~ll, Theorem 1.84 shows that m divides Il. Conversely, if m divides n. then Theorem 2.6 implies that IF," contains IF q"' as a subficld. If a is a root of [in the splitting field of [over IF,!, then [IF ("):IF,,j- m, and so IF (a) ~ IF q ",. Consequently, we have aE IF ", q "., If"
~,_ Theorem. 1[[ is all irreducible po(vnomial ill 1F,lx] o[ degree m. Ihen [has a rOOI a ill IF q"" Funhermore, all the roots 0[[ are simple and are l' given hy the m distincT elements a.a.q.aq~,.",alfm I ofrFq'tl,

{

Z

Let a be a root of [in the splitting field of [ over IF q' Then hence 1F,(a)-lFq"" and in particular "ElF q",. I'\ext we show that if fl c IF"" is a root of [, then fl" is also a roO! of f. Write [( x) ~ il,Y/X +il 1X I au with a,ElF q for O~i~m. Then, using Lt.:mma 2.3 and The(>rem 1.46. we gct Proof

IlFq(,,):lFq]~m. fll

....;..

••

,

[( 13") - am 13 qm ~

(a m f3 m

I

+ a,fl q I

>

+ a,li tau)' = [(13)' ~ O.

00 ~

a'/"f3 qm + ... + aYIi'

I a~

Therdore. the clements a, a q, aq~, ... , a q'" I are rooh of f. It rcmain5. to prove that these clements arc distinct. Suppose, on the contrary, that "q' ~ "q' for somc integers j and k with 0", j < k '" m - I. By raising this identity to the power qm k, we get

It follows then from Lemma 2.12 that [( x) divides x q "' ,., - x. By Lemma 2.13, this is only possible if m divides m - k + j. But we have 0 < m - k T j < m, and so we arrive at a contradiction. 0 2.15, Corollary. 1£1 [ he an irreducihle polynomial ill IFqlx] o[ dewee m. Then the splilting[ield 0[[ over IF, is given bv IF q"" Proof Theorem 2.14 shows that [ splits in IF q "" Furthermore, IF,,('""', ""'. ... ,,,," ')~lFq(")~lFq", for a root" of [in IF q "" where the LJ second identity is taken from the proof of Theorem 2.14. 2.16. Corollary. Any IWO irreducible polYllomials in IF ,Ix J o[ the same degree have isomorphic spliltillg [ields.

49

2. Root:; of Irreducihle Polynomiab

We introduce a convenient terminology for the elements appearing in Theorem 2.14, regardless of whether a E 0' q"' is a root of an irreducible polynomial in fq[x] of degree m or not. 2,~7, Definition, elements a, q ,

Let

fqm

be an extension of f q and let a

E f q ".

Then the 10 IE,

a a< ... ,a.q.':.~-llfe ..calkd..t.heAU2jugal.e.LQU!.~ith.respe<:l

The conjugates of a E 0' q" with respect to 0' q are distinct if and only if the minimal polynomial of a over 0' q has degree m. Otherwise, the degree d of this minimal polynomial is a proper divisor of m, and then the conjugates of a with respect to Fq are the distinct elements a.a q , .••• a q1 repeated m / d times.

' . •

each

2.18, Theorem. The cOl/jugates of a E 0'; wirh respecr to any subfield of 0' q have the .same order In the group 0';' Proof Since 0'; is a cyclic group hy Theorem 2.8. the result follows from Theorem 1.15(ii) and the facl that every power of the characteristic of f q is relatively prime to the order q -I of 0';. 0 2,19. Corollary. If a is a primitive element of 0' q' then so are all its conjugates with respect to any subfield of 0' q'

2.20. Example, Let a EO'" he a root of f(x) = x 4 + X + I E f,[x]. Then the conjugates of a with respect to 0', are a. a', a 4 = a + I. and a" = a' + I. each of them being a primitive clement of IF 16' The conjugates of a with 0 respect to 0'4 are a and a 4 = a + I. There is an intimate relationship between conjugate elements and certain automorphisms of a finite field. Let 0' qm be an extension of 0' q' By an . automorphism
50

StrUl'lurt' of finite Fields

The mappings 00- a I'··· ,01>1_ I arc distinct since they attain distinct values for a primitive element of IF q"" :-iow suppose that a is an arbitrary automorphism of IF q"' over IF q' Let 13 be a primitive clement of IF q ", and let [(x) = x m ~ am_lx m 1+ ... + 0 0 E IF qlx J he its minimal polynomial over IF q' Then O=a(f3m","am_lf3m-1 + '" +ao) = a(f3)m + am la(f3)m 1+ ... + au. so that

a(

13)

is a root of [ in IF q"" It follows from Theorem 2.14 that

a( 13) = f3 q ' for some j. 0", j '" m - 1. Since a is a homomorphism. we get

then a( a) = a q ' for all a E IF q""

D

On the basis of Theorem 2.21 it is evident that the conjugates of a E IF q"' with respect to IF q are obtained hy applying all automorphisms of IF q. over IF q to the element a. The automorphisms of IF q_ over IF q form a group with. the. operation. heing the. usual eompositiolL of mappings. The information provided in Theorem 2.21 shows that this group of llutomorphisms of IF q ., over IF q is a eyelicgroup of order m generated by a,.

3.

TRACES, NORMS. AND BASES

In this section we adopt again the viewpoint of regarding a finite extension F = IF q_ of the finite field K = IF q as a veetor spaee over K (compare with Chapter I, Seetion 4). Then F has dimension mover K, and if (al ..... a m) is a hasis of F over K, each element a E F ean be uniquely represented in the form a

= cIa]

+ ... + ('»Jam with

cj E

K

for 1 ~ j

~

m.

We introduce an important mapping from F to K which will turn out to be linear.

2.22. Definition. For aEF=lF q ", and K=lF q , the trace Tr'/K(a) of a over K is defined hy TrF/K(a)=a+a q + .. · +a q "" If K is the prime suhfield of F. then TrF/K(a) is called the absolute trace of a and simply denoted hy TrF(a).

In other words, the traee of a over K is the sum of the eonjugates of a with respeet to K. Still another deseription of the trace may he obtained as follows. Let[ E K[x] he the minimal polynomial of a over K; its degree d is a divisor of m. Then I(x) = [(x)m/d E K [x] is called the characteristic polynomial of a over K. By Theorem 2.14. the roots of [in F are given by

3.

TrJ~'c~. :'\'orm~.

and

Bas~'~

51

a. a", ... , a"' " and then a remark following Definition 2.17 implies that the

roots of II in F arc precisely the conjugates of a with respect to K. Hence g(x)=xm+am_lxm-I+y.. +a o ~ (x - a)(x - aq) ... (x - a q" '),

(2.1)

and a comparison of coefficients shows that TrF/K(a)~ -am

(2.2)

,.

In particular, TrF/K(a) is always an element of K.

2,23, Theorem, 1£1 K~lFq and Tr F/ K satisfies the followinx properties:

F~lFq"' Then lhe lrace junction

(i) Tr f/K (a + (3) ~ Tr f / K ( a) +Tr F/K ({3) jor all a, (3 E F; (ii) Trf/K(ca) ~ cTrf / K (a) jor all c E K, a E F; (iii) Tr F/ K is a linear transjormation jrom F onl0 K. where both F and K are ciewed as vector spaces over K; (iv) TrF/K(a) ~ rna jor all a E K; (v)

q TrF/K(a ) ~ Trf/K(a) jor all a

E

F.

Proof

(i)

For a, {3 E F we use Theorem 1.46 to get n,'

Tr F/ K (a+{3)=a+{3+(a+{3)q+ ~a

+ a qm

+ {3 + a q + {3q +

I

+(a+{3)q

'-+- (3qm ,

~ Tr f / K (a)+ Tr F/ K ({3).

(ii)

For c E K we have c q ' ~ c for all} '" 0 by Lemma 2.3. Therefore we ohtain for a E F. Trf/K ( co:) = co. + cqa q + t, c 4"" a qm'

= co. + co. q +

+ ca q '"

I

~ cTrF/K(a).

(iii) The properties (i) and (ii), together with the fact that Trf/K(a) E K for all a E F. show that Tr r / K is a linear transformation from F into K. To prove that this mapping is onto, it suffices then to show the existence of an a E F with TrF/K(a) = O. Now TrF/K(a) ~ 0 if and only if a is a root of the polynomial qm x '+ ... + x q + X E K[x] in F. But since this polynomial can have at most qm I roots in F and F has qm clements. we are done. (iv) This follows immediately from the definition of the trace function and Lemma 2.3. qm (v) For a E F we have a ~ a hy Lemma 2.3, and so Tr f / K (a q ) = 42 0. 4 +0. + ... + a q "'= TrF/K(o.). :::J

52

Strul'lUrl' of Finite Field...

The trace function Trf/J\ is not only in itself a linear transformation from F onto K. but serves for a description of all linear transformations from F into K (or. in an equivalent terminology. of all linear functionals on F) that has the advantage of heing independent of a chosen basis. 2.24. Theorem. I.el F be a finile eXlension of Ihe finile field K. hOlh considered as ceclor spaces over K. Then the linear trunsformations from F il/lo K are exaelly Ihe mappings LR• fJ E F. where La(a) ~ Tr F/ K( fJa) for all ex E F. Furthermore, we have Lfl ~ l.y wheneGer f3 Gild yare disTinct eleme1lls of F.

Proof Each mapping 1'8 is a linear transformation from F into K by Theorem 2.23(iii). For fJ. y c F with fJ '" Y. we have 1.8(a)- I./a) ~ Trf/K(fJa)- Tr'/K(ya)'~TrF/K«fJ' y)a)'" for suitable aE F since Tr F/ K maps F onto K. and so the mappings 1'8 and I. y arc different. If K ~ IF q and F ~ IF q"" then the mappings I' R yield qrn different linear transformations from F into K. On the other hand. every linear transformation from F into K can be obtained by assigning arbitrary elements of K to the m clements of a given basis of F over K. Since this can he done in qrn different ways. the mappings L# already exhaust all possihle linear transformations from F into K. 0

°

2.25. Theorem. I.el F he a finile eXlension of K ~ IF q' Then for a E F we ha"e Tr'/K(a) ~ if and only if a ~ fJq - fJ for some fJ E F.

°

Proof The sufficiency of the condition is obvious hy Theorem 2.23(v). To prove the necessity. suppose a E F ~ IF q" with Tr F(K (a) = 0 and let fJ he a root of x q - x - a in some extension field of F. Then fJq - fJ ~ a and

so that

fJ E

fJ) ~ (fJq -

f3) q +

=

(f3q -

=

(fJq - fJ)+ (fJ q' - fJq)+

~

fJqm -

"" (fJq ~

(fl q" -

f3) q" fJq- ')

fJ.

o

F.

In case a chain of extension fields is considered. the composition of trace functions proceeds according to a very simple rule. 2.26. Theorem (Transitivity of Trace). LeI K be a finile field. leI F be a finile eXlension of K and t· a finile eXlension of F. Then

Tr':/K(a)~Tr'/K(Tr"/F(a))

forallaE E.

.i '[ rael". i\nrm.<:" and Ib"l':::

Proof

I.dK

53

~I'"letIF:K]~mand[r::F]~n,sothat[F: K]c

mn by Theorem I.R4. Then for nEt' we have

IrrIA(Tr/(f(n»)~

m

;

.

>n '" I

L Trfll(n)"~ L

;"'0 fI,

I,/

()

" q"

I

/-0\/0

L L: I

! /1

I L: a"'J

I

1

/'/1/

L:

a"I"'~

(J

/..

i

1

a"~TrE;K(I». ()

Anothc.:r inter~~ting function from a finite field to a suhfield IS obtained hy f()rming the.: product of the conjugate:-. of an element of the field with respect to the subfield.

,

2.27. Oefinilion, for over K is defined hy l\fjA"

aE F~I',..

and K -F" the norm l"f/,,(a) of a

(a) "'- a- at{ - ... - a<{~' "= a(It'''

II/III

1).

Ily comparing the constant terms in (2.1). we see that '1// A (I» can be read off from the characteristic polynomial g of a over K -namely, :sIF/K(a) ~

It follows. in particular, that N f /

':'\ 1'/;""

K

(a)

(2.3 )

(-I)"'a". is always an element of K.

2.28. Theorem, Let K ~ 1', and F • 1'," lhen the norm lunl'/ion .'iatisfies the fol/owing properties: (i)

(ii) (iii) (iv)

'IF! A (a{3) ~ N'IK (a)NF! I< (f!) lor all n, {3 N FiA maps F onto K and F* ontO K':

F

F:

l"f/K(a) ~ am lor all a E K: l"FIA(a')~l"f/,,(a)/orallaE

F.

Proof (i) follows immediately from the definition of the norm. We have already noted that l" FlA maps F into K. Since NFl" (a) ~ 0 if and only if a ~ 0, l"FlK maps F' into K'. Property (i) shows that 'If/I< is a group homomorphism hetween these multiplicative groups. Since the elements of the kernel of N FIK are exactly the roots of the polynomial xlqm I)Aq I) - I E K rx I in F, the order d of the kernel satisfies d <; (qm .. I )/( q - I). By Theorem 1.23. the image of N'/K has order (qm -I)jd, which is ;" q-1. Therefore. "'fiK maps F' onto K' and so F ontO K. Property (iii) follows from the definition of the norm and the fact that for a c K the conjugates of a with respect to K arc all equal to a. finally, we have NFIK(a') ~ N flK (a)' .. l" FIK (a) because of (i) and t\ FIK (a) E K. and so (iv) is shown. :J

54

Structure of Finite Fieldf-

2.29. Theorem (Transitivity of :-Jorm). Let K be a finire field. let F be a finite extension of K and E a finite extension of F. Then NI'IK(a) ~ N'IK(N'jF(a»)

forallaE E.

Proof With the same notation as in the proof of Theorem 2.26, we have for a E E. N' fJI<. (,. I"":/F (») a --.'oJ F/K ('q ex ....

·"I'..."')

= (a(4~'~

. l)/(q'"

-

_,q'"

I)/,q

-

u.

1)

,,~

r

qon

... ,

-])/(q

'~t/K

1)

(_)

.......

0

If (a" ... ,a m ) is a basis of the finite field F over a subfield K, the question arises as to the calculation of the coefficients cj(a)E K, 1 ~) ~ m, in the unique representation a=c,(a)a,+'"

+cm(a)a m

(2.4)

of an element aE F. We note that cj : a>-+cj(a) is a linear transformation from F into K, and thus, according to Theorem 2.24, there exists a f3j E F such that c,f a) ~ Tr'IK Wja) for all a E F. Putting a ~ a" I ~ i ~ m, we sec that Tr'!K (f3la,) = 0 for i'" ) and 1 for i ~). Furthermore, (13" ... ,13m) is again a basis of F over K, for if d,f3,+'" +dmf3m~O withd,E K for I~i~m.

then by multiplying by a fixed a, and applying the trace function Tr FIK . one shows that d, ~ O. 2.30. Definition. Let K be a finite field and F a finite extension of K. Then two bases (a, ... .,a",) and (13" ... ,13m) of F over K are said to be dual (or complementary) bases if for I ~ i,) ~ m we bave

fori"'), fori=). In the discussion above we have shown tbat for any basis (a, .... ,a"') of F over K there exists a dual basis {f3" ... ,f3m}. The dual basis is, in fact, uniquely determined since its definition implies that the coefficients c,fa), I~)~m. in (2.4) arc given by ("i(a)~TrFIK(f3ia) for all aEF. and by Theorem 2.24 the element f3i E F is uniquely determined by the linear transformation ci . 2.31.

Example.

Let a E IF, be a root of the irreducible polynomial

+ x' + I in 1F,[x]. Then (a. a'. I + a + a') is a basis of IF, over IF,. One checks easily that its uniquely determined dual basis is again {a, a'. I + a + x'

a'}. Such a basis that is its own dual basis is called a self-dual hasis. The element as E IF Il can he uniquely represented in the form as = cia + Czet:! ......

3. Tran's, ;"'0rms. and

55

Ib~c:-

J

c,(

1 , a -t a·~) with

('I'

('~. c., E=!F ~. and the coefficients are given hy

'" =Tr,)a·a') (' 2 .. Tr r ~ (a"a') c, = so that a'· a' , (I

O. ~

1.

1r,.((1- a+ a.')a')

~ 1.

+ a + a').

0

The number of distinct hases of F over K is rather large (see Exercise 2.37). but there are two special types of bases of particular importance. The first is a polynomial 1)Q.}i.r. {I. U. u 2 •••. , urn I}, made:: up of the powers of a defining clement" elf Faver K. The clement a is often taken to be a primitive clement of F (compare wi)h Theorem 2.10). Another type of basis s a ormal basis defined by a suitahle element of F.

.g

. . Definition. • Let K .. IF q and F ~ IF q"' Then a basis of F over K of the form Ct.ex 4 ••.. ,a 4 }. consisting of a suitabk clement aE F and its conjugates with respect to K, is called a normal hasis of F over K. The basis (a. a.'. I + a - a'} of IF, over IF, discussed in Example 2.31 is a normal basis of!F g over!F~ since I t- ex + a 2 = a 4 . We shall show that a normal hasi."i exists in the general case as well. The proof depends on two lemmas. one on a kind of linear independence property of certain group homomorphisms and one on linear operators. 2.33. Lemma (Artin Lemma). I.el J/.;1 ..... ~>1/ be dislincr homomorphi.wls from a group G infO the mulliplicalice group F* of (In arhilrary field F. and leI a I' .... am he elemenl.'! of F lhal are not all O. Then for some g E G we hace

a,,,,,(g)

1 ••. +am';'m(g)~O.

Proof We proceed by induction on m. The case m ~ I being trivial. we a~~ume that m > 1 and that the statement b shown for any m - I distinct homomorphisms. I'\ow, take ~I, .... J/.;m and a 1..... £J m as in the lemma. If a, ~ O. the induction hypothesis immediately yields the desired result. Thus let a, ~ O. Suppose we had

a,"',(g)+ .. ·• Since

am';'»,(g)~O

forallgEG.

(2.5)

Y. = Ym' there exists h E C with';' ,( h) '" ';'m(h). Then. replacing g hy

hg in (2.5). we get a,y,(h)';',(g)+·'" A fter multiplication hy

ym( h)

amYm(h)';'n,(g)··O

forallgEC.

'we obtain

h,';',(g)+ ... +bm ''''m_,(g)+amYm(g)~O forallgEC. where bl .::..: aiJ/.;i{h)~f'1(h) I for I ~ i ~ m -I. By suhtracting this identity

S6

Slructure of Finite Fit'lds

from (2.5), we arrivc at

where c = a b , for 1 ~ i ~ m 1. But ('I"':' u l - Ul~l(h)o/m(h)-1 we have a contradiction to the induction hypothcsi,. j

j

-

=1=

0, and 0

We recall a few concepts and facts from linear algebra. If T is a linear operator on the finite-dimensional vector space V over the (arbitrary) field K, thcn a polynomial f(x) ~ a,x" ~ ... + a,x + a o C K [xl is said to annihilate T if an Tn + ... -+- Q l T -{ G o 1 = 0, where 1 is the identity operator and 0 the zero operator on V. The uniquely determined monic polynomial of least positivc degree with tbis property is called the millimal polYllomial for T. It divides any other polynomial in K[x] annihilating T. In particular, the minimal polynomial for T divides the characleristic pO(Yllomial x(x) for l' (Cayley-Hamilton theorem), which is given by g(x) ~ det(xl- T) and is a monic polynomial of degree equal to the di mension of V. A vector a E V is callcd a cyclic veClOr for T if the vectors T'a, k ~ 0, I, ... , span V. The following is a standard result from linear algebra.

2.34. I.emma. Let T be a lillear operaTOr all Ihe finile-dimellsiollal vector space V. Then r has a ,yclic veCTOr if and ollly if the characterisric and minimal polynomials for l' are idemical. 2.35. Theorem (:-.Iormal Basis Thcorcm). For any finite field K and any/iniTe eXTension F of K, There exisTs a normal hasis of F over K,

Proof Let K ~ !', and F ~ !'" wi th m ;> 2. hom Theorem 2.21 and the remarks following it, we know that tbe distinct automorphisms of F over K arc given by La. a',. . .,a m '. where e is the identity mapping on F, a( a) - a' for a Eo F. and a power a I refers to the j-fold composition of a with itself. Because of o(a t /3l ~ a(a)+ a(/3) and a(ca) = a(c)a(a) ~ car a) for a, /3 F F and CEo K, the mapping a may also be considered as a linear operator on the vector sptlce F over K. Since am = E, the polynomial x m - I c= K [xl annihilates a. Lemma 2.33, applied to e, a, a 2,. .. , am 'viewed as endomorphisms of P. shows that no nonzero polynomial in K[x] of degree less than m annihilates 0. Consequently, xf'l - I is the minimal polynomial for the linear operator a. Since the characteristic polynomial for a is a monic polynomial of degree m that is divisible by the minimal polynomial for a, it follows that the characteristic polynomial for a is also given by x m - 1. Lemma 2.34 implies thcn the existence of an element a E F such that a.a(a),a'(<»,. .. span F. By dropping repeated elements, we see that It. a( a), 0 '( <» •••• • a m · '( a) span F and thus form a basis of F over K.

Since thi~ ha:-.is consists of ex and its cnnju5atcs with respect to K. it is a normal basis of F over K. ~

3 Tra(t's, Norm" and B,):>t'':>

57

An alternative proof of the normal basis theorem will be provided in Chapter 3. Section 4. by using 50-called linearized polynomials. We introduce an expression that allows us to decide whether a given set of dements forms a basis of an extension field. 2.36.

Definition.

Let K be a finite field and F an extension of K of degree

mover K. Then the discriminQ1lf j,J-;J<(£xl .... 'O':m) of the elements C F is defined hy the determinant of order m given hy

01 ....

'O':m

Tr'/K(a,a,)

Trr/K(ala,)

TrFlK(ala m )

Trr;k(a,a, )

Tr'/K(a,a,)

Tr,/K(a,aml

It follows from the definition that D.F/,(a, ..... an,l is always an clement of K. The following simple characterization of hases can now be

gIven. 1.37. Theorem. Let K he a finite field. F an extension of K of degree mover K. and a l "m E F. ]ilen {a l •...• am} is a basis of F oeer K if and only if D.'/K(a, a m ) '" O. Proof. Let (a, ..... a m ) be a basis of F over K. We prove that D.r/K(a, ..... "ml'" 0 hy showing that the row vectors of the determinant defining D.'/K(a, ..... "m) arc linearly independent. For suppose that c:TrF/K(O':IO':j)- ... .... cmTrr/J«amai)=O

for l:S;;:j~m.

where cl .... .(m E K. Then with f3 = CICt: 1 - ... + cmo m we get TrJ-iI\({3O':j) ~ 0 for 1 <; j <; m. and since a: ..... a .., 'pan F. it follows that Tr F/ K(f3al ~ 0 for all a E F. However. this is only possible if f3 ~ O. and then c:a, + ... + cII/O':m = 0 implies c\ = . . . = (''''1 = O. Conversely. suppose that J. r /1I.·(ol ..... om)==O and cla l - ... ('ma", = 0 for some ': ... "c m E K. Then (,O~O; ...... • cmCt:n/x I 0 fur 1:( j ~ m. >:..

and b\ applying the trace function we get ('~Trr;K(Ct:IOi)..;... ...

+ CmTr,../K(a*1O':j) =0

forl~j~m.

Rut since the row vector:, of the determinant defining Ll OK (ol ..... a*1) are (I ~ . . . ~ c' n ~ O. Therefore. a l ... .. a .., C

linearly independent. it folio",' that ar~ linearly independent over K.

There:.: is another determinant of order m that :-.erve:, the same purpose as the discriminant ,j,flK(Ct:I ..... Ct: m ). The e:.:ntries of this determinant arc. however. elements of the extension field F. For Ct:l ..... o m E F. let

5~

Structuft' (If !:'nill' hl..'lds

A he the m x In matrix. whose entry in the ith row and jth column i~ af-I, where q is the number of elements of K. If AT denotes the transpose of A. then a simple calculation shows that A 'A ~ R. where B is the m x m matrix whose entry in the ith row and jth column is TrF/K(aia J ). By taking determinants. we ohtain

:>J/K(IX, ..... a n ,) ~ det(A)'.

The following result is now implied by Theorem 2.37.

Fq"'

2.38. Corollary. Let a, ....."m Fq if and only if

E

Fq.... Then {a, ..... a m} i, u hu,i, of

Ol,",

a,

a,

al

(Xi

am a mq I

;


a~

I a,

..

O.

Ci~;"

From the criterion above we are led to a relatively simple way of checking whether a given element gives rise to a normal hasis. 2.39. Theorem. For 0: E IF({-. {a. all. Qq! •..•• alj~ '} is a normal ha.~is of F .. Olea F if and only if the polvnomials x m . I and ax m '+ :j 2 q ... ) ...., -1 aqx m -l- . . . . . . . . c/I X + a" in IF l/~,lx I arc rf:'la(iee~v prime.

Proof When (Xl Corollary 2.3R becomes

lX,

-

aq

a q" a

±

a q" aq

4 0':2 = 0. , ...• a"l

I

a q a "

a q'

-a q"

the determinant

aq aq

a q" .. aq

a

a q '"

a

q,

10

" (2.6)

a

after a suitable permutation of the rows. Now consider the resultant R(f. g) of the polynomials fix) ~ x m -I and g(x) - ax m '+ aqx m 2 + ... + a"m' \ "t' a'/"'- I of formal degree m resp. m - I. which is given hy a determinant of order 2m - I in accordance with Definition 1.93. In this determinant. add the (m + 1)st column to the first column, the (m + 2)nd column to the second column. and so on, finally adding the (2m - I)st column to the (m - I)st column. The resulting determinant factorizes into the determinant of the diagonal matrix of order m - I with entries - I along the main diagonal and the determinant in (2.6). Therefore. R(f. g) is. apart from the sign, equal to the determinant in (2.6). The statement of the theorem follows

4. Routs of Unity and (v(:lotolnH.: Polynomiab

59

then from Corollary 2.38 and the fact that R(j.g) = 0 if and'only if f and g 0 are relatively prime.

In connection with the preceding discussion. we mention without proof the following refinement of the normal basis theorem. 2.40 Theorem. Fo,. any fillite ext elISion F of a jinite .field K there exists a normal hasis of F oz.;er K tllat consiSTS (?l primilive elements of f. 4.

ROOTS OF (;!'IITY AND CYCI.OTO:vJ1C POI.YNOMIALS

In this section we investigate the splitting field of the polynomial x" - I over an arhitrary field K. where n is a positivc integer. At the same time we ohtain a generalization of the concept of a root of unity. well kno"n for complex numbers. 2.41. Delinition. Let 11 be a positive integer. The splitting field of x" - I over a field K is called the nIh cvclolomic field over K and denoted by K '" '. The roots of x" - 1 in K Ull are called the 11th roots of unit)' over K and the set of all these roob is denoted by F.(t/). 1\ special case of this general definition is ohtained if K is the field of rational numbers. Then K (111 is a suhfidd of the field of complex numbers and the nth roots of unity have their known geometric interpretation as the vcrtices of a regular polyg~)fi with 1l vertices on the unit circle in the complex plane. For our purposes. the most important case is that of a finite field J(. The hasic properties of rooh of unity can. however. he established without using this restriction. The structure .of 1;.'.{nl .i~ determined hy the relation of 1l to the characteristic of K. as the following thcorem show:... When"we refer to the characteristic"p ~)f k in this discussion. we permit the case p = 0 a~ well.

2.42. Theurem. llCTenstll' p. Then: (i)

(ii)

Lei n he u posilire inleger and K a field uf char-

If P dues nor divide n. then 1:.'('1) is a cyclic group of order n with respect to multiplication in K llll . If P diridt!s n. ',1.·rite n = mp" with positive i1llegers m and e and m not dieisihle hy p. Then K,t/} = K 1no • Ern) = Elm,. and the roots of .\.n - I in KIn) art! the m elements of Elml. ellch a!1ained widE mulliplicily p".

Proof (i) The case n = I is trivial. For n ~ 2. x" - I and its derivative nx') 1 hav~ no common roots. as nx n I only has the root 0 in KIn). Therefore. by Theorem 1.68. x" I cannot have multipic roots. and hence f;'"' has n clements. I'ow if t.~E fY'. then (t~ 1)"=t"(~")-1 =1. thus

60

Structure of Finill' Fil'kls

1;1/ IE E lnl . It follows that 1::'"' is a multiplicative group. Let n ~ pi'p,'" .. p{' be the prime factor decomposition of n. Then one shows by the same argument as in the proof of Theorem 2.8 that for each i, ) <; i <; I, there exists an clement IX, E E'n' that is not a root of the polynomial x"/ p , - 1, that 13/ = Ci.7IP~' has order pf'. and that £1/1) is a cyclic group with generator P ~ PIP,' .. p,. (ii) This follows immediately from x n - I ~ x mp ' - I ~ (x m - I) p' and part (i). 0

2.43. Definition. Let K he a field of characteristic p and n a positive integer not divisible by p. Then a generator of the cyclic group E,n' is called a primitive nth rool of unity over K. By Theorem 1.I5(v) we know that under the conditions of Definition 2.43 there are exactly ¢(n) different primitive nth roots of unity over K. If I; is one of them, then all primitive nth roots of unity over K are given by 1;', where I <; s <; nand gcd(s. n) ~ I. The polynomial whose roots arc prccisely the primitive nth roots of unity over K is of great interest. 2.44. Definition. Let K be a field of characteristic p, n a positive integer not divisible by p, and I; a primitive nth root of unity over K. Then the polynomial n

n

Qn(x)~

(x-I;')

-;-1

gcd(s.n)-l

is called the nth cyclotomic polynomial over K. The polynomial Qn(x) is clearly independent of the choice of 1;. The degree of Qn(x) is (n) and its coefficients obviously belong to the nth cyclotomic field over K. A simple argument will show that they arc actually contained in the prime suhfield of K. We use the product symhol to denotc a product extended over all positive divisors d of a positive integer n.

ndn

2.45. Theorem. Let K be a field of characleristic p and integer nol divisihle by p. Then: (i) x n - I = IldlnQd(x); \ ..•.. \ ;.···.v (ii)

/I

a positive

'

c'· the prime subfield of K. and to

the coefficients of Qnix) heiOf/g 10 Z if the prime subfield of K is the field of rational numbers.

Proof (i) Each nth root of unity over K is a primitive dth root of unity over K for exactly one positive divisor d of n. In detail, if I; is a primitivc /lth root of unity over K and 1;' is an arbitrary nth root of unity over K. then d ~ n /gcd(s, n); that is, d is the order of 1;' in 1::'"'. Since n

xn_)~

n(x-I;'), 5

= 1

..' ,: '-

4. I{O()b

of Lnity and Cvdotomk Polynomiab

61

the formula in (i) is obtained hy collecting those factOrs (x . K') 'for which is a primitive dth root of unity over K. (ii) This is proved by induction on n. "ote that Q,,(x) is a monic polynomial. For n = I we have Q,(x) ~ x - I, and the claim is obviously valid. Now let n > I and suppose the proposition is true for all Qd(X) with I,,;d
K'

2.46.

Example,

Lct r he a prime and k c 1'\1. Then

by Theorem 2.45(i). For k ~ I we simply have Q,(x) ~ I + x xr 1

+ x' + ... + U

An explicit expression for the nth cyclotomic polynomial generalizing the formula for Q,,(x) in Example 2.46 will be given in Chapter 3, Section 2. For applications to finite fields it is useful to know some properties of cyclotomic fields. 2,47.

Theorem.

The lye/otomic field K (", is u simple ulgehraic

ex/ension of K. Moreover: (i)

(ii)

If K ~ 0. Ihen Ihe cye/olOmic polvnomial Q. is irreducihle oDer K und [K'''': Kl ~ ¢(n). If K ~ IF q wirh gcd(q. n) ~ I. Ihen Q"fuctors infO ¢(n)/d disiinci monic irred/lcihle polvnomiu/l in K [x] of Ihe sume degree d. K (., i.1 Ihe SplillillX field of uny such irreducihle fm·tor ODer K. and [K'·': K]= d. where d is Ihe leasl positive integer sucli that qJ", I mod n.

Proof If there exists a primitive nth root of unity rover K. it is clear that K''''· K(KJ, Otherwise. we have the situation described in Theorem 2.42(ii), then K'"'' ~ K'm, and the result follows again. As to the remaining statements, we prove only Oi). the important case for our purposes. Let ~ he a primitive nth root of unitv over IF q • Then ~ E ~ q' if and only if ~q' ~~' and the latter identity is equivalent to q' '" I mod n. Thc smallest positivc integcr for which this holds is k ~ d. and so ~ is in iF q" but

Structure of Finite fields

02

in no proper suhfield thereof. Thus the minimal polynomial of ~ over I'q has degree d, and since ~ is an arhitrary root of Q", the desired results follow. D 2.48. Example. Let K~I'" and Q"(x)~x4-x'-'-I,=I',,lxJ. In the notation of Theorem 2.47(ii) we have d ~ 2. In detail. Q,,(x) factors in the form QI2( x) = (x' + 5x .;. I )(x' - 5x + I), with both factors heing irreduciD ble in I' ,,[x]. The cyclotomic field K I1 " is equal 10 I' "~I' A further connection between cyclotomic fields and finite fields is given hy the following theorem. 2.49, Theorem, The fillite field I' q is Ihe (q - 1 )SI 'J'clolomic field over OilY Olle of itI subfields. Proof The polynomial x q ' - I splits in I' q since its roots are exactly all nonzcro elements of I' q' Ohviously, the polynomial cannot split in any proper subfield of I'q, so that I'q is the splitting field of x q I -lover anyone of its subfields. D

Since IF; is a cyclic group of order q -. I hy Theorem 2.8, there will exist, for any positive divisor n of q - I, a cyclic suhgroup (I, a.... ,an -') of IF; of order n (sec Theorem 1.15(iii)). All elements of this suhgroup are nth roots of unity over any suhfield of IF q and the generating clement a is a primitive nth root of unity over any suhfield of IF q • We conclude this section with a lemma we shall need later on. 2,50, I.emma. If dis 0 dicisnr of Ihe positive integer n willt 1" d < n, Ihen Q"(x) divides (x n -I)j(x d -I) whenever Qn(x) is defined. Proof

From Theorem 2.45(i) we know that Qn(x) divides xn_ I

n

~ (x" .. 1). x - I , x d -I

Since d is a proper divisor of n, the polynomials Qn(x) and x d - 1 have no common root, hence gcd(Qn(x), x d -I) ~ I and the proposition is true. D

5.

REPRESENTATION OF ELEMENTS OF FINITE FIELDS

In this section we describe three different ways of repre;enting the clements of a finite ficld IF q with q ~ pn elements, where p is the characteristic of IF q' The first method is hased on principles expounded in Chapter I, Section 4, and in the present chapter. We note that I' q is a simple algebraic extension of IF p by Theorem 2.10, In fact, if f is an irreducihle polynomial in IFplx] of degree n, thenfhas a root a in I'q according to Theorem 2.14, and so IF q ~ IFp ( a). Then, by Theorem 1.86, every element of IF q can he uniquely

63

~xpresscJ as a polynomial in a over

view Irq as the residue cia'S ring

It fI of degree less than

11.

We may also

~r[xll(f).

251. Example. To represent the elements of 0', in this way. we regard 0', as a simple algebraic extension of 0', of degree 2. which is obtained by adjunction of a root" of an irreducible quadratic polynomial over 0'" say [( x)" x' / IE 0'3[X]. Thus [(a) ~ a' / I ~ 0 in 0'" and the nine elements of IF g are given in the form Go'l ilia with ao' G l E IF j • In detail, IF g = (U. I. 2. o. I + a.2/ a,2a, I / 2a,2 + 2a). The operation tables for 0', may be constructed as in Example 1.62, with a playing the role of the residue e1ass Ixl. 0 If we use Theorems 2.47 and 2.49. we get another possibility of expressing the elements of 0' q' Since 0'" is the (q I)st cyclotomic field over ~r' we can construct it hy finding the decomposition of the (q -I)st cyclotomic polynomial Qq I E= O'rlx[ into irreducible factors in O'r[x], which are all of the same degree. A root of anyone of these factors is then a primitive (q I)st root of unity over 0'r and therefore a primitive element of IF q_ Thus, IF q consish of 0 and appropriate powc:rs of that primitive clement.

2.52. Example. To apply this to the construction of 0'" we note that 0' 9 ~ 0' J". the eighth cyclotomic field over 0:: ,. I\ow Q,(x) ~ x' + 1 ED',[ x I hy Example 2.46. and Q,(x)~ (x' +X~2)(X2

+2x,

2)

is the decomposition of Qx into irn.::ducihlc factors in IF dx J. Let t he a root of x' ~ x ~ 2: then \ is a primitive eighth root of unity over 0'" Thus. all non/.em clements of IF\} can be expressed as powers of and so IF g = (0.1-\'.\'.\4,1'1.\'.\'.\'). We may arrange the nonzero elements of 0', in a "Hailed index tahle. where we list the elements \' according to their exponents i. In order 10 eMablish the connection with the representation in Example 2.51. we ohserve that x' / x I 2 C 0' ,[ x ] has \ ~ 1 + a as a root, where ,,' ~ I ~ 0 as in hample 2.51. Therefor~. the index table for 0', may

r.

ot:'

written as follows:

t

\' I 1-" 2 1 2" 3, 1+2" 4 1 2

5 6

2 - 2a

7

2

8

I

"I " I

We see that we ohtain. of cour~L'. the same clements as in Example 2.51. just in a different order. U A third po~sjhility of representing the dements of IF q is given hy mean:-- of matrices. (n general. the companion maTrix of a monic polynomial

(,4

Structure of Finitt.'

Ficld~

I(x) ~ a o + a,x + ... + an IX" 1+ x n of positive degree n over a field is defined to be the n X n matrix

0 0 I

O·

I 0

0

0

0

(0 A=

0 0

0 0 0

- ao -at

- a,

- an

! It is well known in linear algebra that A satisfies the equation I( A) ~ 0; that is, Go! . . . alA + ... + a n _ IAn 1 + An = O. where I is the n X n identity I

matrix.

Thus, if A is the companion matrix of a monic irreducible polynomial lover IFp of degree n, then I(A) ~ 0, and therefore A can play the role of a root of f. The polynomials in A over IF p of degree less than n yield a representation of the clements of IF q'

2.53. t:xample. As in Example 2.51, let/(x)=x'+IEIF , [x]. The companion matrix of I is

The field IF. can then be represented in the form IF q 21 + A,2A, I +2A,21 +2A). Explicitly:

o = (~ I+A=(: 1+2A=(i

1=

(6

2/+ A

~

{O, 1,21, A, I + A,

n 21~(~ n A~(~

~ (i ~), 2A=(~

2/+2A~(~

~ ),

With IF. given in this way, calculations in this finite field arc then carried out by the usual rules of matrix algebra. For instance,

(2/+A)(J+2A)~(i ~)(~ :)=(~ 6)~2A.

0

In the same way, the method based on the factorization of the cyclotomic polynomial Qq I in IFp[x] can be adapted to yield a representation of the elements of IF q in terms of matrices. 2,54. Example. As in Example 2.52, let h(x) ~ x' + x + 2 E 1F 3 [x] be an irreducible factor of the cyclotomic polynomial Q, E IFJ[x]. The companion matrix of h is

65

The field

~q

can then he represen ted in the form ~" ~ {O. C.

(". ('3. (". c' .C'.

c'. C').

Explicitly:

o· c- '0 I' o ~ (~ 0)' 2). \ I ~

('4

(2

.0)

c'- ~ (2I

c' ~ (:

b)'

:).

-

2

c'- (0 2'- - 2

c'~(; ~ ).

O

(', __ (I

C'.

(6

;).

n-

n

Caleulations proceed hy the rules of matrix algehra. ror instance.

c" -;- C ~ (. 2I 6.

I• (2 2) - 2

D

WEDDERRURYS THEOREM I

All results for finite fields are at the same time also true for all finite division rings hy a famous theorem due to Wedderburn. This theorem states lhat in a finite ring in which all the field properties except commutativity of

multiplication are assumed (i.e.. in a finite division ring), the multiplication must also he commutative. Basically. the first -proof we present of the theorem considers a suhring of the finite divison ring that is a field and establishes a numerical relation between the multiplicative group of the field and the multiplicative group of the whole division ring. Using this relation and information ahout cyclotomic polynomials. one ohtains a contradiction-unless the field is all of the division ring. Before we prove Wedderhurn's theorem in detail. we mention some general principles that will be employed. I.et D be a division ring and Fa subring that is a field (later on, we will express this more hriefly hy saying that F is a subfield of D). Then D can be viewed as a (left) vector space over F (compare with the discussion of the analogous situation for fields in Chapter I. Section 4). If F~ ~q and D is of finite dimension /I over ~q' then D has q" elements. We shall write D* for the multiplicative group of nonzero clements of D. For a group G and a nonempty subset S of G, we defined the normali/.er N(S) of Sin G in Definition 1.24. If S is a singleton {b). we may also refer to N({b}) as the normalizer of the clement h in G. From Theorem This .;,ccti(1n ('iln he omitted without lo~ing nl'~c5~af) information for thl.: following

thaptcf5.

Structurt' of Pinitc fidds

- 66

1.25 we infer that if (; is finite, then the numher of elements in the conjugacy class of h is given hy ICI/IN«b))I. 2.55. is a field.

Theorem (Wedderhurn's Theorem).

f:verv finite dieision ring

Firs! Proof Let () be a finite division ring and let Z ~ {z ED: zd = E D) he the ("emer of D. We omit the ohvious verification that Z is a field. Thus Z = IF q for some prime power q. Now D is a vector space over L of finite dimension n, and so D has q" clements. We shall show that D ~ Z, or. equivalently, that n ": I. Let us suppose, on the contrary. that n > J. "low let a E D and define Nu = {h co D: ab = hal· Then N u is a division ring and N u contains L. Thus N{/ has qr clements, where I ~ r ~ n. We wish to show that r divides n. Since N: is a suhgroup of D*, w~ know that q' - I divides q" 1. If 11 =' rm 1 t with 0" t < r, then qn. I· q'"'q' -I ~ q'(q'''' -I)+(q' -I). "low q'-I divides q" .. I and also q'''' - I, thus it follows that q' - I divides q' - I. But q'. I < q' - I. and so we must have t - O. This implies that r divides n. We consider now the class equation for the group D* (see Theorem 1.27). The center of D* is L*. which has order q - I. For a E D*, the normali/'er of a in D* is exactly N{/*. Therefore, a conjugacy class in D* containing more than one memher has (q" . 1V( q' - I) elements, where r is a divisor of n with I ~ r <. n. Hence the class equation hecomes

dz for all d

q" . I ~ q -I +

Lk

"

i'"' 1

I

q qr, - 1

(2.7)

where r l .... , r A are (not necessarily distinCt} divisors of n with 1 ~ r/

<.

11

for

l:(i~k.

!"ow Jet Q" he the nth cyclotomic polynomial over the field of rational numbers. Then Q,,(q) is an integer by Theorem 2.45(ii). Furthermore. Lemma 2.50 implies that Q,,(q) divide;(q" - I)/(q" - J) for I" i " k. We conclude then from (2.7) that Q,,(q) divides q - I. However. this will lead to a contradiction. By definition. we have n

Q,,(x)~

n

, 1 gcd(5,nl

(x =

1"),

1

r

where the complex numher is a primitive II th root of unity over the field of rationals. Therefore. as compkx numher~. IQ"(q)l~

n"

1 g,CUls.nl

Iq-\'I>

5=

1

n

(q-Ibq-I

5 1 ~cJ(5.n)-1

since n > I and q ~ 2. This inequality is im.:ompatihlc with the

~tatement

6. Wedderhurn\ Theorem

1>7

that Q"(q) divides q - I. Hence we must have theorem is proved.

II ~

I and D = 7, and the 0

Before we start with the second proof of Wedderburn's theorem, we establish some preparatory results. Let D be a finite division ring with center 7, and let F denote a maximal sllbfield of D; that is, F is a subfic1d of D such that thc only subfield of D containing F is F itself. Then F is an extcnsion of Z, for if there were an element: E Z with z ff- F, we could adjoin: to F and obtain a subfield of D properly containing F. from Theorem 2.10 we know that F ~ 7( 0, where ~ E F* is a root of a monic irreducible polynomial f E Z[xl. If we view D as a vector space over F, then for each a ED the assignment Ta ( d) ~ da for d E D defines a linear operator '1~ on this vector space. We consider now the linear operator 7i. If d is an eigenvector of T" then for some A E F* we have d~ = Ad. This implies d~d A and hence dF*d" ~ F*, thus dE N( F*), the normalizer of 1'* in the group D*. Conversely, if d E N( F*), then d~d '~A for some A E F*. and so d is an eigenvector of 7(. This proves the following result.

'=

2.56. Lemma. if dE N( 1'*).

All elemenf d

EO

D* is

1111

eigmGector of 'Ii if alld ollly

Let A be an eigenvalue of 7( with eigenvector d, then d~ ~ Ad. It follows that ~ df(O ~ f(A)d, hence A must he a root of f. If do is another eigenvector corresponding to the eigenvalue A, then dod 'Add o ' ~ A. and so the element b = dod ' commutes with A and, consequently, with every element of F~ 7(A). Let P be the set of all polynomial expressions in h with coefficients in F. Then it is easily checked that P forms a finite integral domain, and so P is a finite field by Theorem 1.31. But P contains F, and thus P ~ F by the maximality of F. In particular, we havc hE F, and since do ~ hd, we conclude that every eigenspace of T( has dimension 1. We use now the following result from linear algebra.

°

2.57. Lemma. Let T be a linear operator on the finite-dimensional veCTOr space V OGer the field K. [hen V has a basis consisting of eigenGectors of T if and onlv if the minimal polynomial for [splits in K infO distincI monic linear factors.

Since frO ~ 0, the polynomial f annihilates the linear operator T". Furthermore, f splits in F into distinct monic linear factors by Theorem 2.14. The minimal polynomial for T" dividesf, and so it also splits in F into distinct monic linear factors. It follows then from Lemma 2.57 that D has a basis as a vector space over' F consisting of eigenvectors of "It. Since every eigenspacc of T( has dimension 1. the dimension m of D ovcr F is equal to the number of distinct eigenvalues of T". Let ~ ~ ~" ~, ..... L bc thc distinct eigenvalues of 7( and let I ~ d" d" .... d m bc corresponding eigenvectors.

Structure of }-in:tc Fields

Because N( F*) is closed under multiplication. it follows from Lemma 2.56 that d,d i must correspond to an eigenvalue~,. say, and hence d,dJ~ ~~, d,dl' Using dJ~ ~ ~idi' we ohtain d'~J ~~, d,. or d/Jd, I.~ ~,. This shows that for eal:h i, 1 ~ i ~ m. the mapping that takes ~j to di~jdl I permutes th~ eigenvalues among themselves. Consequently, the coefficients of K(X) c. (x - ~,) ... (x - ~m) commute with the eigenvectors d" d, ... .,d m of T". Since the coefficients of K ohviously belong to F and thus commute with all the clements of F, they commute with all the elements of D. since these can he written as linear comhinations of d ,. d 2 , ...• d.1I with coefficientt, in F. Thus the coefficients of g arc elements of the center 7. of D. Since g( ~) ~ O. Lemma 2.12 implies that / divides~. On the other hand. we have already observed that every eigenvalue of 1i must he a root of /. and so / ~ g. It follows thatl F: 7. J - lZ( 0: Zj ~ deg( j) ~ m. :"ow m is also the dimension of Dover F, and so the argument in the proof of Theorem I.X4 shows that D is of dimcnsion m~· over 7.. Since thc lattcr dimension is independ~nt of 1". we conclude that every maximal subfield of D ho> the same degree over L. We state this result in the following equivalent form. 1.58.

Lemma.

All maximal suh/ield, a/ D hal'<' Ihe same lirda.

Secolld Proof of Theorem !.55. Let D he a finite division ring. and let 7.. F ~ L( ~). and / Eo 7.[ x [ he as above. Let I:" he an arhitrary maximal subfield of D. Then. hy Lemma 2.5X. L' and F have the sam" order. say q. In view of I.emma 2.4. both F and F arc splitting fields of x' - x over L. It follows then from Theorem 1.91 ,hat there exists an isomorphism from F onto F that keeps the elements of L fixed. The image ~ C". 1,'* of ~ under this isomorphism is therefore a root of / in t·. and so F. ~ 7.( ~). Consider the linear operator T, on the vector space Dover F. Since /( ~) = 0, the polynomial/annihilates T,. But f splits in F, and so there exists a wot A E F of f that is an eigenvalue of TlJ' For a corrc:-oponding eigenvector d we have then d~ ~ Ad. and this implies L* ~ d 'Pd. Thus. f;* is a conjugate of the subgroup F* of D*. For an arbitrary (' C D*, the set of polynomial expressions in (' with coefficients in 7. forms a finite integral domain. and thus a finite field hy Theorem 1.31. Hence. any clement of D* is contained in some suhfield of D, and so in some maximal subfield of D. hom what we have already shown. it follows that any element of D* helongs to some conjugate of F*. fly Theorem 1.25, the numher of distinct conjugates of P is given hy :/)"1 / IN( F*) I. and so it is at most' D* j IF"I. Since each conjugate of F* contains the identity element of D"". the union of the cunjugatc~ of F* has at most

/)"1

- - ( F'

WI

69

Excn:iscs

clements. This number is less than D ~ F. and D is a field.

ID"I

except when D" ~

P.

Hence C

EXERCISES 2.1.

2.2.

2.3. 2.4. 2.5. 2.6. 2.7.

2.R.

2.9. 2.10. 2.11.

2.12. 2.13. 2.14.

2.15. 2.16.

Prove that x 2 + I is irreducible over IF; I and show directly that ~ "~I 'lI(x" I I) has 121 element>. Prove also that x' • x +4 is irreducible over 0: 11 and show that If "Ix \/( x' + I) is isomorphic to If:dxll(x' , x +4). Show that the :-.um of all dements of a finite field is 0, except for IF l . I.et U, be elements of 0: r. n odd. Sbow tbat u' - ab + h" ~ 0 implies a ,. h ~ O. Determine all primitive clements of IF 7' Determine all primitive clements of 1F]7' Determine all primitive clements of IF'!. Write all elements of IF 25 as linear comhinations of hasis clements over 0:,. Then find a primitive clement fJ of 0:" and determine for each 0: E IF Ii the least nonnegative integer II such that a = If the elements of arc represented as powers of a fixed primitive element h <: I' q' then addition in 0: q is facilitated by the introduction of Jarobi's !0l'.arirhm Un) defined by thc cquation I ~ b' ~ b l.",. where the case hn 1 is excluded. Show that we have then bill + hfl = h"" J (n- m) whenever I. is defined. Cot:\struct a tahle of Jacobi's logarithm for 0: 9 and 0: 17' Prove: for any field F. every finite subgroup of the multiplicative group F* is cyclic. Let F be any field. If F' is eyelie, show that F is finite. Prove: if F is a finite field. then H ~ {O} is a subfield of F for every subgroup 11 of the multiplicativc group P if and only if the order of r'" i~ either 1 or a prime number of the form 2 Jl - I with a prime p. For every finite field 0: q -of characteristic p, show that there exists exactly one pth root for each element of 0: q' For a finite field 0:, with q odd, show that an clement a E 0:; has a square root in IF q if and only if a('/ 11(2 = 1. Prove that for given k "N tbe element a is the k th power of Some element of IF II if and only if a(
n

rr.

n=:

EO:;

a"

Strul.:lun: of Finite' fidds

70

2.17. 2.IR. 2.19. 2.20. 2.21.

2.22. 2.23.

2.24.

Prove that [(x)" ~ [(x") for [ en:) x]. Show that any quadratic polynomial in IFq!x] splits over IF. into linear factors. Show that for a E IF q and n f? N the polynomial x q ' - x ;- na is divisible by x q - x /. a over IF". Find all automorphisms of a finite field. If Fis a field and 0/: F ~ F is the mapping defined hy 0/( a) ~ a I if a .r O. 0/( a) ~ 0 if a ~ O. show that 0/ is an automorphism of F if and only if F has at most four dements. Prove: if p is a prime and n a positive integer. then 11 divides (p" -I). (!lim: Usc Corollary 2.19.) Let IFq be a finite field of characteristic p. Prove that [ E IFqlx] satisfies ['(x) ~ 0 if and only if [is the pth power of some polynomial in IFq! x]. Let F be a finite extension of the finite field K with [F: K] ~ m and let [(x) ~ x d + bd' IXd- I + ... + ho E K[x] be the minimal polynomial of aE F over K. Prove that TrF/K(a)~ -(mjd)hd _ 1 and Nf/x(a)

= (_l)mhO'/d

Let F be a finite extension of the finite field K and a E F. The mapping L: fJ E F>-+ afJ E F is a linear transformation of F. considered as a vector space over K. Prove that the characteristic polynomial g( x) of a over K is equal to the characteristic polynomial of the linear transformation L; that is. g(x) = det(xl- L). where 1 is the identity transformation. 2.26. Consider the same situation as in Exercise 2.25. Prove that Trf/x(a) is equal to the trace of the linear transformation L and that NF/K(a) ~ det( I.). 2.27. Prove properties (i) and (ii) of Theorem 2.23 by using the interpretation of Trf/K(a) obtained in Exercise 2.26. 2.28. Prove properties (i) and (iii) of Theorem 2.28 by using the interpretation of NF/K(a) obtained in Exercise 2.26. 2.29. l.et F be a finite extension of the finite field K of characteristic p. Prove that Tr F/. (a P") = (Tr F/.f a»P' for all a E F and 11 EN. 2.30. Give an alternative proof of Theorem 2.25 by viewing F as a vector space over K and showing by dimension arguments that the kernel of the linear transformation Tr F/ K is equal to the range of the linear operator L on F defined by I. f fJ ) ~ fJq - fJ for fJ E F. 2.31. Give an alternative proof of the necessity of the condition in Theorem 2.25 by showing that if aE F with Trfl.(a) = O. y E F with TrflKfY) = - I. and 8, ~ a + a q + ... + a q ' '. then

2.25.

IF: KI

L o,Y"' ,

fJ~ r

satisfies

fJ" - fJ ~ a.

1

2.32. 2.33. 2.34. 2.35.

Let F be a finite extension of K ~ IF q and a = {3q - {3 for some {3 E F. Prove that a ~ yq - y with y E F if and only if {3 - y E K. Let F be a finite extension of K ~ IF q' Prove that for a E F we have ~F/K(a)~l ifandonlyifa~{3q 'forsome{3EF*. Ptove '[7-o'x q ' - c ~ x - a) for all c E K ~ IF q' where the product is extended overall aE F~lFq"' with TrF/K(a)~c. Prove

m

x q "'

..

x

~

n (' mL:' x ,"E-i

2 36,

2.37,

q

q' -

(')'

j=-O

for any m EO N. Consider IF q"' as a vector space over IF q and prove that for every linear operator L on IF q"' there exists a uniquely determined m-tuple ( a", a" " "a m ,) of elements of IF q"' such that

Prove that if the order of basis elements is taken into account, then the number of different bases of IF q"' over IF q is

(qm _ J)(qm- q)(qm _ q').,. (qm _ qm-'), 2.38. 2.39. 2.40.

Prove: if (a"".,a m ) is a basis of F~lFq", over K ~lFq' then Trr/K(a/)=t=O for at least one i, l~i~m.. .., I Prove that there exists a normal basis (a, a q ,,,,, a q ) of F ~ IF q"' over K ~ IF q with TrF/K(a) ~ I. Let K he a finite field, I"~ K(a) u finite simple extension of degree n, and f C K Ix I the minimal polynomial of a over K. Let

f(x)

x-a

2.41.

2.42. 2.43.

~fl0 -{3x+'" +fl 1

II

I

x" 'cF[x]

and

y~f'(a).

Prove that the dual basis of {La"",a" '} is {{joY ',{3,y ' ..... {3" - ,-I . , }. Show that there is a self-dual normal hasis of 1F 4 over IF" hut no self-dual normal basis of ff'16 over IF, (see Example 2.31 for the definition of a self-dual hasis). Construct a self-dual hasis of IF 16 over IF, (see Example 2.31 for the definition of a self-dual hasis). Prove that the dual hasis of a normal hasis of IF q"' over IF q is again a normal hasis of IF qn. over IF/I"

2.44.

Let F he an extension of the finite field K with basis {a" .... a m } over K. Let {3, .. ",{3mEFwith {3,·-"ij_lh,jaj for I«i«m and b,jEK. Let B be the m x m matrix whose U, j) entry is h,j' Prove that !:!. '/K W,,,, "flm) ~ det( 8 )'!J. TIK( a, .. " ,am)'

Structure of I"initl' Fidd"

2.45.

~

Let K

IF q and F ~ IF q"" Prove that for a E F we have A UfiK

(I • a ..... a m

')

n

~

0"

2.46.

2.47. 2.48. 2.49. 2.50.

2.51.

2.53.

2.54.

2.55. 2.56. 2.57.

J ,; m

(.. .

~ I(

""q')' ~

1

Prove that for a F F = IF q" with m " 2 and K ~ IF '/ the diseriminant 6. 1/ K (1. a, .... am - 1) is equal to the discriminant of the characteristic polynomial of a over K. Determine the primitive 4th and Xth roots of unity in IF". Determine the primitive 9th root, of unity in IF ,,,. Let I' bc an nth root of unity over a field K. Prove that 1+ K I 1" f ... + K' ,~O or n according", 1'" I or I' I. For 2 let 1', ..... 1', be all thc (not necessarily distinct) nth roots of unity over an arbitrary field K. Prove that I'f + . " . K;, ~ II for k ." and I'f + ... - 1',; ~ II for k ~ I. 2..... n - I. For an arbitrary field K and an odd positive integer n. show that

n"

°

K(2n)

2.52.

I <

=

K(nl.

Let K be an arbitrary field. Prove that the eyclotomie field Kid' is a subfield of K (,) for any positive divisor d of n E N. Determine thc minimal polynomial over K (4) of a root of unity that can serve as a defining element of K(l2) over K(4). Prove that for p prime the p - I primitive pth roots of unity over Q are linearly independent over Q and therefore form a basis of QIP' over Q. Let K be an arbitrary field and n:;, 2. Prove that the polynomial x n 1+ x l1 - 2 + ... ... x ~ 1 is irrcducihle over K only if n is a prime number. Find the least prime p sueh that x" .,. x" + .. '1 X f I is irreduci· ble over IF p' Find the ten least primes p sueh that XC '+ x P ' - .•• + x ~ I is irreducible over IF,. Prove the following properties of cyelotomic polynomials over a field for which the polynomials exist: (a) Qmpfx) ~ Qm(x P )/Qm(x) if P is prime and meN is not divisible by p: (b) Qmp(x) ~ Qm(x P ) for all mEN divisible by the prime p: (e) Qmp.(x)~Qmp()/ ') if p is a prime and m.kF.N are arbi· trary: (d) Q,,(x) = Qn( - x) if n:;, 3 and II odd: (e) Qn(O)~ I if n:;, 2: (f)

Q,(x ))x·ln)~Q,(x)ifn:;'2;

(g)

O

f Qn(J)~IP ,I

if n ~ I, if n is a power of the prime p, if n has at least two distinct prime faetors:

I·. \L'f<:: 'l'~

(h)

if II ,;, 2. if II ~ I, if n is 2 times a power of the prime p. otherwise.

2.58. 2.59.

Give the matrix representation for the elements of F, uSing the irreducible polynomial .<3 + x + lover F,. Let t he a primitive element of F ~ F 16 with t 4 , t - I ~ O. For k ;;, 0 write ~1.. = L~_ua"mrm with a"m E IF~, and let M k he the 4 x 4 matrix whose (i,i) entry is a k ./ 1,/ I' Sho'N that the 15 matrices M". o~ k ~ 14. and the 4 x 4 zero matrix form a field (with respect to addition and matrix multiplication over F 2) which is isomorphic to F. For 0 ~ k ~ 14 prove that TrF{t') ~ trace of M, ~ a".

Chapter 3

Polynomials over Finite Fields

The th~()ry of pol.vnomials over finik fidd.., i~ important for investigating the algchrail: structure of finik field~ as well as for many applications.

.'\hove all. irreducihle polynomials the prime elements of the polynomial ring over J finite fidJ ar~ indispensable for constructing finite fields and computing with the clements of a finite field.

Scctiml 1

introdllce~

the notion of the order of a polynomial. An

ilnp\)rtant f:.tet i~ the connection hetwecn minimal polynomials of primitive

elemellh (so-called primitive polynomials) and polynomials of thc highest p\)ssihlc orJcr for a given degree. Results ahout irreducihle polynomials going hcyond those di~cus~ed in the prcviou,:; chapters are presented in Sct:tilm 2. The next section i~ devoted In constructive aspects of irreducibil-

ity and deJ.b abll with the prohlcm of cukulating the minimal polynomial of an ckm~nt in an extcn~ion field. Certain ~peciul types of polynomials arc discussed in the last two sections. LineariLed polynomials arc singled out by the property that all the exponent~ occurring in them are powers of the characteristic. The remarkahle theory of these plllynomials enables us. in particular, to give an ultcrnative proof of the normal hasis theorem. Hinomials and trinomials th,.It is, two-term and three-term polynomiab form another class of polynomials for which special results of eonsiderahle interest can he estahlished. We remark that another useful collection of polynomialsnamelv, that of eyclotnmie polynomials-was already considered in Chapter

2. Section 4. and that some additional information on cydotomic polynomi-

al> is contained in Section 2 of the present chapter.

1.

ORDER OF P01.YNOVIIALS ANI> PRIMITIVE I'OLY"OVIIALS

Besides the degree. there is another important integer attached to a nonzero

polynomial over a finite field, namely its order. The definition of the order of a polynomial is based on the following result. 3.1. Lemma. I.el / E IFqlxj he a polynomial 0/ degree m;;, I wilh /(0) '" O. Then Ihere exiSIS a posilive inleger e ,;; q" - I such Ihal /( x) divides

xe

-

J.

Proof The residue class ring IFq[xll(fJ contains q" - I non/.ero residue classes. The q~ residue classes x' I (/).j'=O.L ... ,q~-I, are all nonzero, and so then.: exist intc:gcrs r and .~ with 0 ~ r < s ~ qlll - I such that x' '" x'mod/(x). Since x and /(x) arc relatively prime. it follows that x' '", I mod/(x); that is./(x) divides x' '-I and 0 < S r,;; q" - I. [l

Since a non7ero constant polynomial divides x _. L these polynomials can he included in the following definition. Definition. Let / E IF qlx I he a non7ero polynomial. If /(0), O. then the least positive integer e for which /(x) divides x' - I is called the order of / and denoted by ord( fJ ~ ord(/(x). If /(0) ~ 0, then /(x) ~ x'g(x). where h E 1'\1 and g E IF.rx I with g(O) ""' 0 arc uniquely determined; ord( fJ is then defined to he ord( g).

3.2.

The order of the polynomial/is sometimes also called the period of / or the exponenl of f. The order of an irreducible polynomial/can be characterized in the following alternative fashion. 3.3. Theorem. Lel / E 1F"lx I he an irreducihle polynomial over F q 0/ degree m and wilh /(0) '" O. Then ord(/) is equal 10 order 0/ any rool all in the multiplicative group IF;.".

"Ie"

Proof According to Corollary 2.1 S. IF q" is the splitting field of / over IF q' The roots of / have the same order in the group IF;. hy Theorem 2.1 g. Let" E IF;., be any root of /. Then we ohtwn from I.emma 2.12 that we have ,,' ~ I if and only if /(x) divides x' -I. The result follows now from C the definitions of ord(/) and the order of" in the group IF;... 3.4. Corollary. 1// E IFq[x] is an irreducihle polynomial over degree tn, Ihen ord(/) divides q" - L

IF q 0/

7"

Polynomial... over rinil~' ril'kb

Proof If f(x)~CX with ('Ff;. then ord(f)~'1 and the result is trivial. Othcrwbc. the result follows from Theorem 3.3 and the fact thut lr}, i~ a group of order qm - I. c....; For reducihle polynomials the result of Corollary 3.4 need not he valid (sec Example 3.10). There is another interpretati"n of ordUl hased "n associating a square matrix to J in a canonical fashion and considering the order of this matrix in a certain grnup of matrices (sec LL'mma 6.26). Theorem 3.3 leads to a formula for the numher of monic irrc.:ducihle polynomials of given degree and given order. We usc again $ to denote Euler's function introduced in Theorem 1.I5(iv). The following terminol(')gy will he convenient: if n is a positive integer and the integer h is relatively prime to n. then the least positive integer k for which 1/ .. I mod n is called the multiplicaTive order of h modulo n.

3.5.

Iht!orem.

The numher of

mOlli('

irreducihle polynomial.'\ in

f q["] of degree m and order e is equal 10 $( e)1 m if e " 2 and m is the mulJiplicullre order of q modulo e. l!quallO 2 if m ,- e = I. and equal to 0 in all other cases. In pllrl/l'ular. the degree of WI irreducihle po~rnomial in Irqfxl of order e musT he equal to the nlUITiplicaTit'i' order of q modulo e.

Proof. Let f he an irreducihle polynomial in fqlx] with f(Ol~ O. Then. according to Theorem 3.3. we have ord( f) ~ e if and only if all roots of f ar~ primitive eth mOb of unity over II-(/" In other words. we have ord( f) ~ (' if and only if f divide., the cvclotomic polynomial Q... By Theorem 2.47(ii). any monic irrcducihlc factor of Qe has the same degree m. the Icaq positive integer such that q>n =: 1mod e. and the numher of sW..·h factors i:-. givl:n hy $({')/m. For nl = e.:... 1. we also have 10 take into account the monic irreduejhle polynomial ((.x.) ~ x. U Value, of ord( j) arc availahlc in tahulated form. at least for irreducihle polyn"mials f (sec Chapter 10. Section 2). Since any polynon,ial of positi\'~ degree can he written as a product of irreducible polynomials. the computation of orders of polynomials can he achieved if one kn()\.I;s how to det~rmine the order ()f a power of an irreducihle polynomial anu the order of the product of pairv.'i:-.e relatively prime polynomials. The suhsequent discus~ion i~ devoted to these queqions.

3.6 f

C

I.emma.

I.el

C

he

If- q[ x J "'itll frO) * 0 dit'idel' ., '

(1

positive if1ll!.f!,t'f. Theil the polyll()JlIial I if alld linh' I[ ord( [ ) di"ides "-

If e = ord( [ ) divides "- then f( X) divides x e - I and x" divides XC - I. so that f( x) divides .,' I. Conversely. if f( x) divides x" we have c ~ e. so thut we can write (' . ,., me _. r with 111 L Nand O:S.( r < Since XC 1.= (x me -1)/· (x' - I). it follows that {(x) divide, x' which i, only possihle for r . O. Therefore. e di,ide., (.

Prolif

I

J. (',

I. ~

I Ohkr (If

l)ol~m()mial.' and

Primitivt'

P()l~n()mi..ll"

77

3.7. Corollary. If £'1 alld e:: are posiTi!:e iJ1l('gers. Then The greaTesT common divisor of X"I I and x"·' - 1 in IF qr x 1 is x J - I. where d is The grealeST common dioisor of e] and e:>. Proof I.et f(x) he the (monic) greatest common divisor of x" - I and xt'; - 1. Since x d - I is a common divisor of xt'o - 1, i .:. ;. L 2, it follows that x"" I divides f(x), On the other hand, f(x) is a common divisor of x", I. i ~ 1.2, and so Lemma 3.6 implies that ord(j) divides c, and e,.. Consequently, ord(j) divides d, and hence fIx) divides x" - I by Lemma 3.6. Altogether, we have shown thatf(x) ~ x" I. D Since powers of x arc factored out in advance when determining the order of a polynomial. we need not consider powers of the irreducible polynomials I(x) with g(O) ~ O.

".Ix]

],8. Theorem. rei g <= he irr"du"ihle over "" wilh g(O) ~ 0 and ord(g) ~ e, and lei f ~ gb wilh a posilive inleger h. I.el I he Ihe smallesl inleger wiTh pt ~ b, where p is The characteriSTic of IF IF 171ell ord(f) = ep'. Proof Setting c ~ ordU) and noting that the divisibility of x' - I by f(x) implies the divisibility of x' - I by g(x), we obtain that e divides c by Lemma 3.6, Furthermore, I(x) divides x" -I: therefore, f(x) divides (x" _I)h and, a fortiori, it divides (x' I)r' - x'P' . I. Thus according to I.emma 3.6, c divides ep'. It follows from what we have shown so far that c is of the form c ~ epu with 0 <;; u <;; I. We note now that x" - I has only simple roots, since e is not a multiple of p because of Corollary 3.4, Therefore, all the roots of x,p" - I ~ (x" - I) p" have multiplicity pU, But g(X)h divides x,p" I, whence pU;' h hy comparing multiplicities of roots. and so U;, I. Thus we get u . rand r' ep', 0

3.9. Theorem. l.eT XI'.· .. gA he pairwise rellllieezy prime nvnzero polynumial, over ,,", alld 1"1 f - g,'" g" Theil ord(j) is cqualw Ihe leasl commoll //lUlliplc of ord( g,),.,. ,ord( g, ), Proof It is easily seen that it suffices 10 consider the case where g,(O) ~ 0 for I,,; i <;; k, Set e ~ ord(j) and e, ~ ord(g,) for I,,; i <;; k, and let c ,-: lcm( t' I.·· . . e A ). Then each ,\,U.'(). I ~ i ~ k. divides x<'; - I. and so .~) x) divide~ x( .. 1. Because of the pairwise relative primality of the polynomiab g,,, . .,g,, '-'e <'htain that/Ix) divides x' -I. An application of l.emma 3,(i shows that c divides c. On the other hand, f( x) divides x' .- 1, and so each g,(X), I ~ i ~ k. divides xl" .-1. Again hy I.emma 3.6. it follows that each £',. I ~ i ~ k. divides e. and therefore (' di\'ide~ £'. Thus we ,-=ondude that t' = c.

n By u~ing the same argument as above. nne may. in fact. show that the order of the least common multiple of finitdy many non7ero polynomials is equal to the least'eommon multiple of the orders of the polynomials.

7~

3.10. hample. Let us compute the order of f(X)~XIO+X9+X3~ x' -I '= 6',[x]. The canonical factorization of fiX) over IF, is given hy f(X)~(x'.,-x-I)\x4+x+I). Since ord(x'+x+1)~3. we get ord«x' + x + 1)3) ~ 12 by Theorem 3.8. Furthermore. ord(x 4 + x + 1) = 15_. and so Theorem 3.9 implies that ord(j) is equal to the least common multiple of 12 and 15: that is. ord(j)~60. Note that ord(j) docs not divide 2'" I. which shows that Corollary 3.4 need not hold for reducible polynomials. ':J On the hasis of the information provided above. one arrives then at the following general formula for the order of a polynomial. It suffices to consider polynomials of positive degree and with nonzero constant term. l.ll. Theorem. I.el iF" he a fillile field of eharaeleri,lic p. and ler f E IF q[x] he a po(vnomial of positive degree and Wilh frO) '" O. LeI f = aft'· ··ft·. where aElFq,h, ..... b, EN. alld f" ... ./, are dislinel monic irredueihle polvnomials in IFq[x]. he the callollieal faetorizOlion of fin IFq[x]. '!hen ord(j) ~ ep'. where e i, the leasl common mulliple of ord(j,)..... ord(j,) and t i, the smallest illleger with p' ;, maxi b,..... h, ).

A method of determining lhe order of an irreducible polynomial f in IF) x 1 with frO) ~ () is based on the ohservation that the order e of f is the least positive integer such that X'" I modf(x). Furthermore. hy Corollary 3.4. e divides qffl - I, where m ~ deg(j). Assuming qffl > 2, we stan from the prime factor decomposition qffl_1 ~

n P;"

i _.

I

for 1<; j,,;, we calculate the residues of x'q' "/P'modf(x). This is accomplished by multiplying together a suitable comhi nation of the residues of x, x q. x q', .... x q"' 'mod f( x), If x,qO Ii/p, '" I mod f( x), then e is a multiple of p; . If x lq "' "~ip; '" I mod f( x), then e is not a multiple of p;'. In the latter case we check to sec whether e is a multiple of fI;' '. fI;'-' .... ,Pj by calculating the residues of x(qn,

[liP},

x
[liP,' •••• , X{
f( x).

Thi, computation is repeated for each prime factor of qffl - 1. A key step in the method above is the factorization of the integer qffl _ I. There exist extensive tahles for the complete factorization of num-

hers of this form, especially for the case q ~ 2. We compare now the orders of polynomials obtained from each other hy simple algehraic transformations. The following is a typical example. 3.12.

Oefillition.

Let

79

with a" " O. 1 hen the reciprocal polYllomial 1* of f is defined by

. _." (1) _ "

!(x)-'\!x-aox-ra1x

"' + ... -'-

Q"I X111 '1'

3./3. Theorem. I.el f he a lIoll;era polYllomial ill 1',[ x 1 alld 1* ils renprocall'olvllomial. Theil ord( f) ~ ord( 1*). Proof First consider the casef(O) ~ O. Then the result follows from the fact that f(x) divides x' -I if and only if I*(x) does. If f(O)· O. write f(x) x"g(x) with h EN and g E ~q[xl satisfyingg(O) ~ O. Then from what we have already shown it foll"",s that ord(j) ~ ord(g) ord(g') ~ ordU'). where tht.: last identity is valid since g'" = f"'. 0 0

There is also a close relationship hetween the orders of f(x) and f( x). Since f( x) ~ f( - x) for a field "f eharacteristie 2. it suffices to e"nsidcr finite fields of odd characteristic. 3./4. Theorem. For odd q. leI f (" [1').< I he a polrllomwl of posilit"e degree wilh f(O) * O. LeI e alld F he Ihe orders of f( x) and f( - x). respeclively. lhell £ e if e is a mulliple of 4 and t' ~ 2e if e i, odd. 1/ e i\ [Wice an odd numher. Ihen 1:: ~ e /2 i/ all irreduf"ihle lacrors 011 have evell order and E = e vthenvise. Proof Since ordU(x)) ~ e. f(x) divides x"- I. and so I( x) divides ( X ) ' ' - I ~ x" _. I. Thus L divides 2e by Lemma 3.6. By the same argument. e divides 2£. and so L can only bc 2e. e. or e/2. If e is a multiple of 4. then both e and f; are even. Sincef(x) divides x' 1./( -:<) divides (- x)' - I ~'x' -I. and so £ divides e. Similarly. e divides F. and thus it follows that F ~ e. If e is odd. then I( - x) divides ( - x)' - I ~ - x' - I and so x' + I. But then I( x) cannot divide x' - I. and so we must have E ~ 2e. In the remaining case we have e ~ 2h with an odd integer h. Let I be a power of an irreducible polynomial in 1F,[x]. Then I(x) divides (x h - I)(x h + 1) and I(x) does not divide x h -I since ord(j)~ 2h. But x h - I and x h + I are relatively prime. and this implies that I(x) divides x h - I. Consequently.j( - x) divides (- X)h + I ~ - x" -'- I and so x h - I. 11 follows that £ = e /2. Note that by Theorem 3.8 the power of an irreducihle polynomial has even order if and only if the irreducihle polynomial itself has even order. For general I we have a factorization I ~ g, ... g,. where each g, is a power of an irreducible polynomial and g, ..... g, are pairwise relatively prime. Furthermore. 2h ~ lcm(ord(g,)..... ord(g,)) according to Theorem 3.9. We arrange the g, in such a way that ord(g,) ~ 2h, for I ~ i ~ m and ord(g,) ~ h, for m + I ~ i ~ k. where the h, are odd imegers with lcm(h, ..... h,) ~ h. By what we have already shown. we get "rd(Ii,( x)) ~ h, for I ~ i ~ m and ord(g,( - x))·" 2h, for m' I ~ i ~ k. Then Theorem 3.9 yields ,:: :-:" !/'ITl(

h

h

J 11

') h ,

and . . lJ I. ::.;. II - (' /2 if In = k and F..:..:.. 211 "- (! if m < k. Thcse formula.. . are equivaknl 10 lho:-.e given ill the la:-'1 part of the thcorem. 0 It fnllow ... from I.emma 3.] and Definition 3.2 that the order of a pol;noJllial of degree m:;:] ovcr 0- q is at most qn, --I. Thi~ hound i~ attained fl..)r ;In important c1as", of polynomials-namely, so-called primitive polyn~)mi:ll". I he definition of a primitive polynomial is ha~cd nn the notion of primitive ekment introduced in Definition 2.9.

In"

3.15. Definition. i\ polynomial I c= ~ ,I x] of degree I ;, called a primlflre polynomial over if 4 if it is thc minimal polynomial over IT- 11 of a primitive clement of f '!~. Thu:-.. J primitive polynomi;11 OVCf IF" of degree m may hl: dcscribed a:-- a monic polynomial that is irrl:ducihle ovcr IF Q and has a root ex E IF({- that g.enerate:-- the multiplicative group of [f- " .... Primitive polynomiab can also he charaClerii'.ed a~ follow:"'>. 3./6. Theorem. A polynomial I c= IFqlxl 01 degree m IS a primilive fl0lrnomwl ore< IF" if and on/r il I is mOl/i<. 1(0) ~ O. alld ord( I) q"' - I. " mo/. If J i, primitive over} q' thcn J is monic and 1(0) T O. Sincc I i.. . irrcducihle over :I- l 1 for i - l. 2 h) Lemma 3.1, hence

r

l

l


m,

..

qm_1.

a contradict ion. ThL'rcf~)rL'. J i~ irreducihk over :F,,' and it follows then from Theorcni 3.3 thtlt I i~ a primitive polynomial over [F ". Wc remark that the eonJitioll 1(0) ~ 0 in the tht:0rcm ahove is only nl.'eJcd t~) fllk ~)ul thL' non-primitive p(.llynomial f(x) - x in case q = 2 and nI I. Still
O.

.\Olllt'

I. emma.

J.('1 J "= 0= ,,1).:1 Iw

tl p(l~yl1omi(11 of

pOSiliGe degree wit/z mod I(x) a mod J(x) Kith a ulliqut'~v delermined

I.('! r he IIU' It'us{ [Josili[;e Imeger/or 'r\'/Zich x' is congruent

clemen! of if- q' so that x r

=

Order of Polynomials and Primitive Polynomials

1.

U Eo

1';.

~l

Then ord( /) = hr. where h is the order 0/ a in the multiphcatiDe group

1'*

".

Proof Put e ~ ord(J). Since x' '" I mod/Ix), we must have e;, r. Thu.s we can write e = sr t { with sEN and 0 ::s.;;; { < r. Now

(3.1 ) thus x' '" u 'mod/Ix), and because of the definition of r this is only possihle if t ~ O. The congruence (3.1) yields then II' =;; I mod /( x), thus a' = I, and so s;, hand e", hr. On the other hand, x h, " a' '" I mod f< x), and so e = hr. C ],18, Theorem. The monic polynomial/ E I' qlx I 0/ degree m ;, I is primiti"e polvnomial o"er 1', i/ and only i/ (- I )",/(0) is u primiti"e element 0/1', and the least positive integer r for which x' is congruent mod/Ix) [0 some element o/I'q is r ~ (q"' -I)I(q -I). In case / is primitive o"er I'q. we ha"e x' '" (- 1)m/(o) mod /(x).

II

Proof If / is primitive over I' q' then / has a root a c= I' q.' which is a primitive clement of 1'". By calculating the norm :""./<,,
(3.2)

II.

It follows that the order of (- 1)'''/(0) in 1'; is q - I; that is, (- 1)'''/(0) is a primitive element of I'q. Since/is the minimal polynomial of a over I'q, the identity (3.2) implies that X,,"'-II/tq "", (-I)m/(O)mod/(x),

and so r,;; (qm - 1)1 (q - I). But Theorem 3.16 and Lemma 3.17 yield qm _ I ~ ord(J)';; (q -I)r, thus r ~ (q"' -I)I(q - I). Conversely, suppose the conditions of the theorem arc satisfied. It follows from r ~ (qm - 1)1 (q - I) and Lemma 3.17 that ord(J) is relatively prime to q. Then Theorem 3.11 shows that / has a factorization of the form / = /, ... /" where the J, are distinct monic irreducihle polynomials over I' q' If m,~deg(/,), then ord(J,) divides qm'_1 for I,;;i,;;k according to Corollary 3.4. Now qm, - 1 divides d~(qm'_I)"'(qm'_I)/(q_I)'

"

thus ord(J,) divides d for I,;; i,;; k. It follows from Lemma 3.6 that /,(x) divides x d - I for I,;; i,;; k, and sO/Ix) divides x d - 1:lf k '" 2, then d < (qm"

.. "", - 1)1 (q - I)

~

(qm - 1)/ (q .. I)

a contradiction to the definition of r. Thus k

~

~

r.

I and/is irreducible over 1',.

Polynomial:-. over F:ni{l' I-idJ<;

If fJ E F q"' is a root of f. fJ' ~ ( - I )"'f(O). and so

then the argument leading w (3.2) shows that x' '" ( _. l)"'f(O) mod f (x). Since the order of (- 1)"'f(O) in F; is q - 1, it follows from Lentma 3.17 that ordU) ~ q'" - I. so that f is primitive over Fq by Theorem 3.16. 0

3.19. Example. Consider the polynomial f( x) ~ x' " x' i .x·' ." 2x .\ 2 C F3lx J. Since f is irredueihle over F3' one can use the method outlined after Theorem 3.11 to show that ord(j) ~ RO ~ 34 -1. Consequently. f is primitive over F 3 hy Theorem 3.16. We have x 40 " 2 mod f( x) in accordance with 0 Theorem 3.18.

2.

IRREUUCIBLE POLYNOMIAI.S

We recall that a polynomial f E F"lx J is irreducihle over F q if f has positive degree and every factorization of f in F,,[x] must involve a constant polynomial (see Definition 1.57). Elementary properties of irredueihle polynomials over F" were discussed in Chapter 2. Section 2.

3.20. Theorem. [or eGery finile field F" and merr n C N. Ihe product of all monk irreducible polynomials over IF
Proof According to I.emma 2.13. the monic irredueihle polynomials over IF If occurring in the canoni<.:al factoril.atinn of g( x) ,..., xq~ - x in F"rxl arc precisely. those whose degrees divide n. Since g'(x) .. - 1. Theorem 1.68 implies that g has no multiple roots in its splitting field over Fq • and so each monic irrcducihle polynomial over

IF/I

whose degree divide~

occurs exactly once in the canonical factorization of gin F"lx I.

11

:::J

3.21. Corollary. If Nq(d) is Ihe numher of numi<' irreducihle polynomials in Fq[xl of degree d, Ihen qn~ L.dNq(d)

foral/nEN.

(3.3)

din

where the sum is extended aGer all positir:e di!:-isors d

(~f 11.

Proof The identity (3.3) follows from Theorem 3.20 hy comparing the degree of g(x) ~ x'" - x with the total degree of the canonical factoriLa~ tion of g(x). With a little elementary number theory we can derive from (3.3) an explicit formula for the number of monic irredueihle polynomials in F'/lx I of fixed degree. We need an arithmetic function. called the .\1oebius function. which is defined as follows.

83

2. lrrcducihle Polynomials

3.22.

The Moebius fUllclion p. is the function on 1'\1 defined hy

Definition.

ifll~1.

I

p.( II) ~

{

o( -

I)'

if n is the product of k distinct primes. if II is divisihle hy the square of a prime.

As in (3.3), we usc the summation symbol Lul n to denote a sum extended over all positive divisors d of n E 1'\1. A similar convention applies to the product symholll",".

Lemma.

3,23.

For n Eo 1'\1 Ihe Moehius fUllclion p. salisfies if II ~ I. if n > I.

Proof ror 11 > I we have to take into account only those positive divisors d of II for which p.( d) ~ O-that is, for which d ~ lord is a product of distinct primes. Thus, if p" p, ..... p, are the distinct primc divisors of n. we get

,

L:

LP.(d)~p.(I)+ d

I

1/

L:

p,(p,)+

-1

I.,;;

I

<

p.(p, p,,)+'" +p.(PJ!',"·Pk)

I)";;

-I+(~)(' I)~(~)(-I)'-t

A.

... -(~)(-I)'

~(I+(-I))'~O.

The case

II =

3.24.

(i)

n

I is trivial.

Theorem (Moehius Invcrsion formula) Additivc case: Lei hand H he two funclions from 1'\1 inlo all addilively wrillen ahelian group G. Then H(n) ~

L: h ( d )

for aI/ n E 1'\1

(3.4)

d,n

if and only if

h(n)~ LP.(S)lI(d)~· LP.(d)lI(:) din

(ii)

foral/nEI'\I.

(3.5)

din

Multiplicative case: I.el hand 1I be IWO funclions from 1'\1 inlO a mulliplicatively wrillen aheliall group G. Then H(n) ~

n h(d )

for aI/II '= 1'\1

(3.6 )

d n

If and only if h(n)~ nll(d)"("ld)~ dJn

nil ( -dn )f.!/ dIn

J)

for aI/II EI'\I. (3.7)

Polynvmiah owr Finitc 1-"it'1U:-.

84

Proof

Assuming (3.4) and using Lemma 3.23. we get

LI'(~)H(d)= LI'(d)J[(~)= din

dill

~ L ('In

LI'(d) L din

h(c)

,In/d

I'(d)h(c) ~ Lh(c) L

L d-nle

I'(d) ~ h(n)

din/I'

rln

for all n E N. The converse is derived by a similar calculation. The proof of part (ii) follows immediately from the proof of part (i) if we replace the 0 sums by products and the multiples by powers. 3.25. Theorem. The number Nq( n) of monic irreducible polynomials in !'q[ x I of de?;ree n is gicen by Nq(n)

~ 1. n

L I'(!!-)qd ~

don

d

1. L n

I'(d)q"/d.

din

Proof We apply the additive case of the Moebius inversion formula to the group G~Z.the additive group of integers. Let h(n)=nNq(n) and H( n) = qn for all n E N. Then (3.4) is satisfied because of the identity (3.3). and so (3.5) already gives the desired formula. 0

3.26. Example. The number of monic irreducible polynomials in !'q[x] of degree 20 is given by Nq(20) ~ -'\'(I'(l)q'O + 1'(2)q'O + 1'(4)q' + 1'(5)q4 + l'(lO)q2

+ 1'(20)q)

~2~(q20_qlO_q4+q2).

0

It should be noted that the formula in Theorem 3.25 shows again that for every finite field!' q and every n E N there exists an irreducible polynomial in !'q[xl of degree n (compare with Corollary 2.11). Namely. using 1'(1) ~ I and I'(d)" -I for all dEN. a crude estimate yields N (n);, q

1.n (q' _ qn

1_

qn-2 _ ... _ q) =

1.n (q" _

-.!i) > O.

q" q-l

As another application of the Moebius inversion formula. we establish an explicit formula for the nth cyclotomic polynomial Q" 3.27. Theorem. For a field K of characteristic p and dicisible by p. the nih cyclotomic polynomial Q. acer K satisfies Qn(x) ~

1/ E

N not

n (x d _l)",·/dl n (x·/ d _l)"(d l =

d>1

din

Proof We apply the multiplicative case of the Moebius inversion formula to the multiplicative group G of nonzero rational functions over K. Let h(n) ~ Q"(x) and 1I(n) = x' - I for all n E N. Tben Theorem 2.45(i) shows that (3.6) is satisfied. and so (3.7) yields the desired result. 0

2.

lrrl'uu~ihlc

3.28.

Polynomiab

Example.

H5

For fields K over which Q" is defined. we havc

Q,,(x) ~

n (X"/d .. I)"'dl dl12

~ (x" -I)"«'(x' - 1)"'''(x 4

_

I)"
(x' -1)"(6'(x _1)""21 ~

(x"-I)(x 2 -1) (x' -1)(x 4 -I)

~ x4 - r

,

I- I.

D

The explicit formula in Theorem 3.27 can be uscd to establish the basic properties of cyclotomic polynomials (comparc with Exercise 3.35). In Theorem 3.25 we detcrmined the numher of monic irredueiblc polynomials in !'qlx] of fixed degree. We present now a formula for the product of all monic irreducible polynomials in !' q[ x 1of fixed degree. 3.29. Theorem. The product I(q. n; x) of all monic irreducible polynomials in !' ql x 1of degree II is giGen hv I(q.n;x)~ n(xq'-x)"('/d)~ n(x q··"· X)"'d) d n

Proof

d

11

It follows from Theorem 3.20 that x q"- x= n/(q.d;x). dl'

We apply the multiplicative case of the Moebius inversion formula to tbe multiplicative group (i of nonzero rational functions over IF q' putting

h(n)· I(q. n; x) and H(n) dcsired formula. 3.30.

Example.

=

x q'

x for all nEON. and we obtain the D

For q ~ 2. n = 4 we gct

4 )"(2)( x-x 2 )"(4) 1(2 ,4; x) =(x(6- x )",Il( x-x

X1fl_X

x 15 -1

x4

xJ

--=X

12

-

X

+X

ll

_Xf>+x

1

+1.

D

All monic irreducible polynomials in !'q[x] of degree n can be determined hy factoring I(q. n; x). for this purpose it is advantagcous to have I(q. n; x) availahle in a partially factored form. This is achicved by the following result. 3.31. we have

Theorem.

Let l(q.lI;x) he as 1(". n;

xl =

III

nQm(xl. m

Theorem 3.29. Then for n > I (}.X)

Polynomiah ovcr Finite held!';

where the product is extended over all positive divisors m of qn - I for which n is the multiplicative order of q modulo m, and where Qm(x) is the mth cyclotomic polynomial over IF q' Proof For n > I let S be the set of clements of IF q" that arc of degree n over IF q' Then every a E S has a minimal polynomial over IF q of degree n and is thus a root of I( q, n; x). On the other hand, if fJ is a root of I(q, n; x), then fJ is a root of some monic irreducible polynomial in IFqlx] of degree n, which implies that fJ E S. Therefore,

n (x-a).

I(q,n;x)~

aE:S

If a E S, then a E IF;", and so the order of a in that multiplicative group is a divisor of q" - I. We note that y E IF;" is an element of a proper subfield IF q" of IF q" if and only if yq"~ y-that is, if and only if the order of y divides qd _ I. Thus. the order m of an element a of 8 must be such that n is the least positive integer with q" '" I mod m -that is, such that n is the multiplicative order of q modulo m. For a positive divisor m of qn - I with this property. let 8m be the set of elements of S of order m. Then S is the disjoint union of the subsets Sm' so that we can write

l(q,n;x)=n

n

(x- a).

m al'::Sm

Now Sm contains exactly all clements of IF;" of order m. In other words, 8m is the set of primitive mth roots of unity over IF q' From the definition of cyclotomic polynomials (sec Oefinition 2.44), it follows that

n

(x-a)~Qm(x).

aES",

and so (3.8) is established.

D

3.32. Example. We determine all (monic) irreducible polynomials in 1F,lx] of degree 4. The identity (3.8) yields 1(2,4; x) ~ Q,(x)Q,,(x). By Theorem 2.47(ii),Q,(x)~x'.;-x'+x'+x-rl is irreducible in 1F,lx]. By the same theorem, Q,,(x) factors into two irreducible polynomials in 1F,[x] of degree 4. Since Q,(x+I)=x'+xJ+I is irreducible in 1F,[x], this polynomial must divide Q,,(x), and so

Q,,(x) =x'+x 7+x' +x· + x' +x+ I = (x' +k' + I)(x' Therefore, the irreducible polynomials in 1F,[xj of degree 4 arc + x + I. x' + x) + 1. and x 4 + x + I.

T X

+ I).

x'., x' + x' D

Irreducible polynomials often arise as minimal polynomials of elements of an extension field. Minimal polynomials were introduced in Definition 1.81 and their fundamental properties established in Theorem 1.82. With special reference to finite fields, we summarize now the most useful facts about minimal pOlynomials.

87

.\ Construt'tivn of Irn:ducib1c Polynomiab

3.33. Theorem. LeI" be an elemen/ vf Ihe eXlension field IF q' vf IF q' Suppose (/WI Ihe degree of" ocer IF q is d and Ihal g E IFq[x] is Ihe minimal polYllomial oj " vver IF q' Then: (i) (ii)

g is irreducible acer IF q and ils degree d divides m. A polynomial f ElF q[ x 1salisfies f (,,) = 0 if and only if g divides

j. If f is a monic irreducible polynomial in IF ,[x 1 wi/h f( ,,) = O. (hen j= g. (iv) g(x) divides X,d - x and x q ' - x. (v) The rool,' of g are ". ,,'•...• ",' " and g iI Ihe minimal polynomial acer IF, oj alllhese elemen/s. (vi) If" '" O. Ihen ord(g) is equal 10 Ihe vrder of" in Ihe mulliplica-

(iii)

tive group IF;",. (vii)

g is a primilive polynomial over IF, if and only ij" is of order q d - I 'In IF'q""

Proof (i) The first part follows from Theorem 1.82(i) and the second part from Theorem 1.86. (ii) This follows from Theorem 1.82(ii). (iii) This is an immediate consequence of (ii). (iv) This follows from (i) and Lemma 2.13. (v) The first part follows from (i) and Theorem 2.14 and the second part from (iii). (vi) Since" E IF;, and IF;, is a subgroup of IF;., the result is contained in Theorem 3.3. (vii) If g is primitive over IF" then ord(g) = qd -1, and SO" is of order qd - I in IF;., because of (vi). Conversely, if " is of order qd - I in IF;. and so in IF;d, then a is a primitive element of IF qd. and therefore g is primitive over IF, by Definition 3.15. 0

3. CONSTRUCTION OF IRREDUCIBLE POLYNOMIALS We first deserihe a general principle of obtaining new irreducible polynomials from known ones. It depends on an auxiliary result from number theory. We recall that if n is a positive integer and the integer b is relatively prime to n. then the least positive integer k for which bk I mod n is called the multiplicalive order of b modulo n. We note that tbis multiplicative order divides any other positive integer h for which b h = I mod n.

=

3.34. Lemma. l.el s;" 2 and e ;" 2 be relalively prime inlegers and leI m be Ihe muiliplicalice order of s modulo e. LeI I;" 2 be an in/eger whose prime faclOrs divide e bUI not (sm - I)/e. Assume also Ihal sm I mod4 if I " 0 mod4. Then Ihe mulliplicalive order of s modulo et iI equal to mt.

=

PolXnomials over Finite Fields

Proof We proceed by induction on the number of prime factors of each counted with its multiplicity. First, let 1 bc a prime number. Writing d = (sm -I)/e, we have sm = I + de, and so

I,

s""=(I+de)'

=1+( :)de+U)d'e'+'"

+C~I)d'-'e'

'-rd'e'.

In the last expression, each term except the first and the last is divisible by el because of a property of binomial coefficients noted in the proof of Theorem 1,46. Furthermore, the last term is divisible by el since 1 divides e. Therefore, sm, = I mod el, and so the multiplicative order k of s modulo el divides mI. Also, s' = I mod el implies s' = I mod e, and so k is divisible by m. Since I is a prime number, k can only be m or mI. If k = m, then sm = I mod el, hence de = omod el and I divides d, a contradiction, Thus we must have k = ml, I\ow suppose that 1 has at least two prime factors and write I = rio, where r is a prime factor of I. By what we have already shown, the multiplicative order of s modulo er is equal to mr. If we can prove that each prime factor of 10 divides er but not do = (sm, -I)/er, then the induction hypothesis applied to 10 yields that the multiplicative order of s modulo ert o = el is equal to mrt o = mi. Let ro be a prime factor of 10 , Since every prime factor of I divides e, it is trivial that ro divides er. We write again d=(sm-I)/e.Wehavesm'-I=c(sm-l)withe=sm" "+ .. , +sm+1. thus do = e(sm -Iller = ed/r. Furthermore, since sm = I mode and r divides e, we get sm=lmodr, and so e=r=Omodr, Thus e/r is an integer. Since ro docs not divide d, it suffices to demonstrate that ro docs not divide elr in order to prove that ro does not divide do = edlr. We notc that sm=lmodro, and so e=rmodro' If ro"'r, then elr=lmodro, thus ro does not divide elr. Now let ro = r, Then sm = I + brmod r 2 for some bE Z, hence sml = (I + br)1 = I + }brmod r' for all);> 0, and thus , e=r+br

L

, }=r+br

r(r-I) 2 modr'.

r O

It follows that e r(r-I) -=I+b modr . r 2

If r is odd, then elr = I modr, so that ro = r does not divide elr. In the remaining case we have ro = r = 2. Thcn I = 0 mod 4, and so s m '" I mod 4 by hypothesis. Since e = sm + I in this case, we get e = 2 mod 4, and thus clr = cl2 = I mod2. It follows again that ro does not divide clr. 0 3.35, Theorem, LeI /,(.<), /,(.<), ... ./1"(.<) be alllhe dlwlncl monic Irreducible polynomials in 0= ql'<10/ degree m and order e, and lei I ;> 2 he an

3. Con:-trudion of Irreducihle Polynomial&

R9

integer whme prime factors dicide e but not (qrn -I)/e. Assume also that omod 4. Theil f,(x').f,(x') .... ,fN(x') are all the distinct monic irreducible polynomials ill IF qlx] of degree mt and order el.

qrn" I mod4 if t '"

Proof The condition on e implies e ~ 2. According to Theorem 3.5. monic irreducible polynomials in IF q[x 1 of degree m and order e ~ 2 exist only jf m is the multiplicative order of q modulo e. and then N ~ $(e)/m. By Lemma 3.34. the multiplicative order of q modulo et is equal to mI. and since $(et)/ml ~ (e)/m hy the formula in Exercise 1.4. part (c), it follows that the number of monic irreducible polynomials in IF qlx] of degree ml and order el is also equal to N. Therefore. it remains to show that each of the polynomialsJ;(x'). I" j" N. is irreducible in IFq[x] and of order el. Since the roots of each !,(x) are primitive eth roots of unity over IFq by Theorem 3.3. it follows that !,(x) divides the cyclotomic polynomial Q,(x) over IF q. Then !,(x') divides Q,(x'). and repeated use of the property enunciated in Exercise 2.57. part (b), shows that Q,(x') ~ Q,,(x). Thus J;(x') divides Q,,(x). According to Theorem 2.47(ii). the degree of each irrcducible factor of Q,,(x) in IFq[x] is equal to the multiplicative order of q modulo el. which is mI. Since!,(x') has degree mI. it follows thatJ;(x') is irreducible in IFq[x]. Furthermore, since.!i(x') divides Q,,(x). the order of J;(x') is el. 0

3.36. Example. The irreducible polynomials in IF,lx] of degree 4 and order 15 are x 4 + x + I and x 4 + x 3 + I. Then the irreducible polynomials in 1F,[x] of degree 12 and order 45 are x" + x 3 + I and x" + x 9 + 1. The irreducible polynomials in 1F,[x] of degree 60 and order 225 are x 60 + XIS + I and x 60 + x., + I. The irreducible polynomials in 1F,[x] of degree 100 and order 375 are X'OO + x" + I and x"1O + x" + I. 0 The case in which I " Omod4 and qrn" - I mod 4 is not covered in Theorem 3.35. Here we must have q" - 1 mod4 and m odd. The result referring to this case is somewhat more complicated than Theorem 3.35. 3.37. Theorem. /..el f,(x). f,(x), ... ,fN(x) be alllhe dislincl monic irreducible polynomials in IFq[x] of odd degree m and of order e. LeI q = 2"u - I. I ~ 2bc wilh a, b ~ 2. where u and v are odd and all prime factors of I divide e hUI nOI (qrn - l)j e. LeI k be the smaller of a and b. Then each of Ihe polynomials J; (x') faclors as a producl of 2 k -, monic irreducible polynomials g,,(x) in IFq[x] of degree ml2' - '. The 2 k -'N polynomials g,/x) are alllhe dislincl monic irreducible polynomials in IFq[x] of degree mt2' k and order et. Proof If v ~ 3. then Theorem 3.35 implies that f,(x"). f,(x V ) . . . . , fN(x") are all the distinct monic irreducible polynomials in IFq[x] of odd degree mv and of order ev. Thus we will be done once the special case I = 2 h is settled. Let now 1= 2b • and note that as in the proof of Theorcm 3.35 we obtain that m is the multiplicative order of q modulo e, N = (e)/m. and

YO

Polynomial!:. over Finite Fields

eachfj(x') divides Q,,(x). By Theorem 2.47(ii), Q,.,(x) factors into distinct monic irreducible polynomials in 8'q[x] of degree d, where d is the multiplicative order of q modulo el. Since qd" I mod el, we have qd" I mod e, and so m divides d. Consider first the case a", h. Then q2m - I = (qm _I)(qm + I), and the first factor is divisible bye, whereas the second factor is divisible by I since q" - I mod2" implies q" - I mod I, and thus qm" (_ I)m" - I mod I. Altogether, we get q'm" I mod el, and so d can only be m or 1m. If d ~ m, then qm" I mod el, hence qm" 1mod I, a contradiction. Thus d = 1m = m2 h - ' . , since k ~ b in this case. Now consider the case a < b. We prove hy induction on h that

qm2'" I +.w2 a"h modl a"h"

for all hEN,

(3.9)

where w is odd. For h = 1 we get q2m = (la u _ I )2m =l-l a+'um+

2m

L.

(2;')(_1},m.nlnaun"l+w2a"modla'2

n' 2

with w = - urn. If (3.9) is shown for some hEN, then h

qm2 =

I + w2 ai h + C2 u - hi

I

for some (' E lL.

It follows that

and so the proof of (3.9) is complete. Applying (3.9) with h ~ b - a + I. we get qm2 b ,,~I =. ] mod 2b .... I. Furthermore, qrn.= 1mod e implies qm2 b 0·1 == 1mod e, and so qm2' 0"" I mod L. whcre L is the least common multiple of and e. Now e is even since all prime factors of 1 divide e, but also e"$ Omod4 since qm" 1mod e and qrn " - I mod4. Thcrefore, L ~ el b ~ el, and thus qm2' 0"" 1mod e1. On the other hand, using (3.9) with h = h - a we get

1""'

qm2'

0" I + w2 b '" I mod2 h '

'.

which implies qm2' 0", 1mod el. Consequently, we must have d ~ ml b - a ' , = ml b k+' since k = a in this case. Therefore, the formula d ~ ml b , . , = mlll - k is valid in both cases. Since Q,,(x) factors into distinct monic irreducible polynomials in IFq [x] of degree mIl' -k, each Jj(x') factors into such polynomials. By comparing degrecs, the number of factors is found to be 1" '. Since each irreducihle factor g,/x) of Jj(x') divides Q,,(x), each g'l(x) is of order el. The various polynomials g,/x), I.; i.; 2' " I.;).; N, are distinct, for otherwise one such polynomial, say g(x). would dividcf,,(x') andf,,(x') for )1'" )2' and then any root (3 of g(x) would lead to a common root (3' of Jj-,( x) and Jj,( x). a contradiction. By Theorcm 3.5. the number of monic

3. Construuion of Irreducible Polynomiah

91

irreducible polynomials in IF qlx I of degree m12 1 - k and order el is (el)/mI21.-' = 2' I",(et)/ml = 2k I",(e)/m ~ 2 k - IN. and so the gi/X) yield all such polynomials. D We will show how. from a given irreducible polynomial of order e. all the irreducible polynomials whose orders divide e may be obtained. Since in all cases g(x) ~ x will be among the laller polynomials, we only consider polynomials g with g(O) "" O. Let f be a monic irreducible polynomial in IF ql x] of degree m and order e and with f(O) "" O. Let lX E IFq" be a root of f, and for every I E 1'\1 let g, E IFql x] be the minimal polynomial of lX' over IF q' Let T~ (II' 1" .... 1.) be a set of positive integers such that for each IE 1'\1 there ex.ists a uniquely determined I, I" I" n. with I'" I,qbmod e for some integer b:;, O. Such a set T can. for instance, be constructed as follows. Put 1 1 =1 and. when 11.1, ..... 1;_1 have been constructed. let Ij be the least positive integer such that Ij '" I,qbmod e for 1" I < j and all integers h:;, O. This procedure stops after finitely many steps. With the notation introduced above. we have then the following general result. 3.38. Theorem. The polynomials g", g" .... ,g" are all Ihe dlsllncl monic IrredUcible polynomials in IFqlx] whose orders divide e and whose constant terms are nonzero. Proof Each g" is monic and irreducible in IFqlx] by definition and satisfies g, (0) "" O. Furthermore. since g, has the root lX" whose order in the group 1F;;divides the order of lX, it foll~ws from Theorem 3.3 that ord(g,) divides e. Let g be an arbitrary monic irreducible polynomial in IFqlx] of order d dividing e and with g(O) "" O. If /3 is a root of g. then /3d ~ I implies /3' = 1. and so /3 is an eth root of unity over IFq' Since a is a primitive eth root of unity over IF q' it follows from Theorem 2.42(i) that /3 ~ lX' for some IE 1'\1. Then the definiti0n of the set T implies that I'" liqbmod e for some I. I" i.;; n. and some b:;, O. Hence /3 = a' = (a,,)q', and so /3 is a root of g, because of Theorem 2.14. Since g is the minimal polynomial of /3 over IFq' it follows from Theorem 3.33(iii) that g = g, .. It remains to show that the polyn~mials g,. I.;; I "n, are distinct. Suppose g, = g, for i"" j. Then a" and ,,'j are roots of g,. and so a', ~ (a")"" for ~ome b:;, O. This implies I; '" I,qbmod e. but sin~e we also have I, '" l,qOmod e, we obtain a contradiction to the definition of the set T. D

The minimal polynomial g, of a' E IF q" over IF q is usually calculated by means of the characteristic polynomial f, of a' E IF q" over IFq' From the discussion following Definition 2.22 we know that f, = where r ~ m / k and k is the degree of g,. Since g, is irreducible in IFq[x]. k is the multiplicative order of q modulo d = ord(g,). and d is equal to the order of

gr.

92

Polynomials over Finite Fields

a' in the group IF;"" which is e/gcd(t, e) by Theorem 1.l5(ii), Therefore d, and so k and r, can be determined easily. Several methods are known for calculating!,. One of them is based on a useful relationship between!, and the given polynomial f. 3.39, Theorem, Let J he a monic irreducihle polynomial in IFq[x] oj degree m. Let a E IF q" he a root oj J, and Jor tEN let!, be the characteristic polynomial oj a' E IF q" over IF q' Then

,

!,(x')=(-l)ml'+l)nJ(w;x), 1-"'1

where WI".' \ Wt are the [th roots of unity over IF q counted according to multiplicity.

Proof Let a = a" a, ,, .. ,am be al1 the roots of f. Then a:, are the roots of!, counted according to multiplicity. Thus !,(x')

a~, .. .,a~

n'" (x' -a:)

=

i-'-l

n'" n, (x-a,w

=

/"'-1 J=l

=

n'" n, wJ(w

I ..... ] }

= 1

j

j )

'x-a,).

A comparison of coefficients in the identity x' -I

=

n (x - w) 1-'-1

shows that

, nw;=(-I)"', /=1

and so J,(x')

=

(_1)"'1'-1)

n, n'" (W;-'x - a,) I

= ( -I) ml' -I)

~

,

1i = 1

n J( W 'x) = ( - I) I = I

since

J

,

ml' • I)

n J( w;x) J= 1

w,'" .. ,"',- , run exactly through alit th roots of unity over IF

q'

0

3.40. Example. Consider the irreducihle polynomial J(x) = x + X + I in 1F,[x]. To ealculateJ,. we note that the third roots of unity over IF, are I, w. 4

J. Construction of Irreduc:iblc Polynomial!:.

0.1

and (.,i. where w is a root of x 2 + x + 1 in IF 4' Then

fl(X l ) = (-I)'6 f (x)f(wx)f(w'x) =

(x 4 + X + l)(wx 4 + wx+ 1)(w'x 4 + w2 x + I)

= X 12 + x 9 + X O + x 3 + I, 4 so thatf)(x) = x + xl + x' + x + 1.

o

Another method of calculating I, is hased on matrix theory. Let fIx) ~ x m - am ,x m ' - ... - a,x - a o and let A be the companion matrix of f. which is defined to be the m X m matrix

A=

(o

0

1

0

0

I

o

0

o o o am

'I

Then f is the characteristic polynomial of A in the sense of linear algehra: that is. f( x ) ~ det( x I - A) with I being the m X m identity matrix over IF q' For each tEN, I, is the characteristic polynomial of A', the Ith power of A. Thus, by calculating the powers of A one ohtains the polynomials 1,.

3,4 1. Example. It is of interest to determine which polynomials I, are irreducible in IFq[x]. From the discussion prior to Theorem 3.39 it follows immediately that I, is irreducible in IFq[x] if and only if k ~ m, that is. if and only if m is the multiplicative order of q modulo d ~ e/gcd(l, e). Consider. for instance. the case q ~ 2, m = 6. e ~ 63. Since the multiplicative order of q modulo a divisor of e must be a divisor of m. the only possibilities for the multiplicative order apart from mare k ~ 1.2.3. Then q' - I = I, 3. 7. and q'" I mod d is only possible when d ~ 1.3.7. Thus I, is reducible in 1F,[x] precisely if gcd(l,63) ~ 9.21,63. Since it suffices to consider values of t with 1<;; t <;; 63, it follows that I, is irreducible in 1F 2 [x] except when t = 9,18,21.27,36,42.45.54,63. 0 In practice. irreducihle polynomials often arise as minimal polynomials of elements in an extension field. If in the discussion above we letfbe a primitive polynomial over IF q , so that e ~ qm -I, then the powers of a run through all nonzero elements of IF q" Therefor~. the methods outlined above can be used to calculate the minimal polynomial over IF q of each element of

IF;....

A straightforward method of determining minimal polynomials is the following one. Let 0 be a defining element of IF q" over IF q' so that (I.o,. ... om-') is a basis of IF., over IF q • In order to find the minimal polynomial g of PElF;, over IFq, we express the powers pO,p', ... ,pm in

Polynomials over Finite Fields

94

terms of the basis clements. Let

pi

m

L b,je

1=

I

j

for 1 ~ i .::;; rn

+ 1.

j=l

We write g in the form g(x) ~ cmx m + ... + c,x + co' We wantg to be the monic polynomial of least positive degree with g( (3) ~ O. The condition g(f3) = cmf3 m + ... + c , f3 + Co ~ 0 leads to the homogeneous system of linear equations m-I

E

cl_tb'J=O

forl~j~m

(3.10)

i=1

with unknowns CO,c,'""c m . Let B be the matrix of coefficients of the system-that is, B is the (m + I)Xm matrix whose (i, j) entry is bjj-and let r be the rank of B. Then the dimension of the space of solutions of the system is s = m + 1- r, and since I ~ r ~ m, we have 1 ~ s ~ m. Therefore, we can prescribe values for s of the unknowns co' c,' ... ,cm ' and then the remaining ones are uniquely determined. If s = I, we set C m ~ I, and if s > I, we set em = cm_ 1 = ... = cm-~+2 = 0 and cm_$~ 1 = 1.

3.42. Example. Let 0 E IF64 be a root of the irreducible polynomial x' + x + I in 1F,[x]. For 13 = 0' + O· we have f3o~

1

13 ' ~

0'+0·

f3'~ I +0 + 0'+ 0' f3'~ 0+0'+0'

13·= 13'=1

0 + 0'

+0· +0'+0· f3'~1+0+02 +0·

Therefore, the matrix

B is given by

0 0 0 0 0 I I 0 I I 0 0 B= I I 0 0 1 0 I 0 0 I I 0 I 0 1 o; and its rank is r = 3. Hence., ~ m + I - r = 4, so that we set c, = c, = c. ~ 0, c, = l. The remaining coefficients arc determined from (3.10), and this yields c, ~ I, c, = O. Co = I. Consequently, the minimal polynomial of 13 over I 0 1 0 0 I I

1F2isg(x)~x'+x'+l.

0 0 I I I 0 I

D

Still another method of determininK minimal polynomials is based on Theorem 3.33(v). If we wish to find the minimal polynomial g of 13 E IF qm

95

over IF,. we eakulate the power" {3. fl'. {3". ... until we find the least po,ilJve integer d for which {3'" ". {3. Thi, integer d is the degree of g, and g it'elf i, given by g(x)~(x-{3)(x-{3')",(x-{3q'

'j.

1 he clements {3. fl",· .. ,fJ"" , are the distinct conjugates of {3 with respect to ~ q' anJ g is the minimal polynomial ovcr IF q of all these elements. 3.43. Example. We compute thc minimal polynomials over IF, of all elements of IF ",' Let 0 f' IF 16 he a root of the primitive polynomial x 4 + x + I over !F 2' so that every nonzero clement of IF 16 can be written as a power of O. We have the following index table for IF 16:

0'

0'

o

I

I

0 0' 0' 1+0 0 + 0'

8 9

2 3 4 5 6

0' + 0 3

7

1-0+0 3

1+0' 0 - 03 10 i 1+0 + 0' 110+0',0' 12 I-O~O'+O' 13 1+0'+0 3 14 1+ 0 3

The minimal polynomiab of the clements

{I

(I

fl

~

f3

of IF It> over IF;! are:

0: I:

g,(x)~x.

0:

the Jistinet conjugates of 0 with respect to IF, are 0,0',0 4 ,0'. and the minimal polynomial is

g,(x)~x+1.

g,(x)=(x-O)(x-0')(x-0 4 )(x-0')

= x4 +

fl

~ 0':

+ 1. The di'tinct conjugates of 0' with re'pect to IF, are 0'.0'.0 1',0 240 0'. anJ the minimal polynomial is X

g4(X) = (x - O')(x - 06)(X - 8')(x - 8 1' )

= x4 + X 1

{3

~

8':

x 2 -:- x + 1. Since fJ4 ~ {3. the Jistinct conjugates of this clemcnt with respect to IF, arc 0'. 8 10 • and the minimal polynomial is -

g, (x) ~ (x - 8')( x - 8 10 ) ~ x' + x + I.

/i

0':

The distinct conjugates of 8 7 with respect to IF, arc 0",0'4, 0" ~ 0\.'. 8" ~ 8 11 , and the minimal polynomial is

g, ( , )

(x -

0 7 )(.\ l

=x 4 _x· +l.

-

8 '1 )( X

-

8 13 )( X

-

8 14 )

Polynomial~ over

96

Finite

hcld~

These elements, together with their conjugates with respect to IF ,. exhaust LJ

iF ](>.

I\n important problem is that of the rJelermil1alion of primitive polynomials. Onc approach is hased on the fact that the product of all primitive polynomials over IF q of degree m is equal to the cyclotomic polynomial Q, with e ~ qm - I (sec Theorem 2.47(ii) and Excreise 3.42). Thcrcfore. all primitive polynomials over IF q of dcgree m can he determined by applying one of the factorization algorithms in Chapter 4 to the cyclotomic polynomial Q,. Another method dcpends on constructing a primitive clement of IF q" and then determining the minimal polynomial of this clement ovcr IF q by the mcthods described above. To find a primitive element of IF q_' one sturts from the order qm - I of such an element in thc group IF:., and factors it in the form qm -I ~ h, ... h" where the positive integers h, .... ,h, arc pairwise relatively prime. If for each i. I ~ i ~ k. one can find an element £XI E IF;~. of order hi' then the product 0: 1'" a" has order qm -1 and is thus a primitive element of IF
x' + I = (x 4 - I)' ., x 4 ~(x4-I+x')(x4- I-x'),

and so I(x) = x 4 - x' - I is irredueihle over IF] and with a root 0 of I we have IF,\ ~IF](O). Furthermore, 0 is an element of IF 8' of order 16. To find an elemcnt a of order 5, we write a ~ a _ bO + cO' + dO] with a, h, c, d ElF], and since we must have a lO = I, we get I ~ a'a= (a

+ hO' + cO" + dO")(a + hO + cO' + dO])

~ (a - bO + cO' - dO])(a + bO

~ (a =

+ cO' + dO')

+ CO,)2 _ (bO + dOl)' ~ a' + (2ac - h2 )0 2 + (c' - 2bd) 0 4

-

d'O·

a' + c' - d' + bd + (c' + d 2 - b' - ac + hd )0 2

A comparison of coefficients yields

a' + (', - d' + hd ~ 1. c 2 + d 2 - h' - ac + hd = O. Setting a ~ d = O. we get h' = c' ~ I. Take b ~ c ~ I. and then it is easily checked that a ~ 0 + 0' has order 5. Therefore. t ~ Oa ~ 0' - 0] has order 80 and is thus a primitive clement of IF". The minimal polynomial Ii of t

3.

Con~tructjoll

97

of I rrcducihlc Polynomials

over IF] is g(x) ~

(x - n(x _;-3)(X -;-')(x _;-27)

~ (x -

0' - 03)(x-1 +0 + O')(x- 0' + O')(x -1- ° + 0')

and we have thus obtained a primitive polynomial over 1F 3 of degree 4.

0

3.45. Example. We determine a primitive polynomial over IF, of degree 6. Since 26 -1 ~ 9·7. we first construct two elements of IF:, of ordcr 9 and 7. respectively. The multiplicative ordcr of 2 modulo 9 is 6, and so the cyclotomic polynomial Q,(x) ~ x' + X 3 + I is irreducible over IF,. A root of Q, has order 9 and 1F6,=IF,(0). An element "elF'::' of order 7 satisfies a H = a, thus writing a = L;.. . oa I 8' with OJ E 1F 2 , 0 ~ i ~ 5, we get

°

L5 0,0' ~ (5L /=0

0;0;

)8

/= 0

5

= ,

L0 °i 8R1

~ au

+ 0,0' + 0,07 + 0306 + 0,0 5 + 05°'

~ 00

+ 03 + 0,0 + 0,0' + 0 30 3 + (a, + 05)0' + (a, + a,)OS,

and a comparison of coefficients yields a, ~ 0, a, = a" a, = a, + a,. Choose aO~aJ=a,~0,a,~a,=a5=1, so that ",~O+O'+O' is an clement of order 7. Thus, ;- = 0", ~ 1 + 0' is a primitiVe element of IF... Then ;-' = I + 0',;-3 ~ 0' + 0' + 0',;-' ~ I + 0' + 0 5,;-5 ~ I + ° + 0',;-6 = 1 + 0' + oj + 0' + 0' An application of the method in Example 3.42 yields thc minimal polynomial g(x) = x 6 + x' + x' + x + I of ;- over IF, and thus a primitive polynomial over IF, of degree 6. 0 If a primitivc polynomial g over IF q of dcgree m is known, all other such primitive polynomials can be obtained by considering a root of g in IF q' and detcrmining the minimal polynomials over IF q of all c1cmcnts 0', where I runs through all positive integers,;;; qm - I that are relatively prime to qm - 1. The calculation of these minimal polynomials is carried out by the methods described earlicr in this section. It is useful to be able to decide whcther an irreduciblc polynomial over a finite field remains irreducible over a ccrtain finite extension ficld. The following results address themselves to this question.

°

3.46. Theorem. I.el f be an irreducible polynomial over IFq of degree n and leI keN. fhen f facrors into d irreducible polynomials in IFq.[x] of Ihe same degree n / d, where d ~ gcd( k, n ).

Polynomials over Finite Fields

Proof Since the case frO) ~ 0 is trivial, we can assume frO) ~ o. Let g be an irreducihle factor of fin 0' q' rx]. If ord(f) ~ e. then also ord( g) ~ e hy Theorem 3.3 since the roots of Ii are also roots of f. By Theorem 3.5 the multiplicative ordcr of q modulo e is Ii and the degree of g is equal to the multiplicative order of q' modulo e. Thc powers ql, j ~ 0, I, ... ,considered modulo e, form a cyclic group of order Ii. Thus it follows from Theorem 1.15(ii) that the multiplicative order of q' modulo e is Ii/d, and so the degree of Ii is n! d. 0 3.47. Corollary. An irreducihle po!vnomial over 0' q of deliree remains irreducihle ocer IF <{!- if and on(v if k and 11 are relQiirely prime. Proof

This is an immediate consequcnce of Theorem 3.46.

Ii

0

Example. Consider the primitive polynomial Ii(X) = x' + x 4 ., Xl + X + lover 0', from Example 3.45 as a polynomial over 0' 16' Then. in the notation of Theorem 3.46, we have n ~ 6, k ~ 4. and thus d ~ 2. Therefore. Ii factors in 0' IO[X] into two irreducihle cuhic polynomials. Using thc notation of Examplc 3.45, let g, be thc factor that has I: ~ I + 0' as a root. Thc other roots of g, must he the conjugates 1: 16 and 1:'56 ~ 1: 4 with respect to 0' 10' Since these e1emcnts arc also conjugates with respect to 0'4' it follows that g, is actually in 0'4[X]. Now P = !:," is a primitive third root of unity over 0'" and so 0'4 ~ (O,I.p.p'). Furthermore,

3.48.

x' + (!: + 1: 4 ... 1: 10 )x' -r (!:' + 1: 17 ... l:'O)x '- 1: 21 . We have 1: 4 = I + 0' + 0', 1:" ~ I + 0', and so I: + 1: 4 -r 1: 10 ~ I. Similarly, we ohtain I:s + 1: 17 + I:"' ~ I, so that g,(x) = xJI x' + x 1 p. By dividing g by ~

Ii, we get the second factor and thus the factorization Ii( x) = (Xl

+ x' .\ x I P)( x J + x' + x + P')

in 0'4[X], and hence in O'IO[x]. The two factors of Ii arc primitive polynomials over 0'4' but not over 0' 16' By Corollary 3.47, the polynomial g remains irreducible ovcr certain other extension fields of 0"., such as 0'" and 0' I'"~ 0

4.

LINEARIZED POLYNOMIALS

Both in theory and in applications the spedal class of polynomials to he introduced below is of importance. A useful feature of these polynomials is the structure of the set of roots that facilitates the determination of the roots. Let q, as usual, denote a prime power.

4.

Lincari"l.~d

3.49.

Polynomiab

Definition.

99

A polynomial of the form n

L(x) ~

L

a,x q'

/=0

with coefficients in an extension field I' qm of I' q is called a q-polynomialover I' qm.

If the value of q is fixed once and for all or is clear from the context. it is also customary to speak of a linearized polynomial. This terminology stems from the following property of linearized polynomials. If F is an arbitrary extension field of I'qm and L(x) is a linearized polynomial (Le.• a q-polynomial) over !'qm. then L(p+y)~L(P)+L(y)

L(cP)~cL(P)

forallp,yEF,

forallcEl'q

(3.11)

andallpEF.

(3.12)

The identity (3.11) follows immediately from Theorem 1.46 and (3.12) follows from the fact that cq' ~ c for eEl' q and i;;, O. Thus, if F is considered as a vector space over I'q, then the linearized polynomial L(x) induces a linear operator on F. The special character of the set of roots of a linearized polynomial is shown by the following result. 3.50. Theorem. Let L(x) be a nonzero q-polynomial ocer I'qm and let the extension field I' q' of I'qm contain all the roots of 1.( x). Then each root of L(x) has the same multiplicity, which is either I or a power of q. and the roots form a linear subspace of I' q" where I'q' is regarded as a vector space over IF q" Proof It follows from (3.11) and (3.12) that any linear combination of roots with coefficients in I'q is again a root. and so the roots of L(x) form a linear suhspace of I' q" If n

L(x)~

L

a,x q',

/ ..... 0

then L'(x) ~ ao, so that L(x) has only simple roots in case a o " O. Otherwise, we have a o ~ a, ~ ... = ak 1 = 0, but a k .. 0 for some k ;;, I, and then

L(x)

=

Ln /=1..

a1x q'=

Ln l~k

a~m\q'=

(

Ln

a~(m

If\q'

I

)q'

I=k

which is the q'th power of a linearized polynomial having only simple roOlS. In this case, each root of L(x) has multiplicity qk. 0 There is also a partial converse of Theorem 3.50, which is given hy Theorem 3.52. It depends on a result ahout certain determinants which extends Corollary 2.38.

Polynomiab over Finite hckls

lOa

3.51.

1

1

Lemma.

Let f3 l.f3,•...• f3n be elements oflF q•. Then

13 1 f3r

(3f'

13(-' I

f3i

f3i'

f3r '

13, f3n

13:

~r

13:'

n -I ~f3l

n

j= I c t _

!J".,(f3J ,

1-

,~/,f3,).

1

(3.13)

and so the determinant is '" 0 if and only if 131,13, ..... f3n are linearly independent over IF q' Proof Let Dn be the determinant on the left-hand side of (3.13). We prove (3.13) by induction on n and note that the formula is trivial for n ~ I if the empty product on the right-hand side is intcrpreted as I. Suppose the formula is shown for some n ;;, I. Consider the polynomial

13 1 f3i 13, f3,q

f3( ,

f3( I

f3r '

f3r'

13:"

D(x)=, : f3n

13:

fJnqn

x

xq

x q'

I

x q"

By expansion along the last row we get

D( x)

=

Dnx q" +

n

L

I

a,x q'

,- 0

i"

with a, E IF q", for 0.;; n -I. Assume first that f3 l .... ,f3n are linearly independent over IF q. We have D(f3,) = 0 for I" k" n, and since D(x) is a q-polynomial over IF q ", all linear comhinations c l f3 1 + '" + c"f3n with c, E IFq for I" k" n are roots of D(x). Thus D(x) has qn distinct roots. so that we ohtain a factorization

D(x)

=

D"c,.I)",,( x -

}:, C,f3,).

(3.14)

If f3 l•...• f3n arc linearly dependent over IF q• then Dn = 0 and "[k~lbkf3k = 0 for some hi'" ., bn E IF q' not all of which are O. It follows that

f: b,f3t~ (f: b,f3,)"' =0

k = I

A --1

forj~O.I ..... n.

and so the first n row vectors in the determinant defining D(x) are linearly

Inl

4. Lincari7.cd Polynomiab

dependent OYer IF,,. Thus cases. Consequently.

1)"_,

D(x)~O.

and the identity (3.14) is satisfied in all

~ D(f3".,)~ D",.,.Q"F,(f3".,- '-£,C,f3,). o

and (3.13) is established.

3.52. Theorem. Let U be a linear subspace of IF q"" considered as a Vector space over IF q' Theil for any nOllnegative i/lleger k the polYllomial

n

L(x)~

{x-(3)q

,

PEU

is a q-polynomial over IF q"" Proof Since the q'th power of a q-polynomial over IF q"' is again such a polynomial. it suffices to consider the case k ~ O. Let {f3, •.... f3n) be a basis of U over IF q' Then the determinant Dn on the left-hand side of (3. I3) is "" 0 by Lemma 3.51. and so

L(x)=

n (x-(3)

(jEll

'" DeF'(x- ,tc,f3,)

=

Dn-'D(x)

by (3.14). which shows already that I.(x) is a q-polynomial over IF q""

0

The properties of linearized polynomials lead to the following method of determining roots of such polynomials. Let

L"

L(x) ~

a,x q'

i=O

be a q-polynomial over IF q ",. and suppose we want to find all roots of I.(x) in the finite extension F of IF q"" As we noted above. the mapping I.: f3 E F~ L(f3) E F is a linear operator on the vector space F over IF q . Therefore. L can be represented by a matrix over IF q • Specifically. let {f3 I • • " . f3J be a basis of F over IF q' so that every f3 E F can be written in the form

f3

L cA

= J

with

c, E IF

q

for I

,

then

,

L(f3) ~

L j=l

cI L(f3j

).

"j" ,;

Polynomial.;. OVl'r Finite

102

hdd~

Now let L(f3}) ~

L:

b"f3,

for I"},, s,

k - ,

where bi, ElF q for I " }, k (j, k) entry is h". Then, if

"s, and let H be the s x s matrix over IF

q

whose

(e,,. .. ,c,}B ~ (d,,. ..,d,l, we have L(f3)~

L:

d,f3,.

/';=1

Therefore, the equation L(f3) = 0 is equivalent to

(c" .. .,c,}B ~ (0, ... ,0).

(3.15)

This is a homogeneous system of s linear equation; for c, ..... c,. If r is the rank of the matrix B, then (3.15) has q' 'solution vectors (c,,. . .,c,). Each solution vector (c" ... ,c,) yields a root f3 ~ L~~ of L(x) in F. Thus, the problem of finding the roots of L(x) in F is reduced to the easier prohlem of solving a homogeneous system of linear equations.

,eA

Example, Consider the linearized polynomial I.(x) ~ x' - x 3 - ax 1F 9 [x], where a is a root of the primitive polynomial x' + < -lover IF J • In order to find the roots of I.(x) in IF,;, we choose the hasis {I, l", l"', l"3} of IF"

3.53, E

over 1F 3 , where l" is a root of the primitive polynomial x 4 + x' + x' - x - I over IF, (compare with Example 3.44). Because of the orders involved, we must have a ~ l"'o/ with} ~ I, 3, 5, or 7, and since l"'o + l"'o - I ~ 0, we can take a ~ l"'o ~ - I + l" + l"' -l"J. Next. we calculate I. (I) ~ - a = I -l"-l"' + l"',

t',

L{n ~ l"' -l"3 - al" ~ -l"-l"' L(l') ~ l"" -l"6 - al"' ~ - I + l"'. L(l"3)~l""-l"'-al"'=I-l"3,

and so we get

R~( ~ -I I

-I -I

o o

-I -I

0 0

-

:) I

.

-I,

The system (3.15) has two linearly independent solutions, such as (0,0, I, I) and (-1,1,0, I). All solutions of (3.15) are obtained by forming all linear combinations of these two vectors with coefficients in IF,. The roots of L(x) in 1F 8I arc then 0,=0, O,~l"'+l"J, O,~-l"'-l"j, 04~-1+l"+l"',

103

4. 1.!Oearizl'd I'olynomials

8,=I-K-K'. 8,=-I+K+K'-K 3, 8,=I-K-K'+K 3 • 88 =I-K+K', 8y =-I+K-K'. 0 This method of finding roots can also be applied to a somewhat more general class of polynomials-- namely. affine polynomials. 3.54. Definition. A polynomial of the form A(x) = L(x)- ". where L(x) is a q-polynomial over 0' qm and" E 0'qm, is called an affine q-polynomial over 0' qm. An element (3 E F is a root of A(x) if and only if L({3) = ". In the notation of (3.15), the equation 1.( (3) = 01. is equivalent to (3.16) where "='[;t-,d k{3k' The system (3.16) of linear equations is solved for c" .... c,. and each solution vector (cl .... ,c,) yields a root (3='[;j_ 1cj {3j of A(x) in F. The fact that roots arc easier to determine for affine polynomials suggests the following method of finding the rools of an arbitrary polynomial fIX) over O'.m of positive degree in an extension ficld F of O'qm. First determine a nonzero affine q-polynomial A(x) over 0' qm that is divisible by f(x)-that is, a so-called affine multiple of fIx). Next, obtain all the roots of A(x) in F by the method described above. Since the roots of fIx) in F must be among the roots of A(x) in F. it suffices then to calculatef({3) for all roots (3 of A(x) in F in order to locate the roots of fIx) in F. The only point that remains to be settled is how to determine an affine multiple A(x) of fIx). This can be achieved as follows. Let n:;, I be the degree off(x). For i=O,I, ... ,n-l, calculate the unique polynomial r,(x) of degree <; n -I with x q' " r,(x)modf(x). Then determine elements ", E 0'••. not all 0, such that '[;7~J",r,(x) is a constant polynomial. This involves n .. I conditions concerning the vanishing of the coefficients of xi, 1<; j <; n - I, and thus leads to a homogeneous system of n - I linear equations for the n unknowns "0' "" ... ,,,,. ,. Such a system always has a nontrivial solution. Once a nontrivial solution has been fixed, we have '[;7~J",r,(x) =" for some" E 0' It follows that

.m.

n

I

L i=O

n- 1

",x

q

'"

L

",r,(x)" "modf(x).

i=O

and so

, A (x)

=

L

, ",x q ' -

"

1=0

is a nonzero affinc q-polynomial over O'qm divisible by fIx). It is clear that we may take A(x) to be a monic polynomial.

104

Polynomiab O\"l'r Finite Fields

3.55. Example. Letf(x)~x'+O'x), OxJ. 1x +O E "4[xl,whereOisa root of x' + x+ I eo ",[xl, We want to find the roots of f(x) in We first determine an affine multiple A(x) of fIx) by using the method described above with q ~ 2. Modulo fIx) we havc x ," x', ro(x). x" 7 x' rl(x). x' '" O'x' + Ox' + x + 0 ~ r,( x), x' '" Ox' + 0,' + x ~ 0 ~ r,(x), The condition that ~,,ro(x), a,r,(x'+"I',(x)+,.\r.\(x) should be a eon stant polynomial leads to the system

"04'

0: 0 0: 1

We choose

0: 3

+ +

0'2+

=

=0 0 0-:-

O'IX,+OIX,~().

= I and then obtain

IX

0':3

8a~+8al

0 7 • ex]

0: 2 -

=

IXoro{x)+ IXlrl{x)-r IX,r,{x)

t

02., a o

=

IX"r,( x)

8. Furthermore, =

OJ.,

and so A (x)

= D:3XIl

+ lX 2 X 4 + a 1x 2 ,.

a=

lXOX -

xl'!

+ 0 2 x 4 + fJ2 Xl + Ox + fj'2..

"04'

Next, we caleulate the roots of A(x) in We have to solve the equation L(x) ~ 0' with the 2-polynomial L(x) ~ x' + O'x' + O'x' + Ox over ",. Let be a root of the primitive polynomial x· + x + lover ",. Then (I, t. t'. t'. t'. t') is a basis of over" ,. Sinee 0 is a primitive third root of unity over ",. we can take 0 ~ t" ~ I + t + + t' + t 5 Using 0' = 0 + I ~ t + t' + t' + t'. we obtain

t

L{ I)

Lm

"04

t t

+

t' t2

t'

tS

t'

+

+

r'

+

tJ

t' + t' + t' + t'

+

LW) LW)

r'

+

+

t

tS

L(t')

t'

L(t' )

t'

+

+

t'

Thus the matrix B in (3.16) is given by 10

B=

o 0 o o o

I I

0

0 I I

I 0 I

I 0 I

I I I

I 0

0 0

I 0

I 0

0 I

0

I

I

I

0;

From the representation for 0' given above it follows that the vector (dl,· .. ,d,) in (3.16) is equal to (0,1.0, I. I. I). The general solution of the system (3.16) is then

(1.0.0.0,0.0) + a l (0.1.1.1.0.0) + a,{ I. I. 1.0, 1.0) ~ a,{ I, 1.0.0.0. I)

4. I ,illl::trl/l:J Pohnom:.1h

105

with a,.a,.al"''''' Thus the roots of A(x) in "04 are ~,~I. ~,~I+t'. 4 ~3 ~ I + +1 • ~4 ~ I + I' -CI' + I'.~, ~ I + I +I' + t'. ~6 ~ I' +I' -I'. ~,~ t' _I'. '18 ~ I +I + t' _I' + t' ~ 8. By calculating f(~/) for j ~ 1.2..... R. we find that the roots of f( x) in "04 are ~3' ~,. ~" ~,. 0

r

The method of determining the roots of an affine polynomial shows, in particular. that these roots form an affine subspace-that is. a translate of a linear subspace. This can also be deduced from abstract principles, together with a ~tatcmcnt concerning mUltiplicities.

3.56.

Theorem.

I.et A(x) be an affille q-po(vilomial over "". of

positive df.'gree and let the extension field IF q' of IF q'" contain all The roots of

A(x). Then each root of A(x) has the ,ame multiplicily, which is eilher I or a power of q. and the roOlS form an affine subspace of" q" where" q' is rel{arded as a veCTOr space orer IF cl"

Proof The result about the mUltiplicities is shown in the same way as in the proof of Theorem 3.50. Now let A(x) ~ I.(x)- a. where L(x) is a q-polynomial over" q'" and let {3 be a fixed root of A( x). Then y is a root of A(x) if and only if L(y)~a~ L({3) if and only if L(y-{3)~O if and only if y E {3 + U, where U is the linear suhspace of consisting of the roots of I.(x). Thus the roots of A(x) form an affine subspace of D

E""

"q'

"q"

3.57. Theorem. Let T be an affine subspace of" q'" considered as a cecwr space ocer Theil for any nonnel{atice imeger k Ihe polynomial

"q'

A(x)~

n (x-y)'y~

T

is an affine q-polynomial over",'" Proof

Let T

~ ~

+ U, where U is a linear subspace of "," Then I.(x)=

n (x-{3)'fJ(?-li

is a q-polynomial over" A(x)~

n (x-y)' , ~ n

y€CT

and L(x -

~)

,m according to Theorem 3.52. Furthermore, ,

(x-~-{3)' ~L(x-~).

I3~U

is easily seen to be an affine q-polynomial over"

,m.

C

The ordinary product of linearized polynomials need not be a lineariled polynomial. However, the composition L,( I.,(x)) of two q-polynomials L, (x), L, (x) over",. is again a q-polynomial. Instead of the word composition (or suhstitution) we use the phrase "symbolic multiplication." Thus. we define svmbolic multiplicQlion by

1.,(x)0/.,(x) = 1.,(I.,(x).

Pulynomiab over Finite

rield~

If we consider only q-polynomials over IF", thc.:n a simple investigation shows that symbolic multiplication is commutative, associativc, and distributivc (with rcspect to ordinary addition). In fact. the set of q-polynomials ov~r lFl{ forms an integral domain under the operations of symholic multiplication and ordinary addition. Thc operation of symholie multiplication can he related to the conventional arithmetic of polynomials hy means of the following notion.

3.58.

Ocfinition.

The polynomials

L"

I(x)~ ,

and

",X'

,.

/.(x)

II

t... a,x

,

q'

II

over O'q_ are called q-associales of each other. More specifically. I(x) is the rnnoenlional q-associare of I.(x) and L(x) is thc linearized qwtlssoeiare of I( x).

3.59. Lemma. I.el I.,(x) and 1.,(x) be q-poll'nomials over O'q wilh convelllional q-assoeiales I,(xl alld I,(x). Theil I(xl -/,(x)/,(x) and L(x) I.,(x)@!.,(x) are q-as>ociales of each olher.

Proof

The equations I(x) ~ La,x'~ Lb,X1L",X' ~/,(x)/2(X)

,

,

and

are each true if and only if

al

=

L I • k =

b,c"

for every i.

D

I

If I.,(x) and L(xl are q-polynomials over O'q' we say that I,,(x) svmbolicallv divides L(x) (or that L(x) is symbolically divisible by I,,(x)) if L(x)" L,(x)@!.,(x) for some q-polynomial I.,(x) over O'q. The following criterion is then an immediate consequence of Ll.:mma 3.59. 3.60. Corollary. Lei L,(x) and L(x) be q-polynomials over 0', wilh convelllional q-associares I,(x) and I(x). Ihen I.,(x) symbolically divides L(x) if and ollly if I,(x) divides I(x).

3.61. Example. Let L(x) be a q-polynomial over 0', that symholically x for some mEN. Then there exists a q-polynomial L, (x) divides x" over IF Q such that (3.17)

107

4. [,int'JTl/t'ul'olyllomiJls

This can be applied as follows. Let a be a fixed element of Fq_' Then the affine polynomial L(x)- a has at least one root in Fq"' if and only if L,(a) ~ 0, and if L,(a) ~ 0, then actually all the roots of L(x)- a arc in Fq",. For if fJEF q", is a root of L(x)-a, then L(fJ)= a, and substituting x by fJ in (3.17) yields L,(a)~fJq·-fJ~O. Conversely, suppose L,(a)~O and let y be a root of L(x)- a in some extension field of Fq _; then L( y) = a, and substituting x by y in (3.17) yields yq'" - y ~ I. ,( a) ~ 0, so that y E Fq _. The polynomiall,,(x) can be calculated by Ictting I(x) be the conventional q-associate of L(x), determining I,(x) = (x m -1)II(x), and then taking L,(x) to be the linearized q-associate of I,(x). This application contains Theorem 2.25 as a special case, as one sees easily by choosing L(x) = x q - x. 0 It is an important fact that although symbolic multiplication and ordinary multiplication are quite different operations, the divisibility concepts for linearized polynomials based on these operations are equivalent. 3.62. Theorem. Let L,(x) and l.(x) be q-polynomials over Fq Wilh conventional q-associates I,(x) and I(x). Then the following properties are equivalent; (i) L,(x) symbolically divides L(x); (ii) L,(x) divides l.(x) in the ordinary "ense; (iii) I,(x) divides I(x). Proof Since the equivalence of (i) and (iii) has been established in Corollary 3.60. it suffices to show the equivalence of (i) and (ii). If L,(x) symbolically divides l,(x), then L(x) ~ L,(x )@L,(x) ~ I., (x )@L,(x) ~ 1. 2 (I., (x)) for some q-polynomial L 2 (x) over Fq • Let n

L,(x) =

q a,x ',

L 1"-'

0

then L(x)=aoL,(x)+a,L,(x)q+ .. · +anl.,(x)q',

and so L,(x) divides L(x) in the ordinary sense. Conversely, suppose L,(x) divides L(x) in the ordinary sense, where we can assume that l.,(x) is nonzero. Using the division algorithm, we write I(x) ~ k(x)l,(x)+ r(x), where deg(r(x)) < deg(l,(x», and turning to linearized q-associates we get in an obvious notation L(x) = K(x)@L,(x)+ R(x). By what we have already shown, L,(x) divides K(x)@L,(x) in the ordinary sense, and so I., (x) divides R(x) in the ordinary sense. But since deg( R(x)) < deg( I. ,(x ), R(x) must be the zero polynomial, and this proves that L,(x) symbolically divides L(x). 0 This result can be used to establish an interesting relationship between an irreducible polynomial and the irreducible factors of its linearized q-associate.

10k

Polynomials over finite hdds

3.63. Theorem. I.el j(x) be irreducible in F)x] and lei F(x) be i15 linearized q-as50ciaie. Then Ihe degree oj everv irreducible jaclOr oj F(x)/x in I'"lx] iI' equallU or<1(/(x». I'roo! Since the case j(O) ~ 0 is Irivial, wc can assume flO) '" O. Put e· or<1(/(x» and lei h(x)EFq[x] be an irreducihle factor of F(x)/x of degree d. Then fix) dividcs x' - l. and so hy Theorem 3.62 F(x) divides x q ' - x. It follows that h(x) dividcs x q- - x, hence d divides e by Theorem 3.20. By the division algorithm. we can write x d -I ~ g(x)!(x)+ r(x) wilh g( x). r(x) E= Fqlx] and dcg( r(x)) < deg(/(x ». Turning to linearized q-asso-

ciatcs. we

gC:l

x q "· x~C(x)0F(x).,.R(x),

and since h( x) divides x q ' - x and C(x )®F(x), it follows that h(x) divides R(x). If r(x) is nm the zero polynomiaL then r(x) and!(x) arc relatively prime. and so hy Theorem 1.55 there exist polynomials six). k(x) c F"lx] with s (x) r ( x ) I- k ( x ) j( x )

~

I.

Turning to linearized q.assm.:iatcs, we get S(x)0R(x)-K(x)®F(x)~x.

Since h(x) divides R(x) and F(x), it follows that h(x) divides x. which is impo"ihle. Thus r(x) is the zero polynomial. ><, that fix) divides x d - l. and therefore e divides d by Lemma 3.6. Altogether. we have shown d ~ e. n We say that a q-polynomial L(x) over Fq of degree> I is 5ymbolically irreducihle over Fq if the only symbolic decompositions 1.( x) = I.,(x) ®L,( x) with q-polynomials L, (x), 1.,( x) over F" arc those for which one of the factors has degree I. A symbolically irreducihle polynomial is always reducihle in the ordinary sense since any linearized polynomial of degree > I has the nontrivial factor x. By using l.emma 3.59, one shows immediately that the q-polynomial L(x) is symholically irreducible over Fq if and only if its conventional q-a'Sociate I(x) is irreducible over Fq' hery q-polynomial L(x) over F q of degree> I has a symbolic !aclorizalion into symholically irreducible polynomials over F q and this factorization is essentially unique, in the sense that all other symholic factorilations arc obtained by rearranging factors and by multiplying fac-

tors by nonzero clements of IF q" Using the:

corrcspond~nce

between lin-

earized polynomials and their conventional q-associatcs. DOl: sees that the symholic factorization of I.(x) is obtained by writing down the canonical factorization in !' "Ix J of its conventional q-associate I( x) and then turning to

lineari7.ed q-associatcs.

3.64. Example. Consider the 2-polynomial L(x) ~ x"" x' • x' + x over !F:. Its <.:oo\'cmional 2-associate [(x) = .),..4....;.. x J • x·~ 1 ha~ the canonical

4

109

Lincari/.~rl PolynomJaI~

factorization I(x) ~ (x 2 + X + I)(x

j.

If in 1F,[xJ. Thus.

1. (x) ~ (x 4 1 x' '- x) ® (x' ~ x) ® (x' ~ x) is the symholic factorization of l.(x) into symbolically irredueihle polynomials over !F2. D

For two or more q-polynomials over IF". not all of them 0, we may define their greQ/esr common symbolic divisor to he the monic q-polynomial over IF" of highest degree that symbolically divides all of them. In order to compare this notion with that of the ordinary greatest common divisor, we note first that the roots of the greatest common divisor arc exactly the common roots of the given q-polynomials. Since the intersection of linear subspaees is another linear subspace, it follows that the roots of the greatest common divisor form a linear suhspace of some extension field IF qm, considered as a vector space over IF q' Furthermore, by applying the first part of Theorem 3.50 to the given q-polynomials. we conclude that each root of the greatest common divisor has the same multiplicity, which is either 1 or a power of q. Therefore, Theorem 3.52 implies that the greatest common divisor is a q-polynomial. It follows then from Theorem 3.62 that Ihe , greatest common divisor and the greatest common symholic divisor are identical. An efficient way of calculating the greatest common (symholic) divisor of q-polynomials over IF" is to consider the conventional q-assoeiates and determine their greatest common divisor; then the linearized q-associate of this greatest common divisor is the greatest common (symbolic) divisor of the given q-polynomials. By Theorem 3.50 the roots of a nonzero q-polynomial over IF" form a vector space over IF q • The roots have the additional property that the qth power of a root is again a root. A finite-dimensional vector space Mover!F q that is contained in some extension field of IF q and has the property that the qth power of every element of M is again in M is called a q-modulus. On the basis of this concept we can estahlish the following criterion. 3.65. Theorem. The mOllie polynomial L(x) is a q-polyllomial over IF" if alld only if each roOl of L(x) has Ihe same muliiplicily. which is dlher 1 or a power of q. and The roots form a q-n1odulus. Proof The necessity of the conditions follows from Theorem 3.50 and the remarks ahove. Conversely. the given conditions and Theorem 3.52 imply that I.(x) is a q-polynomial over some extension field of IF". If M is the q-modulus consisting of the roo!> of I.(x). then

L(x)~

n

fi (-

(x-/3)'"

,'vi

for some nonnegative integer k. Since M I.(x)" ."

II fir

11

(x'-!3")"' ~

~

(/I": /I to

n f1~'Af

M). we ohtain

(x'-IJ}'I' ~ I,(x"),

110

Polynomials over hnit(' Fields

If n

L(x) ~

L

a,xq'.

1=0

then n

L

n

L

aixq'·'~l.(x)q~L(xq)~

i=O

so that for 0 ~ i:E; n we have a? = q-polynomial over I' q'

(Xl

and thus

(XI

a,x q" 0

I

E:

IF q" Therefore, /J(x) is a D

Any q-polynomial over I' q of degree q is symholieally irreducible over I' q' for q-polynomials of degree > q, the notion of q-modulus can be used to characterize symbolically irreducible polynomials.

3.66. Theorem. The q-polynomial L (x) over I' q of def!,Yee > q is symholically irreducible over I' q if and only if I. (x) has simple roots and the q-modulus M consisting of the roots of L(x) contains no q-modulus olher than (OJ and M itself. Proof Suppose l.(x) is symbolically irreducible over ff'q. If L(x) had multiple roots, Ihen Theorem 3.65 would imply that we could write l.(x)~ L,(x)q with a q-polynomial L,(x) over ff'q of degree> I. But then l.(x) . xq®L,(x). a contradiction to the symbolic irreducibility of l.(x). Thus l.(x) has only simple roots. Furthermore, if N is a q-modulus contained in M, then Theorem 3.65 shows that 1.,(x)~II#, N(X-f3) is a q-polynomial over ~ q' Since l.,.(x) divides L(x) in the ordinary sense, it symbolically divides L(x) by Theorem 3.62. But L(x) is symbolically irreducible over I'q' and so deg( L,( x» must be either I or deg( L(x)); that is, N is either (OJ or M. To prove the sufficiency of the condition. suppose that L(x) = L,(x) ®I.,(x) is a symbolic decomposition with q-polynomials L,(x). L,(x) over ~ q' Then l.,(x) symholically divides L(x). and so it divides l.(x) in the ordinary sense by Theorem 3.62. It follows that L,(x) has simple roots and that the q-modulus N consisting of the roots of L,(x) is contained in M. Consequently, N is either (OJ or M. and so deg(l.,(x) is either I or deg(L(x)). Thus, either l.,(x) or l.,(x) is of degree I. which means that L(x) is symbolically irreducible over ff'q. D .~.67. Definition. Let L( x) be a nonzero q-polynomial over I' q"' A root l of L(x) is tailed a q-primi(ir~e root over IF t(" if it is not a root of any noo7cro q-polynomial over i'" of lower degree.

Tbis concept may also be viewed as follows. Let g(x) be the minimal polynomial of Z; ovcr 1'." Then Z; is a q-primitive root of l.(x) over ff'q_ if

4. I ,incaril.cd Polynomiab

III

and only if g(x) divides L(x) and g(x) docs not divide any nonzero q-polynomial over I' q. of lower degree. Given an elementl' of a finite extension field of I' q"" one can always find a nonzero q-polynomial over I' q. for which l' is a q-primitive root over I'q"" To sce this. we procced as in the construction of an affine multiple. Lct g(x) bc the minimal polynomial of l' over I'q •• let n be the degree of g(x). and calculate for i ~ O.I ..... n the unique polynomial r,(x) of degree ~ n-I with x q ' = r,(x) mod g(x). Then determine elements a, E I' q.' not all O. such that [7-0 a,r,(x) ~ O. This involves n conditions concerning thc vanishing of the coefficients of Xl. 0 ~ j ~ n -I. and thus leads to a homogeneous system of n linear equations for the n + 1 unknowns aD_ a l •.. •• ll:n' Such a system always has a nontrivial solution. and with such a solution we gct n

L(x)~

L

n

a,x q '=

/"" 0

L

a,r,(x)=Omodg(x).

i=O

so that l.(x) is a nonzero q-polynomial ovcr I'q_ divisible by g(x). By choosing the ", in such a way that L(x) is monic and of the lowest possible degree. one finds that l' is a q-primitive root of L(x) over 1'••. It is easily seen that this monic q-polynomial l.(x) over I'q. of Icast positive degree that is divisible by g(x) is uniquely determined; it is called the minimal q-polynomial of l' over I' qm. 3.68. Theorem. l.ell' be an element of a finile eXlellsion field of 1'". and leI M(x) be its minimal q-po(ynomial ocer I'q-. Then a q-polyllomial K(x) DCer I'q"' has l' as a root if and only if K(x) ~ L(x)0M(x) for some q-polynomial L(x) over I'q •. In particular. for Ihe case m ~ I Ihis means Ihal K(x) has l' as a rOOI if and only if K(x) is symbolically dicisihle hy M(x). Proof If K(x) = L(x)0M(x) that K(l') ~ O. Convcrscly. let

M(x) =

, L

~

L(M(x». it follows immediately

YjX q ,

with Y, = I

"hXq'

with r :;,

J=O

and suppose K (x) ~

L

I

h""'O

has l' as a roo!. Put s ~ r -

I

and Y,

=

0 for j < O. and consider the followir.g

Polynomiab over Finite rields

112

system of s

+I

linear equations in the s + I unknowns {3o'{3" .... {3,: {3u+Ylq-tf3l+yl2f32+'"

" +

PI

q' " v1,-]""2

f3~=o,

+y,q's

+... + vIt

q

' 5 t tPj , , -- U (- t

q' '"" . , + y,-,

f3s =

Or

It i' clear that this system has a unique solution involving elements (3o,{3" ... ,{3,oflF q _. With L(x)~ 'L{3,x q ,

and

R(x)=K(x)-I.(M(x))

0

we get

=

"

"- 0hX h..,.U

q"" "" Y,q' q'" -

"- P, ",-0 J-O

X

t (Uh-t y:~,{3,')Xq,

h=U

1-'-0

It follows from the system ahove that R( x) has degree < q'. But since R(I;) ~ K(I;) - 1.( M(~)) ~ 0, the definition of M(x) implie, that R(x) i' the zero polynomial. Therefore, we have K(x) ~ L( M(x)) ~ L(x)®M(x). LJ

We consider now the problem of determining the number Nl. of q-primitive roots over IF q of a nonzero q-polynomial/.(x) over IF q • If L(x) has multiple root', then by Theorem 3.65 we can write L(x) ~ L,(x)q with a q-polynomial/.,(x) over IF q • Since every root of L(x) is then also a root of L,(x), we have Nl. ~ O. Thus we can assume that L(x) has only simple roots. If I.(x) has degree I, it is obviou, that NJ• ~ L If L(x) has degree qn > I and is monic (without loss of generality), let L(x) ~ L,(x)®'" \

"v

®L,(x) ® ... ® I.,(x)®··· ®L,(x) -'

'-

-y-"-----'

e,

be the symbolic factorization of L(x) with distinct monic symbolically

,, , "

,I

113 ,

irrcducible polynomials L,(x) over IF q' We obtain Nl. by subtracting from th~ .. total number q" of roots the numbcr of roots of L(x) that are already roots' .. of somc nonzero q-polynomial over IFq of degree < q", If I is a root of I,(x) of thc lattcr kind and M(x) is the minimal q-polynomial of lover IF q• then deg(M(x)) < q" and M(x) symbolically divides L(x) by Theorem 3,68. It follows that M(x) symbolically divides one of the polynomials K,(x). I,,; i,,; r. obtaincd from the symbolic factorization of L(x) by omitting thc symbolic factor I,,(x). in which case K,(n ~ 0 by Theorem 3.68. Since every root of K,(x) is automatically a root of L(x). it follows that NL is qn minus the number of I that arc roots of somc K,(x). If qn, is the degree of L,(x). tben the degree. and thus tbe number of roots. of K,(x) is qn n,. If i ,..... i, are distinct subscripts. then thc numbcr of common roots of K, ,(x)..... K,(x) is equal to the degree of the greatest common divisor. , which is thc same as the degree of the greatcst common symbolic divisor (see the discussion following Example 3.64). Using symbolic factorizations. one finds that this degree is equal to qll

11"

..• -11,

Altogether. the inclusion-cxclusion principle of combinatorics yiclds

,

L

N,.=ql1_ Lql1 11,+ 1.,..,1

1~

I

qn 11, I1,T ... +{_l)rqn-nl "'-11,

< J"" r

=qn(I_q-n')· .. (I_q-n,).

This exprcssion can also bc interpreted in a different way. Let l(x) be the conventional q-associate of L(x). Then l(x) = l,(x)"·· ·l,(x)e.

is the canonical factorization of l(x) in IFqlx]. where l,(x) is the conventional q-associate of L,(x). Wc dcfinc an analog of Euler's <[>-function (see Exercise 1.4) for nonzero f E IF qlx 1by letting q(j(x)) ~ /fl denote the number of polynomials in IF ql x 1 that are of smaller degree than f as well as rclatively prime to f. The following rcsult will thcn imply the identity N L ~ q(l(x)) for thc casc undcr considcration. 3.69.

Lemma.

The function q defined for nonzero polynomials in

IF q[ x 1has the following properties: (i) (ii)

(iii)

q(j) = I if dcg(j) = 0: q(jg) ~ /flq( g) wheneGer f and g are relatiGely prime: if deg(j) ~ n.", 1. then q(f) = qn(l_ q-n,)." (1_ q-n,). where the n, are the degrees of the distinct monic irreducible polynomials appearing in the canonical factorization off in IF qlx].

.

,

114

Polynomials over Finite Fields

Proof Property (i) is trivial. For property (ii). lot q(j) ~ sand q(g) ~ I, and let f, .... .[, resp. g, •... ,g, be the polynomials counted by /f) resp. q,q(g). [f h E IFqlx] is a polynomial with deg(h)" deg(jl() and gcd(jl(,h)~I. then gcd(j,h)~gcd(l(.h)=I. and so h=[,modf. h= I(jmod I( for a unique ordered pair (i. j) with [" i "s, [" j " I. On the other hand, given an ordered pair (i, j). the Chinese remainder theorem for IFq[x] (see Exercise 1.37) shows that there exists a unique hE IFq[x] with h [,mod f. h gjmod g, and deg(h) " deg(jl(). This h satisfies gcd(j, h) ~ gcd(l(. h) = I. and so gcd(jl(. h) = I. Therefore. there is a one-to-one correspondence hetween the sl ordered pairs (i, j) and the polynomials hElFq[x] with deg(h)<deg(jg) and gcd(jl(.h)~1. Consequently. q(jg) ~ sl ~ q(f)q(I(). For an irreducible polynomia[ b in IFq[x] of degree m and a positive integer e. we can cakulate q(b') directly. The polynomials h ElF q[x] with deg(h) < deg(h') ~ em that are not relatively prime to h' are exactly those divisible by b, and they are thus of the form h = I(b with dcg(I() < em - m. Since there are q,m-m different choices for g. we get q(b') = q,m _ q,m m = q,m( I - q m). Property (iii) follows now from property (ii). D

=

=

3.70, Theorem, LeI L (x) be a nonzero q-polynomial ocer IF q with conventional q-associale I(x). Then Ihe numher NL of q-primilive rools of I.(x) over IF q is given by NL ~ 0 if L(x) has multiple roots and by N I, = "'q(f(x» if L(x) has simple rOOls.

Proof ~.

This follows from Lemma 3.69 and the discussion preceding D

3.71. Corollary. Every nonzero q-polynomial over IF q wilh "imple rooa has at least one q-primililJe root over IF q"

Earlier in this section we introduced the notion of a q-modu[us. The results about q-primitive roots can be used to construct a special type of basis for a q-modulus. 3,72. Theorem. Let M be a q-modulus of di,!,ensio~ ":' '" I over IFq • Then Ihere exiSlS an element I: E M such thaI {1:.l:q.l:q ,. .. •I:q ) is a basis of M ocer IF q"

Proof According to Theorem 3.65. L(x) ~ np , M(X - (3) is a qpolynomial over IF q • By Corollary 3.7[, L(x) has a q-primitive root I: over IF q • Then 1:,l:q.I:<', ... ,l:qm , are elements of M. If these elements were linearly dependent over IFq • then I: would he a root of a nonzero q-polynomial over IF q of degree less than qm ~ deg( 1.( x a contradiction to the definition of a q-primitive root of 1.( x) over IF q' Therefore. these m elements arc linearly independent over IF q' and so they form a basis of M over IF q' D

n.

5.

Binomiab and

Trinomi31~

liS

3.73. Theorem. I~ IF q " There exisl exacTly $q(x m -I) elemenls qm such Ihal (r.rq.rq·, ...• r ) is a basis oflFqm VGer IF q.

r

Proof . Since IF q" can be viewed as a q-modulus, the argument in the proof of Theorem 3.72 applics. Here

n

L(x)~

f1

F.

(x-fJ)=xqm-x

if lI'"

by Lemma 2.4, and every q-primitive root of L(x) over IFq yields a hasis of the desired type. On the other hand, if E IF q" is not a q-primitive root of qm L(x) over IF q' then r q, r q'.... ,r , are linearly dependent over IFq' and so they do not form a basi~ of IF <" over IF q' Consequently. the number of E IF q"' such that r q•r q·, .. '. q ') is a hasis of IF qm over IFq is equal to the numbcr of q-primitivc roots of L(x) over IF q, which is given by $q(x m -1) according to Theorem 3.70. 0

r. (r,

r

r

r

This result provides a refinement of thc normal basis theorem (compare with Definition 2.32 and Theorcm 2.35). Since each of the clcments r q, r q', ... ,rq" , gcnerates the same normal basis of IF qm over IF q' the numher of diffcrent normal bases of IF qm over IF q is given by (I/m)$q(x m -I).

r,

3.74. Example. We calculate the number of different normal bases of IF.. over IF,. Since 64 ~ 2', this number is given hy t$2(X' - I}. From the canonical factorization

x' - I = (x + 1)'( x 2 + X + 1)2 in 1F,[x] and Lemma 3.69(iii) it follows that

$2(x' -I) ~ 2'(1- t)(1-~) ~ 24, and so there are four different normal hases of 1F'4 over 1F 2.

5.

o

BINOMIALS AND TRINOMIALS

A binomial is a polynomial with two nonzcro terms, one of them being the constanttcrm. Irreducible binomials can bc characterized explicitly. For this purpose it suffices to consider nonlinear. monic binomials. 3.75. Theorem. Lei I" 2 be an intel!er and a ElF;. Then Ihe binomial x' - a is irreducible in IFq[x] if and only if the following two condilions are salisfied: (i) each prime faclOr of I divides Ihe order e of a in IF;, but not (q -I)/e; (ii) q '" 1mod4 if I '" omod 4.

Polynomials over Finite Fields

11(\

Proof Suppose (i) and (ii) are satisfied. Then wc note that f(x) ~ x - a is an irreducible polynomial in IF qf x] of order e, and so fix') ~ x' - a

is irrcducible in IF.[x] by Theorem 3.35. Suppose (i) is violated. Then therc exists a prime factor r of I that either divides (q -Ille or does not divide e. In the first case, we havc rs ~ (q -I)/e for some .. E N. The subgroup of IF; consisting of rth powers has order (q - III r = es and thus contains the subgroup of order e of IF: generated by a. In particular, a ~ b' for some bE IF;, and so x' - a = X'" - b' has the factor x" - h. In the remaining case, r divides neither (q - I)/e nor e, and so r does not divide q -I. Then rlr '" I mod(q -I) for some r l E N, and thus x t

-

a=

XII' -

art' has the factor x tl

-

0'1.

Suppose (i) is satisfied and (ii) is violated. Then 1= 41, for some I, EN and q'" I mod4. But (i) implies that e is even, and since e divides q -I, q must be odd. Hence q '" 3mod4. The fact that x' - a is reducible in IFqlx] is then a consequence of Theorem 3.37. This can also be seen directly as follows. First we note that the information on e and q yields e '" 2mod4. Moreover. a e / 2 = -1, and so Xl - a = Xl + a(e/2)+ 1 = Xl + ad, where d = (e /2)+ I is even. Now ad ~ 4(2- la d/')' ~

4(r la d / ' ) q • I ~ 4c 4 with c ~ (2' lad/' )(q + \)/4,

and this leads to the decomposition ~ (x'" +2cx" +2c')(x'" -2cx" +2c').

o

If q '" 3mod4, we can writc q in the form q ~ 2 u -I with A;, 2 and u odd. Suppose condition (i) in Theorem 3.75 is satisfied and t is divisible by 2 A • We write t ~ Be with B ~ 2 A I and v even. Then k = A in Theorem 3.37, so that with fix) ~ x - a the polynomial fix') ~ x' - a factors as a product of B monic irreducible polynomials in IF qlx 1 of degree 1/ B = c. These irreducible factors can be determined explicitly. Wc note that as in the last part of the proof of Theorem 3.75, d ~ (e/2)+ I is even. Since gcd(2B, q -I) ~ 2, there exists r E N with 2Br '" dmod(q - I). Setting b ~ a' E IF q' we get then the following canonical factorizalion. A

3.76. above, leI

Theorem.

f ·( x )-

Then Ihe rools

'"1, ... ,C 8

Wilh Ihe condilions and Ihe nOlation introduced

~'(B-i-I)!B R', EIF x [ I. ~ -·----x

,oi!(B-2i)!

q

of F(x) are all in IF q, and in IFqlx] we have the

5.

117

l3inomiab and Trinomials

canonical jactorization x' - a ~

8

0

(x" - bc,x'/2 - h 2 ).

, .,j

Proof

For a nonzero element l' in an extension field of IF q we have (x-y)(x+y-I)=x'-f3x-1 withf3=y-y'l

Using the statement and the notation of Waring's formula (see Theorem 1.76). we get ,,(xi' x,) = xf

+ xf '\'

(

£...

-)

)" (i l

+ "I" i, -I)!B ( )" 1 01 XI.Xz

11.12 ~

~2

(B-2'1 )'"./ 2

2

~ 1',

x, =

2,

)"

x,

XI

)/, XIX,.

- 1'-1, we obtain (B'

B/2

I)IB

+ 1'-8 = /~o (-I)' i!(-R'~2i)'1

1"

X I ,X 2

+. )8-",(

(_I),,(B-i,-I)!B(

L..

XI

(

0

i -O

Putting

02

1 1 .1 2 _

11, 2/ 2- 8

f3H "( - I)'

~

If cJ is a root of F(x) in some extension field of IF q and

F(f3). YJ

is such that

then 1'8+1"'= F(c,)~O and so 1"'= -I • Since q+l~ l', -y'I=C ) " J) J' J

2Bu with u odd, we get 1';"+ 1= - I, hence 1';" = - 1'1

I.

Then

cq=(y _y-I)q~yq_y q~_y,-I+y ~c f

and so cj

E

J

J)

J

J

I)'

IFq' Since F( x) is monic, we have B

0

F(x)~

(x-c j

),

)=1

hence B

1" + l' B ~ F( f3) ~

B

0 (f3 - c)J~,0~ 1 (y - y j _ I

I_C) 1 •

It follows that B

y2B+I~

0(y2- c,y-I). j"'" 1

Since this identity holds for any element l' of any eXlension field of IF" (also for l' = 0), we gel the polynomial identity 8

x'B+I~

0 ) = 1

(X' -c1x-!).

Polynomial." over finite Fields

II~

By substituting h -IX'!' for x and multiplying by b 2R , we get a factorization of x Rv + b 2R = Xl + a 2Hr = Xl + ad = x t - a (compare with the final portion of the proof of Theorem 3.75 for the last step). The resulting factors are irreducible in IFq[x] because we know already that the canonical factorization of x' - a involves B irreducible polynomials in IFq[x] of degree v (sec the discussion preceding Theorem 3.76). 3.77. Example. We factor the binomial x 24 - 3 in 1F,!x]. Here q = 23 - 1. so that A = 3, B ~ 4, and c ~ 6. Furthermore, the clement a ~ 3 is of order e ~ 6 in IFj, and so condition (i) in Theorem 3.75 is satisfied and Theorem 3.76 can be applied. We have d ~ 4, and a solution of the congruence 8r" 4mod6 is given by r = 2. Therefore, h ~ 0 2 = 2. Furthermore, F(X)~X4+4x'+2 has the roots ±l and ±3 in IF,. Thus x 24 -3= (X'-2X3_4)(x'+2x'-4)(X6+X3_4)(X6_X3_4) is the canonical factorization in IF,lx]. 0 A trinomial is a polynomial with three nonzero terms, one of them being the constant term. We first consider trinomials that arc also affine polynomials. 3.78, Theorem, Let a E IF q and ler p be the characteristic of IF q' Then the trinomial x P - x - a is irreducihle in IFq[x 1if and only if it has no root in IF q'

Proof If f3 is a root of x P- x - a in some extension field of IFq' then by the proof of Theorem 3.56 the set of roots of x P - x - a is f3 + U, where U is the set of roots of the linearized polynomial x P - x. But U = IFp , and so xP-x-a=

n (x-f3-h).

hEF p

Suppose now that x P - x - a has a factor g E IF q[ x] with I "" r and g monic. Then g{x)~

=

deg( g) < p

n (x-f3-b,)

I-=-\

for certain h, ElF P' A comparison of the coefficients of x' I shows that rf3 + b l + ... + b, is an element of IFq' Since r has a multiplicative inverse in IFq' it follows that f3 E IF q' Thus we have shown that if x P - x - a factors nontrivially in IFq[x], then it has a root in IF q. The converse is trivial. 0 3.79. Corollary. With the nOlOtion of Theorem 3.78, the trinomial x P - x - a is irreducible in IFq[x] if and only ifTrF}a) '" O.

Proof By Theorem 2.25, x P - x - a has a root in IF q if and only if the absolute trace Tr F ,(a) is O. The rest follows from Theorem 3.78. 0

5.

Binomials and Trinomials

119

Since for bE IF; the polynomial fIx) is irreduciblc over IFq if and only if f(bx) is irreduciblc over IFq • the critcria above hold also for trinomials of the form bPx P - bx - a. If we considcr more gcneral trinomials of the above type for which thc degree is a higher powcr of the characteristic. then thesc criteria need not be valid any longer. In fact. the following decomposition formula can be established.

3.80. Theorem. For x q - x - a with a being an element of the subfield K = IF, of F = IF q' we have the decomposition ql,

xLx-a=

n (x'-x-P

1-'

I

(3.18)

)

in IFql x]. where the PI are the distinct elements of IFq with TrFI K(PI) = a. Proof For a givcn P" let y be a root of x' - x extension field of IF q' Then y' - y = PI' and also a = TrFIK(Pj

Pj

in some

)

=TrFIK(y'-y) =(y'-y)+(y'-y)' +(y'-y)" + ... +(y'_y)ql'=yq_y. so that y is a root of x" - x-a. Since x' - x - Pj has only simple roots, x' - x - p] divides x q - x-a. Now the polynomials x' - x - Pi' I (; j (; q/r, are pairwise relatively prime, and so the polynomial on the right-hand side of (3.18) divides x q - x-a. A comparison of degrees and of leading coefficients shows that the two sides of (3.18) are identical.

0

3.81. Example. Consider x' - x -1 in 1F,lx]. Viewing IF, as 1F,(a), where a is a root of the irreducible polynomial x' - x -I in 1F,[x]. we find that the elements of IF, with absolute tracc equal to I are - I, a, I-a. Thus (3.18) yields the decomposition

x' - x -I

=

(x' - x + I)(x' - x - a)(x' - x - I + a).

Since all three factors are irreducible in 1F,lx], we have also ohtained the 0 canonical factorization of x' - x -I in 1F,lx]. The information ahout irreducible trinomials can be applied to the construction of new irreducible polynomials from given ones.

3.82. Theorem. Let f(x)=xm+a m ,x m.'+ ... +a o he an irreducible polynomial over the finite field IFq of characteristic p and let b E IF q' Then the polynomial f(x P - x - b) is irreducihle over IFq if and only if the absolute trace Tr. ,(mb - am ,) is '" O.

Polynomials over Finite

120

Fi~kls

Proof Suppose Tr,}mh - am ,) ~ O. Put K = IF q and let F be tbe splitting field off over K. If a E: F is a root of f. then. according to Theorem 2.14, all the roots of f arc given by a, a q, .... a q" , and F ~ K( a). rurthermore. Trf/K(a) ~ - a m _, by (2.2), and using Theorem 2.26 we get

Tr F (a + h)

=

Tr K(Tr flK (a + b)) ~ Tr K( - a m _, + mb) ~ O.

lJy Corollary 3.79, the trinomial x P - x - (a + b) is irreducible over F. Thus + h). It follows from

[ F(,8); F] ~ p. where ,8 is a root of x P - x - (a Theorem \.R4 that

[F( ,8 ) : K] = [F(,8) : F][ F: K 1~ pm. :>:owa=!JP f3 ·h,sothataEK(,8)andK(,8)~K(a,,8)=F(,8).Hence [K (,8): K ] ~ pill and the minimal polynomial of ,8 over K has degree pm. !lut f(fJP -,8 - b) ~ f(a) ~ 0, and so ,8 is a root of the monic polynomial f(xP-x-h)EK[x] of degree pm. Theorem 3.33(ii) shows that f(x P - x - h) is the minimal polynomial of,8 over K. By Theorem 3.33(i), f(x P - x - b) is irreducible over K = IF q • If Tr, (mb-a m ,)=0, then xP-x-(a+b) is reducible over F, and so [F(,8): FJ< P for any TOot,8 of xP-x-(a+b). The same arguments as above show that,8 is a root of f(x P - x - h) and that [F(,8): K] < pm, heneef(x P - x - b) is reducible over K ~ IF q • D

For certain types of reducible trinomials we can establish the form of the canonical factorization. The hypothesis for this result involves the irreducibility of a binomial, which can be checked by Theorem 3.75. 3.83. Theorem. l£t f(x)~x'-ax-hEI',[x], where r>2 is a power of the characteristic of IF q' and suppose that the binomial x'-' - a is irreducihle ocer IF q • Then fIx) is the product of a linear polynomial and an irreducihle polynomial over 1', of de!?ree r - 1. Proof Since /'( x) = - a '" O,j(x) has only simple roots. If p is the characteristic of IF q• then fIx) is an affine p-polynomial over IF q. Hence, Theorem 3.56 shows that the difference y of two distinct roots of fix) is a root of the p-polynomial x' - ax. and so a root of X,-I - a. From r - I > I and the hypothesis about this binomial, it follows that y is not an element of I'q' and so there exists a root a of fIx) that is not an element of IF q • Then a q ~ a is also a root of fix) and, by what we have already shown, a' - a is a root of the irreducible polynomial X,-l - a over 1'" so that [l'q(aq-a):lFq]=r-\' Since IFq(a'-a)~IF'<")' it follows that m= [IF q( a); IFqJ is a multiple of r - \. On the other hand, a is a root of the polynomial fix) of degree r, so that m", r. Because of r> 2, this is only possible if m ~ r - \. Thus the minimal polynomial of a over IF, is an irrcducible polynomial ovcr IF q of degree r -I that dividesf(x). The result follows now immcdiately. D

~

1~1

Binomial!;, and Trinomials

In the special case of prime fields, one can eharacteri7.c the primitivc polynomials among trinomials of a certain kind. J,84,

Theorem,

rr

For a prime p, the trinomial x P - x - a to IF x Jis a

primitive po~vllomial over IFp if and only if a is a primitive element of IFp and ord(xP-x-I)~(pP-I)/(p-I).

Proof Suppose first that [(x) ~ x P - x - a is a primitive polynomial over IFp • Then a must be a primitive clement of IFp because of Theorem 3.18. If f3 is a root of g(x) ~ x P - x - I in somc extension field of IFp' then

o ~ ag(f3) ~ a({3P -

f3 -I) = a Pf3P - af3 - a ~ [(af3).

,,=

and so af3 is a root of [( x). Conscquently, we have {3' '" I for 0 < r < (pP-I)/(p-I), for othcrwise ,,'(P t'~1 with O
W)· .. (x - W' ').

A comparison of the constant tcrms leads to {3,p'-ItAp-1t

=

I, hence

ord(x P - x - I) ~ (pP - 1)/( p . I) on account of Theorem 3.3.

Conversely, if thc conditions of the theorcm are satisfied, then a and {3 have orders p - I and (pP - 1)/( P -. I), rcspectively, in the multiplicative group IF;,. Now (pP . 1)/( p - I)

= I + P + p' + ... + pp-I

'" I

+ I + I + ... + I

" p" I mod( p - I),

,,=

so that p - I and (pP - 1)/( p " I) are relatively prime. Therefore. a{3 has order (p -1)·( pP -1)/( p - I) = pP -I in IF;,. Hence" is a primitive element of IFp ' and[(x) is a primitive polynomial over IFp • 0 3.85. Example. For p~5 we have (pP-I)/(p-I)=781=11·7J. !'rom the proof of Theorem 3.84 it follows that x"'" I mod(x' - x - I). and since x" .. 1 mod(x' - x -I) and x 7l .. 1mod(x' - x -I), we obtain ord(x' - x-I) ~ 78 J. Now 2 and 3 arc primitive clements of IF,. and so x' - x -2 and x' - x -3 are primitive polynomials over IF, by Theorem ].~

0

for a trinomial x' + x + a over a finite field IF. of odd characteristic. it is easily seen that it is irrcdueible over IF q if and only if a is not of the form a = 4 t - h'. bE IF•. Thus. there are exactly (q - 1)/2 choices for a F IF" that make x' + x + a irreducible over IF q' More generally, the number of a c IF. that make x" + x + a irreducible over IF q is usually asymptotic to q/n. according to the following result.

Polynomials over Finite Fields

122

3.86. Theorem. Let"q be a finite /ield 0/ characteristic p. For an illleger n '" 2 such that 2n(n -I) is not dirisible by p. let T,(q) denote the number 0/ for which the trinomial x' + x + a is irreducible orer Theil there is a constant B" depending only on n, such that

aE".

"q'

IT,(q)-;I",

B,ql/2.

We omit the proof, as it depends on an elaborate investigation of certain Galois groups. In Definition 1.92 we defined the discriminant of a polynomial. The following result gives an explicit formula for the discriminant of a trinomial. 3.87. Theorem. The discriminant ".Ix] with n > k '" I is given by

0/ the

trinomial x' + ax' + bE

D(x'+ax'+b)=(-I)"'-I)/2 b'-1 '(nNb N- K _( -I)N(n _ k)N- KeaN)d, where d = ged(n. k), N

~

nld, K = kid.

EXERCISES

3.1.

Determine the order of the polynomial (x 2 + x + 1)5(x' + X

+ I)

over 0=2'

3.2. 3,3.

Determine the order of the polynomial x' - x· + x 4

-

x2

+ x over

"Determine ,. ord(f) for all monic irreducible polynomials/in ",Ix] of

degree 3. Prove that the polynomial x 8 + x' + x' + x + I is irredueihle over" 2 and determine its order. 3.5. Let / E "qlx] be a polynomial of degree m '" I with /(0) = 0 and suppose that the roots "I'" .,"m of / in the splitting field of / over I' q arc all simple. Prove that ord(f) is equal to the least positive integer e such that af = I for 1 :!S; i ~ m. 3.6. Prove that ord(Q,)= e for all e for which the cyclotomic polynomial Q, E "qlx] is defined. 3.7. Let / be irredueihle over"q with /(0) = O. for e E N relatively prime to q, prove that ord(f) ~ e if and only if / divides the cyclotomic polynomial Q" 3.8. Let / E "qlx] be as in Exercise 3,5 and let bEN. find a general formula showing the relationship between ord(fh) and ord(f). 3.9. Let f q be a finite field of characteristic p, and let / E fqlxl he a 3.4.

123

3.10.

3.11.

3.12. 3.13.

3.14. 3.15.

3.16. 3.17. 3.18. 3.19.

3.20. 3.21.

3.22.

polynomial of positive degree with/(O) * O. Prove that ord(j(x P )) = p ord(j( x)). Let I be an irreducible polynomial in IF qlx] with 1(0) * 0 and ord(j) = e. and let r be a prime not dividing q. Prove: (i) if r divides e. then every irreducible factor of I(x') in IFqlx] has order er; (ii) if r does not divide e. then one irreducible factor of I(x') in IFqlx] has ordcr e and the other faclOrs have ordcr er. Deduce from Exercise 3.10 that if I E IF qlx] is a polynomial of positive degree with 1(0) =O. and if r is a prime not dividing q. then ord(j(x')) = rord(j(x)). Prove that the reciprocal polynomial of an irreducible polynomial I over IF q with 1(0) * 0 is again irreducihle over IF q' A nonzero polynomial I E IF qlx] is called sell-reciprocal if 1= 1*. Prove that if I = gh. whcre g and h are irreducible in IFq[x] and I is self-reciprocal. thcn cithcr (i) h* = ag with a E IF;; or (ii) g* = bg, h* = hh with b = :!: 1. Prove: if I is a self-reciprocal irreducible polynomial in IFqlx] of degree m > I. then m must be even. Prove: if I is a self-reciprocal irredueihle polynomial in IF ql x] of degree> 1 and of order e, then every irreducible polynomial in IFqlxl of degree> 1 whose order divides e is self-reciprocal. Show that x' + x' + x' + x + I is a primitive polynomial over IF,. Show that x' + x' + x' -;- x + I is a primitive polynomial over IF,. Show that x' -x+ 1 is a primitive polynomial over IF,. Let I E IF ql x] be monic of degree m ;, 1. Prove that I is primitive over IF q if and only if I is an irredueihle faclOr over IF q of the cyciOiomic polynomial Qd E IFqlx] with d = qm - I. Determine the number of primitive polynomials over IF" of degree m. If mEN is not a prime. prove that not every monic irreducible polynomial over IF q of degree m can he a primitive polynomial over IF q' If m is a prime. prove that all monic irreducible polynomials over IF q of degree m are primitive over IF q if and only if q = 2 and 2m - I is a prime.

3.23.

If I is a primitive polynomial over IF q' prove that 1(0)

'/*

is again

primitive over IFq.

3.24.

3.25. 3.26.

Prove that the only self-reciprocal primitive polynomials are x I 1 and x' +x+ lover IF, and x+ lover IF, (see Exercise 3.13 for the definition of a self-reciprocal polynomial). Prove: if I(x) is irreducible in IF qlx]. then I( ax + h) is irreducible in IF qlx] for any a, h E IF q with a * O. Prove that Nq(n) '" (ljn)(q" - q) with equality if and only if n is prime.

124

3.27.

Polynomials over Finite Fidds

Prove that N (nb .!.q' - -q._--(q'I' q n n ( q -- I)

3.28. 3.29. 3.30.

-I).

Give a detailed proof of the fact that (3.5) implies (3.4). Prove that the Moebius function p. satisfies p.(mn) = p.(m)p.(n) for all m,nEN with gcd(m.n)=1. Prove the identity

L

p.(d)

din

d

~

$(n)

foralinEN.

n

Prove that '[d"p.(d)$(d) ~ 0 for every even integer n;" 2. Prove the identity '[dl,Ip.(d)1 ~ 2k • where k is the numher of distinct prime factors of n E N. 3.33. Prove that Nq(n) is divisihle hy eq provided that n;" 2, e is a divisor of q - I, and gcd(eq, n) = 1. 3.34. Caleulatc the cyclotomic polynomials QI2 and Q30 from the explicit formula in Theorem 3.27. 3.35. Establish the properties of cyclotomic polynomials listed in Exercise 2.57. Parts (a)-(f), by using the explicit formula in Theorem 3.27. 3.36. Prove that the cyclotomic polynomial Qn with gcd(n,q)=1 is irreducihle over IF q if and only if the multiplicative order of q modulo n is
3.31. 3.32.

b.erci&cs

125

Find the factorization of x J2 - x into irreducible polynomials over IF 2' 3.44. Calculate 1(2,6; x) from the formula in Theorem 3.29. 3.45. Calculate 1(2,6; x) from the formula in Theorem 3.31. 3.46. Prove that 3.43.

1(q,n;x ) =

n( x q' - I -I ),'"/d)

forn>!.

dl"

3.47. 3.48.

3.49.

3.50. 3.51. 3.52. 3.53. 3.54.

3.55. 3.56.

3.57. 3.58. 3.59. 3.60. 3.61. 3.62.

3.63.

Prove that over a finite field of odd order q the polynomial -)(1 + x,q· 1)/' +(1- x),q+ 1)/') is the square of a polynomial. Determine all irreducible polynomials in O=,[x] of degree 6 and order 21 and then all irreducible polynomials in O=,[x] of degree 294 and order 1029. Determine all monic irreducible polynomials in O=,[x] of degree 3 and order 26 and then all monic irreducible polynomials in O=,[x] of degree 6 and order 104. Proceed as in Example 3.41 to determine which polynomials /, are irreducible in 0=q[ x] in the case q = 5, m = 4, e = 78. In the notation of Example 3.41, prove that if I is a prime with I - I dividing m - I, then /, is irreducible in 0=, [x]. Given the irreducible polynomial I(x) ~ x' - x 2 + X + lover 0=" calculate I, and Is by the matrix-theoretic method. Calculate I, and Is in the previous exercise by using the result of Theorem 3.39. Use a root of the primitive polynomial x' - x + lover 0=, to represent all elements of 0=;, and compute the minimal polynomials over 0=, of all elements of 0=27' Let 0 E 0=64 be a root of the irreducible polynomial x· + x + I in O=,[x]. Find the minimal polynomial of f3 ~ I + 0' + 0' over 0=2' Let 0 E 0='4 be a root of the irreducible polynomial x' + x 4 + x' + x + I in o=,[x]. Find the minimal polynomial of f3 = I + 0 + Os over 0=, . Determine all primitive polynomials over 0=, of degree 2. Determine all primitive polynomials over 0=4 of degree 2. Determine a primitive polynomial over O=s of degree 3. Factor the polynomial g E O=,[x] from Example 3.44 in O=q[x] to obtain primitive polynomials over 0=9' ractor the polynomial Ii E O=,[x] from Example 3.45 In O=,[x] to obtain primitive polynomials over 0=,. Find the roots of the following lineari7.ed polynomials in thcir splitting fields: (a) L(x) ~ x'·· x 4 + x' + X E O=,[x]; q (b) I,(x) ~ x + x E o=,[x]. find the roots of the following polynomials in the indicated fields by

126

Polynomials over finite fidds

first determining an affine multiple: (a) !(x)~x'+x'+x'+x'+ I <=IF,[x] in IF,,: (h) fIx) ~ x 4 + Ox' - x' -(0 + I)x + 1- 0 E 1F,[x] in 1F 729 , where 0 is a root of x' - x - I E 1F,lx]. 3.64. Prove that for every polynomial! over IF q"' of positive degree there exists a nonzero q-polynomial over IF qrn that is divisible by!. 3.65. Prove that the greatest common divisor of two or more nonzero q-polynomials over IF ,rn is again a q-polynomial. hut that their least common multiple need not necessarily be a q-polynomial. 3.66. Determine the greatest common divisor of the following linearized polynomials: L,(x) ~ x 64 + x 16 + x' + x 4 + x' + X E IF,fx], l.,(x) ~ x 32 + x' + x' + X E 1F,lx]: (b) l.,(x) ~ X'4' - x" - x' + x' + X E 1F,[x], [.,(x) ~ x" + X E 1F,lx].

(a)

3.67.

Determine the symbolic factorization of the following linearized polynomials into symholieally irreducible polynomials over the given prime fields: (a) [,(x) = x 32 + x 16 + x' + x 4 + x' + x E 1F,lx]: (b) L(x) ~ x" - x' - x' - X E IF,fx]. 3.68. Prove that the q-polynomial L,(x) over IF,rn divides the q-polynomial L(x) over IF q_ if and only if [,(x) ~ [.,(x)0[,,(x) for some q-polynomial [,,(x) over IFq_. 3.69. Prove that the greatest common divisor of two or more affine q-polynomials over IF,"" not all of them 0, is again an affine q-polynomial. 3.70. If A,(x) ~ L,(x)- a, and A,(x) ~ L,(x)- a, are affine q-polynomials over IF,rn and A,(x) divides A,(x), prove that the q-polynomial [,,(x) divides the q-polynomial[,,(x). 3.71. Let fIx) be irreducible in IFqlx] with flO) = and let F(x) be its linearized q-assoeiate. Prove that F(x)/x is irreducible in 1F,[x] if and only if fIx) is a primitive polynomial over IF, or a nonzero constant multiple of such a polynomial. 3.72. Let I be an clement of a finite extension field of IF,rn. Prove that a q-polynomial K(x) over IF,_ has I as a root if and only if K(x) is divisible by the minimal q-polynomial of lover IF q-' 3.73. For a nonzero polynomial !EIF,[x], prove that L,(g)~qd"(j,, where the sum is extended over all monic divisors g E IF ,Ix] of f. 3.74. For a nonzero polynomial! E 1F.r x] and g E IF qlx] with ged(/. g) ~ I. prove that g' '" I mod f. where k - q(/)' 3.75. The function 1', is defined on the set S of nonzero polynomials! over IF, by I'/f) = I if deg(/) ~ 0, 1',(/) ~ if!has at least one multiple root, and I',(f) ~ (-I)' if deg(f) ~ I and! has only simple roots, where k is the numher of irreducible factors in the canonical faetori-

°

°

127

zation of I in IF q[x]. Let L denote a sum extended over all monic divisors g E IFq[x] of j. Prove the following propcrtics: I if deg(j) ~ 0, (a) Ll'q(g)~ { 0 ifdeg(j);d; (b) I'ilg) ~ I'q(j )I'ig) for all I, g E S with gcd(j, g) = I; (c) Lqd"('ll'q(j/g)~~q(j) for alliES; (d) if If is a mapping from S into an additively written abelian group G with If( ej) = If(j) for all e E IF; and IE S, and if '!'(j)=LIf(g) for alliES, thcn If(j)=L.l'q(j/g)'!'(g)~ Ll'q(g)'l'(j/g) for all I E S. 3.76. Prove that the number of different normal bases of IF q" over IF q is

~ m

3.77.

n (qO(dl_l) d!m

provided that gcd( m, q) = I and the multiplicative order of q modulo m is (m). Refer to Example 2.31 for the definition of a self-dual basis and

show that there exists a self-dual normal basis of 1F 2 " over 1F 2 whenever m is odd. (Hinr: Show first that the number of different normal bases of 1F 2" over 1F 2 is odd whenever m is odd.) 3.78. For a prime r and a E IF q' prove that x' - a is either irreducible in IFq[x] or has a root in IF q. 3.79. For an odd primc r, an integer n ;. I, and dE IF q' prove that x," - a is irreducible in IFq[x] if and only if a is not an rth power of an element of IFq' 3.80. Find the canonical factorization of the following binomials over the given prime fields: (a) I(x)= x· + I E 1F 3 [x]; (h) l(x)=x 27 -4EIF 19 [x]; (c) I(x) ~ x" -10 E 1F 23 [x].

3.81. 3.82.

3.83.

3.84. 3.85.

Prove that under the conditions of Theorem 3.76 the roots of the polynomial F( x) introduced there are simple. Prove that the resultant of two binomials x" - a and x m - b in IFq[x] is given by (-I)"(b"/d - am/d)d with d ~ gcd(n, m), where nand m are considered to bc thc formal degrees of the binomials (compare with Definition 1.93). For a nonzero element b of a prime field IF p' prove that the trinomial XC - x - h is irreducible in IFp'[x] if and only if n is not divisible by p. Prove that any polynomial of the form xq-ax-bElFq[x] with a = I has a root in IF q. Prove: if x P - x - a is irredueihle over the field IF q of characteristic p

12~

Polynomials ov('r Finite Hdd...

and fJ is a root of this trinomial in an extension field of IF q' then x P - x - afJP I is irreducible over IFq(fJ). 3.86.

3.87. 3.88.

3.89.

3.90. 3.91. 3.92. 3.93.

3.94. 3.95.

Prove: if f(x) = x m + am IXm 1 + ... + £10 is irreducihlc over the field IF q of charaetcristic p and b E IF q is such that Tr r ,( mb - am. I) = 0, then f(x P - x - h) is the product of p irreducible polynomials over IF q of degree m. If m and p are distinct primes and the multiplicative ordcr of p modulo m is m - I, provc that L;"_o'(X P - x); is irrcdueiblc over IF" Find thc canonical factorization of the given polynomial over thc indicated field: (a) f(x) ~ x' - ax - I E 1F'4[X], where a satisfies ,,' ~ a + I: (b) f(x) ~ x· - ax + a E IFq[x], where ex satisfies ex' ~ ex + 1. Lct A(x)= l.(x)-aElFq[x] be an affine p-polynomial of degrcc r> 2, and suppose the p-polynomial L(x) is such that L(x)jx is irreduciblc over IF q • Prove that A(x) is thc product of a linear polynomial and an irreduciblc polynomial ovcr IF q of degrec r - 1. Prove: tbc trinomial x" + ax' + hE IFq[x], n > k;, I, q evcn, has multiple roots if and only if nand k arc both even. Prove that the degree of cvcry irreducible factor of xl" + x + I in 1F,[x] divides 2n. Prove that the degree of every irreducible factor of x'''· , ~ x + I in 1F,[x] divides 3n. Recall the notion of a self-reciprocal polynomial defined in Exercise 3.13. Prove that if f E 1F,[x] is a self-reciprocal polynomial of positive degree, then f divides a trinomial in IF,lxj only if ord(f) is a multiple of 3. Prove also that the convcrse holds if f is irreducible over !F2' Prove that for odd dEN the cyelotomie polynomial Qd E 1F,[x] divides a trinomial in 1F,[x] if and only if d is a multiple of 3. Let f(x) = x" + ax' + b E IFqlxj, n > k;, I, be a trinomial and let mEN be a multiple of ord(f). Prove thatf(x) divides the trinomial g(x)~xrn '+b-'x"-'+ab

3.96. 3.97.

Prove that only if n = Prove that only if n =

'.

the trinomial x'" + x" + I is irreducible over IF, if and 3' for some nonnegative integer k. the trinomial x 4n + x" + I is irreducible over IF 2 if and 3'5 rn for some nonnegative integers k and m.

Chapter 4

Factorization of Polynomials

Any nonconstant polynomial over a field can bc expressed as a product of irreducible pOlynomials. [n the case of finite fields, somc reasonably efficient algorithms can be devised for the actual calCulation of thc irrcducible factors of a given polynomial of positive degree. The availability of feasible factorization algorithms for polynomials over finite fields is important for coding theory and for the study of linear recurrence relations in finite fields. Beyond the realm of finite fields, there are various computational problems in algebra and number theory that depend in one way or another on the factorization of polynomials over finite fields. We mention the factorization of polynomials over the ring of intcgcrs, the determination of the decomposition of rational primes in algebraic numbcr fields. the calculation of the Galois group of an equation over the rationals, and the construction of field extensions. We shall present several algorithms for the factorization of polynomials over finitc fields. The decision on the choice of algorithm for a specific factorization problem usually depends on whether the underlying finite field is "small" or "large." In Section I we describe those algorithms that are better adapted to "small" finite fields and in the next section those that work belter for "large" finite fields. Some of these algorithms reduce the problem of factoring polynomials to that of finding the roots of certain other polynomials. Therefore. Section 3 is devoted to the discussion of the latter problem from the computational vicwpoint.

Fal:ton1.ation of

130

1.

Pol~momials

FACTORIZATION OVER SMALL FINITE FIELDS

Any polynomialf E IFqlx] of positivc degrce has a canonical factorization in IFqlx] by Theorem 1.59. For the discussion of factorization algorithms it will sufficc to consider only monic polynomials. Our goal is thus to express a monic polynomial f E IF .[x] of positive degrce in the form f~f,"···f:'.

(4.1)

wherc f\ .... ./k are distinct monic irreducible polynomials in IFqlx] and e I" .. , ek are positive integers. First we simplify our task by showing that thc problem can be reduced to that of factoring a polynomial with no repeated faclOrs. which means that the exponents e\ ..... ek in (4.1) are all equal to I (or. equivalently. that the polynomial has no multiplc roots). To this cnd. we calculate d(x) = gcd(f(x).j'(x».

thc greatest common divisor of fix) and its derivative. by thc Euclidean algorithm. If d(x) ~ I. then we know thatf(x) has no repeatcd factors becausc of Thcorem 1.68. If d(x) ~ fix). we must have f'(x) = O. Hcnce fix) = g(x)P. wherc g(x) is a suitable polynomial in IFqlxJ and pis thc characteristic of IFq. If necessary. the reduction process can be continucd by applying the method to g( x). If d(x) = I and d(x) = f(x). thcn d(x) is a nontrivial factor of fix) andf(x)/d(x) has no repeated factors. The factorization off(x) is achievcd by factoring d(x) andf(xJld(x) separately. In casc d(x) still has rcpeated factors. further applications of the rcduction proccss will havc to bc carried out.

By applying this process sufficiently often. the original problcm is reduced to that of factoring a certain number of polynomials with no repeated factors. The canonical factorizations of these polynomials lead directly to the canonical factorization of the original polynomial. Therefore. we may restrict the attention to polynomials with no repeated factors. The following theorem is crucial. 4.1. Theorem. h q ", hmodf. then

If f

fix) ~

E

IFqlx] is monic and

hE

n gcd(f(x). h(x)-c).

IFqlx] is such that (4.2)

eE IF q

Proof Each greatest common divisor on thc right-hand side of (4.2) divides fix). Since the polynomials h(x)- c. c E IF q• are pairwise relatively prime. so arc the greatest common divisors withf(x}. and thus the product of these greatest common divisors divides fix). On the other hand. fix)

l. Factor1lation oVI,.'r Small Finite FiL'1d:,

131

divides

h(x)q-h(x)=

n (h(x)-c),

c E= nq

and sof(x) divides the right-hand side of (4.2). Thus, the two sides of (4.2) are monic polynomials that divide each other, and therefore they must be equal. D In general, (4.2) does not yield the complete factorization of f since ged(f(x), h(x)- c) may be reducible in IFq[x]. If h(x) '" cmodf(x) for some c E IF q' then Theorem 4.1 gives a trivial factorization of f and therefore is of no use. However, if h is such that Theorem 4.1 yields a nontrivial factorization of f, we say that h is an f-reducing polynomial. Any h with II" '" hmodf and 0< deg(h) < deg(f) is obviously f-redueing. In order to obtain factorization algorithms on the basis of Theorem 4.1. we have to find methods of constructing f-reducing polynomials. It should be clear at this stage already that since the factorization provided by (4.2) depends on the calculation of q greatest common divisors, a direct application of this formula will only be feasible for small finite fields IF q • The first method of constructing f-redueing polynomials makes usc of the Chinese remainder theorem for polynomials (see Exercise 1.37). Let us assume that f has no repeated factors, so that f = f, ... fA is a product of distinct monic irreducible polynomials over IF q . If (c" .. .,c,) is any k-tuple of elements of IF", the Chinese remainder theorem implies that there is a unique h E IF qlx] with h(x) '" c, mod j,(x) for I <; i <; k and deg(h) < deg(f). The polynomial h(x) satisfies the condition

h(x)q '"

<' ~ c, '" h(x )modj,(x)

for I <; i <; k,

and therefore

h"=:hmodf,

deg(h) <deg(f}.

(4.3)

On the other hand, if h is a solution of (4.3), then the identity

h(x)"-h(x)=

n (h(x)-c) I" E=

LF q

implies that every irreducible factor of f divides one of the polynomials h(x)- c. Thus, all solutions of (4.3) satisfy h(x) '" c,modj,(x), 1<; i <; k, for some k-tuple (c" ... ,c,) of elements of IF q. Consequently, there are exactly q' solutions of (4.3). We find these solutions by reducing (4.3) to a system of linear equations. With n = deg(f) we construct the II X II matrix B ~ (h,), 0<; i,j <; n ... I, by calculating the powers x,qmodf(x). Specifically, let

x,q",

"

L: t b,;x'modf(x)

r O

forO<;i<;n-1.

(4.4)

Fat:tori/at;on of PolYlh)mial~

Then h(x) ~ only if

00

+ o,x ' ... + n"

,x"

'", IFqlx] is a solution of (4.3) if and

(° 0 ,0, ..... 0" ,)B~ (° 0 .0, ..... 0,,_,).

(4.5)

This follows from the fact that (4.5) holds if and only if

"- ,

L

h(x)·

0rtJ

1=0 1n- I

f/

~

L L } =- ()

I -

"

,

1"-

a

'" L

a,b'J xJ (J

a,x'" ~ h(x)qmodf(x).

The system (4.5) may be written in the equivalent form

(° 0 .0"

.... °"_,)(8 -I) -

(0,0, ... ,0),

(4.6)

where 1 is the n X n identity matrix over 1F1t" By the considerations ahove. the system (4.6) has q' solutions. Thus, the dimellsioll of the Ilull spoce of lhe malrix R - 1 is k. the number of distinct monic irreducible factors of f, and lhe rOllk of 8 . 1 is II - k. Since the constant polynomial h,(x) ~ I is always a solution of (4.3), the vector (1,0, ... ,0) is always a solution of (4.6), as can also he checked directly. There will exist polynomials h,(x).... ,h,(x) of degree "11-' I such that the vectors corresponding to h,(x), h,(x) ,h,(x) form a hasis for the null space of B-1. The polynomials h,(x), ,h,(x) have positive degree and are thus f-reducing. In this approach, an important role is played hy the determination of the rank r of the matrix 8 - I. We have r ~ II - k as noted ahove, so that once the rank r is found. we know that The number of diSTincT monic irredllcih/e faclOrs off is [(ivell hy n - r. On the basis of this information we can then decide when the factorization procedure can he stopped. The rank of B-1 can he determined hy using row and column operations to reduce the matrix to echelon form. However, since we also want to solve the system (4.6), it is advisable to usc only column operations because they leave the null space invariant. Thus. we are allowed to multiply any column of the matrix B-1 by a nonzero clement of IF q and to add any multiple of one of its columns to a different column. The rank r is the numher of nonzero columns in the column echelon form. Having found r, we form k ~ II - r. If k·- I, we know that f is irreducible over IF q and the procedure terminates. In this case, the only solutions of (4.3) are the constant polynomials and the null space of 8 - 1 contains only the vectors of the form (c,O, ... ,0) with C E IF q' If k ~ 2, we take the f-redueing basis polynomial h 2 (x) and calculate

l. facton/.ation

()Wf

133

Small finite Fields

gcd(j(x), h",(x)- c) for all c E IF q' The result will be a nontrivial factorization of f(x) afforded by (4.2). If the usc of h,(x) does not succeed in splittingf(x) into k factors, we calculate ged( g(x), h J(x)- c) for alJ c E IF q and all nontrivial factors K(X) found so far. This procedure is continued until k factors of f(x) arc obtained. Thc process described above must eventualJy yield all the factors. for if we consider two distinct monic irreducible factors of f(x), say f,(x) and f, (x), then hy the argument folJowing (4.3) there exist elements cJ " c j' F IF. such that hj(x) =' cj,modf,(x), h/x) =' cJ,modf,(x) for I",) '" k. Suppose we had C,I = c" for I",) '" k. Then, since any solution h(x) of (4.3) is a linear combination of hl(x)•.... hk(x) with coefficient:s in IF q • there would exist for any such h(x) an element cElF q with h(x)=,cmodf,(x), h(x) '" cmodf,(x). But the argument leading to (4.3) shows, in particular. that there is a solution h(x) of (4.3) with h(x)=Omodf,(x), h(x)=' I modf,(x). This contradiction proves that cJ ' = cj ' for some) with I",) '" k (in fact, since h,(x) = I, we will have) ~ 2). Therefore, hj(x)-c" will be divisible by f,(x), but not by f,(x). Hence any two distinct monic irredueihle factors of f(x) will be separated by some h,(x).

This factorization algorithm based on determining f-redueing polynomials by solving the system (4.6) is ealJed Berlekamp's algorithm. 4.2. Example. factor f(x) = x' + x· + x' + x' + lover IF, hy Berlckamp's algorithm. Since gcd(j(x).j'(x))=J. f(x) has no repeated factors. We have to compute x,qmodf(x) for q = 2 and 0", i '" 7. This yields the following congruences mod f(x):

, =I

XO

x-

x'

:=

x 4 .=:

x'

x~ .=

x·

x' = 1 x lO = I X X

12

=

14

=:

I

+ x·~ + x 4

+ x·

-t-X 2 +X 3 +X 4 +X S 4 7 5 6 x' +X +X +X +X 4 3 5 +x +X +X +X

Therdore, the 8 x 8 matrix FJ is given by I

B=

0 0 0 1 I

0 I

0 0 0 0 0 0 0 1

0 1 0 0 0 1 1 0

0

0 0 1 0 1 1

0 0 0 0 0

0 0 0 I

I

0

0

I

I

1

1 1

I

I

0

0

0

0 0 I I

I

0 0 0 0 0 0

Fal'loriLalion of Polynomiab

1.14

and B - J is given by 0

0

0 0 0

0 0 0 0

I

I I I

I 0

0 I

° ° I I

I

0 0

B-J~

I

I 0 I

0 0 0 0 0

0 0

I

I

I I I

0

°

0 0 0

I I

0 0 0

0 0 0 0

I I

0 0 0 0 0 0 I I

The matrix B-1 has rank 6, and the two vectors (1,0,0,0,0,0,0,0) and (0, I, 1,0,0, I, I, I) form a basis of the null space of B-1. The corresponding polynomials arc h,(x) ~ I and h,(x) = x + x' + x' + x' + x'. We caleulate gcd(j(x), h ,(x)- c) for c E IF, hy the Euelidean algorithm and obtain gcd(j(x), h,(x)) ~ x' + x' + x 4 + X + I, ged(j(x), h,(x)-I) = x' + x -. I. The desired canonical factorization is therefore

I( x l ~ (x' + x' + x 4 + X + 1)( x' + x + I).

D

A second method of obtaining I-reducing polynomials is based on the explicit construction of a family of polynomials among which at least onel-reducing polynomial can he found. Lctlbe again a monic polynomial of degree n with no repeated factors. Let I ~ I, ... Ik he its canonical factorization in IFqlx] with dcg(J;)=n J for I",j",k. If N is the least qN positive integer with x xmod/(x), then it follows from Theorem 3.20 that N = Iem(n" ... ,n,), and it is also easily seen that N is the degree of the splitting field F of lover IF q • Let the polynomial T E IFqlx] be given by , N. . 1;(x)~T(x') fori= 0,I ..... T(x)~x+xq+xq + ... +x q an d defme The following result guarantees that in the case of interest, namely, when I is reducible, there are I-reducing polynomials among the 1;.

=

4.3, polynomials

Theorem. II I is reducible ill IFqlx]. then at least one 1 ~ i ~ n - 1, is I-reducing.

1;,

01 the

Proof It is immediate that any polynomial 1; satisfies 1;q = 1; mod f. Suppose now that for all 1;, I", i '" n -I, the factorization of I afforded by (4.2) were trivial. This means that there exist elements c" ... ,c" ,E IF q such that 1;(x) = c;mod I( x) for I '" i '" n - I, With Co ~ N. viewed as an clcment oflF q, we get T(x;)=c,mod/(x)forO",i",n-1. For any n -,

g(x)=

L a,x;EIF.[x] ,-,

of degree less than n we have then

T(g(X))~T(nt'a,x') = 1-'- ()

nt'a,F(x'l= "t'a,c,mod/(x). 1-0

I - 0

1. Fal'lOrization over Small Finite Fields

135

Putting

n-'

c(g)~ Lu,c,ElF q • 1=0

we obtain T(g(x»=c(g)mod/;(x)

for I «j«k.

(4.7)

Sinec N ~ Icm( n " ... ,n,). at Icast one of the intcgers Nj nJ , say Nj n ,. is not divisible by the charactcristic of IF q' Let 0, be a root of I, in the splitting field F, of I, over IF q . Because of Theorem 2.23(iii) there exists g, ElFq[x] with (4.8) Since k ;;, 2 by assumption, we can apply the Chinese remainder theorem to obtain a polynomial g E IFq[x] of degree < n with g

= g,mod Ii' g = Omod 12'

(4.9)

From (4.8) and (4.9) we deduce that Tr F,/ F,(g(O,»)

=

I.

and Theorems 2.23(iv) and 2.26 imply that Tr F/ F,(g(O,») ~ Njn,. Bceause of thc dcfinitions of the trace and of the element 0,. it follows that

Howcvcr. the second congruence in (4.9) leads to T(g(x» '" Omod I,(x), and since N/n, '" 0 as an element of IF q, we get a contradiction to (4.7). Therefore, at least onc of thc T" 1« i« n -I, is I-reducing. 0 4.4. Example. Factor I(x) ~ x 17 + x 14 + x" + x 12 + x" + x lO + x 9 + x' + x' + x' + x· + x + lover IF,. We have gcd(f(x), j'(x» = x'~ + x' + I, and so/u(x) = I(x)jgcd(f(x),f'(x» = x' + x' + x· + x + I has no repeatcd factors. We factor 10 by finding an fo-reducing polynomial of the type described abovc. To this end, we calculate the powers x, x', x ..... mod/u(x) until we obtain the least positive integcr N with x'" = xmod/o(x). We simplify thc notation by identifying a polynomial l:7:d a;x' with the n-tuple aDa, ... an ,of its coefficients, so that, for instance,fo(x) = 11001101. The calculation of the rcquircd powers of xmod/o(x) is facilitated by thc observation that squaring a polynomial aoa, ... a 6mod/o(x) is the samc as multiplying the vector aDa, ... a 6 by the 7 X 7 matrix of even powers

136

Factorization of Polynomials

xO,x', ... ,x"mod/o(x). This matrix is obtaincd from

=

xO I x' =0 x' =0 x' 0 x' 0 x'o I l2 X == 0

0 0 0 0

= = =

0 I 0 0 1 I 0

I

0 I

0 0 0 0 0 I 0

0 0 1 0 0 0 I

0 0 0 0

0 0 0 I

I

I I

0 0

1

wherc all the congruences are mod/o(x). Thcrefore we gct mod/o(x):

x=o x' =0 x' =0 x' 0 X

X X

I6

l2 64

I2R

X

= =. 1 =I I

I

.= 1

1 0 0 I

=.

256

== 1 == 0 X 1024 == 0 X

512

X

Thus N

~

1 0 0 1 I 0

0 1 0 I 0 1 0 I 0 1 0

0 0 0 0 I 0 0 0 0 I

0

0 0 1 0 0 0 0 I

0 0 0

0 0 0 I 0 0 0 0 1 0 0

0 0 0 I 0 1 I

1 0 1 0

10 and 9

T,lx)=I;x"=1 J

0 0 0

1mod/u(x).

0

Since T,(x) is not congruent to a constant mod/o(x), T,(x) ishrreducing. We have gcd(fo (x).T, (x» ~ gcd(l I 0 0 I 1 0 1, I I I 0 0 0 I) x 5 + x 4 + x 3 + x 2 + 1, gcd(fo(x).T,(x)-I)~ged(1 100 I 10 1.0 I 10001) =

=x' + x + I, and so /u(x) ~ (x' + x'.,.

Xl

+ x' + I)(x' + x + 1).

The second factor is obviously irrcducible in 1F,[x]. Sinec N ~ 10 is the least common multiple of the dcgrees of thc irreducihle factors of /o(x), any nontrivial factorization of the first factor would lead to a value of N diffcrcnt from 10, so that the first factor is also irreducible in 1F,[x].

I. Factorization over Small finite Fil'1d:-.

137

It remains to factor gcd(f(x), !,(X))~XlO+X'+1. We have x 'O + x' + I = (·x' + x 4 + I)', and by checking whether x' + x 4 + I is divisihIe by one of the irreducible factors of lo(x), we find that x' + x 4 + I ~ (x' + x + I )(x' + x + I), with Xl + x + I irreducible in f,lxJ. Hence

I( x)

=

(x' + x 4 + x' -r x' -r 1)( x I + X + 1)'( x'

is the canonical factorization of I(x) in f ,lx J.

-t

x

+ I) I 0

It should be noted that, in general, the I-reducing polynomials T; do not yield the complcte factorization of I since the 7; are not ablc to separate those irreducible factors.li for which Njn, is divisible hy the characteristic of f q • In practice, however, one calculates thc first I-reducing T; and then calculates new T, for each of the resulting factors. In this way, one eventually obtains the complete factorization of f. It is, however, possible to construct a related set of polynomials R i that are capable of separating all the irreducible factors of I at once. We assume, without loss of generality, that 1(0) '" O. Let ord(f(x)) = e, so that I(x) divides x' -1. Since I has no repeated factors, e and q are relatively prime by Corollary 3.4 and Theorem 3.9. ror each i;3 0 let m, be the least positive integer with

x,qm, '" x'mod/(x).

(4.10)

Then we define

R I ( x) =

Xl

+ x lq +

't m , ... + x lq ,

XiII

Since (4.10) is equivalent to iqm, =. imode,

(4.11 )

which is in tum equivalent to qm, '" I mod( e jgcd( e, i)), it follows that m i can also he described as the multiplicativc order of q moduloejgcd(e. i). A comparison with the definition of T;(x) shows that N

T; (x) '" - R, (x )mod/(x). mi It is elear that R~ '" R,modl for all i. so that the R, can he used in (4.2) in place of h. We prove now the claim about the R, made above.

4.5. Theorem. Let I he monic and reducible in f q[x J with no repeated lactors, and suppose that 1(0) =0 and ord(f) ~ e. Then, il all the polynomials R i, I" i" e -I, are used in (4.2), they will separate all irreducible lactors 01

f. Proof Let h(X)~L~~daixiEfqlxJ bc a solution of h(x)q", h(x)mod(x' - I). If we interpret subscripts mod e, then h(x) '" L;:.d a,qxiqmod(x' - I) since iq, i ~ 0,1, ... , e - 1, runs through all residues

ractorintion of Polynomials

mod e as q and e are relatively prime. Since h(x)q = L~-:-ri a,x,q, we get e

L

I

u,x ,q =.

,..,. 0

('

L

I

a,qxlqrnod(x'" -1).

,.,- 0

By considering the exponents mod e, it follows that corresponding coefficients are identical. Thus Q I = a,q for all i, and so a, = a'if = Q ,q 2 = ... for all i. Since tn, is the least positive integer for which (4.11) holds, we ohtain h(x)==

I: a,R,(x)mod(x'-I), I")

where the set J contains exactly one representative from each equivalence class of rcsidues mod e determined by the equivalence relation - which is dcfined by i l - i, if and only if i l == i,q'mod e for some t;" O. Thus, for suitable b, ElF q we havc , - I

h(x)==

I: b,R,(x)mod(x'-I).

(4.12)

1=0

Let now [1(.<) and [,(x) be two distinct monic irreduciblc factors of [(x), and so of x' - I. By the argument leading to (4.3), there is a solution hex) E IFq[x] of hex)" == h(x)mod(x' -I), deg(h(x)) < e, with h(x)==Omod[l(x),

h(x) == I mod[,(x).

(4.13)

Since Ri==R,mod[, the argument subsequent to (4.3) shows that there exist c1cments Cd' C" ElF q with Ri(x) == Cd mod [I(X), R i(X) == cil mod [,(x) for 0 ~ i ~ e - I. If we had cl1 = Ci2 for 0 ~ i ~ e - I, then it would follow from (4.12) that hex) == C mod [,(x), h(x) == c mod [,(x) for some C E IF q' a contradiction to (4.13). Thus cl1 =1= el2 for some i with 0 ~ i ~ e - 1, and since Ro(x) ~ I. we must have i;" I. Then R,(x)- Cd will be divisible by [I(X), but not by [,(x). Hence the use of this R,(x) in (4.2) will separate [,(x) from[,(x). 0 The argument in the proof of Theorem 4.5 shows, of course, that the polynomials R i , with i running through the nonzero elements of the set J, are already separating all irreducible factors of f. However, the determination of the set J depends on knowing the order e, and a direct ealeulation of e (i.e., one that does not have recourse to the canonical factorization of will be lengthy in most eascs. This problem docs not arise in the special cases [(x) ~ x'-I and [(x) ~ Q,(x), the eth cyclotomic polynomial, since it is trivial that ord(x' -I) ~ ord(Q,(x)) ~ e. The polynomials R, are, in fact, well suited for factoring these binomials and cyclotomic polynomials.

n

2. I-"aeloriLalion over Large rinik'l:idds

139

4.6. Example. We dctcrmine the canonical factorization of thc cyclotomic polynomial Q,,(x) in D',[x]. According to Theorcm 3.27 wc have

(x" -I)(x' -·1)

Qs2(X)~ (x"-I)(x 4 -1) = X 24

_ X 22

-x IO

+ x 20

_ XiI!

+X 8 _ X O

x9

X 27

+ x1

X l4 T

(> _

rx 4 -x 2

"1"

x l2

I.

X 243 ,

and since x 2(>:;:::;: - I mod Q,,(x), we get R I(X) '= 0 mod Q,,(x). so that R I is not Q,,-redueing. With R,(x) ~ x' + x' + x" wc get ~ow

RI(x) = x

+

x·~

+

+

gcd( Q" (x). R 2 (x)) ~ x' - x' gcd(Qdx), R,(x)+ I)

= x'

~ x

4

+ Xl(I + +I. x'

-

gcd( Qs2 (x), R 1 (x) - 1) ~ X 12 + x 'O

-

+ I.

x' ~ x' .,. x 4 + x' + 1"' 1(( x),

say, so that (4.2) yields

Q" (x ) ~ (x' - X2 + 1)( x' + x 4

-

X2 + 1) I( (x),

By Theorem 2.47(ii), Q,,(x) is thc product of four irrcducihlc factors in D'3lx] of degree 6. Thus, it remains to factor g(x). Sincc R 3(x) ~ x 3 + x 9 +

+ X 243 + x 72Y == OmodQ52(X)' we next use R 4(x) = x 4 + X I2 + x 30 , We note that x 12 '= - x lO + x' - x 6 - x 4 - x' .. I mod g(x), x l6 '=

x 27 + X!ll

-xlOmodl((x), and so R4(X)'=X'O+X8_X6_X' -I modl((x). Thereforc, gcd( g(x), R 4 (x)) = gcd( 1(( x), x 'O

+ x' - x 6 - x' -

1) ~ 1,

gcd( I((x), R 4(x) + 1) ~ gcd( g( x), x 'O + x' - x 6 - x') ~ x' - x 4 + x' - I. gcd(I((x), R 4(x )-1) ~ gcd(g(x). x 'O + x' - x' - x' + I) ~ x" - x 4 + I. Thus,

Q,,(x) ~ (x' - x' + I)(x' +x 4 - x' + l)(x' - x 4 + x' + l)(x 6 . x 4

,

1)

is the desired canonical factorization.

2.

FACTORIZATION OVER LARGE FINITE FIEI.DS

If D' q is a finitc field with a large number q of c1cmcnts, thc practical implementation of thc methods in the previous section will become morc difficult. We may still be able to find an [-reducing polynomial with a reasonable effort, but a direct application of thc basic formula (4.2) will he

Fnctorizalion of Polynomials

140

prohlematic since it requires the calculation of q greatest common divisors. Thus, to make the usc of j-reducing polynomials feasiblc for large finite fields, it is imperative that we devise ways of reducing the number of elements c E IF q for which the greatest common divisor in (4,2) needs to be calculated, We note that in the context of factorization we consider q to be "large" if q is (suhstantially) bigger than the degree of the polynomial to bc factored. Let j again be a monic polynomial in IF q[x J with no repeated factors, let deg(j) ~ n, and let k be the number of distinct monic irrcducible factors of f. Suppose that hE IFq[x] satisfies h q '" hmodj and 0 < deg(h) < n, so that h is I-reducing. Since the various greatest common divisors in (4.2) arc pairwise relatively prime, it is elcar that at most k of these greatest common divisors will be '" I. The problem is to find an a priori characterization of those cEIF. for which gcd(j(x),h(x)-c)'" I. One such characterization can be obtained hy using thc theory of resultants (see Definition 1.93 and the remarks following it). Let R(j(x), h(x)- c) be the resultant of j(x) and h(x)- c, where the degrees of the two polynomials are taken as the formal degrees in the definition of the resultant. Then gcd(j(x), h(x)- c) '" I if and only if R(j(x), h(x)- c) ~ O. We are thus led to consider F(y)~

R(j(x),h(x)- y),

which, from the representation of the resultant as a detcrminant, is seen to be a polynomial iny of degree", n. Then we bave gcd(j(x), h(x)- c) ~ I if and only if c is a root of F( y) in IF q' The polynomial F(y) may be calculated from the definition, which involves the evaluation of a determinant of order.:::;;; 2n - 1 whose entries arc either elemcnts of IF. or linear polynomials in y. In many eascs it will, however, be preferable to use the following method. Choose n + I distinct clements co' c\, .... C, E IF q and ealeulatc the rcsultantsr, ~ R(j(x). h(x) - c,) for 0", i '" n. Then the unique polynomial F(y) of degree", n with F(c,) ~ r, for 0", i '" n is obtained from the Lagrange interpolation formula (sec Theorem 1.71). This method has the advantage that if any of the r, are 0, we automatically get roOlS of thc polynomial F( y) in IF q' At any rate, the qucstion of isolating the elcments c E IF q with gcd(j(x), h(x)- c) '" I is now reduced to that of finding the roots of a polynomial in IFq' Computational methods for dealing with this problcm will bc discussed in the next section. Factor j(x) = x' - 3x 5 + 5x 4 - 9x' - 5x' + 6x + 7 over IF". Since gcd(j(x),j'(x)) = I.j(x) has no rcpeated factors. We proceed by Berlckamp's algorithm and ealeulatc x";modj(x) for 0 '" i '" 5. This yields

4,7,

Example,

2. I"al.:ton/.ation over Largl.: Finite Field!'>

141

the 6 x 6 matrix

R

I 5 -10

~,

o

0 0 10

o

7

o\

o

0

R

-10

10

o

-3 I

-8

10

7 9

7 2

9 - II 2

-I

II

0

9 - 4

-3

0

-10

-- 9 ;

and thus B-1 is given by

R- I

0 5 -10 0

~

0

0

0

0

-I

-I

8

10

9 9 -4 -10

0 -9

-3 I

0 10 -9

10

-II

7

0 0

2 2 -10 ; Reduction to column echelon form shows that R - I has rank r ~ 3. so that I has k ~ 6 - r ~ 3 distinct monic irreducible factors in IF,,lx 1- A basis for the null space of R - I is given by the vectors h, ~ (1,0,0,0.0,0), h, ~ (0,4,2,1.0,0), h, ~ (0. - 2.9.0, I, I), which correspond to the polynomials h,(x)-Lh,(x)~x3+2x'1-4x.h,(x)~x'+x4+9x'2x. We take the I-reducing polynomial h,(x) and consider II - 3

6

7

9

l-(y)~R{f(x).h,(x)- y)

16 0

~I~

,0

I~

'0

-3 I

5

0 2

-3 I 4

I

2

0 0 0 0

I

0 0 0

9 5 -3

-y 4 2 I 0 0

-5 -9 5 0 -y

6

-5 -9 0 0

4 2 I' 0

Y 4 2 I

7 6

0 7 6

-5 0 0 0 -y 4 2

~I 7

0 0 0

0

0

0

-.I"

4

~l'

-~I

In this case a direct computation of F( y) is feasible. and we obtain F(y)~y'+4y'+3v4-7v·'+IOy'+lly+7.Since I has three distinct monic irreducible factors in 1F,.,[xj. the polynomial Fean have at most three roots in IF". By using either the methods to he discussed in the next section or trial and error, one determines the roots of F in IF" to he - 3. 2. and 6.

Furthc.:rmore. ged{f(x).h,(x)-3) -x -4, ged{f(x),,,,(x)-2)~.\'-.\17.

ged{f( x). ",(x )-6) ~ x'

+ 2x' + 4x - 6,

Fat:torization of Polynomials

so that f{ X) ~ (X .- 4)( x' - x -7)( x) + lx' ~ 4x - 6)

is thc canonical factorization off(x) in IFdxl.

D

Another method of eharaeterizjng thc elements c E IF q for which the greatest common divisors in (4.2) need to bc calculated is hascd on the following considerations. With the notation as ahovc, Ict C be the set of all c '" IF q such that ged(f(x), h(x)- c) * 1. Then (4.2) implies f{x)~

n gcd(f{x),h{x)-c),

(4.14)

( cC

and so fIx) divides nc~c(h(x)- c). We introduec the polynomial G{y)~

n Lv-c).

r" C

Then fIx) dividcs G(h(x» and the polynomial G(y) may be characterized a, follows. 4.11. Theorem. Among 0/1 the polynomial, g E IF q[Y 1such that f(x) divides g(h(x)), the polynomial G(y) is the unique monic polynomial of least degree. Proof We have already shown that the monic polynom;al G(y) is such that fIx) divides G(h(x)). It is easily seen that the polynomials g E IFq[Yl with fIx) dividing g(h(x)) form a nonzero ideal of IFq[y]. By Theorem 1.54, this ideal is a principal ideal generated by a uniquely determined monic polynomial Go E IFq[y]. It follows that Go(y) divides G(y), and so Go{Y)~

n

(Y-c)

ce::C 1

for some subset C, of C. Furthermore, fIx) divides Go(h(x»

n,., c,(h(x)-c), and hence f{x) =

~

n gcd(f{x), h{x)-c). cE:i::C 1

A comparison with (4.14) shows that C, the theorem follows.

~

C. Therefore Go(y) = G(y), and D

This result is applied in the following manner. Let m be the number of clemcots of the set C. Then we write G( y ) =

m

n (y - c) ~ rE C

L I

0

bJ y}

143

2. I-"at'lorilation over 1:ar/z'l' Flnill' Fit::1ds

with coefficient' b,

EO

IF,. !':ow J(x} divides G(h(x». so that we have

L"' J

hjh(x)J:oOmodJ(x). I)

Since bnj = I. this may be viewed as a nomriviallinear dependence relation over IF" of the residues of L h(x). h(x)' ....• h(xj"'modJ(x). Thcorem 4.8 says that with the normalization bm = I this linear dep~ndcnc~ re\(.ltion is unique. and that the rc,idue, of I.h(x}.h(x)' ..... h(x)"' 'modJ(x) are k follows from (4.14). linearly independent over IF,. The bound The polynomial G can thus be determined by calculating the residues modJ(x) of I. h(x). h(x}' .... until wc findthc smalle't power of h(x) that i, linearly depcndcnt (over IF,) on it, predecessors. Thc coefficient' of this first linear dependence relation. in the normalized form, ar~ the coefficients of G. We know that we need not go heyond h(x)' to find this linear depende)1ce relation. and k can he ohtuined from Ikrlekamp's algorithm. The clements of C are now precisely the roots of thc polynomial G. This mcthod of reducing the problem of finding the element' of C to that of calculating the roots of a polynomial in IF q is called the Zussellhulls ulgo-

In"

rifllm.

4.9. Example. Consider again thc polynomial J E 1F2J[x] from Example 4.7. From Berlekamp's algorithm we obtained k ~ 3 and the J-reducing polynomial h(x) ~ xl + 2x' ... 4x E IF ,.,[x]. We apply thc Zassenhaus algorithm in order to determine the clement' (' ElF" for which ged(J( x). h(x}- c} '" L We have h(x) 2

modJ(x).

=: -

4

<

1

h(x) =7x'+7x T2x -2x -6x-7modJ(x).

and '0 it i, clear that h(xl' is not lincarly dependent on I and h(x). Therefore, h(x)" mu,t bc the smallest power of h(x) that is linearly dependent on its predecessors. We have h(x)""'-llx'

Ilx 4 -x'

9x'

5x·-2modJ(x).

and the linear dependence relation is h (x)' - 5h (x)' -'- Ilh (x) - 10 '"

amod J( x).

so that G(y) - v l . 5y'llly' 10. By using either the method, to be discu~s~d in th~ next section or trial and error, one determines the roots of G to he -3,2, and 6. The canonical factorization of J in IF,,,[x] is then ::J ohtained as in thc last part of Examplc 4.7. A method that is conceptually more complicated. but of great theoretical interest, is based on the use of matrices of polynomials. By a

144

nlmrix oj polynomials we I' qlx].

m~an

here a matrix whose entries arc clements of

4.10. Definition. A square matrix of polynomials is called /lOllsingular if its determinant is a nonzero pOlynomial. and it is called unimodular if its determinant is a nonLero clement of IF q"

4.11. Definition. Two square matrices P and Q of polynomials are said to he equivalent if there exists a unimodular matrix U of polynomials and a nonsingular matrix F: with entries in I' q such that P ~ UQf:. It is easily verified that this notion of equivalence is an equivalence

relation, in the scn~c that it is rcncxivc, symmetric, and transitive. We have seen in Section I that there arc polynomials h" .... h, E I'qlx] with 0 < deg(h) < deg(f) for 2 <; i <; k. which together with h, ~ 1 are solutions of h q '= h mod f that arc linearly independent over I' q' Clearly, the polynomials h, may be taken to be monic. The following theorem is fundamental. 4.12, Theorem, Lei f=fl" 'f" where fl, ....f, are distinct monic irreducihle polynomials in I'q[x], and let h, .... ,h, GO I'q[x] he monic pozv/lomials with 0 < deg(h;) < deg(/) for 2 <; i <; k, which together with hi ~ 1 are solution> of h q =: h mod f that are linearly independent over I' q' Then the diagonal matrix of polynomials

f,

0 0

o o

0

[,

()

0

0

, fl

0

0 0 0

D=

f, '

is equivalent to the matrix of polynomials

I f A~

h, h)

"- I

0 0

0

-I

h,

0

()

0

O'

o o -I

Proof By the argument following (4.3) we have h, (x) '= e'l mod f;(x) with e'J C' 1"1 for 1 <:: i,j <:: k. Let F: be the k X k matrix whose (i, j) entry is elj" We shuw first that E is nonsingular. Otherwise. there would exist

145

2. FaCIOri7.ation ovcr I ,i1Tge "ini\t' Fkld:-

clements d" ... ,d, ElF q' not all zero, such that

,

Ld/ejj=O ,~

This implies that

,

forl~j~k.

, L

d;h, '" OmodJ;

for 1,; j", k,

i -1

and '" I:7 ,d,h, '" Omodf. Since deg(h,) < deg(f) for I '" i '" k, it follows that i7_,d,h, ~ O. a contradiction to the linear independence of h, ... .,h,. !':cxt we note that A E is a nonsingular matrix of polynomials. Thus we can write D~(D(AE) ')IIE. so that the theorem is cstahlished once we have shown that U~ D(AE)-' is a unimodular matrix of polynomials. Let b" co IFqlx] he the (i. j) entry of AL·. Then b,; = Ie,} ~ I '" Omod~ ·for 1~.i~A, and for 2~i~k we have bij=h/e1j-e/,= h, - e" ~ OmodJ; for I,; j,; k, so that b" =Omod!,

for

I ';i.j,;k.

(4.15)

Now

"I (AL) ~ dei{AL') (B,}),<,.Jd

(-1)' ,

~ dei(i17(B,J,<;.,<"

where R" is the cofactor of the (j, i) entry in AE, and

. U=D(Al:.)

,

(-1)' ,

~ det(E)!(/,R,;),
Since (4.15) implies that B;J '" 0 mod(j/ /,), it follows that each entry of U is a polynomial over IF q' Furthermore, ( - I)' , det( D) det( U) . d~'tCA I:') ~ det( E) ,

which is a nonzero clement of IF q' Thus, U is a unimodular matrix of polynomials. 0 Theorem 4.12 leads to the theoretical possibility of determining the irreducible factors of I by diagonalizing the matrix A. The number k as well as the entries h" ... .,h, in the first column of II can he obtained with relative case hy Berlckamp's algorithm. The algorithm that achieves the diagonalization of A is, however, quite complicated. The dia!(onalization algnrirhm is hased on the use of the following e1emelltar)' operations: (i) permute any pair of rows (columns): (ii) multiply any row (column) hy an element of IF;: (iii) multiply some row (column) by a monomial (element of IF q) and add the result to any other row (column).

rac!oril.ation of Polynomials

146

The elementary row operations may he performed by multiplying the original matrix from the left by an appropriate unimodular matrix of polynomials. whereas the elementary column operations may be performed hy multiplying the original matrix from the right hy an appropriate nonsingular matrix with entries in I'q. Therefore. the new matrix obtained by any

of these elementary operations is equivalent

to

the original matrix.

One ean show that A is equivalent to a matrix R of polynomials with the property that for each row of R the degree of the diagonal entry is greater than the degrees of the other entries in the row. The matrix R can be computed from A hy performing at most (211 t· k - l)(k - I) elementary operations. where II ~ deg( h 2) 1- ••• + deg( h k)' We note that the diagonal entries of R can be permuted by carrying out suitable row and column permutations. We can thus obtain a matrix S

that. in addition to the property of R stated ahove. satisfies deg(s;,l;> deg(s}/) for I " i " j" k. where the s" are the diagonal entries of S. By multiplying if nece'Sary. we may assume the rows of S hy appropriate elements of that the.\" arc monic polynomials. A matrix S of polynomials with all these properties is called a llormali=ed matrix. The diagonal entries of the matrix D in Theorem 4.12 may also he arranged in such a way that deg(/,)? deg(/,) for I" i" j" k. The resulting equivalent matrix. which we again call D. is then diagonal and normaliled. Using thc fact that the normalized matrix S is equivalent to D. one can then show that deg(s,,) = deg(/,) for I" i " k. Thus. one can read off the degrees of the various irreducihle factors of! from the diagonal entries of S. Furthermore. if d is a positive integer which occurs as the degree of some sJi' and if S(d/ is the square submatrix of .\' whose main diagonal contains exactly all s" of degree d. then one can prove that the determinant of Sed' is equal to the determinant of the corresponding suhmatrix of D. Thus det(S,d')= lid' where gd is the product of all/, of degree d. In this way we are led to the partial faetori'ation

1';.

(4.16) where the product is over all positive integers d that occur as the degree of

some!,. In summary. we see that the matrix S can he used to ohtain the following information ahout the distinct monic irreducible factors of f: the degrees of these factors. the numher of these factors of given degree. and the product of all these factors of given degree. If the /, have distinct degrees. or. equivalently. if the s" have distinct degrees. then (4.16) represents already the canonical factorilation of! in I'qlx]. If (4.16) is not yet the canonical factori7.ation. then one can proceed in various ways. An ohvious option is the application of one of the methods discu'Sed earlier to factor the polynomials Rd' One can also continue with

2. factorization over Largl' Finite hd<.b

147

the diagonalization algorithm in order to obtain the diagonal matrix D equivalent to the normalized matrix S. For the latter purpose, we assume as above that D is put in normalized form. In addition to the properties mentioned above, it is then also true that each of the submatriccs S'd' is equivalent to the corresponding submatrix D,d' of D. It is thcrefore sufficicntlO diagonalizc eaeh of the submatrices S'd' separately. By thc equivalence of S'd' and D'd' wc have S'd} ~ UD'd'/:; for some unimodular matrix U of polynomials and some nonsingular matrix E with entries in IF q • We may then write

where the S;'il. D)dl. and Uf' 0 ~ r ~ d, O:s;; I ~ m, are matrices with entries in IF q , Um "" 0, and 5)/' = D)/l ~ /, the identity matrix of appropriate order. A comparison of thc matrix coefficients of the highest powers of x on both sides of thc equation S,dl = UD,dl/:; yields / ~ UmIE and m ~ O. Thus. U = Uo ~ L'- I and hcnee S,d' ~ E' (D'dlE. Comparing thc matrix coefficicnts of like powers of x in the last identity gives S}J) = E ID}JlF. for O:s;; r:s;; d. Consequently, S)d) and Dr(d) havc the same characteristic polynomial and eigenvalues. and since Did' is diagonal. its eigenvalucs are exactly its diagonal entrics. Therefore, the latter can be determined by finding the roots of the characteristic polynomial of Si dl . which must all be in IF q' As in the earlier methods. we have thus again reduced the factorization problem to that of finding the rOots of certain polynomials in IF q' The partial factorization (4.16) can also be obtained by an entirely different mcthod. To this end, we extend the definition of I:d by letting 1:" i;;, I, be the product of all monic irreducible polynomials in IFq[x] of degrcc i that divide f. In particular. I:,(x) ~ I in case f has no irreducible factor in IFqlx] of dcgrce i. We can thus write

It is trivial that only thosc i with i " dcg( /) need to be considered. Wc calculate now recursively the polynomials ",,(x). '(x),. .. and /'o(x), F(x).... as well as d(x). d,(x), .... Wc start with

148

and for i ~ I we use the formulas

r, (x) "" r, ,(x )qmod F, _,(x ).deg( r,) < deg( F,-,). d,(x) ~ gcdU; ,(x),r,(x)-x). r;(x) ~ r;_

b

)/d,(x).

The algorithm can he stopped when d,(x) 4.13. all i -;, I.

Theorem.

= F,

,(x).

With lhe notation abm;e. we have d,(x) = g,(x)!or

Proof Using the fact that F; divides F; tion shows that r,(x) "" xq'mod /.; ,(x)

I'

a straightforward induc-

for all i -;, I.

(4.17)

We prove now hy induction that

F,

,= n g)

and

d, ~ g,

for all i -;, I.

(4.18)

i"-"'" i

For i have

= I the first identity holds since Fo = f. As to the second identity. we d,(x) ~ ged(ro(x). r,(x )-x) = ged(j(x). x q - x)

by (4.17). and since x q - x is the product of all monic linear polynomials in !'q[xl. it follows that d, is the product of all monic linear polynomials in !' q[xl dividing!. and hence d, = g,. Now assume that (4.18) is shown for some i ~ 1. Then F,~r;_,/di~r;_,/g,~

n / ;;. I I

g).

(4.19)

I

which proves the first identity in (4.18) for i + I. Furthermore. d" ,( x) ~ ged( F, (x). r, , ,( x) - x) = ged( f; (x). x q'" - x)

by (4.17). According to Theorem 3.20. x q '" - x is the product of all monic irreducible polynomials in !'q[xl whose degrees divide i + I. Consequently. d, . , is the product of all monic irreducible polynomials in !' qf x 1that divide F, and whose degrees divide i + I. It follows then from (4.19) that d, _ , = g, , ,. D

In the algorithm above. the most complicated step from the viewpoint of calculation is that of obtaining r, by computing the qth power of r,_,mod r; ,. A common technique of cutting down the amount of ealcula· tion somewhat is based on computing first the residues mod /.; , of r,_I,r/ 1,r/~I.· ... rI2~1 by repeated squaring and reduction mod F; .1' where 2' is the largest power of 2 that is '" q. and then multiplying togcth~r an appropriate combination of these residues mod l; ,to ohtain the residue of

2. Factorization over Large rinite ridds

149

f; _l' For instance, to get the residue of ,/23 1mod F; I' one would multiply together the residues of fjl~ I' r,~ I' "~ I' and r, _Imod F; I. Instead of working with the repeated squaring technique. we could employ the matrix B from Bcrlekamp's algorithm in Section I to calculate r, from r,_,. We write n ~ deg(f) and f,q I mod

,

"

r j

-

1(x)

~ "i..J r'" , 1x' ' /=0

and define

(Sj(O),

s}1) .... ,Sj(n 1») E IF; hy the matrix identity

s(I) ( SIO) , ' I ,

••••

S~lI-I))= (r.(O) ,

/

I'

,.(1) j

,(TI

1····· I-I

I»)B '

(4.20)

where B i' the n x n matrix in (4.5). With

" Sj(x) =

L

,

(4.21 )

S,(Il X l

/ .., 0

we get then r; ,(x)q=s,(x)mod!(x). hence r, ,(x)q=s,(x)modF; ,(x). and thus r;(x) '" s,(x ) mod F; _, (x)

Therefore. once the matrix B has been calculated, we compute r, from r, , in each step by reduction mod f;_, of the polynomial s, obtained from (4.20) and (4.21). 4.14. Example. We consider !(x) ~ x' - 3x' 7 E 1F2J[x] as in Example 4.7. Then

B=

o

+ 5x·

- 9k' - 5x'

I

0

0

o

0

5 -10

0

-I

8

-3

-10

10 7

0

I

-9

-8 7

9

10 7 2

-II 2 -9

II

0

10 9 --4

-3

0

-10

o

+ ox +

We start the algorithm with ro(x) ~ x. f;,(x) ~ !(x). From (4.20) and (4.21) we get s,(x) ~ - lOx' - 3x· + 8x J - x' + 5. and reduction mod f;'(x) yields r,(x) = s,(x). By Theorem 4.13 we have X,(x) ~ d,(x) = gcd(f;'(x), r,(x)-x)~x-4. Furthermore, F,(x)= fo(x)jd,(x)~x'+x·+9x3+ 4x' + Ilx t 4. In the 'econd iteration, we usc again (4.20) and (4.21) to obtain ",(x) ~ 5x' -8x· +9x J -lOx' -II, and reduction mod F,(x) leads to r,(x)~IOx·+IOk'-7x'-9x-'8. By Theorem 4.13 we have g,.<x)= d,(x) ~ gcd(F,(x), r,(x)- x)·= x' - x 17. Furthermore. f,(x) = F,( x )/d,( x) ~. x J + 2x' -t 4x - 6. But, according to the first part of (4.18). all irreducible factors of fi(x) have degree? 3, '0 thot F,.(x) itself must he

ral.:!or.zation of Polynomiab

150

irreducible in n=,',3[X] and

1(3(x)~

1',(x), Thus. we arrive at Ihe partial

factorization !(X)~(X-4)(X2_X+7)(X312x2+4x-6).

which. in this case. is already the canonical factorization of !(x) in n= ,,[ x 1. LJ

3.

CALCULATIO"l OF ROOTS OF POLY"IOVIIALS

We have seen in the preceding section that the problem of determining the canonical factorization of a polynomial can often he reduced to that of finding the roots of an auxiliary polynomial in a finite field. The calculation of roots of a polynomial is. of course. a matter of independent interest as well. In general. on~ will be interested in determining the roots of a polynomial in an extension of the field from which the coefficients arc taken. However, it suffices to consider the situation in which we arc asked to find the roots of a polynomial! '= n= q[x] of positive degree in n=q. since a polynomial over a suhfield can always he viewed as a polynomial over n= q' It is clear that every factorilation algorithm is. in particular. a root-finding algorithm since the roots of! in n= q can he read off from the linear factors that occur in the canonical factorization of! in n=q[x]. Thus. the algorithms presented in the earlier sections of this chapter can also he used for the determination of roots. However. these algorithms will often not be the most efficient procedures for the more specialized task of calculating roots. Therefore. we shall discuss methods that are better suited to this particular purpose. As a first step. one may isolate that part of! which contains the roots of! in n= q' This is achieved by calculating gcd(j(x). x q - x), Since x q - x is the product of all monic linear polynomials in n= qlx I. this greatest common divisor is the product of all monic linear polynomials over n= q dividing!. and so its roots are preciscly the roots of! in n= q' Therefore. we may assume. without loss of generality. that the polynomial for which we want to find the roots in IF
A useful method of finding roots of polynomials was already discussed in Chapter 3. Section 4. It is based on the determination of an affine multiple of the given polynomial. See Example 3.55 for an illustration of this method. In order to arrive at mhcr methods, we consider firM the case of a prime field n=p, As we have seen above. it suffices to deal with polynomials of the form !(x)~

n (x-c,).

,

I

3. Cakulatilln of I{oot:- of Polynomial.;

lSI

where c, ....• c" arc distinct elements of IF p • If p is small. then il is feasible to determine the roots of f by trial and error. that is. hy simply calculating f(O),/( I).... .f( P - I). For large p the following method may be employed. For bE IF p' P odd. we consider f(x-b)~

n" (x-(b+c,»).

,

I

We note thatf(x - b) divides x P - x ~ x(X,p-I)/2 + I)(x" 1)/2 -I). If x is a factor of f( x - b), then f( - b) ~ 0 and a root of f has been found. If x is not a factor of fix - b), then we have

fix - b)

~

ged(f(x - b). x'p-I)/'

+ I)ged(f(x -

b), x,p-1)/2 - I). (4.22)

The identity (4.22) is now used as follows. We calculate the residue modf(x - b) of x'P 'lI'-for example, hy the repeated squaring technique discussed after Theorem 4.13. If X,p-11/2", ± I modf(x - b), then (4.22) yields a nontrivial partial factorization of fix - b). Replacing x by x + b. we get then a nontrivial partial factorization of fix). In the rather unlikely case where x" - 1)/2 '" ± I mod f( x - b), we try another value of b. Thus. by using. if necessary. several choices for b, we will find either a root of f or a nontrivial partial factorization of f. Continuing this process. we will eventually obtain all the roots off. It should be noted that. strictly speaking. this is not a deterministic, but a probabilistic root-finding algorithm, as it depends on the random selection of several elements b E IF p' 4.15. Example. Find the roots of f( x) = x' - 7x' + 3x 4 - 7x' + 4x' x -2EIF,,[x] contained in IF". The roots of f(x) in IF" arc precisely the roots of g(x)~ged(j(x).x"-x) in IF". By the Euclidean algorithm we obtain g(x) ~ x 4 + 6x' - 5x' + 7x - 2. To find the roots of g( x), we usc the algorithm above and first sclect h ~ O. A straightforward calculation yields x' p Ill' ~ x' '" I mod g( x), and so this value of h does not afford a nontrivial partial factorization of g(x). I\ext we choose b = 1. Then g(x - I) = x 4 +2x' -3x - 2 and x' '" -4x' -7x' + 8x -5modg(x - 1). so that b = I yields a nontrivial partial factorization of g(x -I). We have

gcd(g(x-I),x'

+ I) ~ged(x4+2x' -3x -2. -4x' -7x' +8x -4)

and ged(g(x-I).x' -1)~gcd(x4 +2x' -3x-2. -4x'-7x' -r8x-6) ~X2

-8x+8.

152

ractorization (If

Pt)lY:1om~als

hence (4.22) implies g(x-I)~(x'·-7xT4)(x'··8x-,-8),

which leads to the partial factorization g(x)~(x'-5x-2)(x'

6x-I)· 8,(x)g,(x),

say. In order to factor 8,(x) and 8,(X), wc try h ~ 2. We have 8,(x - 2) ~ x' + 8x - 5 and x' = - 8x -t 2mod 8,(x - 2). I'urthermore, gcd( 8, (x - 2), x' -+ I) ~ gcd( x' + 8x - 5, - 8x ~ 3) ~ x -,- 6, and long division yields 8,(X - 2) =

(x ~6)(x ~

8,(X) ~ (x

2), so that

+ 8)(x +4).

Turning to 8,(x), we have g,(x - 2) ~ x' + 7x ~ of 8,(x) and

XIX

+ 7), thus - 2 is a root

8,(X)~(x+2)(x-8).

Comhining these factorizations, we get 8(X)~(x+8)(xT4)(x·,2)(x-8).

Therefore, the roots of 8(x), and thus of f(x), in 1'17 are - 8, - 4. - 2, 8.

D

Next we discuss a root-finding algorithm for large finite fields I' q with small characteristic p. As before, it suffices to consider the case where

"

f(x)=n(x-y,)

,

with distinct clements y,,. . ., y"

E

-,

Fq' Let q = p'" and define the polynomial m

SIx) ~

,

L I ~

x P' 0

We note that for yE:1' II we have S(y)~Tr, a(y)01'", where Tr, q is the ahsolute trace function (sec Definition 2.22). Because of Theorem 2.23(iii), the equation .'i( y) = c has pm' , wluti,ms y '= I' q for every c E Fp' and this ohservation leads to the identity I"

xq

x~

n (S(x)-c). ,'f::

(4.23 )

Fp

Sineef(x) divides x" - x, we get

II

c, l

(S(x)

c)=Omodf(x),

p

and so f(x)c

nr ged(f(x),S(x)-i).

(" (=-

p

(4.24)

.1.

Calculation of Roots of Polvnomiab

Ij.]

This yields a partial factorization of f(x) that calls for thc cakulation of p greatest common divisors. If p is small. this is certainly a feasihle method. It can. howcver. happen that the factorization in (4.24) is trivial- ·-namely. precisely whcn S(x) '" Cmod f(x) for some cEO"f" In this case. othcr auxiliary polynomials relatcd to S(x) have to he used. Let fJ be a defining elemcnt of I'q over I'p, so that (l.fJ.fJ' .... ,flm I) is a hasis of 1'. ovcr I'p. For} ~ O.1.. ... m - I we suhstitute {Jix for x in (4.23) and we get ({Ji)qx"-{Jix~

n (S(fJIX)-C).

c E:.lFp

Sincc WI)q

~

{Ji, we obtain xq-x=fJ

'n (S(fJix)-c). l:' J p

This yields the following gencralization of (4.24): f(x)~

n

gcd(J(x).S(fJJx)-c)

forO.,;;}.;;m-1.

(4.25)

,ElF p

Wc show now that if n ~ dcg( /) '" 2, then there exists at kast one }, O.;;}.;; In -1, for which the partial factorization in (4.25) is nontrivial. For suppose, on the contrary, that all the partial factorizations in (4.25) arc trivial. Then for each}, O.;;}.;; In - I, there exists a cJ E I'p with

.\.( {Jix) '" ci modf( x) .. In particular, we get

S( {JiYI)

=

S( fJly,)

~ ci

for 0 .;; } .;; m - 1.

By the linearity of thc trace it follows that Tr. ,(( YI - y,){Ji) = 0

for O';;},;; m - 1

and

Using the second part of Theorem 2.24, wc concludc that YI - Y2 ~ O. which is a contradiction. Thus, for at least onc} thc partial factorization in (4.25)

is nontrivial. The defining element fJ of I'q ovcr I'p used in .(4.25) is choscn as a root of a known irreducible polynomial in I'p[xl of degree m. Once a nontrivial factorization of the form (4.25) has bccn found. the method is applicd to the nontrivial factors by employing othcr values of j. The argument above shows also that all distinct roOis of f can cventually hc separated by using all the values of} in (4.25).

154

I-actori/i.llion of Polynomials

4.16. Example. Consider IF" '" [F,.Ul). where {3 i, a root of the irredueihle polynomial XU + x -'-I in 1I'.,[x]. and let f( x) = x' -'- ({3' T fl' r fl"

T

{3' )x 3

+ ({3' , {3' + {3' -'- {3 + 1lx'

T(!J'-fl'.,.fl)·,.,.fl't{3<:IF 64 lx]. U~ing

x"", (fl' T fl -t

r

l)x 3 ,(fl' "{3"

{3')x' +(P' -'- {3'1 {3' + 1)x

P'+P'+p'-'-lmodf(x).

we get the following congruences modf(x) hy repeated squaring: x

--;;

x

x·1 ':""(/~~

•

x

(f3~ I /14 ~ /12 .

{f34. /1 3 _fJ))x\'

xl>.., lt

{{4" fJ3 ~ /~:1~3;

l/~)

'J{' , 11x11

N"' -

t

/1)'(. /1J I /~

(fJ~ rfi· 1)'(: '{/1~-r/~' llx -/1~' fJ~

(fi~,/~l_/~lxl_

,.

x 3Z .....

/1' 1)x24-(/f~

(/~)-f3)xJ~fJ~){ltJ':1J~3.{{I-/1'1

(f3~

j

~ 13 2 .

{{4 - {{3

H

t

l)x 2 '(f34

1

fJ'l_t·

W I /1'

, Thus.f(x) divid~s X M

-

x and so has four distinct roots in 1F(J4" We consider

now S(x) =.'(, x 2 +.'(4 t xl!·\

X

l6

-\

x J2 . from the congrucncc:~ above we

ohtuin Six)

=.

({3' + P' + fl' + {3·1 I)x'" {3'x' , ({33

i

{32)X

T fl3 , {3' -r I mOdf(x). and therefore ged(j(x).S(x))~ged(j(x).(fis+fl'I {3'-t {3-'-I)x J +/i'x' ~(fl'

I {32)X

tfl'

i-{3' 11)

~x' t-({34-'-fl'r{32),,'-r(f3'+p'-rI)x+{3'" {3'~K(')

say. and ged(j(x).S(x) -1)" ged(f(x).({3'-r{3'-r{3'

I

{3-I)x'+{3'x'

"(fl' "{32)""'P'+fl')=x

i

{35

Then (4.24) yields f(x)~

g(x)(xTfl')·

To find the roots of g( x). we next usc (4.25) with j

S( flx)

(4.26 ) ~

~

I. We have

{3xr {32 x' + {34 x 4 + fl'x' + p'ox'o -t {3"x" " {3x -'- {3' X 2 T {34 x ' + ( {3' + {32 ) XX ... ( {3"+ {3 +

1) X '0 -'- ({3' + I) x" .

155

and the congruences above yield

S(f3x)=((I'+llx'-(f3'+(I+I)x'-(f3'+f3"

f3'

I

13' .f3+l)x

4

_f3 +f3'-'-f3mndf(x). Since g( x) divides I( x). this congruence holds also mod g(x l. and '" .'i ( f3x ) " ((I'

+ 1).\' - ( 13' + 13 + 1) x' - (f3' I f3' + 13'

+ 13'

f

I-

13) - 13 + I) x

13'1 f3

" (13' + 13' )x' - 13 3X ~ 13 '" f3' . f3 mod g (x). Thu.. . , gcd(g(x).S(f3x))-gcd(g(x).(f3'

I

f3')x'-f3'x-t

~x'-(flJ+I)x-f34-f3''13'

fl'

j

f33'fl)

'f3=h(x).

say. and gcd(g(x).S(f3x)-I)~gcd(g(x).(f!'·f32).\.' .f33 X +f3'-'-f33+f3+l) ~

Then (4.25) with)

~ I

x _ 13' - f3' + J.

yields

(4.27) Tn find the ronts of h(x). we usc (4.25) with) ~ 2. We have

5( 13 'x ) ~ f3'x + f3'x' + f3'x' + f3"x' • f3"x" -t f3"x" 13 'x + (I'x' + (f3J -'- f3')x' +(13' + 13 -I)x' f (13 3 - I )x" _ f3x·\l. and a similar calculation as for S( f3x) yields

S( 13 '.\ ) '" (fl'

f

13 2 + I) x - 13 5 + 13 ' + f3' mod h ( x ).

Therefore. gcd(h(x).S(f3'X»)~gcd(h(x).(f3'+f3)f I)X~f3'+f33 ~x-f3-

i

fl2)

I

and ged(h(x).S(f3'x)-I)~gcd(h(x).(f3'+f3'+I)x4 13'· 13.\+(1'+1)

-x "13 3 -13. so that from (4.25) with)

~

2 we get

h (x) - (x

i

f3

I

1)( x ~ 13' + 13).

(4.2X)

raclOr:z.:lIion of Polynomiab

15(,

Comhining (4.26). (4.27). and (4.28). we arrive at the factorization f( x ) ~ ( x

+ f3 - I) (x

f3' -'- fl)( x - f34 + f3' + 1)( x -'- f3' ).

0

f3 - I. fJ3

and so the roots of f(x) are

-r

f3. f34 + fl' + I. and f35

[]

Finally we consider the root-finding problem for large finite fields" 4 with large characteristic p. As we have seen hefore. it suffices to know how to treat polynomials of the form f(x)~n(x-y,)

E"

i=1

with distinct elements y,,, ... y" q' To cheek whether f(x) has this form. we need only verify the congruence x q " xmodf(x) (compare with the first part of Example 4.16). We can assume that q is the least power of p for which this holds. The polynomial f(x) will. of course. be given by its standard representation

L "/'.

fix)'

i -'- 0

E IF'I for 0 ~ j ~ 1l and an = 1. It will he our first aim to find a nontrivial factor of f(x). To exelude a trivial <..:ase. we can assume 1l ~ 2. I.et q = pt>J and define the polynomials

where

0.,

n

f,(X)~LIX{xi

(4.29)

forO,;;',;;m-l.

i ... 0

so that fo(x) Furthermore.

~

f(x) and each f,(x) is a monic polynomial over {I/

f/

f, ( y,P')

~

L

IX;' YiP'

~I

J' 0

L

\

IX I yi )

pA

~

\ I _. 0

"q.

°

for I ~ i ~ n. O::s; k ::s; m - 1. and so n

f, (x) ~

n (x - Yo")

for 0 ,;; , ,;;

In -

I.

i-I

We ealeulate now the polynomial m-l

F(x)~

[J f,('), /..

-

(4.30)

()

This is a polynomial over IF p sin<..:e

F(x)~

min

n

II (x-y()=

k· O/-"l

n n (x-y'p')~ fU;(x)m1d. I/m-]

/",.1 k

0

n

1-1

where l'~'( x) is the minimal polynomial of y, over" p and d, is its degree (compare with the discussion following Definition 2.22). The F,(x) arc

Y. CaklJl,l1ion of RON" of Polynomials

157

therdore the irreducihle factof' of F(x) in Fplx], hut certain F,(x) could he identical. Thus, the canonical factorization of F(x) in Fplx] has the form F(x) ~G\(x)··· G,(x),

I"

(4.31)

where the C,(x), I" r, arc powers of the distinct ;-;(x). This canonical factorization can he ohtaincd by one of the factorization algorithms in Section 2 of this chapter. Since /(x) ~ /o(x) divides F(x), it follows from (4.31) that

n gcd(f(x),G,(x».

/(x)~

(4.32)

[=1

In most cases, (4.32) will provide a nontrivial panial factorization of /(x). The factorization will be trivial precisely if gcd(/(x),G,(x)) = /(x) for some I, I" r. which is equivalent to r ~ I and/(x) dividing F\(x). A

t"

comparison of degrees shows then n:s:;;: d l = m. Furthermore. the roots of J(x) arc then all conjugate with respect to FP' Thus. hy labelling the roots of /( x) suitably, we can write y,

=

y("

for

I::;;,;

i.:::;;;

n, with 0 -= b l < b 2 < ... < bn < m.

W..: sct hn t I -= m and d~

min (b,. \ - b,). 1";;'1"" n

It is clear that d (A)

(B)

~

111/n. The following two

p()~sibilitics

can occur:

h,. I hi > d for some i. I ~ i ~ n; b,. \ - b, ~ d for all i, I" i " n.

In case (A) we note that the set of roots of [(x) is

(y(',yr" ..... y(.) and the set of roots of /d( x) is

The condition in (A) implies that these two ~Cb of roots are not identical. On thc oth..:r hand. sinl:c hi. I - h, = d for somc i, I ~ i ~ n, the two sets of roots have a common elemcnt. Thus, gcd(/(x), f:/(X»f /(x) and ~ I; that is. gcd
•

=

y{

d,·

1,

for I ~ i ::;;,; n.

I-"actorilation of Polynomial~

15R

hence thc Y, arc exactly all thc conjugatcs of YI with respect to 0'r" Consequently,.!(x) is thc minimal polynomial of y, over O'p' and thus irredudhlc over IF pd.

Thcrefore, corresponding to the cases (A) and (ll) above we have thc following alternativcs: (A)

(ll)

gcd(j(x),j,(x)) is a nontrivial factor of f(x) for some k,l.,;k<m/II: gcd(j(x), f,(x)) ~ I for I .,; k < d = min EN andf(x) = falX) is the minimal polynomial of YI over 0' p"

In alternative (A)

OUf

aim of finding a nontrivial factor of !(x) has

bcen achicved. Furthcr work is needcd in alternative (Il). Let f3 again dcnote a defining element of 0', over 0' P' Then I' p'( f3) ~ 0' q ~ 0'r.' and so /3 is of degrec mid ~ n over lOr" In particular, wc havc f3' '" lOp' for I.,; j.,; n-1. Now let thc coefficients a, of f{x) be such that a" = 0 for some jl1 with I.:::; Jo ~ n -I. Consider f(x)~f3

nf(f3x),

(4.33)

which is a monic polynomial of dcgree n ovcr 0',. Since f3"'j., '" lOr' and a j.l E 1"" it _follows that the coefficicnt of Xl., in f(x) is not an elemcnt of p O'p" Thus fIx) is not a polynomial over lOr" and so .the altcrnative (ll) cannot occur if thc proccdure above is applied to f(x). Since f(x)·' fl"f{ f3 IX), any nontrivial factor of j(x) yields immcdiately a nontrivial factor of f(x). It remains to consider the cast: where alternative (8) is valid and a, ~ 0 for l.,;j";n-1. Thcnf(x) is thc hinomialx" -,-""Cl'p"lx]. !\ow II is not a multiple of p, for otherwise we would havcf{x) ~ (x"/r.,. at )r, which would contradict the irreducihility of f(x) over lOr" We set (4.34 ) and thcn it is casily seen from fl ' 'f lOr' that the coefficicnt of x" I in j(x) is not in IFp'" Thus. the alternative (B) cannot ~ccur if the procedure descrihed ahovc is applied to f(x). Since f{ x) = Wf( fl I( x - I n, any non· trivial factor of j(x) yields immediatcly a nontrivial factor of fIx). This ",,)t·finding algorithm is thus carried l'ut as follows. We first form the polynomials f,(x) according to (4.29) and then thc polynomial F( x) EO 0' pl x] according to (4.30). !\ext, wc apply a factorization algorithm to <)htain thc canonical factorization (4.31) of F(x) in I'p[x]. This leads to thc partial factorization of f(X) given hy (4.32). Should this factorization be trivia\. we calculategcd(j(x),fk(x») for I.;k "OJIl/n. If this also docs nol produce a nontrivial fal·tor of f( x). we trunsform f( x) into j( x) hy eithcr (4.34) or (4.33), depcnding on whether i(x) is a hinomial or no!. As we have shown ahove, an application of the alg,<>rithm to fIX) is bound to yield a

159

EXl.:n.:i~l:s

nontrivial factor of /(x) and thus of I(x). Once a nontrivial factor of I(x) has bccn found, the proccdurc is continucd with the resulting factors in place of I(x), until I(x) is split up completely into lincar factors.

EXERCISES

4.1. 4.2.

4.3. 4.4.

4.5. 4.6. 4.7.

Factor x 12 + x' + x' + x' + x' + x' + lover IF, hy Berlekamp's algorithm. Factor x' + x' + x' - x' + x' - x -I ovcr 1F, by Berlekamp's algorithm. Let IF. = IF,(O) and factor x' + Ox' + x' +(1 + O)x + 0 over IF. by Berlekamp's algorithm. Use Berlekamp's algorithm to prove that x' - x' - x - I is irreducihle in IF ,[x]. Usc Bcrlekamp's algorithm to determine the number of distinct monic irreducible factors of x· + I in IF p[x] for all odd primes p. Usc the polynomials T; in Section I to factor x' + x· + lover IF,. Determine the splitting field of x' + x' + x' + x· + x' -r x' + lover

IF, . 4.8. 4.9.

Determine the splitting field of x' - x· - x' - x.,.. lover IF,. Usc the polynomials R, in Section I to factor the polynomial of

4.10.

Find the canonical factorization of XX + x' + x· + x' + I in 1F,[x] by using the polynomials R, in Section 1. Determine the canonical factorization of the cyclotomic polynomial Q3I(x) in IF,lx]. Factor I(x) = x' + xl + lover IF, and determine ord(j(x )). Factor/(x)=x 9 +x'+ x'+x'+x'+x+1 over IF, and detcrmine ord(j(x)). Prove in detail that if I is a nonzero polynomial over a field and d = gcd(j.j'), then lid has no repeated factors. (Note: Count nonzero constant polynomials among the polynomials with no repeated factors.) Let I be a monic polynomial of positive degrec with integer coefficients. Prove that if I has no repeated factors, then there are only finitely many primcs p such that I, considered as a polynomial over IF p' has repeated factors. Dctermine the numher of monic polynomials in IF .[x 1of degree n ~ I with no repeated factors. Let I be a monic polynomial over IF. and let g" ... ,g, he nonzero polynomials over IF. that are pairwise relatively prime. Prove that if I divides 1(, ••. 1(,. then I = I 1;_, gcd(j, 1(, ). L:se Berlekamp's algorithm to prove the following special case of

Exercise 4.1 over F 2'

4.11. 4.12. 4.13.

4.14.

4.15.

4.\6. 4.17.

4.\ K

J-."iJ~torizJ.li(ln

160

of Poh·nomials

Theorem 3.75: the hinomial x' - a. where I is a prime divisor of q - I and a E IF;. is irreducible in IF ,lx1if and only if a,q ,>/1 * I. 4.19. Let I be an irredueihle polynomial in IF qlx J of degree n and define the n X n matrix B = (b'j) by (4.4). Prove that the characteristic polynomial det(xl - B) of B is equal to x" - I. 4.20. Let 1=1\ I, he a product of k distinct monic irreducihle poly./, in IFq[x] of degree 11\ .... ,11,. respectively. Put nomials 1\ deg(f) ~ II ~ 11\ + ... + II, and define the n X II matrix B ~ (b,) by (4.4). Prove that thc characteristic polynomial det(xl - B) of B is cquallO (x", -I)·" (x n , -- I). 4.21. In the notation of Section 1. prove that the polynomialf.. ~ do not separate thosc irreducible factors 1; of I for which N In I is divisible by the characteristic of IF q' 4.22. Let I E IF ,Ix J bc monic of dcgree n ;, I. Define h E IF ,Ix. y J hy

h (x, y) ~ (y - x)( y - x')( y - x'')· .. (y .. x'' ') - I( y) and write h(X.Y)~5" \(x)y" \+ ... +5\(X)Y+'o(x).

Prove that

I

is irreducihle over IF, if and only if

I

divides " for

O.:E;j~n-I.

4.23.

Usc the criterion in the prceeding exercise to prove that x 7 -I

x" -t

x 3 + x 2 -t 1 is reducihle over 1F 2 "

Prove that the quadratic polynomial I(x) ~ x' T hx I C is irrcducible ovcr IF, if and only if I(x) divides x' + x -'- b. 4.25. Let I be an irrcducible polynomial in IF nIx I of degree m and let I. he a root of lin IF,•. Let g and h he nonzero polynomials in IFq[x]. Prove that h(xjml(g(x)lh(x)) is irreducible in IF)x] if and only if g(x)I.h(x) is irreducible in 1F.,[x]. 4.26. Use the method in Example 4.7 to factor x' -t 3x'; +4x' + 2x - I

4.24.

over IF

4.27. 4.28. 4.29. 4.30.

11"

Use the method in Example 4.7 to factor x' - 6x' 8x" 8 over IF I"~ Usc the Zassenhaus algorithm to factor x' + 3x; + 4x' + 2x - lover IF (J' Lse the Zassenhaus algorithm to factOf x 3 - 6x' - S.x - 8 over IF I"~ Use the Zassenhaus algorithm to factor X'" 3x' + 2x 3 - 6x' - 5 over IF 17.

4.3 I. 4.32. 4.33. 4.34.

Factor x' - 7x) + 4x' -'- 2x j 4 over IF \7' Factor x' 3x 3 -t 4x' .. 6x - 8 over IF ''I' Prove in detail that equivalence of square matrices of polynomials as defined by Definition 4.11 is reflexive. symmetric, and transitive. Lse the method in Example 4.14 10 faelor x 3 - 6x' - 8x - 8 over IF 19.

4.35. 4.36.

Use the method in Example 4.14 to factor x·1 ~3x' -2x-' -6x' + 5 over IF 1";' Usc the method in Example 4.14 to obtain a partial factorilation of x 7 - 2xo - 4.\·4 + 3x 3 - 5x·~ + 3x + 5 over !F; I and t.'omplcte the factorization

4.37. 4.38. 4.39. 4.40.

4.41.

4.42.

4.43.

by

another method.

find the roots of I(x)·- x' - x 4 I 2x' , x)· x -2 E ~\!x] can· tained in IF 5' Find the roots of I( x) ~ x' + 6x 4 + 2x' .. 6x' - 5x- 5 C IF ufx] contained in IF lJ' Prove that all the rool> of l(x)~X"'T8x'+6x-7EIFI9[xJ arc contained in IF It} and find them. Let IF" ~ IF,( Ill, where (j is a root of the irreducible polynomial x' i x' - lover IF,. Prove that all the roots of I(x) ~ x' +(f3' + f3'" I)x' r f32 x + f3' ~ f3' - f3 + I E IF,,[xJ are contained in IF 32 and find them. Let IF" ~ IF,(/l), where (j is a root of the irreducible polynomial x' x + lover IF,. Prove that all the roots of I(x) ~ x' + x 2 (f32 -f3+1)x-f3' -I EIF,,!x] are contained in IF" and find them. I.et IF '" ~ IF ,,( f3), where fI is a root of the irreducible polynomial x'-x-I over IF". Find the roots of l(x)~x'+(3f3-I)x-'-f3­ 5 E IF 169[X J contained in IF 164' If the polynomial I(x - h) in (4.22) is quadratic with constant term t'" 0, prove that the factorization in (4.22) is nontrivial if and only if (' is not the square of an clement of IFf"

4.44.

Let f3 be a defining clement of F~IF,., over IF,. Prove: (a) There exists k, 0", k '" n1 - I. with Tr r ({3') ~ I. (h) For eaeh i ~ 0, 1.. . .,n1 - I there exists an n, E F such that .' a- ,

(c)

({j' =

(Y

,

.-

\f3'+{j'

ifTr F (f3')" O. ifTrF(f3i)~l.

If y~r.';'-U'c,f3', c,EIF" and Trr(Y)~O, then the roots of 1 1 Xl ...... X + Y arc L~ 0 cia' and 1 + 1:;':..0 ('/0'/,

Chapter 5

Exponential Sums

Exponential sums arc important tools in numher theory for solving problems involving integers- and real numhers in general-that are often intractahle hy other means. Analogous sums can be considered in the framework of finite fields and turn out to be useful in various applications of finite fields. A hasie role in setting up exponential sums for finite fields is played hy special group homomorphisms called characters. It is necessary to distinguish between two types of characters-namely, additive and multiplicative characters-depending on whether reference is made 10 the additive or the multiplicative group of the finite field. Exponential sums arc formed hy using the values of one or more characters and possihly eomhining them with weights or with other function values. If we only sum the values of a single character, we speak of a character sum.

In Section I we lay the foundation by first discussing characters offinite abelian groups and then specializing to finite fields. Explicit formulas for additive and multiplicative characters of finite fields can be given. Roth types of characters satisfy important orthogonality relations. Section 2 is devoted to Gaussian sums, which are arguably the most important types of exponential sums for finite fields as they govern the transition from the additive to the multiplicative structure and vice versa. They also appear in many other contexts in algehra and number theory. /Is an illustration oftheir usefulness in number theory, we present a proof of the law of quadratic reciprocity based on properties of Gaussian sums.

j

10J

CharJ.t'ter'

Lxponential sums with the terms of a linear recurring sequence as arguments will be treated in Chapter 6, Section 7. Deep investigations on exponential sums for finite fields have hecn earricd out with thc help of algehraic geomctry, Icading to thc famous results of Weil and Oelignc, hut a prcsentation of this work would lead far beyond the scope of this hook.

I. CHARACTERS

I.et G be a finite ahelian group (written multiplicatively) of order IGlwith identity dement 1(I' A t.:lzaracler X of G is a homomorphism from G into the

multiplicative group U of complex numhers of ahsolute value I-that is, a mapping from G into U with X( g, g,) ~ X( g, )X( g,) for all 15 ,,15, E G. Since X(I,,) ~ x(I" )x(l,,), we must have x(l,,) ~ I. I'urthermore. (X (g))" ~ X (g'" ) ~ X (I,,) ~ I

for every g E G. so that the values of X arc IGlth roots of unity. We note also lhat, X(g)X(g 1)~X(gg ')~x(l,,)~I. and so X(g-')= (X( g»

I

~ X( g) for every 15 E G, where the bar dcnotes complex conjuga-

tion.

Among the eharactcrs of G we have thc tricial character Xo defined hy Xu(,~) ~ I for all 15 E G; all other characters of G arc called nontricial. With cach character X of G there is associated the conjugate character X

x(

defined hy g) ~ X( g) for all 15 E G. Given finitcly many characters x" .... X, of G. one can form the product character X,'" X, hy sctting (XI"'X,)(g)~XI(g)"'X"(g) for all gEG. If XI~"'=X,=X, we write X" for X,'" X,. It is obvious that the set G A of characters of G forms an abelian group under this multiplication of characters. Since the valucs of characters of G can only bc IGlth roots of unity, (i A is finite. After brieny considering the special case of a finitc cyclic group, we establish some basic facts about characters. 5,1. Example. Let G he a finite cyclic group of order n, and let g be a generator of G. For a fixed integer), 0 <; ) <; n - 1, the function XJ ( 15 , ) ~ e"'"

1" "

k

~

0 I •••• , n - I •

defines a character of G. On the other hand. if X is any charactcr of G, then x(g) must he an nth root of unity, say x(g)~e";I/' for some), 0<;)<; n - I. and it follows that X C XI' Therefore. G A consists exactly of the

charactersXO'XI ... .,Xn I'

::l

5.2. Theorem. I,et If he a suhgroul' of the /illite ahelian group (i and let y he u character of If. Then y can be extended to a ('haraerer of G; thar is, there exists a character X of G with X(iI) ~ y(h) for all h E II.

Fxponcnli<.ll Sums

164

Proof We may suppose that !I is a proper suhgraup of C. Choose a Eo G with a'" H. and let H, he the subgroup of C generated hy Hand Q. Let 111 be the kasl PO~ilivc imegcr f\)f which a'" F.1I. Then l..:\"ery ~h.:ment g E HI can be written uniquely in the form g,-:: alh with 0 ~ j < m :.mu hE H. Define a function y, on !I, hy y,(g)~",ly(h). where w is a fixed (,;ompkx numher satisfying v,;o"l = y(01l1). To check that y. is indeed a character of H,.let g, ~ a'h,. 0 <;: k < nJ. h, F /I. he another clement of II,. If j -t k < nJ. then y, ( gg, ) ,- Wi' 'y ( hh, ) ~ y:( g ) y, <.~,). If j - k " nJ, then ggl' llj·l<-m(a'tlhh l ). and ~o

y

(gg,)' w i -' '"y(amhh,) ~ wI'< - my (am)y (hh,) ~ w' "" (hh,) ~

y, (g )", (g,).

It is ohvious that ",(h) ~ ,,(h) for h c- 1/. If II, ~ G. then we art done. Otherwise. we can continue the process above untiL after finitely many

steps. we obtain an extension of y to G. 5.3.

Corollary.

For any

{11:0

:J

distinct

ell!nu:'nl5 .~1'

g::.

E:::

G {here (!xis{s

a characler X of G "-ilh X( g oJ ~ X( g,).

Proof It suffices to show that for h = gig.' 1=1=-1<1 there L'xists a character X of G with X( h) ~ I. This follows. however. from Example 5.\ and Theorem 5.2 hy letting /I he the cyclic suhgroup (If G generated hy h. [' 5.4.

Theorem.

If X is a nontrivial chUfUuer of ,he finf(e uhelil.1f1

group G. Ihel/ (5.1 )

If g

Eo G wilh

g "" Iv- Ihel/

L x ("

Proof

X(g)-O.

(5.2)

(i '

Since X is nontrivial. there exists h Eo G with X( h) "" 1. Then X (h)

L

X (g) ~

L

X (hg ) ~

L

X (g).

g ( (;

heeause if g rullS through G. so does hg. Thus we have

L

(X(h)-I) ~

X(g) ~

o.

t: G

which already implies (5.1). For the second part, we note that the function g defined by ,g(X) ~ X(g) for X eGA is a character of the finite abelian group GA. This character is nontrivial since, hy Corollary 5.3. there exists X E C A with X(g) "" X(1G) = 1. Therefore from (5.1) applied to the group CA.

I.

Charal,:tl'r~

165

o XCG~

The number of characters of a fillite ahelian 1V0uP G

5.5. Theorem. is equa/to IGI· Proof

xE-G~

This follows from

IG'I~

L L

X(g)~

L L

x(g)~IGI.

where we used (5.2) in the first identity and (5.1) in the last identity.

C

The statements of Theorems 5,4 and 5.5 ean be eombined into the orthogonality relations for characters. Let X and", be characters of G. Then

I iGT

-

LX(g)",(g) = 1? E.

{OI

(,

for X "'~. forX~~·

(5.3)

The first part follows. of course. by applying (5.1) to the character x~; the second part is trivial. Furthermore, if g and h are elements of G. then

I

--- (0

iG' x~;. X(g)X(h) ~

,I

for g '" h.

forg~h.

(5,4)

Here. the first part is obtained from (5.2) applied to the clement gh '. whereas the second part follows from Theorem 5.5. Character theory is often used to obtain expressions for the number of solutions of equations in a finite abelian group G. Let f be an arbitrary map from the cartesian produet Gn ~ G X ... x G (n factors) into G. Then. for fixed hc(;, the number N(h) of n-tuples (g" ... ,g,,)EG" with f(g, .... ,g,,)~h is given by I N(h)~jGT

L ... L gllC-G

L

x(f(g, ..... g,,))X(h)

(5.5)

g"cGxE-G"

on account of (5,4). A character X of G may be nontrivial on G, but still annihilate a whole subgroup II of G. in the sense that X(h) ~ I for all h '= H. The set of all characters of G annihilating a given subgroup II is called the annihilator of H in G '. 5.6. Theorem. /.et II he a suhgroup of the finite ahelian group G. Theil the annihilator of'H ill G' is a suhgroup of G' of order! GI / I Hi· Proof Let A be the annihilator in question. Then it is obvious from the definition that A is a subgroup of GA. Let X E A; then p.(gll) = X(g), g E G. is a well-defined character of the faetor group G/H. Conversely, if p.

I-.xpont'ntial Sum:;

166

is a character of GI H, then Xun = Ji(gH), g E G, defines a character of G annihilating H, Distinct clements of A correspond to distinct characters of GIH. Thereforc, A is in one-to-one correspondence with the character group (G/H)", and so thc order of A is equal to the order of (GIH)", which is IGIHI ~ IGI/IIlI according to Theorem 5.5. 0 In a finite field IF q there arc two finite abelian groups that arc of significance-namely, the additivc group and the multiplicativc group of thc field. Therefore, wc will havc to makc an important distinction between the characters pcrtaining to thcse two group structurcs. In hoth cases, cxplicit formulas for the characters can be given. Considcr first the addilice group of IF q' Let p bc the characteristic of IF q: then the prime field contained in IF q is IF p' which we identify with lL/(p). Lct Tr;lFq-lFp be the absolute trace function from IF q to IFp (see Definition 2.22). Then thc function XI dcfined by XI(c) ~ e,.. TWllp

for all c E IF q

(5.6)

is a character of the additive group of IF q' since for c l ' c, E IF q we have Tr(c i + c,) ~ Tr(c l )+ Tr(c,), and so X,(c, + c,) ~ XI(CI)X,(c,). Instead of "character of thc additivc group of IF q," wc shall henceforth usc the term additive character of IF q' Thc eharactcr XI in (5.6) will bc called the canonical additice character of IF q' All additive characters of IF q can he exprcssed in tcrms of XI' 5.7. Theorem, For hElF q , the function Xh with Xh(c)~xl(hc)for all (' EO IF q is an additice character of IF q' and every additive character of IF q is ohtained in this way.

Proof

For c\, ('2

E

IF lj we have

Xb(C, I,·,) = XI(hc I + bc,) = x,(hcllxl(hc,) ~ Xh(cllxb(C,),

and the first part is established. Since Tr maps IF q onto IF p by Theorem 2.23(iii), X, is a nontrivial eharactcr. Thcrefore, if a, b E IF q with a '" h, then x,(c) x,(ac) - - ' ~ - - ~ x,«a - h)c) '" I XI(hc) Xh(C)

for suitable c E IF q' and so Xa and Xh are distinct charactcrs. Hence, if h runs through IF q' wc get q distinct additive charactcrs Xh' On the other hand, IF q has exactly q additive characters hy Thcorem 5.5, and so the list of additive characters of IF q is already complete. 0 By setting h ~ 0 in Theorem 5.7, we obtain the trivial additive character XO' for which Xo( c) = I for all c E IF q' Let E he a finite extension field of IF q' let XI be the canonical

l. Characters

167

additive character of IF q , and let III be the canonical additive character of /:; defined in analogy with (5.6), where Tr is of course replaced hy the ahsolute trace function Tr, from E to IF" Then X, and III are connected hy the identity (5.7)

"I'

is the trace function from E to IF q' This follows from the where Tr transitivity relation Tr,,(fJ) ~ Tr(Tr EI' ,UJ))

for all /l

E

E,

which was shown in Theorem 2.26. Characters of the multiplicati"e group IF: of II- q arc called multiplicati"e characters of IF q' Since IF; is a cyclic group of order q - I by Theorem 2.8, its characters can he easily determined.

5.8. Theorem. Let I( be a fixed primitive element of IF q' For each } ~ 0, I, ... ,q -2. the function.yj with >y,(I(')=e'''''/lq

I)

fork

~0,1, ... ,q-2

defines a multiplicative character of IF q' and every multiplicative character of IF q is obtained in this way. Proof

This follows immediately from Example 5.1.

0

No matter what g is, the character.yo will always represent the trivial multiplicative character, which satisfies .yo(c) ~ I for all c E IF;' 5.9. Corollary. The group of multiplicative characters of IF q is cyclic of order q - 1 with identity element .yo' Proof

Every character.yJ in Theorem 5.8 with} relatively prime to 0

q - I is a generator of the group in question.

5.10. Example, Let q be odd and let 'I be the real-valued function on IF; with '1(c) ~ 1 if c is the square of an element of IF; and '1(c) = -I otherwise. Then 'I is a multiplicative character of IFq' It can also be obtained from the characters in Theorem 5.8 by setting} ~ (q -1)/2. The character 'I annihilates the subgroup of IF; consisting of the squares of elements of IF;, and by Theorem 5.6 it is the only nontrivial character of IF; with this property. This uniquely determined character 'I is called the quadratic character of IFq' If q is an odd prime, then for c E IF; we have '1(e) = (~), the Legendre symbol from elementary number theory. 0 The orthogonality relations (5.3) and (5.4), when applied to additive or multiplicative characters of IFq , yield several fundamental identitie;. We consider first the case of additive characters. in which we use the notation from Theorem 5.7. Then, for additive characters Xu and Xh we have

Exponential SUn1~

16R

fora=h, = b.

(5.H)

fora

In particular, L

Xa(c)=O

(5.9)

fora*O.

cFlF"q

Furthermorc, for e1cmcnts c, d

E

0: q we obtain fOfC"'d, forc=d.

(5.10)

I'or multiplicative characters 'i and T of 0: q we have forY~T,

(5.11 )

for1f=T.

In particular, (5.12)

L'i(c)=Ofor'i"''in' cf:i::

If c, d

E

F;

0:;, then L1f(c)1f(d) =

"

{~_I

forc"'d, forc=d,

wherc thc sum is extended over all multiplicative eharactcrs

2,

(5.13)

y of 0: q'

GAUSSIAN SU:vJS

Lct y be a multiplicativc and X an additive character of 0: q' Then thc Gaussian sum G( 'i, X) is dcfined by

G('i,X)

L

=

Y(c)X(c)·

rEIF:

The ahsolute value of G( 'i, X) can obviously he at most q - 1, hut is in general much smaller, as the following theorem shows. We recall that 'in denotes the trivial multiplicative character and Xo the trivial additive character of 0: q'

5,1/. Theorem, Lei 'i he a rnuliiplicalice and X an addilice character oj 0: q' lhen the Gaussian sum G( 'i, X) salisjies G('i,X)=

{

q-l

jor'i='io,X=Xo'

-I

jor'i='io'X"'Xo'

°

jory'" Yo' X = Xo'

(5.14)

169

If ~ *"

"4- 0

alld X =1"'_

XO'

{hen (5.15 )

Proof The first case in (5.14) is trivial. the third case follows from (5.12). and in the second case we have

L:

G(",o.X)=

X(c)~

("
hy (5.9). for", = >{o and X

~

X(c)-X(O)~-

XI> we get

L: L:

=

L: ("r.F q

,f""lF,;

>{(c) x(e) >{(e,)x(e,)

cfE=iF;

; L: L:

>{(c 'c,h(c,-c). ,;C:..:r; In the inner sum we suhstitute (' 1('1 = d. Then. rt.::lr;

L: L

IG(>{.x)I'=

Cf:-~;

",(d)X(c(d-I»

dE-irq

dF.; f(d{E., X(c(d -1»- X(O») L:

~

",(d)

dc.lf;

L:

X(c(d- I»)

rE::Ir'l

by (5.12). The inner sum has the value q if d = I and the value 0 if d'" I. according to (5.9). Therefore. IG(>{. x)I' = >{(I)q ~ q. and (5.15) is estahCJ lished. The study of the hchavior of Gaussian sums under various transformations of the additive or multiplicative character leads to a number of useful identities.

5.12. Theorem. followinf( properties: (i)

Gaussian sums for the finite field IFq satisfy the

G(>{,Xab)=>{(a) G(f'Xh)foraEIF;.bElFa;

(ii) (iii)

G(f,X)~1f(-I)G(f:X2~

(iv) (v)

G(>{.X)G(.r.X)~"'(-I)qforf=fo.X=x,,; G( ",p. Xh) = (i( >{, XOl bl) for b e: IF q' where p is the charactaisTic

G(.r,X)~Ji;(-I)G(>{.X):

of lF a and a( h) = hP. Proof

(i) for

,'Eo

lFa we have Xah(c)

~ x,(ahc)'~

Xh(ac) hy the

Exponential SUI11:'

17{)

definition in Theorem 5.7. Therefore,

Now set

a(' ~

d. Then G(",Xuh)~

L

>/;(a'd)x,,(d)

den;

. >/;(a ')

L

>/;(d)Xh(d)

d t: n~

~ >/;(a) G(>/;.Xh)'

(ii) We have X~XI, for a suitable hFf. and X«')~Xh(-c)'~ X h( c) for cEo f q' Therefore. by using (i) with a ~ - I and noting that ,,(- I) ~ 1. I. wc gct G(".X)~G(".X ,,)- >/;(-1) G(>/;,Xh)~>/;(-I)G(>/;.X).

,,(

(i~follows from (ii) that G(f. X) ~ f( - I)G(';'. X) = I)G(';'.X)· (iv) By _combining (iii) and (5.15), we obtain G(1J"x)G(f,x)~

y( . I)G(l!q)G(';', X) ·';'(-I)IG(1J"x)['~';'(-I)q. (v) Since Tr(a) ~ Tr(a P ) for a E f q by Theorem 2.23(v), we have :1:,(11) ~ x,(a P) according to (5.6). Thus, for c Eo f q we get Xb(c) ~ x,(bc) = x,lliP('P) ~ X",,,ICP), and '0 G(>/;P,Xh)~

I:

';'P(C)xh(C)~

(" f-IF";

But

('P

run,

through~; as

c runs through

L

1J,(C P )xQ'h'(C P),

(" E:-IF~

f;. and thc dcsircd result follows. o

S.U. Remark. In connection with thc properties above, the value.;,( - I) is of interest. Wc obviously have 1{( -I) ~ ± I. Let m be the order of';': that is, m is th~ least positive integer such that tP rn = t/;o. Then m divides q - 1 ~incl:: \f;iI I Yo' The values of \f are mth roots of unity; in particular, - I can only appear as a value of y if m is even. If X is a primitive clement of IF q' then >/;( K) ~ I. a primitive mth root of unity. If m is even (and so q odd), then >/;( . I) ~ "U".-I,n) ~ I,q '1/'. which is - I precisely if (q - 1)/2 '" m /2 mod m. or. equivalently. (q 1)/ ttl" I mod 2. Therefore. YI - 1) ~ - I if and only if ttl is even and I q .. 1)/ ttl is odd. In all other ca,es we have Y(1)~1.

U

Gaussian sums occur in a variety of contexts, for example in the following. Let y bc a multiplicative character of f q : then, using (5.10). we

may writc

1

(jau~sjan

171

Sum:,>

for any c E IF;. Therefore, I _ ~(c)~ - LG(~,X)x(c) q x

forcEIF;.

(5.16)

where the sum is extended over all additive characters X of IF q' This may be thought of as the Fourier expansion of>} in terms of the additive characters of IF q' with Gaussian sums appearing as Fourier coefficients. Similarly, if X is an additive character of IF q' then. using (5.13), we may write

~_I-I L>}(c) L f(d)x(d) q-

~

forcEIF;'

de:lf-;

Thus we obtain

X(c)

=

-I I "£..,G(>},X)>}(c)

q-

~

for

C

E

IF;,

(5.17)

where the sum is extended over all multiplicative characters ~ of IF q' This can be interpreted as the Fourier expansion of the restriction of X to IF; in terms of the multiplicative characters of IF q' again with Gaussian sums as Fourier coefficients. Therefore, Gaussian sums arc instrumental in the transition from the additive 10 the multiplicative structure (or vice versa) of a finite field. Refore we establish further properties of Gaussian sums, we develop a useful general principle. Let be the set of monic polynomials over IF q' and let A he a complex-valued function on which is multiplicative in the sense that A(gh)~A(g)A(h)

forallg,hE,

(5.18)

and which satisfies IA(g)1 ",I for all gE and A(l)~1. With <1>, denoting the subset of containing the polynomials of degree k, consider the power series

L(z)~

f (L

k -0

A(g))Z'

(5.19)

,l:'1::¢lk

Since there are q' polynomials in <1>" the coefficient of z' is in absolute value", q', and so the power series converges absolutely for Izl < q' '. Because of (5.IR) and unique factorization in lFalx], we may write

t-:..... poncntial Sum"

172

/.(z)~

L

A(g) Z M 8'"

,~

~n(I+A(j)Z'I,,'jl~A(J')Zd"<J"+... ) I

~

n(1 + A(J)Zd,g<J'+A(J)2Z2"O"J'~ ... ). I

where the product is taken over all monic irreducible polynomialsfin IFJx]. It follows that L(z)~ n(I-A(J)Z",,,,,)-1 I

Now apply logarithmic differentiation and multiply the result by z to get

dlogL(z) z-dz'Expansion of (1- A(flz"O"f')

7--1~Tf)zd,g'/J

A(j)deg(fl=",·"f'

= 1

into a geometric series leads to

zdlogL(z) ~ LA(J)deg(J)z'lo"J' dz f

.(1 + A(flz d,g'J'+A(f)'Z2rl"'J'+

... )

~ Ldeg{f)(A(f)z"""I' tA{f)'Z2""'J' I T

A{f)JzJd"'jl_ ... ),

and collecting equal powers of z we ohtain (5.20)

with L, =

L deg( f)A(f) '/do"n.

(5.21 )

I

where the sum is extended over all monic irreducible polynomialsfin with dcg(j) dividing s. Now suppose there exists a positive integer t such that

L

A(g) ~ 0

for all k >

I.

~qlxJ

(5.22)

XC etJ k

Then 1.( Z) is a complex polynomial of degree", t with constant term I. so that we can write

L(z)

~ (1-

"'1=)(1- w2z)'" (1- "',z)

(5.23)

2.

(jau~"ian Sllm~

with complex

Z

number~ wI'

u.:z ....• w,. It follow1:i that

dlog I. (z) ~ _ " dz '-

Wml

,

= -

x

L ," 'X

- L

L

Wml

i-

I i

!

W~,Zj 0 'X

IL

101m-I

/!

\

W~'-')z'-' ~ - L i L W:")"" \" I

\m-l

and comparison with (5.20) yields

1.\

= -

W1 -

w2 -

... -

u:!~

for all s

~

I.

(5.24)

As an application of the principle expressed in (5.24), we consider the following situation. Let X be an additive and if a multiplicative character of I'q. and let F. be a finite extension field of I'q. Then X and y can be "lifted" to F. by setting x'(/3)~x(Tr,./,"({3)) for {3E E and y'({3)= y(N ,/>,<{3)) for {3 E F.". From the additivity of the trace and the multiplieativity of the norm it follows that X' is an additive and if' a multiplicative character of f.-. The following theorem establishes an important relationship between the Gaussian sum G( y. X) in IF q and the Gaussian sum G( y', X') in E. J.I4. Theorem (Davenport-llasse Theorem). I.el X be all addilice and if a mulliplicUlice diGr""ler of IF q' nOI Iimh of Ihem IriDial. Suppose X alld yare lifled 10 charaCler" X' alld if', respeuioe!v. of Ihe fillile exlellsion field E of IF q wilh [ F.: IF q 1= s. Then G(';/·X')

= (

I)' 'G(Y.X)'

Proof It is convenient to extend the definition of if hy setting y(O) = O. We usc the notation of the discussion leading to (5.24); in particular. '" denotes again the set of monic polynomials over IF q • We define A by setting A( I) = I as required. and for I: E '" of positive degree, say g(x)=x' --c,x' ' - ... +(- I)'c" we setA(I:)~Y(c,)X(c,). The multiplicative property (5.IR) i, then easily checked. For k > 1 we split up <1>, according to the values of ('j and c". Each given pair «('1'(''') occurs q"-2

times in

~/..'

and so

L

A(I:) - q'

XL $.

':' (" (

=q'

2( L

n.,

'-cE,n;

y(cll(. (,nL X(c)). q

174

Since one of X and y i, nontrivial. it follows from either (5.9) or (5.12) that

L

A (g) ~ 0 for k > l.

J!€-lP~

Therefore, (5.22) is satisfied with 1 ~ 1. furthermore,
L g0"$:

A(g)=

L

Y(c)X(c)~

(f-'f

L:

"'(clx(c)~ G(y,X)'

'f::.J-;

q

Thu" L(z)=I" G(!f"X)z from (5.19), hence w,~ G(y,X) hy (5.23). Now we consider I"}. which, by (5.21) and the multiplicativity of A, is given by

t, ~

L deg( f)A( f) ,;d'gll, I

.. L • deg( f) A(f,;d'" Ii), (

where the sum is extended over all monic irreducible polynomials I in IF qlx J with deg( I) dividing s, and where the a,terisk indicates that I(x) ~ x is excluded. Faeh sueh/has deg(j) distinct nonzero roots in 1:.", and each root {3 of I has as ito characteristic polynomial over IF q the polynomial , ';0".(/, I(.v)

X.' - ( I X "

-l-

.••

(1)' c~. +_

say. where c, ~ Tr,/,,,<{3) and c, ~ t'lJf}{3) by (2.2) and (2.3). Therefore, I A(f';d'<1 ') ~ ~

and

."k)x(c,) ~ .,,(N u ,}{3))x(Tr,/,"({3)) y'(fl)X'({3),

~o

t, -

L 'deg( f) A(/';."." ") = L ' L (

l(P) If

f

';:(fllx'({3)·

13 ('" F (l

run ... through the fLlnge of summation above. then

f3

runs exactly

through all clements of /:"'. Consequently.

I, -

L

""(fllx'(fl): G(Y·.x·).

Ii ( L'

and an application of (5.24) yields

C(y',X')= -( -C("',X))', which completes the proof.

o

For certain special characters, the assodatcd Gaussian sums can be evaluated explicitly. We therehy obtain formulas that go beyond the trivial

2. (j
175

Sll;T1<:

cases listed in (5.14). A celebrated formula of this kind holds for the quadratic character 1) considered in Example 5.10. 5.15. Theorem. Let IF q he a finite field with q ~ P'. where p is an odd prime and sEN. Let 1) be the quadratic character of IF q and let XI be the canonical additive character of IF q' Then ifp '" I mod4. ifp'" 3mod4. Proof Using Theorem 5.12(iv) and j)~1), we obtain G(1),XI)'~ 1)( - I)q. and since 1)( - I) = I for q'" I mod4 and 1)( -I) ~ - I for q '" 3 mod4 by Remark 5.13. it follows that

if q '" I mod4.

(5.25)

if q '" 3mod4. Thc difficulty of the proof lics in the determination of the correct signs. We first consider the case s = I. Let V be the sct of all complexvalued functions on IF;; it is a (p - I)-dimensional vcctor space over thc complex numbers. A basis for V is formed by the characteristic functions f,.f" ... '/p., of elements of IF;: that is,!,(c)=1 if c~j and 0 othcrwise, wherc j ~ 1.2, ... ,p .. I. From the orthogonality relation (5.11) it follows easily that the multiplicative characters >/;0' ~"",,>/;P 2 of IFp described in Theorem 5.8 also form a basis for V. Let \ ~ e";/p, and define a linear operator T on V by letting Th for h E' V be given by p - 1

(Jh)(c)-

L k

t'h(k)

forc-I.2 ..... p-1.

(5.26 )

1

Then Theorem 5.12(i) implies that T~ = G( >/;. XI),[' for every multiplicative character ~ of IF P' Since ~ ." ~ precisely for the trivial character and the quadratic character. th~ matrix T in the basis Yo. l/;l····.1ft p 2 contains two diagonal entries-namely. G(~",XI)= .. I and G(1).x,)-and a collection of blocks

i

0 \ G(y,X,)

corresponding to pairs

G(;ro'x,l',')

1ft. ~ of conjugate characters that are nontrivial and

nonquadratic. If we compute the determinant of T. then each block contributes

lxponeotia\ Sum,

176

hy'l heorem 5.12(iv). Thu, we obtain (p .. ))/2

II

det(T) ~ - G( '), XI)( - p)IP' 311'

I

l'ow >/-,< -,1) = lp-Jl/2

n

I

",{(-I)~

(-1)1. and so

>/-(-I)~(-I)""-'P 111'

1

(5.27)

>/-i( - I).

I

~

( _ I )IP

I lIP

-31/'

(5.28)

1

Furthermore, si nee

if p" I mod4, if p"' 3mod4, it follows from (5.25) that (5.29) Comhining (5.27), (5.28). and (5.29). we get det(T) ~ ±( -1)'P 1II'iIP 11'/4( -1)'P IIIP 31/'p'P"2)I'

hence (5.30) "low we compute det(T) utilizing the matrix of T in the hasis

II./, .... Jp I' From (5.26) we find det(T)~det(U;')I<j.,

Ii)'
=>I-""'-IP-Ildet((>IU Ii). ) ~ ~ l ..... /,ko;,;p-! ~det(Uj"

II)I<;.,
I)'

whieh is a Vandermonde determinant. Therefore. det(7') =

n

l.,,;;:m<.n ..... p-]

With

(j.

e1://f'

we get

n

det(T)I",

t>1

< n.,,; II

(0'''_8'''') I

(I"-r m ).

d

177

2. Gaussian Sums

II I '"

In <.

8""(8' m-8-'" ml)

n ,;;: f'

n

1,-;: m <

1

n

I)'."

11 '"

I

P

I,;: '" <

(2isin'TT(I/-mJ).

1/ '"-

pIP

Since P

L

(n -

I

1

1/·

L L

mJ-

(nl m)

,,-2m·l

I"'tn
3

P - 1

JP

L

2

1/ (n - I) ~.:.. L (1/' + n) 2 n =2. 2n_1

~.-

~ 3 ( ( I' - 2)( P - 1)( 2 P - 3 J 2 6

\-

(p: 2)( I' - I J) 2

1'( P -1)( I' -2)

~--2

the fiN product is equal to 1)/2

Illf'

furthermore.

A~

[I I,;: m<

'I';;;

P

I

' 'TT(n-m))' - - > 0. (2 smp.

and so det(r)~(-I)(P 1)/2 i

,?

I)/p

21/ l A

withA>O.

Comparison with (5.30) shows that the plus sign always applies in (5.29). and the theorem is estahlished for s ~ I. The general case follows from Theorem 5.14 since the canonical additive character of IFp is lifted to the canonical additive character of IF q by (5.7) and the quadratic character of IF p is lifted to the quadratic character of

Fq •

L

Ikeause of (5.14) and Theorem 5.12(i). a formula for C(11.X) can also be established for any additive character X of'lF q' We turn to another special formula for Gaussian sums which applies to a wider range of multiplkative characters but needs a restriction on the underlying field. We shall have to usc the notion of order of a multiplicative character as introduced in Remark 5.13. 5.16. Theorem (Stiekelberger's Theorem). Lei q be a prime power, Ie/ '" be a nontrivial multiplicative character of IF q' of order m dividing q + I, and let X, be the canonical additive character oflF q ,. Then,

E",pon~'nlial Sum~

I JH

('(' )J

q .,.1 if m odd or ._- even, m q+ 1 if m even and - - odd. m

q

, y, XI - \

\- q

Proof We write F. = IF q' and F = IF q' Let y be a primitive clement of L" and set g = yq- '. Then gq-' = I. so that g E F; furthermore, g is a primitive clement of F. Every IX E E* can be written in the form IX = gly' with 0", j < q -I and 0 '" k < q + 1. Since >{(g) = >{q. I(y) = I, we have q -:.

q

L L

G(";,X,)=

>{(g'y')x,(gly')

,-=Ok ..... O "

cj

L

=

.,;'(y) 0

k

L"

=

-:2

L X,(gly') ,= 0 L

>{'(y)

If T, is the canonical additive character of F, then XI(hyk) by (5.7), Therefore,

L

L

X,(by')=

bcF*

=

TI(TrtjF(hy'»

TI{hTrm(yk»)

hf::.:F"

-I

for 1'r"I" ( y') '" 0,

= { q-I

for Tr"IF( yk) = 0,

Tr tjF( yk) = 0

if and only if y"q-

11 = -

If q is odd, the last condition is equivalent to k = (q (5.32), -I ( q_1

I.

+ 1)12,

fork

=

q+1 -2-'

Together with (5.31) we get G ( y, X, ) =

L"

. A ,

k ((,1-

>{k ( Y) + ( q - I ) >Ii q• 11/' ( Y) 0

1)/2

(5.33) and then by

q+1 for 0 " k < q + I, k '" -2-'

X,(by')=

b0F"

(5.32)

= y' + yk q, and so

because of (5.9). "'ow TrF1F(yk)

L

(5.31)

XI(by').

brP

k=O

q

1: A

';"(y)+ qt/-,q. 11I2(y) 0

~q';"""1I2(y)

,inee y(y)' I and ,;,q;'(y)-I. Now t/-(q-III'(y)-1 if (q+I)/111 is even and· I if ('I + 1)/111 i, odd. and thus for q odd we have .f q -/ I I

--

even.

if

12 I

odd.

111

(5.34)

m

If q is ewn. then the condition in (5.33) is equivalent to ykiq- Ii - I. and the only k with 0" k < q + I satisfying this property is k ~ O. Then hy (5.32).

-I ( q -I

forl:::;;k~q. ~

for k

O.

and (5.31) yields q

(i(Y.XI)~-

1:

'-I

'I

t/-'(y)-q-l~-

1:

1fk(y)_q~q.

k~(l

Comhined with (5.34). this implies the theorem.

We show how to usc G(Jussian sums to estahlish a classical result of number theory. namely the law of quadratic reciprocity. We recall from Example 5.10 that if p is an odd prime and 'I is the quadratic character of ; p' then for c t 0 mod p the I.egendre symbol

(~) is defined hy (~) -11(c).

5.17. Fhmrem (Law of Quadratic Reciprocity). odd primes [J and r we flat'l'

(~)(~) _ ( Proof

I)"

iii'

For any di'tinct

li/4

I.et 11 he the quadratic character of IF p' let X I he the canoni-

cal additive character of IF,. and put G ~ (i( 1/. XI). Then it follows from (5.25) that (i2. ( I)IP lili p = fi. and so

0"

.(G··)"-"/2G~fi"

"1'0'.

(5.35)

Let R he the ring of algehraic integers: that i,. R consists of all complex numbers that arc roots of monic polynomials with integer coefficients. Since the values of (additive and multiplicative) characters of finite fields are complex roots of unity. and since every l.:omplex root of unity is an

algehraic integer, the values of Gaussian sums arc algehraic integers. In particular. G FR. Let (r) he the principal ideal of R generated bv r. Then

IflO

b.poncntial

Sum~

the residue class ring R/(r) has characteristic r. and thus an application of Theorem 1.46 yields

G'~ ( I: ~(cJxI(c)r'" I: ~'(c)Xl(c)mod(r). (,J;

Ct&;

~ow

by Theorem 5.12(i), and so G'=~(r)Gmod(r).

Together with (5.35) we get pi' 1)/2G=~(r)Gmod(r),

and multiplication by G leads to pi' "12p=~(r)pmod(r)

because of G" = p. Since the numbers on both sides of the congruence above arc. in fact. clements of 7L. it follows that fi"

as a congruence in 7L. But

'I/'P:': ~(r)fimod

r

p and r arc relatively prime. hence pi' 1I/)=~(r)modr.

'low fi - (- I)'? "I'p and p' yiclds

'= I mod r.

thus multiplication by pi'

(-I)'? I", "/4=p,,-1)/2~(r)modr. We havc pl'

and the plus sign applies if and only if p is r. Thus.

plr-ll/1

~(r) ~

(5.36)

'>/.' '" :±. I mod r.

congruent to a slJ.uare mod

Sinee

Ill:!

==

(If: )mOdr.

(!..pit we gct from (5.36) (- W ''''

1)/4

= (~)( ~) mod r.

But the integers on both sides of this congruence can only he :: I. and sinee r ? 3, the congruence holds only if the two sides are identical. 0 We consider now charactc:r ~ums involving the quadratic character 11 of f"q odd, and having a quadratic polynomial in the argument. The following cxplicit formula will he needed in Chaptcr 7, Section 2. a2

I

5,/8. Theorem. LeI fix) ~ a,x' + a,x - a o 0I',[x] wilh q odd and O. Pur d ~ af - 4a oa, and leI ~ he Ihe quadraric charader of 'f q' ,hell

Excrci~~~

L '1 (f( c)} ~ (!:IF
L

I -'1(a,) \ ( _. I) ( ) lJ

11 u'2

if d , O. if d

~

0.

Multiplying the sum hy '1(4aj) ~ I, we get

'1 (f(c}) ~'1(a,) L '1(4ajc' +4a,a,c; 4a"a,)

rt:.:.lF"

I

C}1

~'1(a,} L '1((2a)c-a,})-d)~'1(a,} L'1(h'-II). (" .... tr" q

The result for the case d ~

h .... 1 -1

°

follows now immediately. For II '

L '1(h'-II)~ h,n.,. and sinrL' I -t 11( b 2

-

(I

q+ L

we write

f ~(h'-d)).

hr~"

d) is the number of (' r IF with 'I

L

°

(5.37)

'1(h' - d) ~

(,'2 =

h2

q, S(d).

d. we ohtain

(5.38)

be} 1

where S( d) is the numher of ordered pai" (b. c) with h. c E f q and h' - (, ~ d. To sohe this equation. we put h + c ~ u, h - c = c and note that the ordered pairs (h. c) and (u. ,,) are in one-to-one correspondence since q is odd. Thus S( d) is equal to the number of ordered pairs (u. r) with u. I' E f'J and "" ~ d, hence Sid) ~ q - L Together with (5.37) and (5.38), this implies the desired formula. [j

EXERCISES 5. L

Let G be a finite abelian group. H a proper subgroup of G. and g f£ H. Prove that there exists a character X of G that annihilates H. but for which X( g) ~ I. Let H be a subgroup of the finite ahelian group G. Prove that the annihilator II of II in G' is isomorphic to G/ II and that G'/ A is isomorphic to II. Let G he a finite abelian group and mEN. Prove that g '= G is an mth power of an element of G if and only if X(g)~1 for all characters X of (i for which X m is trivial. Let (i , __ . __ G, he finite abelian groups. Define multiplication of k-tuplcs (g,. __ .. g,). (h, __ . __ h,) with g,. h, E G, for I" i" k hy

.~ E G.

5.2.

5.3.

5.4.

( g, . . .. g, )( h, ..... h, ) ~ (g, h, ,. ... g, h, ). Show that with this operation the set of all such k-tup!es forms again a finite abelian group. the so-called direct product (i,0 .. , 0G,.

E"'ponl..'ntial SUI11~

1H2

5.5.

5.6.

5.7. 5.8.

Then prove that (G,0 ... 0G,) A is isomorphic to G( 0 ... 0Ct. Use the structure theorem for finitc ahelian groups. which says in its simplest form that evcry such group is isomorphic to a direct product of finite cyclic groups. to prove that G A is isomorphic to C whcncver C is a finite ahelian group. For additivc characters of IF q in the notation of Theorem 5.7. show that X,Xb ~ X, .. for all a. h E IF q. Thus prove without refcrcncc to Exercise 5.5 that the group of additive characters of IF q is isomorphic to the additive group of IF q' If X, is the canonical additivc character of the finite field IF q of characteristic p. prove that X,( c P ') ~ X, (c) for all c E IF. and j E "". If ti- is a multiplicative character of IF., of order m. provc that the

restriction of y to IF q is a multiplicative character of order m/gcd(m,(q' -I)/(q -I)).

5.9. 5.10.

With the notation of Exercisc 5.8, prove that the restriction of lj. to IF q is the trivial character if and only if m divides (q' -I)/(q -I). Let ti- he a multiplicative charactcr of IF. and let ti-' hc the lifted character of the extension ficld IF q,. Prove that lj.'(c) ~ y'(c) for

cFIF;. Prove that a multiplicative character T of IF q' is equal to a character ti-' lifted from IF. if and only if T q - ' is trivial. 5. I2. If q " I mod m and ti- varies over all multiplicative characters of IF q of order dividing m, prove that the lifted character ti-' of IF., varies over all multiplicative characters of IF., of ordcr dividing m. 5.13. Prove that an additive character X of thc finite cxtension ficld F. of IF q is equal to a character lifted from IF. if and only if X ~ Jlh with hE IF q' whcrc Jl, is the canonical additivc charactcr of L'. 5.14. Prove for c EO IF; that 5.1 I.

L

Jl(d)

Ly'd'(C)

"I',,· Ii ¢(d) ,'"

if (' is a primitive element of IF q' otherwise. where in the outer sum d runs through all positive divisors of q - I and in the inner sum tJ/dl runs through the ¢( d) multiplicative characters of IF" of order d. Here Jl dcnotes the Moebius function (sec Definition 3.22) and Euler's function (see Theorem 1.15 (iv)). 5.15. Show that ~(2) ~ (- I)'" iii', where ~ is the quadratic character of iF" q odd. 5.16. For rEI'\I prove G(ti-P',Xb)~C(Y.Xp'b,), where p(h)~hP' for h F IF q and p is the characteristic of IF q'

IX]

r."'cn;iscs

5.17. 5.18. 5.19.

5.20.

Prove Lx C( y. x) ~ 0 for all multiplicative characters y of IF q • where the sum is extended over all additive characters X of IF q' Prove r..;G(Y.X)~(q -I)X(I) for all additive characters X of IF q • where the sum is extended over all multiplicative characters y of IF q' For the quadratic character ~ of IF,!. q ~ p'. p an odd prime. seN. and an additive character Xb' b <= IF q • in the notation of Theorem 5.7. prove that G ( ~. Xb ) ~ ~ (b )( - I)' q , IlI'i ,,,' . 'p • Sl/4 q 1/2 • If q is odd and

~

is the quadratic character of IF q' prove that

G(~.Xa)C(~,X")- ~(-ah)qfora.hEIF:.

5.21.

ese the law of quadratic reciprocity to evaluate the Legendre symhols ( Po) and (i, ).

5.22.

Determine all primes p such that (

5.23.

~J ) =

I.

Determine all odd prime powers q such that the quadratic character of IF q satisfies ~(3) ~ I. Prove that the polynomial x' t ax - h E 1F"lx]. q odd. is irreducible in IF ql x] if and only if ~(a' - 4b) ~ - I. Determine whether the polynomial X", -12.x +41 is irreducihle in 1F",lx]. Let p and r be distinct odd primes. let sEN he sueh that r' '" 1mod p. and let r he an element of order p in IF,":. For k E Z define

~

5.24. 5.25. 5.26.

C,

p

I

,

I

~ L (~)r"EIF, •. P

Prove the following properties: (i) G, - (; (.

1)lp· 1)/2 p .

)c,;

(ii)

G~ ~

when.~ the last expre~sion is viewed as an clement of

IF,. 5.27.

5.2R.

Use the results of Exercise 5.26 to prove the law of quadratic reciprocity. Prove that

L

5.29.

y(c-a)y(c+b)= ·1

for a. h E= IF If with a ~ b. where V i~ a nontrivial multiplicative character of IF q' Let y he a nontrivial multiplicative character of IF q and let 5 he a suhset of IF q with h clements. Prove that 2

L 'L cC IF

5.30.

,I

(j ("

y(c+a) .I)

=h(q

h).

:

I.et A,. A,. A, be nontrivial multiplicative characters of IF. and let

lX4

Exponential Sums

q2 .. 3q ~ { q'-2q-l 5.31.

if 1"1,, nontriviaL if 1,,1,2 triviaL

Let tf, he a multiplicative character of I' q of ordcr prove

L:

tf,(ac") - {(q -l)y(a) ,err, 0 5.32.

if

In

In>

I. For a E I' q

divides n,

otherwise.

Prove that L".• ~l.rk)) = 0 if q = 3 mod4. ~ is the quadratic character of t", and jErqlx] is an odd polynomial that is, a polynomial with

fl - x) =

fIx).

Chapter 6

Linear Recurring Sequences

Sc.:<..juencc~

in finite fields whose terms depend in a simple manner on their

prcdecc~sors

are of importance for a variety of applications. Such sequences

are ~asy to generate by recursive procedures, which is certainly an advantageous feature from the computational viewpoint, and they also tend to have u~dul structural properties. Of particular interest is the case where the terms depend linearly on a fixed number of predecessors, resuiting in a so-called linear recurring sequence. These sequences arc employed. for instance. in coding theory (sec Chapter g, Section 2), in cryptography (sec Chapter 9, Section 2), and in several branches of electrical engineering. Tn these applications, tbc underlying field is often taken to be l 2' but the theory can be developed quite generally for any finite field. In Section I we show how to implement the generation of linear recurring sequences on special switching circuits called feedback shift registers. We discuss also some basic periodicity properties of such sequences. Section 2 introduces the concept of an impulse response sequence, which is of both practical and theoretical interest. I'urther relations to periodicity properties are found in this way, and also through the use of the so-called characteristic polynomial of a linear recurring sequence. Another application of the characteristic polynomial yields explicit formulas for the terms of

a linear n.::curring sequence. Maximal period sequenc~s arc also defined in this section. These scquem:es will appear in various applications in later chapters. The theory of linear recurring sequences can be approached via linear algebra. ideal theory, or formal power scrics. An approach hased on

l.incaf I'h'curring Sequenccs

the latt~r is pft:sl..:nt~d in Section 3. Thi~ leads to a computation-oriented way of introducing th~ minimal polynomial of a linear recurring sequence in the next section. The minimal polynomial is of crucial importance for the linear recurring sequence, since the order of the minimal polynomial gives the least period of the sequence. In Section 5 we study the collection of all sequences satisfying a given linear recurrence relation. This information is useful in the discussion of opaation~ with lin~ar recurring sequences, such as termwise addition and multiplication for sequences in general finite fields and binary complementation for sequences in F,. We consider also the problem of determining the vari()u~ least periods: of th~ s~qucnces generated by a fixed linear recurrence relation. Section 6 presents some dcterminantal criteria characterizing linear recurring sequences as well as the Berlekamp-Massey algorithm for the calculation of minimal polynomials. Section 7 is devoted to distribution properties of linear rl..:eurring :"lequencl..:s. Exponential sums with linear recurring sequences are the main tools for studying sucb properties.

I.

FEEDBACK SHIFT REGISTERS, PERIODICITY PROPERTIES

I.et k he a positive integer. and let u. QW.;,.•• a k _ 1 be given elements of a finite field IF q' i\ sequence so. s, .... of elcments of IF q satisfying tbe relation ·\'1.k=ak.

ISn·A

I ;·aJ"- 2 .'i'1'J,,

)

•..•

-l-aosn~a

forn~O.l....

(6.1)

is called a (Juh-order) linear recurring st!quence in IF'I' The terms so. I' which determine the: reM of the sequence: uniquely. are referred to as tbe inilial willes. i\ relation of tbe form (6.1) is called a (klh-order) linear recurrenc(' relation. In the older literature one may also find the term "difference: equation:' We ~pea'" of a homogeneous linear recurrence relation if a = 0: othe:rwise th~ linear r~currence rdation is inhomogeneous. The se4u~nce SUo Sl···· it."elf is eall~d a homogeneous. or inhomo/l,eneous. linear recurring "it!quenc(' in IF q' respectively. The generatil)O of linear recurring sequence~ ean he imple:mented on a jeedhack ,hijl regisler. This i, a special kind of electronic switcbing circuit handling inf0rmativn in the form of elements of IF q' which are reprcsentl:d suitahly. Fl)llr types of devices are used. The first is an adder, which ha~ two inputs and one output. the output heing th~ sum in lr,{ of the two inputs. The second is a COlIStWll multiplier, which has one: input and yields as the output tbe prnduct of tbe input witb a constant clement of IF q' Tbe third is a constant udder, which is analogous to a constant multiplier, hut adds a constant element (If IF q to the input. Tbe fourth type of device is a de/av

SI ..... S/..

IR7

e1emell[ ("nip-OOP"), which has one input and one output and is regulated

by an external synchronous clock so that its input at a panicular time appears as its output one unit of time later. We shall not he concerned here with the physical realization of these devices. The representation of the components in circuit diagrams is shown in Figure 6.1.

A feed hack shift register is huilt by interconnecting a finite number of adders, constant multipliers, constant adders, and delay elements along a closed loop in such a way that two outputs are never connected together. Actually, for the purpose of generating linear recurring sequences. it suffices 10 connect the components in a rather special manner. A feedback shift register that gt.:ncrates a linear recurring sequence satisfying (6.1) is shown

in Figure 6.2. At the outset. each delay element {)i' j ~ 0, I, .... k - I, contains the initial value ")' If we think of the arithmetic operations and the transfer along the wires to be performed instantaneously, then after one time unit

each V) will contain ")' \. Continuing in this manner, we sec that the output of the feedback shift register is the string of element so' S\, 5" .... received in intervals of one time unit. In most of the applications the desired linear recurring sequence is homogeneous. in which case the constant adder is not needed. 6.1. Example. In order to generate a linear recurring sequence in .0=, satisfying the homogeneous linear recurrence relation sn+6=sn .

(a) Adder

.s+2s1114-t sn'I+3sn

(bJ C:on<;lant Olultiplier

(c) Consrant adder

for mulTiplying by a

H(;1 IRE 6.1

forn=O,I, ... ,

lbc building block~ of feedback ~hift regiMer<;. (a) Adder. (b) Constant multiplier multiplyin~ b~· a. (c) Con..ta01 adder for addin~ a. (d) Delay element.

for

_--H OUl["lll1

~

FI<;t:RE h.2

(d) LJclay element

for adding a

The

1)1

~en!tral

0 form ofa fel'dback 'ihift

I~l r('gi~ter.

IXX

J.int'ar Recurring, Sr.:qurnccs

one may use the feedback shift register shown in Figure 6.3. Since a 2 O. no connections are necessary at these points. 6.2.

Example.

-

a, :J

Consider the homogeneous linear recurrence relation

S'l.7::'~Sn.4""'Sn'1+S'lt2+Sn'

n=O,l, ... , inlF 2 •

A feedback shift register corresponding to this linear recurrence relation is shown in Figure 6.4. Since multiplication by a constant in IF 2 either preserves or annihilates clements. the effect of a constant multiplier can be simulated by a wire connection or a disconnection. Therefore. a feedback shift register for the generation of binary homogeneous linear recurring sequences requires only delay clements. adders, and wire connections. 0

Let so. 5\, ... be a k th-ordcr linear recurring sequence in IF I{ satisfying (6.1). As we have noted. this sequence can be generated by the feedback shift register in Figure 6.2. If II is a nonnegative integer. then after II time units the delay clement Dr} ~ O.l ..... k -I. will contain So .,' It is therefore natural to call the row vector sn = (sn' sn_ \.... ,5'1 t" I) the nth stale reclOr of the linear recurring sequence (or of the feedback shift register). The state vector so'" (so' s\ ..... s" I) is also referred to as the initial state veCTOr. It is a characteristic feature of linear recurring sequences in finite fields that. after a possibly irregular behavior in the beginning. such sequences arc eventually of a periodic nature (or ultimately periodic in the sense of Definition 6.3 below). Before studying this property in detail. we introduce some terminology and mention a few general facts about ultimately periodic sequences.

Output

FI(;LN.F. 6.3 The feedback 'Shift regiS(Cf for F:xamplc 6.1.

_.-LJ Output

FI (; li RE 6.4

L----'

The feedback shifl regiMer for Example 6.2.

l. I"t'cdhack Shift I{cgislcrs. Pcriodicit.y Propcrtic.. .

1~9

6.3. Definition. Let S be an arhitrary nonempty set, and let 5", 5, •... he a sequence of clements of S. If there exist integers r > 0 and no;;' 0 such that sn I r = sn for all n )- no, then the sequence is called UITimalely periodic and r is called a period of the sequence. The smallest numher among all the possible periods of an ultimately periodic sequence is called the leusl period of the sequence. 6.4. Lemma. Every period of un uilimalelv periodic sequence is divisible hy Ihe leasl period. Proof Let r be an arbitrary period of the ultimately periodic sequence so' SI.· .• and let r\ he its least period. so that we have s1/ {r = sn for alln ~ no and Sn~ r l = SII for all n)- n\ with suitable nonnegative integers no and n,. If r were not divisible by rio we could use the division algorithm for integers to write r ~ mr, + I with integers m;;, 1 and 0 < I < rio Then, for all n;;, max(n o' {/,) we get and so I is a period of the sequence. which contradicts the definition of the least period. 0 6.5. Definition. An ultimately periodic sequence "0,5, period r is called periodic if sn', = 5 n holds for all n = O. I .

with least

The following condition, which is sometimes found in the literature. is equivalent to the definition of a periodic sequence. 6.6.

Lemma.

The sequence so'

exiSTS an inreg,er r > 0 such thaT

SII t

r

5" ..• is periodic if and = s" for all n = O. 1.....

only if Ihere

Proof The necessity of the condition is obvious. Conversely, if the condition is satisfied, then the sequence is ultimately periodic and has a least period rl' Therefore. with a suitable no we have sn+rl = sn for all n)- no' Now let n be an arhitrary nonnegative integer. and choose an integer In ~ no with m == nmod r. Then s""f"r = sm I rl = sm = sn' which shows that the l 0 sequence is periodic in the sense of Definition 6.5. If so. ", .... is ultimately periodic with least period r, then the least nonnegative integer no such that Sn_r = sn for all n ~ 11 0 is called the preperind. The sequence is periodic precisely if the preperiod is O. W~ return now to linear recurring sequences in finite fields and establish the basic results concerning the periodicity behavior of such sequences.

6.7.

Thearem.

Lel I'q he any finile field and k any posilice inreger.

Then every kT