# Finite Fields. Sophie Huczynska. Semester 2, Academic Year

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1 Finite Fields Sophie Huczynska Semester 2, Academic Year

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3 Chapter 1. Introduction Finite fields is a branch of mathematics which has come to the fore in the last 50 years due to its numerous applications, from combinatorics to coding theory. In this course, we will study the properties of finite fields, and gain experience in working with them. In the first two chapters, we explore the theory of fields in general. Throughout, we emphasize results particularly important to finite fields, but allow fields to be arbitrary unless otherwise stated. 1. Group theory: a brief summary We begin by recalling the definition of a group. Definition 1.1. A group is a set G, together with a binary operation, such that the following axioms hold: Closure: G is closed under the operation : x, y G = x y G; Associativity: (x y) z = x (y z) for all x, y, z G; Identity: there exists an element e G (called the identity of G) such that x e = e x = x for all x G; Inverses: for every element x G there exists an element x 1 G (called the inverse of x) such that x x 1 = x 1 x = e. Definition 1.2. A group G is said to be abelian if the binary operation is commutative, i.e. if x y = y x for all x, y G. The operation is often replaced by + for abelian groups, i.e. x y is written x + y Example 1.3. The following are examples of groups: The set Z of integers with the operation of addition; 3

4 4 CHAPTER 1. INTRODUCTION the set of n n matrices with real entries, with the operation of matrix multiplication, forms a group. Let G be the set of remainders of all the integers on division by n, e.g. G = {0, 1,..., n 1}. Let a b be the operation of taking the integer sum a + b and reducing it modulo n. Then (G, ) is a group. Definition 1.4. A multiplicative group G is said to be cyclic if there is an element a G such that for any b G there is some integer j with b = a j. Such an element is called a generator of the cyclic group, and we write G =< a >. Note we may have more than one generator, eg either 1 or 1 can be used to generate the additive group Z. Definition 1.5. For a set S, a subset R of S S is called an equivalence relation on S if it satisfies: (s, s) R for all s S (reflexive) If (s, t) R then (t, s) R (symmetric) If (s, t), (t, u) R then (s, u) R (transitive). An equivalence relation R on S induces a partition of S. If we collect all elements of S equivalent to a fixed s S, we obtain the equivalence class of s, denoted by [s] = {t S : (s, t) R}. The collection of all distinct equivalence classes forms a partition of S, and [s] = [t] (s, t) R. Definition 1.6. For arbitrary integers a,b and positive integer n, we say that a is congruent to b modulo n if the difference a b is a multiple of n, i.e. a = b + kn for some integer k. It is easily checked that congruence modulo n is an equivalence relation on the set Z of integers. Consider the equivalence classes into which the relation partitions Z. These are the sets: [a] = {m Z : m a mod n} = {m Z : m = a + kn for some k Z}.

5 1. GROUP THEORY: A BRIEF SUMMARY 5 E.g. for n = 4 we have: [0] = {,..., 8, 4, 0, 4, 8,...}; [1] = {,..., 7, 3, 1, 5, 9,...}; [2] = {,..., 6, 2, 2, 6, 10,...}; [3] = {,..., 5, 1, 3, 7, 11,...}. We may define on the set {[0], [1],..., [n 1]} a binary operation, which we shall write as + (though it is not ordinary addition) by [a] + [b] = [a + b] where a and b are any elements of the sets [a] and [b] respectively, and a + b is the ordinary sum of a and b. Can show (exercise) that this is well-defined, i.e. does not depend on choice of representatives. Theorem 1.7. The set {[0], [1],..., [n 1]} of equivalence classes modulo n forms a group under the operation + given by [a] + [b] = [a + b]. It is called the group of integers modulo n and is denoted Z n. It is cyclic with [1] as a generator. Proof. See Example Sheet. Definition 1.8. A group is called finite (respectively, infinite) if it contains finitely (respectively, infinitely) many elements. The number of elements of a finite group G is called its order, written G. Definition 1.9. A subset H of the group G is a subgroup of G if H is itself a group with respect to the operation of G. The (cyclic) subgroup consisting of all powers of some element a G is denoted < a > and called the subgroup generated by a. Next, we generalize the notion of congruence, as follows. Theorem If H is a subgroup of G, then the relation R H on G defined by (a, b) R H if and only if a = bh for some h H (additively, a = b+h for some h H) is an equivalence relation. The relation is called left congruence modulo H. Note that when (G, ) = (Z, +) and H =< n >, we get back our previous definition of congruence, since a b mod n a = b + h for some h < n >. The equivalence classes are called the left cosets of H in G; each has size H. Right congruence and right cosets are defined analogously.

6 6 CHAPTER 1. INTRODUCTION Definition The index of H in G (denoted by [G : H]) is the number of left cosets of H in G, and is equal to the number of right cosets of H in G. Theorem The order of a finite group G is equal to the product of the order of any subgroup H and the index of H in G. In particular, the order of H divides the order of G and the order of any element a G divides the order of G. Proof. Exercise We can easily describe subgroups and orders for cyclic groups. In what follows, φ is Euler s function; i.e. φ(n) = the number of integers k with 1 k n which are relatively prime to n. If the integer n has the prime factorization p k 1 1 p k p k r r, then φ(n) = n(1 1 p 1 )(1 1 p 2 ) (1 1 p r ). So, for example, φ(7) = 6 and φ(30) = Sheet for more details = 8. See Example Theorem cyclic. (i) Every subgroup S of a cyclic group G =< a > is (ii) In a finite cyclic group < a > of order m, the element a k generates a subgroup of order m. gcd(k,m) (iii) For any positive divisor d of m, < a > contains precisely one subgroup of order d and precisely one subgroup of index d. (iv) Let f be a positive divisor of m. Then < a > contains φ(f) elements of order f. (v) A finite cyclic group < a > of order m contains φ(m) generators, namely the powers a r with gcd(r, m) = 1. Proof. (i) If S = {1}, then S is cyclic with generator 1. Otherwise, let m be the least positive integer for which a m S. We will show: S =< a m >. Clearly < a m > S. Now, take arbitrary s S; s = a k for some k Z. By the division algorithm for integers, there exist q, r Z with 0 r < m such that k = qm + r. Then a k = a qm+r = (a m ) q.a r, implying a r S. If r > 0, this contradicts the minimality of m, so we must have r = 0 and hence s = a k = (a m ) q < a m >.

7 1. GROUP THEORY: A BRIEF SUMMARY 7 (ii) Set d = gcd(k, m). The order of a k is the least positive integer n such that a kn = e. This identity holds if and only if m divides kn, i.e. if and only if m d divides n. The least positive n with this property is n = m d. (iii) Exercise: see Example Sheet. (iv) Let < a > = m and m = df. By (ii), an element a k is of order f if and only if gcd(k, m) = d. So the number of elements of order f is equal to the number of integers k with 1 k m and gcd(k, m) = d. Equivalently, writing k = dh with 1 h f, the condition becomes gcd(h, f) = 1. There are precisely φ(f) such h. (v) The first part follows from (iv), since the generators of < a > are precisely the elements of order m. The second part follows from (ii). Definition A subgroup H of G is normal its left and right cosets coincide. For a normal subgroup H, the set of (left) cosets of H in G forms a group, denoted G/H. The operation is (ah)(bh) = (ab)h. Definition A mapping f : G H of the group G into the group H is called a homomorphism of G into H if f preserves the operation of G. If f is a bijective homomorphism it is called an isomorphism and we say G and H are isomorphic. An isomorphism of G onto itself is called an automorphism of G. Definition The kernel of the homomorphism f : G H of the group G into the group H is the set (actually, normal subgroup) ker(f) = {a G : af = e H }. The image of f is the set (actually, subgroup) im(f) = {af : a G}. Theorem [First Isomorphism Theorem] Let f : G H be a homomorphism of groups. Then kerf is a normal subgroup of G and Proof. Omitted. G/kerf = imf. Example Take G = Z, H = Z n and f : a [a]. Then f is a homomorphism with ker(f) =< n > and im(f) = Z n, and so the First Isomorphism Theorem says that Z/ < n > and Z n are isomorphic as groups.

8 8 CHAPTER 1. INTRODUCTION 2. Rings and fields Definition 2.1. A ring (R, +, ) is a set R, together with two binary operations, denoted by + and, such that R is an abelian group with respect to +; R is closed under ; is associative, that is (a b) c = a (b c) for all a, b, c R; the distributive laws hold, that is, for all a, b, c R we have a (b+c) = (a b) + (a c) and (b + c) a = (b a) + (c a). Typically, we use 0 to denote the identity element of the abelian group R with respect to addition, and a to denote the additive inverse of a R. Definition 2.2. A ring is called a ring with identity if the ring has a multiplicative identity (usually denoted e or 1). A ring is called commutative if is commutative. A ring is called an integral domain if it is a commutative ring with identity e 0 in which ab = 0 implies a = 0 or b = 0 (i.e. no zero divisors). A ring is called a division ring (or skew field) if the non-zero elements form a group under. A commutative division ring is called a field. Example 2.3. a field; the integers (Z, +, ) form an integral domain but not the rationals (Q, +, ), reals (R, +, ) and complex numbers (C, +, ) form fields; the set of 2 2 matrices with real entries forms a non-commutative ring with identity w.r.t. matrix addition and multiplication. the group Z n with addition as before and multiplication defined by [a][b] = [ab] is a commutative ring with identity [1].

9 2. RINGS AND FIELDS 9 So, in summary: a field is a set F on which two binary operations, called addition and multiplication, are defined, and which contains two distinguished elements e and 0 with 0 e. Moreover, F is an abelian group with respect to addition, having 0 as the identity element, and the non-zero elements of F (often written F ) form an abelian group with respect to multiplication having e as the identity element. The two operations are linked by the distributive laws. Theorem 2.4. Every finite integral domain is a field. Proof. Let R be a finite integral domain, and let its elements be r 1, r 2,..., r n. Consider a fixed non-zero element r R. Then the products rr 1, rr 2,..., rr n must be distinct, since rr i = rr j implies r(r i r j ) = 0, and since r 0 we must have r i r j = 0, i.e. r i = r j. Thus, these products are precisely the n elements of R. Each element of R is of the form rr i ; in particular, the identity e = rr i for some 1 i n. Since R is commutative, we also have r i r = e, and so r i is the multiplicative inverse of r. Thus the non-zero elements of R form a commutative group, and R is a field. Definition 2.5. A subset S of a ring R is called a subring of R if S is closed under + and and forms a ring under these operations. A subset J of a ring R is called an ideal if J is a subring of R and for all a J and r R we have ar J and ra J. Let R be a commutative ring with an identity. Then the smallest ideal containing an element a R is (a) = {ra : r R}. We call (a) the principal ideal generated by a. Definition 2.6. An integral domain in which every ideal is principal is called a principal ideal domain (PID). Example 2.7. Z is a PID. An ideal J of R defines a partition of R into disjoint cosets, residue classes modulo J. These form a ring w.r.t. the following operations: (a + J) + (b + J) = (a + b) + J, (a + J)(b + J) = ab + J. This ring is called the residue class ring and is denoted R/J.

10 10 CHAPTER 1. INTRODUCTION Example 2.8. The residue class ring Z/(n) Here, (n) is the principal ideal generated by the integer n (same set nz as the subgroup < n > but now with two operations). As in the group case, we denote the residue class of a modulo n by [a], as well as by a + (n). The elements of Z/(n) are [0] = 0 + (n), [1] = 1 + (n),..., [n 1] = n 1 + (n). Theorem 2.9. Z/(p), the ring of residue classes of the integers modulo the principal ideal generated by a prime p, is a field. Proof. By Theorem 2.4, it is enough to show that Z/(p) is an integral domain. Now, [a][b] = [ab] = [0] if and only if ab = kp for some k Z. Since p is prime, p divides ab if and only if p divides one of the factors. So, either [a] = [0] or [b] = [0], so Z/(p) contains no zero divisors. These are our first examples of finite fields! Example Here are the addition and multiplication tables for the field Z/(3): (3) 1 + (3) 2 + (3) 0 + (3) 0 + (3) 1 + (3) 2 + (3) 1 + (3) 1 + (3) 2 + (3) 0 + (3) 2 + (3) 2 + (3) 0 + (3) 1 + (3), 0 + (3) 1 + (3) 2 + (3) 0 + (3) 0 + (3) 0 + (3) 0 + (3) 1 + (3) 0 + (3) 1 + (3) 2 + (3) 2 + (3) 0 + (3) 2 + (3) 1 + (3). Remark As you will prove in the Example Sheet, the above result does not hold if p is replaced by a composite n. Definition A mapping φ : R S (R,S rings) is called a ring homomorphism if for any a, b R we have φ(a + b) = φ(a) + φ(b) and φ(ab) = φ(a)φ(b). A ring homomorphism preserves both + and and induces a homomorphism of the additive group of R into that of S. Concepts such as kernel and image are defined analogously to the groups case. We have a ring version of the First Isomorphism Theorem: Theorem 2.13 (First Isomorphism Theorem for Rings) If φ is a ring homomorphism from a ring R onto a ring S then the factor ring R/kerφ and the ring S are isomorphic.

11 2. RINGS AND FIELDS 11 We can use mappings to transfer a structure from an algebraic system to a set without structure. Given a ring R, a set S and a bijective map φ : R S, we can use φ to define a ring structure on S that converts φ into an isomorphism. Specifically, for s 1 = φ(r 1 ) and s 2 = φ(r 2 ), define s 1 + s 2 to be φ(r 1 + r 2 ), and s 1 s 2 to be φ(r 1 )φ(r 2 ). This is called the ring structure induced by φ; any extra properties of R are inherited by S. This idea allows us to obtain a more convenient representation for the finite fields Z/(p). Definition For a prime p, let F p be the set {0, 1,..., p 1} of integers, and let φ : Z/(p) F p be the mapping defined by φ([a]) = a for a = 0, 1,..., p 1. Then F p endowed with the field structure induced by φ is a finite field, called the Galois field of order p. From above, the mapping φ becomes an isomorphism, so φ([a] + [b]) = φ([a])+φ([b]) and φ([a][b]) = φ([a])φ([b]). The finite field F p has zero element 0, identity element 1 and its structure is that of Z/(p). So, computing with elements of F p now means ordinary arithmetic of integers with reduction modulo p. Example F 2 : the elements of this field are 0 and 1. The operation tables are: , We have Z/(5), isomorphic to F 5 = {0, 1, 2, 3, 4}, where the isomorphism is given by [0] 0,..., [4] 4. The operation tables are: , Definition If R is an arbitrary ring and there exists a positive integer n such that nr = 0 for every r R (i.e. r added to itself n times is the zero element) then the least such positive integer n is called the characteristic of R, and R is said to have positive characteristic. If no such positive integer n exists, R is said to have characteristic 0.

12 12 CHAPTER 1. INTRODUCTION Example F 2 and F 5 have characteristic 2 and 5 respectively. Q and R have characteristic 0. Theorem A ring R {0} of positive characteristic with an identity and no zero divisors must have prime characteristic. Proof. Since R contains non-zero elements, R has characteristic n 2. If n were not prime, we could write n = km with k, m Z, 1 < k, m < n. Then 0 = ne = (km)e = (ke)(me), so either ke = 0 or me = 0, since R has no zero divisors. Hence either kr = (ke)r = 0 for all r R or mr = (me)r = 0 for all r R, contradicting the definition of n as the characteristic. Corollary A finite field has prime characteristic. Proof. From Theorem 2.18, we need only show that a finite field F has a positive characteristic. Consider the multiples e, 2e, 3e,... of the identity. Since F contains only finitely many elements, there must exist integers k and m with 1 k < m such that ke = me, i.e. (k m)e = 0, and so F has a positive characteristic. Example The field Z/(p) (equivalently, F p ) has characteristic p. Theorem 2.21 (Freshmen s Exponentiation) Let R be a commutative ring of prime characteristic p. then for a, b R and n N. (a + b) pn = a pn + b pn and (a b) pn = a pn b pn Proof. It can be shown (see Example Sheet) that ( ) p p(p 1) (p i + 1) = 0 mod p i 1 2 i for all i Z with 0 < i < p. By the Binomial Theorem, ( ) ( ) p p (a + b) p = a p + a p 1 b + + ab p 1 + b p = a p + b p 1 p 1 and induction on n establishes the first identity. The second identity follows since a pn = ((a b) + b) pn = (a b) pn + b pn.

13 3. POLYNOMIALS Polynomials Let R be an arbitrary ring. A polynomial over R is an expression of the form f(x) = n a i x i = a 0 + a 1 x + + a n x n, i=0 where n is a non-negative integer, the coefficients a i (0 i n) are elements of R, and x is a symbol not belonging to R, called an indeterminate over R. Definition 3.1. Let f(x) = n i=0 a ix i = a 0 +a 1 x+ +a n x n be a polynomial over R which is not the zero polynomial, so we can suppose a n 0. Then n is called the degree of f(x). By convention, deg(0) =. Polynomials of degree 0 are called constant polynomials. If the leading coefficient of f(x) is 1 (the identity of R) then f(x) is called monic polynomial. Given two polynomials f and g, we can write f(x) = n i=0 a ix i and g(x) = n i=0 b ix i (taking coefficients zero if necessary to ensure same n). We define their sum to be n f(x) + g(x) = (a i + b i )x i and their product to be i=0 f(x)g(x) = 2n c k x k, where c k = a i b j. k=0 i+j=k,0 i n,0 j n Theorem 3.2. With the above operations, the set of polynomials over R forms a ring. It is called the polynomial ring over R and denoted by R[x]. Its zero element is the zero polynomial, all of whose coefficients are zero. Proof. Exercise. Let F denote a (not necessarily finite) field. From now on, we consider polynomials over fields. We say that the polynomial g F [x] divides f F [x] if there exists a polynomial h F [x] such that f = gh. Theorem 3.3 (Division Algorithm) Let g 0 be a polynomial in F [x]. Then for any f F [x], there exist polynomials q, r F [x] such that f = qg + r, where deg(r) < deg(g).

14 14 CHAPTER 1. INTRODUCTION Using the division algorithm, we may show that every ideal of F [x] is principal. Theorem 3.4. F [x] is a principal ideal domain. In fact, for every ideal J (0) of F [x] there is a uniquely determined monic polynomial g F [x] such that J = (g). Proof. Let I be an ideal in F [x]. If I = {0}, then I = (0). If I {0}, choose a non-zero polynomial k(x) I of smallest degree. Let b be the leading coefficient of k(x), and set m(x) = b 1 k(x). Then m I and m is monic. We will show: I = (m). Clearly, (m) I. Now take f(x) I; by the division algorithm there are polynomials q(x), r(x) with f(x) = q(x)m(x) + r(x) where either r(x) = 0 or deg(r) < deg(m). Now, r(x) = f(x) q(x)m(x) I. If r(x) 0, we contradict the minimality of m; so we must have r(x) = 0, i.e. f is a multiple of m and I = (m). We now show uniqueness: if m 1 F [x] is another monic polynomial with I = (m 1 ), then m = c 1 m 1 and m 1 = c 2 m with c 1, c 2 F [x]. Then m = c 1 c 2 m, i.e. c 1 c 2 = 1, and so c 1, c 2 are constant polynomials. Since both m and m 1 are monic, we must have m = m 1. We next introduce an important type of polynomial. Definition 3.5. A polynomial p F [x] is said to be irreducible over F if p has positive degree and p = bc with b, c F [x] implies that either b or c is a constant polynomial. A polynomial which does allow a non-trivial factorization over F is called reducible over F. Note that the field F under consideration is all-important here, e.g. the polynomial x 2 +1 is irreducible in R[x] but reducible in C[x], where it factors as (x + i)(x i). Theorem 3.6 (Unique Factorization) Any polynomial f F [x] of positive degree can be written in the form f = ap e p e k k where a F, p 1,..., p k are distinct monic irreducibles in F [x] and e 1,..., e k are positive integers. This factorization is unique up to the order in which the factors occur; it is called the canonical factorization of f in F [x]. Proof. Omitted. Example 3.7. Find all irreducible polynomials over F 2 of degree 3.

15 3. POLYNOMIALS 15 First, note that a non-zero polynomial in F 2 [x] must be monic. The degree 3 polynomials are of the form x 3 + ax 2 + bx + c, where each coefficient is 0 or 1, i.e. there are 2 3 = 8 of them. Such a polynomial is reducible over F 2 precisely if it has a divisor of degree 1. Compute all products (x + a 0 )(x 2 + b 1 x + b 0 ) to obtain all reducible degree 3 polynomials over F 2. There are 6 of these, leaving 2 irreducibles: x 3 + x + 1 and x 3 + x Theorem 3.8. For f F [x], the residue class ring F [x]/(f) is a field if and only if f is irreducible over F. Proof. Details omitted. For those who know some ring theory this is immediate since, for a PID S, S/(c) is a field if and only if c is a prime element of S. Here, the prime elements of the PID R[x] are precisely the irreducible polynomials. We will be very interested in the structure of the residue class ring F [x]/(f), for arbitrary non-zero polynomial f(x) F [x]. To summarize, F [x]/(f) consists of residue classes g + (f) (also denoted [g]) with g F [x]. Two residue classes g + (f) and h + (f) are identical if and only if g h mod f, i.e. precisely if g h is divisible by f. This is equivalent to: g and h have the same remainder on division by f. Each residue class g + (f) contains a unique representative r F [x] with deg(r) < deg(f), namely the remainder when g is divided by f. The process of moving from g to r is called reduction mod f. (Exercise: uniqueness?) Hence the distinct residue classes comprising F [x]/(f) are precisely the residue classes r + (f), where r runs through all polynomials in F [x] with deg(r) < deg(f). In particular, if F = F p and deg(f) = n, then the number of elements of F p /(f) is equal to the number of polynomials in F p /(f) of degree < n, namely p n. Example 3.9. Let f(x) = x F 2 [x]. The field F p /(x) has 2 1 = 2 elements, namely 0 + (x) and 1 + (x). This field is isomorphic to F 2.

16 16 CHAPTER 1. INTRODUCTION Let f(x) = x 2 + x + 1 F 2 [x]. Then F 2 [x]/(f) is a finite field of 2 2 = 4 elements: {0 + (f), 1 + (f), x + (f), x (f)}. Its behaviour under addition and multiplication is shown below (remember our field has characteristic 2). When performing field operations note that, since we replace each occurrence of f by 0, the polynomial representative for each residue class has degree less than (f) 1 + (f) x + (f) x (f) 0 + (f) 0 + (f) 1 + (f) x + (f) x (f) 1 + (f) 1 + (f) 0 + (f) x (f) 0 + (f) x + (f) x + (f) x (f) 0 + (f) 1 + (f) x (f) x (f) x + (f) 1 + (f) 0 + (f) 0 + (f) 1 + (f) x + (f) x (f) 0 + (f) 0 + (f) 0 + (f) 0 + (f) 0 + (f) 1 + (f) 0 + (f) 1 + (f) x + (f) x (f) x + (f) 0 + (f) x + (f) x (f) 1 + (f) x (f) 0 + (f) x (f) 1 + (f) x + (f) Note that, in the multiplication table, (x + (f))(x + (f)) = x 2 + (f) = f x 1 + (f) = x (f), (x + (f))(x (f)) = x 2 + x + (f) = f 1 + (f) = 1 + (f), (x (f))(x (f)) = x (f) = f x + (f) = x + (f). Comparing these tables to those in Example 2.15, we see that the field F 2 [x]/(f) is isomorphic to F 4. What is the multiplicative order of x + (f) in F 2 [x]/(f)? The multiplicative group of this field has order = 3, so the order must be 1 or 3. Clearly x + (f) 1 + (f), so the order must be 3. Check: (x+(f)) 3 = (x+(f))(x 2 +(f)) = x(x+1)+(f) = x 2 +x+(f) = 1+(f). Let f(x) = x F 3 [x]. We find that F 3 [x]/(f) is a ring of 9 elements which is not even a domain, let alone a field. Its elements are {0 + (f), 1 + (f), 2 + (f), x + (f), x (f), x (f), 2x + (f), 2x (f), 2x (f)}. To see that it is not an integral domain, note that (x (f))(x 1 + (f)) = x (f) = x (f) = 0 + (f), but neither x (f) nor x 1 + (f) are zero. Definition An element a F is called a root (or zero) of the polynomial f F [x] if f(a) = 0..,

17 3. POLYNOMIALS 17 Example (i) The elements 2, 3 Q are roots of x 2 5x + 6 Q[x]. (i) The polynomial x Q[x] has no roots in Q, but two roots ±i C. Definition If f(x) = a 0 + a 1 x + a 2 x a n x n F [x], then the derivative f of f is defined by f = f (x) = a 1 +2a 2 x+ +na n x n 1 F [x]. This obeys the familiar rules: [f(x) + g(x)] = f (x) + g (x) and [f(x)g(x)] = f(x)g (x) + f (x)g(x). Theorem An element a F is a root of the polynomial f F [x] if and only if x a divides f(x). Proof. Using the Division Algorithm, we can write f(x) = q(x)(x a) + c with q F [x] and c F. Substituting x = a, we get f(a) = c, hence f(x) = q(x)(x a) + f(a). The theorem follows from this identity. Definition Let a F be a root of f F [x]. If k is a positive integer such that f(x) is divisible by (x a) k but not (x a) k+1, then k is called the multiplicity of a. If k 2 then a is called a multiple root of f. Theorem An element a F is a multiple root of f F [x] if and only if it is a root of both f and its derivative f. Proof. Exercise Example Consider the polynomial f(x) = x 3 7x 2 +16x 12 Q[x]. It factors as (x 2) 2 (x 3), so its roots are 2 (with multiplicity 2) and 3 (with multiplicity 1). Here, f (x) = 3x 2 14x+16 which factors as (x 2)(3x 8), so we can verify that 2 is also a root of f. The following observation is very important. Theorem If F is a field and f(x) F [x] has degree n, then F contains at most n roots of f(x). Proof. Outline: Suppose F contains n + 1 distinct roots a 1,..., a n+1 of f(x). By Theorem 3.13, we can show that this implies f(x) = (x a 1 )(x a 2 ) (x a n+1 )g(x) for some polynomial g(x), contradicting deg(f) = n.

18 18 CHAPTER 1. INTRODUCTION

19 Chapter 2. Some field theory 4. Field Extensions Definition 4.1. Let F be a field. A subset K that is itself a field under the operations of F is called a subfield of F. The field F is called an extension field of K. If K F, K is called a proper subfield of F. Definition 4.2. A field containing no proper subfields is called a prime field. For example, F p is a prime field, since any subfield must contain the elements 0 and 1, and since it is closed under addition it must contain all other elements, i.e. it must be the whole field. Definition 4.3. The intersection of all subfields of a field F is itself a field, called the prime subfield of F. Remark 4.4. The prime subfield of F is a prime field, as defined above (see Example Sheet). Theorem 4.5. The prime subfield of a field F is isomorphic to Q if F has characteristic 0 and is isomorphic to F p if F has characteristic p. Proof. Denote by P (F ) the prime subfield of F. Let F be a field of characteristic 0; then the elements n1 F (n Z) are all distinct, and form a subring of F isomorphic to Z. The set Q(F ) = {m1 F /n1 F : m, n Z, n 0} is a subfield of F isomorphic to Q. Any subfield of F must contain 1 and 0 and so must contain Q(F ), so Q(F ) P (F ). Since Q(F ) is itself a subfield of F, we also have P (F ) Q(F ), so in fact Q(F ) is the prime subfield of F. If F has characteristic p, a similar argument holds with the set Q(F ) = {0 F, 1 F, 2(1 F ),..., (p 1)1 F }, and this is isomorphic to F p. 19

20 20 CHAPTER 2. SOME FIELD THEORY Definition 4.6. Let K be a subfield of the field F and M any subset of F. Then the field K(M) is defined to be the intersection of all subfields of F containing both K and M; i.e. it is the smallest subfield of F containing both K and M. It is called the extension field obtained by adjoining the elements of M. For finite M = {α 1,..., α n }, we write K(M) = K(α 1,..., α n ). If M = {α}, then L = K(α) is called a simple extension of K and α is called a defining element of L over K. The following type of extension is very important in the theory of fields in general. Definition 4.7. Let K be a subfield of F and α F. If α satisfies a nontrivial polynomial equation with coefficients in K, i.e. if a n α n + a n 1 α n a 1 α 1 + a 0 = 0 for some a i K not all zero, then α is algebraic over K. An extension L of K is called algebraic over K (or an algebraic extension of K) if every element in L is algebraic over K. Example 4.8. The element 3 3 R is algebraic over Q, since it is a root of the polynomial x 3 3 Q[x]. The element i C is algebraic over R, since it is a root of x 2 +1 R[x]. The element π R is not algebraic over Q. An element which is not algebraic over a field F is said to be transcendental over F. Given α F which is algebraic over some subfield K of F, it can be checked (exercise!) that the set J = {f K[x] : f(α) = 0} is an ideal of K[x] and J (0). By Theorem 3.4, it follows that there exists a uniquely determined monic polynomial g K[x] which generates J, i.e. J = (g). Definition 4.9. If α is algebraic over K, then the uniquely determined monic polynomial g K[x] generating the ideal J = {f K[x] : f(α) = 0} of K[x] is called the minimal polynomial of α over K. We refer to the degree of g as the degree of α over K. The key properties of the minimal polynomial are summarised in the next theorem. The third property is the one most useful in practice.

21 5. FIELD EXTENSIONS AS VECTOR SPACES 21 Theorem Let α F be algebraic over a subfield K of F, and let g be the minimal polynomial of α. Then (i) g is irreducible in K[x]; (ii) For f K[x], we have f(α) = 0 if and only if g divides f; (iii) g is the monic polynomial of least degree having α as a root. Proof. (i) Since g has the root α, it has positive degree. Suppose g = h 1 h 2 in K[x] with 1 deg(h i ) < deg(g) (i = 1, 2). This implies 0 = g(α) = h 1 (α)h 2 (α), and so one of h 1 or h 2 must lie in J and hence is divisible by g, a contradiction. (ii) Immediate from the definition of g. (iii) Any monic polynomial in K[x] having α as a root must be a multiple of g by (ii), and so is either equal to g or has larger degree than g. Example The element 3 3 R is algebraic over Q since it is a root of x 3 3 Q[x]. Since x 3 3 is irreducible over Q, it is the minimal polynomial of 3 3 over Q, and hence 3 3 has degree 3 over Q. The element i = 1 C is algebraic over the subfield R of C, since it is a root of the polynomial x R[x]. Since x is irreducible over R, it is the minimal polynomial of i over R, and hence i has degree 2 over R. 5. Field extensions as vector spaces Let L be an extension field of K. An important observation is that L may be viewed as a vector space over K. The elements of L are the vectors and the elements of K are the scalars. We briefly recall the main properties of a vector space. Definition 5.1. A vector space V over F is a non-empty set of objects (called vectors) upon which two operations are defined addition - there is some rule which produces, from any two objects in V, another object in V (denote this operation by +) scalar multiplication - there is some rule which produces, from an element of F (a scalar) and an object in V, another object in V and these objects and operations obey the Vector Space Axioms:

22 22 CHAPTER 2. SOME FIELD THEORY 1. x + y = y + x for all x, y V 2. (x + y) + z = x + (y + z) for all x, y, z V 3. there exists an object 0 V such that x + 0 = x for all x V 4. for every x V there exists an object x such that x + ( x) = 0 5. λ(x + y) = λx + λy for all x, y V and all scalars λ F 6. (λ + µ)x = λx + µx for all x V and all scalars λ, µ F 7. (λµ)x = λ(µx) for all x V and all scalars λ, µ F 8. 1x = x for all x V Definition 5.2. A basis of a vector space V over F is defined as a subset v 1,..., v n of vectors in V that are linearly independent and span V. If v 1,..., v n is a list of vectors in V, then these vectors form a basis if and only if every v V can be uniquely written as v = a 1 v a n v n where a 1,..., a n are elements of the base field F. A vector space will have many different bases, but there are always the same number of basis vectors in each. The number of basis vectors in any basis is called the dimension of V over F. Suppose V has dimension n over F. Then any sequence of more than n vectors in V is linearly dependent. To see that the vector space axioms hold for a field L over a subfield K, note that the elements of L form an abelian group under addition, and that any vector α L may be multiplied by an r K (a scalar ) to get rα L (this is just multiplication in L). Finally, the laws for multiplication by scalars hold since, for r, s L and α, β K we have r(α + β) = rα + rβ, (r + s)α = rα + sα, (rs)α = r(sα) and 1α = α. Example 5.3. Take L = C and let K be its subfield R. Then we can easily check that C is a vector space over R. Since we know from school that C = {a + bi : a, b R}, it is clear that a basis is given by {1, i}.

23 5. FIELD EXTENSIONS AS VECTOR SPACES 23 Definition 5.4. Let L be an extension field of K. If L is finite-dimensional as a vector space over K, then L is said to be a finite extension of K. The dimension of the vector space L over K is called the degree of L over K and written [L : K]. Example 5.5. From above, C is a finite extension of R, of degree 2. Theorem 5.6. If L is a finite extension of K and M is a finite extension of L, then M is a finite extension of K with [M : K] = [M : L][L : K]. Proof. Let [M : L] = m, [L : K] = n; let {α 1,..., α m } be a basis of M over L and let {β 1,..., β n } be a basis of L over K. We shall use them to form a basis of M over K of appropriate cardinality. Every α M can be expressed as a linear combination α = γ 1 α γ m α m for some γ 1,..., γ m L. Writing each γ i as a linear combination of the β j s we get α = m γ i α i = i=1 m n ( r ij β j )α i = i=1 j=1 m n r ij β j α i i=1 j=1 with coefficients r ij K. We claim that the mn elements β j α i form a basis of M over K. Clearly they span M; it suffices to show that they are linearly independent over K. Suppose we have m n s ij β j α i = 0 i=1 j=1 where the coefficients s ij K. Then m n ( s ij β j )α i = 0, i=1 and since the α i are linearly independent over L we must have j=1 n s ij β j = 0 j=1 for 1 i m. Now, since the β j are linearly independent over K, it follows that all the s ij are 0, as required.

24 24 CHAPTER 2. SOME FIELD THEORY Theorem 5.7. Every finite extension of K is algebraic over K. Proof. Let L be a finite extension of K and let [L : K] = m. For α L, the m+1 elements 1, α,..., α m must be linearly dependent over K, i.e. must satisfy a 0 + a 1 α + a m α m = 0 for some α i K (not all zero). Thus α is algebraic over K. Remark 5.8. The converse of Theorem 5.7 is not true, however. See the Example Sheet for an example of an algebraic extension of Q which is not a finite extension. We now relate our new vector space viewpoint to the residue class rings considered previously. Theorem 5.9. Let α F be algebraic of degree n over K and let g be the minimal polynomial of α over K. Then (i) K(α) is isomorphic to K[x]/(g); (ii) [K(α) : K] = n and {1, α,..., α n 1 } is a basis of K(α) over K; (iii) Every β K(α) is algebraic over K and its degree over K is a divisor of n. Proof. (i) Consider the evaluation at α mapping τ : K[x] K(α), defined by τ(f) = f(α) for f K[x]; it is easily shown that this is a homomorphism. Then kerτ = {f K[x] : f(α) = 0} = (g) by the definition of minimal polynomial. Let S be the image of τ, i.e the set of polynomial expressions in α with coefficients in K. By the First Isomorphism Theorem for rings we have S = K[x]/(g). Since g is irreducible, by Theorem 3.8, K[x]/(g) is a field and so S is a field. Since K S K(α) and α S, we have S = K(α) by the definition of K(α), and (i) follows. (ii) Spanning set: Since S = K(α), any β K(α) can be written in the form β = f(α) for some polynomial f K[x]. By the division algorithm, f = qg + r for some q, r K[x] and deg(r) < deg(g) = n. Then β = f(α) = q(α)g(α) + r(α) = r(α),

25 5. FIELD EXTENSIONS AS VECTOR SPACES 25 and so β is a linear combination of 1, α,..., α n 1 with coefficients in K. L.I.: if a 0 + a 1 α + + a n 1 α n 1 = 0 for some a 0,..., a n 1 K, then the polynomial h(x) = a 0 + a 1 x + + a n 1 x n 1 K[x] has α as a root, and is thus a multiple of its minimal polynomial g. Since deg(h) < n = deg(g), this is possible only if h = 0, i.e. a 0 = = a n 1 = 0. Thus the elements 1, α,..., α n 1 are linearly independent over K. (iii) K(α) is a finite extension of K by (ii), and so β K(α) is algebraic over K by Theorem 5.7. Moreover, K(β) is a subfield of K(α). If d is the degree of β over K, then n = [K(α) : K] = [K(α) : K(β)][K(β) : K] = [K(α) : K(β)]d, i.e. d divides n. Remark This theorem tells us that the elements of the simple extension K(α) of K are polynomial expressions in α, and any β K(α) can be uniquely expressed in the form β = a 0 + a 1 α + + a n 1 α n 1 for some a i K. Example Consider the simple extension R(i) of R. We saw earlier that i has minimal polynomial x over R. So R(i) = R[x]/(x 2 + 1), and {1, i} is a basis for R(i) over R. So R(i) = {a + bi : a, b R} = C. Example Consider the simple extension Q( 3 3) of Q. We saw earlier that 3 3 has minimal polynomial x 3 3 over Q. So Q( 3 3) = Q[x]/(x 3 3), and {1, 3 3, ( 3 3) 2 } is a basis for Q( 3 3) over Q. So Q( 3 3) = {a + b c( 3 3) 2 : a, b, c Q}. Note that we have been assuming that both K and α are embedded in some larger field F. Next, we will consider constructing a simple algebraic extension without reference to a previously given larger field, i.e. from the ground up. The next result, due to Kronecker, is one of the most fundamental results in the theory of fields: it says that, given any non-constant polynomial over any field, there exists an extension field in which the polynomial has a root. Theorem 5.13 (Kronecker) Let f K[x] be irreducible over the field K. Then there exists a simple algebraic extension of K with a root of f as a defining element. Proof.

26 26 CHAPTER 2. SOME FIELD THEORY Consider the residue class ring L = K[x]/(f), which is a field since f is irreducible. Its elements are the residue classes [h] = h + (f), with h K[x]. For any a K, think of a as a constant polynomial in K[x] and form the residue class [a]. The mapping a [a] gives an isomorphism from K onto a subfield K of L (exercise: check!), so K may be identified with K. Thus we can view L as an extension of K. For every h(x) = a 0 + a 1 x + + a m x m K[x], we have [h] = [a 0 + a 1 x + + a m x m ] = [a 0 ] + [a 1 ][x] + + [a m ][x] m = a 0 + a 1 [x] + + a m [x] m (making the identification [a i ] = a i ). So, every element of L can be written as a polynomial in [x] with coefficients in K. Since any field containing K and [x] must contain these expressions, L is a simple expression of K obtained by adjoining [x]. If f(x) = b 0 + b 1 x + + b n x n, then f([x]) = b 0 + b 1 [x] + + b n [x] n = [f] = [0], i.e. [x] is a root of f and L is a simple algebraic extension of K. Example Consider the prime field F 3 and the polynomial x 2 +x+2 F 3 [x], irreducible over F 3. Take θ to be a root of f, in the sense that θ is the residue class [x] = x + (f) L = F 3 [x]/(f). Explicitly, we have: f(θ) = f([x]) = f(x + (f)) = (x + (f)) 2 + (x + (f)) + (2 + (f)) = x 2 + x (f) = f + (f) = 0 + (f) = [0]. The other root of f in L is 2θ + 2, since f(2θ + 2) = θ 2 + θ + 2 = 0. By Theorem 5.9, the simple algebraic extension L = F 3 (θ) consists of the nine elements 0, 1, 2, θ, θ + 1, θ + 2, 2θ, 2θ + 1, 2θ + 2.

27 5. FIELD EXTENSIONS AS VECTOR SPACES 27 Example Consider the polynomial f(x) = x 2 + x + 1 F 2 [x], irreducible over F 2. Let θ be the root [x] = x+(f) of f; then the simple algebraic extension L = F 2 (θ) consists of the four elements 0, 1, θ, θ + 1. (The other root is θ + 1). The tables for addition and multiplication are precisely those of Example 3.9, now appropriately relabelled. We give the addition table: θ θ θ θ θ + 1 θ θ θ θ θ + 1 θ + 1 θ 1 0 Note that, in the above examples, adjoining either of two roots of f would yield the same extension field. Theorem Let α and β be two roots of a polynomial f K[x] that is irreducible over K. Then K(α) and K(β) are isomorphic under an isomorphism mapping α to β and keeping the elements of K fixed. Proof. Omitted. Given a polynomial, we now want an extension field which contains all its roots. Definition Let f K[x] be a polynomial of positive degree and F an extension field of K. Then we say that f splits in F if f can be written as a product of linear factors in F [x], i.e. if there exist elements α 1,..., α n F such that f(x) = a(x α 1 ) (x α n ) where a is the leading coefficient of f. The field F is called a splitting field of f over K if it splits in F and if F = K(α 1,..., α n ). So, a splitting field F of a polynomial f over K is the smallest field containing all the roots of f, in the sense that no subfield of F contains all roots of f. The following result answers the questions: can we always find a splitting field, and how many are there? Theorem 5.18 (Existence and uniqueness of splitting field) (i) If K is a field and f any polynomial of positive degree in K[x], then there exists a splitting field of f over K..

28 28 CHAPTER 2. SOME FIELD THEORY (ii) Any two splitting fields of f over K are isomorphic under an isomorphism which keeps the elements of K fixed and maps roots of f into each other. So, we may therefore talk of the splitting field of f over K. It is obtained by adjoining to K finitely many elements algebraic over K, and so we can show (exercise!) that it is a finite extension of K. Example Find the splitting field of the polynomial f(x) = x Q[x] over Q. The polynomial f splits in C, where it factors as (x i 2)(x + i 2). However, C itself is not the splitting field for f. It turns out to be sufficient to adjoin just one of the complex roots of f to Q. The field K = Q(i 2) contains both of the roots of f, and no smaller subfield has this property, so K is the splitting field for F. Splitting fields will be central to our characterization of finite fields, in the next chapter.

29 Chapter 3. Finite fields We have seen, in the previous chapters, some examples of finite fields. For example, the residue class ring Z/pZ (when p is a prime) forms a field with p elements which may be identified with the Galois field F p of order p. The fields F p are important in field theory. From the previous chapter, every field of characteristic p contains a copy of F p (its prime subfield) and can therefore be thought of as an extension of F p. Since every finite field must have characteristic p, this helps us to classify finite fields. 6. Characterizing finite fields Lemma 6.1. Let F be a finite field containing a subfield K with q elements. Then F has q m elements, where m = [F : K]. Proof. F is a vector space over K, finite-dimensional since F is finite. Denote this dimension by m; then F has a basis over K consisting of m elements, say b 1,..., b m. Every element of F can be uniquely represented in the form k 1 b 1 + k m b m (where k 1,..., k m K). Since each k i K can take q values, F must have exactly q m elements. We are now ready to answer the question: What are the possible cardinalities for finite fields? Theorem 6.2. Let F be a finite field. Then F has p n elements, where the prime p is the characteristic of F and n is the degree of F over its prime subfield. Proof. Since F is finite, it must have characteristic p for some prime p (by Corollary 2.19). Thus the prime subfield K of F is isomorphic to F p, by Theorem 4.5, and so contains p elements. Applying Lemma 6.1 yields the result. 29

30 30 CHAPTER 3. FINITE FIELDS So, all finite fields must have prime power order - there is no finite field with 6 elements, for example. We next ask: does there exist a finite field of order p n for every prime power p n? How can such fields be constructed? We saw, in the previous chapter, that we can take the prime fields F p and construct other finite fields from them by adjoining roots of polynomials. If f F p [x] is irreducible of degree n over F p, then adjoining a root of f to F p yields a finite field of p n elements. However, it is not clear whether we can find an irreducible polynomial in F p [x] of degree n, for every integer n. The following two lemmas will help us to characterize fields using root adjunction. Lemma 6.3. If F is a finite field with q elements, then every a F satisfies a q = a. Proof. Clearly a q = a is satisfied for a = 0. The non-zero elements form a group of order q 1 under multiplication. Using the fact that a G = 1 G for any element a of a finite group G, we have that all 0 a F satisfy a q 1 = 1, i.e. a q = a. Lemma 6.4. If F is a finite field with q elements and K is a subfield of F, then the polynomial x q x in K[x] factors in F [x] as x q x = a F(x a) and F is a splitting field of x q x over K. Proof. Since the polynomial x q x has degree q, it has at most q roots in F. By Lemma 6.3, all the elements of F are roots of the polynomial, and there are q of them. Thus the polynomial splits in F as claimed, and cannot split in any smaller field. We are now ready to prove the main characterization theorem for finite fields. Theorem 6.5 (Existence and Uniqueness of Finite Fields) For every prime p and every positive integer n, there exists a finite field with p n elements. Any finite field with q = p n elements is isomorphic to the splitting field of x q x over F p.

31 6. CHARACTERIZING FINITE FIELDS 31 Proof. (Existence) For q = p n, consider x q x in F p [x], and let F be its splitting field over F p. Since its derivative is qx q 1 1 = 1 in F p [x], it can have no common root with x q x and so, by Theorem 3.15, x q x has q distinct roots in F. Let S = {a F : a q a = 0}. Then S is a subfield of F since S contains 0; a, b S implies (by Freshmen s Exponentiation) that (a b) q = a q b q = a b, so a b S; for a, b S and b 0 we have (ab 1 ) q = a q b q = ab 1,so ab 1 S. On the other hand, x q x must split in S since S contains all its roots, i.e its splitting field F is a subfield of S. Thus F = S and, since S has q elements, F is a finite field with q = p n elements. (Uniqueness) Let F be a finite field with q = p n elements. Then F has characteristic p by Theorem 6.2, and so contains F p as a subfield. So, by Lemma 6.4, F is a splitting field of x q x. The result now follows from the uniqueness (up to isomorphism) of splitting fields, from Theorem As a result of the uniqueness part of Theorem 6.5, we may speak of the finite field (or the Galois field) of q elements. We shall denote this field by F q, where q denotes a power of the prime characteristic p of F q. Example 6.6. In Example 5.14, we constructed a field L = F 3 (θ) of 9 elements, where θ is a root of the polynomial x 2 + x + 2 F 3 [x]. By Theorem 6.5, L is the field of 9 elements, i.e. F 9. In Example 5.15, we constructed a field L = F 2 (θ) of 4 elements, where θ is a root of the polynomial x 2 + x + 1 F 2 [x]. By Theorem 6.5, L is the field of 4 elements, i.e. F 4. We can also completely describe the subfields of a finite field F q. Theorem 6.7 (Subfield Criterion) Let F q be the finite field with q = p n elements. Then every subfield of F q has order p m, where m is a positive divisor of n. Conversely, if m is a positive divisor of n, then there is exactly one subfield of F q with p m elements.

32 32 CHAPTER 3. FINITE FIELDS Proof. Clearly, a subfield K of F must have order p m for some positive integer m n. By Lemma 6.1, q = p n must be a power of p m, and so m must divide n. Conversely, if m is a positive divisor of n, then p m 1 divides p n 1, and so x pm 1 1 divides x pn 1 1 in F p [x]. So, every root of x pm x is a root of x q x, and hence belongs to F q. It follows that F q must contain a splitting field of x pm x over F p as a subfield, and (from proof of Theorem 6.5) such a splitting field has order p m. If there were two distinct subfields of order p m in F q, they would together contain more than p m roots of x pm x in F q, a contradiction. So, the unique subfield of F p n of order p m, where m is a positive divisor of n, consists precisely of the roots of x pm x in F p n. Example 6.8. Determine the subfields of the finite field F To do this, list all positive divisors of 30. The containment relations between subfields are equivalent to divisibility relations among the positive divisors of 30. (For diagram, see lectures!) We can also completely characterize the multiplicative group of a finite field. For the finite field F q, we denote the multiplicative group of non-zero elements of F q by F q. Theorem 6.9. For every finite field F q, the multiplicative group F q of nonzero elements of F q is cyclic. Proof. We may assume q 3. Set h = q 1, the order of F q, and let h = p r 1 1 p r p rm m be its prime factor decomposition. For each i, 1 i m, the polynomial x h/p i 1 has at most h/p i roots in F q. Since h/p i < h, it follows that there are nonzero elements of F q which are not roots of this polynomial. Let a i be such an element, and set b i = a h/pr i i i the order of b i divides p r i i the other hand, and so has the form p s i i b pr i 1 i i = a h/p i i 1,. Now, b pri i i = 1, so for some 0 s i r i. On so the order of b i is precisely p r i i. Let b = b 1 b 2... b m. We claim: b has order h(= q 1), i.e. is a generator for the group. Suppose, on the contrary, that the order of b is a proper divisor of h. It is therefore a divisor of at least one of the m integers h/p i, 1 i m; wlog, say of h/p 1. Then 1 = b h/p 1 = b h/p 1 1 b h/p 1 2 b h/p 1 m.

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