We are now ready to answer the question: What are the possible cardinalities for finite fields?

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1 Chapter 3 Fnte felds We have seen, n the prevous chapters, some examples of fnte felds. For example, the resdue class rng Z/pZ (when p s a prme) forms a feld wth p elements whch may be dentfed wth the Galos feld F p of order p. The felds F p are mportant n feld theory. From the prevous chapter, every feld of characterstc p contans a copy of F p (ts prme subfeld) and can therefore be thought of as an extenson of F p. Snce every fnte feld must have characterstc p, ths helps us to classfy fnte felds. 6 Characterzng fnte felds Lemma 6.1 Let F be a fnte feld contanng a subfeld K wth q elements. Then F has q m elements, where m = [F : K]. Proof. F s a vector space over K, fnte-dmensonal snce F s fnte. Denote ths dmenson by m; then F has a bass over K consstng of m elements, say b 1,...,b m. Every element of F can be unquely represented n the form k 1 b 1 + k m b m (where k 1,...,k m K). Snce each k K can take q values, F must have exactly q m elements. We are now ready to answer the queston: What are the possble cardnaltes for fnte felds? Theorem 6.2 Let F be a fnte feld. Then F has p n elements, where the prme p s the characterstc of F and n s the degree of F over ts prme subfeld. Proof. Snce F s fnte, t must have characterstc p for some prme p (by Corollary 2.19). Thus the prme subfeld K of F s somorphc to F p, by Theorem 4.5, and so contans p elements. Applyng Lemma 6.1 yelds the result. So, all fnte felds must have prme power order - there s no fnte feld wth 6 elements, for example. We next ask: does there exst a fnte feld of order p n for every prme power p n? How can such felds be constructed? We saw, n the prevous chapter, that we can take the prme felds F p and construct other fnte felds from them by adjonng roots of polynomals. If f F p [x] s rreducble of degree n over F p, then adjonng a root of f to F p yelds a fnte feld of p n elements. However, t s not clear whether we can fnd an rreducble polynomal n F p [x] of degree n, for every nteger n. The followng two lemmas wll help us to characterze felds usng root adjuncton. 23

2 24 CHAPTER 3. FINITE FIELDS Lemma 6.3 If F s a fnte feld wth q elements, then every a F satsfes a q = a. Proof. Clearly a q = a s satsfed for a = 0. The non-zero elements form a group of order q 1 under multplcaton. Usng the fact that a G = 1 G for any element a of a fnte group G, we have that all 0 a F satsfy a q 1 = 1,.e. a q = a. Lemma 6.4 If F s a fnte feld wth q elements and K s a subfeld of F, then the polynomal x q x n K[x] factors n F[x] as x q x = a F(x a) and F s a splttng feld of x q x over K. Proof. Snce the polynomal x q x has degree q, t has at most q roots n F. By Lemma 6.3, all the elements of F are roots of the polynomal, and there are q of them. Thus the polynomal splts n F as clamed, and cannot splt n any smaller feld. We are now ready to prove the man characterzaton theorem for fnte felds. Theorem 6.5 (Exstence and Unqueness of Fnte Felds) For every prme p and every postve nteger n, there exsts a fnte feld wth p n elements. Any fnte feld wth q = p n elements s somorphc to the splttng feld of x q x over F p. Proof. (Exstence) For q = p n, consder x q x n F p [x], and let F be ts splttng feld over F p. Snce ts dervatve s qx q 1 1 = 1 n F p [x], t can have no common root wth x q x and so, by Theorem 3.15, x q x has q dstnct roots n F. Let S = {a F : a q a = 0}. Then S s a subfeld of F snce S contans 0; a,b S mples (by Freshmen s Exponentaton) 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 splt n S snce S contans all ts roots,.e ts splttng feld F s a subfeld of S. Thus F = S and, snce S has q elements, F s a fnte feld wth q = p n elements. (Unqueness) Let F be a fnte feld wth q = p n elements. Then F has characterstc p by Theorem 6.2, and so contans F p as a subfeld. So, by Lemma 6.4, F s a splttng feld of x q x. The result now follows from the unqueness (up to somorphsm) of splttng felds, from Theorem As a result of the unqueness part of Theorem 6.5, we may speak of the fnte feld (or the Galos feld) of q elements. We shall denote ths feld by F q, where q denotes a power of the prme characterstc p of F q. Example 6.6 In Example 5.14, we constructed a feld L = F 3 (θ) of 9 elements, where θ s a root of the polynomal x 2 + x + 2 F 3 [x]. By Theorem 6.5, L s the feld of 9 elements,.e. F 9. In Example 5.15, we constructed a feld L = F 2 (θ) of 4 elements, where θ s a root of the polynomal x 2 + x + 1 F 2 [x]. By Theorem 6.5, L s the feld of 4 elements,.e. F 4. We can also completely descrbe the subfelds of a fnte feld F q.

3 6. CHARACTERIZING FINITE FIELDS 25 Theorem 6.7 (Subfeld Crteron) Let F q be the fnte feld wth q = p n elements. Then every subfeld of F q has order p m, where m s a postve dvsor of n. Conversely, f m s a postve dvsor of n, then there s exactly one subfeld of F q wth p m elements. Proof. Clearly, a subfeld K of F must have order p m for some postve nteger m n. By Lemma 6.1, q = p n must be a power of p m, and so m must dvde n. Conversely, f m s a postve dvsor of n, then p m 1 dvdes p n 1, and so x pm 1 1 dvdes x pn 1 1 n F p [x]. So, every root of x pm x s a root of x q x, and hence belongs to F q. It follows that F q must contan a splttng feld of x pm x over F p as a subfeld, and (from proof of Theorem 6.5) such a splttng feld has order p m. If there were two dstnct subfelds of order p m n F q, they would together contan more than p m roots of x pm x n F q, a contradcton. So, the unque subfeld of F p n of order p m, where m s a postve dvsor of n, conssts precsely of the roots of x pm x n F p n. Example 6.8 Determne the subfelds of the fnte feld F To do ths, lst all postve dvsors of 30. The contanment relatons between subfelds are equvalent to dvsblty relatons among the postve dvsors of 30. (For dagram, see lectures!) We can also completely characterze the multplcatve group of a fnte feld. For the fnte feld F q, we denote the multplcatve group of non-zero elements of F q by F q. Theorem 6.9 For every fnte feld F q, the multplcatve group F q of nonzero elements of F q s cyclc. Proof. We may assume q 3. Set h = q 1, the order of F q, and let h = pr 1 1 pr prm m be ts prme factor decomposton. For each, 1 m, the polynomal x h/p 1 has at most h/p roots n F q. Snce h/p < h, t follows that there are nonzero elements of F q whch are not roots of. Now, b pr for some 0 s r. On the other hand, ths polynomal. Let a be such an element, and set b = a h/pr dvdes p r and so has the form p s b pr 1 = a h/p 1, = 1, so the order of b so the order of b s precsely p r. Let b = b 1 b 2...b m. We clam: b has order h(= q 1),.e. s a generator for the group. Suppose, on the contrary, that the order of b s a proper dvsor of h. It s therefore a dvsor of at least one of the m ntegers h/p, 1 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. Now, f 2 m, then p r dvdes h/p 1, and so b h/p 1 = 1. Ths forces b h/p 1 1 = 1. Thus the order of b 1 must dvde h/p 1, whch s mpossble snce the order of b 1 s p r 1 1. Thus F q s a cyclc group wth generator b. Defnton 6.10 A generator of the cyclc group F q s called a prmtve element of F q. By Theorem 1.13, F q contans φ(q 1) prmtve elements, where φ s Euler s functon: the number of ntegers less than and relatvely prme to q 1. Recall that, f the nteger n has the prme factorzaton p k 1 1 pk pkr r, then φ(n) = n(1 1 p 1 )(1 1 p 2 ) (1 1 p r ).

4 26 CHAPTER 3. FINITE FIELDS Example 6.11 F 5 has φ(4) = 2 prmtve elements, namely 2 and 3. F 4 has φ(3) = 2 prmtve elements. Expressng F 4 as F 2 (θ) = {0,1,θ,θ + 1}, where θ 2 + θ + 1 = 0, we fnd that both θ and θ + 1 are prmtve elements. We are now ready to prove an mportant result. Theorem 6.12 Let F q be a fnte feld and F r a fnte extenson feld. Then F r s a smple extenson of F q,.e F r = F q (β) for some β F r ; every prmtve element of F r can serve as a defnng element β of F r over F q. Proof. Let α be a prmtve element of F r. Clearly, F q (α) F r. On the other hand, snce F q (α) contans 0 and all powers of α, t contans all elements of F r. So F r = F q (α). So, we can express any fnte feld K wth subfeld F, by adjonng to F a root β of an approprate rreducble polynomal f, whch of course must have degree d = [K : F]. Although the proof of Theorem 6.12 uses a β whch s a prmtve element of K, t s not n fact necessary for β to be a multplcatve generator of K, as the next example shows. Example 6.13 Consder the fnte feld F 9. We can express F 9 n the form F 3 (β), where β s a root of the polynomal x 2 + 1, rreducble over F 3. However, snce β 4 = 1, β does not generate the whole of F 9,.e. β s not a prmtve element of F 9. Corollary 6.14 For every fnte feld F q and every postve nteger n, there exsts an rreducble polynomal n F q [x] of degree n. Proof. Let F r be the extenson feld of F q of order q n, so that [F r : F q ] = n. By Theorem 6.12, F r = F q (α) for some α F r. Then, by propertes of mnmal polynomals, the mnmal polynomal of α over F q s an rreducble polynomal n F q [x] of degree n. 7 Irreducble polynomals In ths secton, we nvestgate rreducble polynomal over fnte felds. Lemma 7.1 Let f F q [x] be an rreducble polynomal over a fnte feld F q and let α be a root of f n an extenson feld of F q. Then, for a polynomal h F q [x], we have h(α) = 0 f and only f f dvdes h. Proof. The mnmal polynomal of α over F q s gven by a 1 f, where a s the leadng coeffcent of f (snce t s a monc rreducble polynomal n F q [x] havng α as a root). The proposton then follows from part () of Theorem Lemma 7.2 Let f F q [x] be an rreducble polynomal over F q of degree m. Then f dvdes x qn x f and only f m dvdes n.

5 7. IRREDUCIBLE POLYNOMIALS 27 Proof. Frst, suppose f dvdes x qn x. Let α be a root of f n the splttng feld of f over F q. Then α qn = α, so α F q n. Thus F q (α) s a subfeld of F q n. Snce [F q (α) : F q ] = m and [F q n : F q ] = n, we have n = [F q n : F q (α)]m, so m dvdes n. Conversely, suppose m dvdes n. Then by Theorem 6.7, F q n contans F q m as a subfeld. Let α be a root of f n the splttng feld of f over F q. Then [F q (α) : F q ] = m, and so F q (α) = F q m. Thus α F q n, hence α qn = α, and so α s a root of x qn x F q [x]. Therefore, by Lemma 7.1, f dvdes x qn x. We are now ready to descrbe the set of roots of an rreducble polynomal. Theorem 7.3 If f s an rreducble polynomal n F q [x] of degree m, then f has a root α n F q m. Moreover, all the roots of f are smple and are gven by the m dstnct elements α,α q,α q2,...,α qm 1 of F q m. Proof. Let α be a root of f n the splttng feld of f over F q. Then [F q (α) : F q ] = m, hence F q (α) = F q m, and so α F q m. We now show that, f β F q m s a root of f, then β q s also a root of f. Wrte f = a m x m + + a 1 x + a 0 (a F q ). Then f(β q ) = a m β qm + + a 1 β q + a 0 = a q mβ qm + + a q 1 βq + a q 0 = (a m β m + + a 1 β + a 0 ) q = f(β) q = 0, usng Lemma 6.3 and Freshmen s Exponentaton. Thus, the elements α,α q,...,α qm 1 are roots of f. We must check that they are all dstnct. Suppose not,.e. α qj = α qk for some 0 j < k m 1. Rasng ths to the power q m k, we get α qm k+j = α qm = α. It then follows from Lemma 7.1 that f dvdes x qm k+j x. By Lemma 7.2, ths s possble only f m dvdes m k + j, a contradcton snce 0 < m k + j < m. Ths result gves us two useful corollares. Corollary 7.4 Let f be an rreducble polynomal n F q [x] of degree m. Then the splttng feld of f over F q s F q m. Proof. Theorem 7.3 shows that f splts n F q m. To see that ths s the splttng feld, note that F q (α,α q,...,α qm 1 ) = F q (α) = F q m. Corollary 7.5 Any two rreducble polynomals n F q [x] of the same degree have somorphc splttng felds. As we shall see later, sets of elements such as those n Theorem 7.3 appear often n the theory of felds. Theorem 7.6 For every fnte feld F q and every n N, the product of all monc rreducble polynomals over F q whose degrees dvde n s equal to x qn x.

6 28 CHAPTER 3. FINITE FIELDS Proof. By Lemma 7.2, the monc rreducble polynomals over F q whch occur n the canoncal factorzaton of g = x qn x n F q [x] are precsely those whose degrees dvde n. Snce g = 1, by Theorem 3.15 g has no multple roots n ts splttng feld over F q. Thus each monc rreducble polynomal over F q whose degree dvdes n occurs exactly once n the canoncal factorzaton of g n F q [x]. Example 7.7 Take q = n = 2; the monc rreducble polynomals over F 2 [x] whose degrees dvde 2 are x, x + 1 and x 2 + x + 1. It s easly seen that x(x + 1)(x 2 + x + 1) = x 4 + x = x 4 x. Corollary 7.8 If N q (d) s the number of monc rreducble polynomals n F q [x] of degree d, then q n = dn q (d) for all n N, where the sum s extended over all postve dvsors d of n. Proof. Ths follows mmedately from Theorem 7.6, upon comparng the degree of g = x qn x wth the total degree of the canoncal factorzaton of g. Ths corollary allows us to obtan an explct formula for the number of monc rreducble polynomals n F q [x] of a gven degree. To do so, we need the followng arthmetc functon, whch wll also prove useful n the next chapter. Defnton 7.9 The Moebus functon µ s the functon on N defned by 1 f n = 1; µ(n) = ( 1) k f n s the product of k dstnct prmes 0 f n s dvsble by the square of a prme. Example 7.10 ()µ(5) = 1; ()µ(35) = 1; ()µ(50) = 0. Lemma 7.11 For n N, the Moebus functon satsfes µ(d) = { 1 f n = 1 0 f n > 1. Proof. The n = 1 case s mmedate. For n > 1 we need only consder the postve dvsors d of n for whch µ(d) s non-zero, namely those d for whch d = 1 or d s a product of dstnct prmes. If p 1,...,p k are the dstnct prme dvsors of n then µ(d) = µ(1) + = 1 + k µ(p ) + =1 ( ) k 1 ( 1) + = (1 + ( 1)) k = 0. ( ) k < 2 k ( 1) µ(p 1 p 2 ) + + µ(p 1 p 2...p k ) ( ) k ( 1) k k

7 7. IRREDUCIBLE POLYNOMIALS 29 Theorem 7.12 (Moebus Inverson Formula) Addtve verson: let h and H be two functons from N nto an addtvely wrtten abelan group G. Then H(n) = h(d) for all n N (3.1) f and only f h(n) = µ( n d )H(d) = µ(d)h( n ) for all n N. (3.2) d Multplcatve verson: let h and H be two functons from N nto a multplcatvely wrtten abelan group G. Then H(n) = h(d) for all n N (3.3) f and only f h(n) = H(d) µ( n d ) = H( n d )µ(d) for all n N. (3.4) Proof. Addtve verson: we prove the forward mplcaton; the converse s smlar and s left as an exercse. Assume the frst dentty holds. Usng Lemma 7.11, we get µ( n d )H(d) = µ(d)h( n d ) = µ(d) h(c) c n d = µ(d)h(c) = h(c) µ(d) = h(n) c n d n c n d n c c for all n N. Multplcatve verson: mmedate upon replacng sums by products and multples by powers. We can now apply ths result as follows. Theorem 7.13 The number N q (n) of monc rreducble polynomals n F q [x] of degree n s gven by N q (n) = 1 µ( n n d )qd = 1 µ(d)q n d. n Proof. Apply the addtve case of the Moebus Inverson Formula to the group G = (Z,+). Take h(n) = nn q (n) and H(n) = q n for all n N. By Corollary 7.8, the dentty (3.1) s satsfed, and so the result follows. Remark 7.14 Snce t s clear from ths formula that N q (n) s greater than zero for all n, ths gves an alternatve proof of Theorem Example 7.15 The number of monc rreducbles n F q [x] of degree 12 s gven by N q (12) = 1 12 (µ(1)q12 + µ(2)q 6 + µ(3)q 4 + µ(4)q 3 + µ(6)q 2 + µ(12)q) = 1 12 (1.q12 + ( 1)q 6 + ( 1)q q q q) = 1 12 (q12 q 6 q 4 + q 2 ).

8 30 CHAPTER 3. FINITE FIELDS We can also obtan a formula for the product of all monc rreducble polynomals n F q [x] of fxed degree. Theorem 7.16 The product I(q,n;x) of all monc rreducble polynomals n F q [x] of degree n s gven by: I(q,n;x) = (x qd x) µ( n d ) = (x q nd x) µ(d). Proof. From Theorem 7.6 we know that x qn x = I(q,d;x). Now apply Moebus Inverson n the multplcatve form to the multplcatve group G of non-zero ratonal functons over F q. Take h(n) = I(q,n;x) and H(n) = x qn x to get the desred formula. Example 7.17 Take q = 2 and n = 4. Then the product of all monc rreducble quartcs n F 2 [x] s: I(2,4;x) = (x 16 x) µ(1) (x 4 x) µ(2) (x 2 x) µ(4) = x16 x x 4 x = x15 1 x 3 1 = x 12 + x 9 + x 6 + x 3 + 1

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