11. FINITE FIELDS. Example 1: The following tables define addition and multiplication for a field of order 4.

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1 11. FINITE FIELDS A Field With 4 Elemets Probably the oly fiite fields which you ll kow about at this stage are the fields of itegers modulo a prime p, deoted by Z p. But there are others. Now although Z 4 is ot a field because 2.2 = 0 i this rig there is a field of order 4. Example 1: The followig tables defie additio ad multiplicatio for a field of order 4. + A B C D A B C D A A B C D A A A A A B B A D C B A B C D C C D A B C A C D B D D C B A D A D B C Clearly the additive idetity is A ad the multiplicative idetity is B. We could write A as 0 ad B as 1. Also, sice D = B + C we could write D = 1 + C. Usig this otatio the additio ad multiplicatio tables become: C 1+C 0 1 C 1+C C 1+C C C C 1+C C C 1+C 0 1 C 0 C 1+C 1 1+C 1+C C C 0 1+C 1 C These table could be recostructed just by kowig just two facts about the field: (1) = 0; (2) C 2 = 1 + C. From property (1) we deduce that x + x = x(1 + 1) = 0 for each of the four values of x. Ad kowig that C 2 = 1 + C eables us to fid the values of C(1 + C) ad (1 + C) 2. Now otice that this field cotais the subfield {0, 1} (see the shaded portios of the tables) which is our old fried Z 2 ad so it is a extesio of Z 2 by a elemet C that satisfies C 2 = 1 + C, or equivaletly C 2 + C + 1 = 0 (remember that subtractio is the same as additio mod 2 sice = 0 implies that 1 = 1). So C is a zero of the quadratic x 2 + x + 1. Ad we recogise this as a prime quadratic over Z 2. It has o zeros withi Z 2 so we have exteded Z 2 by a iveted umber C so as to get a larger system i which the quadratic does factorise. This is the same way i which complex umbers were itroduced. There we had the quadratic x with o real zeros ad we iveted the umber i to exted the reals to give the field of complex umbers i which x ow has zeros The Characteristic of a Field We defie the characteristic of a field F to be the additive order of 1, the multiplicative idetity, except that if 1 has ifiite order, as it does i the field C ad 127

2 all its subfields, we say that the field has characteristic zero. Clearly if a field F has characteristic the x = (1)x = 0x = 0 for all x F. Fiite fields have fiite characteristic, but ote that it is possible to have ifiite fields with fiite characteristic. For example the set of all ratioal fuctios a(x) b(x) with a(x), b(x) Z p[x] is a example of a ifiite field of characteristic p. Example 2: Z p ad all its extesios have characteristic p. Theorem 1: If the characteristic of a field is fiite, it must be prime. Proof: Suppose the characteristic of F is where = ab ad 1 < a, b <. The 1 = (a1)(b1) ad so, sice the Cacellatio Law holds i fields, either a1 = 0 or b1 = 0. Clearly this is a cotradictio. Theorem 2: The order of a fiite field F must be p for some prime p ad some positive iteger. Proof: Let p be the characteristic of F ad let K = {1 Z}. (Note that we write it as 1 rather tha just because is a iteger while 1 is a elemet of the field.) Clearly K Z p. Now F is a fiite-dimesioal vector space over K. Suppose F:K = ad let a 1, a 2,..., a be a basis. The every elemet of F ca be writte uiquely as a liear combiatio of the a i with coefficiets from K ad hece there are p of them. The field K i Theorem 2 is called the prime subfield of F. Polyomials of the form x N x play a importat part i the theory of fiite fields, especially i the case where N is a prime power. Such a polyomial does t usually split ito liear factors, but if N is a multiple of the field characteristic the o prime factor is repeated. Theorem 3: If p divides N the x N x has o repeated prime factors over Z p. Proof: Suppose x N x = a(x) 2 q(x) where the degree of a(x) is at least 1. Differetiatig x N x we get Nx N 1 1 = 0 1 = 1. But, differetiatig a(x) 2 q(x) by the product rule we get a multiple of a(x). Clearly this is a cotradictio. It may seem odd to be usig calculus i the case of fiite fields. Ideed we caot defie derivatives i the usual way as limits of quotiets. But we ca defie the derivative of polyomials i a purely formal way ad prove the product rule directly from this defiitio. Theorem 4: I a field of characteristic p (x + y) p = x p + y p for all x, y ad all 1. Proof: For = 1 we have (x + y) p = x p + y p sice all the other biomial coefficiets are multiples of p ad hece are zero i Z p. Suppose the theorem is true for. The (x + y) p+1 = ((x + y) p ) p = (x p + y p ) p = x p+1 + y p

3 11.3. Costructio of Field Extesios Whe we were dealig with subfields of the complex umbers we could defie F[f (x)] to be the smallest field of C that cotais all the zeros of f (x) because we kow of at least oe field that cotais them all, amely the field of complex umbers. But whe it comes to polyomials over fields, such as Z p, we do t have a obvious field that cotais all the zeros. We have to carry out our field extesios i a differet way. Recall how we defied the rig of itegers mod as Z = {0, 1, 2,..., 1}. These are the possible remaiders whe we divide a iteger by. We make this ito a rig, a mathematical structure with operatios of additio ad multiplicatio, by defiig the operatios i Z to be additio modulo ad multiplicatio modulo. By this we mea that we add or multiply two elemets ad the take the remaider o dividig by. Example 3: Z 7 = {0, 1, 2, 3, 4, 5, 6} ad i Z 7 we have = 2 ad 4.5 = 6. You ll recall that Z is a field if ad oly if is prime. The same idea ca be exteded to prime polyomials to get extesios of ay field. Suppose P(x) is a polyomial of degree over a field F. We create a idetermiate u such that P(u) = 0 i the same way that we created the imagiary umber i such that i = 0 whe extedig the real umbers to the complex umbers. Suppose that P(x) has degree. We defie F[u P(u) = 0] to be the set of all polyomials i u of the form a 1 u a 1 u + a 0 where each a i F. These are the possible remaiders o dividig a polyomial i u by the polyomial P(u). We add these expressios i the usual way ad multiply them modulo P(u). That is, we multiply two expressios of the above form as polyomials but we take the remaider o dividig by P(u). The reaso for chagig the idetermiate from x to u is that we cosider f(x) to be a polyomial i x while f (u) is that polyomial reduced modulo P(u), that is, simplified uder the assumptio that P(u) = 0. Example 4: Take P(x) = x over R. This is prime over R. So R[u u = 0] is {a + bu a, b R}. We add ad multiply as follows: (a 1 + b 1 u) + (a 2 + b 2 u) = (a 1 + a 2 ) + (b 1 + b 2 )u ad (a 1 + b 1 u)(a 2 + b 2 u) = a 1 a 2 + (a 1 b 2 + a 2 b 1 )u + b 1 b 2 u 2 = (a 1 a 2 b 1 b 2 ) + (a 1 b 2 + a 2 b 1 )u. You ll recogise that R[u u = 0] is simply the field of complex umbers where the idetermiate u is more usually writte as i. We shall ow use this techique to costruct a field of order 8. Example 5: The polyomial P(x) = x 3 + x + 1 is prime over Z 2. Z 2 [u u 3 + u + 1 = 0] = {0, 1, u, u + 1, u 2, u 2 + 1, u 2 + u, u 2 + u + 1}. The additio table for this field is: 129

4 + 0 1 u u+1 u 2 u 2 +1 u 2 +u u 2 +u u u+1 u 2 u 2 +1 u 2 +u u 2 +u u+1 u u 2 +1 u 2 u 2 +u+1 u 2 +u u u u u 2 +u u 2 +u+1 u 2 u 2 +1 u+1 u+1 u 1 0 u 2 +u+1 u 2 +u u 2 +1 u 2 u 2 u 2 u 2 +1 u 2 +u u 2 +u u u+1 u 2 +1 u 2 +1 u 2 u 2 +u+1 u 2 +u 1 0 u+1 u u 2 +u u 2 +u u 2 +u+1 u 2 u 2 +1 u u u 2 +u+1 u 2 +u+1 u 2 +u u 2 +1 u 2 u+1 u 1 0 The additio table is idepedet of the prime polyomial P(x). I this case we simply work modulo 2 with the coefficiets. Whe it comes to multiplicatio we will have u 3 ad u 4. It is worth expressig these i the required form before we begi. Sice u 3 + u + 1 = 0 we have u 3 = u + 1. Remember that 1 = 1 because we are usig Z 2 for the coefficiets. The u 4 = u 2 + u. 0 1 u u+1 u 2 u 2 +1 u 2 +u u 2 +u u u+1 u 2 +u+1 u 2 +1 u 2 +u u 2 +u+1 u 0 u u 2 u 2 +u u+1 1 u 2 +u+1 u 2 +1 u+1 0 u+1 u 2 +u u 2 +1 u 2 +u+1 u 2 1 u u 2 0 u 2 u+1 u 2 +u+1 u 2 +u u u u u u 2 u u 2 +u+1 u+1 u 2 +u u 2 +u 0 u 2 +u u 2 +u+1 1 u 2 +1 u+1 u u 2 u 2 +u+1 0 u 2 +u+1 u 2 +1 u 1 u 2 +u u 2 u + 1 If you feel that the above discussio is logically a little usatisfactory let me poit out that there is a perfectly rigorous way of doig all this usig quotiet rigs. You may remember quotiet groups i group theory where the elemets are cosets. Somethig similar ca be doe for rigs, ad F[u P(u) = 0] is simply the quotiet rig F[x]/P(x)F[x]. But sice our excursio ito rig theory will be brief we wo t bother itroducig quotiet rigs here. Theorem 5: P(x) is prime over F if ad oly if F[u P(u) = 0] is a field. Proof: Suppose P(x) = Q(x)R(x) where Q(x) ad R(x) have lower degree tha P(x). The i F[u P(u) = 0], Q(u)R(u) = P(u) = 0. If F[u P(u) = 0 the at least oe of Q(u) ad R(u) would be 0. But this ca oly happe if P(x) divides Q(x) or R(x), which is impossible as they have lower degree tha P(x). Now suppose that P(x) is prime over F. All but oe of the field axioms are obvious ad they work eve if P(x) is t prime. The oe axiom to be checked is the existece of a multiplicative iverse for all o-zero elemets. Suppose that a(u) is a o-zero elemet of F[u P(u) = 0]. The P(x) does ot divide a(x). Sice P(x) is prime over F this meas that a(x) ad P(x) are coprime over F. The greatest commo divisor of a(x) ad P(x) is therefore 1. Just as with greatest commo divisors of itegers it ca be show that the greatest commo divisor of a(x) ad P(x) ca be expressed i the form a(x)h(x) + P(x)k(x) for some h(x), k(x) F[x]. 130

5 [Euclid s algorithm for fidig GCDs works for polyomials just as it does for itegers ad by obtaiig 1 as the GCD ad workig backwards through the algorithm oe ca obtai a equatio of the above form.] So 1 = a(x)h(x) + P(x)k(x) ad substitutig x = u ad we get 1 = a(u)h(u) + P(u)k(u) = a(u)h(u) sice P(u) = 0. It follows that h(u) is the multiplicative iverse of a(u). So give ay prime polyomial P(x) over a field F we ca costruct a extesio of F over which P(x) has a zero. We ca cotiue extedig util we obtai a extesio over which P(x) splits completely ito liear factors. Fially we ca do this for composite polyomials by extedig by zeros from differet prime polyomials. Theorem 6: If F is ay field ad a(x) F[x] there exists a extesio of F over which a(x) splits ito liear factors. Proof: We prove this by iductio o the degree of a(x). We ca write a(x) = P(x)g(x) where P(x), g(x) F[x] ad P(x) is prime over F. We have see how to costruct a extesio K of F over which P(x), ad hece a(x), has a zero, α. Hece we ca write a(x) = (x α)g(x) for some g(x) K[x]. Sice g(x) has smaller degree tha a(x) we ca assume by iductio that there is a extesio H of K over which g(x) splits ito liear factors ad over this field a(x) will split ito liear factors. Theorem 7: For all prime powers p there exists a field of order p. Proof: Let a(x) = x p x ad let K be a extesio of Z p over which a(x) splits ito liear factors. Let H be the set of zeros of a(x) i K. It is easy to check that H is a subfield of K, for if x p = x ad y p = y the (x + y) p = x p + y p = x + y. Closure uder multiplicatio ad iverses are easily checked. Sice a(x) has o repeated zeros H = p. Over a smaller field, such as Z p, x p = x will factorise ito moic prime factors, but these will ot all be liear. The remarkable thig, however, is that every moic prime polyomial whose degree divides will appear exactly oce i this factorisatio The Multiplicative Group of a Fiite Field For ay field F the o-zero elemets form a group F # uder multiplicatio. We call this the multiplicative group of F. If F has order p its multiplicative group has order p 1. Theorem 8: If F is a fiite field the F # is cyclic. Proof: F # is a direct sum of cyclic groups of prime power order. If there is more tha oe direct summad whose order is divisible by p the i F # there are at least p 2 zeros of the polyomial x p 1, a cotradictio. 131

6 Example 6: Fid the order of the smallest field F over which x 14 x splits completely ito liear factors. Solutio: x 14 x = x(x 13 1) so we eed F to cotai a elemet of order 13. If F = p the 13 divides p 1 ad so p = k for some iteger k. The smallest possibility is where p = 27. But will x 14 x split completely over a field of order 27? The aswer is yes because i such a case F # is a cyclic group of order 26 ad so will cotai 12 elemets of order 13 ad these, together with 0 ad 1, will provide 14 distict zeros for x 14 x. Theorem 9: If F is a field of order p the x p = x for all x F. Proof: The order of F # is p 1 ad so if x 0 the x p 1 = 1 ad so x p = x. This is true for x = 0 as well. Corollary: I a field of order p the polyomial x p x has p distict zeros. Theorem 10: Let P(x) be a prime polyomial of degree m over Z p. The P(x) divides x p x if ad oly if m divides. Proof: Suppose m divides. Let F = Z p [u P(u) = 0]. Clearly P(x) is the miimum polyomial of u over Z p. Sice F has order p m we must have u pm = u. So u p2m = (u pm ) pm = u pm = u ad so o. Hece u p = u ad so u is a zero of x p x. This meas that P(x) divides x p x. Suppose ow that P(x) divides x p x. Let F be a fiite degree extesio of Z p over which x p x splits ad let β be a zero of π(x) i F. The Z p [β]:z p = m so F:Z p [β] = /m whece m divides. Corollary : Over Z p, x p x is the product of all the moic prime polyomials over Z p whose degree divides. Proof: We just eed to ote that sice x p x has o repeated zeros each moic prime polyomial oly occurs oce i the above factorisatio. Example 7: The miimum polyomials of the 8 elemets of GF[8] above are as follows: 0 1 α α α 2 + α α + 1 α 2 α 2 + α + 1 x x+ 1 x 3 + x + 1 x 3 + x Example 8: The prime polyomials over Z 2 of degrees 1 ad 3 are: x, x + 1, x 3 + x + 1, x 3 + x ad so x 8 x = x(x + 1)(x 3 + x + 1)(x 3 + x 2 + 1). The prime polyomials over Z 2 of degrees 1, 2 ad 4 are: x, x + 1, x 2 + x + 1, x 4 + x + 1, x 4 + x ad x 4 + x 3 + x 2 + x + 1 ad so x 16 x = x(x + 1)(x 2 + x + 1)(x 4 + x + 1)(x 4 + x 3 + 1)(x 4 + x 3 + x 2 + x + 1). Example 9: Sice x 32 x is the product of all prime polyomials over Z 2 whose degrees divide 5 the (x 32 x)/(x 2 x) must be the product of all prime polyomials of degree 5. Hece there must be 30/5 = 6 of them. 132

7 Example 10: (x 256 x)/(x 16 x) has degree 240. Sice it is the product of all prime polyomials of degree 8 over Z 2 there must be 240/5 = 48 such polyomials. Theorem 11: If F is a field of order p ad π(x) is a prime polyomial of degree over Z p the F Z p [x]/π(x) Z p [x]. Proof: Sice x p x is the product of all prime polyomials over Z p whose degree divides, π(x) x p x. The elemets of F will be zeros of x p x ad so oe of them at least will be a zero of π(x). The α x + π(x) Z p [x] is the required isomorphism. Corollary: All fields of order p are isomorphic to oe aother. Up to isomorphism there s just oe field of order p for ay prime power. We call this GF[p ]. We also kow that there are o fields whose order is ot a prime power. But we caot yet be sure that there is a field for every prime power. Of course if there was o field of order p this would mea that there would be o prime polyomial of degree over Z p. This seems ulikely, but ca we be sure that this is impossible? The Number of Moic Prime Polyomials over Z p. Let P be the umber of moic prime polyomials of degree over Z p. Clearly P 1 = p. Theorem 12: For all ad all primes p, d. P d = p. d Example 11: Fid the umber of prime polyomials of degree 20 over Z 2, that is, fid P 20. P 1 = 2. P 2 = = 1. P 4 = = 3. P 5 = 32 2 = P 10 = = P 20 = 20 = We defie the Möbius fuctio to be: µ(1) = 1; µ(p 1 p 2...p k ) = ( 1) k if p 1,..., p k are distict primes; µ() = 0 if is divisible by the square of a prime. 133

8 Theorem 13: If > 1 the µ ( d) = 0. d Proof: Let = p p k k where p 1,..., p k are distict primes. k k µ ( d ) = µ (1) + µ ( pi ) + µ ( pi p j ) +... = = (1 1) k = 0. d i i, j 1 2 Example 12: µ ( d) = µ(1) + µ(2) + µ(3) + µ(4) + µ(6) + µ(12) d 24 = = 0. 1 d 1 / d Theorem 14: P = µ p = µ ( d) p. d d d / d Proof: µ ( d) p µ ( d) cp = d d c, d cd c d = µ ( d) cpc = cp c µ ( d) = c c= = P. c d c c cp sice µ ( d) = 0 if m > 1, by Theorem 13. d m Example 13: The square-free divisors of 20 are 1, 2, 5 ad 10. Hece the umber of moic prime polyomials over Z 2 of degree 20 is P 20 = = Galois Groups of Fiite Fields The Galois Theory of fiite fields is ot very iterestig. For a start the elemets of fiite fields are roots of uity ad so every polyomial over a fiite field is soluble by radicals. Moreover, as we ow show, the Galois groups of fiite fields are cyclic. Theorem 15: If F is a field of order p the θ(x) = x p is a automorphism of order. Proof: If x, y F the x + y x p + y p = (x + y) p ad xy (xy) p = x p y p. Clearly θ (x) = x p = x for all x so θ = 1, the idetity automorphism. If θ d = 1 for some proper divisor d of the x pd = x for all x F ad so F = p d, a cotradictio. This automorphism is called the Frobeius automorphism. 134

9 Theorem 16: If F is a field of order p ad K is its prime subfield the G(F/K) is a cyclic group of order, geerated by the Frobeius automorphism. Proof: Let a(x) = x p x ad let π(x) be a prime factor of a(x) over K of degree. [Remember that a(x) is the product of all prime polyomials over K whose degree divides so there will be a prime divisor of degree.] Let σ F be a zero of π(x). The 1, σ, σ 2,..., σ -1 is a basis for F over K ad hece every automorphism of F is determied by its effect o σ. But σ must map to oe of the zeros of π(x) ad so there are at most elemets i G(F/K). It follows that the Frobeius automorphism geerates G(F/K). 135

10 EXERCISES FOR CHAPTER 11 Exercise 1: (i) Write dow the additio ad multiplicatio tables for GF[9]. (ii) Fid the miimum polyomial of each of the elemets of GF[9]. (iii) Fid all possible geerators for the multiplicative group of GF[9]. (iv) Fid the Galois group of GF[9] over its prime subfield. Exercise 2: Let a(x) = x 10 + x (i) Show that a(x) splits ito distict liear factors over GF[16]. (ii) Hece or otherwise factorise a(x) ito prime factors over Z 2. Exercise 3: (a) Factorise the polyomial x 8 + x + 1 ito prime polyomials over Z 2. (b) Suppose u 2 = u + 1 i some fiite field F of characteristic 2 ad let K be the prime subfield of F. (i) Show that u 22 = u. (ii) By cosiderig the degree of the miimum polyomial of u over K show that x 2 + x + 1 is composite over Z p for all 3. Exercise 4: Fid a geerator of the multiplicative group of Z 11 [x 2 + 1]. Exercise 5: Fid the umber of prime polyomials of degree 12 over Z 2. Exercise 6: Fid the umber of prime polyomials of degree 18 over Z 2. Exercise 7: Fid the umber of prime polyomials of degree 24 over Z p. Exercise 8: Prove that if p, q are primes the for all m there exists a prime polyomial of degree q m over Z p. Exercise 9: Fid the Galois group G(GF(243)/ Z 3 ). 136

11 SOLUTIONS FOR CHAPTER 11 Exercise 1: Now x is prime over Z 3 so GF[9] = Z 3 [u u = 0]. The elemets of GF[9] are 0, 1, 2, u, u + 1, u + 2, 2u, 2u + 1 ad 2u u u + 1 u + 2 2u 2u + 1 2u u u + 1 u + 2 2u 2u + 1 2u u + 1 u + 2 u 2u + 1 2u + 2 2u u + 2 u u + 1 2u + 2 2u 2u + 1 u u u + 1 u + 2 2u 2u + 1 2u u + 1 u + 1 u + 2 u 2u + 1 2u + 2 2u u + 2 u + 2 u u + 1 2u + 2 2u 2u u 2u 2u + 1 2u u u + 1 u + 2 2u + 1 2u + 1 2u + 2 2u u + 1 u + 2 u 2u + 2 2u + 2 2u 2u u + 2 u u u u + 1 u + 2 2u 2u + 1 2u u u + 1 u + 2 2u 2u + 1 2u u 2u + 2 2u + 1 u u + 2 u + 1 u 0 u 2u 2 u + 2 2u u + 1 2u + 1 u u+1 2u + 2 u + 2 2u 1 2u u u u + 2 2u + 1 2u u u + 1 2u 2 2u 0 2u u 1 2u + 1 u u + 2 u + 2 2u u + 1 u + 2 u u 2u + 2 u 1 2u u + 2 u + 1 2u + 1 u 2 u u (ii) Elemet Miimum Polyomial 0 x 1 x x + 1 u x u u + 1 x 2 + x + 2 2u + 1 u + 2 x 2 + 2x + 2 2u + 2 (iii) (u + 1) 2 = 2u, (2u) 2 = 1, so u + 1 has order 8 ad so u + 1 geerates the multiplicative group of GF[9]. So do (u + 1) 3, (u + 1) 5 ad (u + 1) 7. Hece the geerators are u + 1, 2u + 1, 2u + 2 ad u + 2. (iv) G(GF[9]/ Z 3 ) C 2 ad is geerated by the automorphism x x

12 Exercise 2: (i) I GF[16], x 16 x = x(x 5 1)a(x) splits ito 16 distict liear factors ad hece so does a(x). (ii) x 16 x = x(x 1)(x 4 + x 3 + x 2 + x + 1)a(x) is the product of all the prime polyomials over Z 2 whose degree is 1, 2 or 4. Hece the prime factors of a(x) have degrees 2 or 4. Sice x 2 + x + 1 is the oly prime quadratic it must divide a(x) ad the other prime factors must be two of the three prime quartics: x 4 + x + 1, x 4 + x 3 + 1, x 4 + x 3 + x 2 + x + 1. But x 4 + x 3 + x 2 + x + 1 already occurs so a(x) = (x 2 + x + 1)(x 4 + x + 1)(x 4 + x 3 + 1). Exercise 3: (a) (i) x 2 + x + 1 ad x 4 + x + 1 have o zeros i Z 2 ad hece they must be prime. (ii) x 8 + x + 1 = (x 2 + x + 1)(x 6 + x 5 + x 3 + x 2 + 1). (b) (i) u 22 = (u 2 ) 2 = (u + 1) 2 = u = u. (ii) Suppose that x 2 + x + 1 is prime over K. This will be the miimum polyomial of u over K. Sice x 22 x is the product of all prime polyomials whose degree divides 2 it follows that 2 2, that is 2-1 divides. If 3 this is a cotradictio. Exercise 4: We ca easily show that x is prime over Z 11 ad so Z 11 [x 2 + 1] GF(121). Let i be a idetermiate such that i 2 = 1. We are lookig for a + bi of order 120, where a, b Z 11. We ca systematically try various combiatios of a, b util we fid a geerator. Sice ϕ(120) = 32, about 1 i 3 of the elemets of Z 11 # will be a geerator. Clearly b = 0 will ot work, so let s try α = 1 + i. α 2 = 2i so α 3 = 2i(1 + i) = 2 + 2i, α 4 = 4, α 5 = 4 4i ad α 6 = 8i = 3i. Hece α 15 = ( 2 + 2i)( 4) 3 = 128(1 i) = 7 7i ad so α 60 = 7 4 (1 i) 4 = 5 2 ( 2i) 2 = 12 = 1. So the order of α does t divide 60 ad so must be a multiple of 8. α 40 = ( 4) 10 = 5 5 = = 3.4 = 12 = 1 so α has order 40. Now we might try α = 2 + i, but surely there s a easier way! If we fid elemets of order 3, 5 ad 8 ad multiply them we will get a elemet of order 120. I C a elemet of order 3 is ω = 1 + 3i 2. Let s adapt this to Z 11 [i]. Squarig the elemets of Z 11 we soo fid that 5 2 = 3. Ad ½ i Z 11 is clearly 6. So let s try ( 1 + 5i).6 = i = 5 + 8i. Ideed this does have order 3. Fidig a elemet of order 5 is easy because there is oe iside Z 11 itself sice Z 11 # = 10. I fact, sice 2 5 = 1, 4 5 = 1. So 4 has order 5. For a elemet of order 8 we eed a square root of i. Let (a + bi) 2 = i. Therefore a 2 b 2 = 0 ad 2ab = 1. So b = 1 2a = 6 a. Thus a 2 = 36 a 2 ad so a4 = 36, whece a 2 = ± 6 = 5 or 6. Sice 5 = 16, we ca take 138

13 a = ± 4 = 4, 7. Let s take a = 7. The b = 6 7 = 6.8 = 48 = 4. So 7 + 4i has order 8. Alteratively we could observe that i C, 1 + i 2 has order 8. But x2 2 has o zeros i Z 11. However we ca write 1 + i 2 = 1 + i ad sice 3 2 = 2 the 3 ca play the role 2i of 2i i this field. Fially, 1 3 = 4 i Z 11, so we try 4( 1 + i) = 7 + 4i. Ideed 7 + 4i does have order 8. Hece (5 + 8i).4.(7 + 4i) = 4(3 +76i) = 4(3 i) = 1 4i has order 120 ad so is a geerator of the multiplicative group. Exercise 5: Fid the umber of prime polyomials of degree 12 over Z 2. Let P be the umber of (moic) prime polyomials of degree over Z 2. The P 1 = 2, P 2 = 1 ad P 3 = 2. 1.P P P 4 = 2 4 so P 4 = = P P P P 6 = 2 6 so P 6 = P P P P P P 12 = so P 12 = 12 = 335. = 9. Alteratively π(12) = µ(1)212 + µ(2)2 6 + µ(3)2 4 + µ(4)2 3 + µ(6)2 2 + µ(12) = 335. = Exercise 6: The umber of prime polyomials of degree 18 over Z 2 is P 18 = 1 18 [ ] = 1 18 [ ] = Exercise 7: P 24 = 1 24 [µ(1)p24 + µ(2)p 12 + µ(3)p 8 + µ(4)p 6 + µ(6)p 4 + µ(8)p 3 + µ(12)p 2 + µ(24)p] = 1 24 [p24 p 12 p 8 + p 4 ]. This is the umber of moic prime polyomials of degree 24. The total umber of (p 1) prime polyomials is therefore 24 [p 24 p 12 p 8 + p 4 ]. 139

14 Exercise 8: The umber of moic prime polyomials of degree q m over Z p is 1 q [µ(1)pq + µ(q)p q 1 1 ] = q [pq p q 1 ] > 0 Exercise 9: 243 = 3 5 so G(GF(243)/ Z 3 ) C