MATH 371 Class notes/outline October 15, 2013

MATH 371 Class notes/outline October 15, 2013 More on olynomials We now consider olynomials with coefficients in rings (not just fields) other than R and C. (Our rings continue to be commutative and have multilicative identities). The formal definition of a olynomial with coefficients in a ring R is that it is a function : N Ñ R such that n 0 for all but finitely many n. We tend to write n rather than n, and instead of writing 0, 1,... we write 0 1x 2 x 2 3 x 3. So x is the function 0, 1, 0, 0,..., x 2 is the function 0, 0, 1, 0, 0,... and we can identify an element r of R with the function r, 0, 0, 0,.... Given two olynomials and we form by letting n n n n and n n ņ i0 i n i. In this way we make the ring of olynomials (in one variable) with coefficients in R into a ring, denoted Rrxs. The degree of a olynomial is the largest value of n for which n 0, the leading coefficient is n, and the leading term of is n x n. Two olynomials and are eual if and only if n n for all n 0. There is a natural inclusion R Ñ Rrxs that sends r P R to the constant olynomial r this is a ring homomorhism. There s some weirdness that can haen for olynomials with coefficients in an arbitrary commutative ring R. For instance, let R Z{x4y and consider 2x 1. Then deg deg 2 but 1 in this ring, so deg 0. But if the leading coefficient of or is not a zero divisor then deg deg deg. Also, if R is an integral domain, then Rrxs R (where we identify R with the olynomials of degree zero in Rrxs). Just as we have to be careful with the degree of a roduct, we have to be a little bit careful with the division algorithm. The most general statement one can make is that if the leading coefficient of the olynomial d P Rrxs is not a zero-divisor, then given f P Rrxs, there exist olynomials, r P Rrxs such that f d r where either r 0 or none of the terms in r is divisible by the leading term of d. Note the care with which we have to say this, and the odd things that can haen in the division algorithm, which now goes as follows: 1. Given f and d, where the leading coefficient d n of d is not a zero divisor, begin by setting 0, r 0 and s f. Note that f d r s. 2. If s 0, then we re done, outut and r. 3. Let s m x m be the leading term of s. 4. If d n x n divides s m x m, then m n and s m cd n for a uniue c P R and so s m x m cx m n d n x n. In this case, ut : cx m n and s : x cx m n d.

2 5. On the other hand, of d n x n does not divide s m x m, then ut r : r s m x m and s : s s m x m. 6. After all of this, we still have f d r s. 7. Go ack to ste 2. Because the degree of s decreases each time through the loo, the rocess will sto after at most degs 1 times and yield the result described above. If the leading coefficient of d is a unit in R, then we have the standard result that f d r with degr degd. Note that this is true for monic olynomials (leading coefficient is 1) and that a monic olynomial of degree 1 is never a unit in Rrxs (roof?). Roots. Given any ρ P R, we have the evaluation homomorhism ϕ ρ : Rrxs Ñ R (note which way it goes): ϕ ρ ρ 0 1 ρ n ρ n. Borrow the notation from affine varieties and set V tρ P R ρ 0u to be the set of roots of P Rrxs. We have that ρ P R is a root of P Rrxs if and only if x ρ divides. (The roof uses the division algorithm to write x ρ r with degr 0, i.e., r P R, so ρ r and ρ is a root if and only if r 0, i.e., x ρ.) The multilicity of a root ρ of is denoted ν ρ and is the largest value of n such that x ρ n. A little weirdness: Let R Z{x6y and let x 2 3x 2 P Rrxs. Here s a table of ρ for ρ P R: ρ 0 1 2 3 4 5 ρ 2 0 0 2 0 0 So V t1, 2, 4, 5u and has four roots even though its degree is only 2. It s certainly not true that x 1x 2x 4x 5, although x 1x 2 x 4x 5 (since 3 3 in R etc). On the other hand, if R is an integral domain, and, P Rrxs then V V Y V. This in turn imlies that if 0 and V tρ 1,..., ρ s u then x Qxx ρ 1 νρ 1 x ρ s νρs, where Q P Rrxs and V Q H. The number of roots of, counted with multilicities, is bounded by the degree of. (Prove this by induction on the degree of ). An interesting examle: Consider the olynomial x x P F rxs. Then V x x F by Fermat s little theorem, therefore x x xx 1x 2 x 1 in F rxs. Comare the coefficients of degree 1 on both sides and get 1! 1 in F, which gives another (easier? more natural?) roof of Wilson s theorem. Derivatives: In the context of a general commutative ring R, we can t use calculus (limits and such) to define the derivative of a olynomial. But we can just aroriate the formula from there, and define 1 na n x n 1 n 1a n 1 x n 2 a 1 if a n x n a n 1 x n 1 a 1 x a 0. Then you can formally rove the sum and roduct rules for derivatives.

3 It s easy to rove that if 2 then 1 and an element ρ P R is a multile root of (i.e., ν ρ 1) if and only if ρ is a root of both and 1. One fact about derivatives that doesn t carry over from calculus is the mean-value theorem. So there are non-constant olynomials with derivative zero for instance x P F rxs. Cyclotomic olynomials: Let s go back to Crxs for a bit and consider the nth roots of unity, i.e., the comlex numbers ξ that satisfy ξ n 1. As is well known, the nth roots of unity are ξ e 2πki{n for k 0, 1,..., n 1. The number ξ is called a rimitive nth root of unity if ξ n 1 but ξ k 1 for 0 k n. We have that e 2πki{n is a rimitive nth root of unity if and only if gcdk, n 1. Thus there are ϕn rimitive nth roots of unity (where ϕ is Euler s ϕ-function). Moreover, if ζ is a rimitive nth root of unity and ζ m 1. then n m (because then e 2πmki{n 1 so mk{n is an integer; n mk and gcdk, n 1 imlies n m). The nth cyclotomic olynomial Φ n x is defined to be the monic olynomial whose roots are recisely the rimitive nth roots of unity. So The first few Φ n are Φ 1 x x 1 Φ n x Φ 2 x x 1 Φ 3 x x 1 2 ¹ 1 k n, gcdk,n1? 3 i 2 Φ 4 x x ix i x 2 1 x x e 2πki{n. 1 2 i? 3 2 x 2 x 1 It is remarkable that the cyclotomic olynomials seem to (and do) all have integer coefficients, which allows us to define them as olynomials over any ring, and the following is true: Proosition: For all n 1, (i) x n 1 ± d n Φ dx, and (ii) Φ n x P Zrxs, i,e., the cyclotomic olynomials have integer coefficients. Proof. The roots of x n 1 are all the nth roots of unity. The roots of the Φ d x are the rimitive dth roots of unity, where d n, so all the roots of the roduct on the right side of (i) are roots of x n 1. But each root of x n 1 must be a rimitive dth root of unity for some d n for which d n. Thus the olynomials on the left and right sides of (i) have the same roots, and they are both monic, so they are eual (since C is a field). To rove ± Φ n x P Zrxs we use induction on n. We know the first few cases are true. For n 1, set f d n, d n Φ d, so that x n 1 Φ n f. By induction (since f is the roduct of Φ d s for d n), we know that f is a monic integer olynomial. Division of olynomials in Zrxs gives x n 1 f r where r 0 or degr degf and P Zrxs. Since f is monic, we have that and r are uniue in Zrxs as well as in Crxs, so we must have Φ n and r 0. Therefore Φ n P Zrxs. The identity (i) above is true in any Rrxs, via the canonical homomorhism from Z to R, extended to be a homomorhism from Zrxs to Rrxs. So we generalize the notion of rimitive nth root of unity to any commutative ring R: α P R is a rimitive nth root of unity if α n 1 and α k 1 for 1 k n.

4 Lemma: Suose R is an integral domain, and let α P R. If Φ n α 0 and if α is not a multile root of x n 1 P Rrxs, then α is a rimitive nth root of unity in R. ± Proof. The identity x n 1 d n Φ dx in Rrxs means there is a factorization Φ n x n 1 for some P Rrxs. Therefore α n 1 αφ n α 0 and so α n 1. If α is a rimitive dth root of unity for some 1 ± d n, then we must have d n by the arenthetical remark above. In this case, we have x n 1 c d Φ cx by (i) again, and since R is an integral domain ± we ll have Φ c α 0 for some c d. But now α is a root of at least two of the factors in x n 1 d n Φ dx, namely Φ n and Φ c for some c d n, so α is a multile root of x n 1, a contradiction. Using this lemma, we can rove an imortant result due to Gauss: Theorem: Let F be a field and let G F be a finite subgrou of the grou of units in F. Then G is cyclic. Proof. Let N G and consider the olynomial x N 1 ± d N Φ dx P F rxs. The roots of x N 1 are recisely the elements of G, since every element of G is a root, and there are at most N, and hence exactly N such roots. This tells us that none of the roots of x N 1 are multile roots. But then Φ N must have degφ N ϕn roots, which are rimitive Nth roots of unity by the lemma above, and hence are generators of G. A corollary of this theorem is that F is a cyclic grou. An integer a such that ras generates F is called a rimitive root mod. For instance, 2 is a rimitive root mod 13 (try it!). There doesn t seem to be any way to identify the ϕ 1 rimitive roots among the elements of F (the roortion of them can be arbitrarily small). Another alication of cyclotomic olynomials: Theorem: There are infinitely many rime numbers 1 mod n for any n 2. Proof. It is enough to show that there exists a rime number 1 mod n for every n 2 (why?). From the definition of Φ n, we have for n 2 that Φ n n 1. So there is a rime such that Φ n n. Now the constant term of Φ n is 1 since Φ n 0 1 and Φ n 0 P Z, which shows that n (since if n then would divide every term of Φ n n excet the constant term 1, but we re assuming Φ n n). Therefore rns is not a multile root of x n 1 P F rxs (since does not divide the derivative of x n 1 evaluated at x n). Since Φ n rns 0 in F, this imlies by the lemma above that the order of rns is n in F. Therefore n divides F 1 and so 1 mod n. More on ideals in olynomial rings. We already know that if F is a field, then F rxs is a Euclidean domain (the degree of a olynomial is the Euclidean function). Therefore F rxs is a rincial ideal domain and a uniue factorization domain and the division algorithm works in F rxs We illustrate this by finding gcdx 5 x 1, x 4 x 3 x 1 in F 2 rxs. i 1 0 1 2 3 r i x 5 x 1 x 4 x 3 x 1 x 3 x 2 x x 2 x 1 0 i x 1 x x λ i 1 0 1 x µ i 0 1 x 1 x 2 x 1 So the gcd is x 2 x 1 and x 2 x 1 xx 5 x 1 x 2 x 1x 4 x 3 x 1 in F 2 rxs.

5 Recall that the units in F rxs are the non-zero constants, and if is not irreducible then there are olynomials 1 and 2 such that 1 2 and 0 deg 1, deg 2 deg. So the following are direct conseuences of things we already know: Proosition: For P F rxs, (i) The ideal xy is maximal if and only if is irreducible, in which case F rxs{xy is a field. (ii) is a unit if and only if deg 0. (iii) If deg 1 then is irreducible (and F rxs{xy F ). (iv) If is irreducible and deg 1 then does not have any roots. (v) If deg 2 or 3 then is irreducible if and only if it has no roots. Examles: The olynomial x 3 x 1 P F 5 rxs is irreducible since it is degree 3 and has no roots: x 0 1 2 3 4 x 1 3 1 1 4 But x 4 x 2 1 P F 2 rxs has no roots since 0 1 and 1 1, but x 4 x 2 1 x 2 x 1 2 in F 2 rxs. Gauss roved (and we might rove one of these days) that the cyclotomic olynomials are irreducible in Qrxs. In the homework we ll exlore which cyclotomic olynomials Φ n are irreducible in F rxs. In Galois theory, one studies the situation where there is a field F and a olynomial P F rxs with no roots in F, along with an extension field E F containing an element α for which α 0 (we view also as an element of Erxs). The most familiar case of this is F R, E C, x 2 1 and α i. There is a natural construction of such an E, given F and. For instance Rrxs{xx 2 1y C. Because it s really no harder, we ll do this construction in the general case Rrxs where the coefficients come from a ring that is not necessarily a field. First a remark: Suose I is an ideal in Rrxs such that R X I x0y (where we consider R to be the subring of constant olynomials in Rrxs, so the only constant olynomial in I is the zero olynomial). If r 1, r 2 P R and rr 1 s rr 2 s P R{I, then r 1 r 2 P R X I and so r 1 r 2. So if R X I x0y we can simly write r to denote the element rrs in Rrxs{I. Proosition: Let R be a ring and x n a n 1 x n 1 a 1 x a 0 P Rrxs be a monic olynomial of degree n. Then R X xy x0y. Each element rs xy in the uotient ring Rrxs{xy can be exressed uniuely as a olynomial of degree less than n in rxs: b n 1 α n 1 b 1 α b 0, where b 0,..., b n 1 P R and α rxs. In Rrxs{xy we have the identity α n a n 1 α n 1 a 1 α a 0. It is essential that is a monic olynomial so that the considerations about degree on age 1 of these notes aly. Note that the natural ring homomorhism ϕ: R Ñ Rrxs{xy given by ϕr rrs is injective, so we can view R as a subring of Rrxs{xy. In the secial case that R F, a field and is an irreducible olynomial, then xy is a maximal ideal and F rxs{xy is an extension field E of F, and α rxs P E is actually a root of.

6 Examle. Let x 2 x 1 P F 2 rxs, which is irreducible since it has no roots. By the roosition, the uotient ring E F 2 rxs{xx 2 x 1y is a field, whose elements are of the form a bα, where a, b P F 2 and α 2 1 α 1 α determines the multilication rule: a bαc dα ac ad bcα bdα 2 ac bd ad bc bdα (it doesn t matter whether we use lus or minus signs since the characteristic of the field is 2). Note that E is an extension field of F 2 having 4 elements. The law of uadratic recirocity. Before the break, we were concerned with which in F are uadratic residues, i.e., which half of the non-zero elements of F can be exressed as the suares of elements of F. We introduced the Legendre symbol: a $ & % 0 if a 1 if a is a uadratic residue modulo 1 if a is a uadratic non-residue modulo. Recall that the Legendre symbol satisfies a a k for any k P Z, and if is an odd rime and a is an integer not divisible by, then we have Euler s formula a a 1{2 mod. This allows us to conclude that if is an odd rime, then the Legendre symbols satisfy: ab a b and we noted that 1 1 1{2 tells us that if is an odd rime, then 1 is a uadratic residue mod if 1 mod 4 and 1 is a uadratic non-residue mod if 3 mod 4. We can get a little more information in an elementary way by following in Gauss s footstes. We start as follows: For odd rimes, we re used to writing the numbers in F as 0, 1,..., 1, but we could just as easily write them as 1 2, 3 3,..., 2, 1, 0, 1, 2,..., 2 2, 1 2. For any integer a such that a, we consider the list of numbers a, 2a, 3a,..., 1 2 a. None of these numbers is divisible by, and no air of these are congruent to each other mod. We set µ a (or just µa if is clear from the context) eual to the number of elements of this list that are congruent to negative numbers in the above listing of F (or to numbers bigger than {2 in the standard listing of F ). For instance, if 11 then µ6 3, since 6, 12, 18, 24, 30 are

7 congruent to 5, 1, 4, 2, 3 mod 11. Using the µ function, we can give another characterization of Legendre symbols: Lemma (Gauss): With the above notation, if a, then a 1 µa. Idea of roof: Each number ka for k 1,..., 1{2 is congruent to m k for 1 m k 1{2. When 1 j, k 1{2 and j k, we cannot have ja ka mod (since F is a field), and by the definition of µ we conclude that a 1{2 1 2! 1 µa 1 2! mod and so Gauss s result follows from Euler s after canceling off the 1{2!. Using this, we can determine when 2 is a uadratic residue mod for an odd rime. Namely, 2 is a uadratic residue mod of 1 mod 8 or 7 mod 8, and 2 is a uadratic non-residue mod if 3, 5 mod 8. To see this, we need to comute µ 2, i.e., how many of the numbers 2, 4,..., 1 are greater than {2. And if 1 mod 4 then this number is 1{4, where if 3 mod 4 it s 1{4. Therefore 2 $ '& '% 1 if 1 mod 8 1 if 3 mod 8 1 if 5 mod 8 1 if 7 mod 8 To do much more, we need the owerful law of uadratic recirocity, due to Gauss. It states that if and are odd rimes then Another way to say this is $ '& '% 1 1 1{4. It is remarkable that the two congruences if 3 mod 4 otherwise x 2 mod and x 2 mod should have any connection. But here s an examle that shows the usefulness of the law of uadratic recirocity in comuting Legendre symbols: 19 43 5 19 4 2 2 1 43 19 19 5 5 5 5 and so the congruence x 2 19 mod 43 has no solutions. To rove Gauss s law of uadratic recirocity we will work in the ring R F rxs{x1 x x 1 y.

8 From the roosition on age 5, an element in R can be written uniuely in terms of α rxs as where c 0,..., c 2 P F. c 0 c 1 α c 1 α 2 Lemma: The element α is a rimitive th root of unity in R. Moreover, if l and β α l then in R. 1 β β 1 0 Proof. We know from the roosition that α, α 2,..., α 2 1 and α 1 1 α α 2 1. But α αα 1 1, and so α is a rimitive th root of unity. If l then gcd, l 1, and so t1, α,..., α 1 u t1, β,..., β 1 u, which gives the euation in the lemma. Gauss sums. We define the Gauss sum in R to be G 1 k1 k α k. Because we re working in R (where α 1), the individual terms satisfy k k m α k α k m for every m P Z. We ll use this often to rove two imortant roerties of G: 1. G 2 1 1{2. 2. If, then G is an invertible element in the ring R. Proof. The invertibility of G follows from (1) since P F R is invertible in R since it is invertible in F for. To rove (1), we start calculating: 1 1 k k G 2 α k α k k1 1 j1 j j1 k1 1 k1 1 α j k1 k (where we reversed the second sum and used that 1 1 j k G 2 1 1 j1 k1 1 1{2 1 jk 1 j1 k1 k α j k α k α k α j k j 2 k α j1 k k α k ). Next,

1 where in the last euality we used the fact about from near the bottom of age 6 and we relaced k with jk, since as k runs through 1,..., 1 the remainders of jk mod also run through j 2 1,..., (though not necessarily in the same order). Since 1 by definition, we get because 1 k1 1 G 2 1 1{2 k1 1 1{2 1 k1 k k 1 α j1 k j1 1 α j1 k k 0 (half the numbers between 1 and 1 are uadratic residues mod ). From the lemma above, we have that the formula for G 2 in (1) above. 1 j0 j0 α j1 k 0 unless k 1, in which case the sum is. This gives Proof of the law of uadratic recirocity. Raise G to the th ower in R and get G G 2 1{2 G 1 1 1{4 1{2 G 1 1 1{4 G using Euler s formula for the Legendre symbol. On the other hand, we can calculate G from the definition and use the freshman dream in the ring R to get G 1 j1 1 j1 j α j j 1 j1 α j j α j Since G is invertible, we can cancel G from the two exressions for G and get the law of uadratic recirocity: G 1 1 1{4. The above is one of the half-dozen or so roofs that Gauss gave of the law of uadratic recirocity. He was so taken with the theorem that he called it his Theorema Aureum. Finite fields. Next we turn to the remarkable fact that for every rime and every n 1 there exists a uniue field with n elements (we constructed a field with 2 2 elements above). Lemma: Suose F is a finite field, then F n, where is a rime number, n 1, and there exists an irreducible olynomial f P F rxs of degree n such that F F rxs{xfy. 9

10 Proof. Start with the uniue ring homomorhism κ: Z Ñ F, which is not injective since F is finite. Therefore the characteristic (generator of the kernel of κ) of F is a rime number and F, being the image of κ, is a subring of F. By the first theorem on age 4, we have that F is a cyclic grou, so let σ be a generator of F. Thus, every element in F is either 0 or else some ower σ n of σ. Since ϕ σ x σ, and so ϕ σ x n σ n, the ring homomorhism ϕ σ : F rxs Ñ F is surjective, and in fact, since x P F rxs F rxs, we can restrict ϕ σ to F rxs and get a surjective homomorhism ϕ: F rxs Ñ F. The kernel of ϕ is a rincial ideal xfy F rxs, and F rxs{xfy F, so xfy is a maximal ideal. Therefore f is an irreducible olynomial (by (i) of the Proosition on age 5). And F n, where n degf by the other roosition on age 5. Our goal now is to rove the main result of this subsection: Theorem: There exists a finite field with n elements, where is a rime number and n 1. More recisely: (i) There exists an irreducible olynomial in F rxs of degree n. (ii) If F and F 1 are finite fields with n elements, then there is a ring isomorhism F Ñ F 1. Proof. To rove (i), we are going to use cyclotomic olynomials since the cyclotomic olynomial Φ k has integer coefficients, we can use the homomorhism κ: Z Ñ F to consider Φ k as an element of F rxs. We are going to show that if f is an irreducible olynomial dividing Φ n 1 in F rxs, then degf n. To do this, suose degf d, then we know that E F rxs{xfy is a field with d elements and α rxs is a root of f P F rxs Erxs. Since f Φ n 1 we have gf Φ n 1 for some g P F rxs and we get that Φ n 1α gαfα 0. The derivative of x n 1 1 P F rxs is x n 2, therefore α is not a multile root of x n 1 and so α is a rimitive n 1th root of unity. But α d 1 1 (that s the order of the grou E ), and so n 1 d 1. On the other hand, let R tξ P E ξ n ξu, which is a subring of E (use the freshman dream to get additivity). Since α n 1 1, we must have α P R, and since E ta 0 a 1 α a d 1 α d 1 a i P F u, it follows that R E (since R contains 1 and all owers of α and is a subring of E). Now we know there is a rimitive d 1th root of unity ζ in E, and since E R we have ζ P R and so ζ n 1 1. But then d 1 n 1 and combining this with the receding aragrah tell us that d 1 n 1, or d n. This comletes the roof of (i). To rove (ii), suose F and F 1 are finite fields with n elements. By the lemma above, F F rxs{xfy for some irreducible olynomial f of degree n, and fα 0, where α rxs P F. The set I tg P F rxs gα 0u F rxs is an ideal in F rxs, and f P I. Therefore xfy I, but xfy is a maximal ideal (because F is a field) and so I xfy. Now F is a finite grou with n 1 elements, therefore β n 1 1 0 for every β P F, which imlies that x n x P I and therefore f x n x in F rxs. On the other hand, in F 1 rxs we have that x n x ¹ γpf 1 x γ,

11 since every γ P F 1 satisfies γ n γ 0 as well. Therefore f P F rxs F 1 rxs must have a root α 1 P F 1 since f divides x n x. So consider the ring homomorhism ϕ α 1 : F rxs Ñ F 1. Chearly xfy kerϕ α 1, but since kerϕ α 1 is a roer ideal and xfy is a maximal ideal in F rxs, we must have xfy kerϕ α 1. Therefore there is an injective ring homomorhism F rxs{xfy Ñ F 1 which must also be surjective since F 1 has the same number of elements as F rxs{xfy F. Thus F F 1 and we are done. We know that x n x xx n 1 1 x ¹ d n 1 Φ d in F rxs. And by the theorem on the receding age, we know that x n x is divisible by an irreducible olynomial of degree n. But we can say a bit more about this, in articular we can calculate the comlete irreducible factorization of x n x in F rxs. For instance in F 2 rxs, and in F 3 rxs, x 22 x x 4 x xx 1x 2 x 1 x 32 x x 9 x xx 1x 2x 2 1x 2 x 2x 2 2x 2. In general we have the following: Theorem. The olynomial x n x P F rxs is the roduct x n x f 1 f 2 f k of all the monic irreducible olynomials f 1,..., f k in F rxs of all degrees d for which 1 d n and d n. Proof. We can restate the theorem as follows: For d such that 1 d n and f P F rxs an irreducible monic olynomial of degree d, f x n x if and only if d n. Furthermore x n x is not divisible by the suare of any irreducible olynomial. So we suose d satisfies 1 d n and f P F rxs is an irreducible monic olynomial of degree d. Then we have E F rxs{xfy is a field with d elements, and α rxs P E satisfies α d α (because E is a cyclic grou of order d 1). Now if d n, then raising both sides of α d α to the d ower times, where n d, gives us that α n α in E. And this means that α n α rx n xs r0s P E F rxs{xfy, in other words, x n x P xfy, in other words f x n x. Now let s assume that the monic irreducible olynomial f P F rxs of degree d divides x n x and we wish to show that d n. Once again consider the field E F rxs{xfy, and let gx x n x P Erxs. Clearly 1 P E satisfies g1 0, and α rxs P E satisfies gα 0, since f g and fα 0 in E. Now use the freshman s dream to conclude that the set of elements e of E which satisfy ge 0 is a subring of E, and hence it is all of E. But E has d elements, so E is a cyclic grou of order d 1. And if σ is a generator of E then σ d 1 1, and also σ n 1 1 since this is true for all elements of E. Thus d 1 n 1. We claim that this imlies d n and will rove this below. U to this oint, we ve shown that the x n x is the roduct of the monic irreducible olynomials of degrees d which divide n. Now we have to show that none of these irreducible olynomials occur

12 to a ower higher than 1 in the factorization of x n x. But if f is an irreducible factor of x n x, then f 2 cannot divide evenly into x n x, since the derivative of x n x n x n 1 1 1 in F rxs (and use the first sentence on age 3). So the last detail we have to take care of is a roof that if t, d and n are ositive integers, with t 1, then t d 1 t n 1 if and only if d n. Start by writing n d r with 0 r d. Then But 0 t r 1 theorem. t n 1 t d 1 td t r 1 t d td t r t r t r 1 1 t d 1 t r td 1 t d 1 t r 1 t d 1 t r 1 t d t d 1 t r 1 t d 1 t d 1, so the division works if and only if r 0. This comletes the roof of the If we take the degree of both sides of the factorization x n x f 1 f k from the theorem, we get the euation n dn d where N d is the number of monic irreducible olynomials of degree d in F rxs. Since we know that there are monic irreducible olynomials of degree 1 in F rxs, namely x, x 1, x 2,..., x 1 we have N 1. So if is a rime number, then d n N N 1 N and we can conclude that More generally, we have N n 1 n N {. n d n,d n dn d. Another imortant conseuence of the theorem above is the following lemma: Lemma: Let f P F rxs be an irreducible olynomial of degree d. Then f x d x and f does not divide x c x if c d. Using this result we can find factors of a given olynomial f P F rxs using the Euclidean algorithm. Suose that g P F rxs, degg d and g g 1 g 2 g d where g i is the roduct of all the irreducible olynomials of degree i that divide g. It then follows from the theorem that gcdx i x, g is the roduct of all the g j for j i. So we can find the g j by successively inserting i 1, 2,... into gcdx i x, g and using the Euiclidean algorithm to comute the gcd. Factoring in F rxs: We can use linear algebra to hel decide whether a olynomial in F rxs of degree 4 is irreducible. To do this, we consider the Frobenius ma F : F Ñ F where F λ λ

13 (this is a ring homomorhism because of the freshman s dream ). Given a olynomial f P F rxs we extend F to the ring R F rxs{xfy, and we ll still call this ma F : R Ñ R. But we can view R as a vector sace over F, and because λ λ for λ P F, the ma F (extended to R) is a linear maing of vector saces. It might hel to do an examle of this. Examle: Let f x 5 x 1 P F 2 rxs. Then R F 2 rxs{xfy is a vector sace over F 2 with basis t1, α, α 2, α 3, α 4 u where α rxs. Since fα 0 in R, we have that α 5 α 1. What doe the Frobenius ma F λ λ 2 do to this basis? Well, F 1 1, F α α 2, F α 2 α 4, F α 3 α 6 αα 5 αα 1 α 2 α, and F α 4 α 8 α 3 α 5 α 3 α 1 α 4 α 3. Therefore the matrix of the ma F with resect to this basis is M F 1 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 1 0 1 Note that this matrix is invertible, since if we aly the ermutation 2453 to it, it becomes uer triangular with 1s on the diagonal (so det M F 1). Now if M F were not invertible, then we could find a non-constant olynomial g P F rxs such that degg degf and rgs 0. And if were an irreducible olynomial such that f then we would have g. Therefore gcdf, g is a non-trivial divisor of f (i.e., 0 deggcdf, g degf). Next, suose g P F rxs is a olynomial such that 0 degg degf and rgs rgs 0 in R F {xfy. In other words, rgs is in the kernel of the linear ma F I : R Ñ R (viewing R as a vector sace over F ). Since x x xx 1 x 1 in F rxs, we also have the factorization. g g gg 1 g 1 in F rxs. If is an irreducible factor of f, and since f g g (because rg gs 0 P R F {xfy), we obtain that will divide one of g, g 1,...,g 1. And so one of gcdf, g, gcdf, g 1,...,gcdf, g 1 is a non-trivial factor of f (since degg degf). Examle (continued): The matrix of F I for the examle above is M F I 0 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 0 0 1 1 0 0 1 0 0 Now r1, 0, 0, 0, 0s T P kerm F I, but we knew that would haen since a a 0 for all a P F. But there is a second, linearly indeendent element of kerm F I, namely r1, 1, 0, 1, 1s T. This means that the olynomial g 1 x x 3 x 4 satisfies f g 2 g. Using the Euclidean algorithm, we can comute that gcdx 5 x 1, x 4 x 3 x 1 x 2 x 1.

14 and so x 2 x 1 is a nontrivial factor of x 5 x 1. So we have a way to find non-trivial factors of olynomials in F rxs. It might be a bit surrising to know that if the method given above doesn t work to find a factor of f, then f is irreducible: Theorem: Suose f P F rxs is a non-constant olynomial and let F : R Ñ R be the Frobenius ma, where R F rxs{xfy. Then f is irreducible if and only if kerf 0 and kerf I F. Proof. We have seen above that kerf 0 and kerf I F if f is irreducible, since otherwise we can use the method above to find a non-trivial factor of f. So conversely, assume that kerf 0 and kerf I F, and let r be a non-zero element of R. We re going to show that r is invertible in R, which will imly that R is a field, and thus that f is irreducible. Consider the F -linear ma A: R Ñ R given by Ax rx, and suose that x P kera X ima. Then x ry for some y P R and rx 0. But then F x F ry r y r 2 y 1 rx 0, and so x P kerf. Therefore x 0 and so kera X ima 0. But since dimkera dimima dimr (the dimensions are taken as vector saces over F ), we have kera ima R Now, if x P kera then so is F x, since AF x rx rxx 1 0. Likewise, if x P ima then so is F x, since if x Ay ry then F x x ry rr 1 y P ima. We can exress 1 P R uniuely as x y where x P kera and y P ima. But then F 1 1 F x F y, and so F x x and F y y. But since kerf I F we have x P F and y P F. The only way x can also be in kera is for x 0 (since x is a scalar ), and so y 1. But now 1 P im A so there is a z P R such that rz Az 1, and we are done.