Math 121 Homework 3 Solutions Problem 13.4 #6. Let K 1 and K 2 be finite extensions of F in the field K, and assume that both are splitting fields over F. (a) Prove that their composite K 1 K 2 is a splitting field over F. (b) Prove that K 1 K 2 is a splitting field over F. (Use Exercise 5 in 13.3, which was in the previous assignment.) Solution. By hypothesis K 1 is the splitting field of a polynomial, which we may assume to be monic: f(x) = (x α 1 ) (x α n ). Then K 1 = F (α 1,, α n ), since by the definition of the splitting field, it is the smallest field containing F and all the roots of f. Note that we are not assuming f to be irreducible. Similarly suppose that K 2 = F (β 1,, β m ) is the splitting field of g(x) = (x β 1 ) (x β m ). Then K 1 K 2 is the smallest field containing both K 1 and K 2, and clearly this is F (α 1,, α n, β 1,, β m ). This is the splitting field of f(x)g(x), proving (a). Part (b) requires more careful reasoning. Let γ K 1 K 2, and let h be the irreducible polynomial over F. Since K 1 is a splitting field, by Exercise 5 in 13.3, h factors into linear factors in K 1 [x], so Similarly it has a factorization into h(x) = (x γ 1 ) (x γ p ), γ = γ 1. (1) h(x) = (x γ 1) (x γ p), γ = γ 1 (2) 1
in K 2 [x]. But (working in K) we may substitute each root γ i from the factorization (3) into equation (4) and since h(γ i ) = 0, we see that each γ i is one of the γ i. In other words, each γ i is in K 2 as well as K 1. Since γ i K 1 K 2, we see that the irreducible polynomial splits into linear factors in (K 1 K 2 )[x]. This shows that K 1 K 2 is a splitting field, by the criterion of Exercise 5 in 13.3. Problem 13.5 #1. Prove that the derivative D x of a polynomial satisfies and D x (f(x) + g(x)) = D x (f(x)) + D x (g(x)) D x (f(x)g(x)) = D x (f(x))g(x) + f(x)d x (g(x)) for any two polynomials f and g. Solution. Of course we cannot use calculus to prove this, but rather the definition of the derivative from Page 546 of Dummit and Foote. That is, if f(x) = n a k x k, then D x (f(x)) = k=0 n k=1 k a k 1 k. The maps f D x f is obviously linear, so the first identity is trivial. For the second identity, let us check this first when f and g are monomials, say f(x) = x n and g(x) = x m. Then D x (fg) = (n + m)x n+m 1 = nx n 1 x m + x n mx m 1 = D x (f)g + fd x (g). If f and g are general, we note that both sides of the equation D x (f(x)g(x)) = D x (f(x))g(x) + f(x)d x (g(x)) are bilinear in f and g, reducing to the case where f and g are monomials. Problem 13.5 #2. Find all irreducible polynomials of degree 1, 2 and 4 over F 2 and prove that their product is x 16 x. Observation. A polynomial over F 2 has no root in F 2 if and only if it has an odd number of terms (so 1 is not a root) and constant term 1 (so that 0 is not a root). 2
Solution. The polynomials x and x + 1 of degree 1 are clearly irreducible. Remember that a polynomial of degree 2 is irreducible in F [x] if and only if it does not have a root in F.) By the observation, there is one irreducible of degree 2: x 2 + x + 1. The irreducibles of degree 4 may be found as follows. First, an irreducible cannot have a root in F 2, and the observation gives us four such polynomials. One is ruled out: x 4 +x 2 +1 = (x 2 +x+1) 2 has no root in F 2, but is reducible. So here is the result. The product of these six polynomials is x 16 + x. degree irreducible polynomials x 1 x + 1 2 x 2 + x + 1 x 4 + x 3 + x 2 + x + 1 4 x 4 + x 3 + 1 x 4 + x + 1 Problem 13.5 # 5. For any prime p and any nonzero a F p, show that the polynomial x p x + a is irreducible and separable over F p. Solution. First we check the following fact. Lemma 1. If α is a root of f(x) = x p x + a, then so is α + t. Proof. If α is a root of f and t F p then α + t is also a root of f since t p = t and so (α + t) p (α + t) + a = (α p + t) (α + t) + a = α p α + a = 0. Thus the roots of f are the p distinct values α + t with t F p. Now suppose that f is reducible. Note that f has no roots in F p since t F p satisfies t p t = 0 and a is assumed nonzero. Therefore every irreducible factor of f has degree 2. Let h be such an irreducible factor. Then h has at least two roots, which are among the roots of f. If α is one then α + t is another, for some t F p, t 0. By Theorem 8, there is a homomorphism θ : F q (α) F q (α + t) that sends α α + t. Of course F q (α + t) = F q (α), so θ is actually an automorphism. Not θ(α + t) = θ(α) + θ(t) = (α + t) + t = α + 2t. Then 3
h(α + 2t) = h(θ(α + t)) = θ(h(α + t)) = 0, so α + 2t is also a root. Similarly, α + 3t, α + 4t, are all roots of h. We see that α, α + t, α + 2t, are all roots of h. Since every root of f is a root of h, f = h is irreducible. For separability, the derivative f = 1 is obviously prime to f, so f is separable. Problem 13.5 # 6. Prove that Conclude that x pn 1 1 = a F p n (x a). (3) α = ( 1) pn, (4) α F p n so that product of the nonzero elements of a finite field of characteristic p is 1 if p = 2 and 1 if p is odd. Solution. The relevant ideas are contained in the Example on pages 549-550 of Dummit and Foote. So much of our solution will consist of reviewing what they do there. They consider the polynomial f(x) = x pn x. They show it is separable, so it has p n distinct roots, and they show that the set of roots in a splitting field is the splitting field. Thus it follows from what they prove in this example that x pn x = (x α). α F p n Now note that α = 0 is one of these roots, and we may divide both sides of the identity by x to obtain (8). Evaluating both sides of (3) at x = 0 gives 1 = α F p n ( α) which implies (9). Problem 13.5 # 8. Prove that f(x) p = f(x p ) for any polynomial f F p [x]. Solution. Let φ : F p [x] F p [x] be the Frobenius map φ(f) = f p. We have the identities φ(f + g) = φ(f) + φ(g), φ(fg) = φ(f)φ(g). 4
The second identity is true in any ring, and the first just depends on the binomial coefficients ( p k) = 0 when 1 k p 1, so it is true in any ring of characteristic p. In other words, Proposition 35 on page 548 of Dummit and Foote, which is stated for a field of characteristic p, is actually true in a ring of characteristic p, in this case F p. Using this, let We have f(x) = a n x n + a n 1 x n 1 +... + a 0. φ(f(x)) = φ(a n )φ(x) n + φ(a n 1 )φ(x) n 1 +... + φ(a 0 ). (5) Now we make use of a property of F p. Lemma 2 (Fermat). If a F p then a p = a. Proof. This is obvious if a = 0, and if a 0, then a p 1 = 1 since a F p, a group of order p 1. Multiplying a p 1 = 1 by a gives a p = a. So we see that φ(a i ) = a i and φ(x) = x p. Applying this in (10) gives f(x) p = a n x pn + a n 1 x p(n 1) +... + a 0 = f(x p ). Problem 13.6 # 1. Suppose that m and n are relatively prime integers. Let ζ m be a primitive m-th root of unity, and let ζ n be a primitive n-th root. Show that ζ m ζ n is a primitive nm-th root of unity. Solution. Let ζ = ζ m ζ n. To show that ζ is a primitive mn-th root of unity, We have to show that: ζ mn = 1, If ζ a = 1 then a is a multiple of mn. For the first claim, ζ mn = (ζm) m n (ζn) n m = 1 n 1 m = 1. For the second claim, suppose that ζ a = 1. Then 1 = 1 n = ζm an ζn an = ζ an m. Since ζ m is a primitive m-th root of unity, this implies that an is a multiple of m, and since m, n are coprime, it follows that a is a multiple of m. Similarly it is a multiple of n. Since m and n are coprime, it follows that a is a multiple of mn, as required. 5
Problem 13.6 # 10. Let ϕ denote the Frobenius map x x p on the finite field F p n. Show that ϕ is an isomorphism of F p n with itself. ( Automorphism. ) Show that ϕ n is the identity map of F p n but no smaller power of ϕ is the identity. Solution. The fact that ϕ is a ring homomorphism follows from the identities ϕ(a + b) = ϕ(a) + ϕ(b), ϕ(ab) = ϕ(a)ϕ(b) from Proposition 35 on page 548 of Dummit and Foote. To see that it is injective, note that ϕ(a) = 0 means a p = 0 so a = 0. To see that it is surjective, use the Pigeonhole Principle: if X is a finite set then a map X X is surjective if and only if it is injective. We have shown that ϕ is a bijective ring homomorphism, so it is an automorphism. To show that ϕ is the identity map on F p n note that if a F p n then a pn = a. This fact is contained in the Example in 549-550, but let us give a quick proof of it. If a = 0 then clearly a pn = 0 also, so we may assume that a 0. Then a lies in the F p which is an abelian group of order n pn 1. Hence a pn 1 = 1 and multiplying this identity by a gives a pn = a. Now ϕ n (a) = a pn so we have proved that ϕ n is the identity on F p n. We must also show that n is the smallest integer for which this is true. If ϕ d is the identity then a pd = a for all a F p n. This means that every element of F p n is a root of the equation x pd x = 0. The degree of the polynomial must therefore be at least p n = F p n, so d n. 6