Math 121 Homework 3 Solutions


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1 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
2 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 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
3 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 x 2 + x + 1 x 4 + x 3 + x 2 + x x 4 + x 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
4 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 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
5 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 a 0. φ(f(x)) = φ(a n )φ(x) n + φ(a n 1 )φ(x) n 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 mth root of unity, and let ζ n be a primitive nth root. Show that ζ m ζ n is a primitive nmth root of unity. Solution. Let ζ = ζ m ζ n. To show that ζ is a primitive mnth 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 mth 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
6 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 , 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
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