Math 11 Homework 6 Solutions Problem 14. # 17. Let K/F be any finite extension and let α K. Let L be a Galois extension of F containing K and let H Gal(L/F ) be the subgroup corresponding to K. Define the norm of α from K to F to be N K/F (α) = σ σ(α), where the product is taken over all embeddings of K into an algebraic closure of F (so over a set of coset representatives for H in Gal(L/F ) by the Fundamental Theorem of Galois Theory.) In particular if K/F is Galois this is σ Gal(K/F ) σ(α). (a) Prove that N K/F (α) F. (b) Prove that N K/F (αβ) = N K/F (α)n K/F (b), so the norm is a multiplicative map from( K to F. D ) (c) Let K = F be a quadratic extension of K. Show that N K/F (a + b ) D = a Db. (d) Let m α (x) = x d +a d 1 x d 1 +...+a 1 x+a 0 be the minimal polynomial of α K over F. Let n = [K : F ]. Prove that d n, that there are d distinct Galois conjugates of α which are all repeated n/d times in the product above, and conclude that N K/F (α) = ( 1) n a n/d 0. Solution. We review some points that are made in the proof of Theorem 14 (The Fundamental Theorem of Galois Theory), page 575-6. We may take the algebraic closure F of F to contain L. We will denote G = Gal(L/K). Lemma 1. Let G = Gal(L/F ) and H = Gal(L/K) G. Then every embedding σ : K F extends to an automorphism σ 1 G, and σ 1, σ G 1
have the same restriction to K if and only if they represent the same coset σ 1 H = σ H. Therefore the distinct embeddings of K into F over F are in bijection with the cosets σh. Proof. Since L is a splitting field, every embedding σ : K F has image in L, and extends to an automorphism of L. If σ 1 and σ Gal(L/K) are two such extensions, then they are the same if σ 1 (x) = σ (x) for all x K, in other words, σ1 1 σ (x) = x which means σ1 1 al(l/k) = H. This is equivalent to σ 1 H = σ H. So we can write N K/F (α) = cosets σh σ(α). (1) Now let us show N K/F (α) F. Since F is the fixed field of G, it is sufficient to show that τn K/F (α) = N K/F (α) for τ G. This is because σh τσh is a permutation of the cosets and so N K/F (α) = cosets σh σ(α) = cosets τσh τσ(α) = τ cosets σh σ(α) = τn K/F (α). This proves (a). As for (b), σ(αβ) = σ(α)σ(β) and taking the product over all embeddings σ of K into L over F gives N K/F (αβ) ( = N K/F (α)n K/F (β). D ) For (c), the assumption that F is a quadratic extension of K implies that the polynomial x D is irreducible over F. Its roots are D and D, so N K/F (α + b ) ( D = a + b ) ( D a b ) D = a Db. Before we prove (d), we will prove another important property of the norm. Proposition. Let E K F and let α E. Then N E/F (α) = N K/F (N E/K (α)). This property is sometimes called the transitivity of the norm map.
Proof. We may choose the Galois extension L of F in the definition of the norm so that it contains both K and E. Let H = Gal(L/K) as before, and let M = Gal(L/E) so M H. Now we can choose coset representatives for M in G as follows. First, let σ 1,, σ k be left coset representatives for H = Gal(L/K) in G = Gal(L/F ). This means G = σ i H (disjoint). i Similarly, let τ 1,, τ l be left coset representatives for M in H, so H = j τ j M (disjoint). Then G = i σ i H = i,j σ i τ j M, so the σ i τ j are a set of coset representatives for M in G. This means that N E/F (α) = σ i τ j (α) = ( ) σ i τ j α = σ i (N E/K (α)) = N K/F (N E/K (α)). i,j i j i Now let us prove (d). We have inclusions K F (α) F. Let [F (α) : F ] = d so [K : F (α)] = n/d. By the transitivity of the norm, N K/F (α) = N F (α)/f N K/F (α) (α). Since α F (α), τ(α) = α for every embedding of K into L over F (α), and there are n/d such embeddings, so N K/F (α) (α) = τ τ(α) = α n/d. Now let us compute N F (α)/f (α). This is the product of the conjugates σ(α) of α. These are the distinct roots of the minimal polynomial d (x σ i α). i=1 3
(The roots of this polynomial are all distinct since L/F is Galois and therefore separable, and so any intermediate extension such as F (α)/f is separable.) The constant term a 0 of this polynomial is then ( 1) d σ i (α) = ( 1) d N F (α)/f (α). Therefore N K/F (α) = N F (α)/f N K/F (α) (α) = N F (α)/f (α n/d ) = (N F (α)/f (α)) n/d = (( 1) d a 0 ) n/d = ( 1) n a n/d 0. Problem 14. #18. With notation as in the previous problem, define the trace of α from K to F to be Tr K/F (α) = σ σ(α), a sum of Galois conjugates of α. (a) Prove that Tr K/F (α) F. (b) Prove that Tr K/F (α + β) = Tr K/F (α) + Tr K/F (β), so that the trace is an additive map from ( K to F. D ) ( (c) Let K = F be a quadratic extension of K. Show that Tr K/F a + b ) D a. (d) Let m α (x) be as in the previous problem. Prove that Tr K/F (α) = n a d d 1. Solution: This is so similar to the previous problem that we won t write out solutions to (a) and (b). For (c), as in the previous problem the conjugates of a + b D over F are a + b D and a b D, so the trace is the sum a of these. For (d), this is also similar to the previous problem. We may use the transitivity property of the trace: Proposition 3. Let E K F and let α E. Then Tr E/F (α) = Tr K/F (Tr E/K (α)). Problem 14. #1. Use the linear independence of characters to show that for any Galois extension K of F there is an elment α K with Tr K/F (α) 0. Solution: The linear independence of characters is Theorem 7 on page 569. It is due to Artin. Since K/F is Galois, then Tr K/F (α) = χ(α). χ Gal(K/F ) If this is always zero, then χ = 0, contradicting Theorem 7 with all a i = 1. 4 =
Problem 14. #. Suppose K/F is a Galois extension of F and let σ be an element of Gal(K/F ). (a) Suppose that α K is of the form α = β for some nonzero β K. σβ Prove that N K/F (α) = 1. (b) Suppose that α K is of the form α = β σβ for some β K. Prove that Tr K/F (α) = 0. This problem sets up Hilbert s Theorem 90 (Exercise 3) which we will be discussing later. Solution: Since K/F is Galois, N(β) = τ(β), N(σβ) = τ Gal(K/F ) τ Gal(K/F ) τσ(β), But τ τσ just permutes the elements of Gal(K/F ). Therefore N(β) = N(σ(β)). Thus N(β/σβ) = N(β)/N(σβ) = 1. Part (b) is similar. Problem 14.3 #9. Let q = p m be a power of the prime p and let F q = F p m be the finite field with q elements. Let σ q = σp m be the m-th power of the Frobenius automorphism σ p, called the q-frobenius automorphism. (a) Prove that σ q fixes F q. (b) Prove that every finite extension of F q of degree n is the splitting field of x qn x over F q, and hence there is a unique such extension. (c) Prove that every finite extension of F q of degree n is cyclic with σ q as a generator. (d) Prove that the subfields of the unique extension of F q of degree n are in bijection with the divisors d of n. Solution. This problem is very similar to Proposition 15 on page 586. We could argue some points alternatively by making more use of Proposition 15. (a) We may interpret σ q as an automorphism of any field containing F p, in particular of F q. But as an automorphism of F q, it is trivial, and this is the content of (a). To see this, we must show that σ q (a) = a if a F q, that is, a q = a. If a = 0 this is obvious, so assume that a F q. This is a finite group of order q 1, so a q 1 = 1. Multiplying this equation by a gives a q = a. (b) If [E : F q ] = n, then E is a vector space of dimension n over F q, so E has cardinality q n. Now if a E, then we claim a qn = a. We may prove this the same way as (a): if x = 0 this is clear, and if a 0, then a lies in a group E of order q n 1, so a qn 1 = 1, so a qn = a. Now the polynomial 5
f(x) = x qn x is separable, since f (x) = 1, so f and f are coprime. Thus its roots are distinct, but we have shown that the q n elements of E are roots, so the elements of E are precisely the roots of this polynomial in an algebraic closure of E. It is now clear that E is the splitting field of f. (c) With F = F q and E = F q n, the field E is the splitting field of a separable polynomial, so it is Galois over E. The group G = Gal(E/F ) has order n = [E : F ], and σ q is an element. To show that it is cyclic with generator σ q sufficient to show that if m is a divisor of n = G and σq m = 1 E then m = n. Indeed, for all a E, we have a = σq m (a) = a qm, so the polynomial x qm x has q n roots, namely all elements of E. Because a polynomial of degree q m cannot have more than q m roots, we must have m = n. (d) This now follows from the fundamental theorem of Galois theory, since a cyclic group of order n has one subgroup for each divisor of n. Problem 14.3 #10. Prove that n divides ϕ(p n 1). [Hint: observe that ϕ(p n 1) is the order of the group of automorphisms of a cyclic group of order p n 1. Solution. By Proposition 16 on page 135 of Dummit and Foote, the group of automorphisms of the cyclic group Z m of order m is (Z/mZ), which has order φ(m). We apply this with m = p n 1. The group F p is cyclic n of order p n 1 so Aut(F p n) has order φ(pn 1). Now we may exhibit an automorphism of order exactly n, namely the Frobenius map x x p. By Lagrange s theorem, every element of Aut(F pn) must have order dividing Aut(F p n) and so n pn 1. Problem 14.5 #3. Determine the quadratic equation satisfied by the period α = ζ 5 + ζ5 1 of the fifth root of unity ζ 5. Determine the quadratic equation satisfied by ζ 5 over Q(α) and use this to explicitly solve for the fifth root of unity. Solution. Let al(q(ζ 5 )/Q). Then σ(ζ 5 ) = ζ5 a where a can be 1,, 1 or. So σ(ζ 5 ) = ζ5 a + ζ5 a. α± α 4 This means that the conjugates of α are α = ζ 5 + ζ5 1 and β = ζ5 + ζ5. Now 1 + α + β = 1 + ζ 5 + ζ 1 5 + ζ 5 + ζ 5 = 0. Also αβ = (ζ 5 + ζ 1 5 )(ζ 5 + ζ 5 ) = ζ 3 5 + ζ 1 5 + ζ 5 + ζ 3 5 = 1. 6
Since α + b = αβ = 1 we have (x α)(x β) = x (α + β)x + αβ = x + x 1. The f(x) = x + x 1 is the quadratic equation satisfied by α, β and so α, β are 1 ± 5. With ζ 5 = e πi/5 we have α = cos ( ) π 5 > 0 since 0 < π < π, while 5 β = cos ( ) 4π 5 < 0. So α = 5 1 5 + 1, β =. α± α 4 Now we are asked to find the quadratic equation satisfied by ζ = ζ 5 over Q(α). Since α is real, complex conjugation σ 1 Gal(Q(ζ 5 )/Q(α)). Thus the conjugates of ζ over Q(α) are ζ and ζ 1, and the irreducible polynomial they satisfy is (x ζ)(x ζ 1 ) = x αx + 1. Using the quadratic equation again, ζ = α ± α 4. Here α 4 < 0 so if we interpet the square root as ( 4 α ) i then we want the positive sign because ζ has positive imaginary part sin ( ) π 5. Thus ζ = α + α 4. Problem 14.5 #4. Let σ a Gal(Q(ζ n )/Q) denote hte automorphism of the cyclotomic field of n-th roots of unity such that σ a (ζ n ) = ζ a n. Show that σ a (ζ) = ζ a for every n-th root of unity ζ. Solution. For some m we have ζ = ζ m n. Therefore σ a (ζ) = σ a (ζ n ) m = ζ am n = (ζ m n ) a = ζ a. 7