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ALGEBRA HW 5 CLAY SHONKWILER 510.5 Let F = Q(i). Prove that x3 − 2 and x3 − 3 are irreducible over F . Proof. If x3 − 2 is reducible over F then, since it is a polynomial of degree 3, it must reduce into a product with at least one linear factor. This, in turn, implies that at least one root of x3 − 2 must be in F . Now, if we let √ √ √ √ −3 1 ζ = 2 + 2 , then the roots of x3 − 2 are 3 2, 3 2ζ, 3 2ζ, where ζ is the complex √ conjugate of ζ. Now, elements of F can be written in the form a + b −1, where a, b ∈ Q. Clearly, then, none of the roots of x3 − 2 is in F , so x3 − 2 is irreducible over F . Similarly, if x3 − 3 is not irreducible over F , then one of its roots must lie √ √ √ 3 3 3 3 in F . However, the roots of x − 3 are 3, 3ζ, 3ζ, so, again, none of its roots lie in F , meaning x3 − 3 is also irreducible over F . 510.8 Let F be a field of characteristic 6= 2. Let D1 and 2 be elements of F , √ D√ neither of which is a square in F . Prove that F ( D1 , D2 ) is of degree 4 over F if √ D1 D2√is not a square in F and is of degree 2 over F otherwise. When F ( D1 , D2 ) is of degree 4 over F the field is called a biquadratic extension of F . √ √ √ √ Proof. D1 is of degree 2 over F , the degree of the extension F ( D1 , D2 )/F ( D1 ) 2 is at √ most 2 and is precisely 2 if and only if x − D2 is irreducible over F ( D1 ). Since this √ polynomial is of degree 2,√it is reducible √ if and only if it has a root √ on F ( D1 ), √ which is to say that D2 ∈ F ( D1 ). If this is the case, then D2 = a + b D1 for some a, b ∈ F . Squaring both sides of this expression, we get p D2 = (a2 + D1 b2 ) + 2ab D1 . √ If ab 6= 0, then we could solve for D1 in terms of elements of F ; since D1 is not a square, this is not possible. Hence, we see that either a = 0 or b = 0. If b = 0, then we see that D2 = a2 ; since D2 is not a square, this is impossible. On the other hand, if a = 0, then D2 = D1 b2 , 1 2 CLAY SHONKWILER meaning p p D2 = b2 D1 . √ Now, multiplying both sides by D1 , we see that this can only be the case if p p p D1 D2 = D1 D2 = b2 D1 ∈ F. If D1 D2 is not a square in F , then, since all √ of our implications go √ both√ways, we √see that x2 − D2 is irreducible over F ( D1 ), meaning [F ( D1 , D2 ) : F ( D1 )] = 2. In this case, p p p p p p [F ( D1 , D2 ) : F ] = [F ( D1 , D2 ) : F ( D1 )][F ( D1 ) : F ] = 2 · 2 = 4. √ √ √ On the other hand, if D1 D2 is a square in F , then the degree of F ( D1 , D2 )/F ( D1 ) is less than two; that is, the degree of this extension is 1. In this case, we see that p p p p p p [F ( D1 , D2 ) : F ] = [F ( D1 , D2 ) : F ( D1 )][F ( D1 ) : F ] = 1 · 2 = 2. 510.9 Let F be a field of characteristic 6= 2. Let a, b be elements of the field F withpb not a square in F . Prove that a necessary and sufficient condition √ √ √ for a + b = m + n for some m, np ∈ F is that a2 − b is a square in F . √ Use this to determine when the field Q( a + b) is biquadratic over Q. √ Proof. (⇒) Suppose a2 − b is a square in F . Then a2 − b ∈ F . Now, let √ √ a + a2 − b a − a2 − b m= n= . 2 2 Then n, m ∈ F since char(F ) 6= 2. Now, p p √ √ √ √ !2 √ √ a + a2 − b (a + b) + 2 a2 − b + (a − b) a+ b+ a− b m= = = , 2 4 2 meaning √ p m= a+ √ b+ 2 p a− √ b . On the other hand √ √ √ √ a − a2 − b (a + b) − 2 a2 − b + (a − b) n= = = 2 4 meaning √ p a+ √ b− 2 p a− p p √ √ a+ b− a− b n= . 2 Hence, √ m+ √ p p p p √ √ √ √ q √ a+ b+ a− b a+ b− a− b n= + = a + b. 2 2 √ !2 b , ALGEBRA HW 5 3 Now, with this proof in hand, we have the exact conditions under which p √ √ √ √ √ √ √ Q( a + b) = Q( m + n). Since Q( mp + n) = Q( m, n), we can √ use problem 510.8 above to conclude that Q( a + b) is biquadratic if and only if a2 − b is a square in F and √ √ a + a2 − b a − a2 − b b mn = = 2 2 4 p √ is not a square in F . Since b is not a square in F , we see that Q( a + b) is biquadratic precisely when a2 − b is a square in F . ♣ 510.13 Suppose F = Q(α1 , α2 , . . . , αn ) where αi2 ∈ Q for i = 1, 2, . . . , n. Prove √ 3 / F. that 2 ∈ Proof. Our basic tactic is to demonstrate that the degree of the extension √ F/Q is a power of 2 and then to demonstrate thatany field containing 3 2 must be of degree divisible by 3 over Q. To that end, first we prove the following lemma: Lemma 0.1. If F = Q(α1 , α2 , . . . , αn ) where αi2 ∈ Q for i = 1, . . . , n, then the degree of the extension F/Q is 2k for some k ∈ N. Proof. By induction. Suppose n = 1. Then F = Q(α1 ) for some α1 , where / Q, α12 ∈ Q. If, in fact, α1 ∈ Q, then F = Q and so [F : Q] = 1 = 20 . If α1 ∈ then mα1 ,Q (x) = x2 − α1 , so [F : Q] = 2 = 21 . Now, suppose [Q(α1 , . . . , αn ) : Q] = 2k for all n < j, j ≥ 2 where the αi are as above. Let F = Q(α1 , . . . , αj ). Since αj2 ∈ Q, αj is of degree 1 or 2 over Q, depending on whether or not αj ∈ Q. Hence, [F : Q(α1 , . . . , αj−1 )] ≤ 2. Since this leaves only two choices for [F : Q(α1 , . . . , αj−1 )], 1 and 2, we see that [F : Q] = [F : Q(α1 , . . . , αj−1 )][Q(α1 , . . . , αj−1 ) : Q] = [Q(α1 , . . . , αj−1 ) : Q] or [F : Q] = [F : Q(α1 , . . . , αj−1 )][Q(α1 , . . . , αj−1 ) : Q] = 2[Q(α1 , . . . , αj−1 ) : Q]. In either case, since [Q(α1 , . . . , αj−1 ) : Q] = 2k for some k ∈ N, we see that [F : Q] = 2m , where m = k or m = k + 1. 4 CLAY SHONKWILER √ √ Now, suppose 3 2 ∈ F . Since any field containing 3 2 must contain √ Q( 3 2), we see that the following tower of containments holds: F | √ Q( 3 2) | Q √ Now, since [Q( 3 2) : Q] = 3, we see that √ √ √ 3 3 3 [F : Q] = [F : Q( 2)][Q( 2) : Q] = 3[F : Q( 2)] meaning 3 divides [F : Q]. However, by the lemma, [F √ : Q] = 2k for some k ∈ N. From this contradiction, then, we conclude that 3 2 ∈ / F. 510.18 Let k be a field and let k(x) be the field of rational functions in x with P (x) coefficients from k. Let t ∈ k(x) be the rational function Q(x) with relatively prime polynomials P (x), Q(x) ∈ k[x], with Q(x) 6= 0. Then k(x) is an extension of k(t) and to compute its degree it is necessary to compute the minimal polynomial with coefficients in k(t) satisfied by x. (a) Show that the polynomial P (X) − tQ(X) in the variable X and coefficients in k(t) is irreducible over k(t) and has x as a root. Proof. P (X) − tQ(X) certainly has x as a root, since P (x) − tQ(x) = P (x) − P (x) Q(x) = P (x) − P (x) = 0. Q(x) Now, k(t) is the field of fractions of the UFD k[t], so, by Gauss’ Lemma, P (X) − tQ(X) is irreducible in (k(t))[X] if and only if it is irreducible over (k[t])[X]. Now, (k[t])[X] = (k[X])[t], so we can check for irreducibility in (k[X])[t]. Since P (X) − tQ(X) is linear in (k[X])[t] it is certainly irreducible in (k[X])[t] and, hence, in (k(t))[X]. Therefore, we conclude that P (X) − tQ(X) is irreducible over k(t). (b) Show that the degree of P (X) − tQ(X) as a polynomial in X with coefficients in k(t) is the maximum of the degrees of P (x) and Q(x). Proof. Let n = max{deg(P (x)), deg(Q(x))}. Then P (x) = pn xn + . . . + p1 x + p0 and Q(x) = qn xn + . . . + q1 x + q0 for pi , qj ∈ k for i = 0, . . . , n where at least one of pn , qn is non-zero. Now, certainly deg(P (X) − tQ(X)) ≤ n; we need only show that the coefficient on ALGEBRA HW 5 5 X n is non-zero. However, the coefficient on X n in the polynomial P (X) − tQ(X) is pn − tqn . Now, pn , qn ∈ k and it is not the case that both are zero. Furthermore, t ∈ k(x) but t ∈ / k so it cannot be the case that pn = tqn . Hence, pn − tqn 6= 0. Therefore, we see that deg(P (X) − tQ(X)) ≥ n. Since we’ve shown inequalities in both directions, we conclude that deg(P (X) − tQ(X)) = n = max{deg(P (x)), deg(Q(x))}. P (x) (c) Show that [k(x) : k(t)] = [k(x) : k( Q(x) )] = max(deg P (x), deg Q(x)). Proof. In part (a) above we demonstrated that P (X) − tQ(X) is irreducible over k(t) and has x as a root, so it must be the case that P (X) − tQ(X) is the minimal polynomial of x over k(t). Hence, k(x) ' (k(t)[X])/(P (X) − tQ(X)) and, using our result from part (b), we see that [k(x) : k(t)] = deg(P (X) − tQ(X)) = max{deg(P (x)), deg(Q(x))}. 510.22 Let K1 and K2 be two finite extensions of a field F contained in the field K. Prove that the F -algebra K1 ⊗F K2 is a field if and only if [K1 K2 : F ] = [K1 : F ][K2 : F ]. Proof. (⇐) Suppose K1 ⊗F K2 is a field. Then the dimension of K1 ⊗F K2 as a free module over F is equal to the degree of the extension K1 ⊗F K2 /F . However, we know the dimension of K1 ⊗F K2 as a free module: it is the product of the dimensions of the Ki as F -modules for i = 1, 2. By the definition of the degree of an extension, we know that the dimension of Ki as an F -module (i.e. an F -vector space) is simply [Ki : F ]. Hence, if K1 ⊗F K2 is a field, then [K1 ⊗F K2 : F ] = [K1 : F ][K2 : F ]. Now, define φ : K1 × K2 → K1 K2 , where φ(a, b) = ab. Then, if a, a1 , a2 ∈ K1 , b, b1 , b2 ∈ K2 and r ∈ F , φ((a1 , b)+(a2 , b)) = φ(a1 +a2 , b) = (a1 +a2 )b = a1 b+a2 b = φ(a1 , b)+φ(a2 , b), φ((a, b1 )+(a, b2 )) = φ(a, b1 +b2 ) = a(b1 +b2 ) = ab1 +ab2 = φ(a, b1 )+φ(a, b2 ) and φ(ar, b) = (ar)b = a(rb) = φ(a, rb) 6 CLAY SHONKWILER since r ∈ F ⊆ K1 ∩ K2 . Hence, φ is bilinear, so, by the universal property of the tensor product, it induces a module homomorphism Φ : K1 ⊗F K2 → K1 K2 . Now, we know K1 ⊗F K2 is a field, so we must check that this homomorphism preserves products. Let a ⊗ b, c ⊗ d ∈ K1 ⊗F K2 . Then Φ((a ⊗ b)(c ⊗ d)) = Φ(ac ⊗ bd) = acbd = abcd = Φ(a ⊗ b)φ(c ⊗ d). Hence, Φ is a field homomorphism. Since field homomorphisms are necessarily injective, we need only check that Φ is surjective to confirm that it is a field isomorphism. However, this is clear: if ab ∈ K1 K2 , where a ∈ K1 , b ∈ K2 , then a ⊗ b ∈ K1 ⊗F K2 and Φ(a ⊗ b) = ab. Therefore, we conclude that [K1 K2 : F ] = [K1 ⊗F K2 : F ] = [K1 : F ][K2 : F ]. (⇒) On the other hand, suppose [K1 K2 : F ] = [K1 : F ][K2 : F ]. Then, since [K1 K2 : F ] = [K1 K2 : K1 ][K1 : F ], we see that [K1 K2 : K1 ] = [K2 : F ]. That is to say, for any a ∈ K2 , either a ∈ F or a ∈ / K1 . Now, we know that K1 ⊗F K2 is an F -algebra. Furthermore, just as above, we can construct the F -algebra homomorphism Φ : K1 ⊗F K2 → K1 K2 . Now, since K1 ∩ K2 = F , every element in K1 K2 can be written uniquely as either a product of an element in K1 with an element of K2 or as an element of F . Now, we define Ψ : K1 K2 → K1 ⊗F K2 in the following way. Let a ∈ K1 K2 . Then either a = bc for b ∈ K1 and c ∈ K2 or a ∈ F . If a ∈ / F , define Ψ(a) = Ψ(bc) = b ⊗ c. If a ∈ F , then define Ψ(a) = a ⊗ 1. Then Ψ is certainly an F -algebra homomorphism; we want to demonstrate that it is the inverser of Φ. To that end, let ab ∈ K1 K2 . Then a ⊗ b ab ∈ /F Φ ◦ Ψ(ab) = Φ(Ψ(ab)) = Φ = ab. ab ⊗ 1 ab ∈ F On the other hand, let a ⊗ b ∈ K1 ⊗F K2 . Then a ⊗ b ab ∈ /F Ψ ◦ Φ(a ⊗ b) = Ψ(Φ(a ⊗ b)) = Ψ(ab) = ab ⊗ 1 ab ∈ F. Now, if ab ∈ / F , then we have nothing to worry about. If ab ∈ F and b ∈ F , then ab ⊗ 1 = a ⊗ b, so again we have no problems. However, it cannot be the case that ab ∈ F and b ∈ / F , since K1 ∩ K2 = F . Therefore, we conclude that Φ ◦ Ψ ≡ Id and Ψ ◦ Φ ≡ Id, ALGEBRA HW 5 7 meaning that Ψ is an F -algebra isomorphism. Since K1 K2 is a field, this implies that K1 ⊗F K2 is a field as well. 525.4 Determine the splitting field and its degree over Q for x6 − 4. Answer: First, note that x6 − 4 = (x3 − 2)(x3 + 2). Hence, the splitting field K for x6 − 4 must contain the splitting fields for x3 − 2 and x3 + 2. Now, using the same notation as in problem 510.5 √ √ √ 3 3 3 3 above, the roots of x − 2 are 2, 2ζ, 2ζ and the roots of x3 + 2 are √ √ √ 3 3 3 −2, −2ζ, −2ζ. It is clear that √ √ 3 3 −2 = − 2, so we see that the roots of x2 +2 are contained in the splitting field of x3 −2. Hence, all of the roots of x6 − 4 are contained in the splitting field of x3 − 2, meaning K is contained in the splitting field of x3 − 2. Since we already knew that K must contain the splitting field of x3 − 2, we see that, in fact, K is equal to the splitting field of x3 − 2. Now, as we’ve seen in class, the √ 3 3 splitting field of x √− 2 is simply Q( 2, ζ), which is of degree 6 over Q. Therefore, K = Q( 3 2, ζ) and [K : Q]6. ♣ 536.10 Let φ denote the Frobenius map x 7→ xp on the finite field Fpn . Prove that φ gives an isomorphism of Fpn to itself. Prove that φn is the identity map and that no lower power of φ is the identity. Proof. Since Fpn is a field of characteristic p, we know that the Frobenius map is necessarily an injective homomorphism. However, since Fpn is finite, any injective endomorphism must also be surjective, so we see that φ gives an automorphism of Fpn . Now, to see that φn is the identity map, let a ∈ Fpn . If a = 0, then φn (a) = φn (0) = 0. × n If a 6= 0, then a ∈ F× pn . Now, Fpn is an abelian group of order p −1, meaning n p −1 = 1. Therefore, that for all b ∈ F× pn , b n n −1 φn (a) = ap = aap = a(1) = a. Therefore, we see that φn is the identity. Now, suppose there exists some k < n such that φk is also the identity on k Fpn . Then, for all b ∈ Fpn , φk (b) = bp = b, which implies that each element k k of Fpn is a root of the polynomial xp − x over Fp . However, xp − k has only pk distinct roots; since k < n and |Fpn | = pn , this is clearly impossible. 8 CLAY SHONKWILER From this contradiction, then, we conclude that there is no such k, so no lower power of φ is the identity. DRL 3E3A, University of Pennsylvania E-mail address: [email protected]