# ALGEBRA HW 5 CLAY SHONKWILER

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1 ALGEBRA HW 5 CLAY SHONKWILER Let F = Q(i). Prove that x 3 and x 3 3 are irreducible over F. Proof. If x 3 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 x 3 must be in F. Now, if we let ζ = 1 + 3, then the roots of x3 are 3, 3 ζ, 3 ζ, 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 x 3 is in F, so x 3 is irreducible over F. Similarly, if x 3 3 is not irreducible over F, then one of its roots must lie in F. However, the roots of x 3 3 are 3 3, 3 3ζ, 3 3ζ, so, again, none of its roots lie in F, meaning x 3 3 is also irreducible over F Let F be a field of characteristic. Let D 1 and D be elements of F, neither of which is a square in F. Prove that F ( D 1, D ) is of degree 4 over F if D 1 D is not a square in F and is of degree over F otherwise. When F ( D 1, D ) is of degree 4 over F the field is called a biquadratic extension of F. Proof. D 1 is of degree over F, the degree of the extension F ( D 1, D )/F ( D 1 ) is at most and is precisely if and only if x D is irreducible over F ( D 1 ). Since this polynomial is of degree, it is reducible if and only if it has a root on F ( D 1 ), which is to say that D F ( D 1 ). If this is the case, then D = a + b D 1 for some a, b F. Squaring both sides of this expression, we get D = (a + D 1 b ) + ab D 1. If ab 0, then we could solve for D 1 in terms of elements of F ; since D 1 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 D = a ; since D is not a square, this is impossible. On the other hand, if a = 0, then D = D 1 b, 1

2 CLAY SHONKWILER meaning D = b D 1. Now, multiplying both sides by D 1, we see that this can only be the case if D1 D = D 1 D = b D 1 F. If D 1 D is not a square in F, then, since all of our implications go both ways, we see that x D is irreducible over F ( D 1 ), meaning [F ( D 1, D ) : F ( D 1 )] =. In this case, [F ( D 1, D ) : F ] = [F ( D 1, D ) : F ( D 1 )][F ( D 1 ) : F ] = = 4. On the other hand, if D 1 D is a square in F, then the degree of F ( D 1, D )/F ( D 1 ) is less than two; that is, the degree of this extension is 1. In this case, we see that [F ( D 1, D ) : F ] = [F ( D 1, D ) : F ( D 1 )][F ( D 1 ) : F ] = 1 = Let F be a field of characteristic. Let a, b be elements of the field F with b not a square in F. Prove that a necessary and sufficient condition for a + b = m + n for some m, n F is that a b is a square in F. Use this to determine when the field Q( a + b) is biquadratic over Q. Proof. ( ) Suppose a b is a square in F. Then a b F. Now, let m = a + a b n = a a b. Then n, m F since char(f ). Now, m = a + a b = (a + b) + a b + (a ( b) a + b + a ) b =, 4 meaning On the other hand n = a a b meaning Hence, a + b + m + n = a + b + a b m =. = (a + b) a b + (a b) 4 a + b a b n =. a b a + b + ( a + b a ) b =, a b = a + b.

3 ALGEBRA HW 5 3 Now, with this proof in hand, we have the exact conditions under which Q( a + b) = Q( m + n). Since Q( m + n) = Q( m, n), we can use problem above to conclude that Q( a + b) is biquadratic if and only if a b is a square in F and mn = a + a b a a b = b 4 is not a square in F. Since b is not a square in F, we see that Q( a + b) is biquadratic precisely when a b is a square in F Suppose F = Q(α 1, α,..., α n ) where α i Q for i = 1,,..., n. Prove that 3 / F. Proof. Our basic tactic is to demonstrate that the degree of the extension F/Q is a power of and then to demonstrate thatany field containing 3 must be of degree divisible by 3 over Q. To that end, first we prove the following lemma: Lemma 0.1. If F = Q(α 1, α,..., α n ) where αi Q for i = 1,..., n, then the degree of the extension F/Q is k for some k N. Proof. By induction. Suppose n = 1. Then F = Q(α 1 ) for some α 1, where α 1 Q. If, in fact, α 1 Q, then F = Q and so [F : Q] = 1 = 0. If α 1 / Q, then m α1,q(x) = x α 1, so [F : Q] = = 1. Now, suppose [Q(α 1,..., α n ) : Q] = k for all n < j, j where the α i are as above. Let F = Q(α 1,..., α j ). Since α j Q, α j is of degree 1 or over Q, depending on whether or not α j Q. Hence, [F : Q(α 1,..., α j 1 )]. Since this leaves only two choices for [F : Q(α 1,..., α j 1 )], 1 and, we see that [F : Q] = [F : Q(α 1,..., α j 1 )][Q(α 1,..., α j 1 ) : Q] = [Q(α 1,..., α j 1 ) : Q] or [F : Q] = [F : Q(α 1,..., α j 1 )][Q(α 1,..., α j 1 ) : Q] = [Q(α 1,..., α j 1 ) : Q]. In either case, since [Q(α 1,..., α j 1 ) : Q] = k for some k N, we see that [F : Q] = m, where m = k or m = k + 1.

4 4 CLAY SHONKWILER Now, suppose 3 F. Since any field containing 3 must contain Q( 3 ), we see that the following tower of containments holds: F Q( 3 ) Q Now, since [Q( 3 ) : Q] = 3, we see that [F : Q] = [F : Q( 3 )][Q( 3 ) : Q] = 3[F : Q( 3 )] meaning 3 divides [F : Q]. However, by the lemma, [F : Q] = k for some k N. From this contradiction, then, we conclude that 3 / F Let k be a field and let k(x) be the field of rational functions in x with coefficients from k. Let t k(x) be the rational function P (x) Q(x) with relatively prime polynomials P (x), Q(x) k[x], with Q(x) 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) = p n x n p 1 x + p 0 and Q(x) = q n x n q 1 x + q 0 for p i, q j k for i = 0,..., n where at least one of p n, q n is non-zero. Now, certainly deg(p (X) tq(x)) n; we need only show that the coefficient on

5 ALGEBRA HW 5 5 X n is non-zero. However, the coefficient on X n in the polynomial P (X) tq(x) is p n tq n. Now, p n, q n 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 p n = tq n. Hence, p n tq n 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))}. (c) Show that [k(x) : k(t)] = [k(x) : k( P (x) 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))} Let K 1 and K be two finite extensions of a field F contained in the field K. Prove that the F -algebra K 1 F K is a field if and only if [K 1 K : F ] = [K 1 : F ][K : F ]. Proof. ( ) Suppose K 1 F K is a field. Then the dimension of K 1 F K as a free module over F is equal to the degree of the extension K 1 F K /F. However, we know the dimension of K 1 F K as a free module: it is the product of the dimensions of the K i as F -modules for i = 1,. By the definition of the degree of an extension, we know that the dimension of K i as an F -module (i.e. an F -vector space) is simply [K i : F ]. Hence, if K 1 F K is a field, then [K 1 F K : F ] = [K 1 : F ][K : F ]. Now, define φ : K 1 K K 1 K, where φ(a, b) = ab. Then, if a, a 1, a K 1, b, b 1, b K and r F, φ((a 1, b)+(a, b)) = φ(a 1 +a, b) = (a 1 +a )b = a 1 b+a b = φ(a 1, b)+φ(a, b), φ((a, b 1 )+(a, b )) = φ(a, b 1 +b ) = a(b 1 +b ) = ab 1 +ab = φ(a, b 1 )+φ(a, b ) and φ(ar, b) = (ar)b = a(rb) = φ(a, rb)

6 6 CLAY SHONKWILER since r F K 1 K. Hence, φ is bilinear, so, by the universal property of the tensor product, it induces a module homomorphism Φ : K 1 F K K 1 K. Now, we know K 1 F K is a field, so we must check that this homomorphism preserves products. Let a b, c d K 1 F K. 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 K 1 K, where a K 1, b K, then a b K 1 F K and Therefore, we conclude that Φ(a b) = ab. [K 1 K : F ] = [K 1 F K : F ] = [K 1 : F ][K : F ]. ( ) On the other hand, suppose [K 1 K : F ] = [K 1 : F ][K : F ]. Then, since [K 1 K : F ] = [K 1 K : K 1 ][K 1 : F ], we see that [K 1 K : K 1 ] = [K : F ]. That is to say, for any a K, either a F or a / K 1. Now, we know that K 1 F K is an F -algebra. Furthermore, just as above, we can construct the F -algebra homomorphism Φ : K 1 F K K 1 K. Now, since K 1 K = F, every element in K 1 K can be written uniquely as either a product of an element in K 1 with an element of K or as an element of F. Now, we define Ψ : K 1 K K 1 F K in the following way. Let a K 1 K. Then either a = bc for b K 1 and c K or a F. If a / F, define If a F, then define Ψ(a) = Ψ(bc) = b c. 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 K 1 K. Then ({ }) a b ab / F Φ Ψ(ab) = Φ(Ψ(ab)) = Φ = ab. ab 1 ab F On the other hand, let a b K 1 F K. 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 K 1 K = F. Therefore, we conclude that Φ Ψ Id and Ψ Φ Id,

7 ALGEBRA HW 5 7 meaning that Ψ is an F -algebra isomorphism. Since K 1 K is a field, this implies that K 1 F K is a field as well Determine the splitting field and its degree over Q for x 6 4. Answer: First, note that x 6 4 = (x 3 )(x 3 + ). Hence, the splitting field K for x 6 4 must contain the splitting fields for x 3 and x 3 +. Now, using the same notation as in problem above, the roots of x 3 are 3, 3 ζ, 3 ζ and the roots of x 3 + are 3, 3 ζ, 3 ζ. It is clear that 3 = 3, so we see that the roots of x + are contained in the splitting field of x 3. Hence, all of the roots of x 6 4 are contained in the splitting field of x 3, meaning K is contained in the splitting field of x 3. Since we already knew that K must contain the splitting field of x 3, we see that, in fact, K is equal to the splitting field of x 3. Now, as we ve seen in class, the splitting field of x 3 is simply Q( 3, ζ), which is of degree 6 over Q. Therefore, K = Q( 3, ζ) and [K : Q] Let φ denote the Frobenius map x x p on the finite field F p n. Prove that φ gives an isomorphism of F p n to itself. Prove that φ n is the identity map and that no lower power of φ is the identity. Proof. Since F p n is a field of characteristic p, we know that the Frobenius map is necessarily an injective homomorphism. However, since F p n is finite, any injective endomorphism must also be surjective, so we see that φ gives an automorphism of F p n. Now, to see that φ n is the identity map, let a F p n. If a = 0, then φ n (a) = φ n (0) = 0. If a 0, then a F p n. Now, F p n is an abelian group of order pn 1, meaning that for all b F p n, bpn 1 = 1. Therefore, φ n (a) = a pn = aa pn 1 = a(1) = a. Therefore, we see that φ n is the identity. Now, suppose there exists some k < n such that φ k is also the identity on F p n. Then, for all b F p n, φ k (b) = b pk = b, which implies that each element of F p n is a root of the polynomial x pk x over F p. However, x pk k has only p k distinct roots; since k < n and F p n = p n, this is clearly impossible.

8 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 address:

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