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Math 545: Problem Set 6 Math 545: Problem Set 6 Due Wednesday, March 4 1. Suppose that K is an arbitrary field, and consider K(t), the rational function field in one variable over K. Show that if f(X) 2 K[t][X] is reducible over K(t), then in fact f(X) is reducible over K[t]. [Hint: Note that just as Q is the fraction field of Z, so is K(t) the fraction field of K[t]. Carefully adapt the proof given in class for the analogous statement concerning f(X) 2 Z[X]. You will need to define the notion of a primitive polynomial in this new context, and prove a version of Gauss's Lemma.] 2. (A Generalized Eisenstein Criterion) Let R be an integral domain, and p ⊂ R a prime ideal. Suppose that n n−1 p(X) = X + rn−1X + ··· + r1X + r0 2 R[X] 2 satisfies r0; r1; : : : ; rn−1 2 p, and r0 2= p (i.e. p(X) is Eisenstein for p). Prove that p(X) is irreducible in R[X]. 3. This problem is designed to convince you that the Primitive Element Theorem fails in characteristic p > 0. Let K = Fp(S; T ) be the field of rational functions in two variables S; T with coefficients in the finite field with p elements. a) Prove that Xp −S is irreducible in K[X]. [Hint: Use the previous two problems and the fact that K = Fp(T )(S).] Hence p p p Fp( S; T ) = K[X]=(X − S) pp is a field extension of K of degree p (here S denotes thep coset p p p X+(X −S)). Then show that Y −T is irreducible in Fp( S; T )[Y ]. Thus, p p p p p p p Fp( S; T ) = Fp( S; T )[Y ]=(Y − T ) pp is a field extension of Fp( S; T ) of degree p. p p p p 2 b) Setting L = Fp( S; T ), show [L : K] = p and that p f p SiT j j 0 ≤ i; j < pg is a basis for L as a K-vector space. 1 c) Show that L is not a simple extension of K by showing that degK (γ)jp for every γ 2 L. [Hint: If F is a field of characteristic p > 0 and a; b 2 F , compute (a+b)p using the binomial theorem.] d) Show that Xp −S has a single root of multiplicity p in L. That is, show that Xp − S factors as the pth power of a linear polynomial in L[X]. An irreducible polynomial with multiple rootsp is called pinseparable, and the fact that the minimal polynomials of p S and p T over K are inseparable is what causes the problem above. We are motivated to make the following definition: if K is an arbitrary field, then an irreducible polynomial p(X) 2 K[X] is separable if it has no multiple roots in any extension field of K. An algebraic field extension LjK is called separable if mα,K (X) is a separable polynomial for all α 2 L. A field K is called perfect if all irreducible polynomials in K[X] are separable. We have seen that characteristic zero fields are perfect. It turns out that finite fields are perfect as well. p 4. ([S] 33.9) Prove that if a is a positive constructible number, then a is constructible. (See [S] for a hint.) 5. ([S] 26.4) Let K be a subfield of C and N a finite normal field extension of K. Prove that NjK is a splitting field extension without using the fact that N is a simple extension of K. (See [S] for a hint.) 6. ([S] 26.5) Determine the Galois group of p(X) = X4 − 2X2 − 4 over Q. Prove that this group is isomorphic to D4, the dihedral group of order 8. Recall that D4 is the symmetry group of a square, and it has a presentation in terms of generators and relations as follows: 4 2 −1 −1 D4 = fx; y j x = e; y = e; yxy = x g ∼ Hence, to show that Gal(p) = D4, you just need to show that jGal(p)j = 8 and that there exist σ; τ 2 Gal(p) such that σ has order 4, τ has order 2, and τστ −1 = σ−1. 7. ([S] 26.12) Suppose F is a subfield of C and LjF is a splitting field extension with Galois group G. Suppose that K = F (α) is an inter- mediate field of LjF , and set H = Gal(LjK). If N is the splitting field T −1 of mα,F over F , prove that Gal(LjN) = g2G gHg . 2.
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