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) ∈ 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) ∈ 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 ∈ R[X]

2 satisﬁes r0, r1, . . . , rn−1 ∈ p, and r0 ∈/ 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 Fp( S,T ) = K[X]/(X − S)

√p is a ﬁeld extension of K of degree p (here S denotes the√ coset p p p X+(X −S)). Then show that Y −T is irreducible in Fp( S,T )[Y ]. Thus, √ √ √ p p p p Fp( S, T ) = Fp( S,T )[Y ]/(Y − T ) √p is a ﬁeld extension of Fp( S,T ) of degree p. √ √ p p 2 b) Setting L = Fp( S, T ), show [L : K] = p and that √ { p SiT j | 0 ≤ i, j < p}

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)|p for every γ ∈ L. [Hint: If F is a ﬁeld of characteristic p > 0 and a, b ∈ 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 roots√ is called √inseparable, 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) ∈ K[X] is separable if it has no multiple roots in any extension field of K. An algebraic field extension L|K is called separable if mα,K (X) is a separable polynomial for all α ∈ 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. √ 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 N|K 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 = {x, y | x = e, y = e, yxy = x } ∼ Hence, to show that Gal(p) = D4, you just need to show that |Gal(p)| = 8 and that there exist σ, τ ∈ 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 L|F is a splitting field extension with Galois group G. Suppose that K = F (α) is an inter- mediate field of L|F , and set H = Gal(L|K). If N is the splitting field T −1 of mα,F over F , prove that Gal(L|N) = g∈G gHg .