M 380 D 511380 Final Exam

Name:

Do three out of the four questions below and please indicate here which questions you chose:

1. Let F be a field and σ_{1}, …, σ_{n} distinct automorphisms of F. Suppose that a_{1}, …, a_{n} are elements of F such that ∑ a_{i}σ_{i}(x) = 0 for all x ∈ F. Prove that a_{1} = ⋯ = a_{n} = 0.

[This is lemma (22.6) of Isaacs]

Induction on n. For n = 1, a_{1} = a_{1}σ_{1}(1) = 0. For n > 1, if ∑ a_{i}σ_{i}(x) = 0, x ∈ F and a ∈ F then 0 = ∑ a_{i}σ_{i}(ax) = ∑ a_{i}σ_{i}(x)σ_{i}(a). On the other hand ∑ a_{i}σ_{i}(x)σ_{1}(a) = 0. Subtracting the last two equations gives ∑ a_{i}σ_{i}(x)(σ_{i}(a) − σ_{1}(a)) = 0. This equation has n−1 terms so, by the induction hypothesis, either a_{i} = 0, for all i > 1 or σ_{i}(a)−σ_{1}(a) = 0 for some i > 1. In the first case we also get a_{1} = 0 and we are done. In the second case, since a was arbitrary, we get σ_{1} = σ_{i}, contrary to the hypothesis, and we are also done.

2. Let f(x) = x^{8} + 1.

a) Let K be the splitting field of f(x) over Q, the field of rational numbers. Determine the Galois group of K/Q.

b) How many subfields of K are of degree 4 over Q? How many of these are Galois over Q? Explain.

c) Let L be the splitting field of f(x) over GF (41), the field of 41 elements. Determine the Galois group of L/GF (41).

If ϵ is a root of f, then ϵ^{8} = −1, so ϵ^{16} = 1, so ϵ is a primitive 16-th root of unity, and conversely, every primitive 16-th root of unity is a root of f, thus f = Φ_{16}, the 16-th cyclotomic polynomial. So Gal(K/Q) is isomorphic to (Z/16Z)^{∗}, which is isomorphic to Z/2Z ⊕ Z/4Z. This solves a).

The subfields of K degree 4 over Q correspond, by the fundamental theorem of Galois theory, to the subgroups of index 4 of Gal(K/Q), that is, subgroups of order 2. Hence there are three such subgroups. Since Gal(K/Q) is abelian, every subgroup of it is normal, so every subfield of K is Galois over Q.

For c) note that 40 is divisible by 8 but not by 16 and we know that GF (41)^{∗} is cyclic of order 40. If ϵ is a root of f, then ϵ^{2} is an 8-th root of unity so is in GF (41), but ϵ is a primitive 16-th root of unity, so is not in GF (41). Therefore L/GF (41) is a quadratic extension, so its Galois group is Z/2Z.

3. Let f(x) be an irreducible polynomial in Q[x]. Let α ∈ C be a root of f(x). Let n be the degree of f(x) and r the number of roots of f(x) in Q(α).

a) Show that r is independent of the choice of a root of f(x).

b) Show that r divides n.

c) Show that if the Galois group of the splitting field of f(x) over Q is abelian then r = n.

d) Determine r if f(x) = x^{6} − 2.

We know that the Galois group of the splitting field F of f(x) over Q acts transitively on the roots of f(x). Let us denote this group by G. Let α = σ_{1}(α), σ_{2}(α), …, σ_{r}(α) be the roots of f in Q(α). If τ(α) is another root of f in C, then τ is an isomorphism between Q(α) and Q(τ(α)) so τ ∘ σ_{1}(α), τ ∘ σ_{2}(α), …, τ ∘ σ_{r}(α) are the the roots of f in Q(τ(α)), so r is independent of α.

Now, the set of τ in G for which Q(α) = Q(τ(α)) is a subgroup H of G which contains Gal(F/Q(α)). The image of α under σ ∈ H is a root of f in Q(α) and every root arises in this way. Also σ(α) = α if and only if σ ∈ Gal(F/Q(α)). Therefore r is the number of cosets of Gal(F/Q(α)) in H and thus divides the number of cosets of Gal(F/Q(α)) in G, which is equal to n, showing b).

If G is abelian, then any subfield of F is Galois, hence normal, over Q. So Q(α)/Q is normal and the irreducible polynomial f has a root in Q(α), so it splits in Q(α) and therefore Q(α) contains all n roots of f, that is, r = n.

Let α be the positive real sixth root of 2, then Q(α) is real. Now the roots of f in C are ζα, ζ^{6} = 1 and the only real sixth roots of unity are ±1, so the roots of f in Q(α) are ±α, so r = 2.

4.a) Let R be a Noetherian ring and R[[x]] the ring of power series with coefficients in R. Show that R[[x]] is Noetherian. (Hint: Mimic the proof of the Hilbert basis theorem using the least non-zero coefficient of a power series).

b) Show that if F is a field then F [[x]] is a principal ideal domain.

If α = ∑ a_{i}x^{i} ∈ R[[x]] is such that a_{i} = 0, i < n we will write α = a_{n}x^{n} + · · ·. Let I be an ideal of R[[x]]. For n ≥ 0 define I_{n} = {r ∈ R|∃α = rx^{n} + ⋯ ∈ I}. Then I_{n} is an ideal of R, since given α = rx^{n} + ⋯ ∈ I and β = sx^{n} + ⋯ ∈ I, we have α + β = (r + s)x^{n} + ⋯ ∈ I, so r + s ∈ I_{n} if r, s ∈ I_{n}. Likewise ar ∈ I_{n} if r ∈ I_{n} and a ∈ R. If α = rx^{n} + ⋯ ∈ I then xα = rx^{n+1} + ⋯ ∈ I so I_{n} ⊂ I_{n+1}. Since R is Noetherian, there exists m_{0} such that I_{m} ⊂ I_{m+1}, m ≥ m_{0}. Also I_{n} is finitely generated for all n. Let r_{n1}, …, r_{nk_{n}} be generators for I_{n} and choose α_{nj} = r_{nj}x^{n} + ⋯, for j = 1, …, k_{n}. We will prove that α_{nj}, n = 0, …, m_{0}, j = 1, …, k_{n} generate I. If α = rx^{n} + ⋯ ∈ I and n ≤ m_{0}, then r ∈ I_{n} so r = ∑ s_{nj}r_{nj}, s_{nj} ∈ R, so α − ∑ s_{nj}α_{nj} = r′x^{n+1} + ⋯ ∈ I. If α is as above but n > m_{0}, then r ∈ I_{n} = I_{m_{0}}, so r = ∑ s_{nj}r_{m_{0}j}, s_{nj} ∈ R and α′ = α−∑ s_{nj}x^{n−m_{0}}α_{m_{0}j} = r′x^{n+1}+⋯ ∈ I. Replacing α by α′ and continuing inductively we get α = ∑(∑ s_{nj}x^{n−m_{0}})α_{m_{0}j}, so I is generated by the α_{nj}.

If F is a field, using the same notation as above, we have either I = (0) or I_{n} = (0), n < m_{0}, I_{n} = F, n ≥ m_{0}, since the only ideals of F are (0) and F, so the above proof shows that α_{m_{0}1} = x^{m_{0}} + ⋯ generates I.