M 380 D 511380 Midterm Name: Do three out of the four questions below and please indicate here which questions you chose: Notation: A^{∗} is the group of invertible elements of a ring A. Φ_{n} is the n-th cyclotomic polynomial. GF (q) is the field of q elements.

1. Let R be a domain and a, b ∈ R, (a, b) = R. Prove that, if c ∈ R is such that bc ∈ (a), then c ∈ (a).

Since (a, b) = R, there exists x, y ∈ R, ax + by = 1. Then c = acx + bcy and acx ∈ (a), clearly and bcy ∈ (a), by hypothesis. Thus c ∈ (a).

2. Show that Φ_{n} is irreducible over GF (q) if and only if q generates (Z/nZ)^{∗}. Show that Φ_{15} is never irreducible over GF (q).

If the characteristic of GF (q) divides n, then Φ_{n} is not separable and q is not prime to n, so this case is trivial. Assume then that the characteristic of GF (q) doesn’t divide n. Let E be the splitting field of Φ_{n} over GF (q). We know by general theory of cyclotomic extensions that Gal(E/GF (q)) is a subgroup of (Z/nZ)^{∗} via the homomorphism that takes σ ∈ Gal(E/GF (q)) to a ∈ (Z/nZ)^{∗}, where σ(ζ) = ζ^{a}, for ζ a primitive n-th root of unity. On the other hand, by the theory of finite fields, Gal(E/GF (q)) is generated by x ↦ x^{q}, which corresponds therefore to q ∈ (Z/nZ)^{∗}. Finally, Φ_{n} is irreducible over GF (q) if and only if [E : GF (q)] = deg Φ_{n} = ϕ(n) = |(Z/nZ)^{∗}|. Since, on the other hand [E : GF (q)] = | Gal(E/GF (q))| = | < q > |, the result follows.

For the second part just notice that (Z/15Z)^{∗} is not cyclic, so can’t be generated by any q.

3. Let E/F be a separable, normal, algebraic extension, not necessarily finite. Prove that Fix(Gal(E/F)) = F.

We have that F ⊂ Fix(Gal(E/F)), so we need to prove the reverse inclusion. Let α ∈ E, α ∉ F and let f(x) be the minimal polynomial of α over F (recall that E, thus α ∈ E is algebraic over F.). Since E/F is separable, f has distinct roots. Let K be the splitting field of f over F, then since E/F is normal, K ⊂ E. Now, K is the splitting field of a separable polynomial over F so K/F is Galois. Hence Fix(Gal(K/F)) = F, that is, there exists σ ∈ Gal(K/F) with σ(α) ≠ α. Since E/F is normal, we know that σ extends to an automorphism of E, that is, an element of Gal(E/F) and σ does not fix α, so α ∉ Fix(Gal(E/F)), completing the proof.

4. Let E/F be a Galois extension of fields and f(x) ∈ F [x] be an irreducible polynomial. Prove that Gal(E/F) acts transitively on the set of irreducible factors of f(x) in E[x].

Let g(x) be an irreducible factor of f(x) in E[x] and σ ∈ Gal(E/F). Then σ(g(x)) is also an irreducible factor of f(x) in E[x]. So, since both g(x) and σ(g(x)) are irreducible, they are either equal or coprime. let Γ be the orbit of g(x) under Gal(E/F) and h(x) the product of all elements of Γ. By the above, the elements of Γ are pairwise coprime factors of f(x), hence h(x)|f(x). On the other hand h(x) is invariant under Gal(E/F) so h(x) ∈ F [x]. Since f(x) is irreducible over F, this yields h(x) = f(x). So any other factor of f(x) in E[x] divides h(x) and is therefore in Γ, the orbit of g(x) under Gal(E/F), which completes the proof.