Mathematics 352 – Abstract Algebra II Solutions for Problem Set 8 March 31, 2008

Section 11.1

11. K = ℚ(\sqrt{2}, i) is the splitting field over ℚ of the polynomial f(x) = (x^{2} − 2)(x^{2} + 1). The same “two-step” argument done in class and in Example 2.A on page 374-375 of the text shows that there are elements of Gal_{ℚ}ℚ(\sqrt{2}, i) acting on the generators as follows:

id : \sqrt{2} ↦ \sqrt{2}
i ↦ i
α : \sqrt{2} ↦ −\sqrt{2}
i ↦ i
β : \sqrt{2} ↦ \sqrt{2}
i ↦ −i
γ : \sqrt{2} ↦ −\sqrt{2}
i ↦ −i

(Recall this works because for each action on \sqrt{2}, K is still the splitting field of x^{2} + 1 over ℚ(\sqrt{2}) so we can extend either action on \sqrt{2} to each possible action on i using Corollary 10.8 (the “Key Corollary” of Theorem 10.7). This yields a Galois group isomorphic to the Klein 4-group (ℤ_{2} × ℤ_{2}, +) since γ = β ∘ α and α, β, γ all have order 2.

12. The idea is similar to what happened above in Exercise 11 and Example 2.A. Applying a “three-step” argument, we have elements of Gal_{ℚ}ℚ(\sqrt{2}, \sqrt{3}, \sqrt{5}) as follows:

id : \sqrt{2} ↦ \sqrt{2}
\sqrt{3} ↦ \sqrt{3}
\sqrt{5} ↦ \sqrt{5}
α_{100} : \sqrt{2} ↦ −\sqrt{2}
\sqrt{3} ↦ \sqrt{3}
\sqrt{5} ↦ \sqrt{5}
α_{010} : \sqrt{2} ↦ \sqrt{2}
\sqrt{3} ↦ −\sqrt{3}
\sqrt{5} ↦ \sqrt{5}
α_{001} : \sqrt{2} ↦ \sqrt{2}
\sqrt{3} ↦ \sqrt{3}
\sqrt{5} ↦ −\sqrt{5}

α_{110} : \sqrt{2} ↦ −\sqrt{2}
\sqrt{3} ↦ −\sqrt{3}
\sqrt{5} ↦ \sqrt{5}
α_{101} : \sqrt{2} ↦ −\sqrt{2}
\sqrt{3} ↦ \sqrt{3}
\sqrt{5} ↦ −\sqrt{5}
α_{011} : \sqrt{2} ↦ \sqrt{2}
\sqrt{3} ↦ −\sqrt{3}
\sqrt{5} ↦ −\sqrt{5}
α_{111} : \sqrt{2} ↦ −\sqrt{2}
\sqrt{3} ↦ −\sqrt{3}
\sqrt{5} ↦ −\sqrt{5}

By examining the composition table for these ℚ-automorphisms, we see that they form an abelian group of order 8 in which every element other than id has order 2. By the structure theorem for finite abelian groups this means that Gal_{ℚ}ℚ(\sqrt{2}, \sqrt{3}, \sqrt{5}) ≃ ℤ_{2} × ℤ_{2} × ℤ_{2}.

15. (a) If σ : ℝ → ℝ is an automorphism, then for all x ∈ ℝ, σ(x^{2}) = σ(x · x) = σ(x) · σ(x) = (σ(x))^{2}. If x ≠ 0, then σ(x) ≠ 0 since σ is one-to-one. Therefore if a > 0, then we can write σ(a) = σ(x^{2}) = (σ(x))^{2} in ℝ. Therefore σ(a) is the square of a nonzero real number, hence is strictly positive. It follows that σ takes positive elements of ℝ to positive elements of ℝ.

(b) If a, b ∈ R and σ ∈ Gal_{ℚ}ℝ, suppose a < b. Then b − a > 0, so by part (a), we have σ(b − a) > 0. But σ(b − a) = σ(b) − σ(a) > 0 since σ is an automorphism of ℝ. It follows that σ(b) > σ(a), or equivalently σ(a) < σ(b). (In other words, σ is order-preserving.)

(c) Now suppose that c < r < d with c, d ∈ ℚ and r ∈ ℝ. If σ ∈ Gal_{ℚ}ℝ, then part (b) shows σ(c) < σ(r) < σ(d). But c, d ∈ ℚ, so σ(c) = c and σ(d) = d. Therefore c < σ(r) < d, and we have:

c < r < d ⇒ c < σ(r) < d  (1)

whenever c, d ∈ ℚ. We claim that this implies σ(r) = r for all r ∈ ℝ, and hence σ = id. The reason this is true is that given any ε > 0 and r ∈ R, there exist rational numbers d ∈ (r, r + ε) and c ∈ (r − ε, r). If we apply (1) with such c, d, then we see r − ε < σ(r) < r + ε. Since we can do this for all ε > 0, σ(r) = r.

16. When ζ satisfies the condition in this problem (that the powers of ζ are n distinct solutions of x^{n} − 1 = 0), we say that ζ is a primitive nth root of one. In the irreducible factorization of x^{n} − 1 in ℚ[x], suppose that ζ is a root of the irreducible polynomial p(x). Then by Theorem 11.3, if σ ∈ Gal_{ℚ}ℚ(ζ), then σ(ζ) is also a root of p(x). However this means that σ(ζ) = ζ^{k} for some k ∈ {1, …, n}. Now suppose τ ∈ Gal_{ℚ}ℚ(ζ) is another element of the Galois group. By the same reasoning τ(ζ) = ζ^{ℓ} for some ℓ ∈ {1, …, n}. But now consider the composition σ ∘ τ:

(σ ∘ τ)(ζ) = σ(τ(ζ)) = (ζ^{ℓ})^{k} = ζ^{k·ℓ} = (ζ^{k})^{ℓ} = τ(σ(ζ)) = (τ ∘ σ)(ζ).

By Theorem 11.4, this means that σ ∘τ = τ ∘σ on all of ℚ(ζ). Since σ, τ are two arbitrary elements, Gal_{ℚ}ℚ(ζ) is an abelian group.

Comment: Since σ(ζ) = ζ^{k} and τ(ζ) = ζ^{ℓ} must also be primitive nth roots of one when σ and τ are automorphisms of ℚ(ζ), the possible values for ℓ and k here are the elements of the set {1, 2, …, n} that are relatively prime to n. This observation shows that Gal_{ℚ}ℚ(ζ) ≃ ℤ_{n}^{∗}, the multiplicative group of units in the ring ℤ_{n}, which is of course an abelian group.

Section 11.2

7b. The Galois correspondence here is the correspondence between subgroups of the Galois group and subfields of ℚ(\sqrt{2}, i) containing ℚ. From Exercise 11 in Section 11.1 above, we have G = Gal_{ℚ}ℚ(\sqrt{2}, i) = \{id, α, β, γ\} ≃ ℤ_{2} × ℤ_{2}. Since each non-identity element of G has order 2, there are exactly 5 subgroups of G, including the trivial subgroups {id} and G itself. The following diagram gives the inclusions between these subgroups (with the direction of the arrows indicating the direction of inclusion):

\xymatrix{ & \{id\} \ar[dl] \ar[d] \ar[dr] \\ \{id, α\} \ar[dr] & \{id, β\} \ar[d] & \{id, γ\} \ar[dl] \\  & G & .}

Now we show the fixed fields of each of these subgroups in a parallel diagram. For each subgroup H above, the subfield in the corresponding location is the fixed field

E_{H} = \{u ∈ ℚ(\sqrt{2}, i) : σ(u) = u for all σ in H \}.

For instance, if H = \{id, α\}, then by definition, u = u_{0} + u_{1}\sqrt{2} + u_{2}i + u_{3}i\sqrt{2} (u_{i} ∈ ℚ) is in E_{H} if and only if α(u) = u_{0} − u_{1}\sqrt{2} + u_{2}i − u_{3}i\sqrt{2} = u. This implies u_{1} = u_{3} = 0 and u_{0}, u_{2} are arbitrary. Therefore

E_{H} = {u_{0} + u_{2}i : u_{0}, u_{2} ∈ ℚ} = ℚ(i).

Similarly, if H = \{id, β\}, then by definition, u = u_{0} + u_{1}\sqrt{2} + u_{2}i + u_{3}i\sqrt{2} (u_{i} ∈ ℚ) is in E_{H} if and only if β(u) = u_{0} + u_{1}\sqrt{2} − u_{2}i − u_{3}i\sqrt{2} = u. This implies u_{2} = u_{3} = 0 and u_{0}, u_{1} are arbitrary. Therefore

E_{H} = \{u_{0} + u_{1}\sqrt{2} : u_{0}, u_{1} ∈ ℚ\} = ℚ(\sqrt{2})

in this case. Finally if H = \{id, γ = α ∘ β\}, then by definition, u = u_{0} + u_{1}\sqrt{2} + u_{2}i + u_{3}i\sqrt{2} (u_{i} ∈ ℚ) is in E_{H} if and only if γ(u) = u_{0} − u_{1}\sqrt{2} − u_{2}i + u_{3}i\sqrt{2} = u. This implies u_{1} = u_{2} = 0 and u_{0}, u_{3} are arbitrary. Therefore

E_{H} = \{u_{0} + u_{3}i\sqrt{2} : u_{0}, u_{3} ∈ ℚ\} = ℚ(i\sqrt{2}).

Every element of ℚ(\sqrt{2}, i) is fixed by the identity subgroup. The only elements of ℚ(\sqrt{2}, i) fixed by all elements of G are the rational numbers. Hence we obtain the corresponding diagram of fixed fields and inclusions:

\xymatrix{ & ℚ(\sqrt{2}, i) \\ ℚ(i) \ar[ur] & ℚ(\sqrt{2}) \ar[u] & ℚ(i\sqrt{2}) \ar[ul] \\  & ℚ \ar[ul] \ar[u] \ar[ur]}

(Note that the arrows are all reversed from the first diagram!)