Mathematics 352 – Abstract Algebra II Solutions for Problem Set 9 April 7, 2008

Section 11.2

8. First, Gal_{E}K is a subgroup of the abelian group Gal_{F} K, so it is an abelian group. Second, since E is normal over F, by Theorem 11.11 (the Fundamental Theorem of Galois Theory), we have

Gal_{F} E ≃ Gal_{F} K/Gal_{E}K.

Any time G is an abelian group and H is a normal subgroup, the quotient group G/H is abelian. (Proof: For all a, b ∈ G,

(Ha)(Hb) = H(ab) (definition of coset product)
= H(ba) (since ab = ba in G)
= (Hb)(Ha).)

Therefore, Gal_{F} E is also abelian.

9. (a) By the result of Exercise 8, we know that Gal_{E}K and Gal_{F} E are both abelian. Now, we use some standard facts about cyclic groups. First, every subgroup of a cyclic group is cyclic (see Theorem 7.16 in Hungerford, or your text from Algebraic Structures). Then every quotient group G/H of a cyclic group is cyclic. This follows since if a is a generator of G, then every b ∈ G satisfies b = a^{k} for some k ∈ ℤ. But then Hb = Ha^{k} = (Ha)^{k}, which shows that G/H is cyclic, generated by Ha.

(b) Since we assume K is Galois over F, the Galois correspondence is a one-to-one correspondence by Theorem 11.11. In the cyclic group G = Gal_{F} K ≃ ℤ_{n}, there is exactly one subgroup of order d for each divisor d of n. (If a generates G, then this subgroup is generated by a^{n/d}; see Exercise 40 in Section 7.3 of Hungerford.) Hence there is one intermediate field for each divisor d of n, and these are the only intermediate fields.

11. (a) The roots of x^{4} − 2 = 0 are

x = \sqrt[4]{2}, i\sqrt[4]{2}, −\sqrt[4]{2}, −i\sqrt[4]{2}.

Hence the splitting field over ℚ is

K = ℚ(\sqrt[4]{2}, i\sqrt[4]{2}, −\sqrt[4]{2}, −i\sqrt[4]{2}).

We claim that K = ℚ(\sqrt[4]{2}, i). First, it is clear that K ⊆ ℚ(\sqrt[4]{2}, i), since all the roots of x^{4} − 2 = 0 are contained in this field. On the other hand ℚ(\sqrt[4]{2}, i) ⊆ K since \sqrt[4]{2} ∈ K and i = \frac{i \sqrt[4]{2}}{\sqrt[4]{2}} ∈ K. Therefore the splitting field is equal to ℚ(\sqrt[4]{2}, i).

(b) From the tower of field extensions

ℚ ⊂ ℚ(\sqrt[4]{2}) ⊂ ℚ(\sqrt[4]{2}, i),

we see that

[ℚ(\sqrt[4]{2}, i) : ℚ] = [ℚ(\sqrt[4]{2})(i) : ℚ(\sqrt[4]{2})][ℚ(\sqrt[4]{2}) : ℚ].

The second factor on the right is 4 by Theorem 10.7, since the minimal polynomial of \sqrt[4]{2} over ℚ is the irreducible polynomial x^{4} − 2 ∈ ℚ[x]. The first factor on the right is 2 by Theorem 10.7, since the polynomial x^{2} + 1 has i as a root and it is still irreducible in ℚ(\sqrt[4]{2})[x]. (This follows since x^{2} + 1 has no roots in ℚ(\sqrt[4]{2}) ⊂ ℝ.) Hence [ℚ(\sqrt[4]{2}, i) : ℚ] = 8.

(c) Consider the extension K/ℚ(i). K is still the splitting field of x^{4} − 2 over ℚ(i). The polynomial x^{4} − 2 is still irreducible in ℚ(i)[x] since there are no roots in ℚ(i) and there are no polynomials of degree 2 with coefficients in ℚ(i) that divide x^{4} − 2. (In other words there are no pairs r, s of roots of x^{4} − 2 such that (x − r)(x − s) ∈ ℚ(i)[x]. So by Theorem 11.3, there is a ℚ(i)-automorphism σ of K that satisfies σ(\sqrt[4]{2}) = i\sqrt[4]{2}. Since σ is a ℚ(i)-automorphism, it satisfies σ(i) = i. But σ(a) = a for all a ∈ ℚ as well, so σ defines an element of Gal_{ℚ}K.

(d) Using Corollary 11.13, the restriction of the complex conjugation mapping a + ib ↦ a − ib is also an element τ of Gal_{ℚ}K. Now by part (1) of Theorem 11.11, |Gal_{ℚ}K| = [K : ℚ] = 8. Consider the action of the following combinations of σ and τ on the generators of K. We have

id : \sqrt[4]{2} ↦ \sqrt[4]{2}
i ↦ i
σ : \sqrt[4]{2} ↦ i\sqrt[4]{2}
i ↦ i
σ^{2} : \sqrt[4]{2} ↦ −\sqrt[4]{2}
i ↦ i
σ^{3} : \sqrt[4]{2} ↦ −i\sqrt[4]{2}
i ↦ i
τ : \sqrt[4]{2} ↦ \sqrt[4]{2}
i ↦ −i
στ : \sqrt[4]{2} ↦ i\sqrt[4]{2}
i ↦ −i
σ^{2}τ : \sqrt[4]{2} ↦ −\sqrt[4]{2}
i ↦ −i
σ^{3}τ : \sqrt[4]{2} ↦ −i\sqrt[4]{2}
i ↦ −i.

These give 8 distinct mappings on K by Theorem 11.4. Hence this shows

Gal_{ℚ}K = {id, σ, σ^{2}, σ^{3}, τ, στ, σ^{2}τ, σ^{3}τ}.

(e) We claim Gal_{ℚ}K ≃ D_{4}, the dihedral group of order 8. To see this note that the generators σ and τ satisfy σ^{4} = id, τ^{2} = id and

τσ : \sqrt[4]{2} ↦ −i\sqrt[4]{2}
i ↦ −i.

Hence τσ = σ^{3}τ = σ^{−1}τ. This shows that the generators satisfy the same relations as the generators of D_{4}, and both groups have order 8. It follows that Gal_{ℚ}K ≃ D_{4}. (Another way to say this is that, viewing D_{4} as the group of symmetries of the square in ℝ^{2}, the mapping

φ : Gal_{ℚ}K → D_{4}
σ ↦ r_{1} (the \frac{π}{2} rotation)
τ ↦ v (one of the reflections)

is an isomorphism of groups – it is a one-to-one and onto group homomorphism.)

12. By Theorem 11.11, Gal_{ℚ(i)}K is the subgroup of Gal_{ℚ}K fixing all elements of ℚ(i). By examining the list in 11 (d), we see that this subgroup is H = {id, σ, σ^{2}, σ^{3}}, which is cyclic of order 4. The other four elements map i to −i so they are not elements of Gal_{ℚ(i)}K. Therefore Gal_{ℚ(i)}K ≃ ℤ_{4}.

13. The Galois correspondence is given by the following diagrams of subgroups and intermediate fields. D_{4} contains 5 elements of order 2, so there are 5 subgroups of order 2. There are also 3 subgroups of order 4, one cyclic (namely, ⟨σ⟩), and the other two isomorphic to ℤ_{2} × ℤ_{2}. Here are the subgroups, with arrows showing the inclusions:

\xymatrix{ &  & \{id\} \ar[dl] \ar[dll] \ar[d] \ar[dr] \ar[drr] \\ ⟨τ⟩ \ar[dr] & ⟨σ^{2}τ⟩ \ar[d] & ⟨σ^{2}⟩ \ar[d] \ar[dl] \ar[dr] & ⟨στ⟩ \ar[d] & ⟨σ^{3}τ⟩ \ar[dl] \\  & \{id, τ, σ^{2}, σ^{2}τ\} \ar[dr] & ⟨σ⟩ \ar[d] & \{id, στ, σ^{2}, σ^{3}τ\} \ar[dl] \\  &  & Gal_{ℚ}K &  & .}

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[4]{2}, i) : α(u) = u for all α in H\}.

For instance, the fixed field of the subgroup ⟨στ⟩ would be determined like this: If a = a_{0} + a_{1} \sqrt[4]{2} + a_{2}\sqrt{2} + a_{3} \sqrt[4]{8} + a_{4}i + a_{5}i\sqrt[4]{2} + a_{6}i\sqrt{2} + a_{7}i\sqrt[4]{8}, satisfies σ(τ(a)) = a, then a_{5} = a_{1}, a_{2} = 0, a_{3} = −a_{7}, and a_{4} = 0. a_{0}, a_{6}, a_{1}, a_{3} are arbitrary and the general element of the fixed field looks like

a_{0} + a_{1}(1 + i)\sqrt[4]{2} + a_{6}i\sqrt{2} + a_{3}(1 − i)\sqrt[4]{8}.

We claim this is a general element of the field ℚ((1 + i)\sqrt[4]{2}). To see this note that if we write β = (1 + i)\sqrt[4]{2}, then

β^{2} = −2i\sqrt{2}
β^{3} = 2(1 − i)\sqrt[4]{8}
β^{4} = −8 ∈ ℚ.

This shows that the fixed field of H = ⟨τ⟩ is E_{H} = ℚ((1 − i)\sqrt[4]{2}) as claimed. We also see that ℚ(i\sqrt{2}) will be contained in ℚ(β) and ℚ(γ) = ℚ((1 − i)\sqrt[4]{2}) from this computation. (Note that β^{2} and γ^{2} are rational multiples of i\sqrt{2}.) That is the fixed field of the subgroup H = \{id, στ, σ^{2}, σ^{3}τ\}.

\xymatrix{ &  & ℚ(\sqrt[4]{2}, i) \\ ℚ(\sqrt[4]{2}) \ar[urr] & ℚ(i\sqrt[4]{2}) \ar[ur] & ℚ(\sqrt{2}, i) \ar[u] & ℚ((1 + i)\sqrt[4]{2}) \ar[ul] & ℚ((1 − i)\sqrt[4]{2}) \ar[ull] \\  & ℚ(\sqrt{2}) \ar[ur] \ar[u] \ar[ul] & ℚ(i) \ar[u] & ℚ(i\sqrt{2}) \ar[ul] \ar[u] \ar[ur] \\  &  & ℚ \ar[ul] \ar[u] \ar[ur]}

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