Brian Bi
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## Section 16.7. The Main Theorem

Exercise 16.7.1 We assume that $$\sqrt{b} \notin F(\sqrt{a})$$ and $$\sqrt{a} \notin F(\sqrt{b})$$. This implies that $$[F(\sqrt{a}, \sqrt{b}) : F] = 4$$. Any intermediate fields must therefore be quadratic extensions of $$F$$. Clearly, $$F(\sqrt{a})$$, $$F(\sqrt{b})$$, and $$F(\sqrt{ab})$$ are three such intermediate fields. If there is any other intermediate field besides these three, it must be generated by an element of the form $$\zeta = p\sqrt{a} + q\sqrt{b} + r\sqrt{ab}$$, where $$p, q, r \in F$$. We have that $$\zeta^2 = (p^2 a + q^2 b + r^2 ab) + 2qrb\sqrt{a} + 2pra\sqrt{b} + 2pq\sqrt{ab}$$. In order for $$\zeta$$ to have degree 2, we need $$[p : q : r] = [2qrb : 2pra : 2pq]$$. If any of $$p, q, r$$ vanish, it is easy to see that this condition forces at least one other to vanish, so one of the three intermediate fields previously described is obtained. Otherwise, write $$p^2ra = q^2rb$$, implying $$p\sqrt{a} = \pm q\sqrt{b}$$. If $$p, q \ne 0$$, this implies that $$\sqrt{a} \in F(\sqrt{b})$$, contradiction. So there are no other intermediate fields.

Exercise 16.7.2

1. According to Corollary 16.7.2(c), intermediate fields with $$[L : F] = 4$$ correspond to subgroups of $$\Gal(K/F)$$ with index 4, that is, order 6. There are three such subgroups, namely $$\{0\} \times C_6, C_2 \times C_3$$, and $$\{(x, 2x) \mid x \in \{0, \ldots, 5\}\}$$. So there are three intermediate fields with the desired property.

2. None, because a group of order 24 can't have a subgroup of index 9.

3. By the fundamental theorem, intermediate fields $$L$$ with $$\Gal(K/L) \cong C_4$$ correspond to copies of $$C_4$$ in $$\Gal(K/F)$$. There are two such subgroups, namely the one generated by $$(0, 3)$$ and the one generated by $$(1, 3)$$. So there are two such intermediate fields.

Exercise 16.7.3

1. According to Corollary 16.7.2(c), intermediate fields with $$[L : F] = 2$$ correspond to subgroups of $$A_4$$ with index 2. No such subgroup exists, so there are no such intermediate fields.

2. The group $$D_4 = \langle x, y \mid x^4 = y^2 = 1, yx = x^{-1}y\rangle$$ has three subgroups of index 2, namely $$\{1, x, x^2, x^3\}, \{1, xy, x^2, x^3 y\}, \{1, y, x^2, x^2 y\}$$. Thus, there are three such intermediate fields.

Exercise 16.7.4 We argued in Exercise 16.6.2 that $$\Gal(\Q(\sqrt{2}, \sqrt{3}, \sqrt{5})/\Q) \cong C_3^2$$. The intermediate fields correspond to the subgroups of $$C_3^2$$. Besides the obvious trivial subgroup and $$C_3^2$$ itself, which, by the fundamental theorem, correspond to $$\Q(\sqrt{2}, \sqrt{3}, \sqrt{5})$$ and $$\Q$$, respectively, there should be 7 intermediate fields $$L$$ with $$[L : F] = 4$$, corresponding to the 7 subgroups of $$C_2^3$$ of index 4 (order 2), each generated by one of its nontrivial elements. These correspond to the 7 intermediate fields of degree 4 over $$\Q$$, namely $$\Q(\sqrt{2}, \sqrt{3})$$, $$\Q(\sqrt{2}, \sqrt{5})$$, $$\Q(\sqrt{3}, \sqrt{5})$$, $$\Q(\sqrt{2}, \sqrt{15})$$, $$\Q(\sqrt{3}, \sqrt{10})$$, $$\Q(\sqrt{5}, \sqrt{6})$$, and $$\Q(\sqrt{6}, \sqrt{10})$$. The subgroups of $$C_2^3$$ of index 2 (order 4) must all be isomorphic to $$V_4$$, and correspond to 2-dimensional subspaces of $$C_2^3$$ viewed as a vector space over $$\mathbb{F}_2$$, so there are 7 such subgroups, namely $$\{\{(x, y, z) \mid x, y, z \in \mathbb{F}_2, px + qy + rz = 0\} \mid (p, q, r) \in \mathbb{F}_2^3 \setminus \{0\}\}$$. These correspond to the 7 intermediate fields of degree 2 over $$\Q$$, namely $$\Q(\sqrt{2})$$, $$\Q(\sqrt{3})$$, $$\Q(\sqrt{5})$$, $$\Q(\sqrt{6})$$, $$\Q(\sqrt{10})$$, $$\Q(\sqrt{15})$$, $$\Q(\sqrt{30})$$.

Exercise 16.7.5 Let $$F$$ denote the splitting field of $$f$$, and let $$K$$ denote the splitting field of $$(x^3 - 1)f(x)$$. Let $$\alpha, \beta, \gamma$$ denote the roots of $$f$$ in $$F$$, and let $$\omega$$ denote a complex cube root of unity. It is known that $$[F : \Q] = |S_3| = 6$$. The possibilities are as follows:

• Case 1: $$\omega \in F$$. Then $$K = F$$, and $$\Gal(K/\Q) = \Gal(F/\Q) = S_3$$.

• Case 2: $$\omega \notin F$$. Then $$K = F(\omega)$$ and $$[K : F] = 2$$ since $$\omega$$ has degree 2 over $$\Q$$. Consequently $$[K : \Q] = 12$$, and $$\Gal(K/\Q)$$ is one of the five groups of order 12 determined in section 7.8. Since $$F$$ is an intermediate field and $$F/\Q$$ is a Galois extension, $$\Gal(K/\Q)$$ possesses a normal subgroup isomorphic to $$S_3$$, although since such a subgroup would be of index 2, it is automatically normal anyway. In any case, this immediately rules out the abelian groups $$C_{12}$$ and $$C_2 \times C_6$$. It is well-known that $$A_4$$ also doesn't have a subgroup isomorphic to $$S_3$$. Finally, $$\operatorname{Dic}_3$$ only has one element of order 2, so it can't contain $$S_3$$, which has three elements of order 2, as a subgroup. Therefore $$\Gal(K/\Q)$$ can only be $$D_6$$.

Exercise 16.7.6 Since $$S_3$$ has subgroups of index 3, there is an intermediate field $$L$$ with degree 3 over $$F$$. Since 3 is prime, it follows that $$L = F(\alpha)$$ where $$\alpha$$ is any element of $$L\setminus F$$. The minimal polynomial of $$\alpha$$ over $$F$$ is therefore cubic; call it $$f$$. As the subgroups of $$S_3$$ of index 3 are not normal, $$L/F$$ is not a Galois extension; therefore, $$f$$ does not split completely in $$L$$. But since $$K$$ is a splitting field over $$F$$ and $$f$$ a root in $$K$$, it follows that $$f$$ splits completely in $$K$$. The field generated over $$F$$ by the roots of $$f$$ is strictly larger than $$L$$, but since $$[K : L] = 2$$, this field can only be $$K$$. Therefore $$K$$ is the splitting field of the irreducible cubic $$f$$.

Exercise 16.7.7

1. Since $$\Q \subseteq \Q(i + \sqrt{2}) \subseteq \Q(i, \sqrt{2})$$, and $$[\Q(i, \sqrt{2}) : \Q] = 4$$, the degree of $$\Q(i + \sqrt{2})$$ over $$\Q$$ can only be 2 or 4. Our argument in the solution to Exercise 16.7.1 implies that $$\sqrt{-1} + \sqrt{2}$$ can't have degree 2 over $$\Q$$, so it must have degree 4. The minimal polynomial is therefore $$(x - i - \sqrt{2}) (x - i + \sqrt{2})(x + i - \sqrt{2})(x + i + \sqrt{2}) = x^4 - 2x^2 + 9$$.

2. $$[\Q(i, \sqrt{2}) : \Q] = 4$$ and clearly the given set spans $$\Q(i, \sqrt{2})$$ over $$\Q$$ and has cardinality 4, so it is also a basis.

Exercise 16.7.8

1. The polynomial $$x^4 - 2$$ is irreducible over $$\Q$$ by the Eisenstein criterion.

2. Obviously it factors as $$(x^2 - \sqrt{2})(x^2 + \sqrt{2})$$. Since $$\Q(\sqrt{2})$$ doesn't contain square roots of $$\pm\sqrt{2}$$, these factors are irreducible.

3. The factorization into irreducibles over $$\Q(i, \sqrt{2})$$ is $$(x^2 - \sqrt{2})(x^2 + \sqrt{2})$$ for the same reasons as in (b).

4. Over $$\Q(\sqrt[4]{2})$$ we can evidently take the factorization further than in (b), namely to $$(x - \sqrt[4]{2})(x + \sqrt[4]{2})(x^2 + \sqrt{2})$$. The third factor remains irreducible since $$-\sqrt{2}$$ has no square roots in any subfield of the reals.

5. Over $$\Q(\sqrt[4]{2}, i)$$ the polynomial $$x^4 - 2$$ splits completely as $$(x - \sqrt[4]{2})(x + \sqrt[4]{2})(x - i\sqrt[4]{2}) (x + i\sqrt[4]{2})$$.

Exercise 16.7.9 By construction $$\Q(\zeta)$$ can't be larger than the splitting field, and $$x^5 - 1$$ does split completely in $$\Q(\zeta)$$ as $$\prod_{i=0}^4 (x - \zeta^i)$$. Therefore $$\Q(\zeta)$$ is the splitting field for $$x^5 - 1$$ over $$\Q$$. The cyclotomic polynomial $$(x^5 - 1)/(x - 1) = x^4 + x^3 + x^2 + x + 1$$ is irreducible over $$\Q$$, so $$[K : \Q] = 4$$. Since $$K$$ is a splitting field over $$\Q$$, Theorem 16.6.4 implies that $$K/\Q$$ is a Galois extension. Lemma 16.6.3 implies that $$\Gal(K/\Q)$$ contains an automorphism $$\sigma$$ that sends $$\zeta$$ to $$\zeta^2$$. Evidently $$\sigma^2(\zeta) = \zeta^4 \ne \zeta$$, so $$\Gal(K/\Q)$$ is not the group $$V_4$$, but must be $$C_4$$.

Exercise 16.7.10 The fixed field $$K^H$$ is, by definition, left invariant by $$H$$, so $$H$$ is a subgroup of the stabilizer of every element of $$K^H$$. By Theorem 15.8.1, the extension $$K^H/F$$ has a primitive element $$\beta$$. Any $$F$$-automorphism of $$K$$ fixing $$\beta$$ would fix $$F(\beta) = K^H$$, so $$\beta$$ cannot be left invariant by any $$F$$-automorphism not in $$H$$; that is, the stabilizer of $$\beta$$ is precisely $$H$$.

Exercise 16.7.11

1. Since $$\Q \subseteq \Q(\gamma) \subseteq L$$, and $$[L : \Q] = 6$$, the degree of $$\gamma$$ over $$\Q$$ can only be 2, 3, or 6. Now $$\gamma^2 = 3 + \sqrt[3]{4} + 2\sqrt[3]{2}\sqrt{3}$$ and $$\gamma^3 = 2 + 3\sqrt[3]{4}\sqrt{3} + 9\sqrt[3]{2} + 3\sqrt{3}$$, so if any $$\Q$$-linear combination involving $$1, \gamma, \gamma^2, \gamma^3$$ vanishes, it must have zero coefficient for $$\gamma^3$$, which uniquely contains a $$\sqrt[3]{4}\sqrt{3}$$ factor, and $$\gamma^2$$, which uniquely contains a $$\sqrt[3]{4}$$ factor. Evidently this implies that the other two coefficients must be zero as well. So the degree of $$\gamma$$ over $$\Q$$ exceeds 3; it must be 6. Thus $$\gamma$$ is a primitive element for the extension $$\Q(\alpha, \beta) / \Q$$. By inspection, a polynomial with $$\gamma$$ as a root is \begin{equation*} f(x) = \prod_{j=0}^2 \prod_{k=0}^1 (x - e^{2\pi i j/3}\alpha - (-1)^k \beta) = x^6 - 9x^4 - 4x^3 + 27x^2 - 36x - 23 \end{equation*} Since this has degree 6 and rational coefficients, it is the minimal polynomial sought.

2. $$K$$ evidently contains $$L$$, but $$K$$ also contains the complex cube roots of 2, so $$K$$ is strictly larger than $$L$$; it contains the quadratic factor $$x^2 + \alpha x + \alpha^2$$ which remains irreducible over $$L$$. So $$[K : L] = 2$$, and $$[K : \Q] = 12$$. As in Exercise 16.7.5, we use the fact that there are only five possible isomorphism groups for $$\Gal(K/\Q)$$. Since there is an intermediate field $$\Q(\beta)$$ of degree 2 over $$\Q$$, it follows that $$\Gal(K/\Q)$$ contains a subgroup of index 2, which rules out $$A_4$$. Furthermore, the intermediate field $$L$$ corresponds to a subgroup of index 6 and order 2. But $$L$$ isn't a splitting field over $$\Q$$, since the irreducible polynomial $$f$$ has the root $$\gamma \in L$$ but doesn't split completely in $$L$$. So $$L/\Q$$ isn't a Galois extension and the subgroup corresponding to $$L$$ isn't normal in $$\Gal(K/\Q)$$. This rules out the two abelian groups $$C_3 \times C_4$$ and $$C_2 \times C_2 \times C_3$$, as well as $$\operatorname{Dic}_3$$, which has only one subgroup of order 2. Therefore $$\Gal(K/\Q) \cong D_6$$ is the only possibility.