*Algebra*

## 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

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.

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

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

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.

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

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 \((x - i - \sqrt{2}) (x - i + \sqrt{2})(x + i - \sqrt{2})(x + i + \sqrt{2}) = x^4 - 2x^2 + 9\).

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

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

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.

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).

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.

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

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.

\(K\) 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.