Let \(F \subseteq K \subseteq L\) be finite-degree field extensions. For each of the following assertions, prove the statement or provide a counterexample with full justification:
If \(L/F\) is Galois, then \(K/F\) is Galois.
If \(L/F\) is Galois, then \(L/K\) is Galois.
If \(K/F\) and \(L/K\) are both Galois, then \(L/F\) is Galois.
Solution
Goal: Adjudicate the three transitivity/sub-extension assertions for Galois extensions.
1. Part (a): False.
1.1. Counterexample: Let \(F = \mathbb{Q}\), \(K = \mathbb{Q}(\sqrt[3]{2})\), and \(L = \mathbb{Q}(\sqrt[3]{2}, \omega)\), where \(\omega = e^{2\pi i / 3}\).
1.2. \(L/F\) is Galois: The polynomial \(f(x) = x^3 - 2 \in \mathbb{Q}[x]\) is irreducible by Eisenstein’s criterion at \(p = 2\). Its roots in \(\mathbb{C}\) are \(\sqrt[3]{2}, \sqrt[3]{2}\omega, \sqrt[3]{2}\omega^2\). The field \(L = \mathbb{Q}(\sqrt[3]{2}, \omega)\) is the splitting field of \(f(x)\) over \(\mathbb{Q}\). Since \(\operatorname{char}(\mathbb{Q}) = 0\), \(f(x)\) is separable, so \(L/\mathbb{Q}\) is a Galois extension.
1.3. \(K/F\) is not Galois: The intermediate field \(K = \mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}\) contains one root of the irreducible polynomial \(x^3 - 2\), but contains neither of the non-real roots \(\sqrt[3]{2}\omega, \sqrt[3]{2}\omega^2 \notin \mathbb{R}\).
1.4. Thus \(K/\mathbb{Q}\) is not a normal extension, hence is not Galois.
1.5. Therefore, \(L/F\) Galois does not imply \(K/F\) Galois.
2. Part (b): True.
2.1. Since \(L/F\) is a finite Galois extension, it is both normal and separable.
2.2. \(L/K\) is separable: Let \(\alpha \in L\). Since \(L/F\) is separable, the minimal polynomial \(m_{\alpha, F}(x) \in F[x]\) has distinct roots in an algebraic closure \(\overline{F}\). Since \(F \subseteq K\), the minimal polynomial \(m_{\alpha, K}(x) \in K[x]\) divides \(m_{\alpha, F}(x)\) in \(K[x]\). Any divisor of a separable polynomial is separable, so \(m_{\alpha, K}(x)\) has distinct roots. Thus every element of \(L\) is separable over \(K\).
2.3. \(L/K\) is normal: Since \(L/F\) is a finite normal extension, \(L\) is the splitting field of some polynomial \(g(x) \in F[x]\) over \(F\). Since \(F \subseteq K\), \(g(x) \in K[x]\), so \(L\) is also generated over \(K\) by the roots of \(g(x)\). Thus \(L\) is the splitting field of \(g(x)\) over \(K\), which implies \(L/K\) is normal.
2.4. Since \(L/K\) is finite, normal, and separable, \(L/K\) is a Galois extension.
3. Part (c): False.
3.1. Counterexample: Let \(F = \mathbb{Q}\), \(K = \mathbb{Q}(\sqrt{2})\), and \(L = \mathbb{Q}(\sqrt[4]{2})\).
3.2. \(K/F\) is Galois: \(K = \mathbb{Q}(\sqrt{2})\) is an extension of degree 2 over \(\mathbb{Q}\), which is the splitting field of \(x^2 - 2\) over \(\mathbb{Q}\), hence Galois.
3.3. \(L/K\) is Galois: \(L = K(\sqrt[4]{2})\) is obtained by adjoining a root of \(x^2 - \sqrt{2} \in K[x]\). Since \(\sqrt[4]{2} \notin K\) (as \([\mathbb{Q}(\sqrt[4]{2}) : \mathbb{Q}] = 4\) and \([K : \mathbb{Q}] = 2\)), \([L : K] = 2\). Every degree 2 extension in characteristic zero is Galois, so \(L/K\) is Galois.
3.4. \(L/F\) is not Galois: The polynomial \(x^4 - 2 \in \mathbb{Q}[x]\) is irreducible over \(\mathbb{Q}\) by Eisenstein at \(p = 2\). The field \(L = \mathbb{Q}(\sqrt[4]{2}) \subset \mathbb{R}\) contains the real root \(\sqrt[4]{2}\), but does not contain the complex roots \(\pm i \sqrt[4]{2} \notin \mathbb{R}\).
3.5. Thus \(L/\mathbb{Q}\) is not normal, and therefore is not Galois.
3.6. Hence, transitivity of Galois extensions fails in general.
4. Conclusion:
(a) is false, (b) is true, and (c) is false.