Let \(K \subset L \subset M\) be a tower of finite degree field extensions. In each of the following parts, either prove the assertion or give a counterexample (with justification).
If \(M/K\) is Galois, then \(L/K\) is Galois
If \(M/K\) is Galois, then \(M/L\) is Galois.
Solution
1. The assertion “if \(M/K\) is Galois, then \(L/K\) is Galois” is false.
Take \[ K=\QQ,\qquad L=\QQ(\sqrt[3]{2}),\qquad M=\QQ(\sqrt[3]{2},\zeta_3), \] where \(\zeta_3\) is a primitive cube root of unity. Then \(M\) is the splitting field of \(x^3-2\) over \(\QQ\), so \(M/\QQ\) is finite Galois.
However, \(L/\QQ\) is not normal: \(x^3-2\) is irreducible over \(\QQ\), has a root \(\sqrt[3]{2}\in L\), and its other two roots are nonreal, hence not in the real field \(L\). Thus \(L/\QQ\) is not Galois.
2. The assertion “if \(M/K\) is Galois, then \(M/L\) is Galois” is true.
Since \(M/K\) is finite Galois, it is separable and normal. Separability passes to the larger base field \(L\): the minimal polynomial over \(L\) of any element of \(M\) divides its separable minimal polynomial over \(K\).
For normality, every \(L\)-embedding \(\sigma:M\to\overline K\) is in particular a \(K\)-embedding. Normality of \(M/K\) forces \(\sigma(M)=M\). Hence \(M/L\) is normal.
Thus \(M/L\) is finite, separable, and normal, so it is Galois.