Tower of Galois Extensions [duplicate]

Let $ K/ L /F $ be fields. If $K / L$ is Galois and $ L / F $ is Galois, then $ K / F$ is Galois. We mentioned this very quickly in today's class without justifying. But I have trouble seeing this. Any help is appreciated.

Our definition for $ K / F$ being Galois is the $F$-automorphism group $G = G( K / F) : = \{ \sigma \in aut (G) : \sigma_F = id_F \}$ fixes only $F$, i.e. $ \{ x \in K : \sigma (x) = x \text { for all } \sigma \in G \} = F$.

But we also showed that $ K /F$ is Galois if and only if $ K/F$ is normal and separable.

Solution 1:

This is not true. E.g. $\mathbb Q(\sqrt{1+2i})$ is Galois over $\mathbb Q(i)$, which is Galois over $\mathbb Q$. But $\mathbb Q(\sqrt{1+2i})$ is not Galois over $\mathbb Q$.

Another example, maybe easier to check: $\mathbb Q(\sqrt{1+\sqrt{2}}) \supset \mathbb Q(\sqrt{2})\supset \mathbb Q$.

The Galois closure will of the first field over $\mathbb Q$ will contain $\sqrt{1-\sqrt{2}}$ as well; but this is not a real number, whereas $\mathbb Q(\sqrt{1+\sqrt{2}})$ consists entirely of real numbers.

Solution 2:

In the book Abstract Algebra, third edition, by David S. Dummit and Richard M. Foote, they mention the example $\mathbb{Q}\subset\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}(\sqrt[4]{2})$ as a pair of Galois extensions which are not transitive:

The field $\mathbb{Q}(\sqrt[4]{2})$ is not Galois over $\mathbb{Q}$ since any automorphism is determined by where it sends $\sqrt[4]{2}$ and of the four possibilities $\{\pm\sqrt[4]{2},\pm i\sqrt[4]{2}\}$, only two are elements of the field (the two real roots).