Does $E$ a finite field and $F\subset E$ imply that $E$ is Galois over $F$?

Question

Is this the case? I don't know whether to go fishing for a counterexample or to try to prove it.

Answer 1

Yes, it is the case. The reason why comes out of the construction of finite fields.

Since $E$ is finite, we have $E = \mathbb{F}_{p^n}$ for some prime $p$. It is a standard argument of abstract algebra that $E$ is then the splitting field of $x^{p^n}-x$ over $\mathbb{F}_p$. This is a separable polynomial, hence $E$ is Galois over $\mathbb{F}_p$.

Moreover, $\mathbb{F}_p \subset F \subset E$, so this implies that $E$ is Galois over $F$ as well.