Converse of Galois Theorem.

Question

If $F\subset K\subset E$ field extension such that $K\subset E$ and $F\subset K$ is both finite and Galois extensions then $F\subset E$ is Galois extension.

My intuition says that this is false but I cannot find a counterexample for this.

Answer 1

Indeed, it is false. The general intuition from this comes from the fundamental theorem of Galois theory. By the correspondence between Galois extensions and normal subgroups, this would suggest that normality is also transitive, which it isn't.

Here is an explicit construction using quadratic extensions. Letting $F = \mathbb{Q}$, we note that $K = \mathbb{Q}(\sqrt{2})$ is a quadratic extension of $F$, as the minimal polynomial over $\mathbb{Q}$ for $\sqrt{2}$ is $x^{2}-2$. Hence, as all quadratic extensions are Galois, we note that $K$ is Galois over $F$. Furthermore, let $E = K(\sqrt[4]{2})$. Then $E$ is a degree $2$ extension of $K$, since the minimal polynomial for $\sqrt[4]{2}$ over $K$ is $x^{2}-\sqrt{2}$, so this extension is again quadratic and therefore Galois over $K$.

However, $E = F(\sqrt{2}, \sqrt[4]{2}) = F(\sqrt[4]{2})$ is NOT Galois over $F$. The minimal polynomial for $\sqrt[4]{2}$ over $\mathbb{Q}$ is $x^{4}-2$, which has two imaginary roots!