Construct the Galois closure of $\mathbb{Q} \subset \mathbb{Q}(\sqrt[4]2)$.
I think the closure is M = $\mathbb{Q}(\sqrt[4]2, i)$. This is Galois over $\mathbb{Q}$ because it is the splitting field of the separable polynomial $x^4 - 2$. But how to show that $M$ is the smallest Galois extension?
The key is that every irreducible polynomial with one root in a Galois extension splits in that extension.
So, let $K$ be any Galois extension containing a 4th root of $2$. Then $K$ has (by definition) a root of $x^4 - 2$ in it, so it has all of them; in particular it has $i\sqrt[4]{2}$ and $\sqrt[4]{2}$ and since it's a field it has $i$.