Are the extensions $\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $\mathbb{Q}(\sqrt[3]{5})$ normal over $\mathbb{Q}$

Question

are the extensions $\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $\mathbb{Q}(\sqrt[3]{5})$ normal over $\mathbb{Q}$

I am not really sure how to start. I tried to construct a counterexample for the first extension. The second seems to be true. But I did not succeed.

Any hints? Thanks in advance.

Answer 1

$\mathbb{Q}(\sqrt[3]{5})$ is not normal over $\mathbb{Q}$:

The polynomial $X^3-5$ is irreducible over $\mathbb{Q}$. (Eisenstein criterion with $p=5$) And $X=\sqrt[3]{5}$ is a root. But $X^3-5=(X-\sqrt[3]{5})(X-(-1)^{2/3}\sqrt[3]{5})(X+\sqrt[3]{-5})$. Hence it can not be factored into linear factors over $\mathbb{Q}(\sqrt[3]{5})$.

$\mathbb{Q}(\sqrt{2},\sqrt{3})$ is normal, since $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are field extensions of degree 2 and therefore normal. So the composition $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is normal.