I'm trying to prove whether or not $\mathbb{Q}(\sqrt[4]{-3})$ is a normal extension. I'm pretty sure it isn't, but I haven't gotten very far with my method...
First of all, $\sqrt[4]{-3}$ has minimal polynomial $x^4+3$ over $\mathbb{Q}$. This has roots $\sqrt[4]{-3}, -\sqrt[4]{-3},i\sqrt[4]{-3}$ and $-i\sqrt[4]{-3}$. So, supposing that $\mathbb{Q}(\sqrt[4]{-3})$ is in fact normal, these would all lie in $\mathbb{Q}(\sqrt[4]{-3})$. From this I have shown that then $\mathbb{Q}(\sqrt[4]{-3})=\mathbb{Q}(\sqrt{3},i)$ and then I don't know where else to go. Any help would be really appreciated!
Consider the intermediate field $\mathbb{Q}(\sqrt{-3})$. Over this field, one has $X^4 + 3 = (X^2 - \sqrt{-3})(X^2 + \sqrt{-3})$ and both quadratic factors are irreducible. Over $\mathbb{Q}(\sqrt[4]{-3}) = \mathbb{Q}(\sqrt{\sqrt{-3}})$, the first factor becomes a product of two linear polynomials. But then the second factor has to remain irreducible, since $\sqrt{-3}$ and $-\sqrt{-3}$ lie in a different square class ($-1$ is not a square in $\mathbb{Q}(\sqrt{-3})$). Hence, $X^4 + 3$ does not split into a product of linear factors in $\mathbb{Q}(\sqrt[4]{-3})$, whereby the extension is not normal.