# Prove $\mathbb{Q}(2^{1/4})/ \mathbb{Q}$ is NOT a Galois extension.

I'm trying to show that, if $K/L$ and $L/F$ are Galois extensions, this doesn't necessarily imply that $K/F$ is also a Galois extension.

I have already shown that $\mathbb{Q}(2^{1/4})/ \mathbb{Q}(\sqrt 2)$ and $\mathbb{Q}(\sqrt 2) / \mathbb Q$ are Galois extensions. To finish the proof, I just need to show that $\mathbb{Q}(2^{1/4})/ \mathbb{Q}$ is NOT a Galois extension. But how do I show it?

• Isn't $i2^{1/4}$ a Galois conjugate of $2^{1/4}$? May 22, 2018 at 20:15

## 2 Answers

Galois extensions are normal. This means that every polynomial either splits completely in the extension or has no roots in the extension. For your case, take the polynomial $x^4-2$. It can't split completely, because it has imaginary roots and your field is contained in $\mathbb{R}$. But it also has one root.

The polynomial $x^4-2$ has a root in $\mathbb{Q}\left(\sqrt[4]2\right)$, but not all of its roots belong to it.

• I have observed this already. But how does this imply that it is not a Galois extension? May 22, 2018 at 20:17
• ah i guess we posted the same thing at the same time May 22, 2018 at 20:18
• @Pascal'sWager An extension $E$ over $F$ is a Galois extension if and only if very irreducible polynomial in $F[x]$ with at least one root in $E$ splits over $E$ and is separable. May 22, 2018 at 20:19