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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?

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    $\begingroup$ Isn't $i2^{1/4}$ a Galois conjugate of $2^{1/4}$? $\endgroup$
    – Angina Seng
    May 22, 2018 at 20:15

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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.

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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.

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  • $\begingroup$ I have observed this already. But how does this imply that it is not a Galois extension? $\endgroup$
    – Pascal's Wager
    May 22, 2018 at 20:17
  • $\begingroup$ ah i guess we posted the same thing at the same time $\endgroup$
    – Sheel Stueber
    May 22, 2018 at 20:18
  • $\begingroup$ @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. $\endgroup$
    – José Carlos Santos
    May 22, 2018 at 20:19

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