I want to show that $\sqrt[4]{2}$ is not contained in any field $K$ that is Galois over $\mathbb{Q}$ with $G(K/\mathbb{Q}) \cong S_n$, for any positive integer $n$.

The statement is obvisouly true when $n = 1,2,3$. Indeed, if $n \leqslant 3$ and $\sqrt[4]{2} \in K$, then $x^4 - 2$ splits completely in $K$. Hence, $K$ contains $\mathbb{Q}(\sqrt[4]{2},i)$, which implies $8 \mid n!$, a contradiction. However, I have no idea how to prove this for $n \geqslant 4$. I also know that the condition $G(K/\mathbb{Q}) \cong S_n$ implies $K$ is the splitting field of some irreducible polynomial $f$ over $Q$. But I don't know whether this fact is useful here.

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    $\begingroup$ Let $L=\Bbb{Q}(\root4\of2,i)$. So in the interesting case $L\subseteq K$. As $L/\Bbb{Q}$ is normal we have $G(K/L)\lhd G(K/\Bbb{Q})$. What do you know about the normal subgroups of $S_n$? $\endgroup$ Aug 7, 2022 at 3:11

1 Answer 1


Let $K_0=\mathbb Q(\sqrt[4]{2},i)$ and suppose $K/K_0/\mathbb Q$ such that $Gal(K/\mathbb Q)\cong S_n$. Since $Gal(K_0/\mathbb Q)\cong D_8$ is normal, by Galois correspondence this must mean that $D_8$ is a quotient of $S_n$.

Then for $n\geq 5$ this cannot be true since the only normal subgroup of $S_n$ is $A_n$ so the quotient cannot be $D_8$. For $S_4$ you can proceed similarly by looking at all its normal subgroups and resulting quotients.


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