# Certain Galois extension over $\mathbb{Q}$ not contain $\sqrt{2}$

I want to show that $$\sqrt{2}$$ 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 $$\sqrt{2} \in K$$, then $$x^4 - 2$$ splits completely in $$K$$. Hence, $$K$$ contains $$\mathbb{Q}(\sqrt{2},i)$$, 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.

• 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 $S_n$? Aug 7, 2022 at 3:11

Let $$K_0=\mathbb Q(\sqrt{2},i)$$ 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.