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Suppose $p(x)$ is a irreducible polynomial over a field $K\subseteq \mathbb{C}$ with degree n such that the Galois group $Gal(p(x)/K)\cong S_n, n\geq 5$. Take $\alpha $ a root of $p(x)$. Show that $Gal(K(\alpha)/K)$ is solvable but that there is no radical extension which contains $K(\alpha)$.
Since the Galois group is isomorphic to the symmetric group, with $n\geq 5$, we know it is not solvable and as such there is no radical extension that contains the splitting field of $p(x)$. But I do not know how to get started with this one apart from this.

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  • $\begingroup$ It doesn't really make sense to use the notation $\text{Gal}$ here, because $K(\alpha)/K$ is not a Galois extension. $\endgroup$
    – Mark
    Commented Apr 5 at 12:20
  • $\begingroup$ It's the notation used in the exercise, but if there was no other root of $p(x)$ in $K(\alpha)$ this would just be the trivial group. $\endgroup$ Commented Apr 5 at 12:37

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Note that $\alpha$ is the only root of $p$ in $K(\alpha)$. Indeed, if there was some other root $\beta$ then it would mean that $\beta=f(\alpha)$ for some $f\in K[x]$. But then if we let $L$ be the splitting field of $p$ over $K$ then every automorphism $\sigma\in\text{Gal}(L/K)$ that fixes $\alpha$ must also fix $\beta=f(\alpha)$. This is a contradiction to the Galois group being $S_n$ with $n>2$.

Thus, the group $\text{Aut}(K(\alpha)/K)$ (the notation $\text{Gal}$ is usually used only for Galois extensions, at least in most books) is trivial, and in particular solvable.

Now assume there is some radical extension $M/K$ that contains $K(\alpha)$. It can be shown that $M$ is contained in some normal radical extension of $K$. Indeed, assume $M=K(\gamma_1,...,\gamma_r)$ where $\gamma_1,...,\gamma_r$ is a radical sequence. Let $m_i$ be the minimal polynomial of $\gamma_i$ over $K$, and let $T$ be the splitting field of $m_1m_2...m_r$ over $M$. Then clearly $M\subseteq T$, and $T$ is also the splitting field over $K$. Thus $T/K$ is a normal extension. We now show this extension is radical. If $\delta_i$ is a root of $m_i$ then there is an isomorphism $\varphi:K(\gamma_i)\to K(\delta_i)$ over $K$, and it can be extended to a homomorphism $\psi:T\to\overline{T}$. But since $T/K$ is a normal, $\psi$ is actually an automorphism of $T$. Then the sequence:

$\psi(\gamma_1),\psi(\gamma_2),...,\psi(\gamma_i)=\delta_i,...,\psi(\gamma_r)$

is a radical sequence of $T$. In particular, $\delta_i$ belongs to a tower of simple radical extensions over $K$. By taking the union of such sequences for all roots of $m_1m_2...m_r$, we get that $T$ is radical over $K$.

So $T/K$ is normal and radical, and so $p$ must split in $T$. This means $p$ is a solvable by radicals over $K$, a contradiction.

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  • $\begingroup$ thanks, why is a radical extension also a normal extension? $\endgroup$ Commented Apr 5 at 12:55
  • $\begingroup$ @riescharlison Sorry, it doesn't have to be. But it is contained in some normal radical extension. I added the idea of a proof of this not very trivial fact. $\endgroup$
    – Mark
    Commented Apr 5 at 13:11
  • $\begingroup$ We do have seen that for a radical extension $E/K$ there exists a radical extension $E'/K$ such that $E'$ Galois is over K and thus normal. $\endgroup$ Commented Apr 5 at 13:21

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