# prove Galois group is solvable but that there is no radical expansion

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.

• It doesn't really make sense to use the notation $\text{Gal}$ here, because $K(\alpha)/K$ is not a Galois extension.
– Mark
Commented Apr 5 at 12:20
• 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. Commented Apr 5 at 12:37

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.

• thanks, why is a radical extension also a normal extension? Commented Apr 5 at 12:55
• @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.
– Mark
Commented Apr 5 at 13:11
• 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. Commented Apr 5 at 13:21