# Solvability of polynomials over fields of characteristic zero

1) Let $K$ be a field, $\operatorname{char}(K)= 0$, and $f ∈ K [x]$ with $\deg(f)\le4$. Then $f$ is solvable by radicals.

Proof:

$\operatorname{Gal} (F/K) \cong S_4$ then $\operatorname{Gal}(F/K)$ is solvable by radicals, where $F$ is a splitting field of $f$ over $K$.

2) If $\deg(f)=p \ge 5$, ($p$ prime ?), then $f$ is not solvable by radicals.

Can somebody explain to me these two results?

## 1 Answer

• If $K$ is a field and $\operatorname{char}(K)= 0$, $f \in K [x]$ with $\deg(f)\le4$, then $f$ is solvable by radicals.

Well, $S_4$ is a solvable group, where the appropriate chain of subgroups are as follows:

$$\{e\} \subset \mathbb{Z}_2 \times \mathbb{Z}_2 \subset A_4 \subset S_4$$

You can check by hand that each $G_j$ is normal in $G_{j+1}$ and that each factor group $G_{j+1}/G_j$ is abelian. Now, it is a theorem that $Gal(f)$ can be viewed as a permutation group acting on the $\deg(f)$ roots of $f$, and in particular, $Gal(f)$ is a subgroup of $S_{\deg(f)}$. Noting that $S_2$, $S_3$, and trivially $S_4$ are all subgroups of $S_4$, we can finish up by applying the fact that every subgroup of a solvable group is solvable.

• If $\deg(f)=p ≥5$, ($p$ prime?), then $f$ is not solvable by radicals.

This is not always true. I.e. there are irreducible quintic polynomials that are solvable (e.g. with $Gal(f) \cong \mathbb{Z}_5$). What we can say is that if $\deg(f) = p \geq 5$, then $f$ is not necessarily solvable by radicals.

For example, such a polynomial will not be solvable if $Gal(f)$ contains a $2$-cycle when we look at it as a subgroup of $S_p$ acting on the roots of $f$. This is justified as follows:

Since $f$ is irreducible, then if we adjoin one of its real roots to $\mathbb{Q}$, we get the tower of fields $\mathbb{Q} \subset \mathbb{Q}[\alpha] \subset K$ where $\mathbb{Q}[\alpha]$ is a degree $5$ extension. From here, we have $|Gal(f)| = [K:\mathbb{Q}] = 5 \cdot [K:\mathbb{Q}[\alpha]]$. In particular, $5$ divides $|Gal(f)|$, and so $Gal(f)$ must contain an element of order $5$ by Cauchy's theorem. This element necessarily corresponds to a $5$ cycle. From here, it is a theorem that any $2$-cycle together with any $p$-cycle will generate the entire permutation group $S_p$. Hence, in this situation, $Gal(f) \cong S_5$. Finally, we know that $S_n$ is not solvable for all $n \geq 5$.

Footnote: A case where $Gal(f)$ would contain a $2$-cycle would be if $f$ had two complex roots. Recall that, if $a+bi$ is a root of some polynomial, then so too is $a-bi$. Hence, complex conjugation is a legitimate $\mathbb{Q}$-automorphism of order $2$ that swaps the two complex roots.

• Glad I could help! – Kaj Hansen Jul 18 '14 at 20:22