# Is this polynomial solvable by radicals?3

Suppose you have a field $\mathbb{F}$.

Show that the polynomial $x^n-n\cdot1_{\mathbb{F}}\in \mathbb{F}[x]$, where $n\geq 2$ is solvable by radicals.

• Are we working in characteristic zero? If char$(\mathbb{F})\ | \ n$, then this becomes trivial. – Ken Duna Jun 23 '16 at 4:21
• @KenDuna there is no mention of the characteristic of the field so I think you have to prove it in general. – richarddedekind Jun 23 '16 at 4:52

Let $f(x) = x^n - n\cdot 1_\mathbb{F}$.

As I mentioned in the comments, if char$(\mathbb{F}) \ | \ n$ then $f(x) = x^n$ is solvable by radicals as its roots are only $0_\mathbb{F}$.

Assume that char$(\mathbb{F}) \not \mid \ n$.

If $n\cdot 1_{\mathbb{F}}$ has an $n^{th}$ root in $\mathbb{F}$ (I confess that I don't know whether this case ought actually to arise) then the splitting field of $x^n - n\cdot 1_{\mathbb{F}}$ is just a cyclotomic extension. Thus its Galois group is abelian and hence solvable.

Assume that $n\cdot 1_{\mathbb{F}}$ does not have an $n^{th}$ root in $\mathbb{F}$.

We want to show that the Galois group of the splitting field is solvable. We will use the fact that if $G$ is a finite group and $H$ is a normal subgroup, then $$G \text{ is solvable} \iff H \text{ and } G/H \text{ are solvable}.$$

Let $K = \mathbb{F}(\omega,\sqrt[n]{n})$ be the splitting field of $f(x)$ (here $\omega$ is a primitive $n^{th}$ root of unity), $L = \mathbb{F}(\omega)$, $G=Gal(K \backslash \mathbb{F})$, and $H \subseteq G$ be the subgroup of $G$ corresponding to $L$ under the Galois correspondence. That is, the fixed field of $H$ is $L$.

Note that $L$ is a Galois extension of $\mathbb{F}$ (it is the splitting field of $x^n-1_\mathbb{F}$). In addition $Gal(L\backslash\mathbb{F})$ is a finite abelian group (in fact it is isomorphic to a subgroup of $\left(\mathbb{Z}_n\right)^*$) and hence solvable.

By the Galois correspondence, $H$ is normal and $$G/H \cong Gal(L\backslash \mathbb{F}).$$ So $G/H$ is solvable.

Again by the Galois correspondence, $H \cong Gal(K\backslash L)$. The roots of $f(x)$ are all of the form $\omega^k \sqrt[n]{n}$. If $\sigma \in Gal(K\backslash L)$, then $\sigma(\sqrt[n]{n}) = \omega^{r_\sigma}\sqrt[n]{n}$ for some $r_\sigma \in \mathbb{N}$. This defines a map from $Gal(K \backslash L) \to \mathbb{Z}_n$ taking $\sigma$ to $\overline{r_\sigma} \in \mathbb{Z}_n$. It is not hard to show that this is an isomorphism.

Therefore $H \cong Gal(K \backslash L) \cong \mathbb{Z}_n$ is solvable.