# Are there irreducible polynomials that are partially solvable by radicals?

In general, polynomials of degree 5 or higher, with rational coefficients, are not solvable by radicals. There are some exceptions and trivial cases, for example $x^5-5x+12$ and $x^{100}+x^{25}+3$, respectively.

All irreducible polynomials with rational coefficients, that I am aware of, have either none or all roots that can be expressed by radicals.

Do partially-solvable irreducible polynomials exist? If they do, are there any examples known?

I mean, for example, an irreducible polynomial of $n$th degree that can be split over $Q(\sqrt{2})$ into two polynomials of 3rd and $(n-3)$th degrees, where the polynomial of $(n-3)$th is not solvable by radicals.

At https://math.stackexchange.com/q/662972 Alexander Gruber wrote:

Polynomials which are not solvable by radicals have (at least one) root that cannot be written by any combination of the operations of addition, multiplication, and the taking of nth roots.

I am not sure if he meant only reducible polynomials, and/or with non-rational coefficients, as polynomials that have "at least one" root that cannot be expressed by radicals.

• If you mean real radicals see Does one real radical root imply they all are? for an example of such. – dxiv Dec 30 '16 at 20:40
• $\Phi_{13}(x)=\frac{x^{13}-1}{x-1}$ is an irreducible polynomial completely solvable through radicals. It is an interesting exercise in Galois theory I saw here (somewhere). – Jack D'Aurizio Dec 30 '16 at 20:52
• I'm not sure to well understand, but, if $P(x)$ is not solvable by radicals and $Q(x)$ is solvable by radicals, does $P(x)Q(x)$ answers your question? – Emilio Novati Dec 30 '16 at 21:18
• @dxiv - No, I mean both real and complex radicals. – klajok Dec 30 '16 at 21:52
• @JackD'Aurizio Yes, all $x^p=1$ are solvable by radicals. – klajok Dec 30 '16 at 21:52

Proof. Let $L/K$ be a radical extension, and induct on the degree $n = [L : K]$. For $n = 1$ the claim is obvious. Pick a proper subfield $L'$ of $L$ such that $L = L'(\alpha)$ for some $\alpha \in L$ such that $\alpha^k \in L'$ for some $k > 1$, such a subfield exists since $n > 1$ and $L/K$ is a radical extension. Then, the normal closure of $L/K$ is $N/K$, where $N$ is the compositum of the splitting field of the minimal polynomial of $\alpha$ over $K$, and the normal closure $N'/K$ of $L'/K$. The extension $N'/K$ is a radical extension by induction hypothesis, and since being radical is transitive, we are done if we prove that $N/N'$ is a radical extension.
$N$ is generated over $N'$ by the $K$-conjugates of $\alpha$. Let $a = \alpha^k \in N'$, and take $\sigma \in \textrm{Aut}(N/K)$. Any conjugate of $\alpha$ is $\sigma(\alpha)$ for some $\sigma$, and we have $\sigma(a) = \sigma(\alpha^k) = \sigma(\alpha)^k$, where $\sigma(a)$ is a conjugate of $a$, and thus $\sigma(a) \in N'$ by normality of $N'/K$. Thus, every conjugate of $\alpha$ is a root of some element of $N'$, and since $N$ is generated over $N'$ by these conjugates, $N/N'$ is a radical extension, and we are done.
To conclude, let $L/K$ be any extension which is contained in a radical extension $M/K$. Then, the normal closure of $L/K$ is contained in the normal closure of $M/K$, and the latter is a radical extension by the proposition. Thus, the normal closure of $L/K$ is also contained in a radical extension of $K$. Now, note that if $L = K(\alpha)$ where $\alpha$ has minimal polynomial $p \in K[X]$, then the normal closure is the splitting field of $p$, and thus contains every root of $p$.