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.

  • $\begingroup$ If you mean real radicals see Does one real radical root imply they all are? for an example of such. $\endgroup$
    – dxiv
    Dec 30, 2016 at 20:40
  • $\begingroup$ $\Phi_{13}(x)=\frac{x^{13}-1}{x-1}$ is an irreducible polynomial completely solvable through radicals. It is an interesting exercise in Galois theory I saw here (somewhere). $\endgroup$ Dec 30, 2016 at 20:52
  • $\begingroup$ 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? $\endgroup$ Dec 30, 2016 at 21:18
  • $\begingroup$ @dxiv - No, I mean both real and complex radicals. $\endgroup$
    – klajok
    Dec 30, 2016 at 21:52
  • $\begingroup$ @JackD'Aurizio Yes, all $x^p=1$ are solvable by radicals. $\endgroup$
    – klajok
    Dec 30, 2016 at 21:52

1 Answer 1


If you allow complex radicals (such as roots of unity), then the answer is no. This follows from the following:

Proposition. The normal closure of a finite radical extension is a radical extension.

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 $.

  • $\begingroup$ Thank you very much for the elaborative answer. To summarize it: if any root of irreducible polynomial can be expressed by radicals then all of them can be too. I am writing this comment before accepting your answer because your first sentence contains "the answer is TRUE" which means for me: "there DO exist irreducible polynomials such that only some of their roots can be expressed by radicals". $\endgroup$
    – klajok
    Dec 30, 2016 at 22:01
  • $\begingroup$ Yes, that's correct. I edited the answer - the answer to the question "Do partially-solvable irreducible polynomials exist?" is "no". $\endgroup$
    – Ege Erdil
    Dec 30, 2016 at 22:07

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