I'm wondering precisely what is meant by the phrase "solution in radicals". We know that general quintic (5th degree) polynomials don't have a solution in radicals. I'm wanting a "formal" description of what that phrase means in the symbol-manipulation/computer-algorithms sense. I've seen lots of imprecise and ambiguous wording in English that glosses over what that means, and all of those definitions seem to be lacking. Take for example the entry at Britannica:

At stake was the existence of a formula that expresses the roots of a quintic equation in terms of the coefficients. This formula, moreover, must involve only the operations of addition, subtraction, multiplication, and division, together with the extraction of roots, since that was all that had been required for the solution of quadratic, cubic, and quartic equations. If such a formula were to exist, the quintic would accordingly be said to be solvable by radicals.

Notice the omission of exponentation, which I assume is a typo, but it of course can be recovered by taking reciprocal roots (e.g. $ x^3 = \sqrt[^1/_3]{x} $ ). Ignoring that issue for the moment, this would seem to allow terms like:

  • $ \pi^a $
  • $ e^{a \pi i} $
  • $ \sqrt{2}^{\sqrt{2}} $
  • $ \sqrt[\pi]{a^{b^c}} $
  • $ i^a $

...where $ a,b,... $ (etc.) are coefficients of the original polynomial (i.e. $ ax^5 + bx^4 + cx^3 + dx^2 + ex + f $ ). This seems pretty typical of the information out there, where there are a lot of potential ambiguities.

Wikipedia's entry for Algebraic Solution, is different (more accurate?), stating:

An algebraic solution or solution in radicals is a closed form expression, and more specifically a closed-form algebraic expression, that is the solution of an algebraic equation in terms of the coefficients, relying only on addition, subtraction, multiplication, division, raising to integer powers, and the extraction of roots (square roots, cube roots, etc.).

...differences include "raising to integer powers", and extraction of roots, but not being exactly clear if the n of nth root can be irrational, imaginary, or a coefficient of the equation. The mention of "raising to integer powers" seems a little odd, since I can effectively raise to rational powers by combining exponentation and root extraction $ a^{^7/_{16}} = \sqrt[16]{a^7} $

Also missing is the mention of a finite number of terms, which is present in a lot of presentations. I also assume that we're talking about a fixed number of terms, so there would be no number of terms depending on an intermediate calculation, or towers of exponents based on a coefficient. And there is no mention if the use of transcendental constants are allowed.

To that end, I'd like an executable computer algorithm that describes the form of a "solution in radicals". Here's one attempt at a "solution in radicals" checker using Mathematica notation:

radicalsQ[_Symbol] := True
radicalsQ[_?NumericQ] := True
radicalsQ[x_ + y_] := radicalsQ[x] && radicalsQ[y]
radicalsQ[x_ - y_] := radicalsQ[x] && radicalsQ[y]
radicalsQ[x_ * y_] := radicalsQ[x] && radicalsQ[y]
radicalsQ[x_ / y_] := radicalsQ[x] && radicalsQ[y]
radicalsQ[x_ ^ _Integer] := radicalsQ[x]
radicalsQ[root[_Integer,x_]] := radicalsQ[x]
radicalsQ[_] := False

...which you can execute on-line here with the mathics.net webpage. Can anyone point me in the direction of a reference that might verify or contradict the above? Some day I'd like to learn how to prove Abel's Impossibility Theorem, but I'm not there yet. But in the mean time, I'm just interested in this small aspect.

Here are some tests, showing a few terms that pass and fail the above criteria.

trusies = Map[radicalsQ,{123, a, 2*b, (-b + root[2,b^2-4*a*c])/(2*a), 
                         root[16,a^7],root[16,a]^7, Pi*b, 
falsies = Map[radicalsQ,{2^a,root[a,b],a^I}]

allTests = (And @@ trusies) && !(Or @@ falsies)
  • $\begingroup$ Do you know what a field extension is? $\endgroup$ May 7, 2014 at 0:13
  • 7
    $\begingroup$ We do not want exponentiation, this is not a typo. Your example of $x^3$ we still have, since we have multiplication. We most definitely do not want to allow terms such as $2^{\sqrt2}$. In particular, all the numbers we obtain through the allowed operations and starting with the rationals are algebraic, while $2^{\sqrt2}$ is not. $\endgroup$ May 7, 2014 at 0:13
  • $\begingroup$ Also, $\pi$ and $e$ are generally not allowed (though we can write $i=\sqrt{-1}$)—both definitions specify that we must express the answer in terms of the coefficients. So we can only use this if $\pi$ is one of the coefficients of our polynomial, or otherwise expressible in terms of the coefficients. $\endgroup$ May 7, 2014 at 1:14

1 Answer 1


In the language of field extensions:

A polynomial $f$ has a solution in radicals if there is some tower of fields $$\mathbb Q=K_1\subset K_2\subset\cdots K_n$$ where $f$ has a root in $K_n$, and for each $i$ we have some $\alpha_i\in K_{i+1}$ and some integer $m_i>1$ such that $K_{i+1}=K_i(\alpha_i)$ and $\alpha_i^{m_i}\in K_i$.

We can give the same definition without referring to fields, but it requires more work:

A polynomial $f$ has a solution in radicals if we have some sequence of numbers $x_1,\ldots,x_n$ such that $x_n$ is a root of $f$, and for each $x_i$ one of the following holds:

  1. $x_i\in \mathbb Q$
  2. We have some $j,k<i$ such that $x_i=x_j+x_k$
  3. We have some $j,k<i$ such that $x_i=x_j-x_k$
  4. We have some $j,k<i$ such that $x_i=x_jx_k$
  5. We have some $j,k<i$ such that $x_i=x_j/x_k$
  6. We have some integer $m_i>1$ and some $j<i$ such that $x_i^{m_1}=x_j$
  • $\begingroup$ We actually want a slightly more general notion, don't we? Where K_1 rather than $\mathbb Q$ is $\mathbb Q(a_0,\dots,a_n)$, where the $a_i$ are the coefficients of the polynomial under consideration. $\endgroup$ May 7, 2014 at 0:26
  • $\begingroup$ @AndresCaicedo The definition is certainly clearer that way. Along the same lines, for the second one, (1) should read "$x_i$ is a coefficient of $f$". $\endgroup$
    – Jack M
    May 7, 2014 at 0:27
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    $\begingroup$ @GregBuchholz Only one of the 6 needs to hold. For $x^2-2$, we could have $x_1=2$ and $x_2=\sqrt{2}$, since $x_1$ satisfies 1 and $x_2$ satisfies 6. $\endgroup$ May 7, 2014 at 1:32
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    $\begingroup$ @GregBuchholz I am making the assumption for this answer that $f$ is over $\mathbb Q$; it simplifies notation and is more relevant to the historical usage of the phrase and the context of your question IMHO. $\endgroup$ May 7, 2014 at 3:09
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    $\begingroup$ @GregBuchholz The first definition I give is quite standard; it should be in any algebra textbook which introduces Galois Theory. In Dummit & Foote 3rd Ed. it is on page 627. The second is just a rephrasing of the first in non-field theoretic terms, which I don't have a reference for. $\endgroup$ May 7, 2014 at 3:25

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