I know the Abel-Ruffini theorem, which states that a general polynomial equation in one variable with degree $\geq 5$ is not solvable in radicals. I wonder whether there is a similar result for systems of polynomial equations? Say I have a system \begin{cases} f_1(x_1,x_2,\ldots,x_n)=0\\ f_2(x_1,x_2,\ldots,x_n)=0\\ \vdots\\ f_n(x_1,x_2,\ldots,x_n)=0\\ \end{cases} where $f_i(x_1,x_2,\ldots,x_n)=0$ is a polynomial of some degree in the variables $x_1,\ldots,x_n$. (Just to clarify my notation, I am not restricting to homogeneous systems, as $f_i$ may contain also the constant term as well as mixed terms such as $x_1^2x_2^3x_8$ and so on.) My question is, what can I say about the solvability of this system in radicals?

I would be tempted to say that if the total degree of the system is $\geq5$ then the system is not solvable in radicals, because a system of degree $\alpha$ corresponds be a single equation in one variable of degree $\alpha$, but I am not sure if it is right. How would you comment this issue?

  • $\begingroup$ I guess you're assuming that the system has finitely many (complex) solutions. Also your expectation sounds weird, maybe you mean "is not always solvable by radicals" to at least fit with the $n=1$ case, or better define "general". $\endgroup$
    – YCor
    Jul 18, 2019 at 14:20
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    $\begingroup$ You should read about elimination theory. The relevant point is that generically, the degees of polynomials are multiplied, rather than added, under elimination. So a $2\times 2$ polynomial system with equations of degrees $m$ and $n$ generically has $mn$ solutions, with each individual (first or second) variable satisfying a polynomial of degree $mn$. $\endgroup$ Jul 18, 2019 at 17:51
  • $\begingroup$ @VictorProtsak This (plus some additional reasoning) shows that the equations which are generically solvable are one equation of degree $\leq 4$, or two equations of degree $2$, plus any number of equations fo degree $1$. $\endgroup$
    – Will Sawin
    Jul 18, 2019 at 19:41
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    $\begingroup$ See arxiv.org/abs/1405.1252 $\endgroup$ Jul 18, 2019 at 23:05

1 Answer 1


(@VictorProtsak points out I was assuming the "total degree" $n$ referred to the maximum of the degrees of the polynomials. I'm not sure what the actual definition of "total degree" is but perhaps the only sensible definition is to take the degree in the scheme theoretic sense, i.e. the number of solutions if it is reduced. In which case the example below still applies:)

To show equations are not "generally" solvable in radicals it suffices to show that there exists at least one specialization which is not solvable. So surely you immediately see that these equations cannot be solved for $d \ge 5$ by taking $f_1 = x^5_1 - x_1 - 1$ and then $f_i = x_i$ for $i \ge 1$.

If the degree is $\le 4$ then the Galois group acts on the roots via a subgroup of $S_n$ for $n \le 4$ and so it will be solvable in that case.

(If you bound the maximal degree) you can do better; if $n \ge 3$, then you can take

$$f_1 = x_1 - x^2_2, \ f_2 = x_2 - x^2_3, \ f_3 = x^2_1 - x_3 - 1,$$ and $f_i = x_i$ for $i > 3$. Since this implies that $x^8_3 - x_3 - 1 = 0$, which is not solvable, so there can be no general solution in radicals for $d \ge 2$ if $n \ge 3$.

For the remaining cases, $d = 1$ is always solvable using linear algebra. For $n = 2$ there are not general solutions for $d \ge 3$; you use the same idea as above by taking $f_1 = x_1 -x^3_2$ and $f_2 = x^3_1-x_2-1$.

The last case is $n = d = 2$ which is solvable as one can see in a number of ways, for example, by using resultants to get equations in each single variable of degree $4$.

  • $\begingroup$ FWIW, the question referenced the total degree of the system, so that for three quadratic equations in three variables ("you can do better" paragraph), the total degree is $2+2+2=6$. $\endgroup$ Jul 18, 2019 at 17:40
  • $\begingroup$ @VictorProtsak oh good point, I confess I haven't heard the word "total degree" for a system of polynomials before. I guess (as in your comment) you mean $2 \times 2 \times 2 = 8$? (Certainly adding the degrees doesn't make sense since then a system of linear equations would have degree $n$?) $\endgroup$
    – user143272
    Jul 18, 2019 at 19:38
  • $\begingroup$ I agree that the total (additive) degree of the system is not a particularly relevant quantity for this property, which is why I wrote "FWIW". But it is a reasonable, even if crude, measure of complexity of a system in a general sense. $\endgroup$ Jul 18, 2019 at 19:51

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