3
$\begingroup$

Inspired by the top answer to this MO question, I would like to push the limit of the Hermite-Brioschi-Kronecker theorem. Suppose we only allow solutions to be expressed in terms of basic arithmetic operations in the complex field and a universal collection of finite number of univariate meromorphic functions $\mathfrak{F}_6$. Is the sextic equation still unsolvable?

If we allow some element of $\mathfrak{F}_6$ to be bivariate (for instance the Lauricella functions), then an arbitrary sextic equation becomes solvable, since one can bring polynomial equations to the Bring-Jerrard normal form. More generally what's the lowest maximum valency of $\mathfrak{F}_n$ possible for general $n$-th degree polynomial equations? I conjecture it's $n-4$, i.e., the trivial upper bound. Is that correct?

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ This is an old question, but a comment might be useful. One can use a Tschirnhausen transformation in radicals to eliminate the $x^{n-1}, x^{n-2}, x^{n-3}$ terms of the general equation of deg $n\geq4$. For $n=6$, this leaves us with $$x^6+x^2+ax+b = 0$$ but this is qualitatively different from the quintic version which doesn't have the $x^2$ term. For $n=6$, if we wish to get rid of the $x^{n-4}$ term as well, then the required Tschirnhausen transformation is no longer in radicals. $\endgroup$
    – Tito Piezas III
    Commented Sep 22, 2019 at 2:05
  • $\begingroup$ @TitoPiezasIII Out of curiosity, if we remove the $x^{6-4}=x^2$ term, would one need a quintic Tschirnhausen transformation or which type? Thanks $\endgroup$
    – Тyma Gaidash
    Commented Jul 2, 2023 at 0:30
  • $\begingroup$ @TymaGaidash Remove 4 terms from the sextic, including the $x^2$ term? A deg-4 Tschirnhaus transformation would suffice, but its coefficients will be deg $4!=24$ so it wont be in radicals. $\endgroup$
    – Tito Piezas III
    Commented Jul 2, 2023 at 0:56
  • 2
    $\begingroup$ @TymaGaidash I just remembered. There is a special quintic Tschirnhausen transformation for the general sextic, but its function is to knock off the three odd terms $(x^5, x^3, x)$ so it becomes the "cubic" $y^6+Ay^4+By^2+C$. The strange thing is one needs to solve another sextic to do this. (A version of this sextic resolvent was known to Hilbert and Joubert.) I'll make a question about it in MSE or MO soon. $\endgroup$
    – Tito Piezas III
    Commented Jul 4, 2023 at 13:15
  • 1
    $\begingroup$ @SidharthGhoshal No, the method just gives birth to a second sextic which needs to be solved too. Might as well use those non-radicals to solve the first sextic directly. I gotta rest. Later. $\endgroup$
    – Tito Piezas III
    Commented Oct 28 at 17:16

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .