Timeline for Can roots of any polynomial be expressed using Eulerian function?
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13 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 6, 2012 at 23:53 | vote | accept | Anixx | ||
| Feb 23, 2012 at 14:45 | comment | added | B R | Thank you, François and Noam for clarifying this for me. | |
| Feb 23, 2012 at 10:31 | comment | added | François Brunault | In his Traité p. 378, Jordan proves the following theorem : The solution of the general equation of degree > 5 cannot be reduced to that of equations arising from circular or elliptic functions. As far as I can see the proof boils down to show that the alternating group $\mathcal{A}_n$ with $n \geq 6$ is not isomorphic to $\mathrm{PSL}_2(\mathbf{Z}/p\mathbf{Z})$ for any prime $p$. After that, Jordan also remarks that any equation can be solved by bisecting periods of hyperelliptic functions, as Noam said above. | |
| Feb 22, 2012 at 19:59 | comment | added | Noam D. Elkies | @D.Speyer: because the kernel is not a congruence subgroup. | |
| Feb 22, 2012 at 16:25 | comment | added | David E Speyer | @Noam There is a point that is confusing me in the other direction. $S_6$ is generatable by two elements, so there is a surjection $\Gamma(2) \to S_6$. So there is some function $\phi$ on the upper half plane, invariant for a finite index non-congruence subgroup, which generates an $S_6$ extension of $\mathbb{C}(j)$. Why can't I use $\phi(j^{-1}(z))$ to solve an arbitrary sextic? | |
| Feb 22, 2012 at 6:36 | comment | added | Noam D. Elkies | I guess what happens for Siegel modular functions is that the roots of any polynomial $P(x)$ generate the $2$-torsion field of the Jacobian of the hyperelliptic curve $C: y^2=P(x)$, so once you can compute modular functions for abelian varieties of genus $g(C)$ you can factor $P$. | |
| Feb 22, 2012 at 3:06 | comment | added | B R | In his appendix to Mumford's Tata II (using google books), Umemura writes, "Jordan [(Traité des substitutions et des équations algébriques)] showed that we can solve any algebraic equation of higher degree by modular functions. Jordan's idea is clarified by Thomae's formula[.] In this appendix, we show how we can deduce from Thomae's formula the resolution of algebraic equations by a Siegel modular function which is explicitly expressed by theta constants." I can't find what Jordan actually says in "Traité", though. | |
| Feb 22, 2012 at 2:16 | comment | added | Noam D. Elkies | Looks right. The $\eta$ function gives access to classical modular functions, which give field extensions with Galois group contained in a quotient of some $\text{GL}_2({\bf Z}/N{\bf Z})$. That's enough to deal with the generic quintic, but not sextics and beyond (though the sextics only barely fail: $A_6$ is of $\text{GL}_2$ type, but over the field of $9$ elements!). | |
| Feb 22, 2012 at 2:09 | history | edited | David E Speyer | CC BY-SA 3.0 |
deleted 40 characters in body
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| Feb 22, 2012 at 1:35 | comment | added | David E Speyer | Should be. But there are a lot of caveats in the above. I'm hoping someone will give a clear reference to clean this all up. | |
| Feb 22, 2012 at 0:40 | comment | added | J.C. Ottem | So basically, the answer should be 'yes' for degree 5 or less and 'no' for higher degree? | |
| Feb 22, 2012 at 0:21 | history | answered | David E Speyer | CC BY-SA 3.0 |