Timeline for Can every curve be written as $f(x)=g(y)$?
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| Jan 5, 2013 at 19:58 | comment | added | ACL | @pranavk: if $A$ is a (non-zero) Abelian variety over $k=\bar{\mathbf Q}$, the rank of $A(k)$ is infinite. Consequently, if $C$ is a curve over $k$, one may find effective divisors of arbitrary degree $D,E$ such that $D-E$ has degree zero, but is not torsion. If $\deg(D)>2g$, Riemann-Roch implies that $D$ is the divisor of zeroes of some function $x$, and same for $E={\rm div}^-(y)$. Then $x$ and $y$ are not related by an equation of the form $f(x)=g(y)$. | |
| Jan 5, 2013 at 19:57 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Jan 5, 2013 at 19:28 | comment | added | Jason Starr | Perhaps this is the same objection raised by Donu and Mike, but there are several problems with this answer. First of all, a generic curve depends on 3g-3 parameters, not g parameters. More importantly, although the arithmetic genus of the image curve in $\mathbb{P}^2_{\mathbb{C}}$ is quadratic in $d$, the image may be very singular. Thus the geometric genus of the normalization may be much smaller than $d$. So I do not see any very easy "parameter count" that rules this out. Probably a better approach is to argue that such curves have non-simple Jacobians. | |
| Jan 5, 2013 at 19:07 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Jan 5, 2013 at 19:05 | comment | added | Victor Protsak | Correction: generic curves of genus $g$ depend on $3g-3$ parameters. | |
| Jan 5, 2013 at 18:59 | comment | added | Alexandre Eremenko | Michael, on your second comment: if $f$ is a rational function, not necessary a polynomial, then $(f(x))^-\sim d(x)^-$ where $d$ is the degree. Here $\sim$ is the equivalence, not equality. Donu, and Michael (first comment): I agree, I gave a solution of a slightly different problem. I edited my answer. | |
| Jan 5, 2013 at 18:55 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Jan 5, 2013 at 18:31 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Jan 5, 2013 at 17:00 | comment | added | Michael Zieve | Also, although you said that $f$ and $g$ are rational functions, your identity of pole divisors only follows if $f$ and $g$ are polynomials. But under additional hypotheses one can obtain your conclusion for rational functions $f$ and $g$. For instance, this follows if $C$ contains points $\alpha,\beta$ such that $\phi(\alpha)=\phi(\beta)$, $\pi(\alpha)=\pi(\beta)$, and the ratio of the ramification indices of $\alpha$ and $\beta$ under $\pi$ is different from the analogous ratio under $\phi$. Variants of this argument are in my paper with Beals and Wetherell in Israel J.Math. | |
| Jan 5, 2013 at 16:51 | comment | added | Michael Zieve | Thanks Alex. But I agree with Donu: you showed that for most choices of $\pi\colon C\to\mathbb{P}^1$ and $\phi\colon C\to\mathbb{P}^1$, there do not exist nonconstant polynomials $f$ and $g$ such that $f\circ\phi=g\circ\pi$. Whereas my question is whether this can happen for every choice of $\phi$ and $\pi$, subject to the additional constraint that $f(x)=g(y)$ is irreducible. | |
| Jan 5, 2013 at 16:32 | comment | added | Donu Arapura | Hi Alex, you seem to be showing that if $g>0$, $x,y$ can be chosen so that $f(x)=g(y)$ cannot hold. But the question is whether $x,y$ can be chosen so that such a relation does hold. It is certainly possible sometimes, e.g. when $C$ is hyperelliptic. I suspect it is false for generic $C$, however. | |
| Jan 5, 2013 at 16:23 | comment | added | user30379 | How about over countable ground fields like $\overline{\mathbf{Q}}$? | |
| Jan 5, 2013 at 15:54 | history | answered | Alexandre Eremenko | CC BY-SA 3.0 |