Timeline for Can every curve be written as $f(x)=g(y)$?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 8, 2013 at 20:31 | comment | added | Michael Zieve | Incidentally, in the literature, curves of the form $g(x)=h(y)$ are called "variables separated curves". | |
| Jan 8, 2013 at 20:03 | comment | added | Michael Zieve | @jason: your "inclusions" idea works in the hyperelliptic case, namely, the curve $y^2=f(x)$ is isomorphic to $y^2=f(x)h(x)^2$ for any $h$. It seems conceivable that all examples where $\max(d,e)$ is sufficiently large compared to $g$ might behave similarly to the hyperelliptic ones. | |
| Jan 8, 2013 at 13:27 | comment | added | Jason Starr | @michael: Yes, this would imply that there are only finitely many choices of $d$ and $e$. So perhaps the guess is wrong. There are two possible explanations consistent with Lang-Vojta. First, there could be "inclusions" among the subvarieties of $M_g$, i.e., it could happen that every curve that is Zieve for some $(d,e)$ is automatically Zieve for infinitely many other choices. The second explanation is that there is some other (non-generic) algebraic property (perhaps a condition on extra endomorphisms of the Jacobian) such that every Zieve curve has this property. | |
| Jan 7, 2013 at 19:11 | comment | added | Michael Zieve | @jason: I don't understand how there could be finitely many sequences of multiplicities for a fixed genus. Wouldn't this imply that there are only finitely many possibilities for $d$ and $e$ for a fixed genus? | |
| Jan 7, 2013 at 17:25 | comment | added | Jason Starr | @paul: I just remembered the Lang and Vojta conjectures about rational curves and rational points on varieties of general type. According to the conjectures, there is a proper, closed subvariety of $M_g$ that contains <B>all</B> rational curves in $M_g$. This strongly suggests that there are only finitely many sequences of multiplicities $(m_i)$ that give Zieve curves of genus $g$. I bet we could bound these sequences, and then use that to find a curve of genus $g$ defined over some number field that is not a Zieve curve. | |
| Jan 7, 2013 at 14:34 | comment | added | Jason Starr | @paul: I don't know, and that is a good question. I have seen examples due to Zarhin of curves over $\mathbb{Q}$ that have simple Jacobian. However, I do not see how to prove that Zieve curves have non-simple Jacobian. Careful bookkeeping of the relation between the multiplicities $m_i$ above, the genus $g$, and the dimension of the corresponding parameter space might help; for instance, if for each genus $g$ there are only finitely many possibilities for the sequence of multiplicities. Then the Zieve curves would live in a proper, closed subvariety of $M_g$. | |
| Jan 6, 2013 at 16:38 | comment | added | paul Monsky | Suppose you restrict attention to curves defined over Q? Is the result still true, and can you exhibit an example? What about the Klein quartic for example? I believe its Jacobian is a factor of the Jacobian of the Fermat curve of degree 7, but it's not clear to me whether it is a model of some f(x)=g(y). | |
| Jan 6, 2013 at 1:09 | comment | added | Jason Starr | There is a geometric question (I believe inspired by arithmetic questions of Nick Katz): what is the maximal-dimensional uniruled or rationally connected subvariety of $M_g$. So far the biggest subvarieties are moduli spaces of trigonal curves. Perhaps moduli of Zieve curves are a contender(?). | |
| Jan 5, 2013 at 23:06 | comment | added | JSE | I'm totally going to start saying "Harris and Mumford proved that the generic genus g curve is not Zieve." | |
| Jan 5, 2013 at 21:14 | vote | accept | Michael Zieve | ||
| Jan 5, 2013 at 20:51 | history | edited | Jason Starr | CC BY-SA 3.0 |
Addressed issue of rational functions rather than polynomial functions.
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| Jan 5, 2013 at 20:08 | history | answered | Jason Starr | CC BY-SA 3.0 |