Does every irreducible curve admit an equation of the form $f(x)=g(y)$, where $f$ and $g$ are polynomials? What if we allow $f$ and $g$ to be rational functions?

Actually, I'd like to understand this in the presence of an additional constraint: if we're given a finite cover of curves $\pi\colon C\to\mathbb{P}^1$, do we expect there to be a cover $\phi\colon C\to\mathbb{P}^1$ and rational functions $f(x)$ and $g(x)$ such that $C$ is isomorphic to $f(x)=g(y)$ and also $f\circ\phi=g\circ\pi$? In other words, not only is $C$ isomorphic to $f(x)=g(y)$, but this isomorphism can be chosen so that $\pi$ is the projection onto the $y$ coordinate.

This is reminiscent of Chad Schoen's paper "Varieties dominated by product varieties", but I don't see a precise connection between the two.

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    $\begingroup$ what do you mean by "curve"? since you mention a cover $C \to \mathbb{P}^1$, then it is probably a projective curve, but the nature of equation als suggests that you are talking about a plane affine curve. $\endgroup$ – Dima Sustretov Jan 5 '13 at 12:52
  • $\begingroup$ I mean a smooth projective curve. When I refer to "$f(x)=g(y)$", I mean the normalization of its projective closure. $\endgroup$ – Michael Zieve Jan 5 '13 at 14:00
  • $\begingroup$ Is $C$ meant to be geometrically irreducible? Is the ground field algebraically closed? Of characteristic 0? $\endgroup$ – user30180 Jan 5 '13 at 14:30
  • $\begingroup$ Yes, my question is for curves over the complex numbers. $\endgroup$ – Michael Zieve Jan 5 '13 at 14:32
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    $\begingroup$ Welcome, Mike!! $\endgroup$ – Lubin Jan 5 '13 at 14:35

This would contradict the Harris-Mumford(-Eisenbud) theorem that $M_g$ is non-uniruled for $g$ at least $23$. Let $C$ be a general curve of genus $g$. If $C$ is in "Zieve form", then it is the normalization of the (almost certainly) singular curve in $\mathbb{CP}^1 \times \mathbb{CP}^1$, $$D = \{ ([x_0,x_1],[y_0,y_1]) \in \mathbb{CP}^1\times \mathbb{CP}^1 \vert y_0^e f(x_0,x_1) - x_0^dg(y_0,y_1) \}, $$ where $f(x_0,x_1)$, respectively $g(y_0,y_1)$, is a homogeneous polynomial of degree $d$, resp. $e$, such that $f(0,1)$ and $g(0,1)$ are nonzero (or else the defining polynomial factors to a simpler form). By direct computation, the singular points occur where $[x_0,x_1]$ is a multiple root of $f(x_0,x_1)$ and $[y_0,y_1]$ is a multiple root of $g(y_0,y_1)$ or the point is $([0,1],[0,1])$. Moreover, at each point, the local analytic type of the singularity is the same as the plane curve with equation $y^n-x^m$, where $m$, resp. $n$, is the vanishing order of $f(x_0,x_1)$, resp. $g(y_0,y_1)$ at that point. In particular, the "delta invariant" depends only on $(m,n)$. Thus, if you "deform" $f(x_0,x_1)$ and $g(y_0,y_1)$ so that the number and type of multiple roots remains constant, then the normalizations of the corresponding curves in $\mathbb{CP}^1\times \mathbb{CP}^1$ remain of genus $g$. However, the family of such deformations of $(f,g)$ is a rational variety. Precisely, if you write $$ f(x_0,x_1) = (x_1-a_1x_0)^{m_1}(x_1-a_2x_0)^{m_2}\cdots (x_1-a_rx_0)^{m_r}, $$ with $(a_1,\dots,a_r)$ pairwise distinct, then the deformation space for $f$ is just a Zariski open subset of the affine space with coordinates $(a_1,\dots,a_r)$, and similarly for $g(x_0,x_1)$. Since $M_g$ is non-uniruled, this is a contradiction: there is only the constant morphism from a rational variety to $M_g$ whose image contains the general point parameterizing $C$.

Edit. Mike also asks whether this could be true if $f$ and $g$ are rational functions rather than polynomial functions. This is equivalent to replacing the defining equation above in $\mathbb{CP}^1 \times \mathbb{CP}^1$ by the more general equation $$ g_0(y_0,y_1)f_1(x_0,x_1) - f_0(x_0,x_1)g_1(y_0,y_1), $$ where $f_0$, $f_1$ are homogeneous of degree $d$ with no common factor, and where $g_0$, $g_1$ are homogeneous of degree $d$ with no common factor. The same observations apply: the number and types of singularities depend only on the number and multiplicities of the roots of $f_0$, $f_1$, $g_0$ and $g_1$. By varying those (distinct, likely repeated) roots as in the previous paragraph, one gets a morphism from a rational, quasi-projective variety to $M_g$. By Harris-Mumford(-Eisenbud), the only such morphism is constant if the image contains a general point of $M_g$.

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    $\begingroup$ I'm totally going to start saying "Harris and Mumford proved that the generic genus g curve is not Zieve." $\endgroup$ – JSE Jan 5 '13 at 23:06
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    $\begingroup$ 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(?). $\endgroup$ – Jason Starr Jan 6 '13 at 1:09
  • $\begingroup$ 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). $\endgroup$ – paul Monsky Jan 6 '13 at 16:38
  • $\begingroup$ @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 $M_g$. $\endgroup$ – Jason Starr Jan 7 '13 at 14:34
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    $\begingroup$ @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. $\endgroup$ – Jason Starr Jan 7 '13 at 17:25

I think that the answer is no.

Here is a somewhat related problem. Every curve curve $C$ of genus >0, has meromorphic functions $x,y$ on it which are not related by any equation of the form $f(x)=g(y)$. I denote by $(x)=(x)^+-(x)^-$ the principal divisor of an element $x$, zeros minus poles. If $x$ and $y$ are two elemets of the field of meromorphic functions on $C$, related by $f(x)=g(y)$, where $f,g$ are ratonal functions, then the divisors of poles of $x$ and $y$ are related as follows: $$m(x)^-\sim n(y)^-,$$ where $\sim$ means the usual equivalence of divisors. (Two dividors $d$ and $e$ are equivalent if $d=e+(z)$). And $m,n$ are degrees of $f,g$.

Now the factor of the set of all divisors over this equivalent equation is a torus of dimension $g$ ($g$ is the genus of $C$). We only need the fact that it is uncountable for $g>0$. So we can always find incommensurable divisors of the form $(x)^-$ and $(y)^-$. These $x$ and $y$ are related by some polynomial relation $F(x,y)=0$, but cannot be related by an equation of the form $f(x)=g(y)$.

This solution was explained me by Drinfeld in 1980 when I asked him more general question: Can every algebraic relation $F(x,y)=0$ be obtained from a chain $x=x_1,x_2,x_3,\ldots,x_n=y$ where $x_i$ and $x_{i+1}$ are related by $f_i(x_i)=f_{i+1}(x_{i+1})$, with some rational functions $f_i$, by elimination of $x_2,...x_{n-1}$? The answer is no, for the same reason).

  • $\begingroup$ How about over countable ground fields like $\overline{\mathbf{Q}}$? $\endgroup$ – user30379 Jan 5 '13 at 16:23
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    $\begingroup$ 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. $\endgroup$ – Donu Arapura Jan 5 '13 at 16:32
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    $\begingroup$ 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. $\endgroup$ – Michael Zieve Jan 5 '13 at 16:51
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    $\begingroup$ 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. $\endgroup$ – Jason Starr Jan 5 '13 at 19:28
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    $\begingroup$ @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 $f(x)=g(y)$. $\endgroup$ – ACL Jan 5 '13 at 19:58

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