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.



Can every curve be written as $f(x)=g(y)$?
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 $M\_g$.


