Let $P(x)$ and $Q(x)$ be two general polynomial of the same degree $d$ (d can be arbitrary large). Consider the surface $S : z^2 = P(x) Q(y)$ in the projective space $\mathbb{P}^3$. I want to prove the following assertion (which I'm almost sure that should be true):

If the coefficients of the polynomials $P$ and $Q$ are chosen generically, then the surface $S$ is a hyperbolic surface in $\mathbb{P}^3$. Equivalently, I wish to show that there is no rational/elliptic curve on the surface $S$.

By hyperbolicity I mean there is no non-constant holomorphic function $f: \mathbb{C} \rightarrow S$.

Note that the surface $S$ is a singular surface with singularities at $(a_i, b_j, 0)$ where $a_i$ and $b_j$ are roots of the polynomials $P$ and $Q$ respectively.\

Jason, thanks, I think in your answer you mean the hyperbolicity of $D(zw)$ rather than $D(y)$. Also what is the generic condition on the polynomials $P$ and $Q$? Is the affine surface D(zw) in the above surface $S$ always hyperbolic?

If we consider the equation $S$ on a number field $K$. Can we deduce that the there are finite number of $K$-rational points on the affine surface $D(zw)$ by Lang conjecture?

  • $\begingroup$ It seems that the components of the ramification locus are all rational. $\endgroup$ Mar 31, 2014 at 6:03
  • $\begingroup$ What are the coordinates? $\endgroup$
    – abx
    Mar 31, 2014 at 6:06
  • $\begingroup$ if we homogenize the equation of the surface $S$ with respect to the variable $w$ it becomes of the form $(\frac{z}{w})^2 = P(\frac{x}{w})Q(\frac{y}{w})$ in $\mathbb{P}^3$ with coordinates $x, y, z, w$. More precisely, by hyperbolicity I mean the affine part of the surface $S$, i.e. the part $\{w\neq 0\}$ is hyperbolic. Equivalently the affine part of the surface $S$ doesn't contain any rational/elliptic curves. ( the image of any non constant meromorphic functions is contained in the part $\{w=0\}$. $\endgroup$
    – Mehdi
    Mar 31, 2014 at 10:16
  • $\begingroup$ Mehdi: can you explain the word "equivalently" in your question? I thought that equivalence of hyperbolicity to the absence of rational/elliptic curves is a major unsolved conjecture. $\endgroup$ Mar 31, 2014 at 13:17
  • $\begingroup$ Yes , Alexandre, this is one of the conjectures made by Lang $\endgroup$
    – Mehdi
    Apr 1, 2014 at 7:22

1 Answer 1


I am just writing as an answer the point raised by Alex Degtyarev in the comments: for every root $a_i$ of $P(x)$, resp. for every root $b_j$ of $Q(y)$, the surface contains the rational curve $\{(a_i,y,0):y\in \mathbb{C}\}$, resp. the rational curve $\{(x,b_j,0):x\in \mathbb{C}\}$.

Edit. The OP clarifies that the question is whether the Zariski open subset $D(z)$ is hyperbolic. This open subset is a finite, unbranched cover of the affine hyperbolic surfaces $$[\mathbb{C}\setminus Z(P(x))]\times [\mathbb{C}\setminus Z(Q(y))],$$ the product of two affine hyperbolic curves.

  • $\begingroup$ I guess that these are all kinds of rational curves in the surface $S$, it means that there is no rational/elliptic curve in the part $\{z\neq 0 \}\cup \{w\neq0\}$. $\endgroup$
    – Mehdi
    Mar 31, 2014 at 13:29
  • $\begingroup$ Just to clarify, are you amending your question to ask about hyperbolicity of the affine surface where $zw$ is nonzero? $\endgroup$ Mar 31, 2014 at 13:34
  • $\begingroup$ Yes, Jason, I ask about the hyperbolicity of the affine surface where $zw$ is nonzero. $\endgroup$
    – Mehdi
    Mar 31, 2014 at 13:47
  • $\begingroup$ This is a way of thinking about this problem: we can define the map $S \righarrow \mathbb{P}^1$ by $(x, y, z, w) \mapsto (x, y)$ which is defined everywhere except the finite set $\Sigma$ of the singularities of the surface $S$. This set can be blown up to yield a holomorphic map from the blowup of $S$, i.e. $g: \overline{S} \rightarrow \mathbb{P}^1$. Now if we could prove that the map $g$ can be factorized as $\overline{S} \rightarrow C \rightarrow \mathbb{P}^1$ where $C$ is a hyper elliptic curve, then we can almost prove the hyperbolicity of the surface $S$ by a standard argument. $\endgroup$
    – Mehdi
    Mar 31, 2014 at 13:55
  • $\begingroup$ Jason, you mean the Zariski open subset $D(zw)$? $\endgroup$
    – Mehdi
    Mar 31, 2014 at 17:16

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