Cross-posted from MSE.

I am not good at algebraic geometry and almost surely am misunderstanding something.

Got an alleged argument against Bombieri-Lang conjecture and would like to know what the mistake is.

One of the most simplest formulations of Bombieri-Lang conjecture is in Joe Silverman's answer on MO, paraphrasing

...For a surface of general type ... the Bombieri-Lang conjecture says that the solutions in rational numbers ...are.. not Zariski dense (lie on a finite set of curves).

Take the affine surface over the rationals:

$$ z^6 + x^4 - y^2=0$$

According to Magma it is of general type.

Fix $z$ squarefree integer $k$. This gives quartic model of elliptic curve $k^6+x^4-y^2=0$.

This is birationally equivalent to Weierstrass $$v^2=-4k^6u+u^3 $$

As $k$ varies, this gives infinitely many positive rank elliptic curves.

What is wrong with this alleged contradiction?

Comment suggests it might be not of general type.

In Magma online: http://magma.maths.usyd.edu.au/calc/

 p:=z^6 + x^4*t^2 - y^2*t^4;
 // 2 0 General type
 // 2

Added The main misunderstanding was caused by incorrect usage of Magma function. The correct way to check is KodairaEnriquesType(S : CheckADE := true); because of certain assumptions, but this might take much longer.

With this change I get error and no result.

  • 4
    $\begingroup$ I suspect the issue is that the KodairaDimension function is only meant for mildly singular surfaces. The webpage magma.maths.usyd.edu.au/magma/handbook/text/1354 lists the related function KodairaEnriquesDimension as only working for surfaces with ADE singularities. $\endgroup$
    – dhy
    Aug 16, 2015 at 8:58
  • 5
    $\begingroup$ A general type surface (in characteristic 0) cannot be swept out by a family of elliptic curves (this follows from a short argument with pluricanonical forms.) This implies that the number of elliptic curves on a general type surface is countable; a deep conjecture (closely related to Bombieri-Lang) predicts that there are only finitely many elliptic curves. $\endgroup$
    – dhy
    Aug 16, 2015 at 9:02
  • $\begingroup$ @dhy likely you are right, thank you. After RTFM this fails: KodairaEnriquesType(S : CheckADE := true); $\endgroup$
    – joro
    Aug 16, 2015 at 9:05
  • 3
    $\begingroup$ It is worse than it might seem: your surface is unirational, and hence rational. Probably you can find a reference somewhere, but you can also prove it directly. For $(v,w)\in \mathbb{G}_m^2$, basically $(k^*)^2$, consider $x=w^3(v-v^{-1})/2$, $y=w^6(v^3-v-v^{-1} + v^{-3})/8$, and $z=w^2(v-v^{-1})/2$. In other words, set $z=x/w$, and then parameterize the resulting rational curve over $k(w)$. $\endgroup$ Aug 16, 2015 at 11:35
  • $\begingroup$ @JasonStarr Thanks, will check your argument tomorrow. Two people on MSE claim it is elliptic surface, btw. $\endgroup$
    – joro
    Aug 16, 2015 at 13:05

2 Answers 2


I am only posting this as an answer because it annoys me to see a question like this listed as "unanswered", thus "hovering" near the top of the list of unanswered questions. If dhy wants to write up his comment as an answer, then I will delete this answer.

The surface given by the OP is as far as possible from being "general type". Just to remind, a projective variety $S$ over a characteristic $0$ field is of general type if one, and hence any, smooth, projective model of the variety has big dualizing sheaf. This is a birational property. In particular, since the dualizing sheaf on $\mathbb{P}^n$, namely $\mathcal{O}_{\mathbb{P}^n}(-n-1)$, is anti-ample, rational varieties are not of general type. As dhy correctly points out, much more is known: for a diagram $$\begin{array}{lcc} T & \xrightarrow{f} & S \\ \downarrow \pi \\ R \end{array},$$ such that $f$ is dominant and generically finite and such that $\pi$ is projective and flat of positive fiber dimension, then the geometric generic fiber of $\pi$ is of general type (this can easily fail in positive characteristic). In particular, if there exists a diagram as above such that the geometric generic fiber of $\pi$ is rational or an Abelian variety, then also $S$ is not of general type. (Once upon a time, Serge Lang had a conjecture that was roughly the converse of this statement, but nobody seems to believe that conjecture anymore.)

Back to the surface $S$ of the OP, i.e., the zero locus in $\mathbb{A}^3$ of $$F(x,y,z) = z^6 + x^4 - y^2,$$ this is an irreducible, affine hypersurface with an isolated singularity at the origin. By the way, such a singularity is called quasi-homogeneous, in this case with weights $(3,6,2)$. This came up a little while ago, because for a quasi-homogeneous function $$g:\mathbb{A}^r \to \mathbb{A}^1,$$ the critical locus of $g$ is $\{0\}\subset \mathbb{A}^1$. Anyway, the open subset $U= D(xz)\cap S$ of $S$ is dense. Since being of general type is birational, it suffices to work with $U$. Denote $w = x/z\in \mathcal{O}^{\times}_{\mathbb{A}^3}(D(xz))$. Then the defining equation of $S$ becomes, $$ \widetilde{F}(w,x,y) = x^4(w^{-6}x^2+1) - y^2.$$ Since $x$ is invertible on $D(xz)$, also $y/x^2$ is in $\mathcal{O}_{\mathbb{A}^3}(D(xz))$. Thus the defining equation is the same as, $$F(w,x,y) = x^4\left( 1 - \left(\frac{y}{x^2}\right)^2 + (w^{-3}x)^2 \right).$$ Now denote $$ v = \frac{y}{x^2} + \frac{x}{w^3}.$$ Then there is an isomorphism of $k$-algegras $$\frac{k[x,y,z][1/xz]}{\langle F(x,y,z) \rangle} \xrightarrow{\cong} k[v,w]\left[\frac{1}{vw}\right],$$ $$x = \frac{w^3(v-v^{-1})}{2}, \ y = \frac{w^6(v-v^{-1})^2(v+v^{-1})}{8}, \ z = \frac{w^2(v-v^{-1})}{2}.$$ In particular, $U$ is isomorphic to $\mathbb{G}_m^2$. Therefore $S$ is a rational surface. So $S$ is not of general type.

One commenter points out that for general rational functions $a(z)$, $b(z)$ and $c(z)$, for the element $$ F(x,y,z) = x^4 +a(z)x^2 + b(z)x + c(z) - y^2,$$ of $k(z)[x,y]$, the zero locus is an elliptic surface, hence not of general type.

  • $\begingroup$ Thank you. Isn't much simpler argument: treat one of the variables as parameter. If the genus is $1$ it is elliptic surface? $\endgroup$
    – joro
    Aug 17, 2015 at 6:49

This is more or less what Jason has done, but maybe a bit more direct, and it is so elementary that it's hard not to call it an exercise that possibly does not belong on MO. Start with $y^2=x^4+z^6$. Change variables $(x,y,z)=(zx_1,z^2y_1,z)$ to get $y_1^2=x_1^4+z^2$. Next change variables $(x_1,y_1,z)=(x_1,x_1^2y_2,x_1^2z_2)$ to get $y_2^2+1=z_2^2$. This eliminates a variable, so your surface is birational to $\mathbb P^1$ times the curve $Y^2+1=Z^2$. This latter curve is also trivially a copy of $\mathbb P^1$, so your surface is birational to $\mathbb P^1\times\mathbb P^1$. (Jason, don't delete your answer, it's great having a perspective that's more intrinsic than just "change variables and voila!".)


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