3
$\begingroup$

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/

 K<x,y,z,t>:=ProjectiveSpace(Rationals(),3);
 p:=z^6 + x^4*t^2 - y^2*t^4;
 S:=Surface(K,p);
 KodairaEnriquesType(S);
 KodairaDimension(S);
 //returns:
 // 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.

$\endgroup$
8
  • 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

10
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
9
$\begingroup$

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!".)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.