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Looks like I found a counterexample to a theorem assuming Lang's conjecture, but not sure it is correct.

Boundedness of Mordell–Weil ranks of certain elliptic curves and Lang’s conjecture p. 2

Theorem 1.3. Let $K$ be a finitely generated field over $\mathbf{Q}$. Let $n \ge 8$ be an integer and let $\alpha_i, (i=0,\ldots,n)$ be fixed elements of $K$. Suppose Conjecture 1.2 holds for $k$ a finitely generated field over $\mathbf{Q}$ (cf. [2]). Then there are only finitely many elliptic curves of the form $y^2=a x^4 + b x^2 +c \; (a,b,c \in K)$ which have $\alpha_i$ as the $x$-coordinates of some $K$-rational points. In particular the Mordell–Weil ranks of such elliptic curves are bounded.

Let $K=\mathbb{Q}[\sqrt{21}], a=c=\frac{-\frac{112}{75} s^{4} + \frac{6647}{4725} s^{2} - \frac{83521}{33339600}}{s^{2}} ,b=\frac{\frac{1799}{75} s^{4} - \frac{171377}{37800} s^{2} + \frac{21464897}{533433600}}{s^{2}}, s \in K$.

Let $P(x)=a x^4 + b x^2 + a$. The discriminant depends on $s$ and $P(x)=P(-x)$.

Consider the elliptic curve $y^2=P(x)$ for $s$ for which the discriminant doesn't vanish.

Let $n=9$ and $\alpha_i=\{1,2,4,1/2,1/4,-1,-2,-4,-1/2,-1/4\}$

$$P(1)= \left(21\right) \cdot s^{-2} \cdot (s - \frac{17}{84})^{2} \cdot (s + \frac{17}{84})^{2}$$ $$P(2)= \left(\frac{1764}{25}\right) \cdot s^{-2} \cdot (s^{2} + \frac{289}{7056})^{2}$$ $$P(4)= 289$$ $$P(1/2)= \left(\frac{441}{100}\right) \cdot s^{-2} \cdot (s^{2} + \frac{289}{7056})^{2}$$ $$P(1/4)= 289/256$$

All of the above are squares as are $P(-x)$.

For all infinitely many admissible choices of $s$, $\alpha_i$ are $x$-coordinates.

The $j$ invariant of the Jacobian depends on $s$.

Is this really a counterexample to Theorem 1.3?

$a,b$ in machine readable form:

 a=c=-1/33339600*(289+7056*s^2)^2/s^2+289/189
 b=257/533433600*(289+7056*s^2)^2/s^2-4913/756
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4  
Of course, if this is a counterexample to Theorem 1.3, then the most likely explanation is that there's an error in the proof of Theorem 1.3 and the condition $n\ge8$ needs to be replaced with $n\ge9$ or $n\ge10$. It's much less likely that this would contradict the Bombieri-Lang conjecture. But still, you've found a nice example. (Hmmm... Another possibility is that Theorem 1.3 is misstated and actually only holds on a Zariski open subset of the parameter space, which would allow for a finite number of families of exceptions.) –  Joe Silverman Apr 27 at 15:00

1 Answer 1

up vote 28 down vote accepted

Yes, this is a counterexample to Theorem 1.3. But it looks like the issue is with the proof of Theorem 1.3, and is not relevant to Lang's conjecture. Namely, your example contradicts Theorem 4.2 of that paper, which does not rely on Lang's conjecture. Then the proof of Theorem 1.3 relies on Theorem 4.2, in addition to relying on Lang's conjecture. The only proof given for Theorem 4.2 is "by the argument in [4], we obtain the following theorem", so it is not completely surprising if an error occurs there.

Added later: In fact, the problem is that the author needed to assume that $\alpha_i^2\ne\alpha_j^2$ for $i\ne j$, rather than just that $\alpha_i\ne\alpha_j$. Otherwise everything he says about $W_n$ is wrong. Namely, given elements $\alpha_0,\alpha_1,\dots,\alpha_n$ of a number field $K$, he defines $W_n$ to be the subvariety of $\mathbb{P}^n$ defined by the equations $$ \left| \begin{array}{cccc} 1&1&1&1 \\ \alpha_0^2&\alpha_1^2&\alpha_2^2&\alpha_i^2 \\ \alpha_0^4&\alpha_1^4&\alpha_2^4&\alpha_i^4 \\ Y_0 & Y_1 & Y_2&Y_i \end{array} \right| = 0\,\,\,(i=3,4,\dots,n). $$ Theorem 4.2 asserts that if $n\ge 8$ then the only curves on $W_n$ of genus $0$ or $1$ are the lines $$ (Y_0,\dots,Y_9)=\bigl(\pm(s+t\alpha_0^2),\pm(s+t\alpha_1^2),\dots,\pm(s+t\alpha_n)^2\bigr).$$ Of course, your example produces further genus-$0$ curves on $W_9$, and hence contradicts Theorem 4.2. But as I said, the real issue is that the author should have assumed that the $\alpha_i^2$ are pairwise distinct, since otherwise everything he says about $W_n$ is wrong.

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Thank you. So adding $\alpha_i^2\ne\alpha_j^2$ probably makes the paper correct? –  joro Apr 28 at 5:57
    
That's what I would guess, but I didn't check all the proofs (mainly I didn't check the ones which rely on carrying over arguments from Vojta's paper). –  Michael Zieve Apr 28 at 7:46
2  
At some point, I did check it (including Vojta's paper) and indeed, one just want evaluate the quadrics az^2+bz+c at 8 given points. In this paper, the 8 points are the squares of the alphas. –  Pasten Apr 28 at 15:08
    
Thanks for the info! The result of this paper is neat, so I'm happy to hear that someone can vouch for it (after the slight modification). –  Michael Zieve Apr 28 at 15:13
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The closest to such a thing is Math.Reviews (although they always say that journals are responsible for this, rather than them -- but then apparently nobody is responsible for bugs in papers published in conference proceedings). Anyway just now I emailed Math.Reviews pointing out the mistakes in the paper, they will give the reviewer the option to revise his review in light of my remarks. I also emailed the editor who handled this paper; based on prior experience, I'm certain that he will proceed with the greatest professionalism. –  Michael Zieve Apr 29 at 15:02

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