A conjecture by Goldfeld says that half of all elliptic curves have rank zero (i.e. their Mordell-Weil group has finite order.)

Are there any known infinite families of elliptic curves (over $\mathbb{Q}$) with only finitely many rational points?

For example, in Silverman/Tate, there is an computation which shows that $y^2=x^3+x$ has exactly one rational point and $y^2=x^3+4x$ has exactly three rational points (not counting the point at infinity). I'm wondering if there are any known parameterizations giving an infinite number of such curves.


This is Corollary 6.2.1 of Chapter X in Silverman's "The Arithmetic of Elliptic Curves": Let $p$ be a prime such that $p\equiv 7$ or $11\bmod 16$. Then the Mordell-Weil group of $E_p: y^2=x^3+px$ has rank $0$ (moreover, the $2$-torsion of Sha is also trivial!). Hence, $E(\mathbb{Q})$ has only finitely many points.

In fact, in Proposition 6.1 of Silverman (same chapter X), it is shown that the torsion subgroup of $E_p(\mathbb{Q})$ is $\mathbb{Z}/2\mathbb{Z}$, so $E_p(\mathbb{Q})$ has only $2$ points.

The only drawback to this example is that all these curves $E_p$ have constant $j$-invariant $1728$.


EDIT: From the Wikipedia essay on congruent numbers, http://en.wikipedia.org/wiki/Congruent_number#Congruent_number_problem, we get "a positive rational number $n$ is congruent if and only if the equation $y^2 = x^3 - n^2x$ has a rational point with $y$ not equal to $0$ ... the existence of a rational point with $y$ nonzero is equivalent to saying the elliptic curve has positive rank ... if $p$ is a prime number then if $p \equiv 3 \pmod8$, then $p$ is not a congruent number...."

Putting this together, if $p=8k+3$ is prime, then $y^2=x^3-p^2x$ has rank zero.

It is safe to ignore the following paragraphs, left over from an earlier attempt.

Ken Ono, Rank zero quadratic twists of modular elliptic curves, http://www.mathcs.emory.edu/~ono/publications-cv/pdfs/014.pdf, proves that if $E$ is defined by $y^2=x^3-1$, and $r$ is 1, 2, 5, 10, 13, 14, 17, or 22, then there are infinitely many positive squarefree integers $D\equiv r\pmod{24}$ such that $E_D$, the $D$-quadratic twist of $E$, has rank zero.

I guess this isn't exactly what you want, as it doesn't specify which integers $D$ work and which do not, but still there may be something in that paper which will be of use to you.


For any rational number $x$, there exists an elliptic curve $E$ over $\mathbf{Q}$ without any rational points whose $j$-invariant is $x$.

I can't give you the precise details at the moment, but the idea is that you can twist your elliptic curve and get rid of any rational point.

I can't give you any equations either.

At least we know there exists such a family now.

  • $\begingroup$ With all due respect, I'm pretty sure this doesn't really answer the question. You provide a way to construct infinitely many rank zero curves, not a parametrization... $\endgroup$ – Bruno Joyal Feb 25 '12 at 21:48
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    $\begingroup$ You're right, I slightly misunderstood the question. I guess it was already known to the OP that there are infinitely many ell curves with finitely many rational points over $\mathbf{Q}$. The question of parametrizing such a family is a more difficult question. $\endgroup$ – Hoedan Feb 25 '12 at 22:28

But I have a very little knowledge about these things, but as far as my knowledge is concerned , there is some recent celebrated work by Manjul Bhargava and Arul Shankar, where in-fact they prove that a positive proportion of elliptic curves have rank $0$.

The have also used the recent results of Dokchitser, and proved it. So you can find their article here. To give a bird-eye view of what they have done, one can say that in fact, they are able to construct families such that exactly half of them have positive sign in their functional equation, and with average $3$-Selmer rank bounded by $7/6$. (This can be achieved by imposing appropriate conditions on the coefficients of the elliptic curve, and computing the root number as a product of local roots numbers.) Now work of the Dokchitser brothers on the parity conjecture implies that for elliptic curves with sign $+1$, the rank of the $3$-Selmer group is even. When combined with the bound of $7/6$, they deduce that the $3$-Selmer groups of the curves with sign $+1$ that lie in their family must be trivial.

Now (under some additional assumptions about the $3$-torsion, and some other technical assumptions, which they are able to impose on their family) by applying the results of Skinner and Urban on the Main Conjecture (which lets one pass from triviality of a Selmer group to non-vanishing of the $L$-function) they deduce that the curves in their family having sign $+1$ also have non-vanishing $L$-value at $s = 1$.

Now a positive proportion of elliptic curves overall lie in their family, and so putting all this together, one finds that a positive proportion of elliptic curves have both $3$-Selmer rank zero (and in particular, Mordell--Weil rank ( $g$ ) zero) and also analytic rank zero. Thus, a positive proportion of elliptic curves satisfy (the rank part of) BSD.

On the other hand, works by Kolyvagin and Don Zagier are also appreciable and gave rise to many theorems in this direction.

And regarding your parameterization, there is some parameterization done in the case of positive rank, so I can point the database maintained by Andrej, which is here, and it is well known that certain elliptic curves of the form $y^2= x^3-n^2x$ has a rank $0$, so there is a nice article by Keqin Feng and Maosheng Xiong which is here.

I am still searching for some nice articles, I will surely re-edit the post once I get them.

Thank you.


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