For elliptic curve $y^{2}=x(x+a^{2})(x+(1-a)^{2})$,($a$ is a rational number and does not equal 0,1,1/2),is its rank always 0?
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2$\begingroup$ Without further explanation or motivation, this looks to me like a homework exercise. Perhaps math.stackexchange.com would be a better venue for this question. $\endgroup$– David Roberts ♦Commented May 3, 2011 at 2:25
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$\begingroup$ The elliptic curves in my question,is related to a number theory problem I considered.If for some elliptic curves like these,ranks are not zero,the infomation given by the rational points on the elliptic curves of this kind is not so useful as to the problem I considered. I have the impression that to determine the ranks of elliptic curves is very hard,even for some specific elliptic curve.I wonder if someone has considered similar questions for certain class of elliptic curves. Thanks to comments made by David Roberts. $\endgroup$– leiweixiCommented May 3, 2011 at 6:12
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5$\begingroup$ For $a=7$, the rank is nonzero, as it is for $a=31$. A third example is $a=751$. $\endgroup$– JunkieCommented May 3, 2011 at 13:02
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$\begingroup$ Thanks a lot.Junkie have just give me a negative answer. $\endgroup$– leiweixiCommented May 3, 2011 at 13:41
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7$\begingroup$ Junkie, your comment is definitely good enough to go in the answer box. $\endgroup$– S. Carnahan ♦Commented May 3, 2011 at 14:58
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3 Answers
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Your family of elliptic curves are exactly those elliptic curves with torsion subgroup containing $\Bbb Z / 2 \Bbb Z \times \Bbb Z / 4 \Bbb Z$. To see this make the linear change of variables $t = a/(1-a)$ and look at the parameterizations on page 101 of Husemöller's book "Elliptic Curves".
So in particular there are values of $a$ for which your elliptic curve has Mordell-Weil rank 8, the first of which was discovered by Elkies in 2005. See Dujella's website. Moreover, Eroshkin found that there are infinitely many such curves with rank at least 5.
These happen to be my favorite elliptic curves. I'd be very interested to know why you were interested in them.
answered May 10, 2011 at 6:29
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1$\begingroup$ An example with rank 9 can be found in dx.doi.org/10.1112/S1461157014000023 $\endgroup$– dujeCommented Jul 31, 2014 at 12:19
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Although Junkie has answered the question, I'd like to point out that in the case of parametrized families of elliptic curves (such as this) it is often easy to find an explicit subfamily with positive rank. In the present case, let us take $x=2a^2$ and see what condition on $a$ forces this to give a point on the elliptic curve. Making this substitution we obtain. $$ y^2=6 a^4 (3 a^2-2a+1). $$ We can simplify by defining $b=y/6 a^2$. Thus $$ 6b^2=3 a^2-2a+1. $$ This is a conic with the point $(a,b)=(-1,1)$ and so we can parametrize all the solutions: $$ a=(-t^2 + 4t - 10/3)/(t^2 - 2), \qquad b=(t^2 - 8t/3 + 2)/(t^2 - 2) $$ where $t$ is rational. The argument gives that $$ (x,y)=(2a^2,6a^2 b)$$ is a rational point on the elliptic curve provided $a$, $b$ have the above shape. It should be possible through a slightly tedious computer algebra calculation to determine all rational numbers $t$ where the point above is torsion (using Mazur's Theorem), and for all other values of $t$ the rank is positive.
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2$\begingroup$ How did it occur to you to take x = 2a^2? $\endgroup$– KConradCommented May 4, 2011 at 7:17
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1$\begingroup$ Actually, we can set $x = ka^2$ for more or less any $k$. This will give $$y^2 = k(k+1)a^4 ((k+1)a^2 - 2a + 1)$$. After setting $y = k(k+1)b$ you get $$ k(k+1)b^2 = (k+1)a^2 - 2a +1. $$ Multiply by $k+1$ and set $b = c/(k+1)$ to get $$ k c^2 = (k+1)^2 a^2 - 2(k+1)a + (k+1) = ((k+1)a - 1)^2 + k. $$ and setting $a = 1 + kd/(k+1)$ you get $$ c^2 = kd^2 + 1 $$ which has one, and therefore infinitely many rational points. Most of them will give non-torsion points on the original elliptic curve. I can't read Samir's mind, but setting x = ka^2 seems like an easy way to get rid of the $x(x+a^2)$. $\endgroup$ Commented May 5, 2011 at 4:01
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I'm probably going out on a limb here, but I think that the following conjecture is reasonable.
Conjecture. Let $E:y^2=x^3+a(T)x+b(T)$ be an elliptic curve defined over $\mathbb{Q}(T)$ with the property that its $j$-invariant is not in $\mathbb{Q}$. Then there are infinitely many rational numbers $t$ such that $E_t(\mathbb{Q})$ has positive rank.
Of course, I have no idea how one might prove this. As Samir pointed out, in some cases one can find a covering $C\to \mathbb{P}^1$ such that the pullback of $E$ has positive rank over $C$ and such that $C$ has genus 0 (or 1) with infinitely many rational points. But in general this won't be possible.
One final note. It's possible to create families for which the sign of the functional equation is always even, in which case the fibers have even rank (if we believe Birch-Swinnerton-Dyer). But even in this case, I'd expect infinitely many fibers to have rank 2 (or greater).
answered May 5, 2011 at 3:02
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$\begingroup$ I wonder if your conjecture follows from the 'rank distribution conjecture' (due to Goldfeld, Katz-Sarnak?) that (on average) half of all elliptic curves have rank 0, and half have rank 1. I've just taken a look at a preprint of Bhargava, and one of his theorems is that, assuming finiteness of sha, a positive proportion of all elliptic curves have rank 1. I would think that such a result could be used to prove that a positive proportion of curves in a family (like in your conjecture) have positive rank. $\endgroup$ Commented May 9, 2011 at 9:07
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$\begingroup$ @Joe Silverman: What is the expected rank distribution in a family with constant parity? I guess the minimalist philosophy would say 100% of specializations $t$ would have either rank $r$ or rank $r+1$, depending on which parity is obtained where $r$ is the Mordell-Weil rank of $E$ over $\Bbb Q(T)$. Are there examples of these constant parity families over $\Bbb Q$? I know of twist families over number fields, the so called "lawful curves" of Dokchitser & Dokchitser, but not of any over $\Bbb Q$. If so could you point me to an example? $\endgroup$ Commented May 10, 2011 at 6:01
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$\begingroup$ @Jamie Weigandt: I agree that the minimalist philosophy is reasonable. I don't know a reference for constant parity families over $\mathbb{Q}$, but the following paper says that for any open interval $I$ in $[-1,1]$, there is a twist families such that the average rank lies in $I$. MR1641093 Rizzo, Ottavio G., Average root numbers in families of elliptic curves. Proc. Amer. Math. Soc. 127 (1999), no. 6, 1597–1603. $\endgroup$ Commented May 10, 2011 at 13:15
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1$\begingroup$ Rohrlich's paper for twist families is: archive.numdam.org/article/CM_1993__87_2_119_0.pdf He quotes (0.5) Cassels and Schinzel for $7(1+t^4)y^2=x^3-x$, the idea resting on Stephens, that the 2-selmer rank is odd for twists by 6,7 (mod 8) of the congruent number curve. blms.oxfordjournals.org/content/14/4/345.full.pdf $\endgroup$– JunkieCommented May 11, 2011 at 22:31