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Let $K/\mathbb{Q}$ be an algebraic extension, and let $E_1,E_2/\mathbb{Q}$ be elliptic curves. Is it possible that the Mordell-Weil rank of $E_1(K)$ is finite while that of $E_2(K)$ is infinite?

Pablo
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3 Answers3

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The answer is yes, at least if you assume that Tate-Shafarevich conjecture.

Let $E$ be an elliptic curve over a number field $k$. Under some mild hypothesis (see Corollary 1.10 of http://arxiv.org/abs/0904.3709) there exist quadratic twists of $E$ which have trivial Mordell-Weil rank. In other words, there exists quadratic extensions in which the rank of $E$ does not go up. Assuming in addition the Tate-Shafarevich conjecture there are also quadratic twists of $E$ whose Mordel-Weil rank is $1$ (see Corollary 1.12 loc. cit.), and hence quadratic extensions in which the rank of $E$ goes up. Generalizing the methods of loc. cit. along the line of http://arxiv.org/abs/1504.02343 one can show that (assuming the Tate-Shafaretich conjecture) if $E_1$ and $E_2$ are elliptic curves over $k$ satisfying certain independence conditions then one can find a quadratic extension in which the rank of $E_1$ stays the same and the rank of $E_2$ goes up. Applying this construction successively one can construct an infinite extension $K/k$ such that the rank of $E_1(K)$ is finite and the rank of $E_2(K)$ is infinite.

Yonatan Harpaz
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    Thanks for this. It’s what I would have expected, but what do I know? – Lubin Aug 06 '15 at 13:17
  • @YonatanHarpaz This sounds amazing! Can you point out some properties of the arising field like how many rational primes ramify there, or what is the Galois group over $\mathbb{Q}.$ – Pablo Aug 06 '15 at 14:31
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    It seems that this kind of construction should even work unconditionally, replacing Tate-Shafarevich by the Heegner construction (for positive rank) and Kolyvagin etc. or even an explicit 2-descent (for rank zero). – Noam D. Elkies Aug 06 '15 at 14:59
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    to Pablo's question: the field will be the compositum of infinitely many quadratic extensions, and thus will have infinitely many ramified prime and a Galois group that's abelian of exponent 2. – Noam D. Elkies Aug 06 '15 at 15:06
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    @NoamD.Elkies Please note that if we confine ourselves to compistums of quadratic extensions, then by a 1974 result of Frey and Jarden it is not possible for all primes to ramify (they prove that an elliptic curve over the compositum of all quadratic extensions of $\mathbb{Q}$ has infinite rank). In light of that I conjecture that if many primes ramify (say with respect to some density) then all elliptic curves should have rank $\infty$. How big can you make the set primes ramifying here? Some related questions are: – Pablo Aug 06 '15 at 15:20
  • http://mathoverflow.net/questions/213033/an-elliptic-curve-trivial-over-any-extension-unramified-outside-7-and-infinity http://mathoverflow.net/questions/212906/is-the-compositum-of-all-quadratic-extensions-of-the-rationals-an-ample-field – Pablo Aug 06 '15 at 15:22
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If you're willing to replace $\mathbb Q$ by a finite (quadratic) extension $k$, then my recollection is that there are $\mathbb Z_p$ extensions $K$ of $k$ such that some elliptic curves over $k$ have finite Mordell-Weil group over $K$ and some have infinite rank. In any case, it would be worth looking up what's known about rank growth in $\mathbb Z_p$-extensions, including for $k=\mathbb Q$, starting with the work of Mazur.

Joe Silverman
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  • It follows from CFT over $\mathbb{Q}$ that there is a unique $\mathbb{Z}_p$-extension for every prime number $p$. Furthermore, this extension is ramified at only $p$ which implies that every elliptic curve over it has finite rank (see Theorem 1.2 in http://alozano.clas.uconn.edu/wp-content/uploads/sites/490/2014/01/lozano-robledo_abelian_revised_web.pdf). – Pablo Aug 06 '15 at 14:22
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    @Pablo Thanks. I of course did know that $\mathbb Q$ has a unique $\mathbb Z_p$ extension for every $p$, but I didn't remember what happens for elliptic curves over such towers. But in any case, that's why I suggested looking at the families of $\mathbb Z_p$ extensions available over a quadratic field. There are certainly results about ranks up the anti-cycoltomic tower, but I don't know offhand what happens with the other choices of towers. – Joe Silverman Aug 06 '15 at 14:49
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    If this is of interest to you, I have found the following http://annals.math.princeton.edu/wp-content/uploads/annals-v162-n1-p01.pdf which states (Corollary 5) that if $K$ is imaginary quadratic then all elliptic curves have finite rank over an anti-cyclotomic tower. – Pablo Aug 06 '15 at 15:26
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    @Pablo: what Joe writes works well. Let $M$ be an imaginary quadratic extension of ${\mathbb Q}$ and $K$ the anticyclotomic ${\mathbb Z}_p$-extension of $M$. Then if $E_1$ has say square-free conductor with an odd number of prime divisors inert in $M$, then $E_1(K)$ will be finitely generated (as in the article you quoted). But if we take another elliptic curve $E_2$ whose square-free conductor has an even number of primes factors which are inert in $M$, then $E_2(K)$ will be infinitely generated (as there will be Heegner points present up the anticyclotomic tower). – sibilant Aug 10 '15 at 04:08
  • @RobertPollack how many rational primes ramify in $K$? – Pablo Aug 10 '15 at 08:11
  • @Pablo: I was implicitly assuming none --- that the conductor of the curve in both cases was relatively prime to the discriminant of M. Probably the ramified case can be handled as well, but I don't know the details of that. – sibilant Aug 10 '15 at 14:48
  • @RobertPollack I am asking the following: Let $M$ be an imaginary quadratic extension of $\mathbb{Q}$, let $p$ be a prime number, and let $K$ be the anticyclotomic $\mathbb{Z}_p$-extension of $M$. What prime numbers ramify in $K$? Are there infinitely many of them? Thanks! – Pablo Aug 10 '15 at 15:16
  • @Pablo: Sorry, I was confusing $K$ and $M$. In a ${\mathbf Z}_p$-extension, the only primes that can ramify are the ones over $p$. – sibilant Aug 13 '15 at 01:36
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Are you thinking of the case $K=\overline{\mathbb{Q}}$? If $K$ is a finite extension, then the rank is always finite. And I think that for $K=\overline{\mathbb{Q}}$ it is always infinite. Maybe I'm missing something...

loglog
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    The question is whether or not the finite generation of $E(K)$, as $K \subset \bar{\mathbb{Q}}$ is fixed but $E$ varies, is an intrinsic property of the field $K$. This is so for instance if $K$ has finite degree over $\mathbb{Q}$, or if $K = \bar{\mathbb{Q}}$, or if $K$ is the maximal totally real or totally $p$-adic extension of $\mathbb{Q}$. – Vesselin Dimitrov Aug 06 '15 at 12:34