Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

According to a well-known theorem of Mordell, the group of rational points $E(\mathbf{Q})$ of an elliptic curve $E/\mathbf{Q}$ is finitely generated. Weil generalized this theorem to abelian varieties over number fields.

Less well-known is the following generalization, due I believe to Néron : if $K$ is a field of finite type (that is, finitely generated over its prime field) and $A$ is an abelian variety over $K$, then $A(K)$ is finitely generated. There is an even more general statement, the Lang-Néron theorem, for relative field extensions which are finitely generated (see Brian Conrad's article for the precise statement and a proof of this theorem).

Q1. Are there other fields $K$ for which the group of $K$-rational points of an abelian variety over $K$ is always finitely generated?

In the other direction, there exist fields $K$ for which $A(K)$ is clearly never finitely generated whenever $\operatorname{dim}(A) \geq 1$. For example $K=\mathbf{C}$, in which case it follows from the description af abelian varieties as complex tori. If $K$ is a finite extension of $\mathbf{Q}_p$, then $A(K)$ contains a finite-index subgroup isomorphic to $\mathcal{O}_K^{\operatorname{dim} A}$, so $A(K)$ is again never finitely generated. Other examples I can think of are complete discretely valued fields and algebraically closed fields. Note that we often have the stronger result that $A(K) \otimes \mathbf{Q}$ is infinite-dimensional (except when $K=\overline{\mathbf{F}}_p$, in which case $A(K)$ is a torsion group).

Q2. Are there other fields $K$ for which the group of $K$-rational points of a non-trivial abelian variety over $K$ is never finitely generated?

share|improve this question
1  
For Q2, you could take $K$ to be the maximal unramified extension of ${\bf Q}_p$ (so $K$ is not a finite extension of ${\bf Q}_p$, it is not complete for a discrete valuation, and it is not algebraically closed). –  Chandan Singh Dalawat Oct 1 '12 at 13:29
3  
There is a conjecture a Mazur to the effect that for any number field F and abelian variety A over F, the group $A(F_{p^\infty})$ is finitely generated; here $p$ is a fixed prime number and $F_{p^\infty}$ is the union of the largest $p$-power subextensions of fields of the form $F(\mu_{p^n})$, for all $n\geq 1$. See for instance Lang's "Survey of Diophantine geometry", I, par. 4 (p. 29). –  Damian Rössler Oct 1 '12 at 16:20
1  
@Damian : This is interesting, this means that the field $F_{p^\infty}$ will (conjecturally) answer Q1. This reminds me Kato proved the following theorem (in Astérisque 295) : if $E/\mathbf{Q}$ is an elliptic curve and $m>1$ is any integer, then $E(\mathbf{Q}(\mu_{m^\infty}))$ is finitely generated. More generally, he proves it for any abelian variety which is a quotient of $J_0(N)$, so it applies to any abelian variety over $\mathbf{Q}$ with GL_2-type. –  François Brunault Oct 1 '12 at 17:40
2  
@François: trivial generalizations of your examples for Q2: replace $Q_p$ by any (algebraic extension of) complete discrete valuation fields; replace $\mathbb C$ by any algebraically closed field $K$ (you don't need to use complex uniformization, just observe that $K$ is the algebraic closure of a purely transcendental extension $L$ of its prime field. Then any $A$ over $K$ is defined over a finite extension of $L$. If $A(K)$ was of finite type, then there exists a finite extension $L'$ of $L$, containing the defintion field of $A$, such that $A(K)=A(L')$. This is impossible because $K\ne L'$ –  Qing Liu Oct 3 '12 at 7:14
1  
and there are plenty of $K$-points in $A$ as $K$ is algebraically closed (use a finite surjective morphism from $A$ to $\mathbb P^d$ if you want). –  Qing Liu Oct 3 '12 at 7:15

3 Answers 3

Here is an [INCOMPLETE, POSSIBLY INCORRECT] answer to question 1. Yes. Let $C_n/k,n=1,2,\ldots$ be a sequence of curves of increasing genus defined over a finite field $k$ with maps $C_{n+1} \to C_n$ for all $n$. Let $K = \bigcup k(C_n)$. Assume further that $Jac(C_{n+1})/Jac(C_n)$ is simple for all $n$, where $Jac$ is the Jacobian (this is will be the typical case in such a tower). Then $A(K)$ is finitely generated for any abelian variety $A$, as $A(K) = A(k(C_n))$ where $n$ is the largest integer for which $A$ occurs as a factor of $Jac(C_n)$.

EDIT: As pointed out by Will in the comment below, this only works if $A$ is defined over $k$.

Here is an answer for question 2. Yes. Let $K$ be an infinite subfield of the algebraic closure of a finite field. It follows easily from the Weil bound that $A(K)$ is an infinite torsion group so is not finitely generated.

share|improve this answer
1  
$A$ is a variety over $K$, not a variety over $k$. $Jac(C_n)$ is a variety over $k$. Why should they be the same? –  Will Sawin Oct 1 '12 at 19:14
    
Thanks for your answer. I don't know if the following would work, but what happens with the simpler case $K=\cup_n k(T^{1/n})$? Or to mimick the situation in Mazur's conjecture, one could try to take an inverse system of multiplication-by-$n$ maps on a given elliptic curve $E/k$. –  François Brunault Oct 1 '12 at 21:12
2  
@François: The first field that you suggest has been used by Ulmer to construct elliptic curves with arbitrary large rank over function fields, using that $k(T^{1/n})$ are all abstractly the same field, so that won't work. Don't know about the second one. –  Felipe Voloch Oct 1 '12 at 22:16

[Edited November 2 for brevity]

(1) An extension of the finitely generated case for Q1: let $K_0$ be finitely generated over the prime field, and let $K=K_0((x_i)_{i\in I})$ be a purely transcendental extension of $K_0$. Then $K$ "satisfies Q1". Indeed, any abelian variety $A/K$ is defined over some intermediate $K_1:=K_0((x_i)_{i\in J})$, $J\subset I$ finite. Then $A(K_1)$ is finitely generated, but $A(K)=A(K_1)$ since $K/K_1$ is purely transcendental.

(2) Another "easy" case for Q2: if $K/\mathbb{F}_p$ is infinite algebraic, then for any $A$ the group $A(K)$ is torsion, but must be infinite by Weil's estimates, hence is not finitely generated.

(3) A general result on Q2: Say a field $K$ is fertile if for every smooth irreducible $K$-variety $X$, if $X(K)$ is nonempty, then it is Zariski-dense.
(Pop, who invented the concept, called these fields "large"; others say "ample").

I claim that every fertile field $K$ satisfies Q2. This includes in particular:

(3a) all Henselian valued fields (already mentioned by Pete, but there is no restriction on the rank here, except the valuation must be nontrivial).

(3b) Pseudo-algebraically closed fields (i.e. such that every geometrically irreducible variety has a rational point). This includes example (2) above.

Proof of claim: Let $A$ be an abelian $K$-variety of dimension $g>0$, with origin $e$. We may assume $g\geq2$ (if $g=1$, consider $A\times A$). Let $t_1,\dots,t_g$ be a regular system of parameters at $e$. Consider the rational map $(t_1:\dots:t_g):A\dots\to\mathbb{P}^{g-1}_K$. It induces a morphism $f:U\smallsetminus\{e\}\to\mathbb{P}^{g-1}_K$ where $U\subset A$ is a neighborhood of $e$. Let $\widetilde{U}$ be the blow-up of $e$ in $U$. By the assumption on $t_1,\dots,t_g$, we get a morphism $\widetilde{f}:\widetilde{U}\to \mathbb{P}^{g-1}_K$ which induces an isomorphism $E\to\mathbb{P}^{g-1}_K$where $E$ is the exceptional divisor. Moreover, $\widetilde{f}$ is smooth along $E$. Shrinking $U$, we may assume $\widetilde{f}$ smooth.
For every $y\in\mathbb{P}^{g-1}(K)$, $\widetilde{f}^{-1}(y)$ is a smooth curve with a rational point on $E$. Since $K$ is fertile, $\widetilde{f}^{-1}(y)$ also has rational points on $U\smallsetminus\{e\}$. Hence $f:U(K)\smallsetminus\{e\}\to\mathbb{P}^{g-1}(K)$ is surjective.
On the other hand, if $A(K)$ were finitely generated there would be a finitely generated subfield of definition $K_0\subset K$ for $A$, $U$ and $f$ such that $A(K)=A(K_0)$, which would imply $f(U(K))\subset\mathbb{P}^{g-1}(K_0)$. This is a contradiction because $K_0\neq K$ (finitely generated fields are not fertile).

share|improve this answer
    
I was thinking of the example (1) as well, but I couldn't decide whether $A(K) = A(K_1)$: why is this? –  Pete L. Clark Nov 1 '12 at 1:37
2  
@Pete: enough to prove $A(K_1(x))=A(K_1)$. An element of $A(K_1(x))$ is the same thing as a $K_1$-rational map from the affine line to $A$. These are all constant. –  Laurent Moret-Bailly Nov 1 '12 at 8:08
    
@Laurent: Thanks again. That one at least I should have seen for myself. –  Pete L. Clark Nov 4 '12 at 6:31
    
Thank you very much for your answer and this interesting result that fertile fields satisfy Q2. Looking for references on fertile fields, I found out the recent article "Large implies henselian" by Pop, who proves that fraction fields of henselian rings are fertile. So for example, fields like $K((x_1,\ldots,x_n))$ or more generally fraction fields of $I$-adically complete rings will satisfy Q2. –  François Brunault Nov 6 '12 at 12:40
1  
@François: infinite algebraic extensions of finite fields are counterexamples ($A(K)\otimes\mathbb{Q}$ is zero). But they are a bit too trivial. Whether they are the only ones is a good question. –  Laurent Moret-Bailly Nov 6 '12 at 13:52

This is an attempt at a relatively mild generalization of what others have said:

Let $K$ be a field and $|\cdot|: K \rightarrow \mathbb{R}$ be a nontrivial absolute value on $K$.

$\bullet$ If $K$ is complete for $|\cdot|$, then $E(K)$ has the structure of a $K$-analytic Lie group in the sense of Serre. In particular it is a $K$-analytic manifold so has at least continuum cardinality.

$\bullet$ When $|\cdot|$ comes from a rank one valuation $v$, I suspect that even if $K$ is merely Henselian for $v$, then $E(K)$ cannot be finitely generated.

Here is a proof in the case that the valuation is discrete and the residue field $k$ is infinite: standard arguments involving the formal group still give a filtration

$E(K) \supset E^0(K) \supset E^1(K) \supset E^2(K) \supset \ldots$

such that (by Hensel's Lemma) for all $n \geq 1$, $E^n(K)/E^{n+1}(K) \cong (k,+)$. (Just last night I noticed that Cassels's Lectures on Elliptic Curves has a beautiful, elementary take on this. He works with the case $K = \mathbb{Q}_p$ but the argument holds much more generally.) If $k$ is infinite, then its additive group is not finitely generated and thus $E(K)$, having a subquotient which is not finitely generated, is itself not finitely generated.

share|improve this answer
    
Pete, if I'm understanding correctly, this would be an(other) example of a theorem proved using second-order axioms (completeness) but actually follows from first-order consequences (Henselianness). Is this at least on the right track? –  François G. Dorais Oct 30 '12 at 1:45
    
@François: yes, you are understanding correctly: at least, this is a useful way to view what I am talking about. And indeed I was thinking along these lines myself. A more explicitly model-theoretic question along these lines is: is the class of fields for which the Mordell-Weil Theorem holds for all elliptic curves elementary? (But I suspect that the answer to this is "no"...) –  Pete L. Clark Oct 30 '12 at 3:17
    
@Pete: the answer is indeed no, consider (ultraproducts of) finite fields! –  Laurent Moret-Bailly Oct 30 '12 at 7:23
    
@Laurent: but a (non)principal ultraproduct of finite fields is not elementarily equivalent to a finite field. Indeed, such a field is PAC, whereas finite fields are not. Or am I just not understanding you properly? –  Pete L. Clark Oct 30 '12 at 7:26
2  
@Pete: right, then take an elliptic curve $E$ of rank one (say) over $\mathbb{Q}$. If $K$ is a nontrivial ultrapower of $\mathbb{Q}$, then $E(K)$ contains the corresponding ultrapower of $\mathbb{Z}$, which is not finitely generated. –  Laurent Moret-Bailly Oct 30 '12 at 8:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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