Over which fields does the Mordell-Weil theorem hold?

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?

• 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
• 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
• @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
• @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
• 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

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.

• $A$ 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
• @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).

• 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
• @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
• @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.

• 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
• @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