Over which fields does the Mordell-Weil theorem hold? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T03:29:03Z http://mathoverflow.net/feeds/question/108543 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108543/over-which-fields-does-the-mordell-weil-theorem-hold Over which fields does the Mordell-Weil theorem hold? François Brunault 2012-10-01T13:01:11Z 2012-11-02T11:05:34Z <p>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.</p> <p>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 <a href="http://math.stanford.edu/~conrad/papers/Kktrace.pdf" rel="nofollow">Brian Conrad's article</a> for the precise statement and a proof of this theorem).</p> <p>Q1. Are there other fields $K$ for which the group of $K$-rational points of an abelian variety over $K$ is always finitely generated?</p> <p>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).</p> <p>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?</p> http://mathoverflow.net/questions/108543/over-which-fields-does-the-mordell-weil-theorem-hold/108548#108548 Answer by Felipe Voloch for Over which fields does the Mordell-Weil theorem hold? Felipe Voloch 2012-10-01T14:35:58Z 2012-10-01T19:27:14Z <p>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 $Jac(C_n)$. </p> <p>EDIT: As pointed out by Will in the comment below, this only works if $A$ is defined over $k$. </p> <p>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. </p> http://mathoverflow.net/questions/108543/over-which-fields-does-the-mordell-weil-theorem-hold/111024#111024 Answer by Pete L. Clark for Over which fields does the Mordell-Weil theorem hold? Pete L. Clark 2012-10-29T19:37:31Z 2012-10-30T03:18:51Z <p>This is an attempt at a relatively mild generalization of what others have said:</p> <p>Let $K$ be a field and $|\cdot|: K \rightarrow \mathbb{R}$ be a nontrivial absolute value on $K$. </p> <p>$\bullet$ If $K$ is <strong>complete</strong> 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.</p> <p>$\bullet$ When $|\cdot|$ comes from a rank one valuation $v$, I suspect that even if $K$ is merely <strong>Henselian</strong> for $v$, then $E(K)$ cannot be finitely generated. </p> <p>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 </p> <p>$E(K) \supset E^0(K) \supset E^1(K) \supset E^2(K) \supset \ldots$ </p> <p>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 <em>Lectures on Elliptic Curves</em> 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.</p> http://mathoverflow.net/questions/108543/over-which-fields-does-the-mordell-weil-theorem-hold/111067#111067 Answer by Laurent Moret-Bailly for Over which fields does the Mordell-Weil theorem hold? Laurent Moret-Bailly 2012-10-30T10:28:48Z 2012-11-02T11:05:34Z <p>[Edited November 2 for brevity] </p> <p>(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 <em>purely transcendental</em> 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. </p> <p>(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. </p> <p>(3) A general result on Q2: Say a field $K$ is <em>fertile</em> if for every smooth irreducible $K$-variety $X$, if $X(K)$ is nonempty, then it is Zariski-dense.<br> (Pop, who invented the concept, called these fields "large"; others say "ample"). </p> <p>I claim that <strong>every fertile field $K$ satisfies Q2</strong>. This includes in particular: </p> <p>(3a) all Henselian valued fields (already mentioned by Pete, but there is no restriction on the rank here, except the valuation must be nontrivial). </p> <p>(3b) Pseudo-algebraically closed fields (i.e. such that every geometrically irreducible variety has a rational point). This includes example (2) above. </p> <p><strong>Proof of claim</strong>: 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 <code>$(t_1:\dots:t_g):A\dots\to\mathbb{P}^{g-1}_K$</code>. It induces a morphism <code>$f:U\smallsetminus\{e\}\to\mathbb{P}^{g-1}_K$</code> 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 <code>$\widetilde{f}:\widetilde{U}\to \mathbb{P}^{g-1}_K$</code> which induces an isomorphism <code>$E\to\mathbb{P}^{g-1}_K$</code>where $E$ is the exceptional divisor. Moreover, $\widetilde{f}$ is smooth along $E$. Shrinking $U$, we may assume $\widetilde{f}$ smooth.<br> For every <code>$y\in\mathbb{P}^{g-1}(K)$</code>, <code>$\widetilde{f}^{-1}(y)$</code> is a smooth curve with a rational point on $E$. Since $K$ is fertile, <code>$\widetilde{f}^{-1}(y)$</code> also has rational points on <code>$U\smallsetminus\{e\}$</code>. Hence <code>$f:U(K)\smallsetminus\{e\}\to\mathbb{P}^{g-1}(K)$</code> is surjective.<br> 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 <code>$f(U(K))\subset\mathbb{P}^{g-1}(K_0)$</code>. This is a contradiction because $K_0\neq K$ (finitely generated fields are not fertile).</p>