Hello, Suppose $A$ is an Abelian variety of dimension $g$ over a number field $k$. Then using height functions one can show that there are nontorsion points in $A(\bar k)$. This looks like an overkill. Is there an easy, elementary way to see this? Thanks! Ramin

2$\begingroup$ I like this question! It's a good way of making the point that countable unions of closed subschemes are weird. I blogged about this and related problems: quomodocumque.wordpress.com/2008/12/15/… (note  that post is out of date.) $\endgroup$– JSEFeb 19, 2011 at 14:39

$\begingroup$ Just read the blog post. Incidentally, this question came up in Davesh's colloquium talk yesterday. He and I both thought that the formal group argument that Jared suggests is a good natural way to do this, except that formal groups are not exactly elementary, and that there are no formal groups for general algebraic dynamical systems (the latter remark is due to A. Medvedev). $\endgroup$– RaminFeb 19, 2011 at 15:59

1$\begingroup$ Is it right to say that SGP's padic log argument and Jared's Silvermanstyle argument and the MaulikPoonenVoisin argument are all, in some sense, cousins? I think all show, in some sense, that the torsion points are nowhere dense adelically. $\endgroup$– JSEFeb 19, 2011 at 23:56

$\begingroup$ Maybe there is some kind of weak approximation lurking in the background? $\endgroup$– RaminFeb 20, 2011 at 3:24
4 Answers
Let's take a page from Silverman's book, VII.3. Let $\mathfrak{p}$ be one of the primes of good reduction of $A$. Let $K/k$ be any extension, and let $\mathfrak{P}$ be a prime of $K$ above $\mathfrak{p}$. The reduction map $A(K)\to A(\mathcal{O}_K/\mathfrak{P})$ becomes injective when you restrict to torsion points of order prime to the residue characteristic of $\mathfrak{p}$  this is proved using an appeal to formal groups.
Now choose two such primes $\mathfrak{p}$ and $\mathfrak{p}'$ with distinct residue characteristics. Convince yourself that there exists a $K/k$ and primes $\mathfrak{P},\mathfrak{P}'$ of $K$ for which $A(K)\to A(\mathcal{O}_K/\mathfrak{P})\times A(\mathcal{O}_K/\mathfrak{P}')$ has nontrivial kernel. Nontrivial points in the kernel must not be torsion.

$\begingroup$ @Jared: There is also a schemetheoretic proof of the injectivity (see for example exercise C.10 on page 294 of HindrySilverman). $\endgroup$– RaminFeb 20, 2011 at 2:41
An argument due to T. Saito goes as follows: Let p be a prime of good reduction for the abelian variety $A$ over a number field $K$. Consider the padic logarithm on $A(\bar{K_p})$; this vanishes precisely on the torsion points. Since $A(\bar{K})$ is dense in $A(\bar{K}_p)$ and the padic logarithm is not identically zero, $A(\bar{K})$ contains nontorsion points.

2$\begingroup$ As a fan of the $p$adic logarithm, I find this argument particularly appealing. $\endgroup$– LubinFeb 19, 2011 at 18:27

Let $l$ be the field cut out by the action of the Galois group on all the torsion points of $A$, and let $\mathfrak{g} = \operatorname{Gal}(l/k)$. Then $\mathfrak{g}$ is a closed subgroup of $\operatorname{GL}_{2 \operatorname{dim} A}(\widehat{\mathbb{Z}})$. It is relatively easy to see that $\mathfrak{g}$ is much smaller than the full absolute Galois group of $\mathbb{Q}$  for instance, I believe basic group theory shows that there are only finitely many $n$ for which the symmetric group $S_n$ can occur as a quotient of $\mathfrak{g}$ (please let me know if I am wrong or if this turns out to be hard to show). On the other hand by Hilbert Irreducibility we have for each $n$ a Galois extension $k_n/k$ with Galois group $S_n$.
Take an affine open subset $A^{\circ}$ of $A$ and by Noether Normalization choose a finite $k$morphism $\varphi: A^{\circ} \rightarrow \mathbb{A}^n$. Let $P$ be a point on $\mathbb{A}^n$ whose coordinates generate the field $k_n$, and let $P' \in A^{\circ}(\overline{k})$ be any point with $\varphi(P') = P$. Then $k(P') \supset k_n$. So $P'$ does not lie in $A(l)$ and is thus a nontorsion point.
If $A = E$ is an elliptic curve, you can choose $\varphi$ just to be the $x$coordinate function, and one should be able to use this argument to construct an explicit nontorsion point on $E(\overline{k})$.
Added: now let $k$ be any field which is not algebraic over a finite field. Then if $A$ is an abelian variety defined over $k$ it is also defined over a subfield $k_0$ which is finitely generated either over $\mathbb{Q}$ or over $\mathbb{F}_p(t)$. In particular the field $k_0$ is Hilbertian, and the above argument goes through to show that $A(\overline{k_0})$  and hence also $A(\overline{k})$  has nontorsion points. This is the best possible result, since if $k$ is algebraic over a finite field, $A(\overline{k}) = A(\overline{k})[\operatorname{tors}]$.

$\begingroup$ @Ramin: no worries. (By the way, I believe that Jared's answer can also be adapted to prove the "general case" treated in my answer.) $\endgroup$ Feb 21, 2011 at 0:53
Jan Denef once pointed out to me that this is a simple consequence of the ManinMumford Conjecture, i.e., Raynaud's Theorem that if $A$ is an Abelian variety defined over a number field and $C$ is a curve on $A$ that is not a coset of an abelian subvariety then $C$ contains only finitely many torsion points.
In this problem we may assume that $A$ has no proper abelian subvarieties. If $A$ has dimension at least 2, take $C$ any curve on $A$ defined over $\bar k$, then $C(\bar k)$ is infinite, but contains only finitely many torsion points. If $A$ is an elliptic curve, let $C$ be any curve of genus at least 2 on $A\times A$. Again, $C(\bar k)$ contains only finitely many torsion points.

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