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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 non-torsion points in $A(\bar k)$. This looks like an overkill. Is there an easy, elementary way to see this? Thanks! Ramin

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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 I blogged about this and related problems: quomodocumque.wordpress.com/2008/12/15/… (note -- that post is out of date.) –  JSE Feb 19 '11 at 14:39
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). –  Ramin Feb 19 '11 at 15:59
Is it right to say that SGP's p-adic log argument and Jared's Silverman-style argument and the Maulik-Poonen-Voisin argument are all, in some sense, cousins? I think all show, in some sense, that the torsion points are nowhere dense adelically. –  JSE Feb 19 '11 at 23:56
Maybe there is some kind of weak approximation lurking in the background? –  Ramin Feb 20 '11 at 3:24
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4 Answers

up vote 10 down vote accepted

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.

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@Jared: Thanks! –  Ramin Feb 19 '11 at 15:59
@Jared: There is also a scheme-theoretic proof of the injectivity (see for example exercise C.10 on page 294 of Hindry-Silverman). –  Ramin Feb 20 '11 at 2:41
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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 p-adic 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 p-adic logarithm is not identically zero, $A(\bar{K})$ contains non-torsion points.

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As a fan of the $p$-adic logarithm, I find this argument particularly appealing. –  Lubin Feb 19 '11 at 18:27
@SGP: Thank you! –  Ramin Feb 20 '11 at 2:24
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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, $A(\overline{k}) = A(\overline{k})[\operatorname{tors}]$.

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@Pete: Thank you! –  Ramin Feb 20 '11 at 2:20
Too bad mathoverflow doesn't let me accept more than one answer. –  Ramin Feb 20 '11 at 22:34
@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.) –  Pete L. Clark Feb 21 '11 at 0:53
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Jan Denef once pointed out to me that this is a simple consequence of the Manin-Mumford 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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+1 for most elegant overkill. –  Pete L. Clark Feb 21 '11 at 18:23
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