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

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. 


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. 


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, $A(\overline{k}) = A(\overline{k})[\operatorname{tors}]$. 


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. 

