Rational points over completions of a number field

Let $X$ be a smooth geometrically irreducible $k$-variety over a number field $k$. I do not assume that $X$ has a $k$-point. Is it true that $X$ has $k_v$-points for almost all places $v$ of $k$?

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This ought to be the Lang-Weil estimate to get points over almost all residue fields, then Hensel's lemma. So if the reductions are almost all smooth, where is the obstruction? – Charles Matthews May 13 '11 at 18:09
The reductions are almost all smooth if $X$ is smooth, basically because singularity means that some matrix of partial derivatives has zero det, but this can only happen mod $p$ for finitely many $p$. Note that you do need geom irred though, because e.g. the spectrum of a number field $L$ Galois over $k$ doesn't have $k_v$-points when $v$ doesn't split. – Kevin Buzzard May 13 '11 at 18:32
@Charles: doesn't Lang-Weil say something about points over bigger and bigger finite fields of the same characteristic? Does it also work if you're changing the characteristic? – Kevin Buzzard May 13 '11 at 18:35
@Kevin: Lang-Weil does apply but, if you are bothered by it, apply Deligne instead. – Felipe Voloch May 13 '11 at 19:09
@Felipe Voloch: sorry but, may I ask which result of Deligne? Thanks. – shenghao May 13 '11 at 20:18

Without loss of generality $X$ is affine, so embed it in projective space and apply the Bertini Theorem to conclude that $X$ contains a smooth, geometrically integral affine curve $C^{\circ}$ missing $N$ points from its projective completion $C$. Such a guy will remain smooth modulo $v$ for almost all places of $v$ -- this follows immediately from the Jacobian condition -- and now you are reduced to Weil's theorem: you have a family of smooth affine curves $C^{\circ}$ over finite fields $\mathbb{F}_v$ of fixed genus and missing $N_v \leq N$ points from its projective completion (in fact $N_v = N$ for sufficiently large $v$). Now the Weil bounds tell you that the number of $\mathbb{F}_v$-rational points goes to infinity with $\# \mathbb{F}_v$. Finally, apply Hensel's Lemma.