# Rational points on varieties over local fields

In his expanded lecture notes Rational points on varieties, Bjorn Poonen writes the following:

REMARK 2.5.3: There is an algorithm that, given a local field $k$ of characteristic $0$ and a $k$-variety $X$, decides whether $X(k)$ is nonempty. (...)

I would like to see if this algorithm can be applied to determine the existence of rational points on a very specific family of varieties over $\mathbb{Q}_p$. Unfortunately, I could not find anything related on the web so far. Could you point to some literature on this matter, or explain some of the ideas?

Let us assume that $X$ is smooth and projective for simplicity, given by a number of polynomial equations with coefficients in the ring of integers $\mathcal O$ of $k$. Let $\kappa$ denote the residue class field of $k$. Reduce the equation moduloe the maximal ideal to get the reduced variety $\bar X$. Enumerate the $\kappa$-points of $\bar X$ (this is a finite problem). If $\bar X(\kappa) = \emptyset$, then $X(k)$ is empty. If there is a smooth point $\bar P \in \bar X(\kappa)$ (and $\dim \bar X = \dim X$), then it lifts by Hensel's Lemma to a point on $X(k)$, so $X(k)$ is non-empty. Otherwise, consider the lifts of each point to points modulo the square of the maximal ideal. This comes down to looking at equations $f(\dots, a_i + \pi x_i, \dots)$, where $f$ runs through the equations defining an affine patch of $X$ containing a lift of $\bar P$, $a_i$ are lifts of the coordinates of $\bar P$ and $x_i$ are variables; divide the equations by the highest possible power of $\pi$, which is a generator of the maximal ideal, and reduce mod $\pi$. This gives equations for the reduction mod $\pi$ of a different model of $X$. Smoothness of $X$ guarantees that the process will terminate after finitely many lifting steps.
If $X$ is not smooth, consider its singular locus $X'$ first (recursively). If $X'(k)$ is non-empty, so is $X(k)$. Otherwise, run the previous procedure on $X$. Since $X'(k)$ is empty, all points that lift indefinitely will end up on the smooth part of $X$, which again gives termination.
• Great, thanks very much! I was aware of Hensel's Lemma when $X$ has a smooth integral model over $\mathcal{O}$, but I didn't know it could essentially work in the same way without a smooth integral model. I'll try to run this computation! – thierry stulemeijer Oct 7 '17 at 11:24
• Dear Michael, thanks again for your explanation. Unfortunately, there seems to be a gap in the given algorithm. Indeed, $\bar{X}(\kappa )$ empty only implies that $X(\mathcal{O})$ is empty, but $X(k)$ could still be nonempty. A trivial example of this situation is given by the point $X =$ Spec $\mathbb{Q}_p[T]/(pT-1)$. This phenomenon also arises in the cases I'm interested in. Do you know how to get around that? – thierry stulemeijer Oct 8 '17 at 20:20
• You are right. It does work for projective varieties, though. I'll fix that in the answer. So you could look at the projective closure and switch to another affine patch. There may be problems if the part you add is very singular (which one should be able to remedy by choosing a better compactification); otherwise a $k$-point on the boundary will have $k$-points of $X$ nearby (so when you find a point on the boundary, it is good enough). – Michael Stoll Oct 8 '17 at 20:56