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Let $X$ be as smooth variety over a field $k$ of characteristic $0$.

Consider the following statements:

The variety $X$ has no $k((t))$-rational points.
No smooth compactification of $X$ has a $k$-rational point.

Are these equivalent? If not, what additional assumptions on $X$ would make them equivalent? I'm particularly interested in the case where $X$ is a homogenous space of a "nice" algebraic group over $k$.

edited May 3 2012 at 19:19

asked May 3 2012 at 0:18

Wanderer
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Assume that $X$ is itself smooth and proper. Then the first condition just says that $X$ doesn't contain any rational curves defined over $k$. That seems much stronger than not having $k$-rational points. – Keerthi Madapusi Pera May 3 2012 at 2:09

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@Keerthi Madapusi Pera: I'm guessing you read $k(t)$ for the proposer's $k((t))$. Even for $k(t)$ it's not quite true because any $k$-rational point is automatically $k(t)$-rational (the map from the $t$-line to $X$ can be constant). – Noam D. Elkies May 3 2012 at 4:56

@Noam--Of course. That was very silly of me. – Keerthi Madapusi Pera May 3 2012 at 5:23

ulrich · Accepted Answer · 2012-05-03 03:35:57Z UTC

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Yes, this is true.

One implication is immediate: if $X$ has a $k((t))$ point then by the valuative criterion of properness there is a map $Spec(k[[t]])$ to any compactification of $X$, so the image of the closed point gives a $k$-point of the compactification.

For the converse, if a smooth compactification has a $k$-point then choose a general curve $C$ through that point (which is smooth at that point). Since $C$ is general, it is not contained in the boundary. By completing the local ring of the curve at the smooth point you get a map $Spec(k[[t]])$ to the compactification. Since $C$ is not contained in the boundary, the map retricted to the generic point (which is sent to the generic point of the curve) gives a $k((t))$-point of $X$.

answered May 3 2012 at 3:35

ulrich
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Nice answer. Where do we use the smoothness of $X$ and the fact that $k$ is of characteristic zero? I think the "obvious" implication you mention using the val. crit. holds without these hypotheses. Moreover, I don't see how you use these hypotheses in the second implication either. Also, why does there always exist a "general" curve on $\overline X$? Looking at the dimension 1 case this doesn't seem to be the case (but of course the dimension 1 case could be handled separately). – Harry May 3 2012 at 7:04

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Charateristic zero is used implicitly in order that there exists a smooth compactification of $X$. Smoothness is used to ensure that there exists a curve passing through any $k$-rational point that is smooth at that point. – ulrich May 3 2012 at 7:59

Great answer, thanks a lot. Could you just explain a bit more why the existence of such a curve follows from smoothness? – Wanderer May 3 2012 at 16:28

Let $\dim(X) = n$. For $x$ a $k$-rational point let $f_1,f_2,\dots,f_n$ be generators of the maximal ideal at $x$. Then we get an etale morphism $F$ from an open neighbourhood $U$ of $X$ to $\mathbb{A}^n$ given by $(f_1,f_2,\dots,f_n)$ such that $F(x) = 0$. Let $L$ be any line in $\mathbb{A}^n$ passing through $0$ and not contained in the image of the boundary. The irreducible component of $F^{-1}(L)$ containing $x$ can be taken to be the curve $C$. – ulrich May 4 2012 at 3:13

Rational points on smooth compactifications

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