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 $k$.



Rational points on smooth compactifications
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 $X$.


