most recent 30 from http://mathoverflow.net 2013-05-22T22:33:54Z http://mathoverflow.net/feeds/question/95819 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95819/rational-points-on-smooth-compactifications Wanderer 2012-05-03T00:18:56Z 2012-05-03T19:19:37Z

<p>Let $X$ be as smooth variety over a field $k$ of characteristic $0$.</p> <p>Consider the following statements:</p> <ul> <li>The variety $X$ has no $k((t))$-rational points.</li> <li>No smooth compactification of $X$ has a $k$-rational point.</li> </ul> <p>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 $k$.</p>

http://mathoverflow.net/questions/95819/rational-points-on-smooth-compactifications/95824#95824 ulrich 2012-05-03T03:35:57Z 2012-05-03T03:35:57Z

<p>Yes, this is true.</p> <p>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.</p> <p>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 $X$.</p>