Rational points on smooth compactifications - MathOverflow 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 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 Answer by ulrich for Rational points on smooth compactifications 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>