Canonical divisor of a curve base point free (if g>0) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T06:34:01Z http://mathoverflow.net/feeds/question/80988 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80988/canonical-divisor-of-a-curve-base-point-free-if-g0 Canonical divisor of a curve base point free (if g>0) GiulioP 2011-11-15T15:42:17Z 2011-11-15T15:42:17Z <p>Is there a way to prove that the canonical divisor $W$ of an algebraic function field in one variable $F$ over a field $K$ (that is the function field of an algebraic curve) of genus $g>0$ is base point free, <em>without</em> using Clifford´s theorem?<br> Note that K is not necessarily algebraically closed! in that case I know how to solve the problem. </p> <p>Equivalently, I have to show that for any place $P$ of $F/K$ there exists an holomorphic differential $\omega$ of $F/K$ such that $v_P(\omega)=0$, that is the support of the associated canonical divisor $(\omega)$ to $\omega$ does not contain the place $P$.</p>