most recent 30 from http://mathoverflow.net 2013-05-19T05:44:36Z http://mathoverflow.net/feeds/question/70181 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70181/why-is-it-a-non-basepoint phil 2011-07-12T21:47:31Z 2011-07-12T22:17:31Z

<p>In the proof of the Castelnuovo theorem for curves in $\mathbb{P}^3$ (Hartshorne IV, 6.4.) the following is done: One considers a smooth, complete curve $C$ in the projective space $\mathbb{P}^3$ over an algebraically closed field. The degree of this curve is denoted by $d$. Then one takes a hyperplane section $D=P_1+...+P_d$ of the curve such that $P_1$,..., $P_d$ are different and no three of them are collinear. Then one wants to show that $P_i$ is not a basepoint of the linear system $|nD-P_1-P_2-...-P_{i-1}|$ if $i \leq \mbox{min}(d, 2n+1)$. At this point the following argument is given, which I do not understand: "To show that $P_i$ is not a basepoint it suffices to find a surface of degree $n$ in $\mathbb{P}^3$ that contains $P_1$, $P_2$,..., $P_{i-1}$ but not $P_i$". I do not get why this suffices. What is the reason that $P_i$ is not a basepoint, if there exists such a surface?</p>

http://mathoverflow.net/questions/70181/why-is-it-a-non-basepoint/70183#70183 Jack Huizenga 2011-07-12T21:56:44Z 2011-07-12T21:56:44Z

<p>Every degree $n$ surface $S$ passing through $P_1,\ldots,P_{i-1}$ gives rise to a member of the series $|nD - P_1-\cdots -P_{i-1}|$ by looking at the points cut on $C$ by $S$ that are residual to $P_1,\ldots,P_{i-1}$.</p>