# Why is it a non-basepoint?

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?

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Didn't you say that $P_i$ does not lie on the surface? Your final question seems to contradict the previous statement... Or I don't understand correctly? –  diverietti Jul 12 '11 at 21:58
Thank you. I've edited it –  phil Jul 12 '11 at 22:17

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 $P_1,\ldots,P_{i-1}$.
Thank you for your answer. You mean that the points cut on $C$ by $S$ and different from $P_1,..., P_{i-1}$ form a divisor linearly equivalent to $nD-P_1-...-P_{i-1}$. But can you explain me, why these two are linearly equivalent. I do not get it immediately. –  phil Jul 12 '11 at 22:49
By the very definition of linear equivalence, $n$ times a hyperplane and $S$ are linearly equivalent in $\mathbb P^3$, and so their intersections with $C$ are linearly equivalent too. Subtracting $P_1,\ldots,P_{i−1}$ from each of these intersections then also gives linearly equivalent divisors. –  Emerton Jul 12 '11 at 23:49