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 $nDP_1P_2...P_{i1}$ 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 $P_1$, $P_2$,..., $P_{i1}$ 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?

$\begingroup$ 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? $\endgroup$ – diverietti Jul 12 '11 at 21:58
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Every degree $n$ surface $S$ passing through $P_1,\ldots,P_{i1}$ gives rise to a member of the series $nD  P_1\cdots P_{i1}$ by looking at the points cut on $C$ by $S$ that are residual to $P_1,\ldots,P_{i1}$.

$\begingroup$ Thank you for your answer. You mean that the points cut on $C$ by $S$ and different from $P_1,..., P_{i1}$ form a divisor linearly equivalent to $nDP_1...P_{i1}$. But can you explain me, why these two are linearly equivalent. I do not get it immediately. $\endgroup$ – phil Jul 12 '11 at 22:49

2$\begingroup$ 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. $\endgroup$ – Emerton Jul 12 '11 at 23:49