Let $C$ be a reduced curve in a smooth degree $d$ ($d \ge 5$) surface in $\mathbb{P}^3$. Suppose $C=C_1 \cup ... \cup C_r$ with $C_i$ irreducible and $C_i^2<0$ for all $i$. Then is the linear system $|C|$ base-point free?
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5$\begingroup$ Do you mean $C_i^2>0$ ? If $r=1$, the answer if of course no: any curve in $|C|$ is equivalent to $C$. $\endgroup$– Jérémy BlancCommented Sep 22, 2012 at 11:12
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As Jérémy points out, there are easy examples where $|C|$ is $\textbf{not}$ base point free. Perhaps the OP is asking if it is always the case that $|C|$ is not base point free. However, there are examples where $|C|$ is base point free. For instance, using homogeneous coordinates $x_0,x_1,x_2,x_3$ on $\mathbb{P}^3$, let $L_1,\dots,L_d$ be generic linear polynomials in $x_0,x_1,x_2$. Let $G$ be the degree $d$ polynomial $L_1\cdot \dots \cdot L_d$. Let $C$ be the singular plane curve $Z(G,x_3)$, the union of the $d$ irreducible curves $C_i = Z(L_i,x_3)$. By Bertini's Theorem, for a generic homogeneous polynomial $H$ of degree $d-1$ in $x_0,\dots,x_3$, the polynomial $F = G + x_3H$ is smooth away from the base locus $C$ of the linear system. If $H$ is nonzero at each of the intersection points of $C_i\cap C_j$, $i\neq j$, then $F$ is everywhere smooth. Every curve $C_i$ has negative intersection on the smooth surface $Z(F)$. Yet $C$ is a hyperplane section, $Z(F,x_3)$, hence $|C|$ is base point free.