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?

  • 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 Blanc Sep 22 '12 at 11:12

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.