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Fix a Hilbert polynomial $P$ of a non-plane curve in $\mathbb{P}^3$. By a curve we mean a reduced scheme of pure dimension $1$ i.e., it can be reducible but is reduced. Suppose that the degree of these curves is $e$. For a general curve $C$ in $Hilb_P$ (the Hilbert scheme corresponding to $P$) and a general smooth degree $d$ ($d \ge 5$) surface $X$ in $\mathbb{P}^3$ for $d \ge e+2$ containing $C$, should we expect that the dimension of the linear series $|C|$ which is equal to $h^0(\mathcal{O}_X(C))-1$ to be equal to zero? (Intuitively I would expect this because in this case if $C' \subset C$ irreducible then $C'^2<0$ implying $\dim |C'|=0$)

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e=1, d=2 gives a counterexample. – quim Sep 18 '12 at 20:27
Maybe you want to assume the curves are nondegenerate, ie, not contained in planes? – quim Sep 18 '12 at 20:29
The hilbert-space tag didn't fit here, so I removed it. – user5117 Sep 18 '12 at 20:51
@quim: I have edited the question to non-plane curves – Naga Venkata Sep 18 '12 at 20:53
By Bertini (at least if the surface is nonsingular) the moving part of |C| is irreducible, which combined with your statement that the selfintersection of every component is negative would give that there is no moving part. However, there are a couple of things to check: 1) that the general surface containing C is nonsingular, 2) that the components of every element D∈|C| also have negative selfintersection. I don't know how you argue this selfintersection, and how the argument interacts with the choices of general C -> general S -> general D – quim Sep 19 '12 at 9:34

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