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Let $C$ be a curve in $\mathbb{P}^3$, possibly non-reduced. Assume, there exists a smooth surface in $\mathbb{P}^3$ containing $C$. Is it true that for $d \gg 0$, a generic element of $I_d(C)$ defines a smooth surface in $\mathbb{P}^3$?

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up vote 5 down vote accepted

I hope this is not a homework exercise, but I do not recall this from the standard textbooks.

This is already false for planar double lines. Let $\mathbb{P}^3$ have homogeneous coordinates $[T_0,T_1,T_2,T_3]$. Let $S$ be the plane $Z(T_3)$. Let $C$ be the curve $Z(T_2^2,T_3)$ with induced reduced curve $L=Z(T_2,T_3)$. Let $R$ be a smooth hypersurface of degree $d$ in $\mathbb{P}^3$ that contains $L$. Since $C$ is a local complete intersection scheme, if $R$ contains $C$, then $C$ is a Cartier divisor on $R$. Thus $[C]$ equals $2[L]$ as an effective Cartier divisor on $R$. In particular, the kernel $\mathcal{J}$ of the reduction homomorphism, $$ \mathcal{O}_C \to \mathcal{O}_L, $$ equals $\mathcal{O}_R(-L)|_L$, i.e., the dual of the normal sheaf of $L$ in $R$ (the conormal sheaf).

Considering the special case when $R$ equals $S$, we compute that $\mathcal{J}$ is isomorphic to $\mathcal{O}_L(-1)$. On the other hand, by adjunction, Chern class computations, etc., for general $R$ of degree $d$ that contains $L$, $\mathcal{O}_R(-L)|_L$ is isomorphic to $\mathcal{O}_L(2-d)$. Thus, if $R$ contains $C$, then $d$ must equal $1$.

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EDIT. The first version of this answer claimed that the result was true for any $C$, but it was uncorrect as remarked by J. Starr. In fact, Jason's answer shows that the result may be false for a non-reduced $C$.

However, the claim is true when $C$ is reduced; let me give a sketch of the proof.

Since $\mathcal{O}_{\mathbb{P}^3}(1)$ is ample, there exists $d_0 \in \mathbb{N}$ such that for any $d \geq d_0$ the sheaf $\mathcal{I}_C(d)$ is generated by its global sections. Hence the base locus of $|\mathcal{I}_C(d)|$ consists of the curve $C$ only. In particular, the linear system $|\mathcal{I}_C(d)|$ has not fixed components, so Bertini theorem implies that

the general element of the linear system $|\mathcal{I}_C(d)|$ is smooth outside $C$. $(*)$

Let now $S$ be a smooth surface of degree $n$ containing $C$ (we may assume $d_0 > n$) and consider the elements in $|\mathcal{I}_C(d)|$ of the form $$X=S +H_1+H_2+ \cdots + H_{d-n},$$ where the $H_i$ are hyperplanes.

If $p$ is any point of $C$ and we choose as $H_i$ hyperplanes not containing $p$, then $X$ is a surface of degree $d$ which is smooth at $p$.

Since $p \in C$ is arbitrary, by using $(*)$ it follows that the general element of $|\mathcal{I}_C(d)|$ is smooth everywhere for $d \geq d_0$.

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Your argument is incorrect. Let $S$ be a linear plane, and let $C$ be a planar curve of very high degree. By Bezout's theorem, every quadric surface that contains $C$ must also contain the plane $S$. Bertini's theorem only says that the general member of the linear system is smooth away from the base locus. However, $S$ is in the base locus of the system of surfaces obtained as $S+H$ by varying $H$. –  Jason Starr Apr 2 '13 at 17:30
    
Right, this argument must be fixed. Unfortunately I have no time now, I will do it later. Thank you! –  Francesco Polizzi Apr 2 '13 at 18:15
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@Olivier: it seems to me that the argument proving that the base locus of $I_C(d)$ consists of $C$ only (for $d >>0$) assumes that $C$ is considered with its reduced structure, and that it cannot be carried out if instead $C$ has some nilpotent structure (as Jason's example shows). I'm missing something? –  Francesco Polizzi Apr 3 '13 at 8:52
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I don't follow either. In Jason's example, the singularities of $R$ lie on the base locus (and for every $d>1$, the base locus of $\mathcal{I}_C(d)$ is just $C$). But these singularities move with $R$. That is, for every $p\in C$ there is indeed a surface smooth at $p$ and containing $C$, but it must be singular at some other point $p\in C$. –  quim Apr 3 '13 at 11:47
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Rather, the other point must be $q\in C$ ;) –  quim Apr 3 '13 at 11:48
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