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 $\mathbb{P}^3$?



Existence of smooth surfaces containing a curve
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 $1$. 


