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Given a projective curve $C$, is it possible that $C$ can degenerate into union of lines i.e., does there exist a family of curves $\pi:\mathcal{C} \to B$ such that $\pi^{-1}(0)=C$ and there exists $a \in B$ such that $\pi^{-1}(a)$ is the union of lines?

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Yes, the Hessian family of cubics is one example. –  Felipe Voloch Apr 10 '12 at 21:49
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Are you asking whether some projective curve admits such a degeneration, or whether every projective curve does? –  Charles Staats Apr 10 '12 at 22:12
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Take the family of plane conics $xy+tz^2=0$ (homogeneous coordinates $x$, $y$, $z$, parameter $t$). –  Laurent Moret-Bailly Apr 11 '12 at 6:53
    
@charles: I was asking about any projective curve –  Naga Venkata Apr 11 '12 at 15:50

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up vote 10 down vote accepted

Yes, in the following sense. Pick a trivalent graph $G$ with $v$ vertices and regard it as the dual complex of the stable curve $E$ consisting of one copy of $\mathbb P^1$ for each vertex of $G$ and one node for each edge. The genus $g$ of $E$ is given by $2g-2=v$. The stack of stable curves of genus $g$ is irreducible, so any smooth curve $C$ of genus $g$ can be joined to the degenerate curve $E$ (the base $B$ can be taken to be any irreducible variety that dominates the stack; for example, some Hilbert scheme of pluricanonically embedded curves).

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If the question is about degenerations inside a fixed projective space, then the answer in general is NO, at least if one requires that the union of lines be reduced and have only nodes as singularities. A counterexample was found by Hartshorne and is described in [R. Hartshorne: Families of curves in P3 and Zeuthen’s problem. Mem. Amer. Math. Soc. 130 (1997), no. 617].

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If two one-dimensional closed subschemes have the same degree and arithmetic genus, then they lie in the same connected component of the Hilbert scheme. Thus, it looks to me like finding this counterexample amounts to finding a genus $g$ and degree $d$ such that no arrangement of $d$ lines with genus $g$ can be embedded as a closed subscheme of $\mathbb P^3$. (Such an arrangement necessarily exists, as shown in inkspot's answer.) Is this correct? –  Charles Staats Apr 11 '12 at 18:52

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