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Let $X$ be a nodal curve (over $\mathbb{C}$) consisting of two components $C$ and $E$ where $C$ is hyperelliptic of genus three and $E$ is an elliptic curve. The inverse image of the node on $C$ is not an Weierstrauss point. Now let $L$ be a line bundle on $X$ such that $\deg L|_C =2$ and $h^0(C, L|_C)=2$ and $L|_E = \mathcal{O}(x)$ for some point $x \in E$.

The question is, is there a stable curve $f: \mathcal{X} \to Spec R$ with $R$ a DVR whose residue field is $\mathbb{C}$ and a line bundle $\mathcal{L}$ on $\mathcal{X}$ such that the generic fiber is smooth, and it gives $X$ and $L$ when restricted to the special fiber and $rank\ f_* \mathcal{L} = 2$?

It seems that (from the theory of admissible covers) the answer is YES. But how to argue rigorously?

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I believe you probably want to add the condition that the generic fiber of $\mathcal{X}$ is smooth. – Jason Starr Apr 23 '13 at 15:42
Thanks! I've just updated it. Sorry for my careless... – marker Apr 23 '13 at 19:42
up vote 1 down vote accepted

Let $\pi:\mathcal{X} \to T$ be a versal family of stable curves over $T$ such that the fiber over $t_0\in T$ equals $X$. By unobstructedness of deformations of stable curves, $T$ is smooth of dimension $3(4)-3=9$.

I do not claim that the relative Picard functor is representable, but there a family $(T'\to T,\mathcal{L})$ consisting of a smooth morphism $T'\to T$ of relative dimension $4$, and an invertible sheaf $\mathcal{L}$ on the pullback $\mathcal{X}' = T'\times_T \mathcal{X}$ such that for a point $t'_0\in T'$ mapping to $t_0$, the fiber $\mathcal{L}_{t'_0}$ equals $L$ and such that the pair $(T'\to T,\mathcal{L})$ is versal.

Define $W^2_3(T')\subset T'$ to be the closed subscheme parameterizing points $t'$ such that $\mathcal{L}_{t'}$ has $h^0\geq 2$. As explained in AGCH or Harris-Morrison, $W^2_3(T')$ is a determinantal subset, every component of which has codimension $\leq 4$. In particular, there exists a component $W_i$ of codimension $\leq 4$ in $T'$ containing $t'_0$. Moreover, the intersection of $W_i$ with the fiber over $t_0$ has dimension $1$. Thus, denoting by $Z_i \subset T$ the closure of the image of $W_i$ in $T$, the generic fiber dimension of $W_i$ is $\leq 1$. Thus, $Z_i$ has codimension $\leq 1$ in $T$.

If either $Z_i$ has codimension $0$, or if $Z_i$ is not contained in the boundary of $T$, then the inverse image of the interior in $W_i$ is a nonempty Zariski open subset of $W_i$, hence a dense open subset. Thus, there exists a smooth morphism from a DVR (or even just a normal curve) to $W_i$ whose generic point maps to the interior,i.e., parameterizes an invertible sheaf on a smooth curve, and whose special point maps to $t'_0$, i.e., parameterizes $(X,L)$.

Thus, by way of contradiction, assume that $Z_i$ has codimension $1$ and is contained in the boundary. Then $Z_i$ contains the component $\Delta_{3,1}$ of the boundary. But now just deform the genus $3$ component $C$ to a non-hyperelliptic curve $C_\eta$. There is no $\mathfrak{g}^1_2$ on $C_\eta$. Thus, there is no deformation $L_\eta$ of $L$ to $X_\eta$. This contradiction proves that $(X,L)$ does deform to a $\mathfrak{g}^1_3$ on some smoothing $X_\eta$ of $X$.

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