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?



line bundle on a nodal curve coming degeneration of smooth ones
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 $X$. 


