0
$\begingroup$

Let $C$ be a smooth, projective, geometrically irreducible curve of genus at least $2$ over a complete discrete valued field $F$ of characteristic zero (not necessarily algebraically closed). Let $R$ be the ring of integers of $F$. Assume that the residue field of $R$ is algebraically closed. Does there exist a model of $C$ over $R$ such that the special fiber is irreducible with at worst nodal singularities?

I know that by Deligne-Mumford, there always exists a semi-stable model, which only guarentees a reducible curve with at worst nodal singularities, but I am asking if we can do better?

|cite|improve this question
$\endgroup$
  • $\begingroup$ No. If you have a model where the special fiber is stable and reducible (for instance, the union of 2 smooth curves meeting transversally, both of genus $>0$), you cannot find a semi-stable model with an irreducible fiber. $\endgroup$ – abx Jul 2 '15 at 20:51
1
$\begingroup$

By Lemma 1.12 in the Deligne-Mumford paper, the answer is no. Any isomorphism of generic fibers extends uniquely to an isomorphism over $R$. This is one way of saying that the stack $\mathcal{M}_g$ of genus $g$ stable curves is proper.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ note that the uniqueness requires stability, which includes a condition on not having certain irreducible components. However I guess having one irreducible component always implies this $\endgroup$ – Will Sawin Jul 4 '15 at 1:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.