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Let $O$ be a Dedekind domain with fraction field $K$. Let $C$ be a smooth projective geometrically connected curve of genus $g>1$ over $K$.

Let $p:X \to \mathrm{Spec} \ O $ be the canonical model of $X$. Assuming that $C$ has semi-stable reduction over $O$, we have that $p$ is a stable curve.

When is $X$ regular?

One possible answer would be when the minimal regular model and the canonical model coincide.

But when does this happen? When there aren't any curves to be contracted on the minimal regular model. But I have a feeling this actually never happens...Why?

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up vote 3 down vote accepted

The minimal regular model coincides with the canonical one if and only if there is no vertical irreducible curve $C$ such that $C$ is smooth, rational, of self-intersection $C^2=-2$. When the minimal regular model is semi-stable, the condition $C^2=-2$ is equivalent to $C$ meets the other irreducible components at exactly two points (rational over the field of definition of $C$) or one point (quadratic over the field of definition of $C$).

In the other direction, given the stable model $X$, at any singular point $x$ of a close fiber $X_s$, the local ring $O_{X,x}$ is rather simple. Suppose $x$ is rational over $k(s)$, then $$\widehat{O}_{X,x}\simeq \widehat{O}_{S,s}[[u,v]]/(uv-t^e), $$ where $t$ is a uniformizing element of the DVR $O_{S,s}$. Then $x$ is a regular point of $X$ if and only if $e=1$. Regular points of $X_s$ are regular in $X$.

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Thanks for this. I'm still wondering though. Does it happen a lot that a semi-stable regular model is stable? So as you mention this means that the semi-stable regular model does not contain a vertical irreducible curve C such that C is smooth, rational and of self-intersection $C^2=-2$. I have a feeling that such a curve actually exists "almost always". Is my feeling right or wrong here? – Fozad Nov 11 '11 at 18:44
@Fozad: the problem is how do you qualify "almost always" ? If you have a singular stable model, then you can construct a regular stable model (with a different generic fiber) whose close fiber is isomorphic to the given one. The converse holds true also. In fact, in some sense, regular stable model is the generic situation (in some moduli space, they correspond to sections over the base scheme which cut transversally the boundary). – user18119 Nov 12 '11 at 11:22

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