# Stabilization of semistable curves in a concrete case

Some days ago I learned for the first time about the stabilization of a semistable curve (from Knudsen's article), but I am still quite confused. If $C/k$ is a semistable curve (i.e. we allow rational components with two nodes, but not less) we blow down all the unstable components (in whatever order) and get a stable curve $C'$. This operation may be performed in families: consider $\omega=\omega_{X/S}$ the canonical sheaf of $\pi:X\to S$ (a semistable curve over $S$), then $\pi^*\pi_*\omega$ surjects onto $\omega$ and the map $X\to \mathbb{P}(\pi_*\omega^n)$ where $n=3$ or greater (other references need $n=4$ for some reason).

I am having trouble understanding this map, apart from the case where $S=\operatorname{Spec} k$ in which case I just forget about the sheaves and see it in the intuitive way.

Let us consider the next simplest case: take $S=\operatorname{Spec}k[\epsilon]$ and $X/S$ such that its geometric fiber is $Y\cup Z$ where $Y$ is of genus $1$, $Z$ is of genus $0$ and $Y\cap Z=\{p_1,p_2\}$ such that $$\widehat{\mathcal{O}}_{p_i}\cong k[\epsilon][\![x,y]\!]/(xy-a_i\epsilon), a_i\in k$$

Concrete question: what is $\widehat{\mathcal{O}}_p$, where $p$ is the unique node of the stabilization of $X/S$ (whose inverse image is $Z$)?

My computations tell me that it should $k[\epsilon][\![x,y]\!]/(xy)$, but are not rigorous (I could elaborate on this, but I believe they do not make enough sense), independently from the $a_i$s. Is it right? If yes, I find it difficult to interpret it, can someone explain why every infinitesimal structure is collapsed?

I gave an example just to be specific, but if you have more general suggestions (behaviour at the nodes in more general families, for instance), of course they are very well welcome.

## 1 Answer

Consider the total space $S$ of your family over $k[\epsilon]$. Note that $S$ is a smooth surface, provided $a_i \ne 0$ (it could be singular at the singular points of $C$, but the first order conditions ensure that it is smooth). The normal bundle of $C$ in $S$ is trivial, hence $$O_Z(Z) = O_Z(-Y) = O_Z(-2).$$ Thus $Z$ is a smooth $(-2)$-curve on a smooth surface $S$. Contracting it you get a surface $S'$ with an ordinary double point. The curve $C'$ is a divisor on $S'$ passing through the singularity, hence it has an ordinary double point too. Thus, around this point it looks as $xy = \epsilon^2$. In particular, if you mod out $\epsilon^2$, you get $xy = 0$.