The Stacks project

Proof. We have already proven the hard part of the lemma which is the existence of a minimal model (whose proof relies on resolution of surface singularities), see Proposition 55.8.6. To prove uniqueness, suppose that $X$ and $Y$ are two minimal models. By Resolution of Surfaces, Lemma 54.17.2 there exists a diagram of $S$-morphisms

where each morphism is a blowup in a closed point. The exceptional fibre of the morphism $X_ n \to X_{n - 1}$ is an exceptional curve of the first kind $E$. We claim that $E$ is contracted to a point under the morphism $X_ n = Y_ m \to Y$. If this is true, then $X_ n \to Y$ factors through $X_{n - 1}$ by Resolution of Surfaces, Lemma 54.16.1. In this case the morphism $X_{n - 1} \to Y$ is still a sequence of contractions of exceptional curves by Resolution of Surfaces, Lemma 54.17.1. Hence by induction on $n$ we conclude. (The base case $n = 0$ means that there is a sequence of contractions $X = Y_ m \to \ldots \to Y_1 \to Y_0 = Y$ ending with $Y$. However as $X$ is a minimal model it contains no exceptional curves of the first kind, hence $m = 0$ and $X = Y$.)

Proof of the claim. We will show by induction on $m$ that any exceptional curve of the first kind $E \subset Y_ m$ is mapped to a point by the morphism $Y_ m \to Y$. If $m = 0$ this is clear because $Y$ is a minimal model. If $m > 0$, then either $Y_ m \to Y_{m - 1}$ contracts $E$ (and we're done) or the exceptional fibre $E' \subset Y_ m$ of $Y_ m \to Y_{m - 1}$ is a second exceptional curve of the first kind. Since both $E$ and $E'$ are irreducible components of the special fibre and since $g_ C > 0$ by assumption, we conclude that $E \cap E' = \emptyset $ by Lemma 55.9.11. Then the image of $E$ in $Y_{m - 1}$ is an exceptional curve of the first kind (this is clear because the morphism $Y_ m \to Y_{m - 1}$ is an isomorphism in a neighbourhood of $E$). By induction we see that $Y_{m - 1} \to Y$ contracts this curve and the proof is complete. $\square$