Lemma 55.10.1. Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_ C) = K$ and genus $> 0$. There is a unique minimal model for $C$.
55.10 Uniqueness of the minimal model
If the genus of the generic fibre is positive, then minimal models are unique (Lemma 55.10.1) and consequently have a suitable mapping property (Lemma 55.10.2).
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
\[ X = X_0 \leftarrow X_1 \leftarrow \ldots \leftarrow X_ n = Y_ m \to \ldots \to Y_1 \to Y_0 = Y \]
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. $\square$
Lemma 55.10.2. Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_ C) = K$ and genus $> 0$. Let $X$ be the minimal model for $C$ (Lemma 55.10.1). Let $Y$ be a regular proper model for $C$. Then there is a unique morphism of models $Y \to X$ which is a sequence of contractions of exceptional curves of the first kind.
Proof. The existence and properties of the morphism $X \to Y$ follows immediately from Lemma 55.8.5 and the uniqueness of the minimal model. The morphism $Y \to X$ is unique because $C \subset Y$ is scheme theoretically dense and $X$ is separated (see Morphisms, Lemma 29.7.10). $\square$
Example 55.10.3. If the genus of $C$ is $0$, then minimal models are indeed nonunique. Namely, consider the closed subscheme defined by $T_1T_2 - \pi T_0^2 = 0$. More precisely $X$ is defined as $\text{Proj}(R[T_0, T_1, T_2]/(T_1T_2 - \pi T_0^2))$. Then the special fibre $X_ k$ is a union of two exceptional curves $C_1$, $C_2$ both isomorphic to $\mathbf{P}^1_ k$ (exactly as in Lemma 55.9.11). Projection from $(0 : 1 : 0)$ defines a morphism $X \to \mathbf{P}^1_ R$ contracting $C_2$ and inducing an isomorphism of $C_1$ with the special fiber of $\mathbf{P}^1_ R$. Projection from $(0 : 0 : 1)$ defines a morphism $X \to \mathbf{P}^1_ R$ contracting $C_1$ and inducing an isomorphism of $C_2$ with the special fiber of $\mathbf{P}^1_ R$. More precisely, these morphisms correspond to the graded $R$-algebra maps In Lemma 55.12.4 we will study this phenomenon.
\[ X \subset \mathbf{P}^2_ R \]
\[ R[T_0, T_1] \longrightarrow R[T_0, T_1, T_2]/(T_1T_2 - \pi T_0^2) \longleftarrow R[T_0, T_2] \]
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