Lemma 55.8.1. Let $V_1 \to V_2$ be a closed immersion of algebraic schemes over $K$. If $X_2$ is a model for $V_2$, then the scheme theoretic image of $V_1 \to X_2$ is a model for $V_1$.

## 55.8 Models

In this chapter $R$ will be a discrete valuation ring and $K$ will be its fraction field. If needed we will denote $\pi \in R$ a uniformizer and $k = R/(\pi )$ its residue field.

Let $V$ be an algebraic $K$-scheme (Varieties, Definition 33.20.1). A *model* for $V$ will mean a flat finite type^{1} morphism $X \to \mathop{\mathrm{Spec}}(R)$ endowed with an isomorphism $V \to X_ K = X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(K)$. We often will identify $V$ and the generic fibre $X_ K$ of $X$ and just write $V = X_ K$. The special fibre is $X_ k = X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(k)$. A *morphism of models $X \to X'$ for $V$* is a morphism $X \to X'$ of schemes over $R$ which induces the identity on $V$.

We will say *$X$ is a proper model of $V$* if $X$ is a model of $V$ and the structure morphism $X \to \mathop{\mathrm{Spec}}(R)$ is proper. Similarly for separated models, smooth models, and add more here. We will say *$X$ is a regular model of $V$* if $X$ is a model of $V$ and $X$ is a regular scheme. Similarly for normal models, reduced models, and add more here.

Let $R \subset R'$ be an extension of discrete valuation rings (More on Algebra, Definition 15.111.1). This induces an extension $K'/K$ of fraction fields. Given an algebraic scheme $V$ over $K$, denote $V'$ the base change $V \times _{\mathop{\mathrm{Spec}}(K)} \mathop{\mathrm{Spec}}(K')$. Then there is a functor

sending $X$ to $X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(R')$.

**Proof.**
Using Morphisms, Lemma 29.6.3 and Example 29.6.4 this boils down to the following algebra statement. Let $A_1$ be a finite type $R$-algebra flat over $R$. Let $A_1 \otimes _ R K \to B_2$ be a surjection. Then $A_2 = A_1 / \mathop{\mathrm{Ker}}(A_1 \to B_2)$ is a finite type $R$-algebra flat over $R$ such that $B_2 = A_2 \otimes _ R K$. We omit the detailed proof; use More on Algebra, Lemma 15.22.11 to prove that $A_2$ is flat.
$\square$

Lemma 55.8.2. Let $X$ be a model of a geometrically normal variety $V$ over $K$. Then the normalization $\nu : X^\nu \to X$ is finite and the base change of $X^\nu $ to the completion $R^\wedge $ is the normalization of the base change of $X$. Moreover, for each $x \in X^\nu $ the completion of $\mathcal{O}_{X^\nu , x}$ is normal.

**Proof.**
Observe that $R^\wedge $ is a discrete valuation ring (More on Algebra, Lemma 15.43.5). Set $Y = X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(R^\wedge )$. Since $R^\wedge $ is a discrete valuation ring, we see that

where $K^\wedge $ is the fraction field of $R^\wedge $. Since $V$ is geometrically normal, we find that this is a normal scheme. Hence the first part of the lemma follows from Resolution of Surfaces, Lemma 54.11.6.

To prove the second part we may assume $X$ and $Y$ are normal (by the first part). If $x$ is in the generic fibre, then $\mathcal{O}_{X, x} = \mathcal{O}_{V, x}$ is a normal local ring essentially of finite type over a field. Such a ring is excellent (More on Algebra, Proposition 15.52.3). If $x$ is a point of the special fibre with image $y \in Y$, then $\mathcal{O}_{X, x}^\wedge = \mathcal{O}_{Y, y}^\wedge $ by Resolution of Surfaces, Lemma 54.11.1. In this case $\mathcal{O}_{Y, y}$ is a excellent normal local domain by the same reference as before as $R^\wedge $ is excellent. If $B$ is a excellent local normal domain, then the completion $B^\wedge $ is normal (as $B \to B^\wedge $ is regular and More on Algebra, Lemma 15.42.2 applies). This finishes the proof. $\square$

Lemma 55.8.3. Let $X$ be a model of a smooth curve $C$ over $K$. Then there exists a resolution of singularities of $X$ and any resolution is a model of $C$.

**Proof.**
We check condition (4) of Lipman's theorem (Resolution of Surfaces, Theorem 54.14.5) hold. This is clear from Lemma 55.8.2 except for the statement that $X^\nu $ has finitely many singular points. To see this we can use that $R$ is J-2 by More on Algebra, Proposition 15.48.7 and hence the nonsingular locus is open in $X^\nu $. Since $X^\nu $ is normal of dimension $\leq 2$, the singular points are closed, hence closedness of the singular locus means there are finitely many of them (as $X$ is quasi-compact). Observe that any resolution of $X$ is a modification of $X$ (Resolution of Surfaces, Definition 54.14.1). This will be an isomorphism over the normal locus of $X$ by Varieties, Lemma 33.17.3. Since the set of normal points includes $C = X_ K$ we conclude any resolution is a model of $C$.
$\square$

Definition 55.8.4. Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_ C) = K$. A *minimal model* will be a regular, proper model $X$ for $C$ such that $X$ does not contain an exceptional curve of the first kind (Resolution of Surfaces, Section 54.16).

Really such a thing should be called a minimal regular proper model or even a relatively minimal regular projective model. But as long as we stick to models over discrete valuation rings (as we will in this chapter), no confusion should arise.

Minimal models always exist (Proposition 55.8.6) and are unique when the genus is $> 0$ (Lemma 55.10.1).

Lemma 55.8.5. Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_ C) = K$. If $X$ is a regular proper model for $C$, then there exists a sequence of morphisms

of proper regular models of $C$, such that each morphism is a contraction of an exceptional curve of the first kind, and such that $X_0$ is a minimal model.

**Proof.**
By Resolution of Surfaces, Lemma 54.16.11 we see that $X$ is projective over $R$. Hence $X$ has an ample invertible sheaf by More on Morphisms, Lemma 37.50.1 (we will use this below). Let $E \subset X$ be an exceptional curve of the first kind. See Resolution of Surfaces, Section 54.16. By Resolution of Surfaces, Lemma 54.16.8 we can contract $E$ by a morphism $X \to X'$ such that $X'$ is regular and is projective over $R$. Clearly, the number of irreducible components of $X'_ k$ is exactly one less than the number of irreducible components of $X_ k$. Thus we can only perform a finite number of these contractions until we obtain a minimal model.
$\square$

Proposition 55.8.6. Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_ C) = K$. A minimal model exists.

**Proof.**
Choose a closed immersion $C \to \mathbf{P}^ n_ K$. Let $X$ be the scheme theoretic image of $C \to \mathbf{P}^ n_ R$. Then $X \to \mathop{\mathrm{Spec}}(R)$ is a projective model of $C$ by Lemma 55.8.1. By Lemma 55.8.3 there exists a resolution of singularities $X' \to X$ and $X'$ is a model for $C$. Then $X' \to \mathop{\mathrm{Spec}}(R)$ is proper as a composition of proper morphisms. Then we may apply Lemma 55.8.5 to obtain a minimal model.
$\square$

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