Lemma 89.2.1. Let $(A, \mathfrak m, \kappa )$ be a $2$-dimensional Noetherian local domain such that $U = \mathop{\mathrm{Spec}}(A) \setminus \{ \mathfrak m\} $ is a normal scheme. Then any modification $f : X \to \mathop{\mathrm{Spec}}(A)$ is a morphism as in (89.2.0.1).
89.2 Modifications
Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring. We set $S = \mathop{\mathrm{Spec}}(A)$ and $U = S \setminus \{ \mathfrak m\} $. In this section we will consider the category
89.2.0.1
\begin{equation} \label{spaces-resolve-equation-modification} \left\{ f : X \longrightarrow S \quad \middle | \quad \begin{matrix} X\text{ is an algebraic space}
\\ f\text{ is a proper morphism}
\\ f^{-1}(U) \to U\text{ is an isomorphism}
\end{matrix} \right\} \end{equation}
A morphism from $X/S$ to $X'/S$ will be a morphism of algebraic spaces $X \to X'$ compatible with the structure morphisms over $S$. In Algebraization of Formal Spaces, Section 88.30 we have seen that this category only depends on the completion of $A$ and we have proven some elementary properties of objects in this category. In this section we specifically study cases where $\dim (A) \leq 2$ or where the dimension of the closed fibre is at most $1$.
Proof. Let $f : X \to S$ be a modification. We have to show that $f^{-1}(U) \to U$ is an isomorphism. Since every closed point $u$ of $U$ has codimension $1$, this follows from Spaces over Fields, Lemma 72.3.3. $\square$
Lemma 89.2.2. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring. Let $g : X \to Y$ be a morphism in the category (89.2.0.1). If the induced morphism $X_\kappa \to Y_\kappa $ of special fibres is a closed immersion, then $g$ is a closed immersion.
Proof. This is a special case of More on Morphisms of Spaces, Lemma 76.49.3. $\square$
Lemma 89.2.3. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local domain of dimension $\geq 1$. Let $f : X \to \mathop{\mathrm{Spec}}(A)$ be a morphism of algebraic spaces. Assume at least one of the following conditions is satisfied
$f$ is a modification (Spaces over Fields, Definition 72.8.1),
$f$ is an alteration (Spaces over Fields, Definition 72.8.3),
$f$ is locally of finite type, quasi-separated, $X$ is integral, and there is exactly one point of $|X|$ mapping to the generic point of $\mathop{\mathrm{Spec}}(A)$,
$f$ is locally of finite type, $X$ is decent, and the points of $|X|$ mapping to the generic point of $\mathop{\mathrm{Spec}}(A)$ are the generic points of irreducible components of $|X|$,
add more here.
Then $\dim (X_\kappa ) \leq \dim (A) - 1$.
Proof. Cases (1), (2), and (3) are special cases of (4). Choose an affine scheme $U = \mathop{\mathrm{Spec}}(B)$ and an étale morphism $U \to X$. The ring map $A \to B$ is of finite type. We have to show that $\dim (U_\kappa ) \leq \dim (A) - 1$. Since $X$ is decent, the generic points of irreducible components of $U$ are the points lying over generic points of irreducible components of $|X|$, see Decent Spaces, Lemma 68.20.1. Hence the fibre of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ over $(0)$ is the (finite) set of minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ r$ of $B$. Thus $A_ f \to B_ f$ is finite for some nonzero $f \in A$ (Algebra, Lemma 10.122.10). We conclude $\kappa (\mathfrak q_ i)$ is a finite extension of the fraction field of $A$. Let $\mathfrak q \subset B$ be a prime lying over $\mathfrak m$. Then
\[ \dim (B_\mathfrak q) = \max \dim ((B/\mathfrak q_ i)_{\mathfrak q}) \leq \dim (A) \]
the inequality by the dimension formula for $A \subset B/\mathfrak q_ i$, see Algebra, Lemma 10.113.1. However, the dimension of $B_\mathfrak q/\mathfrak m B_\mathfrak q$ (which is the local ring of $U_\kappa $ at the corresponding point) is at least one less because the minimal primes $\mathfrak q_ i$ are not in $V(\mathfrak m)$. We conclude by Properties, Lemma 28.10.2. $\square$
Lemma 89.2.4. If $(A, \mathfrak m, \kappa )$ is a complete Noetherian local domain of dimension $2$, then every modification of $\mathop{\mathrm{Spec}}(A)$ is projective over $A$.
Proof. By More on Morphisms of Spaces, Lemma 76.43.6 it suffices to show that the special fibre of any modification $X$ of $\mathop{\mathrm{Spec}}(A)$ has dimension $\leq 1$. This follows from Lemma 89.2.3. $\square$
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