The Stacks project

86.11 Application to modifications

Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. We set $S = \mathop{\mathrm{Spec}}(A)$ and $U = S \setminus V(I)$. In this section we will consider the category

86.11.0.1
\begin{equation} \label{restricted-equation-modification} \left\{ f : X \longrightarrow S \quad \middle | \quad \begin{matrix} X\text{ is an algebraic space} \\ f\text{ is locally of finite type} \\ 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 $S$.

Let $A \to B$ be a homomorphism of Noetherian rings and let $J \subset B$ be an ideal such that $J = \sqrt{I B}$. Then base change along the morphism $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ gives a functor from the category (86.11.0.1) for $A$ to the category (86.11.0.1) for $B$.

Lemma 86.11.1. Let $(A, I)$ be a pair consisting of a Noetherian ring and an ideal $I$. Let $A^\wedge $ be the $I$-adic completion of $A$. Then base change defines an equivalence of categories between the category (86.11.0.1) for $A$ with the category (86.11.0.1) for the completion $A^\wedge $.

Proof. Set $S = \mathop{\mathrm{Spec}}(A)$ as in (86.11.0.1) and $T = V(I)$. Similarly, write $S' = \mathop{\mathrm{Spec}}(A^\wedge )$ and $T' = V(IA^\wedge )$. The morphism $S' \to S$ defines an isomorphism $S'_{/T'} \to S_{/T}$ of formal completions. Let $\mathcal{C}_{S, T}$, $\mathcal{C}_{S_{/T}}$, $\mathcal{C}_{S'_{/T'}}$, and $\mathcal{C}_{S', T'}$ be the corresponding categories as used in (86.10.3.1). By Theorem 86.10.9 (in fact we only need the affine case treated in Lemma 86.10.8) we see that

\[ \mathcal{C}_{S, T} = \mathcal{C}_{S_{/T}} = \mathcal{C}_{S_{/T'}'} = \mathcal{C}_{S', T'} \]

Since $\mathcal{C}_{S, T}$ is the category (86.11.0.1) for $A$ and $\mathcal{C}_{S', T'}$ the category (86.11.0.1) for $A^\wedge $ this proves the lemma. $\square$

Lemma 86.11.2. Notation and assumptions as in Lemma 86.11.1. Let $f : X \to \mathop{\mathrm{Spec}}(A)$ correspond to $g : Y \to \mathop{\mathrm{Spec}}(A^\wedge )$ via the equivalence. Then $f$ is quasi-compact, quasi-separated, separated, proper, finite, and add more here if and only if $g$ is so.

Proof. You can deduce this for the statements quasi-compact, quasi-separated, separated, and proper by using Lemmas 86.10.10 86.10.11, 86.10.12, 86.10.11, and 86.10.13 to translate the corresponding property into a property of the formal completion and using the argument of the proof of Lemma 86.11.1. However, there is a direct argument using fpqc descent as follows. First, note that $\{ U \to \mathop{\mathrm{Spec}}(A), \mathop{\mathrm{Spec}}(A^\wedge ) \to \mathop{\mathrm{Spec}}(A)\} $ is an fpqc covering with $U = \mathop{\mathrm{Spec}}(A) \setminus V(I)$ as before. The base change of $f$ by $U \to \mathop{\mathrm{Spec}}(A)$ is $\text{id}_ U$ by definition of our category (86.11.0.1). Let $P$ be a property of morphisms of algebraic spaces which is fpqc local on the base (Descent on Spaces, Definition 72.9.1) such that $P$ holds for identity morphisms. Then we see that $P$ holds for $f$ if and only if $P$ holds for $g$. This applies to $P$ equal to quasi-compact, quasi-separated, separated, proper, and finite by Descent on Spaces, Lemmas 72.10.1, 72.10.2, 72.10.18, 72.10.19, and 72.10.23. $\square$

Lemma 86.11.3. Let $A \to B$ be a local map of local Noetherian rings such that

  1. $A \to B$ is flat,

  2. $\mathfrak m_ B = \mathfrak m_ A B$, and

  3. $\kappa (\mathfrak m_ A) = \kappa (\mathfrak m_ B)$

(equivalently, $A \to B$ induces an isomorphism on completions, see More on Algebra, Lemma 15.42.9). Then the base change functor from the category (86.11.0.1) for $(A, \mathfrak m_ A)$ to the category (86.11.0.1) for $(B, \mathfrak m_ B)$ is an equivalence.

Proof. This follows immediately from Lemma 86.11.1. $\square$

Lemma 86.11.4. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring. Let $f : X \to S$ be an object of (86.11.0.1). Then there exists a $U$-admissible blowup $S' \to S$ which dominates $X$.

Proof. Special case of More on Morphisms of Spaces, Lemma 74.39.4. $\square$


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