Lemma 87.7.1. Let $S$ be a scheme. Let $Y$ be a Noetherian integral algebraic space over $S$. Assume there exists an alteration $f : X \to Y$ with $X$ regular. Then the normalization $Y^\nu \to Y$ is finite and $Y$ has a dense open which is regular.

## 87.7 Implied properties

In this section we prove that for a Noetherian integral algebraic space the existence of a regular alteration has quite a few consequences. This section should be skipped by those not interested in “bad” Noetherian algebraic spaces.

**Proof.**
By étale localization, it suffices to prove this when $Y = \mathop{\mathrm{Spec}}(A)$ where $A$ is a Noetherian domain. Let $B$ be the integral closure of $A$ in its fraction field. Set $C = \Gamma (X, \mathcal{O}_ X)$. By Cohomology of Spaces, Lemma 67.20.2 we see that $C$ is a finite $A$-module. As $X$ is normal (Properties of Spaces, Lemma 64.25.4) we see that $C$ is normal domain (Spaces over Fields, Lemma 70.4.6). Thus $B \subset C$ and we conclude that $B$ is finite over $A$ as $A$ is Noetherian.

There exists a nonempty open $V \subset Y$ such that $f^{-1}V \to V$ is finite, see Spaces over Fields, Definition 70.8.3. After shrinking $V$ we may assume that $f^{-1}V \to V$ is flat (Morphisms of Spaces, Proposition 65.32.1). Thus $f^{-1}V \to V$ is faithfully flat. Then $V$ is regular by Algebra, Lemma 10.162.4. $\square$

Lemma 87.7.2. Let $(A, \mathfrak m, \kappa )$ be a local Noetherian domain. Assume there exists an alteration $f : X \to \mathop{\mathrm{Spec}}(A)$ with $X$ regular. Then

there exists a nonzero $f \in A$ such that $A_ f$ is regular,

the integral closure $B$ of $A$ in its fraction field is finite over $A$,

the $\mathfrak m$-adic completion of $B$ is a normal ring, i.e., the completions of $B$ at its maximal ideals are normal domains, and

the generic formal fibre of $A$ is regular.

**Proof.**
Parts (1) and (2) follow from Lemma 87.7.1. We have to redo part of the proof of that lemma in order to set up notation for the proof of (3). Set $C = \Gamma (X, \mathcal{O}_ X)$. By Cohomology of Spaces, Lemma 67.20.2 we see that $C$ is a finite $A$-module. As $X$ is normal (Properties of Spaces, Lemma 64.25.4) we see that $C$ is normal domain (Spaces over Fields, Lemma 70.4.6). Thus $B \subset C$ and we conclude that $B$ is finite over $A$ as $A$ is Noetherian. By Resolution of Surfaces, Lemma 54.13.2 in order to prove (3) it suffices to show that the $\mathfrak m$-adic completion $C^\wedge $ is normal.

By Algebra, Lemma 10.96.8 the completion $C^\wedge $ is the product of the completions of $C$ at the prime ideals of $C$ lying over $\mathfrak m$. There are finitely many of these and these are the maximal ideals $\mathfrak m_1, \ldots , \mathfrak m_ r$ of $C$. (The corresponding result for $B$ explains the final statement of the lemma.) Thus replacing $A$ by $C_{\mathfrak m_ i}$ and $X$ by $X_ i = X \times _{\mathop{\mathrm{Spec}}(C)} \mathop{\mathrm{Spec}}(C_{\mathfrak m_ i})$ we reduce to the case discussed in the next paragraph. (Note that $\Gamma (X_ i, \mathcal{O}) = C_{\mathfrak m_ i}$ by Cohomology of Spaces, Lemma 67.11.2.)

Here $A$ is a Noetherian local normal domain and $f : X \to \mathop{\mathrm{Spec}}(A)$ is a regular alteration with $\Gamma (X, \mathcal{O}_ X) = A$. We have to show that the completion $A^\wedge $ of $A$ is a normal domain. By Lemma 87.6.2 $Y = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A^\wedge )$ is regular. Since $\Gamma (Y, \mathcal{O}_ Y) = A^\wedge $ by Cohomology of Spaces, Lemma 67.11.2. We conclude that $A^\wedge $ is normal as before. Namely, $Y$ is normal by Properties of Spaces, Lemma 64.25.4. It is connected because $\Gamma (Y, \mathcal{O}_ Y) = A^\wedge $ is local. Hence $Y$ is normal and integral (as connected and normal implies integral for separated algebraic spaces). Thus $\Gamma (Y, \mathcal{O}_ Y) = A^\wedge $ is a normal domain by Spaces over Fields, Lemma 70.4.6. This proves (3).

Proof of (4). Let $\eta \in \mathop{\mathrm{Spec}}(A)$ denote the generic point and denote by a subscript $\eta $ the base change to $\eta $. Since $f$ is an alteration, the scheme $X_\eta $ is finite and faithfully flat over $\eta $. Since $Y = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A^\wedge )$ is regular by Lemma 87.6.2 we see that $Y_\eta $ is regular (as a limit of opens in $Y$). Then $Y_\eta \to \mathop{\mathrm{Spec}}(A^\wedge \otimes _ A \kappa (\eta ))$ is finite faithfully flat onto the generic formal fibre. We conclude by Algebra, Lemma 10.162.4. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)