Lemma 115.22.1. Let (A, \mathfrak m, \kappa ) be a Noetherian local ring. The category of Algebraization of Formal Spaces, Equation (88.30.0.1) for A is equivalent to the category Algebraization of Formal Spaces, Equation (88.30.0.1) for the henselization A^ h of A.
115.22 Modifications
Here is a obsolete result on the category of Algebraization of Formal Spaces, Equation (88.30.0.1). Please visit Algebraization of Formal Spaces, Section 88.30 for the current material.
Proof. This is a special case of Algebraization of Formal Spaces, Lemma 88.30.3. \square
The following lemma on rational singularities is no longer needed in the chapter on resolving surface singularities.
Lemma 115.22.2. In Resolution of Surfaces, Situation 54.9.1. Let M be a finite reflexive A-module. Let M \otimes _ A \mathcal{O}_ X denote the pullback of the associated \mathcal{O}_ S-module. Then M \otimes _ A \mathcal{O}_ X maps onto its double dual.
Proof. Let \mathcal{F} = (M \otimes _ A \mathcal{O}_ X)^{**} be the double dual and let \mathcal{F}' \subset \mathcal{F} be the image of the evaluation map M \otimes _ A \mathcal{O}_ X \to \mathcal{F}. Then we have a short exact sequence
0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{Q} \to 0
Since X is normal, the local rings \mathcal{O}_{X, x} are discrete valuation rings for points of codimension 1 (see Properties, Lemma 28.12.5). Hence \mathcal{Q}_ x = 0 for such points by More on Algebra, Lemma 15.23.3. Thus \mathcal{Q} is supported in finitely many closed points and is globally generated by Cohomology of Schemes, Lemma 30.9.10. We obtain the exact sequence
0 \to H^0(X, \mathcal{F}') \to H^0(X, \mathcal{F}) \to H^0(X, \mathcal{Q}) \to 0
because \mathcal{F}' is generated by global sections (Resolution of Surfaces, Lemma 54.9.2). Since X \to \mathop{\mathrm{Spec}}(A) is an isomorphism over the complement of the closed point, and since M is reflexive, we see that the maps
M \to H^0(X, \mathcal{F}') \to H^0(X, \mathcal{F})
induce isomorphisms after localization at any nonmaximal prime of A. Hence these maps are isomorphisms by More on Algebra, Lemma 15.23.13 and the fact that reflexive modules over normal rings have property (S_2) (More on Algebra, Lemma 15.23.18). Thus we conclude that \mathcal{Q} = 0 as desired. \square
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