Lemma 113.20.1. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring. The category of Restricted Power Series, Equation (86.11.0.1) for $A$ is equivalent to the category Restricted Power Series, Equation (86.11.0.1) for the henselization $A^ h$ of $A$.

## 113.20 Modifications

Here is a obsolete result on the category of Restricted Power Series, Equation (86.11.0.1). Please visit Restricted Power Series, Section 86.11 for the current material.

**Proof.**
This is a special case of Restricted Power Series, Lemma 86.11.3.
$\square$

The following lemma on rational singularities is no longer needed in the chapter on resolving surface singularities.

Lemma 113.20.2. 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

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

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

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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