The Stacks project

Lemma 54.8.9. Let $p$ be a prime number. Let $A$ be a regular local ring of dimension $2$ and characteristic $p$. Let $A_0 \subset A$ be a subring such that $\Omega _{A/A_0}$ is free of rank $r < \infty $. Set $\omega _ A = \Omega ^ r_{A/A_0}$. If $X \to \mathop{\mathrm{Spec}}(A)$ is the result of a sequence of blowups in closed points, then there exists a map

\[ \varphi _ X : (\Omega ^ r_{X/\mathop{\mathrm{Spec}}(A_0)})^{**} \longrightarrow \omega _ X \]

extending the given identification in the generic point.

Proof. Observe that $A$ is Gorenstein (Dualizing Complexes, Lemma 47.21.3) and hence the invertible module $\omega _ A$ does indeed serve as a dualizing module. Moreover, any $X$ as in the lemma has an invertible dualizing module $\omega _ X$ as $X$ is regular (hence Gorenstein) and proper over $A$, see Remark 54.7.7 and Lemma 54.3.2. Suppose we have constructed the map $\varphi _ X : (\Omega ^ r_{X/A_0})^{**} \to \omega _ X$ and suppose that $b : X' \to X$ is a blowup in a closed point. Set $\Omega ^ r_ X = (\Omega ^ r_{X/A_0})^{**}$ and $\Omega ^ r_{X'} = (\Omega ^ r_{X'/A_0})^{**}$. Since $\omega _{X'} = b^!(\omega _ X)$ a map $\Omega ^ r_{X'} \to \omega _{X'}$ is the same thing as a map $Rb_*(\Omega ^ r_{X'}) \to \omega _ X$. See discussion in Remark 54.7.7 and Duality for Schemes, Section 48.19. Thus in turn it suffices to produce a map

\[ Rb_*(\Omega ^ r_{X'}) \longrightarrow \Omega ^ r_ X \]

The sheaves $\Omega ^ r_{X'}$ and $\Omega ^ r_ X$ are invertible, see Divisors, Lemma 31.12.15. Consider the exact sequence

\[ b^*\Omega _{X/A_0} \to \Omega _{X'/A_0} \to \Omega _{X'/X} \to 0 \]

A local calculation shows that $\Omega _{X'/X}$ is isomorphic to an invertible module on the exceptional divisor $E$, see Lemma 54.3.6. It follows that either

\[ \Omega ^ r_{X'} \cong (b^*\Omega ^ r_ X)(E) \quad \text{or}\quad \Omega ^ r_{X'} \cong b^*\Omega ^ r_ X \]

see Divisors, Lemma 31.15.13. (The second possibility never happens in characteristic zero, but can happen in characteristic $p$.) In both cases we see that $R^1b_*(\Omega ^ r_{X'}) = 0$ and $b_*(\Omega ^ r_{X'}) = \Omega ^ r_ X$ by Lemma 54.3.4. $\square$

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