Lemma 54.9.7. In Situation 54.9.1 assume $A$ has a dualizing complex $\omega _ A^\bullet $ and is not regular. Let $X$ be the blowup of $\mathop{\mathrm{Spec}}(A)$ in $\mathfrak m$ with exceptional divisor $E \subset X$. Let $\omega _ X$ be the dualizing module of $X$. Then
$\omega _ E = \omega _ X|_ E \otimes \mathcal{O}_ E(-1)$,
$H^1(X, \omega _ X(n)) = 0$ for $n \geq 0$,
the map $f^*\omega _ A \to \omega _ X$ of Lemma 54.9.6 is surjective.
Proof. We will use the results of Lemma 54.9.5 without further mention. Observe that $\omega _ E = \omega _ X|_ E \otimes \mathcal{O}_ E(-1)$ by Duality for Schemes, Lemmas 48.14.2 and 48.9.7. Thus $\omega _ X|_ E = \omega _ E(1)$. Consider the short exact sequences
By Algebraic Curves, Lemma 53.6.4 we see that $H^1(E, \omega _ E(n + 1)) = 0$ for $n \geq 0$. Thus we see that the maps
are surjective. Since $H^1(X, \omega _ X(n))$ is zero for $n \gg 0$ (Cohomology of Schemes, Lemma 30.16.2) we conclude that (2) holds.
By Algebraic Curves, Lemma 53.6.7 we see that $\omega _ X|_ E = \omega _ E \otimes \mathcal{O}_ E(1)$ is globally generated. Since we seen above that $H^1(X, \omega _ X(1)) = 0$ the map $H^0(X, \omega _ X) \to H^0(E, \omega _ X|_ E)$ is surjective. We conclude that $\omega _ X$ is globally generated hence (3) holds because $\Gamma (X, \omega _ X) = \omega _ A$ is used in Lemma 54.9.6 to define the map. $\square$
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