Lemma 47.25.6. Let $\varphi : R \to A$ be a finite type homomorphism of Noetherian rings. Assume $R$ local and let $\mathfrak m \subset A$ be a maximal ideal lying over the maximal ideal of $R$. If $\omega _ R^\bullet$ is a normalized dualizing complex for $R$, then $\varphi ^!(\omega _ R^\bullet )_\mathfrak m$ is a normalized dualizing complex for $A_\mathfrak m$.

Proof. We already know that $\varphi ^!(\omega _ R^\bullet )$ is a dualizing complex for $A$, see Lemma 47.24.3. Choose a factorization $R \to P \to A$ with $P = R[x_1, \ldots , x_ n]$ as in the construction of $\varphi ^!$. If we can prove the lemma for $R \to P$ and the maximal ideal $\mathfrak m'$ of $P$ corresponding to $\mathfrak m$, then we obtain the result for $R \to A$ by applying Lemma 47.16.2 to $P_{\mathfrak m'} \to A_\mathfrak m$ or by applying Lemma 47.17.2 to $P \to A$. In the case $A = R[x_1, \ldots , x_ n]$ we see that $\dim (A_\mathfrak m) = \dim (R) + n$ for example by Algebra, Lemma 10.111.7 (combined with Algebra, Lemma 10.113.1 to compute the dimension of the fibre). The fact that $\omega _ R^\bullet$ is normalized means that $i = -\dim (R)$ is the smallest index such that $H^ i(\omega _ R^\bullet )$ is nonzero (follows from Lemmas 47.16.5 and 47.16.11). Then $\varphi ^!(\omega _ R^\bullet )_\mathfrak m = \omega _ R^\bullet \otimes _ R A_\mathfrak m[n]$ has its first nonzero cohomology module in degree $-\dim (R) - n$ and therefore is the normalized dualizing complex for $A_\mathfrak m$. $\square$

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