Lemma 47.25.9. Let $R \to A$ be a finite type homomorphism of Noetherian rings. Let $\mathfrak p \subset R$ be a prime ideal. Assume

$R_\mathfrak p$ is Cohen-Macaulay, and

for any minimal prime $\mathfrak q \subset A$ we have $\text{trdeg}_{\kappa (R \cap \mathfrak q)} \kappa (\mathfrak q) \leq r$.

Then

\[ H^ i(\omega _{A/R}^\bullet )_\mathfrak p \not= 0 \Rightarrow - r \leq i \]

and $H^{-r}(\omega _{A/R}^\bullet )_\mathfrak p$ is $(S_2)$ as an $A_\mathfrak p$-module.

**Proof.**
We may replace $R$ by $R_\mathfrak p$ by Lemma 47.25.1. Thus we may assume $R$ is a Cohen-Macaulay local ring and we have to show the assertions of the lemma for the $A$-modules $H^ i(\omega _{A/R}^\bullet )$.

Let $R^\wedge $ be the completion of $R$. The map $R \to R^\wedge $ is flat and $R^\wedge $ is Cohen-Macaulay (More on Algebra, Lemma 15.42.3). Observe that the minimal primes of $A \otimes _ R R^\wedge $ lie over minimal primes of $A$ by the flatness of $A \to A \otimes _ R R^\wedge $ (and going down for flatness, see Algebra, Lemma 10.38.19). Thus condition (2) holds for the finite type ring map $R^\wedge \to A \otimes _ R R^\wedge $ by Morphisms, Lemma 29.27.3. Appealing to Lemma 47.25.1 once again it suffices to prove the lemma for $R^\wedge \to A \otimes _ R R^\wedge $. In this way, using Lemma 47.22.4, we may assume $R$ is a Noetherian local Cohen-Macaulay ring which has a dualizing complex $\omega _ R^\bullet $.

Let $\mathfrak m \subset A$ be a maximal ideal. It suffices to show that the assertions of the lemma hold for $H^ i(\omega _{A/R}^\bullet )_\mathfrak m$. If $\mathfrak m$ does not lie over the maximal ideal of $R$, then we replace $R$ by a localization to reduce to this case (small detail omitted).

We may assume $\omega _ R^\bullet $ is normalized. Setting $d = \dim (R)$ we see that $\omega _ R^\bullet = \omega _ R[d]$ for some $R$-module $\omega _ R$, see Lemma 47.20.2. Set $\omega _ A^\bullet = \varphi ^!(\omega _ R^\bullet )$. By Lemma 47.24.11 we have

\[ \omega _{A/R}^\bullet = R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ R[d] \otimes _ R^\mathbf {L} A, \omega _ A^\bullet ) \]

By the dimension formula we have $\dim (A_\mathfrak m) \leq d + r$, see Morphisms, Lemma 29.50.2 and use that $\kappa (\mathfrak m)$ is finite over the residue field of $R$ by the Hilbert Nullstellensatz. By Lemma 47.25.6 we see that $(\omega _ A^\bullet )_\mathfrak m$ is a normalized dualizing complex for $A_\mathfrak m$. Hence $H^ i((\omega _ A^\bullet )_\mathfrak m)$ is nonzero only for $-d - r \leq i \leq 0$, see Lemma 47.16.5. Since $\omega _ R[d] \otimes _ R^\mathbf {L} A$ lives in degrees $\leq -d$ we conclude the vanishing holds. Finally, we also see that

\[ H^{-r}(\omega _{A/R}^\bullet )_\mathfrak m = \mathop{\mathrm{Hom}}\nolimits _ A(\omega _ R \otimes _ R A, H^{-d - r}(\omega _ A^\bullet ))_\mathfrak m \]

Since $H^{-d - r}(\omega _ A^\bullet )_\mathfrak m$ is $(S_2)$ by Lemma 47.17.5 we find that the final statement is true by More on Algebra, Lemma 15.23.11.
$\square$

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