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
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 15.43.3). 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 10.39.19). Thus condition (2) holds for the finite type ring map $R^\wedge \to A \otimes _ R R^\wedge $ by Morphisms, Lemma 29.28.3. 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
By the dimension formula we have $\dim (A_\mathfrak m) \leq d + r$, see Morphisms, Lemma 29.52.2 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
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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)