The Stacks project

Lemma 47.25.8. Let $R \to A$ be a flat finite type homomorphism of Noetherian rings. Let $\mathfrak q \subset A$ be a prime ideal lying over $\mathfrak p \subset R$. Then

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

where $d$ is the dimension of the fibre of $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(R)$ over $\mathfrak p$ at the point $\mathfrak q$. If all fibres of $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(R)$ have dimension $\leq d$, then $\omega _{A/R}^\bullet $ has tor amplitude in $[-d, 0]$ as a complex of $R$-modules.

Proof. The lower bound has been shown in Lemma 47.25.7. Choose a factorization $R \to P \to A$ with $P = R[x_1, \ldots , x_ n]$ as in Section 47.24 so that $\omega _{A/R}^\bullet = R\mathop{\mathrm{Hom}}\nolimits (A, P)[n]$. The upper bound means that $\mathop{\mathrm{Ext}}\nolimits ^ i_ P(A, P)$ is zero for $i > n$. This follows from More on Algebra, Lemma 15.77.5 which shows that $A$ is a perfect $P$-module with tor amplitude in $[-n, 0]$.

Proof of the final statement. Let $R \to R'$ be a ring homomorphism of Noetherian rings. Set $A' = A \otimes _ R R'$. Then

\[ \omega _{A'/R'}^\bullet = \omega _{A/R}^\bullet \otimes _ A^\mathbf {L} A' = \omega _{A/R}^\bullet \otimes _ R^\mathbf {L} R' \]

The first isomorphism by Lemma 47.25.1 and the second, which takes place in $D(R')$, by More on Algebra, Lemma 15.61.2. By the first part of the proof (note that the fibres of $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(R')$ have dimension $\leq d$) we conclude that $\omega _{A/R}^\bullet \otimes _ R^\mathbf {L} R'$ has cohomology only in degrees $[-d, 0]$. Taking $R' = R \oplus M$ to be the square zero thickening of $R$ by a finite $R$-module $M$, we see that $R\mathop{\mathrm{Hom}}\nolimits (A, P) \otimes _ R^\mathbf {L} M$ has cohomology only in the interval $[-d, 0]$ for any finite $R$-module $M$. Since any $R$-module is a filtered colimit of finite $R$-modules and since tensor products commute with colimits we conclude. $\square$

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