The Stacks project

Proof. Follows from Lemmas 47.24.4, 47.24.6, and 47.24.5 and the definitions. $\square$

Proof. Choose $R \to P \to A$ as in the definition of $\varphi ^!$. Recall that $R \to A$ is a perfect ring map (More on Algebra, Lemma 15.82.4) and hence $A$ is perfect as a $P$-modue (More on Algebra, Lemma 15.82.2). This shows that $\omega _{A/R}^\bullet $ is in $D^ b_{\textit{Coh}}(A)$ by Lemma 47.24.2. To show $\omega _{A/R}^\bullet $ is $R$-perfect it suffices to show it has finite tor dimension as a complex of $R$-modules. This is true because $\omega _{A/R}^\bullet = \varphi ^!(R) = R\mathop{\mathrm{Hom}}\nolimits (A, P)[n]$ maps to $R\mathop{\mathrm{Hom}}\nolimits _ P(A, P)[n]$ in $D(P)$, which is perfect in $D(P)$ (More on Algebra, Lemma 15.74.15), hence has finite tor dimension in $D(R)$ as $R \to P$ is flat. This proves (1).

Proof of (2). The object $R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _{A/R}^\bullet , \omega _{A/R}^\bullet )$ of $D(A)$ maps in $D(P)$ to

This is equal to $A$ by the already used More on Algebra, Lemma 15.74.15.

Proof of (3). By Lemma 47.25.1 there is an isomorphism

and the right hand side is a dualizing complex by Lemma 47.24.3. $\square$

Proof. By the uniqueness of dualizing complexes, it doesn't matter which dualizing complex we pick for $A$; we omit the detailed proof. Denote $\varphi : k \to A$ the algebra structure. We may take $\omega _ A^\bullet = \varphi ^!(k[0])$ by Lemma 47.24.3. We conclude by Lemma 47.25.2. $\square$

Proof. Recall that a local complete intersection homomorphism is a perfect ring map by More on Algebra, Lemma 15.82.6. Hence the final statement holds by Lemma 47.24.10. By More on Algebra, Definition 15.33.2 we can write $A = R[x_1, \ldots , x_ n]/I$ where $I$ is a Koszul-regular ideal. The construction of $\varphi ^!$ in Section 47.24 shows that it suffices to show the lemma in case $A = R/I$ where $I \subset R$ is a Koszul-regular ideal. Checking $\omega _{A/R}^\bullet $ is invertible in $D(A)$ is local on $\mathop{\mathrm{Spec}}(A)$ by More on Algebra, Lemma 15.126.4. Moreover, formation of $\omega _{A/R}^\bullet $ commutes with localization on $R$ by Lemma 47.24.4. Combining More on Algebra, Definition 15.32.1 and Lemma 15.30.7 and Algebra, Lemma 10.68.6 we can find $g_1, \ldots , g_ r \in R$ generating the unit ideal in $A$ such that $I_{g_ j} \subset R_{g_ j}$ is generated by a regular sequence. Thus we may assume $A = R/(f_1, \ldots , f_ c)$ where $f_1, \ldots , f_ c$ is a regular sequence in $R$. Then we consider the ring maps

and we use Lemma 47.24.7 (and the final statement already proven) to see that it suffices to prove the lemma for each step. Finally, in case $A = R/(f)$ for some nonzerodivisor $f$ we see that the lemma is true since $\varphi ^!(R) = R\mathop{\mathrm{Hom}}\nolimits (A, R)$ is invertible by Lemma 47.13.10. $\square$

Proof. If (2) holds, then the fibre rings $A \otimes _ R \kappa (\mathfrak p)$ have invertible dualizing complexes, and hence are Gorenstein. See Lemmas 47.25.2 and 47.21.4.

For the converse, assume (1). Observe that $\omega _{A/R}^\bullet $ is in $D^ b_{\textit{Coh}}(A)$ by Lemma 47.24.2 (since flat finite type homomorphisms of Noetherian rings are perfect, see More on Algebra, Lemma 15.82.4). Take a prime $\mathfrak q \subset A$ lying over $\mathfrak p \subset R$. Then

Applying Lemmas 47.25.2 and 47.21.4 and assumption (1) we find that this complex has $1$ nonzero cohomology group which is a $1$-dimensional $\kappa (\mathfrak q)$-vector space. By More on Algebra, Lemma 15.77.1 we conclude that $(\omega _{A/R}^\bullet )_ f$ is an invertible object of $D(A_ f)$ for some $f \in A$, $f \not\in \mathfrak q$. This proves (2) holds. $\square$

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 10.112.7 (combined with Algebra, Lemma 10.114.1 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 $A_\mathfrak m$. $\square$

Proof. Choose a factorization $R \to P \to A$ with $P = R[x_1, \ldots , x_ n]$ as in Section 47.24 so that $\omega _{A/R}^\bullet = R\mathop{\mathrm{Hom}}\nolimits (A, P)[n]$. We have to show that $R\mathop{\mathrm{Hom}}\nolimits (A, P)_\mathfrak q$ has vanishing cohomology in degrees $< n - d$. By Lemma 47.13.3 this means we have to show that $\mathop{\mathrm{Ext}}\nolimits _ P^ i(P/I, P)_{\mathfrak r} = 0$ for $i < n - d$ where $\mathfrak r \subset P$ is the prime corresponding to $\mathfrak q$ and $I$ is the kernel of $P \to A$. We may rewrite this as $\mathop{\mathrm{Ext}}\nolimits _{P_\mathfrak r}^ i(P_\mathfrak r/IP_\mathfrak r, P_\mathfrak r)$ by More on Algebra, Lemma 15.65.4. Thus we have to show

by Lemma 47.11.1. By Lemma 47.11.5 we have

The two expressions on the right hand side agree by Algebra, Lemma 10.116.4. $\square$

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

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. $\square$

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. $\square$