The Stacks project

47.25 Relative dualizing complexes in the Noetherian case

Let $\varphi : R \to A$ be a finite type homomorphism of Noetherian rings. Then we define the relative dualizing complex of $A$ over $R$ as the object

\[ \omega _{A/R}^\bullet = \varphi ^!(R) \]

of $D(A)$. Here $\varphi ^!$ is as in Section 47.24. From the material in that section we see that $\omega _{A/R}^\bullet $ is well defined up to (non-unique) isomorphism.

Lemma 47.25.1. Let $R \to R'$ be a homomorphism of Noetherian rings. Let $R \to A$ be of finite type. Set $A' = A \otimes _ R R'$. If

  1. $R \to R'$ is flat, or

  2. $R \to A$ is flat, or

  3. $R \to A$ is perfect and $R'$ and $A$ are tor independent over $R$,

then there is an isomorphism $\omega _{A/R}^\bullet \otimes _ A^\mathbf {L} A' \to \omega ^\bullet _{A'/R'}$ in $D(A')$.

Lemma 47.25.2. Let $\varphi : R \to A$ be a flat finite type map of Noetherian rings. Then

  1. $\omega _{A/R}^\bullet $ is in $D^ b_{\textit{Coh}}(A)$ and $R$-perfect (More on Algebra, Definition 15.78.1),

  2. $A \to R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _{A/R}^\bullet , \omega _{A/R}^\bullet )$ is an isomorphism, and

  3. for every map $R \to k$ to a field the base change $\omega _{A/R}^\bullet \otimes _ A^\mathbf {L} (A \otimes _ R k)$ is a dualizing complex for $A \otimes _ R k$.

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 15.77.4) and hence $A$ is perfect as a $P$-modue (More on Algebra, Lemma 15.77.2). 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 15.70.14), 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

\begin{align*} R\mathop{\mathrm{Hom}}\nolimits _ P(\omega _{A/R}^\bullet , R\mathop{\mathrm{Hom}}\nolimits (A, P)[n]) & = R\mathop{\mathrm{Hom}}\nolimits _ P(R\mathop{\mathrm{Hom}}\nolimits _ P(A, P)[n], P)[n] \\ & = R\mathop{\mathrm{Hom}}\nolimits _ P(R\mathop{\mathrm{Hom}}\nolimits _ P(A, P), P) \end{align*}

This is equal to $A$ by the already used More on Algebra, Lemma 15.70.14.

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

\[ \omega _{A/R}^\bullet \otimes _ A^\mathbf {L} (A \otimes _ R k) \cong \omega ^\bullet _{A \otimes _ R k/k} \]

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

Lemma 47.25.3. Let $K/k$ be an extension of fields. Let $A$ be a finite type $k$-algebra. Let $A_ K = A \otimes _ k K$. If $\omega _ A^\bullet $ is a dualizing complex for $A$, then $\omega _ A^\bullet \otimes _ A A_ K$ is a dualizing complex for $A_ K$.

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. $\square$

Lemma 47.25.4. Let $\varphi : R \to A$ be a local complete intersection homomorphism of Noetherian rings. Then $\omega _{A/R}^\bullet $ is an invertible object of $D(A)$ and $\varphi ^!(K) = K \otimes _ R^\mathbf {L} \omega _{A/R}^\bullet $ for all $K \in D(R)$.

Proof. Recall that a local complete intersection homomorphism is a perfect ring map by More on Algebra, Lemma 15.77.6. Hence the final statement holds by Lemma 47.24.10. By More on Algebra, Definition 15.32.2 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 15.115.4. Moreover, formation of $\omega _{A/R}^\bullet $ commutes with localization on $R$ by Lemma 47.24.4. Combining More on Algebra, Definition 15.31.1 and Lemma 15.29.7 and Algebra, Lemma 10.67.6 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

\[ R \to R/(f_1) \to R/(f_1, f_2) \to \ldots \to R/(f_1, \ldots , f_ c) = A \]

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. $\square$

Lemma 47.25.5. Let $\varphi : R \to A$ be a flat finite type homomorphism of Noetherian rings. The following are equivalent

  1. the fibres $A \otimes _ R \kappa (\mathfrak p)$ are Gorenstein for all primes $\mathfrak p \subset R$, and

  2. $\omega _{A/R}^\bullet $ is an invertible object of $D(A)$, see More on Algebra, Lemma 15.115.4.

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 15.77.4). Take a prime $\mathfrak q \subset A$ lying over $\mathfrak p \subset R$. Then

\[ \omega _{A/R}^\bullet \otimes _ A^\mathbf {L} \kappa (\mathfrak q) = \omega _{A/R}^\bullet \otimes _ A^\mathbf {L} (A \otimes _ R \kappa (\mathfrak p)) \otimes _{(A \otimes _ R \kappa (\mathfrak p))}^\mathbf {L} \kappa (\mathfrak q) \]

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 15.72.5 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. $\square$

The following lemma is useful to see how dimension functions change when passing to a finite type algebra over a Noetherian ring.

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 $A_\mathfrak m$. $\square$

Lemma 47.25.7. Let $R \to A$ be a finite type homomorphism of Noetherian rings. Let $\mathfrak q \subset A$ be a prime ideal lying over $\mathfrak p \subset R$. Then

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

where $d$ is the dimension of the fibre of $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(R)$ over $\mathfrak p$ at the point $\mathfrak q$.

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, Remark 15.62.21. Thus we have to show

\[ \text{depth}_{IP_\mathfrak r}(P_\mathfrak r) \geq n - d \]

by Lemma 47.11.1. By Lemma 47.11.5 we have

\[ \text{depth}_{IP_\mathfrak r}(P_\mathfrak r) \geq \dim ((P \otimes _ R \kappa (\mathfrak p))_\mathfrak r) - \dim ((P/I \otimes _ R \kappa (\mathfrak p))_\mathfrak r) \]

The two expressions on the right hand side agree by Algebra, Lemma 10.115.4. $\square$

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 15.72.8 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 15.59.2. 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. $\square$

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

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

  2. 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. $\square$


Comments (0)


Post a comment

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0E9M. Beware of the difference between the letter 'O' and the digit '0'.