## 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 $\omega _ A^\bullet = \varphi ^!(k)$ 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. $\square$

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).