## 47.15 Dualizing complexes

In this section we define dualizing complexes for Noetherian rings.

Definition 47.15.1. Let $A$ be a Noetherian ring. A dualizing complex is a complex of $A$-modules $\omega _ A^\bullet$ such that

1. $\omega _ A^\bullet$ has finite injective dimension,

2. $H^ i(\omega _ A^\bullet )$ is a finite $A$-module for all $i$, and

3. $A \to R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet )$ is a quasi-isomorphism.

This definition takes some time getting used to. It is perhaps a good idea to prove some of the following lemmas yourself without reading the proofs.

Lemma 47.15.2. Let $A$ be a Noetherian ring. If $\omega _ A^\bullet$ is a dualizing complex, then the functor

$D : K \longmapsto R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet )$

is an anti-equivalence $D_{\textit{Coh}}(A) \to D_{\textit{Coh}}(A)$ which exchanges $D^+_{\textit{Coh}}(A)$ and $D^-_{\textit{Coh}}(A)$ and induces an anti-equivalence $D^ b_{\textit{Coh}}(A) \to D^ b_{\textit{Coh}}(A)$. Moreover $D \circ D$ is isomorphic to the identity functor.

Proof. Let $K$ be an object of $D_{\textit{Coh}}(A)$. Pick an integer $n$ and consider the distinguished triangle

$\tau _{\leq n}K \to K \to \tau _{\geq n + 1}K \to \tau _{\leq n}K[1]$

see Derived Categories, Remark 13.12.4. Since $\omega _ A^\bullet$ has finite injective dimension we see that $R\mathop{\mathrm{Hom}}\nolimits _ A(\tau _{\geq n + 1}K, \omega _ A^\bullet )$ has vanishing cohomology in degrees $\geq n - c$ for some constant $c$. On the other hand, we obtain a spectral sequence

$\mathop{\mathrm{Ext}}\nolimits _ A^ p(H^{-q}(\tau _{\leq n}K), \omega _ A^\bullet ) \Rightarrow \mathop{\mathrm{Ext}}\nolimits _ A^{p + q}(\tau _{\leq n}K, \omega _ A^\bullet ) = H^{p + q}(R\mathop{\mathrm{Hom}}\nolimits _ A(\tau _{\leq n}K, \omega _ A^\bullet ))$

which shows that these cohomology modules are finite. Since for $n > p + q + c$ this is equal to $H^{p + q}(R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet ))$ we see that $R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet )$ is indeed an object of $D_{\textit{Coh}}(A)$. By More on Algebra, Lemma 15.91.2 and the assumptions on the dualizing complex we obtain a canonical isomorphism

$K = R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet ) \otimes _ A^\mathbf {L} K \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ A(R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet ), \omega _ A^\bullet )$

Thus our functor has a quasi-inverse and the proof is complete. $\square$

Lemma 47.15.3. Let $A$ be a Noetherian ring. Let $K \in D^ b_{\textit{Coh}}(A)$. Let $\mathfrak m$ be a maximal ideal of $A$. If $H^ i(K)/\mathfrak m H^ i(K) \not= 0$, then there exists a finite $A$-module $E$ annihilated by a power of $\mathfrak m$ and a map $K \to E[-i]$ which is nonzero on $H^ i(K)$.

Proof. Let $I$ be the injective hull of the residue field of $\mathfrak m$. If $H^ i(K)/\mathfrak m H^ i(K) \not= 0$, then there exists a nonzero map $H^ i(K) \to I$. Since $I$ is injective, we can lift this to a nonzero map $K \to I[-i]$. Recall that $I = \bigcup I[\mathfrak m^ n]$, see Lemma 47.7.2 and that each of the modules $E = I[\mathfrak m^ n]$ is of the desired type. Thus it suffices to prove that

$\mathop{\mathrm{Hom}}\nolimits _{D(A)}(K, I) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _{D(A)}(K, I[\mathfrak m^ n])$

This would be immediate if $K$ where a compact object (or a perfect object) of $D(A)$. This is not the case, but $K$ is a pseudo-coherent object which is enough here. Namely, we can represent $K$ by a bounded above complex of finite free $R$-modules $K^\bullet$. In this case the $\mathop{\mathrm{Hom}}\nolimits$ groups above are computed by using $\mathop{\mathrm{Hom}}\nolimits _{K(A)}(K^\bullet , -)$. As each $K^ n$ is finite free the limit statement holds and the proof is complete. $\square$

Let $R$ be a ring. Recall that an object $L$ of $D(R)$ is invertible if it is an invertible object for the symmetric monoidal structure on $D(R)$ given by derived tensor product. In More on Algebra, Lemma 15.115.4 we have seen this means $L$ is perfect, $L = \bigoplus H^ n(L)[-n]$, this is a finite sum, each $H^ n(L)$ is finite projective, and there is an open covering $\mathop{\mathrm{Spec}}(R) = \bigcup D(f_ i)$ such that $L \otimes _ R R_{f_ i} \cong R_{f_ i}[-n_ i]$ for some integers $n_ i$.

Lemma 47.15.4. Let $A$ be a Noetherian ring. Let $F : D^ b_{\textit{Coh}}(A) \to D^ b_{\textit{Coh}}(A)$ be an $A$-linear equivalence of categories. Then $F(A)$ is an invertible object of $D(A)$.

Proof. Let $\mathfrak m \subset A$ be a maximal ideal with residue field $\kappa$. Consider the object $F(\kappa )$. Since $\kappa = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(\kappa , \kappa )$ we find that all cohomology groups of $F(\kappa )$ are annihilated by $\mathfrak m$. We also see that

$\mathop{\mathrm{Ext}}\nolimits ^ i_ A(\kappa , \kappa ) = \text{Ext}^ i_ A(F(\kappa ), F(\kappa )) = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(F(\kappa ), F(\kappa )[i])$

is zero for $i < 0$. Say $H^ a(F(\kappa )) \not= 0$ and $H^ b(F(\kappa )) \not= 0$ with $a$ minimal and $b$ maximal (so in particular $a \leq b$). Then there is a nonzero map

$F(\kappa ) \to H^ b(F(\kappa ))[-b] \to H^ a(F(\kappa ))[-b] \to F(\kappa )[a - b]$

in $D(A)$ (nonzero because it induces a nonzero map on cohomology). This proves that $b = a$. We conclude that $F(\kappa ) = \kappa [-a]$.

Let $G$ be a quasi-inverse to our functor $F$. Arguing as above we find an integer $b$ such that $G(\kappa ) = \kappa [-b]$. On composing we find $a + b = 0$. Let $E$ be a finite $A$-module wich is annihilated by a power of $\mathfrak m$. Arguing by induction on the length of $E$ we find that $G(E) = E'[-b]$ for some finite $A$-module $E'$ annihilated by a power of $\mathfrak m$. Then $E[-a] = F(E')$. Next, we consider the groups

$\mathop{\mathrm{Ext}}\nolimits ^ i_ A(A, E') = \text{Ext}^ i_ A(F(A), F(E')) = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(F(A), E[-a + i])$

The left hand side is nonzero if and only if $i = 0$ and then we get $E'$. Applying this with $E = E' = \kappa$ and using Nakayama's lemma this implies that $H^ j(F(A))_\mathfrak m$ is zero for $j > a$ and generated by $1$ element for $j = a$. On the other hand, if $H^ j(F(A))_\mathfrak m$ is not zero for some $j < a$, then there is a map $F(A) \to E[-a + i]$ for some $i < 0$ and some $E$ (Lemma 47.15.3) which is a contradiction. Thus we see that $F(A)_\mathfrak m = M[-a]$ for some $A_\mathfrak m$-module $M$ generated by $1$ element. However, since

$A_\mathfrak m = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(A, A)_\mathfrak m = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(F(A), F(A))_\mathfrak m = \mathop{\mathrm{Hom}}\nolimits _{A_\mathfrak m}(M, M)$

we see that $M \cong A_\mathfrak m$. We conclude that there exists an element $f \in A$, $f \not\in \mathfrak m$ such that $F(A)_ f$ is isomorphic to $A_ f[-a]$. This finishes the proof. $\square$

Lemma 47.15.5. Let $A$ be a Noetherian ring. If $\omega _ A^\bullet$ and $(\omega '_ A)^\bullet$ are dualizing complexes, then $(\omega '_ A)^\bullet$ is quasi-isomorphic to $\omega _ A^\bullet \otimes _ A^\mathbf {L} L$ for some invertible object $L$ of $D(A)$.

Proof. By Lemmas 47.15.2 and 47.15.4 the functor $K \mapsto R\mathop{\mathrm{Hom}}\nolimits _ A(R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet ), (\omega _ A')^\bullet )$ maps $A$ to an invertible object $L$. In other words, there is an isomorphism

$L \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , (\omega _ A')^\bullet )$

Since $L$ has finite tor dimension, this means that we can apply More on Algebra, Lemma 15.91.2 to see that

$R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , (\omega '_ A)^\bullet ) \otimes _ A^\mathbf {L} K \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ A(R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet ), (\omega _ A')^\bullet )$

is an isomorphism for $K$ in $D^ b_{\textit{Coh}}(A)$. In particular, setting $K = \omega _ A^\bullet$ finishes the proof. $\square$

Lemma 47.15.6. Let $A$ be a Noetherian ring. Let $B = S^{-1}A$ be a localization. If $\omega _ A^\bullet$ is a dualizing complex, then $\omega _ A^\bullet \otimes _ A B$ is a dualizing complex for $B$.

Proof. Let $\omega _ A^\bullet \to I^\bullet$ be a quasi-isomorphism with $I^\bullet$ a bounded complex of injectives. Then $S^{-1}I^\bullet$ is a bounded complex of injective $B = S^{-1}A$-modules (Lemma 47.3.8) representing $\omega _ A^\bullet \otimes _ A B$. Thus $\omega _ A^\bullet \otimes _ A B$ has finite injective dimension. Since $H^ i(\omega _ A^\bullet \otimes _ A B) = H^ i(\omega _ A^\bullet ) \otimes _ A B$ by flatness of $A \to B$ we see that $\omega _ A^\bullet \otimes _ A B$ has finite cohomology modules. Finally, the map

$B \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet \otimes _ A B, \omega _ A^\bullet \otimes _ A B)$

is a quasi-isomorphism as formation of internal hom commutes with flat base change in this case, see More on Algebra, Lemma 15.92.2. $\square$

Lemma 47.15.7. Let $A$ be a Noetherian ring. Let $f_1, \ldots , f_ n \in A$ generate the unit ideal. If $\omega _ A^\bullet$ is a complex of $A$-modules such that $(\omega _ A^\bullet )_{f_ i}$ is a dualizing complex for $A_{f_ i}$ for all $i$, then $\omega _ A^\bullet$ is a dualizing complex for $A$.

Proof. Consider the double complex

$\prod \nolimits _{i_0} (\omega _ A^\bullet )_{f_{i_0}} \to \prod \nolimits _{i_0 < i_1} (\omega _ A^\bullet )_{f_{i_0}f_{i_1}} \to \ldots$

The associated total complex is quasi-isomorphic to $\omega _ A^\bullet$ for example by Descent, Remark 35.3.10 or by Derived Categories of Schemes, Lemma 36.8.4. By assumption the complexes $(\omega _ A^\bullet )_{f_ i}$ have finite injective dimension as complexes of $A_{f_ i}$-modules. This implies that each of the complexes $(\omega _ A^\bullet )_{f_{i_0} \ldots f_{i_ p}}$, $p > 0$ has finite injective dimension over $A_{f_{i_0} \ldots f_{i_ p}}$, see Lemma 47.3.8. This in turn implies that each of the complexes $(\omega _ A^\bullet )_{f_{i_0} \ldots f_{i_ p}}$, $p > 0$ has finite injective dimension over $A$, see Lemma 47.3.2. Hence $\omega _ A^\bullet$ has finite injective dimension as a complex of $A$-modules (as it can be represented by a complex endowed with a finite filtration whose graded parts have finite injective dimension). Since $H^ n(\omega _ A^\bullet )_{f_ i}$ is a finite $A_{f_ i}$ module for each $i$ we see that $H^ i(\omega _ A^\bullet )$ is a finite $A$-module, see Algebra, Lemma 10.22.2. Finally, the (derived) base change of the map $A \to R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet )$ to $A_{f_ i}$ is the map $A_{f_ i} \to R\mathop{\mathrm{Hom}}\nolimits _ A((\omega _ A^\bullet )_{f_ i}, (\omega _ A^\bullet )_{f_ i})$ by More on Algebra, Lemma 15.92.2. Hence we deduce that $A \to R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet )$ is an isomorphism and the proof is complete. $\square$

Lemma 47.15.8. Let $A \to B$ be a finite ring map of Noetherian rings. Let $\omega _ A^\bullet$ be a dualizing complex. Then $R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )$ is a dualizing complex for $B$.

Proof. Let $\omega _ A^\bullet \to I^\bullet$ be a quasi-isomorphism with $I^\bullet$ a bounded complex of injectives. Then $\mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet )$ is a bounded complex of injective $B$-modules (Lemma 47.3.4) representing $R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )$. Thus $R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )$ has finite injective dimension. By Lemma 47.13.4 it is an object of $D_{\textit{Coh}}(B)$. Finally, we compute

$\mathop{\mathrm{Hom}}\nolimits _{D(B)}(R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet ), R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )) = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet ), \omega _ A^\bullet ) = B$

and for $n \not= 0$ we compute

$\mathop{\mathrm{Hom}}\nolimits _{D(B)}(R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet ), R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )[n]) = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet ), \omega _ A^\bullet [n]) = 0$

which proves the last property of a dualizing complex. In the displayed equations, the first equality holds by Lemma 47.13.1 and the second equality holds by Lemma 47.15.2. $\square$

Lemma 47.15.9. Let $A \to B$ be a surjective homomorphism of Noetherian rings. Let $\omega _ A^\bullet$ be a dualizing complex. Then $R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )$ is a dualizing complex for $B$.

Proof. Special case of Lemma 47.15.8. $\square$

Lemma 47.15.10. Let $A$ be a Noetherian ring. If $\omega _ A^\bullet$ is a dualizing complex, then $\omega _ A^\bullet \otimes _ A A[x]$ is a dualizing complex for $A[x]$.

Proof. Set $B = A[x]$ and $\omega _ B^\bullet = \omega _ A^\bullet \otimes _ A B$. It follows from Lemma 47.3.10 and More on Algebra, Lemma 15.66.5 that $\omega _ B^\bullet$ has finite injective dimension. Since $H^ i(\omega _ B^\bullet ) = H^ i(\omega _ A^\bullet ) \otimes _ A B$ by flatness of $A \to B$ we see that $\omega _ A^\bullet \otimes _ A B$ has finite cohomology modules. Finally, the map

$B \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ B(\omega _ B^\bullet , \omega _ B^\bullet )$

is a quasi-isomorphism as formation of internal hom commutes with flat base change in this case, see More on Algebra, Lemma 15.92.2. $\square$

Proposition 47.15.11. Let $A$ be a Noetherian ring which has a dualizing complex. Then any $A$-algebra essentially of finite type over $A$ has a dualizing complex.

Lemma 47.15.12. Let $A$ be a Noetherian ring. Let $\omega _ A^\bullet$ be a dualizing complex. Let $\mathfrak m \subset A$ be a maximal ideal and set $\kappa = A/\mathfrak m$. Then $R\mathop{\mathrm{Hom}}\nolimits _ A(\kappa , \omega _ A^\bullet ) \cong \kappa [n]$ for some $n \in \mathbf{Z}$.

Proof. This is true because $R\mathop{\mathrm{Hom}}\nolimits _ A(\kappa , \omega _ A^\bullet )$ is a dualizing complex over $\kappa$ (Lemma 47.15.9), because dualizing complexes over $\kappa$ are unique up to shifts (Lemma 47.15.5), and because $\kappa$ is a dualizing complex over $\kappa$. $\square$

## Comments (1)

Comment #4850 by Weixiao Lu on

In the proof of 0A7C, it should be "$\ge c - n$" instead of "$\ge n - c$" and "$p + q + n > n$" instead of "$n > p + q + c$".

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