## 47.27 Relative dualizing complexes

For a finite type ring map $\varphi : R \to A$ of Noetherian rings we have the relative dualizing complex $\omega _{A/R}^\bullet = \varphi ^!(R)$ considered in Section 47.25. If $R$ is not Noetherian, a similarly constructed complex will in general not have good properties. In this section, we give a definition of a relative dualizing complex for a flat and finitely presented ring maps $R \to A$ of non-Noetherian rings. The definition is chosen to globalize to flat and finitely presented morphisms of schemes, see Duality for Schemes, Section 48.28. We will show that relative dualizing complexes exist (when the definition applies), are unique up to (noncanonical) isomorphism, and that in the Noetherian case we recover the complex of Section 47.25.

The Noetherian reader may safely skip this section!

Definition 47.27.1. Let $R \to A$ be a flat ring map of finite presentation. A relative dualizing complex is an object $K \in D(A)$ such that

1. $K$ is $R$-perfect (More on Algebra, Definition 15.83.1), and

2. $R\mathop{\mathrm{Hom}}\nolimits _{A \otimes _ R A}(A, K \otimes _ A^\mathbf {L} (A \otimes _ R A))$ is isomorphic to $A$.

To understand this definition you may have to read and understand some of the following lemmas. Lemmas 47.27.3 and 47.27.2 show this definition does not clash with the definition in Section 47.25.

Lemma 47.27.2. Let $R \to A$ be a flat ring map of finite presentation. Any two relative dualizing complexes for $R \to A$ are isomorphic.

Proof. Let $K$ and $L$ be two relative dualizing complexes for $R \to A$. Denote $K_1 = K \otimes _ A^\mathbf {L} (A \otimes _ R A)$ and $L_2 = (A \otimes _ R A) \otimes _ A^\mathbf {L} L$ the derived base changes via the first and second coprojections $A \to A \otimes _ R A$. By symmetry the assumption on $L_2$ implies that $R\mathop{\mathrm{Hom}}\nolimits _{A \otimes _ R A}(A, L_2)$ is isomorphic to $A$. By More on Algebra, Lemma 15.98.3 part (3) applied twice we have

$A \otimes _{A \otimes _ R A}^\mathbf {L} L_2 \cong R\mathop{\mathrm{Hom}}\nolimits _{A \otimes _ R A}(A, K_1 \otimes _{A \otimes _ R A}^\mathbf {L} L_2) \cong A \otimes _{A \otimes _ R A}^\mathbf {L} K_1$

Applying the restriction functor $D(A \otimes _ R A) \to D(A)$ for either coprojection we obtain the desired result. $\square$

Lemma 47.27.3. Let $\varphi : R \to A$ be a flat finite type ring map of Noetherian rings. Then the relative dualizing complex $\omega _{A/R}^\bullet = \varphi ^!(R)$ of Section 47.25 is a relative dualizing complex in the sense of Definition 47.27.1.

Proof. From Lemma 47.25.2 we see that $\varphi ^!(R)$ is $R$-perfect. Denote $\delta : A \otimes _ R A \to A$ the multiplication map and $p_1, p_2 : A \to A \otimes _ R A$ the coprojections. Then

$\varphi ^!(R) \otimes _ A^\mathbf {L} (A \otimes _ R A) = \varphi ^!(R) \otimes _{A, p_1}^\mathbf {L} (A \otimes _ R A) = p_2^!(A)$

by Lemma 47.24.4. Recall that $R\mathop{\mathrm{Hom}}\nolimits _{A \otimes _ R A}(A, \varphi ^!(R) \otimes _ A^\mathbf {L} (A \otimes _ R A))$ is the image of $\delta ^!(\varphi ^!(R) \otimes _ A^\mathbf {L} (A \otimes _ R A))$ under the restriction map $\delta _* : D(A) \to D(A \otimes _ R A)$. Use the definition of $\delta ^!$ from Section 47.24 and Lemma 47.13.3. Since $\delta ^!(p_2^!(A)) \cong A$ by Lemma 47.24.7 we conclude. $\square$

Lemma 47.27.4. Let $R \to A$ be a flat ring map of finite presentation. Then

1. there exists a relative dualizing complex $K$ in $D(A)$, and

2. for any ring map $R \to R'$ setting $A' = A \otimes _ R R'$ and $K' = K \otimes _ A^\mathbf {L} A'$, then $K'$ is a relative dualizing complex for $R' \to A'$.

Moreover, if

$\xi : A \longrightarrow K \otimes _ A^\mathbf {L} (A \otimes _ R A)$

is a generator for the cyclic module $\mathop{\mathrm{Hom}}\nolimits _{D(A \otimes _ R A)}(A, K \otimes _ A^\mathbf {L} (A \otimes _ R A))$ then in (2) the derived base change of $\xi$ by $A \otimes _ R A \to A' \otimes _{R'} A'$ is a generator for the cyclic module $\mathop{\mathrm{Hom}}\nolimits _{D(A' \otimes _{R'} A')}(A', K' \otimes _{A'}^\mathbf {L} (A' \otimes _{R'} A'))$

Proof. We first reduce to the Noetherian case. By Algebra, Lemma 10.168.1 there exists a finite type $\mathbf{Z}$ subalgebra $R_0 \subset R$ and a flat finite type ring map $R_0 \to A_0$ such that $A = A_0 \otimes _{R_0} R$. By Lemma 47.27.3 there exists a relative dualizing complex $K_0 \in D(A_0)$. Thus if we show (2) for $K_0$, then we find that $K_0 \otimes _{A_0}^\mathbf {L} A$ is a dualizing complex for $R \to A$ and that it also satisfies (2) by transitivity of derived base change. The uniqueness of relative dualizing complexes (Lemma 47.27.2) then shows that this holds for any relative dualizing complex.

Assume $R$ Noetherian and let $K$ be a relative dualizing complex for $R \to A$. Given a ring map $R \to R'$ set $A' = A \otimes _ R R'$ and $K' = K \otimes _ A^\mathbf {L} A'$. To finish the proof we have to show that $K'$ is a relative dualizing complex for $R' \to A'$. By More on Algebra, Lemma 15.83.5 we see that $K'$ is $R'$-perfect in all cases. By Lemmas 47.25.1 and 47.27.3 if $R'$ is Noetherian, then $K'$ is a relative dualizing complex for $R' \to A'$ (in either sense). Transitivity of derived tensor product shows that $K \otimes _ A^\mathbf {L} (A \otimes _ R A) \otimes _{A \otimes _ R A}^\mathbf {L} (A' \otimes _{R'} A') = K' \otimes _{A'}^\mathbf {L} (A' \otimes _{R'} A')$. Flatness of $R \to A$ guarantees that $A \otimes _{A \otimes _ R A}^\mathbf {L} (A' \otimes _{R'} A') = A'$; namely $A \otimes _ R A$ and $R'$ are tor independent over $R$ so we can apply More on Algebra, Lemma 15.61.2. Finally, $A$ is pseudo-coherent as an $A \otimes _ R A$-module by More on Algebra, Lemma 15.82.8. Thus we have checked all the assumptions of More on Algebra, Lemma 15.83.6. We find there exists a bounded below complex $E^\bullet$ of $R$-flat finitely presented $A \otimes _ R A$-modules such that $E^\bullet \otimes _ R R'$ represents $R\mathop{\mathrm{Hom}}\nolimits _{A' \otimes _{R'} A'}(A', K' \otimes _{A'}^\mathbf {L} (A' \otimes _{R'} A'))$ and these identifications are compatible with derived base change. Let $n \in \mathbf{Z}$, $n \not= 0$. Define $Q^ n$ by the sequence

$E^{n - 1} \to E^ n \to Q^ n \to 0$

Since $\kappa (\mathfrak p)$ is a Noetherian ring, we know that $H^ n(E^\bullet \otimes _ R \kappa (\mathfrak p)) = 0$, see remarks above. Chasing diagrams this means that

$Q^ n \otimes _ R \kappa (\mathfrak p) \to E^{n + 1} \otimes _ R \kappa (\mathfrak p)$

is injective. Hence for a prime $\mathfrak q$ of $A \otimes _ R A$ lying over $\mathfrak p$ we have $Q^ n_\mathfrak q$ is $R_\mathfrak p$-flat and $Q^ n_\mathfrak p \to E^{n + 1}_\mathfrak q$ is $R_\mathfrak p$-universally injective, see Algebra, Lemma 10.99.1. Since this holds for all primes, we conclude that $Q^ n$ is $R$-flat and $Q^ n \to E^{n + 1}$ is $R$-universally injective. In particular $H^ n(E^\bullet \otimes _ R R') = 0$ for any ring map $R \to R'$. Let $Z^0 = \mathop{\mathrm{Ker}}(E^0 \to E^1)$. Since there is an exact sequence $0 \to Z^0 \to E^0 \to E^1 \to Q^1 \to 0$ we see that $Z^0$ is $R$-flat and that $Z^0 \otimes _ R R' = \mathop{\mathrm{Ker}}(E^0 \otimes _ R R' \to E^1 \otimes _ R R')$ for all $R \to R'$. Then the short exact sequence $0 \to Q^{-1} \to Z^0 \to H^0(E^\bullet ) \to 0$ shows that

$H^0(E^\bullet \otimes _ R R') = H^0(E^\bullet ) \otimes _ R R' = A \otimes _ R R' = A'$

as desired. This equality furthermore gives the final assertion of the lemma. $\square$

Lemma 47.27.5. Let $R \to A$ be a flat ring map of finite presentation. Let $K$ be a relative dualizing complex. Then $A \to R\mathop{\mathrm{Hom}}\nolimits _ A(K, K)$ is an isomorphism.

Proof. By Algebra, Lemma 10.168.1 there exists a finite type $\mathbf{Z}$ subalgebra $R_0 \subset R$ and a flat finite type ring map $R_0 \to A_0$ such that $A = A_0 \otimes _{R_0} R$. By Lemmas 47.27.2, 47.27.3, and 47.27.4 there exists a relative dualizing complex $K_0 \in D(A_0)$ and its derived base change is $K$. This reduces us to the situation discussed in the next paragraph.

Assume $R$ Noetherian and let $K$ be a relative dualizing complex for $R \to A$. Given a ring map $R \to R'$ set $A' = A \otimes _ R R'$ and $K' = K \otimes _ A^\mathbf {L} A'$. To finish the proof we show $R\mathop{\mathrm{Hom}}\nolimits _{A'}(K', K') = A'$. By Lemma 47.25.2 we know this is true whenever $R'$ is Noetherian. Since a general $R'$ is a filtered colimit of Noetherian $R$-algebras, we find the result holds by More on Algebra, Lemma 15.83.7. $\square$

Lemma 47.27.6. Let $R \to A \to B$ be a ring maps which are flat and of finite presentation. Let $K_{A/R}$ and $K_{B/A}$ be relative dualizing complexes for $R \to A$ and $A \to B$. Then $K = K_{A/R} \otimes _ A^\mathbf {L} K_{B/A}$ is a relative dualizing complex for $R \to B$.

Proof. We will use reduction to the Noetherian case. Namely, by Algebra, Lemma 10.168.1 there exists a finite type $\mathbf{Z}$ subalgebra $R_0 \subset R$ and a flat finite type ring map $R_0 \to A_0$ such that $A = A_0 \otimes _{R_0} R$. After increasing $R_0$ and correspondingly replacing $A_0$ we may assume there is a flat finite type ring map $A_0 \to B_0$ such that $B = B_0 \otimes _{R_0} R$ (use the same lemma). If we prove the lemma for $R_0 \to A_0 \to B_0$, then the lemma follows by Lemmas 47.27.2, 47.27.3, and 47.27.4. This reduces us to the situation discussed in the next paragraph.

Assume $R$ is Noetherian and denote $\varphi : R \to A$ and $\psi : A \to B$ the given ring maps. Then $K_{A/R} \cong \varphi ^!(R)$ and $K_{B/A} \cong \psi ^!(A)$, see references given above. Then

$K = K_{A/R} \otimes _ A^\mathbf {L} K_{B/A} \cong \varphi ^!(R) \otimes _ A^\mathbf {L} \psi ^!(A) \cong \psi ^!(\varphi ^!(R)) \cong (\psi \circ \varphi )^!(R)$

by Lemmas 47.24.10 and 47.24.7. Thus $K$ is a relative dualizing complex for $R \to B$. $\square$

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).