Proof. For any $X,Y\in D^b_{Coh}(A)$ , we have

$$\begin{align*}S^{-1}\operatorname{Hom}_{D(A)}(X,Y)=\operatorname{Hom}_{D(S^{-1}A)}(S^{-1}X,S^{-1}Y),\end{align*}$$

by [25, Tag 0A6A]. Thus, we can spread out a quasi-isomorphism and its inverse and the results are inverse quasi-isomophisms.

Proof. K is represented by a bounded above complex $F^{\bullet }$ of finite free A-modules such that $F^m=0$ for all $m>b$ . Let $K^{\bullet }$ be the complex of A-modules with $K^m=F^m\ (m> a), K^{a}=\operatorname {coker}(F^{a-1}\to F^a), K^m=0\ (m<a)$ , that is, $K^{\bullet }$ is the canonical truncation $\tau _{\geq a}F^{\bullet }$ . Then $K^{\bullet }$ also represents K, and $K^{\bullet }$ is a finite complex of finitely presented A-modules such that $K^m=0$ for all $m\not \in [a,b]$ and $K^m$ free for all $a<m\leq b$ . By [25, Tag 05N7], there exists an index i and a complex of finitely presented $A_i$ -modules $K_i^{\bullet }$ such that $K_i^m=0$ for all $m\not \in [a,b]$ and $K_i^m$ free for all $a<m\leq b$ , and that $K_i^{\bullet }\otimes _{A_i}A$ is isomorphic to $K^{\bullet }$ as cochain complexes. Since $A_i\to A$ is flat, $K\cong K_i^{\bullet }\otimes _{A_i}^L A$ .

Proof. Choose a basis of $E^{m-1},E^m,E^{m+1}$ , so $d^{m-1}$ and $d^m$ are represented by matrices with coefficients in A. The statement of the lemma is a first-order statement of the coefficients and f, so it suffices to show the lemma for a Noetherian local A.

The condition $E^{\bullet }/fE^{\bullet }$ is exact at degree m implies $\ker (d^m)=\operatorname {im}(d^{m-1})+(fE^m\cap \ker (d^m))$ . Since f is a nonzerodivisor on $E^{m+1}$ , $fE^m\cap \ker (d^m)=f\ker (d^m)$ . Thus, $\ker (d^m)=\operatorname {im}(d^{m-1})+f \ker (d^m)$ . By Nakayama’s Lemma, $\ker (d^m)=\operatorname {im}(d^{m-1})$ .

Proof. Since $A_\natural /\mathfrak {m} A_\natural \neq 0$ , it suffices to show $B\to A_\natural $ is flat.

We know $A\to A_\natural $ is flat [Reference Schoutens23, Corollary 3.3.3]. By [25, Tag 051C], we may base change to the reduction of A and assume A reduced. If ${\dim A=0}$ , then $A=k$ , so $B=l$ is a field. Thus, we may assume $\dim A>0,$ so $\operatorname {depth} A>0.$

We need to show $N\otimes _B^L A_\natural $ is concentrated in degree $0$ for all finite B-modules N. If $N=l$ , then since $\mathfrak {n}=\mathfrak {m} B$ , $N\otimes _B^L A_\natural =k\otimes _A^L A_\natural $ is concentrated in degree $0$ as $A\to A_\natural $ is flat. Thus, $N\otimes _B^L A_\natural $ is concentrated in degree $0$ for all N of finite length.

For a general N, we may assume, by Noetherian induction, that $\overline {N}\otimes _B^L A_\natural $ is concentrated in degree $0$ for all proper quotients $\overline {N}$ of N. Since $N[\mathfrak {n}^\infty ]$ is of finite length, we may thus assume $N[\mathfrak {n}^\infty ]=0$ , i.e., $\operatorname {depth} N>0$ .

Since $\operatorname {depth} A>0$ and since $\mathfrak {m} B=\mathfrak {n}$ , there exists an $f\in \mathfrak {m}$ that is both a nonzerodivisor in A (thus a noninvertible nonzerodivisor in both B and $A_\natural $ ) and a nonzerodivisor in N. Thus, $N\otimes _B^L A_\natural /fA_\natural \cong N/fN \otimes _B^L A_\natural $ is concentrated in degree $0$ . Since N and thus $N\otimes _B^L A_\natural $ can be represented by a complex of finite free modules, Lemma 3.1 implies that $N\otimes _B^L A_\natural $ is concentrated in degree $0$ .

Proof. We show by induction on $h=\dim \operatorname {Supp} N$ that $\operatorname {Ext}^m_{A}(N,M)=0$ for all $b+1\leq m\leq b+d+1-h$ , for all finite A-modules N. This implies the result, since then $\operatorname {Ext}^{b+1}_{A}(N,M)=0$ for all finite A-modules N, so dimension shifting and the fact $M\in D^{\leq b}(A)$ shows $\operatorname {Ext}^{m}_{A}(N,M)=0$ for all finite A-modules N and all $m>b$ .

If $h=0$ , then N has finite length, thus is a successive extension of k and the vanishing holds by the assumption. Assume $h>0$ , so there exists $f\in \mathfrak {m}$ such that $\dim \left ( V(fA)\cap \operatorname {Supp} N\right )<h$ . Let $C=\operatorname {Cone}(N\xrightarrow {\times f}N)$ , so we have a distinguished triangle $N\xrightarrow {\times f}N\to C\to +1$ . We note that $C\in D^{[-1,0]}_{Coh}(A)$ and that the dimension of the support of the cohomology modules of C is less than h. Thus, the induction hypothesis tells us $\operatorname {Ext}^m_{A}(C,M)=0$ for all $b+2\leq m\leq b+d+2-h$ . The distinguished triangle above and Nakayama’s Lemma tells us $\operatorname {Ext}^m_{A}(N,M)=0$ for all $b+1\leq m\leq b+d+1-h$ , as desired.

Proof. By Lemma 3.3, $\operatorname {Ext}^m_A(N,M)=0$ for all finite A-modules N and all $m>0$ . Thus, M has injective amplitude $[a,0]$ and $R\operatorname {Hom}_{A}(k,M)\cong k$ . Since $M\in D^{[a,0]}(A)$ , we see $R\operatorname {Hom}_{A}(M,M)\in D^{[a,-a]}(A)$ . Thus, $R\operatorname {Hom}_{A}(M,M)=A$ , so M is a normalized dualizing complex.

Proof. By Popescu’s theorem [25, Tag 07GC], B is the colimit of a direct system of smooth A-algebras $A_i$ . The composition $A_i\to B\to B/IB\cong A/I$ gives an A-algebra map $A_i\to A/I$ , thus $A_i$ admits a section $A_i\to A$ since $(A,I)$ is Henselian, see [25, Tag 07M7]. Thus [Reference Schoutens23, Theorem 7.1.1] applies (see also [Reference Lyu17, Lemma 3.4]), so there exists an A-algebra map $B\to A_\natural $ where $A_\natural =A^X/\mathcal {U}=S_{\mathcal {U}}^{-1}A^X$ is an ultrapower of A. Here X is a set, $\mathcal {U}$ is an ultrafilter on X, and $S_{\mathcal {U}}=\{e_U\mid U\in \mathcal {U}\}$ , where $e_U$ is defined by $(e_U)_x=1$ when $x\in U$ and $(e_U)_x=0$ when $x\not \in U$ . This map is flat by Theorem 3.2.

Let $K\in D^b_{Coh}(B)$ be a normalized dualizing complex. Then $F:=K\otimes _B^L A_\natural \in D(A_\natural )$ is pseudo-coherent and bounded. Writing $k_\natural =k\otimes _A^L A_\natural =k\otimes _B^L A_\natural $ , we have

$$ \begin{align*} R\operatorname{Hom}_{A_\natural}(k_\natural,F)&=R\operatorname{Hom}_{B}(k,K)\otimes_B^L A_\natural\cong k_\natural,\\ R\operatorname{Hom}_{A_\natural}(F,F)&=R\operatorname{Hom}_{B}(K,K)\otimes_B^L A_\natural= A_\natural, \end{align*} $$

by [25, Tag 0A6A] and by the fact K is a normalized dualizing complex.

Let $d=\dim A$ . Then $K\in D^{[-d, 0]}(B)$ and thus $F\in D^{[-d, 0]}(A_\natural )$ . By Lemma 2.2, there exists an $e\in S_{\mathcal {U}}$ and an $M\in D((A^X)_e)$ represented by a complex $M^{\bullet }$ of finitely presented $(A^X)_e$ -modules such that $M\otimes ^L_{(A^X)_e}A_\natural \cong F$ , that $M^m=0$ for all $m>0$ and $m<-d$ , and that $M^m$ is free for all $-d<m\leq 0$ . Note that if $e=e_U$ , then $(A^X)_e=A^U$ , thus after replacing X by a subset in $\mathcal {U}$ we may assume $e=1$ .

We note that another application of Lemma 2.2 with $a=-2d$ shows that, after replacing X by a subset, M is also represented by a complex of free A-modules $P^{\bullet }$ such that $P^m=0$ for all $m>0$ and that $P^m$ is finite for all $m\geq -2d-1$ . Thus, $R\operatorname {Hom}_{A^X}(M,M)$ is represented by the Hom complex $E^{\bullet }:=\operatorname {Hom}^{\bullet }_{A^X}(P^{\bullet },M^{\bullet })$ , and $E^m$ is finitely presented for all $m\leq d+1$ . Also note that $R\operatorname {Hom}_{A^X}(k^X,M)$ is represented by a complex of finitely presented modules in degrees $\geq -d$ , since $k^X=A^X\otimes ^L_A k$ is represented by a complex of finite free modules in degrees $\leq 0$ .

Let $T^{-1}\to T^0\to T$ be a three-term complex of finitely presented $A^X$ -modules. Fix presentations $P^{i,-1}\to P^{i,0}\to T^{i}\to 0\ (i=-1,0,1)$ , where $P^{i,j}$ are finite free, and maps $P^{-1,0}\to P^{0,0}\to P^{1,0}$ lifting $T^{-1}\to T^0\to T^1$ . We do not require the composition $P^{-1,0}\to P^{1,0}$ to be zero. Fix bases of $P^{i,j}$ , so the maps between $P^{i,j}$ are represented by matrices. The statement that the cohomology $\ker (T^0\to T^1)/\operatorname {im}(T^{-1}\to T^0)$ is $0$ (resp. $A^X,k^X$ ) is a first-order statement of the coefficients of the matrices. This observation and the observation on Hom complexes above tells us that after replacing X by a subset in $\mathcal {U}$ , we have

$$ \begin{align*} \tau_{\leq d+1}R\operatorname{Hom}_{A^X}(k^X,M)&\cong k^X,\\ \tau_{\geq -d}\tau_{\leq d}R\operatorname{Hom}_{A^X}(M,M)&=A^X. \end{align*} $$

Apply the projection $A^X\to A$ to some coordinate, we get an object $M_1\in D^{[-d,0]}_{Coh}(A)$ such that

$$ \begin{align*} \tau_{\leq d+1}R\operatorname{Hom}_{A}(k,M_1)&\cong k,\\ \tau_{\geq -d}\tau_{\leq d}R\operatorname{Hom}_{A}(M_1,M_1)&=A. \end{align*} $$

Then $M_1$ is a dualizing complex for A by Corollary 3.4.

Proof. Let $A^h$ be the Henselization of A, so $A^h$ is a Henselian G-ring [25, Tag 07QR]. Thus, $A^h$ admits a dualizing complex ${K\in D(A^h)}$ by Theorem 3.5 (and [25, Tag 0BFR] or our Theorem 1.3). Since $A\to A^h$ is the filtered colimit of elementary étale-local ring maps $A\to A'$ [25, Tag 04GN] and since each $A'\to A^h$ is flat (cf. [25, Tag 08HS]), by Lemma 2.2 there exists an $A'$ and a $K'\in D^b_{Coh}(A')$ such that $K'\otimes ^L_{A'}A^h=K$ . Since K is a dualizing complex for $A^h$ , by flatness again we see $K'$ is a dualizing complex for $A'$ , see [25, Tag 0E4A].

Proof. For all $\mathfrak {p}\in \operatorname {Spec}(A)$ , we have

$$\begin{align*}\operatorname{ht}\mathfrak{p}-\operatorname{depth} A_{\mathfrak{p}}=\max \{b-a\mid a\leq b, H^a(K_{\mathfrak{p}})\neq 0, H^b(K_{\mathfrak{p}})\neq 0\}. \end{align*}$$

By [Reference Grothendieck8, Proposition 6.10.6], the function on the left-hand side is constructible on $\operatorname {Spec}(A)$ . Let N be the maximum of this function, and for each minimal prime $\mathfrak {q}$ of A let $c_{\mathfrak {q}}$ be the unique integer such that $H^{c_{\mathfrak {q}}}(K_{\mathfrak {q}})\neq 0$ . Then it is clear that K is concentrated in degrees $[-N+\min _{\mathfrak {q}} c_{\mathfrak {q}},\max _{\mathfrak {q}} c_{\mathfrak {q}}+N]$ . (In fact, $[\min _{\mathfrak {q}} c_{\mathfrak {q}},\max _{\mathfrak {q}} c_{\mathfrak {q}}+N]$ .)

Proof. Since K is a pseudo-dualizing complex, we have, for all $\mathfrak {p}\in \operatorname {Spec}(A)$ ,

$$\begin{align*}\operatorname{ht}\mathfrak{p}-\operatorname{depth} A_{\mathfrak{p}}=\max \{b-a\mid a\leq b, H^a(K_{\mathfrak{p}})\neq 0, H^b(K_{\mathfrak{p}})\neq 0\}. \end{align*}$$

Since $K\in D^b_{Coh}(A)$ , the function on the right-hand side is constructible. Therefore, [Reference Grothendieck8, Proposition 6.10.6] shows that for every $\mathfrak {p}\in \operatorname {Spec}(A)$ , $\operatorname {Spec}(A/\mathfrak {p})$ contains a nonempty Cohen–Macaulay open subscheme. Since B is finite over A the same is true for B. Since K is bounded, $L\in D_{Coh}(B)$ . It is clear that $L_{\mathfrak {q}}$ is a dualizing complex for $B_{\mathfrak {q}}$ for all $\mathfrak {q}\in \operatorname {Spec}(B)$ . We conclude by Lemma 4.3.

Proof. Recall that $f^!$ is well defined for all separated morphisms of finite type [25, Tag 0AA0], compatible with composition [25, Tag 0ATX], flat base change [25, Tag 0E9U], and open immersion [25, Tag 0AU0].

Let K be a pseudo-dualizing complex for Y. Since the result is true after base change to $\operatorname {Spec}(\mathcal {O}_{Y,y})$ for all $y\in Y$ [25, Tag 0AA3], we see $(f^!K)_x$ is a dualizing complex for $\mathcal {O}_{X,x}$ for all $x\in X$ . Thus it suffices to show $f^!K\in D^b_{Coh}(X)$ . We may assume X and Y affine by compatibility with open immersion. By compatibility with composition, it suffices to treat the cases $X=\mathbf {A}^1_Y$ and f is finite (or even closed immersion). If $X=\mathbf {A}^1_Y$ the result is trivial, see [25, Tag 0AA1]. If f is finite, the result is true by [25, Tag 0AA2] and Lemma 4.4. Therefore, the result is true for any f.

Proof. The “in particular” statement follows from the fact that A is a pseudo-dualizing complex for a Gorenstein A.

Every finite type A-algebra admits a pseudo-dualizing complex by Corollary 4.5, so A is universally catenary by Remark 4.2. To see A is Gor-2, we may assume by Lemma 4.4 that A is an integral domain, and we must show $A_f$ is Gorenstein for some nonzero f. This is clear: a pseudo-dualizing complex $K\in D^b_{Coh}(A)$ is a shift of the fraction field at the generic point of A, thus for some $f\neq 0$ , $K_f$ is a shift of $A_f$ by Lemma 2.1.

Proof. By definition, we may assume A Gorenstein, and we conclude by Corollary 4.7.

Proof. By Lemma 4.4, $D_K(A/\mathfrak {p})\in D^b_{Coh}(A)$ for all $\mathfrak {p}\in \operatorname {Spec}(A)$ . Thus $D_K$ maps $D^b_{Coh}(A)$ into $D^b_{Coh}(A)$ . The fact that $\operatorname {id}\to D_K\circ D_K$ is an isomorphism can be checked locally by [25, Tag 0A6A], and the local case is well-known, cf. [25, Tag 0A7C].

Proof. Let $L=R\operatorname {Hom}_A(K,K')$ . Then $L=D_{K'}\circ D_{K}(A)$ is an invertible object by Corollary 4.9 and [25, Tag 0A7E]. The statement that the canonical map $K\otimes ^L_A L\to K'$ is an isomorphism can be checked locally by [25, Tag 0A6A], and when A is local it follows from [25, Tag 0A69].

Proof. “If” follows from Lemma 4.4.

We proceed to show “only if.” Every finite type A-algebra admits a pseudo-dualizing complex, Corollary 4.5, so we may replace A by any finite type A-algebra that admits a section.

By Remark 4.2, Corollary 4.7, and [25, Tag 0AWY], A is universally catenary, has a codimension function, is Gor-2, and has Gorenstein formal fibers. By [Reference Kawasaki14, p. 2738, proof of Theorem 1.3], there exists a finite type A-algebra B that is Cohen–Macaulay and admits a section. Replace A by B, we may assume our A is Cohen–Macaulay.

Assume A is Cohen–Macaulay. We may assume $\operatorname {Spec}(A)$ is connected, and K is concentrated in degree $0$ . Then the square-zero extension $A\oplus H^0(K)$ is Gorenstein by [Reference Reiten22] (see also [Reference Aoyama1, Corollary 2.12]), finishing the proof.

Proof. We may assume $(A,\mathfrak {m},k)$ local. We have

$$ \begin{align*}R\operatorname{Hom}_{A}(k,K)=R\operatorname{Hom}_{A/I}(k,R\operatorname{Hom}_{A}(A/I,K))\end{align*} $$

is a shift of k, so K is a dualizing complex by [Reference Hartshorne10, Chapter V, Proposition 3.4].

Proof. Note that $R\operatorname {Hom}_A(A/fA,M)\in D^+(A/fA)$ since $M\in D^+(A)$ . If (i) holds, then $R\operatorname {Hom}_A(A/fA,M)\in D^+_{Coh}(A/fA)$ , so for any $X\in D^b_{Coh}(A/fA)$ ,

$$\begin{align*}R\operatorname{Hom}_A(X,M)=R\operatorname{Hom}_{A/fA}(X,R\operatorname{Hom}_A(A/fA,M)), \end{align*}$$

is in $D_{Coh}(A/fA)$ , in particular this holds for $X=C:=\operatorname {Cone}(A\xrightarrow {\times f}A)$ . Thus (ii) holds.

Now assume (ii), and assume A is f-adically complete and M is derived f-adically complete. From the distinguished triangle $A\xrightarrow {\times f} A\to C\to +1$ We know $R\operatorname {Hom}_A(C,M)\cong \operatorname {Cone}(M\xrightarrow {\times f}M)[-1]$ , so $\operatorname {Cone}(M\xrightarrow {\times f}M)\in D_{Coh}(A/fA)$ . Let b be an integer such that $M\in D^{\leq b}(A)$ . Then $H^b(M)/fH^b(M)=H^b(\operatorname {Cone}(M\xrightarrow {\times f}M))$ is finite. Thus there exists a finite free A-module F and a map $F[-b]\to M$ in $D(A)$ inducing a surjective map on $H^b(-)/fH^b(-)$ . Since $F[-b]$ is also derived f-adically complete [25, Tag 091T], we see $F[-b]\to M$ induces a surjective map on $H^b(-)$ by [25, Tag 09B9]. Thus $\operatorname {Cone}(F[-b]\to M)\in D^{\leq b-1}(A)$ , and we see inductively $H^c(M)$ is finite for all c.

Proof. For (i), we may assume $I=fA$ principal. Let a be the smallest integer such that $H^a(K)\neq 0$ . Then $H^a(K)[f]=H^a(L)$ is finite. The exact sequences $0\to H^a(K)[f]\to H^a(K)[f^{n+1}]\xrightarrow {\times f} H^a(K)[f^n]$ tells us $H^a(K)$ is finite. Apply dimension shifting to the distinguished triangle

$$\begin{align*}H^a(K)[-a]\to K\to \tau_{>a}K\to +1, \end{align*}$$

we see inductively $H^c(K)$ is finite for all $c\in \mathbf {Z}$ .

For (ii), from Corollary 4.7 we know $A/I$ is Gor-2, so A is Gor-2 by definition, since I is nilpotent. By Lemma 4.3, it suffices to show (ii) when $(A,\mathfrak {m},k)$ is local. Then

$$ \begin{align*}R\operatorname{Hom}_{A}(k,K)=R\operatorname{Hom}_{A/I}(k,R\operatorname{Hom}_{A}(A/I,K)),\end{align*} $$

is a shift of k, so K has finite injective dimension [25, Tag 0AVJ], in particular bounded. Since $K\in D_{Coh}(A)$ by (i), (ii) follows from Lemma 5.3.

Proof. Let $N\in \mathbf {Z}_{\geq 1}$ be such that $J:=A[f^\infty ]=A[f^N]$ . After replacing A by some $A_g$ , we may assume $A/\mathfrak {p}$ Cohen–Macaulay and our inequality holds for all $\mathfrak {q}\in V(\mathfrak {p})$ and all $n< N$ by [Reference Grothendieck8, Proposition 6.10.6]. By the same proposition may also assume that, for all $\mathfrak {q}\in V(\mathfrak {p})$ , $\operatorname {depth} J_{\mathfrak {q}}\geq \operatorname {ht}(\mathfrak {q}/\mathfrak {p})$ and $\operatorname {depth} (A_{\mathfrak {q}}/J_{\mathfrak {q}})\geq \operatorname {ht}(\mathfrak {q}/\mathfrak {p})+1$ . Since f is a nonzerodivisor on $A/J$ we have $\operatorname {depth} (A_{\mathfrak {q}}/(f^nA+J)_{\mathfrak {q}})\geq \operatorname {ht}(\mathfrak {q}/\mathfrak {p})$ for all $n\in \mathbf {Z}_{\geq 1}$ .

For $n\geq N$ , the sequence

is exact. Thus the depth inequalities above imply $\operatorname {depth} (A_{\mathfrak {q}}/f^nA_{\mathfrak {q}})\geq \operatorname {ht}(\mathfrak {q}/\mathfrak {p})$ for all ${n\in \mathbf {Z}_{\geq N}}$ . Now for all $n\geq N$ we have $ \operatorname {ht}(\mathfrak {q}/f^nA)-\operatorname {depth} \left (A_{\mathfrak {q}}/f^nA_{\mathfrak {q}}\right )\leq \operatorname {ht}(\mathfrak {q})-\operatorname {ht}(\mathfrak {q}/\mathfrak {p})$ . The right-hand side equals $\operatorname {ht}(\mathfrak {p})$ after localization by [Reference Grothendieck8, Proposition 6.10.6].

Proof of Theorem 5.1

A pseudo-dualizing complex is a dualizing complex if and only if the ring is finite-dimensional, see Remark 4.2. Since I is in the Jacobson radical of A, A is finite-dimensional if and only if $A/I$ is. Thus it suffices to prove the result for pseudo-dualizing complexes.

We may assume $I=fA$ principal. Let $A_n=A/f^nA$ .

Let $(J_1^{\bullet },d_1^{\bullet })$ be a bounded below complex of injective $A_1$ -modules that represents $K_1$ . Fix $a\in \mathbf {Z}$ so that $J_1^m=0$ for all $m<a$ . Further, fix $c_0\in \mathbf {Z}_{\geq a}$ such that for all minimal primes $\mathfrak {q}$ of $fA$ , there exists $c\leq c_0$ such that $H^c(J_1^{\bullet })_{\mathfrak {q}}\neq 0$ .

For each m, let $J^m$ be an injective hull of $J_1^m$ as an A-module. Thus $J^m$ is an essential extension of $J_1^m$ and $J_1^m=J^m[f]$ . Let $J^m_\infty =J^m[f^\infty ]$ . Then $J^m_\infty $ is an injective A-module by [25, Tag 08XW].

The proof is given after the main argument. Granting Claim 5.7, $J^{\bullet }_\infty $ is now a bounded below complex of $f^\infty $ -torsion injective A-modules. We next show that $J^{\bullet }_\infty \in D(A)$ is bounded. Note that if $\dim A_1$ is finite, then $K_1$ has finite injective dimension, so we could choose $J_1^{\bullet }$ so that $J_1^m=0$ for $m\gg 1$ , and $J_\infty $ is automatically bounded. Without this assumption, writing $J^m_n=J^m[f^n]$ , we have that $R\operatorname {Hom}_{A_n}(A_1,J^{\bullet }_n)=J^{\bullet }_n[f]=J^{\bullet }_1\in D(A_1)$ , since $J^{\bullet }_n$ is a bounded below complex of injectives. Thus $J^{\bullet }_n$ is a pseudo-dualizing complex for $A_n$ by Lemma 5.5. Let $\mathfrak {q}$ be a minimal prime of $fA$ . Then $(J^{\bullet }_n)_{\mathfrak {q}}$ is exact except at a single degree c, and applying $R\operatorname {Hom}_{A_{\mathfrak {q}}/f^n A_{\mathfrak {q}}}(A_{\mathfrak {q}}/f A_{\mathfrak {q}},-)$ we see $(J^{\bullet }_1)_{\mathfrak {q}}$ is exact except at degree c, so $c\leq c_0$ . Thus for all $\mathfrak {p}\in V(fA)$ , $H^c(J^{\bullet }_n)_{\mathfrak {p}}\neq 0$ for some $c\leq c_0$ .

Note that $A/fA$ is Gor-2 by Corollary 4.7, so Lemma 5.6 shows that there exists $b\in \mathbf {Z}_{\geq 1}$ such that for all $\mathfrak {p}\in V(fA)$ and all $n\in \mathbf {Z}_{\geq 1}$ ,

$$\begin{align*}\operatorname{ht}(\mathfrak{p}/f^nA)-\operatorname{depth} \left(A_{\mathfrak{p}}/f^nA_{\mathfrak{p}}\right)\leq b. \end{align*}$$

Therefore, $(J^{\bullet }_n)_{\mathfrak {p}}\in D^{[a,b+c_0]}(A)$ , so $(J^{\bullet }_\infty )_{\mathfrak {p}}\in D^{[a,b+c_0]}(A)$ for all $\mathfrak {p}\in V(fA)$ . Since f is in the Jacobson radical of A we have $J^{\bullet }_\infty \in D^{[a,b+c_0]}(A)$ .

Let $K\in D(A)$ be the derived f-adic completion of $J^{\bullet }_\infty $ , so $K\in D^b(A)$ by for example [25, Tag 091Z]. By [25, Tag 0A6Y] (and [25, Tags 091T and 0A6R]) we have

$$\begin{align*}R\operatorname{Hom}_A(A/fA,K)=R\operatorname{Hom}_A(A/fA,J^{\bullet}_\infty), \end{align*}$$

and the right-hand side is just $J^{\bullet }_\infty [f]=J^{\bullet }_1$ since $J^{\bullet }_\infty $ is a bounded below complex of injectives. Thus $K\in D^b_{Coh}(A)$ by Lemma 5.4. We conclude that K is a pseudo-dualizing complex for A by Lemma 5.3.

Proof of Claim 5.7

We first show that there exist maps $d^m_2:J^m_2\to J^{m+1}_2$ extending $d^m_1$ such that $(J^{\bullet }_2,d^{\bullet }_2)$ is a complex. This follows from the dual version of [25, Tag 0DYR]. We give the proof in our case for the reader’s convenience.

Let $\delta ^m_2:J^m_2\to J^{m+1}_2$ be arbitrary maps extending $d^m_1$ . Composing, we get maps $\delta ^{m+1}_2\circ \delta ^m_2:J^m_2\to J^{m+2}_2$ . Since $(J^{\bullet }_1,d^{\bullet }_1)$ is a complex, $\delta ^{m+1}_2\circ \delta ^m_2$ is zero on $J^m_1=J^m_2[f]$ . Since ${f^2=0\in A_2}$ , we have $fJ^m_2\subseteq J^m_1$ , so $\operatorname {im}(\delta ^{m+1}_2\circ \delta ^m_2)\subseteq J^{m+2}_1.$ This tells us $(J^{\bullet }_2/J^{\bullet }_1,\delta ^{\bullet }_2)$ is a complex, and that $\delta ^{m+1}_2\circ \delta ^m_2$ induces a map $J^m_2/J^m_1\to J^{m+2}_1$ . It is clear that $\delta ^{\bullet +1}_2\circ \delta ^{\bullet }_2:J^{\bullet }_2/J^{\bullet }_1\to J^{\bullet +2}_1$ is a map of complexes.

Since $J^m_2$ is injective, we have canonical isomorphisms

$$\begin{align*}J^m_2/J^m_1=\operatorname{Hom}_{A_2}(fA_2,J^m_2)=\operatorname{Hom}_{A_1}(fA_2,J^m_1). \end{align*}$$

Therefore, $J^{\bullet }_2/J^{\bullet }_1$ represents $R\operatorname {Hom}_{A_1}(fA_2,J^{\bullet }_1)$ , so $R\operatorname {Hom}_{A_1}(J^{\bullet }_2/J^{\bullet }_1,J^{\bullet +2}_1)=R\operatorname {Hom}_{A_1}(R\operatorname {Hom}_{A_1}(fA_2,J^{\bullet }_1),J^{\bullet }_1[2])=fA_2[2]$ by Corollary 4.9. Thus $\delta ^{\bullet +1}_2\circ \delta ^{\bullet }_2:J^{\bullet }_2/J^{\bullet }_1\to J^{\bullet +2}_1$ is zero in $D(A_1),$ hence homotopic to zero (see [25, Tag 05TG]). Let $g^{\bullet }_2:J^{\bullet }_2/J^{\bullet }_1\to J^{\bullet +1}_1$ be a homotopy between $\delta ^{\bullet +1}_2\circ \delta ^{\bullet }_2$ and $0$ . View each $g^m_2$ as a map $g^m_2:J^m_2\to J^{m+1}_2$ , we see $d^m_2=\delta ^m_2-g^m_2$ is what we want.

Now $J^{\bullet }_2$ is a bounded below complex of injective $A_2$ -modules, so

$$\begin{align*}R\operatorname{Hom}_{A_2}(A_1,J^{\bullet}_2)=J^{\bullet}_2[f]=J^{\bullet}_1,\end{align*}$$

thus $J^{\bullet }_2$ is a pseudo-dualizing complex for $A_2$ by Lemma 5.5. The same argument as above tells us we can extend $d^{\bullet }_2$ to $d^{\bullet }_3$ , ad infinitum, showing Claim 5.7.

Proof. We may assume $A/\mathfrak {p}$ Gorenstein. Let B be the $\mathfrak {p}$ -adic completion of A, so $A\to B$ is a flat ring map with $A/\mathfrak {p}=B/\mathfrak {p} B$ , thus open subsets of $V(\mathfrak {p})$ in $\operatorname {Spec}(A)$ are in one-to-one correspondence with open subsets of $V(\mathfrak {p} B)$ in $\operatorname {Spec}(B)$ . Note that for all $\mathfrak {P}\in V(\mathfrak {p})$ , $A_{\mathfrak {P}}^\wedge =B_{\mathfrak {P} B}^\wedge $ since $A/\mathfrak {p}^n=B/\mathfrak {p}^n B$ for all n. Thus, the base change of a dualizing complex for (resp. canonical module of) $A_{\mathfrak {p}}$ to $B_{\mathfrak {p} B}$ is a dualizing complex for (resp. canonical module of) $B_{\mathfrak {p} B}$ , and we may apply flat descent ([25, Tag 0E4A] and [Reference Aoyama1, Theorem 4.2]) for $\mathfrak {P}\in V(\mathfrak {p})$ . Thus it suffices to prove the lemma in the case A is $\mathfrak {p}$ -adically complete and $A/\mathfrak {p}$ is Gorenstein.

In this case, A admits a pseudo-dualizing complex E by Theorem 5.1, and we have $K_{\mathfrak {p}}\cong E_{\mathfrak {p}}$ (resp. $M_{\mathfrak {p}}\cong H^0(E_{\mathfrak {p}})$ and $E_{\mathfrak {p}}\in D^{\geq 0}(A_{\mathfrak {p}})$ ; [Reference Aoyama1, (1.5)]) after a shift. After localizing we have $K\cong E$ by Lemma 2.1 (resp. $M\cong H^0(E)$ and $E\in D^{\geq 0}(A)$ ), as desired.

Proof. We know (ii) implies (i) and (iv) implies (iii) by Lemma 2.2. On the other hand, (ii) and (iv) follows from Lemma 6.1 and general topology [25, Tag 0541].

Proof. Let $\mathfrak {p}\in \operatorname {Spec}(A)$ . By Corollary 3.6, there exists an étale ring map $A\to B$ and a prime $\mathfrak {q}\in \operatorname {Spec}(B)$ lying above $\mathfrak {p}$ such that $B_{\mathfrak {q}}$ admits a dualizing complex. Note that B is Gor-2 since it is of finite type over A, Corollary 4.8. By Theorem 6.2, localizing B near $\mathfrak {q}$ we may assume B admits a pseudo-dualizing complex. If A is finite-dimensional then so is B, so B admits a dualizing complex. Now we take $A'$ to be a finite product of such B so that $\operatorname {Spec}(A')\to \operatorname {Spec}(A)$ is surjective.

Proof. $R\operatorname {Hom}_A(k,K)$ is a shift of k, so $R\operatorname {Hom}_B(B/\mathfrak {m} B,K\otimes _A^L B)$ is a shift of $B/\mathfrak {m} B$ by [25, Tag 0A6A]. We conclude by Lemma 5.3.

Proof. Let a be a parameter of A. The $aA$ -adic completion $A^\wedge $ of A is a complete local ring, thus admits a normalized dualizing complex M.

Let E be a dualizing complex for the Artinian ring $A_a$ concentrated in degree $-1$ . The map $A_a\to (A^\wedge )_a$ is a flat map between Artinian rings with Gorenstein fibers by our assumptions, thus $E\otimes ^L_{A_a}(A^\wedge )_a$ is a dualizing complex by Lemma 7.1. On the other hand $M_a$ is another dualizing complex of $(A^\wedge )_a$ concentrated in degree $-1$ by [25, Tag 0A7V]. Thus $ E\otimes ^L_{A_a}(A^\wedge )_a\cong M_a $ as $(A^\wedge )_a$ is Artinian. By [Reference Bhatt2, Theorem 1.4], there exists $K\in D(A)$ such that $K\otimes ^L_A A^\wedge \cong M$ . Then K is a dualizing complex for A [25, Tag 0E4A].

Proof. We may assume $(U,K)$ cannot be enlarged to any $(W,K_W)$ , and we shall show $\dim \mathcal {O}_{X,x}>1$ for all $x\not \in U$ .

Let $x\not \in U$ , and assume $\dim \mathcal {O}_{X,x}\leq 1$ . Let $V=\operatorname {Spec}(A)$ be an affine open neighborhood of $x\in X$ such that A has a pseudo-dualizing complex L, which exists by Lemma 7.2 and Theorem 6.2. We may assume that all generic points of V specialize to x. Since $\dim \mathcal {O}_{X,x}=1$ , we see from [25, Tag 0A7V] that, after shifting, we may assume $L_{\mathfrak {q}}$ is concentrated at degree $0$ for all minimal primes $\mathfrak {q}$ of A.

If $U\times _X \operatorname {Spec}(\mathcal {O}_{X,x})$ is empty, then we may assume $U\cap V=\emptyset $ , so we can enlarge U to $U\cup V$ . Otherwise $U\times _X \operatorname {Spec}(\mathcal {O}_{X,x})$ has dimension zero, so it is affine. Thus, we may assume $U\cap V$ is affine by [25, Tag 01Z6]. Write $B=\mathcal {O}(U\cap V)$ , so B admits a dualizing complex $K_0$ given by the restriction of K to $U\cap V$ . Then $L\otimes ^L_{A}B_{\mathfrak {p}}\cong K_0\otimes ^L_B B_{\mathfrak {p}}$ , where $\mathfrak {p}\in \operatorname {Spec}(A)$ corresponds to x, since $\dim B_{\mathfrak {p}}=0$ and both sides are dualizing complexes concentrated in degree $0$ . Thus there exists $g\in A\setminus \mathfrak {p}$ such that $L\otimes ^L_{A} B_{g}\cong K_0\otimes ^L_B B_{g}$ by Lemma 2.1, so we can enlarge U to $U\cup \operatorname {Spec}(A_g)$ .

Proof. “Only if” follows from [25, Tag 0AWY] and Corollary 4.7. “If” follows from Theorem 7.4 with $U=\emptyset $ .

Proof. Since $c(-)$ is a codimension function of $A/I$ , for $Q\subseteq Q'\in V(I+\mathfrak {p})$ , we have $c(\mathfrak {p},Q')-c(\mathfrak {p},Q)=\operatorname {ht}(Q'/Q)+\operatorname {ht}(Q/\mathfrak {p})-\operatorname {ht}(Q'/\mathfrak {p})$ . Since A is catenary, the number on the right-hand side is $0$ . Thus for $Q\subseteq Q'\in V(I+\mathfrak {p})$ , we have $c(\mathfrak {p},Q')=c(\mathfrak {p},Q)$ . Since $(A,I)$ is Henselian, $V(I+\mathfrak {p})$ is connected, see [25, Tag 09Y6]. This shows (i). For (ii), note that $V(I+\mathfrak {p})$ is nonempty for all $\mathfrak {p}\in \operatorname {Spec}(A)$ , so our function is well defined. To show it is a codimension function, it suffices to show for $\mathfrak {p}\subseteq \mathfrak {p}'\subseteq Q$ where $Q\in V(I)$ , we have $c(\mathfrak {p}',Q)-c(\mathfrak {p},Q)=\operatorname {ht}(\mathfrak {p}'/\mathfrak {p})$ . This is clear as A is catenary.

Proof. By Lemma 8.5, it suffices to show A is universally catenary. We may, therefore, assume A is an integral domain and there exists a finite injective ring map $A\to B$ such that B is universally catenary. We may also assume B is an integral domain. Let $c:\operatorname {Spec}(A/I)\to \mathbf {Z}$ be a codimension function. Let $\delta (Q)=c((Q\cap A)/I)-\operatorname {ht}(Q)$ for $Q\in V(IB)$ . For $Q\subseteq Q'\in V(IB)$ , we have

$$ \begin{align*} c((Q'\cap A)/I)-c((Q\cap A)/I)&=\operatorname{ht}((Q'\cap A)/(Q\cap A))\\ &=\operatorname{ht}(Q'/Q)\\ &=\operatorname{ht}(Q')-\operatorname{ht}(Q), \end{align*} $$

where the first identity is because c is a codimension function, the second identity follows from the dimension formula [25, Tag 02IJ] as $A/I$ is universally catenary, and the third identity is because B is a catenary domain. Therefore, $\delta (Q')=\delta (Q)$ . Since the pair $(B,IB)$ is Henselian [25, Tag 09XK], $V(IB)$ is connected, see [25, Tag 09Y6]. Thus, $\delta $ is a constant function on $V(IB)$ . Therefore, for all maximal ideals $\mathfrak {n}$ of B, which necessarily contains $IB$ , $\operatorname {ht}(\mathfrak {n})$ depends only on $\mathfrak {n}\cap A$ , and thus must equal to $\operatorname {ht}(\mathfrak {n}\cap A)$ . As discussed in Remark 6.7, we see A is universally catenary, as desired.

Proof of Theorem 8.4

Assume that A is I-adically complete, $A/I$ is a quotient of a Cohen–Macaulay ring, and $A/I$ is either quasi-excellent or semilocal and Nagata. Note that $(A,I)$ is a Henselian pair [25, Tag 0ALJ].

If $A/I$ is quasi-excellent, then A is quasi-excellent [Reference Kurano and Shimomoto15], in particular CM-quasi-excellent. If $A/I$ is semilocal and Nagata, then A is semilocal and Nagata [Reference Marot18], and thus has Cohen–Macaulay formal fibers by [Reference Takumi19, Theorem C]. Then A is CM-quasi-excellent, see [Reference Grothendieck8, Proposition 7.3.18]. Consequently, in both cases, A is CM-quasi-excellent.

Now A satisfies condition (i) in Lemma 8.6 by the discussions in Remark 6.7, thus is universally catenary and admits a codimension function by Lemma 8.6. By Theorem 8.3, A is a quotient of a Cohen–Macaulay ring.