Formal lifting of dualizing complexes and consequences

For any $X,Y\in D^{b}_{Coh}(A)$, we have

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

$K$ 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 [Stacks, 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, $K\cong K_{i}^{\bullet}\otimes_{A_{i}}^{L}A$. ∎

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, $\ker(d^{m})=\operatorname{im}(d^{m-1})$. ∎

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 [Sch10, Corollary 3.3.3]. By [Stacks, 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 $0ptA>0.$

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., $0ptN>0$.

Since $0ptA>0$ 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 $0$. ∎

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. ∎

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. ∎

By Popescu’s theorem [Stacks, 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 [Stacks, Tag 07M7]. Thus [Sch10, Theorem 7.1.1] applies (see also [Lyu, 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

by [Stacks, 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

Apply the projection $A^{X}\to A$ to some coordinate, we get an object $M_{1}\in D^{[-d,0]}_{Coh}(A)$ such that

Then $M_{1}$ is a dualizing complex for $A$ by Corollary 3.4. ∎

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

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

By [EGA ̵IV2, 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, $[\min_{\mathfrak{q}}c_{\mathfrak{q}},\max_{\mathfrak{q}}c_{\mathfrak{q}}+N]$.) ∎

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

Since $K\in D^{b}_{Coh}(A)$, the function on the right hand side is constructible. Therefore [EGA ̵IV2, 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. ∎

Recall that $f^{!}$ is well-defined for all separated morphisms of finite type [Stacks, Tag 0AA0], compatible with composition [Stacks, Tag 0ATX], flat base change [Stacks, Tag 0E9U], and open immersion [Stacks, 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 $y\in Y$ [Stacks, 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 [Stacks, Tag 0AA1]. If $f$ is finite, the result is true by [Stacks, Tag 0AA2] and Lemma 4.4. Therefore the result is true for any $f$. ∎

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. ∎

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

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 [Stacks, Tag 0A6A], and the local case is well-known, cf. [Stacks, Tag 0A7C]. ∎

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

“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 [Stacks, Tag 0AWY], $A$ is universally catenary, has a codimension function, is Gor-2, and has Gorenstein formal fibers. By [Kaw08, 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 [Rei72] (see also [Aoy83, Corollary 2.12]), finishing the proof. ∎

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

is a shift of $k$, so $K$ is a dualizing complex by [Har66, Chapter V, Proposition 3.4]. ∎

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 $X\in D^{b}_{Coh}(A/fA)$,

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 [Stacks, Tag 091T], we see $F[-b]\to M$ induces a surjective map on $H^{b}(-)$ by [Stacks, 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 $c$. ∎

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