hcju/stacksqa huggingface dataset:
Suppose that $J = (g_1, \ldots, g_m)$. Then $R\Gamma_Z(M)$ is computed by the complex $$ M \to \prod M_{g_{j_0}} \to \prod M_{g_{j_0}g_{j_1}} \to \ldots \to M_{g_1g_2\ldots g_m} $$ by Lemma \ref{lemma-local-cohomology-adjoint}. By More on Algebra, Lemma \ref{more-algebra-lemma-when-derived-completion-is-completion} the derived $I$-adic completion of this complex is given by the complex $$ \lim M/I^nM \to \prod \lim (M/I^nM)_{g_{j_0}} \to \ldots \to \lim (M/I^nM)_{g_1g_2\ldots g_m} $$ of usual completions. Since $R\Gamma_Z(M/I^nM)$ is computed by the complex $ M/I^nM \to \prod (M/I^nM)_{g_{j_0}} \to \ldots \to (M/I^nM)_{g_1g_2\ldots g_m}$ and since the transition maps between these complexes are surjective, we conclude that (1) holds by More on Algebra, Lemma \ref{more-algebra-lemma-compute-Rlim-modules}. Part (2) then follows from More on Algebra, Lemma \ref{more-algebra-lemma-break-long-exact-sequence-modules}.
Since $M$ is a finite $A$-module, we have that $M$ is $I$-adically complete. The proof of Lemma \ref{lemma-completion-local} shows that $$ H^0(R\Gamma_Z(M)^\wedge) = \Ker(M^\wedge \to \prod M_{g_j}^\wedge) = \Ker(M \to \prod M_{g_j}^\wedge) $$ where on the right hand side we have usual $I$-adic completion. The kernel $K_j$ of $M_{g_j} \to M_{g_j}^\wedge$ is $\bigcap I^n M_{g_j}$. By Algebra, Lemma \ref{algebra-lemma-intersection-powers-ideal-module} for every $\mathfrak p \in V(IA_{g_j})$ we find an $f \in A_{g_j}$, $f \not \in \mathfrak p$ such that $(K_j)_f = 0$. \medskip\noindent Let $s \in H^0(R\Gamma_Z(M)^\wedge)$. By the above we may think of $s$ as an element of $M$. The support $Z'$ of $s$ intersected with $D(g_j)$ is disjoint from $D(g_j) \cap V(I)$ by the arguments above. Thus $Z'$ is a closed subset of $\Spec(A)$ with $Z' \cap V(I) \subset V(J)$. Then $Z' \cup V(J) = V(J')$ for some ideal $J' \subset J$ with $V(J') \cap V(I) \subset V(J)$ and we have $s \in H^0_{V(J')}(M)$. Conversely, any $s \in H^0_{V(J')}(M)$ with $J' \subset J$ and $V(J') \cap V(I) \subset V(J)$ maps to zero in $M_{g_j}^\wedge$ for all $j$. This proves the lemma.
We have to show: if $K \in D^+(A)$ has finite cohomology modules, then the complex $R\Hom(B, K)$ has finite cohomology modules too. This follows for example from Lemma \ref{lemma-RHom-ext} if we can show the ext modules $\Ext^i_A(B, K)$ are finite $A$-modules. Since $K$ is bounded below there is a convergent spectral sequence $$ \Ext^p_A(B, H^q(K)) \Rightarrow \text{Ext}^{p + q}_A(B, K) $$ This finishes the proof as the modules $\Ext^p_A(B, H^q(K))$ are finite by Algebra, Lemma \ref{algebra-lemma-ext-noetherian}.
This is true because $R\Hom(E, -)$ is the right adjoint to $- \otimes^\mathbf{L}_R E$, see Differential Graded Algebra, Lemma \ref{dga-lemma-tensor-hom-adjoint}. Hence this functor plays the same role as the functor $R\Hom(A, -)$ for the map $R \to A$ (Lemma \ref{lemma-right-adjoint}), whence these functors must correspond via the equivalence $- \otimes^\mathbf{L}_E A : D(E, \text{d}) \to D(A)$.
We apply Divided Power Algebra, Lemma \ref{dpa-lemma-tate-resoluton-pseudo-coherent-ring-map} to choose a Tate resolution $(E, \text{d})$ of $A$ over $R$. Note that $E^i = 0$ for $i > 0$, $E^0 = R[x_1, \ldots, x_n]$ is a polynomial algebra, and $E^i$ is a finite free $E^0$-module for $i < 0$. It follows that $E$ viewed as a complex of $R$-modules is a bounded above complex of free $R$-modules. We check the assumptions of Lemma \ref{lemma-RHom-is-tensor}. The first holds because $A$ is perfect (hence compact by More on Algebra, Proposition \ref{more-algebra-proposition-perfect-is-compact}) and the second by More on Algebra, Lemma \ref{more-algebra-lemma-RHom-out-of-projective}. From the lemma conclude that $K \mapsto R\Hom(E, K)$ is isomorphic to $K \mapsto K \otimes_R^\mathbf{L} N$ for some differential graded $E$-module $N$. Observe that $$ (R \otimes_R E) \otimes_E^\mathbf{L} A = R \otimes_E E \otimes_E A $$ in $D(A)$. Hence by Differential Graded Algebra, Lemma \ref{dga-lemma-compose-tensor-functors-general-algebra} we conclude that the composition of $- \otimes_R^\mathbf{L} N$ and $- \otimes_R^\mathbf{L} A$ is of the form $- \otimes_R M$ for some $M \in D(A)$. To finish the proof we apply Lemma \ref{lemma-RHom-dga}.
Since $A$ has the finite projective resolution $$ 0 \to I \to R \to A \to 0 $$ we see that $A$ is a perfect $R$-module. By Lemma \ref{lemma-RHom-is-tensor-special} it suffices to prove that $R\Hom(A, R)$ is represented by $N[-1]$ in $D(A)$. This means $R\Hom(A, R)$ has a unique nonzero cohomology module, namely $N$ in degree $1$. As $\text{Mod}_A \to \text{Mod}_R$ is fully faithful it suffice to prove this after applying the restriction functor $i_* : D(A) \to D(R)$. By Lemma \ref{lemma-RHom-ext} we have $$ i_*R\Hom(A, R) = R\Hom_R(A, R) $$ Using the finite projective resolution above we find that the latter is represented by the complex $R \to I^{\otimes -1}$ with $R$ in degree $0$. The map $R \to I^{\otimes -1}$ is injective and the cokernel is $N$.
To see that the map is an isomorphism, it suffices to prove it is an isomorphism after applying $\varphi'_*$. Applying the functor $\varphi'_*$ to (\ref{equation-base-change}) and using that $A' = A \otimes_R^\mathbf{L} R'$ we obtain the base change map $R\Hom_R(A, K) \otimes_R^\mathbf{L} R' \to R\Hom_{R'}(A \otimes_R^\mathbf{L} R', K \otimes_R^\mathbf{L} R')$ for derived hom of More on Algebra, Equation (\ref{more-algebra-equation-base-change-RHom}). Unwinding the left and right hand side exactly as in the proof of More on Algebra, Lemma \ref{more-algebra-lemma-base-change-RHom} and in particular using More on Algebra, Lemma \ref{more-algebra-lemma-upgrade-adjoint-tensor-RHom} gives the desired result.
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 \ref{derived-remark-truncation-distinguished-triangle}. Since $L$ has finite injective dimension we see that $R\Hom_A(\tau_{\geq n + 1}K, L)$ has vanishing cohomology in degrees $\geq c - n$ for some constant $c$. Hence, given $i$, we see that $\Ext^i_A(K, L) \to \Ext^i_A(\tau_{\leq n}K, L)$ is an isomorphism for some $n \gg - i$. By Derived Categories of Schemes, Lemma \ref{perfect-lemma-coherent-internal-hom} applied to $\tau_{\leq n}K$ and $L$ we see conclude that $\Ext^i_A(K, L)$ is a finite $A$-module for all $i$. Hence $R\Hom_A(K, L)$ is indeed an object of $D_{\textit{Coh}}(A)$.
Let $K$ be an object of $D_{\textit{Coh}}(A)$. From Lemma \ref{lemma-finite-ext-into-bounded-injective} we see $R\Hom_A(K, \omega_A^\bullet)$ is an object of $D_{\textit{Coh}}(A)$. By More on Algebra, Lemma \ref{more-algebra-lemma-internal-hom-evaluate-isomorphism-technical} and the assumptions on the dualizing complex we obtain a canonical isomorphism $$ K = R\Hom_A(\omega_A^\bullet, \omega_A^\bullet) \otimes_A^\mathbf{L} K \longrightarrow R\Hom_A(R\Hom_A(K, \omega_A^\bullet), \omega_A^\bullet) $$ Thus our functor has a quasi-inverse and the proof is complete.
Let $\mathfrak m \subset A$ be a maximal ideal with residue field $\kappa$. Consider the object $F(\kappa)$. Since $\kappa = \Hom_{D(A)}(\kappa, \kappa)$ we find that all cohomology groups of $F(\kappa)$ are annihilated by $\mathfrak m$. We also see that $$ \Ext^i_A(\kappa, \kappa) = \text{Ext}^i_A(F(\kappa), F(\kappa)) = \Hom_{D(A)}(F(\kappa), F(\kappa)[i]) $$ 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 $F(\kappa) = \kappa[-a]$. \medskip\noindent 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 $$ \Ext^i_A(A, E') = \text{Ext}^i_A(F(A), F(E')) = \Hom_{D(A)}(F(A), E[-a + i]) $$ 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 $E$ (More on Algebra, Lemma \ref{more-algebra-lemma-detect-cohomology}) 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 $$ A_\mathfrak m = \Hom_{D(A)}(A, A)_\mathfrak m = \Hom_{D(A)}(F(A), F(A))_\mathfrak m = \Hom_{A_\mathfrak m}(M, M) $$ 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.
By Lemmas \ref{lemma-dualizing} and \ref{lemma-equivalence-comes-from-invertible} the functor $K \mapsto R\Hom_A(R\Hom_A(K, \omega_A^\bullet), (\omega_A')^\bullet)$ maps $A$ to an invertible object $L$. In other words, there is an isomorphism $$ L \longrightarrow R\Hom_A(\omega_A^\bullet, (\omega_A')^\bullet) $$ Since $L$ has finite tor dimension, this means that we can apply More on Algebra, Lemma \ref{more-algebra-lemma-internal-hom-evaluate-isomorphism-technical} to see that $$ R\Hom_A(\omega_A^\bullet, (\omega'_A)^\bullet) \otimes_A^\mathbf{L} K \longrightarrow R\Hom_A(R\Hom_A(K, \omega_A^\bullet), (\omega_A')^\bullet) $$ is an isomorphism for $K$ in $D^b_{\textit{Coh}}(A)$. In particular, setting $K = \omega_A^\bullet$ finishes the proof.
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 \ref{lemma-localization-injective-modules}) 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 $$ B \longrightarrow R\Hom_A(\omega_A^\bullet \otimes_A B, \omega_A^\bullet \otimes_A B) $$ is a quasi-isomorphism as formation of internal hom commutes with flat base change in this case, see More on Algebra, Lemma \ref{more-algebra-lemma-base-change-RHom}.
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 \ref{descent-remark-standard-covering} or by Derived Categories of Schemes, Lemma \ref{perfect-lemma-alternating-cech-complex-complex-computes-cohomology}. 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 \ref{lemma-localization-injective-modules}. 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 \ref{lemma-injective-flat}. 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 \ref{algebra-lemma-cover}. Finally, the (derived) base change of the map $A \to R\Hom_A(\omega_A^\bullet, \omega_A^\bullet)$ to $A_{f_i}$ is the map $A_{f_i} \to R\Hom_A((\omega_A^\bullet)_{f_i}, (\omega_A^\bullet)_{f_i})$ by More on Algebra, Lemma \ref{more-algebra-lemma-base-change-RHom}. Hence we deduce that $A \to R\Hom_A(\omega_A^\bullet, \omega_A^\bullet)$ is an isomorphism and the proof is complete.
Let $\omega_A^\bullet \to I^\bullet$ be a quasi-isomorphism with $I^\bullet$ a bounded complex of injectives. Then $\Hom_A(B, I^\bullet)$ is a bounded complex of injective $B$-modules (Lemma \ref{lemma-hom-injective}) representing $R\Hom(B, \omega_A^\bullet)$. Thus $R\Hom(B, \omega_A^\bullet)$ has finite injective dimension. By Lemma \ref{lemma-exact-support-coherent} it is an object of $D_{\textit{Coh}}(B)$. Finally, we compute $$ \Hom_{D(B)}(R\Hom(B, \omega_A^\bullet), R\Hom(B, \omega_A^\bullet)) = \Hom_{D(A)}(R\Hom(B, \omega_A^\bullet), \omega_A^\bullet) = B $$ and for $n \not = 0$ we compute $$ \Hom_{D(B)}(R\Hom(B, \omega_A^\bullet), R\Hom(B, \omega_A^\bullet)[n]) = \Hom_{D(A)}(R\Hom(B, \omega_A^\bullet), \omega_A^\bullet[n]) = 0 $$ which proves the last property of a dualizing complex. In the displayed equations, the first equality holds by Lemma \ref{lemma-right-adjoint} and the second equality holds by Lemma \ref{lemma-dualizing}.
Set $B = A[x]$ and $\omega_B^\bullet = \omega_A^\bullet \otimes_A B$. It follows from Lemma \ref{lemma-injective-dimension-over-polynomial-ring} and More on Algebra, Lemma \ref{more-algebra-lemma-finite-injective-dimension} 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 $$ B \longrightarrow R\Hom_B(\omega_B^\bullet, \omega_B^\bullet) $$ is a quasi-isomorphism as formation of internal hom commutes with flat base change in this case, see More on Algebra, Lemma \ref{more-algebra-lemma-base-change-RHom}.
This is true because $R\Hom_A(\kappa, \omega_A^\bullet)$ is a dualizing complex over $\kappa$ (Lemma \ref{lemma-dualizing-quotient}), because dualizing complexes over $\kappa$ are unique up to shifts (Lemma \ref{lemma-dualizing-unique}), and because $\kappa$ is a dualizing complex over $\kappa$.
By Lemma \ref{lemma-dualizing-finite} the complex $\omega_B^\bullet$ is dualizing for $B$. We have $$ R\Hom_B(\kappa', \omega_B^\bullet) = R\Hom_B(\kappa', R\Hom(B, \omega_A^\bullet)) = R\Hom_A(\kappa', \omega_A^\bullet) $$ by Lemma \ref{lemma-right-adjoint}. Since $\kappa'$ is isomorphic to a finite direct sum of copies of $\kappa$ as an $A$-module and since $\omega_A^\bullet$ is normalized, we see that this complex only has cohomology placed in degree $0$. Thus $\omega_B^\bullet$ is a normalized dualizing complex as well.
By induction on the length of $N$ we see that $R\Hom_A(N, \omega_A^\bullet)$ is a module of finite length sitting in degree $0$. Thus $R\Hom_A(-, \omega_A^\bullet)$ induces an anti-equivalence on the category of finite length modules. Since the same is true for $\Hom_A(-, E)$ by Proposition \ref{proposition-matlis} we see that $$ N \longmapsto \Hom_A(R\Hom_A(N, \omega_A^\bullet), E) $$ is an equivalence as in Lemma \ref{lemma-equivalence-finite-length}. Hence it is isomorphic to the identity functor. Since $\Hom_A(-, E)$ applied twice is the identity (Proposition \ref{proposition-matlis}) we obtain the statement of the lemma.
We prove this by induction on $d$. If $d = 0$, this follows from Lemma \ref{lemma-dualizing-finite-length} and Matlis duality (Proposition \ref{proposition-matlis}) which guarantees that $\Hom_A(M, E)$ is nonzero if $M$ is nonzero. \medskip\noindent Assume the result holds for modules with support of dimension $< d$ and that $M$ has depth $> 0$. Choose an $f \in \mathfrak m$ which is a nonzerodivisor on $M$ and consider the short exact sequence $$ 0 \to M \to M \to M/fM \to 0 $$ Since $\dim(\text{Supp}(M/fM)) = d - 1$ (Algebra, Lemma \ref{algebra-lemma-one-equation-module}) we may apply the induction hypothesis. Writing $E^i = \Ext^i_A(M, \omega_A^\bullet)$ and $F^i = \Ext^i_A(M/fM, \omega_A^\bullet)$ we obtain a long exact sequence $$ \ldots \to F^i \to E^i \xrightarrow{f} E^i \to F^{i + 1} \to \ldots $$ By induction $E^i/fE^i = 0$ for $i + 1 \not \in \{-\dim(\text{Supp}(M/fM)), \ldots, -\text{depth}(M/fM)\}$. By Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) and Algebra, Lemma \ref{algebra-lemma-depth-drops-by-one} we conclude $E^i = 0$ for $i \not \in \{-\dim(\text{Supp}(M)), \ldots, -\text{depth}(M)\}$. Moreover, in the boundary case $i = - \text{depth}(M)$ we deduce that $E^i$ is nonzero as $F^{i + 1}$ is nonzero by induction. Since $E^i/fE^i \subset F^{i + 1}$ we get $$ \dim(\text{Supp}(F^{i + 1})) \geq \dim(\text{Supp}(E^i/fE^i)) \geq \dim(\text{Supp}(E^i)) - 1 $$ (see lemma used above) we also obtain the dimension estimate (2). \medskip\noindent If $M$ has depth $0$ and $d > 0$ we let $N = M[\mathfrak m^\infty]$ and set $M' = M/N$ (compare with Lemma \ref{lemma-divide-by-torsion}). Then $M'$ has depth $> 0$ and $\dim(\text{Supp}(M')) = d$. Thus we know the result for $M'$ and since $R\Hom_A(N, \omega_A^\bullet) = \Hom_A(N, E)$ (Lemma \ref{lemma-dualizing-finite-length}) the long exact cohomology sequence of $\Ext$'s implies the result for $M$.
We will use the results of Lemma \ref{lemma-sitting-in-degrees} without further mention. Fix a finite module $M$. If $M$ is Cohen-Macaulay, then only $\Ext^{-d}_A(M, \omega_A^\bullet)$ can be nonzero, hence (1) $\Rightarrow$ (3). The implication (3) $\Rightarrow$ (2) is immediate. Assume (2) and let $N = \Ext^{-\delta}_A(M, \omega_A^\bullet)$ be the nonzero $\Ext$ where $\delta = \text{depth}(M)$. Then, since $$ M[0] = R\Hom_A(R\Hom_A(M, \omega_A^\bullet), \omega_A^\bullet) = R\Hom_A(N[\delta], \omega_A^\bullet) $$ (Lemma \ref{lemma-dualizing}) we conclude that $M = \Ext_A^{-\delta}(N, \omega_A^\bullet)$. Thus $\delta \geq \dim(\text{Supp}(M))$. However, since we also know that $\delta \leq \dim(\text{Supp}(M))$ (Algebra, Lemma \ref{algebra-lemma-bound-depth}) we conclude that $M$ is Cohen-Macaulay. \medskip\noindent To prove the final statement, it suffices to show that $N = \Ext^{-d}_A(M, \omega_A^\bullet)$ is in $CM_d$ for $M$ in $CM_d$. Above we have seen that $M[0] = R\Hom_A(N[d], \omega_A^\bullet)$ and this proves the desired result by the equivalence of (1) and (3).
Since $A_\mathfrak p$ has dimension zero, there exists an integer $n > 0$ such that $\mathfrak p^nA_\mathfrak p$ is zero. Set $B = A/\mathfrak p^n$ and $\omega_B^\bullet = R\Hom_A(B, \omega_A^\bullet)$. Since $B_\mathfrak p = A_\mathfrak p$ we see that $$ (\omega_B^\bullet)_\mathfrak p = R\Hom_A(B, \omega_A^\bullet) \otimes_A^\mathbf{L} A_\mathfrak p = R\Hom_{A_\mathfrak p}(B_\mathfrak p, (\omega_A^\bullet)_\mathfrak p) = (\omega_A^\bullet)_\mathfrak p $$ The second equality holds by More on Algebra, Lemma \ref{more-algebra-lemma-base-change-RHom}. By Lemma \ref{lemma-normalized-quotient} we may replace $A$ by $B$. After doing so, we see that $\dim(A) = e$. Then we see that $H^i(\omega_A^\bullet)_\mathfrak p$ can only be nonzero if $i = -e$ by Lemma \ref{lemma-sitting-in-degrees} parts (1) and (2). On the other hand, since $(\omega_A^\bullet)_\mathfrak p$ is a dualizing complex for the nonzero ring $A_\mathfrak p$ (Lemma \ref{lemma-dualizing-localize}) we see that the remaining module has to be nonzero.
The complex $\omega_A^\bullet \otimes_A A_\mathfrak p$ is a dualizing complex for $A_\mathfrak p$ (Lemma \ref{lemma-dualizing-localize}). The dimension of $A_\mathfrak p$ is zero as $\mathfrak p$ is minimal. Hence the result follows from Lemma \ref{lemma-dualizing-artinian}.
Let $\mathfrak p \subset \mathfrak q$ be an immediate specialization. We have to show that $\delta(\mathfrak p) = \delta(\mathfrak q) + 1$. We may replace $A$ by $A/\mathfrak p$, the complex $\omega_A^\bullet$ by $\omega_{A/\mathfrak p}^\bullet = R\Hom(A/\mathfrak p, \omega_A^\bullet)$, the prime $\mathfrak p$ by $(0)$, and the prime $\mathfrak q$ by $\mathfrak q/\mathfrak p$, see Lemma \ref{lemma-quotient-function}. Thus we may assume that $A$ is a domain, $\mathfrak p = (0)$, and $\mathfrak q$ is a prime ideal of height $1$. \medskip\noindent Then $H^i(\omega_A^\bullet)_{(0)}$ is nonzero for exactly one $i$, say $i_0$, by Lemma \ref{lemma-nonvanishing-generically}. In fact $i_0 = -\delta((0))$ because $(\omega_A^\bullet)_{(0)}[-\delta((0))]$ is a normalized dualizing complex over the field $A_{(0)}$. \medskip\noindent On the other hand $(\omega_A^\bullet)_\mathfrak q[-\delta(\mathfrak q)]$ is a normalized dualizing complex for $A_\mathfrak q$. By Lemma \ref{lemma-nonvanishing-generically-local} we see that $$ H^e((\omega_A^\bullet)_\mathfrak q[-\delta(\mathfrak q)])_{(0)} = H^{e - \delta(\mathfrak q)}(\omega_A^\bullet)_{(0)} $$ is nonzero only for $e = -\dim(A_\mathfrak q) = -1$. We conclude $$ -\delta((0)) = -1 - \delta(\mathfrak q) $$ as desired.
Because $\Spec(A)$ has a dimension function by Lemma \ref{lemma-dimension-function} it is catenary, see Topology, Lemma \ref{topology-lemma-dimension-function-catenary}. Hence $A$ is catenary, see Algebra, Lemma \ref{algebra-lemma-catenary}. It follows from Proposition \ref{proposition-dualizing-essentially-finite-type} that $A$ is universally catenary. \medskip\noindent Because any dualizing complex $\omega_A^\bullet$ is in $D^b_{\textit{Coh}}(A)$ the values of the function $\delta_{\omega_A^\bullet}$ in minimal primes are bounded by Lemma \ref{lemma-nonvanishing-generically}. On the other hand, for a maximal ideal $\mathfrak m$ with residue field $\kappa$ the integer $i = -\delta(\mathfrak m)$ is the unique integer such that $\Ext_A^i(\kappa, \omega_A^\bullet)$ is nonzero (Lemma \ref{lemma-find-function}). Since $\omega_A^\bullet$ has finite injective dimension these values are bounded too. Since the dimension of $A$ is the maximal value of $\delta(\mathfrak p) - \delta(\mathfrak m)$ where $\mathfrak p \subset \mathfrak m$ are a pair consisting of a minimal prime and a maximal prime we find that the dimension of $\Spec(A)$ is bounded.
We will use Lemma \ref{lemma-sitting-in-degrees} without further mention. By Lemma \ref{lemma-nonvanishing-generically-local} the support of $\omega_A$ contains the irreducible components of dimension $d$. Let $\mathfrak p \subset A$ be a prime. By Lemma \ref{lemma-dimension-function} the complex $(\omega_A^\bullet)_{\mathfrak p}[-\dim(A/\mathfrak p)]$ is a normalized dualizing complex for $A_\mathfrak p$. Hence if $\dim(A/\mathfrak p) + \dim(A_\mathfrak p) < d$, then $(\omega_A)_\mathfrak p = 0$. This proves the support of $\omega_A$ is the union of the irreducible components of dimension $d$, because the complement of this union is exactly the primes $\mathfrak p$ of $A$ for which $\dim(A/\mathfrak p) + \dim(A_\mathfrak p) < d$ as $A$ is catenary (Lemma \ref{lemma-universally-catenary}). On the other hand, if $\dim(A/\mathfrak p) + \dim(A_\mathfrak p) = d$, then $$ (\omega_A)_\mathfrak p = H^{-\dim(A_\mathfrak p)}\left( (\omega_A^\bullet)_{\mathfrak p}[-\dim(A/\mathfrak p)] \right) $$ Hence in order to prove $\omega_A$ has $(S_2)$ it suffices to show that the depth of $\omega_A$ is at least $\min(\dim(A), 2)$. We prove this by induction on $\dim(A)$. The case $\dim(A) = 0$ is trivial. \medskip\noindent Assume $\text{depth}(A) > 0$. Choose a nonzerodivisor $f \in \mathfrak m$ and set $B = A/fA$. Then $\dim(B) = \dim(A) - 1$ and we may apply the induction hypothesis to $B$. By Lemma \ref{lemma-divide-by-nonzerodivisor} we see that multiplication by $f$ is injective on $\omega_A$ and we get $\omega_A/f\omega_A \subset \omega_B$. This proves the depth of $\omega_A$ is at least $1$. If $\dim(A) > 1$, then $\dim(B) > 0$ and $\omega_B$ has depth $ > 0$. Hence $\omega_A$ has depth $> 1$ and we conclude in this case. \medskip\noindent Assume $\dim(A) > 0$ and $\text{depth}(A) = 0$. Let $I = A[\mathfrak m^\infty]$ and set $B = A/I$. Then $B$ has depth $\geq 1$ and $\omega_A = \omega_B$ by Lemma \ref{lemma-divide-by-finite-length-ideal}. Since we proved the result for $\omega_B$ above the proof is done.