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 \input{preamble} % OK, start here. % \begin{document} \title{Dualizing Complexes} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Introduction} \label{section-introduction} \noindent A reference is the book \cite{RD}. \medskip\noindent The goals of this chapter are the following: \begin{enumerate} \item Define what it means to have a dualizing complex $\omega_A^\bullet$ over a Noetherian ring $A$, namely \begin{enumerate} \item we have $\omega_A^\bullet \in D^{+}(A)$, \item the cohomology modules $H^i(\omega_A^\bullet)$ are all finite $A$-modules, \item $\omega_A^\bullet$ has finite injective dimension, and \item we have $A \to R\Hom_A(\omega_A^\bullet, \omega_A^\bullet)$ is a quasi-isomorphism. \end{enumerate} \item List elementary properties of dualizing complexes. \item Show a dualizing complex gives rise to a dimension function. \item Show a dualizing complex gives rise to a good notion of a reflexive hull. \item Prove the finiteness theorem when a dualizing complex exists. \end{enumerate} \section{Essential surjections and injections} \label{section-essential} \noindent We will mostly work in categories of modules, but we may as well make the definition in general. \begin{definition} \label{definition-essential} Let $\mathcal{A}$ be an abelian category. \begin{enumerate} \item An injection $A \subset B$ of $\mathcal{A}$ is {\it essential}, or we say that $B$ is an {\it essential extension of} $A$, if every nonzero subobject $B' \subset B$ has nonzero intersection with $A$. \item A surjection $f : A \to B$ of $\mathcal{A}$ is {\it essential} if for every proper subobject $A' \subset A$ we have $f(A') \not = B$. \end{enumerate} \end{definition} \noindent Some lemmas about this notion. \begin{lemma} \label{lemma-essential} Let $\mathcal{A}$ be an abelian category. \begin{enumerate} \item If $A \subset B$ and $B \subset C$ are essential extensions, then $A \subset C$ is an essential extension. \item If $A \subset B$ is an essential extension and $C \subset B$ is a subobject, then $A \cap C \subset C$ is an essential extension. \item If $A \to B$ and $B \to C$ are essential surjections, then $A \to C$ is an essential surjection. \item Given an essential surjection $f : A \to B$ and a surjection $A \to C$ with kernel $K$, the morphism $C \to B/f(K)$ is an essential surjection. \end{enumerate} \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-union-essential-extensions} Let $R$ be a ring. Let $M$ be an $R$-module. Let $E = \colim E_i$ be a filtered colimit of $R$-modules. Suppose given a compatible system of essential injections $M \to E_i$ of $R$-modules. Then $M \to E$ is an essential injection. \end{lemma} \begin{proof} Immediate from the definitions and the fact that filtered colimits are exact (Algebra, Lemma \ref{algebra-lemma-directed-colimit-exact}). \end{proof} \begin{lemma} \label{lemma-essential-extension} Let $R$ be a ring. Let $M \subset N$ be $R$-modules. The following are equivalent \begin{enumerate} \item $M \subset N$ is an essential extension, \item for all $x \in N$ there exists an $f \in R$ such that $fx \in M$ and $fx \not = 0$. \end{enumerate} \end{lemma} \begin{proof} Assume (1) and let $x \in N$ be a nonzero element. By (1) we have $Rx \cap M \not = 0$. This implies (2). \medskip\noindent Assume (2). Let $N' \subset N$ be a nonzero submodule. Pick $x \in N'$ nonzero. By (2) we can find $f \in$ with $fx \in N$ and $fx \not = 0$. Thus $N' \cap M \not = 0$. \end{proof} \section{Injective modules} \label{section-injective-modules} \noindent Some results about injective modules over rings. \begin{lemma} \label{lemma-product-injectives} Let $R$ be a ring. Any product of injective $R$-modules is injective. \end{lemma} \begin{proof} Special case of Homology, Lemma \ref{homology-lemma-product-injectives}. \end{proof} \begin{lemma} \label{lemma-injective-flat} Let $R \to S$ be a flat ring map. If $E$ is an injective $S$-module, then $E$ is injective as an $R$-module. \end{lemma} \begin{proof} This is true because $\Hom_R(M, E) = \Hom_S(M \otimes_R S, E)$ by Algebra, Lemma \ref{algebra-lemma-adjoint-tensor-restrict} and the fact that tensoring with $S$ is exact. \end{proof} \begin{lemma} \label{lemma-injective-epimorphism} Let $R \to S$ be an epimorphism of rings. Let $E$ be an $S$-module. If $E$ is injective as an $R$-module, then $E$ is an injective $S$-module. \end{lemma} \begin{proof} This is true because $\Hom_R(N, E) = \Hom_S(N, E)$ for any $S$-module $N$, see Algebra, Lemma \ref{algebra-lemma-epimorphism-modules}. \end{proof} \begin{lemma} \label{lemma-hom-injective} Let $R \to S$ be a ring map. If $E$ is an injective $R$-module, then $\Hom_R(S, E)$ is an injective $S$-module. \end{lemma} \begin{proof} This is true because $\Hom_S(N, \Hom_R(S, E)) = \Hom_R(N, E)$ by Algebra, Lemma \ref{algebra-lemma-adjoint-hom-restrict}. \end{proof} \begin{lemma} \label{lemma-essential-extensions-in-injective} Let $R$ be a ring. Let $I$ be an injective $R$-module. Let $E \subset I$ be a submodule. The following are equivalent \begin{enumerate} \item $E$ is injective, and \item for all $E \subset E' \subset I$ with $E \subset E'$ essential we have $E = E'$. \end{enumerate} In particular, an $R$-module is injective if and only if every essential extension is trivial. \end{lemma} \begin{proof} The final assertion follows from the first and the fact that the category of $R$-modules has enough injectives (More on Algebra, Section \ref{more-algebra-section-injectives-modules}). \medskip\noindent Assume (1). Let $E \subset E' \subset I$ as in (2). Then the map $\text{id}_E : E \to E$ can be extended to a map $\alpha : E' \to E$. The kernel of $\alpha$ has to be zero because it intersects $E$ trivially and $E'$ is an essential extension. Hence $E = E'$. \medskip\noindent Assume (2). Let $M \subset N$ be $R$-modules and let $\varphi : M \to E$ be an $R$-module map. In order to prove (1) we have to show that $\varphi$ extends to a morphism $N \to E$. Consider the set $\mathcal{S}$ of pairs $(M', \varphi')$ where $M \subset M' \subset N$ and $\varphi' : M' \to E$ is an $R$-module map agreeing with $\varphi$ on $M$. We define an ordering on $\mathcal{S}$ by the rule $(M', \varphi') \leq (M'', \varphi'')$ if and only if $M' \subset M''$ and $\varphi''|_{M'} = \varphi'$. It is clear that we can take the maximum of a totally ordered subset of $\mathcal{S}$. Hence by Zorn's lemma we may assume $(M, \varphi)$ is a maximal element. \medskip\noindent Choose an extension $\psi : N \to I$ of $\varphi$ composed with the inclusion $E \to I$. This is possible as $I$ is injective. If $\psi(N) \subset E$, then $\psi$ is the desired extension. If $\psi(N)$ is not contained in $E$, then by (2) the inclusion $E \subset E + \psi(N)$ is not essential. hence we can find a nonzero submodule $K \subset E + \psi(N)$ meeting $E$ in $0$. This means that $M' = \psi^{-1}(E + K)$ strictly contains $M$. Thus we can extend $\varphi$ to $M'$ using $$M' \xrightarrow{\psi|_{M'}} E + K \to (E + K)/K = E$$ This contradicts the maximality of $(M, \varphi)$. \end{proof} \begin{example} \label{example-reduced-ring-injective} Let $R$ be a reduced ring. Let $\mathfrak p \subset R$ be a minimal prime so that $K = R_\mathfrak p$ is a field (Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}). Then $K$ is an injective $R$-module. Namely, we have $\Hom_R(M, K) = \Hom_K(M_\mathfrak p, K)$ for any $R$-module $M$. Since localization is an exact functor and taking duals is an exact functor on $K$-vector spaces we conclude $\Hom_R(-, K)$ is an exact functor, i.e., $K$ is an injective $R$-module. \end{example} \begin{lemma} \label{lemma-characterize-injective} Let $R$ be a ring. Let $E$ be an $R$-module. The following are equivalent \begin{enumerate} \item $E$ is an injective $R$-module, and \item given an ideal $I \subset R$ and a module map $\varphi : I \to E$ there exists an extension of $\varphi$ to an $R$-module map $R \to E$. \end{enumerate} \end{lemma} \begin{proof} The implication (1) $\Rightarrow$ (2) follows from the definitions. Thus we assume (2) holds and we prove (1). First proof: The lemma follows from More on Algebra, Lemma \ref{more-algebra-lemma-characterize-injective-bis}. Second proof: Since $R$ is a generator for the category of $R$-modules, the lemma follows from Injectives, Lemma \ref{injectives-lemma-characterize-injective}. \medskip\noindent Third proof: We have to show that every essential extension $E \subset E'$ is trivial, see Lemma \ref{lemma-essential-extensions-in-injective}. Pick $x \in E'$ and set $I = \{f \in R \mid fx \in E\}$. The map $I \to E$, $f \mapsto fx$ extends to $\psi : R \to E$ by (2). Then $x' = x - \psi(1)$ is an element of $E'$ whose annihilator in $E'/E$ is $I$ and which is annihilated by $I$ as an element of $E'$. Thus $Rx' = (R/I)x'$ does not intersect $E$. Since $E \subset E'$ is an essential extension it follows that $x' \in E$ as desired. \end{proof} \begin{lemma} \label{lemma-sum-injective-modules} Let $R$ be a Noetherian ring. A direct sum of injective modules is injective. \end{lemma} \begin{proof} Let $E_i$ be a family of injective modules parametrized by a set $I$. Set $E = \bigcup E_i$. To show that $E$ is injective we use Lemma \ref{lemma-characterize-injective}. Thus let $\varphi : I \to E$ be a module map from an ideal of $R$ into $E$. As $I$ is a finite $R$-module (because $R$ is Noetherian) we can find finitely many elements $i_1, \ldots, i_r \in I$ such that $\varphi$ maps into $\bigcup_{j = 1, \ldots, r} E_{i_j}$. Then we can extend $\varphi$ into $\bigcup_{j = 1, \ldots, r} E_{i_j}$ using the injectivity of the modules $E_{i_j}$. \end{proof} \begin{lemma} \label{lemma-localization-injective-modules} Let $R$ be a Noetherian ring. Let $S \subset R$ be a multiplicative subset. If $E$ is an injective $R$-module, then $S^{-1}E$ is an injective $S^{-1}R$-module. \end{lemma} \begin{proof} Since $R \to S^{-1}R$ is an epimorphism of rings, it suffices to show that $S^{-1}E$ is injective as an $R$-module, see Lemma \ref{lemma-injective-epimorphism}. To show this we use Lemma \ref{lemma-characterize-injective}. Thus let $I \subset R$ be an ideal and let $\varphi : I \to S^{-1} E$ be an $R$-module map. As $I$ is a finitely presented $R$-module (because $R$ is Noetherian) we can find find an $f \in S$ and an $R$-module map $I \to E$ such that $f\varphi$ is the composition $I \to E \to S^{-1}E$ (Algebra, Lemma \ref{algebra-lemma-hom-from-finitely-presented}). Then we can extend $I \to E$ to a homomorphism $R \to E$. Then the composition $$R \to E \to S^{-1}E \xrightarrow{f^{-1}} S^{-1}E$$ is the desired extension of $\varphi$ to $R$. \end{proof} \begin{lemma} \label{lemma-injective-module-divide} Let $R$ be a Noetherian ring. Let $I$ be an injective $R$-module. \begin{enumerate} \item Let $f \in R$. Then $E = \bigcup I[f^n] = I[f^\infty]$ is an injective submodule of $I$. \item Let $J \subset R$ be an ideal. Then the $J$-power torsion submodule $I[J^\infty]$ is an injective submodule of $I$. \end{enumerate} \end{lemma} \begin{proof} We will use Lemma \ref{lemma-essential-extensions-in-injective} to prove (1). Suppose that $E \subset E' \subset I$ and that $E'$ is an essential extension of $E$. We will show that $E' = E$. If not, then we can find $x \in E'$ and $x \not \in E$. Let $J = \{a \in R \mid ax \in E'\}$. Since $R$ is Noetherian we can choose $x$ with $J$ maximal. Since $R$ is Noetherian we can write $J = (g_1, \ldots, g_t)$ for some $g_i \in R$. Say $f^{n_i}$ annihilates $g_ix$. Set $n = \max\{n_i\}$. Then $x' = f^n x$ is an element of $E'$ not in $E$ and is annihilated by $J$. By maximality of $J$ we see that $R x' = (R/J)x' \cap E = (0)$. Hence $E'$ is not an essential extension of $E$ a contradiction. \medskip\noindent To prove (2) write $J = (f_1, \ldots, f_t)$. Then $I[J^\infty]$ is equal to $$(\ldots((I[f_1^\infty])[f_2^\infty])\ldots)[f_t^\infty]$$ and the result follows from (1) and induction. \end{proof} \begin{lemma} \label{lemma-injective-dimension-over-polynomial-ring} Let $A$ be a Noetherian ring. Let $E$ be an injective $A$-module. Then $E \otimes_A A[x]$ has injective-amplitude $[0, 1]$ as an object of $D(A[x])$. In particular, $E \otimes_A A[x]$ has finite injective dimension as an $A[x]$-module. \end{lemma} \begin{proof} Let us write $E[x] = E \otimes_A A[x]$. Consider the short exact sequence of $A[x]$-modules $$0 \to E[x] \to \Hom_A(A[x], E[x]) \to \Hom_A(A[x], E[x]) \to 0$$ where the first map sends $p \in E[x]$ to $f \mapsto fp$ and the second map sends $\varphi$ to $f \mapsto \varphi(xf) - x\varphi(f)$. The second map is surjective because $\Hom_A(A[x], E[x]) = \prod_{n \geq 0} E[x]$ as an abelian group and the map sends $(e_n)$ to $(e_{n + 1} - xe_n)$ which is surjective. As an $A$-module we have $E[x] \cong \bigoplus_{n \geq 0} E$ which is injective by Lemma \ref{lemma-sum-injective-modules}. Hence the $A[x]$-module $\Hom_A(A[x], I[x])$ is injective by Lemma \ref{lemma-hom-injective} and the proof is complete. \end{proof} \section{Projective covers} \label{section-projective-cover} \noindent In this section we briefly discuss projective covers. \begin{definition} \label{definition-projective-cover} Let $R$ be a ring. A surjection $P \to M$ of $R$-modules is said to be a {\it projective cover}, or sometimes a {\it projective envelope}, if $P$ is a projective $R$-module and $P \to M$ is an essential surjection. \end{definition} \noindent Projective covers do not always exist. For example, if $k$ is a field and $R = k[x]$ is the polynomial ring over $k$, then the module $M = R/(x)$ does not have a projective cover. Namely, for any surjection $f : P \to M$ with $P$ projective over $R$, the proper submodule $(x - 1)P$ surjects onto $M$. Hence $f$ is not essential. \begin{lemma} \label{lemma-projective-cover-unique} Let $R$ be a ring and let $M$ be an $R$-module. If a projective cover of $M$ exists, then it is unique up to isomorphism. \end{lemma} \begin{proof} Let $P \to M$ and $P' \to M$ be projective covers. Because $P$ is a projective $R$-module and $P' \to M$ is surjective, we can find an $R$-module map $\alpha : P \to P'$ compatible with the maps to $M$. Since $P' \to M$ is essential, we see that $\alpha$ is surjective. As $P'$ is a projective $R$-module we can choose a direct sum decomposition $P = \Ker(\alpha) \oplus P'$. Since $P' \to M$ is surjective and since $P \to M$ is essential we conclude that $\Ker(\alpha)$ is zero as desired. \end{proof} \noindent Here is an example where projective covers exist. \begin{lemma} \label{lemma-projective-covers-local} Let $(R, \mathfrak m, \kappa)$ be a local ring. Any finite $R$-module has a projective cover. \end{lemma} \begin{proof} Let $M$ be a finite $R$-module. Let $r = \dim_\kappa(M/\mathfrak m M)$. Choose $x_1, \ldots, x_r \in M$ mapping to a basis of $M/\mathfrak m M$. Consider the map $f : R^{\oplus r} \to M$. By Nakayama's lemma this is a surjection (Algebra, Lemma \ref{algebra-lemma-NAK}). If $N \subset R^{\oplus R}$ is a proper submodule, then $N/\mathfrak m N \to \kappa^{\oplus r}$ is not surjective (by Nakayama's lemma again) hence $N/\mathfrak m N \to M/\mathfrak m M$ is not surjective. Thus $f$ is an essential surjection. \end{proof} \section{Injective hulls} \label{section-injective-hull} \noindent In this section we briefly discuss injective hulls. \begin{definition} \label{definition-injective-hull} Let $R$ be a ring. A injection $M \to I$ of $R$-modules is said to be an {\it injective hull} if $I$ is a injective $R$-module and $M \to I$ is an essential injection. \end{definition} \noindent Injective hulls always exist. \begin{lemma} \label{lemma-injective-hull} Let $R$ be a ring. Any $R$-module has an injective hull. \end{lemma} \begin{proof} Let $M$ be an $R$-module. By More on Algebra, Section \ref{more-algebra-section-injectives-modules} the category of $R$-modules has enough injectives. Choose an injection $M \to I$ with $I$ an injective $R$-module. Consider the set $\mathcal{S}$ of submodules $M \subset E \subset I$ such that $E$ is an essential extension of $M$. We order $\mathcal{S}$ by inclusion. If $\{E_\alpha\}$ is a totally ordered subset of $\mathcal{S}$, then $\bigcup E_\alpha$ is an essential extension of $M$ too (Lemma \ref{lemma-union-essential-extensions}). Thus we can apply Zorn's lemma and find a maximal element $E \in \mathcal{S}$. We claim $M \subset E$ is an injective hull, i.e., $E$ is an injective $R$-module. This follows from Lemma \ref{lemma-essential-extensions-in-injective}. \end{proof} \begin{lemma} \label{lemma-injective-hull-unique} Let $R$ be a ring. Let $M$, $N$ be $R$-modules and let $M \to E$ and $N \to E'$ be injective hulls. Then \begin{enumerate} \item for any $R$-module map $\varphi : M \to N$ there exists an $R$-module map $\psi : E \to E'$ such that $$\xymatrix{ M \ar[r] \ar[d]_\varphi & E \ar[d]^\psi \\ N \ar[r] & E' }$$ commutes, \item if $\varphi$ is injective, then $\psi$ is injective, \item if $\varphi$ is an essential injection, then $\psi$ is an isomorphism, \item if $\varphi$ is an isomorphism, then $\psi$ is an isomorphism, \item if $M \to I$ is an embedding of $M$ into an injective $R$-module, then there is an isomorphism $I \cong E \oplus I'$ compatible with the embeddings of $M$, \end{enumerate} In particular, the injective hull $E$ of $M$ is unique up to isomorphism. \end{lemma} \begin{proof} Part (1) follows from the fact that $E'$ is an injective $R$-module. Part (2) follows as $\Ker(\psi) \cap M = 0$ and $E$ is an essential extension of $M$. Assume $\varphi$ is an essential injection. Then $E \cong \psi(E) \subset E'$ by (2) which implies $E' = \psi(E) \oplus E''$ because $E$ is injective. Since $E'$ is an essential extension of $M$ (Lemma \ref{lemma-essential}) we get $E'' = 0$. Part (4) is a special case of (3). Assume $M \to I$ as in (5). Choose a map $\alpha : E \to I$ extending the map $M \to I$. Arguing as before we see that $\alpha$ is injective. Thus as before $\alpha(E)$ splits off from $I$. This proves (5). \end{proof} \begin{example} \label{example-injective-hull-domain} Let $R$ be a domain with fraction field $K$. Then $R \subset K$ is an injective hull of $R$. Namely, by Example \ref{example-reduced-ring-injective} we see that $K$ is an injective $R$-module and by Lemma \ref{lemma-essential-extension} we see that $R \subset K$ is an essential extension. \end{example} \begin{definition} \label{definition-indecomposable} An object $X$ of an additive category is called {\it indecomposable} if it is nonzero and if $X = Y \oplus Z$, then either $Y = 0$ or $Z = 0$. \end{definition} \begin{lemma} \label{lemma-indecomposable-injective} Let $R$ be a ring. Let $E$ be an indecomposable injective $R$-module. Then \begin{enumerate} \item $E$ is the injective hull of any nonzero submodule of $E$, \item the intersection of any two nonzero submodules of $E$ is nonzero, \item $\text{End}_R(E, E)$ is a noncommutative local ring with maximal ideal those $\varphi : E \to E$ whose kernel is nonzero, and \item the set of zerodivisors on $E$ is a prime ideal $\mathfrak p$ of $R$ and $E$ is an injective $R_\mathfrak p$-module. \end{enumerate} \end{lemma} \begin{proof} Part (1) follows from Lemma \ref{lemma-injective-hull-unique}. Part (2) follows from part (1) and the definition of injective hulls. \medskip\noindent Proof of (3). Set $A = \text{End}_R(E, E)$ and $I = \{\varphi \in A \mid \Ker(f) \not = 0\}$. The statement means that $I$ is a two sided ideal and that any $\varphi \in A$, $\varphi \not \in I$ is invertible. Suppose $\varphi$ and $\psi$ are not injective. Then $\Ker(\varphi) \cap \Ker(\psi)$ is nonzero by (2). Hence $\varphi + \psi \in I$. It follows that $I$ is a two sided ideal. If $\varphi \in A$, $\varphi \not \in I$, then $E \cong \varphi(E) \subset E$ is an injective submodule, hence $E = \varphi(E)$ because $E$ is indecomposable. \medskip\noindent Proof of (4). Consider the ring map $R \to A$ and let $\mathfrak p \subset R$ be the inverse image of the maximal ideal $I$. Then it is clear that $\mathfrak p$ is a prime ideal and that $R \to A$ extends to $R_\mathfrak p \to A$. Thus $E$ is an $R_\mathfrak p$-module. It follows from Lemma \ref{lemma-injective-epimorphism} that $E$ is injective as an $R_\mathfrak p$-module. \end{proof} \begin{lemma} \label{lemma-injective-hull-indecomposable} Let $\mathfrak p \subset R$ be a prime of a ring $R$. Let $E$ be the injective hull of $R/\mathfrak p$. Then \begin{enumerate} \item $E$ is indecomposable, \item $E$ is the injective hull of $\kappa(\mathfrak p)$, \item $E$ is the injective hull of $\kappa(\mathfrak p)$ over the ring $R_\mathfrak p$. \end{enumerate} \end{lemma} \begin{proof} As $R/\mathfrak p \subset \kappa(\mathfrak p)$ we can extend the embedding to a map $\kappa(\mathfrak p) \to E$. Hence (2) holds. For $f \in R$, $f \not \in \mathfrak p$ the map $f : \kappa(\mathfrak p) \to \kappa(\mathfrak p)$ is an isomorphism hence the map $f : E \to E$ is an isomorphism, see Lemma \ref{lemma-injective-hull-unique}. Thus $E$ is an $R_\mathfrak p$-module. It is injective as an $R_\mathfrak p$-module by Lemma \ref{lemma-injective-epimorphism}. Finally, let $E' \subset E$ be a nonzero injective $R$-submodule. Then $J = (R/\mathfrak p) \cap E'$ is nonzero. After shrinking $E'$ we may assume that $E'$ is the injective hull of $J$ (see Lemma \ref{lemma-injective-hull-unique} for example). Observe that $R/\mathfrak p$ is an essential extension of $J$ for example by Lemma \ref{lemma-essential-extension}. Hence $E' \to E$ is an isomorphism by Lemma \ref{lemma-injective-hull-unique} part (3). Hence $E$ is indecomposable. \end{proof} \begin{lemma} \label{lemma-indecomposable-injective-noetherian} Let $R$ be a Noetherian ring. Let $E$ be an indecomposable injective $R$-module. Then there exists a prime ideal $\mathfrak p$ of $R$ such that $E$ is the injective hull of $\kappa(\mathfrak p)$. \end{lemma} \begin{proof} Let $\mathfrak p$ be the prime ideal found in Lemma \ref{lemma-indecomposable-injective}. Say $\mathfrak p = (f_1, \ldots, f_r)$. Pick a nonzero element $x \in \bigcap \Ker(f_i : E \to E)$, see Lemma \ref{lemma-indecomposable-injective}. Then $(R_\mathfrak p)x$ is a module isomorphic to $\kappa(\mathfrak p)$ inside $E$. We conclude by Lemma \ref{lemma-indecomposable-injective}. \end{proof} \begin{proposition}[Structure of injective modules over Noetherian rings] \label{proposition-structure-injectives-noetherian} Let $R$ be a Noetherian ring. Every injective module is a direct sum of indecomposable injective modules. Every indecomposable injective module is the injective hull of the residue field at a prime. \end{proposition} \begin{proof} The second statement is Lemma \ref{lemma-indecomposable-injective-noetherian}. For the first statement, let $I$ be an injective $R$-module. We will use transfinite induction to construct $I_\alpha \subset I$ for ordinals $\alpha$ which are direct sums of indecomposable injective $R$-modules $E_{\beta + 1}$ for $\beta < \alpha$. For $\alpha = 0$ we let $I_0 = 0$. Suppose given an ordinal $\alpha$ such that $I_\alpha$ has been constructed. Then $I_\alpha$ is an injective $R$-module by Lemma \ref{lemma-sum-injective-modules}. Hence $I \cong I_\alpha \oplus I'$. If $I' = 0$ we are done. If not, then $I'$ has an associated prime by Algebra, Lemma \ref{algebra-lemma-ass-zero}. Thus $I'$ contains a copy of $R/\mathfrak p$ for some prime $\mathfrak p$. Hence $I'$ contains an indecomposable submodule $E$ by Lemmas \ref{lemma-injective-hull-unique} and \ref{lemma-injective-hull-indecomposable}. Set $I_{\alpha + 1} = I_\alpha \oplus E_\alpha$. If $\alpha$ is a limit ordinal and $I_\beta$ has been constructed for $\beta < \alpha$, then we set $I_\alpha = \bigcup_{\beta < \alpha} I_\beta$. Observe that $I_\alpha = \bigoplus_{\beta < \alpha} E_{\beta + 1}$. This concludes the proof. \end{proof} \section{Duality over Artinian local rings} \label{section-artinian} \noindent Let $(R, \mathfrak m, \kappa)$ be an artinian local ring. Recall that this implies $R$ is Noetherian and that $R$ has finite length as an $R$-module. Moreover an $R$-module is finite if and only if it has finite length. We will use these facts without further mention in this section. Please see Algebra, Sections \ref{algebra-section-length} and \ref{algebra-section-artinian} and Algebra, Proposition \ref{algebra-proposition-dimension-zero-ring} for more details. \begin{lemma} \label{lemma-finite} Let $(R, \mathfrak m, \kappa)$ be an artinian local ring. Let $E$ be an injective hull of $\kappa$. For every finite $R$-module $M$ we have $$\text{length}_R(M) = \text{length}_R(\Hom_R(M, E))$$ In particular, the injective hull $E$ of $\kappa$ is a finite $R$-module. \end{lemma} \begin{proof} Because $E$ is an essential extension of $\kappa$ we have $\kappa = E[\mathfrak m]$ where $E[\mathfrak m]$ is the $\mathfrak m$-torsion in $E$ (notation as in More on Algebra, Section \ref{more-algebra-section-formal-glueing}). Hence $\Hom_R(\kappa, E) \cong \kappa$ and the equality of lengths holds for $M = \kappa$. We prove the displayed equality of the lemma by induction on the length of $M$. If $M$ is nonzero there exists a surjection $M \to \kappa$ with kernel $M'$. Since the functor $M \mapsto \Hom_R(M, E)$ is exact we obtain a short exact sequence $$0 \to \Hom_R(\kappa, E) \to \Hom_R(M, E) \to \Hom_R(M', E) \to 0.$$ Additivity of length for this sequence and the sequence $0 \to M' \to M \to \kappa \to 0$ and the equality for $M'$ (induction hypothesis) and $\kappa$ implies the equality for $M$. The final statement of the lemma follows as $E = \Hom_R(R, E)$. \end{proof} \begin{lemma} \label{lemma-evaluate} Let $(R, \mathfrak m, \kappa)$ be an artinian local ring. Let $E$ be an injective hull of $\kappa$. For any finite $R$-module $M$ the evaluation map $$M \longrightarrow \Hom_R(\Hom_R(M, E), E)$$ is an isomorphism. In particular $R = \Hom_R(E, E)$. \end{lemma} \begin{proof} Observe that the displayed arrow is injective. Namely, if $x \in M$ is a nonzero element, then there is a nonzero map $Rx \to \kappa$ which we can extend to a map $\varphi : M \to E$ that doesn't vanish on $x$. Since the source and target of the arrow have the same length by Lemma \ref{lemma-finite} we conclude it is an isomorphism. The final statement follows on taking $M = R$. \end{proof} \noindent To state the next lemma, denote $\text{Mod}^{fg}_R$ the category of finite $R$-modules over a ring $R$. \begin{lemma} \label{lemma-duality} Let $(R, \mathfrak m, \kappa)$ be an artinian local ring. Let $E$ be an injective hull of $\kappa$. The functor $D(-) = \Hom_R(-, E)$ induces an exact anti-equivalence $\text{Mod}^{fg}_R \to \text{Mod}^{fg}_R$ and $D \circ D \cong \text{id}$. \end{lemma} \begin{proof} We have seen that $D \circ D = \text{id}$ on $\text{Mod}^{fg}_R$ in Lemma \ref{lemma-evaluate}. It follows immediately that $D$ is an anti-equivalence. \end{proof} \begin{lemma} \label{lemma-duality-torsion-cotorsion} Assumptions and notation as in Lemma \ref{lemma-duality}. Let $I \subset R$ be an ideal and $M$ a finite $R$-module. Then $$D(M[I]) = D(M)/ID(M) \quad\text{and}\quad D(M/IM) = D(M)[I]$$ \end{lemma} \begin{proof} Say $I = (f_1, \ldots, f_t)$. Consider the map $$M^{\oplus t} \xrightarrow{f_1, \ldots, f_t} M$$ with cokernel $M/IM$. Applying the exact functor $D$ we conclude that $D(M/IM)$ is $D(M)[I]$. The other case is proved in the same way. \end{proof} \section{Injective hull of the residue field} \label{section-hull-residue-field} \noindent Most of our results will be for Noetherian local rings in this section. \begin{lemma} \label{lemma-quotient} Let $R \to S$ be a surjective map of local rings with kernel $I$. Let $E$ be the injective hull of the residue field of $R$ over $R$. Then $E[I]$ is the injective hull of the residue field of $S$ over $S$. \end{lemma} \begin{proof} Observe that $E[I] = \Hom_R(S, E)$ as $S = R/I$. Hence $E[I]$ is an injective $S$-module by Lemma \ref{lemma-hom-injective}. Since $E$ is an essential extension of $\kappa = R/\mathfrak m_R$ it follows that $E[I]$ is an essential extension of $\kappa$ as well. The result follows. \end{proof} \begin{lemma} \label{lemma-torsion-submodule-sum-injective-hulls} Let $(R, \mathfrak m, \kappa)$ be a local ring. Let $E$ be the injective hull of $\kappa$. Let $M$ be a $\mathfrak m$-power torsion $R$-module with $n = \dim_\kappa(M[\mathfrak m]) < \infty$. Then $M$ is isomorphic to a submodule of $E^{\oplus n}$. \end{lemma} \begin{proof} Observe that $E^{\oplus n}$ is the injective hull of $\kappa^{\oplus n} = M[\mathfrak m]$. Thus there is an $R$-module map $M \to E^{\oplus n}$ which is injective on $M[\mathfrak m]$. Since $M$ is $\mathfrak m$-power torsion the inclusion $M[\mathfrak m] \subset M$ is an essential extension (for example by Lemma \ref{lemma-essential-extension}) we conclude that the kernel of $M \to E^{\oplus n}$ is zero. \end{proof} \begin{lemma} \label{lemma-union-artinian} Let $(R, \mathfrak m, \kappa)$ be a Noetherian local ring. Let $E$ be an injective hull of $\kappa$ over $R$. Let $E_n$ be an injective hull of $\kappa$ over $R/\mathfrak m^n$. Then $E = \bigcup E_n$ and $E_n = E[\mathfrak m^n]$. \end{lemma} \begin{proof} We have $E_n = E[\mathfrak m^n]$ by Lemma \ref{lemma-quotient}. We have $E = \bigcup E_n$ because $\bigcup E_n = E[\mathfrak m^\infty]$ is an injective $R$-submodule which contains $\kappa$, see Lemma \ref{lemma-injective-module-divide}. \end{proof} \noindent The following lemma tells us the injective hull of the residue field of a Noetherian local ring only depends on the completion. \begin{lemma} \label{lemma-compare} Let $R \to S$ be a flat local homomorphism of local Noetherian rings such that $R/\mathfrak m_R \cong S/\mathfrak m_R S$. Then the injective hull of the residue field of $R$ is the injective hull of the residue field of $S$. \end{lemma} \begin{proof} Set $\kappa = R/\mathfrak m_R = S/\mathfrak m_S$. Let $E_R$ be the injective hull of $\kappa$ over $R$. Let $E_S$ be the injective hull of $\kappa$ over $S$. Observe that $E_S$ is an injective $R$-module by Lemma \ref{lemma-injective-flat}. Choose an extension $E_R \to E_S$ of the identification of residue fields. This map is an isomorphism by Lemma \ref{lemma-union-artinian} because $R \to S$ induces an isomorphism $R/\mathfrak m_R^n \to S/\mathfrak m_S^n$ for all $n$. \end{proof} \begin{lemma} \label{lemma-endos} Let $(R, \mathfrak m, \kappa)$ be a Noetherian local ring. Let $E$ be an injective hull of $\kappa$ over $R$. Then $\Hom_R(E, E)$ is canonically isomorphic to the completion of $R$. \end{lemma} \begin{proof} Write $E = \bigcup E_n$ with $E_n = E[\mathfrak m^n]$ as in Lemma \ref{lemma-union-artinian}. Any endomorphism of $E$ preserves this filtration. Hence $$\Hom_R(E, E) = \lim \Hom_R(E_n, E_n)$$ The lemma follows as $\Hom_R(E_n, E_n) = \Hom_{R/\mathfrak m^n}(E_n, E_n) = R/\mathfrak m^n$ by Lemma \ref{lemma-evaluate}. \end{proof} \begin{lemma} \label{lemma-injective-hull-has-dcc} Let $(R, \mathfrak m, \kappa)$ be a Noetherian local ring. Let $E$ be an injective hull of $\kappa$ over $R$. Then $E$ satisfies the descending chain condition. \end{lemma} \begin{proof} If $E \subset M_1 \subset M_2 \ldots$ is a sequence of submodules, then $$\Hom_R(E, E) \to \Hom_R(M_1, E) \to \Hom_R(M_2, E) \to \ldots$$ is sequence of surjections. By Lemma \ref{lemma-endos} each of these is a module over the completion $R^\wedge = \Hom_R(E, E)$. Since $R^\wedge$ is Noetherian (Algebra, Lemma \ref{algebra-lemma-completion-Noetherian-Noetherian}) the sequence stabilizes: $\Hom_R(M_n, E) = \Hom_R(M_{n + 1}, E) = \ldots$. Since $E$ is injective, this can only happen if $\Hom_R(M_n/M_{n + 1}, E)$ is zero. However, if $M_n/M_{n + 1}$ is nonzero, then it contains a nonzero element annihilated by $\mathfrak m$, because $E$ is $\mathfrak m$-power torsion by Lemma \ref{lemma-union-artinian}. In this case $M_n/M_{n + 1}$ has a nonzero map into $E$, contradicting the assumed vanishing. This finishes the proof. \end{proof} \begin{lemma} \label{lemma-describe-categories} Let $(R, \mathfrak m, \kappa)$ be a Noetherian local ring. Let $E$ be an injective hull of $\kappa$. \begin{enumerate} \item For an $R$-module $M$ the following are equivalent: \begin{enumerate} \item $M$ satisfies the ascending chain condition, \item $M$ is a finite $R$-module, and \item there exist $n, m$ and an exact sequence $R^{\oplus m} \to R^{\oplus n} \to M \to 0$. \end{enumerate} \item For an $R$-module $M$ the following are equivalent: \begin{enumerate} \item $M$ satisfies the descending chain condition, \item $M$ is $\mathfrak m$-power torsion and $\dim_\kappa(M[\mathfrak m]) < \infty$, and \item there exist $n, m$ and an exact sequence $0 \to M \to E^{\oplus n} \to E^{\oplus m}$. \end{enumerate} \end{enumerate} \end{lemma} \begin{proof} We omit the proof of (1). \medskip\noindent Let $M$ be an $R$-module with the descending chain condition. Let $x \in M$. Then $\mathfrak m^n x$ is a descending chain of submodules, hence stabilizes. Thus $\mathfrak m^nx = \mathfrak m^{n + 1}x$ for some $n$. By Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) this implies $\mathfrak m^n x = 0$, i.e., $x$ is $\mathfrak m$-power torsion. Since $M[\mathfrak m]$ is a vector space over $\kappa$ it has to be finite dimensional in order to have the descending chain condition. \medskip\noindent Assume that $M$ is $\mathfrak m$-power torsion and has a finite dimensional $\mathfrak m$-torsion submodule $M[\mathfrak m]$. By Lemma \ref{lemma-torsion-submodule-sum-injective-hulls} we see that $M$ is a submodule of $E^{\oplus n}$ for some $n$. Consider the quotient $N = E^{\oplus n}/M$. By Lemma \ref{lemma-injective-hull-has-dcc} the module $E$ has the descending chain condition hence so do $E^{\oplus n}$ and $N$. Therefore $N$ satisfies (2)(a) which implies $N$ satisfies (2)(b) by the second paragraph of the proof. Thus by Lemma \ref{lemma-torsion-submodule-sum-injective-hulls} again we see that $N$ is a submodule of $E^{\oplus m}$ for some $m$. Thus we have a short exact sequence $0 \to M \to E^{\oplus n} \to E^{\oplus m}$. \medskip\noindent Assume we have a short exact sequence $0 \to M \to E^{\oplus n} \to E^{\oplus m}$. Since $E$ satisfies the descending chain condition by Lemma \ref{lemma-injective-hull-has-dcc} so does $M$. \end{proof} \begin{proposition}[Matlis duality] \label{proposition-matlis} Let $(R, \mathfrak m, \kappa)$ be a complete local Noetherian ring. Let $E$ be an injective hull of $\kappa$ over $R$. The functor $D(-) = \Hom_R(-, E)$ induces an anti-equivalence $$\left\{ \begin{matrix} R\text{-modules with the} \\ \text{descending chain condition} \end{matrix} \right\} \longleftrightarrow \left\{ \begin{matrix} R\text{-modules with the} \\ \text{ascending chain condition} \end{matrix} \right\}$$ and we have $D \circ D = \text{id}$ on either side of the equivalence. \end{proposition} \begin{proof} By Lemma \ref{lemma-endos} we have $R = \Hom_R(E, E) = D(E)$. Of course we have $E = \Hom_R(R, E) = D(R)$. Since $E$ is injective the functor $D$ is exact. The result now follows immediately from the description of the categories in Lemma \ref{lemma-describe-categories}. \end{proof} \section{Deriving torsion} \label{section-bad-local-cohomology} \noindent Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal (if $I$ is not finitely generated perhaps a different definition should be used). Let $Z = V(I) \subset \Spec(A)$. Recall that the category $I^\infty\text{-torsion}$ of $I$-power torsion modules only depends on the closed subset $Z$ and not on the choice of the finitely generated ideal $I$ such that $Z = V(I)$, see More on Algebra, Lemma \ref{more-algebra-lemma-local-cohomology-closed}. In this section we will consider the functor $$H^0_{I} : \text{Mod}_A \longrightarrow I^\infty\text{-torsion},\quad M \longmapsto M[I^\infty] = \bigcup M[I^n]$$ which sends $M$ to the submodule of $I$-power torsion. \medskip\noindent Let $A$ be a ring and let $I$ be a finitely generated ideal. Note that $I^\infty\text{-torsion}$ is a Grothendieck abelian category (direct sums exist, filtered colimits are exact, and $\bigoplus A/I^n$ is a generator by More on Algebra, Lemma \ref{more-algebra-lemma-I-power-torsion-presentation}). Hence the derived category $D(I^\infty\text{-torsion})$ exists, see Injectives, Remark \ref{injectives-remark-existence-D}. Our functor $H^0_I$ is left exact and has a derived extension which we will denote $$R\Gamma_I : D(A) \longrightarrow D(I^\infty\text{-torsion}).$$ {\bf Warning:} this functor does not deserve the name local cohomology unless the ring $A$ is Noetherian. The functors $H^0_I$, $R\Gamma_I$, and the satellites $H^p_I$ only depend on the closed subset $Z \subset \Spec(A)$ and not on the choice of the finitely generated ideal $I$ such that $V(I) = Z$. However, we insist on using the subscript $I$ for the functors above as the notation $R\Gamma_Z$ is going to be used for a different functor, see (\ref{equation-local-cohomology}), which agrees with the functor $R\Gamma_I$ only (as far as we know) in case $A$ is Noetherian (see Lemma \ref{lemma-local-cohomology-noetherian}). \begin{lemma} \label{lemma-adjoint} Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. The functor $R\Gamma_I$ is right adjoint to the functor $D(I^\infty\text{-torsion}) \to D(A)$. \end{lemma} \begin{proof} This follows from the fact that taking $I$-power torsion submodules is the right adjoint to the inclusion functor $I^\infty\text{-torsion} \to \text{Mod}_A$. See Derived Categories, Lemma \ref{derived-lemma-derived-adjoint-functors}. \end{proof} \begin{lemma} \label{lemma-local-cohomology-ext} Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. For any object $K$ of $D(A)$ we have $$R\Gamma_I(K) = \text{hocolim}\ R\Hom_A(A/I^n, K)$$ in $D(A)$ and $$R^q\Gamma_I(K) = \colim_n \text{Ext}_A^q(A/I^n, K)$$ as modules for all $q \in \mathbf{Z}$. \end{lemma} \begin{proof} Let $J^\bullet$ be a K-injective complex representing $K$. Then $$R\Gamma_I(K) = J^\bullet[I^\infty] = \colim J^\bullet[I^n] = \colim \Hom_A(A/I^n, J^\bullet)$$ By Derived Categories, Lemma \ref{derived-lemma-colim-hocolim} we obtain the first equality. The second equality is clear because $H^q(\Hom_A(A/I^n, J^\bullet)) = \text{Ext}^q_A(A/I^n, K)$ and because filtered colimits are exact in the category of abelian groups. \end{proof} \begin{lemma} \label{lemma-bad-local-cohomology-vanishes} Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Let $K^\bullet$ be a complex of $A$-modules such that $f : K^\bullet \to K^\bullet$ is an isomorphism for some $f \in I$, i.e., $K^\bullet$ is a complex of $A_f$-modules. Then $R\Gamma_I(K^\bullet) = 0$. \end{lemma} \begin{proof} Namely, in this case the cohomology modules of $R\Gamma_I(K^\bullet)$ are both $f$-power torsion and $f$ acts by automorphisms. Hence the cohomology modules are zero and hence the object is zero. \end{proof} \noindent Let $A$ be a ring and $I \subset A$ a finitely generated ideal. By More on Algebra, Lemma \ref{more-algebra-lemma-I-power-torsion} the category of $I$-power torsion modules is a Serre subcategory of the category of all $A$-modules, hence there is a functor \begin{equation} \label{equation-compare-torsion} D(I^\infty\text{-torsion}) \to D_{I^\infty\text{-torsion}}(A) \end{equation} see Derived Categories, Section \ref{derived-section-triangulated-sub}. \begin{lemma} \label{lemma-not-equal} Let $A$ be a ring and let $I$ be a finitely generated ideal. Let $M$ and $N$ be $I$-power torsion modules. \begin{enumerate} \item $\Hom_{D(A)}(M, N) = \Hom_{D({I^\infty\text{-torsion}})}(M, N)$, \item $\text{Ext}^1_{D(A)}(M, N) = \text{Ext}^1_{D({I^\infty\text{-torsion}})}(M, N)$, \item $\text{Ext}^2_{D({I^\infty\text{-torsion}})}(M, N) \to \text{Ext}^2_{D(A)}(M, N)$ is not surjective in general, \item (\ref{equation-compare-torsion}) is not an equivalence in general. \end{enumerate} \end{lemma} \begin{proof} Parts (1) and (2) follow immediately from the fact that $I$-power torsion forms a Serre subcategory of $\text{Mod}_A$. Part (4) follows from part (3). \medskip\noindent For part (3) let $A$ be a ring with an element $f \in A$ such that $A[f]$ contains a nonzero element $x$ annihilated by $f$ and $A$ contains elements $x_n$ with $f^nx_n = x$. Such a ring $A$ exists because we can take $$A = \mathbf{Z}[f, x, x_n]/(fx, f^nx_n - x)$$ Given $A$ set $I = (f)$. Then the exact sequence $$0 \to A[f] \to A \xrightarrow{f} A \to A/fA \to 0$$ defines an element in $\text{Ext}^2_A(A/fA, A[f])$. We claim this element does not come from an element of $\text{Ext}^2_{D(f^\infty\text{-torsion})}(A/fA, A[f])$. Namely, if it did, then there would be an exact sequence $$0 \to A[f] \to M \to N \to A/fA \to 0$$ where $M$ and $N$ are $f$-power torsion modules defining the same $2$ extension class. Since $A \to A$ is a complex of free modules and since the $2$ extension classes are the same we would be able to find a map $$\xymatrix{ 0 \ar[r] & A[f] \ar[r] \ar[d] & A \ar[r] \ar[d]_\varphi & A \ar[r] \ar[d]_\psi & A/fA \ar[r] \ar[d] & 0 \\ 0 \ar[r] & A[f] \ar[r] & M \ar[r] & N \ar[r] & A/fA \ar[r] & 0 }$$ (some details omitted). Then we could replace $M$ by the image of $\varphi$ and $N$ by the image of $\psi$. Then $M$ would be a cyclic module, hence $f^n M = 0$ for some $n$. Considering $\varphi(x_{n + 1})$ we get a contradiction with the fact that $f^{n + 1}x_n = x$ is nonzero in $A[f]$. \end{proof} \section{Local cohomology} \label{section-local-cohomology} \noindent Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Set $Z = V(I) \subset \Spec(A)$. We will construct a functor \begin{equation} \label{equation-local-cohomology} R\Gamma_Z : D(A) \longrightarrow D_{I^\infty\text{-torsion}}(A). \end{equation} which is right adjoint to the inclusion functor. For notation see Section \ref{section-bad-local-cohomology}. The cohomology modules of $R\Gamma_Z(K)$ are the {\it local cohomology groups of $K$ with respect to $Z$}. In fact, we will show $R\Gamma_Z$ computes cohomology with support in $Z$ for the associated complex of quasi-coherent sheaves on $\Spec(A)$. By Lemma \ref{lemma-not-equal} this functor will in general {\bf not} be equal to $R\Gamma_I( - )$ even viewed as functors into $D(A)$. In Section \ref{section-local-cohomology-noetherian} we will show that if $A$ is Noetherian, then the two agree. \begin{lemma} \label{lemma-local-cohomology-adjoint} Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. There exists a right adjoint $R\Gamma_Z$ (\ref{equation-local-cohomology}) to the inclusion functor $D_{I^\infty\text{-torsion}}(A) \to D(A)$. In fact, if $I$ is generated by $f_1, \ldots, f_r \in A$, then we have $$R\Gamma_Z(K) = (A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r}) \otimes_A^\mathbf{L} K$$ functorially in $K \in D(A)$. \end{lemma} \begin{proof} Say $I = (f_1, \ldots, f_r)$ is an ideal. Let $K^\bullet$ be a complex of $A$-modules. There is a canonical map of complexes $$(A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r}) \longrightarrow A.$$ from the extended {\v C}ech complex to $A$. Tensoring with $K^\bullet$, taking associated total complex, we get a map $$\text{Tot}\left( K^\bullet \otimes_A (A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r})\right) \longrightarrow K^\bullet$$ in $D(A)$. We claim the cohomology modules of the complex on the left are $I$-power torsion, i.e., the LHS is an object of $D_{I^\infty\text{-torsion}}(A)$. Namely, we have $$(A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r}) = \colim K(A, f_1^n, \ldots, f_r^n)$$ by More on Algebra, Lemma \ref{more-algebra-lemma-extended-alternating-Cech-is-colimit-koszul}. Moreover, multiplication by $f_i^n$ on the complex $K(A, f_1^n, \ldots, f_r^n)$ is homotopic to zero by More on Algebra, Lemma \ref{more-algebra-lemma-homotopy-koszul}. Since $$H^q\left( LHS \right) = \colim H^q(\text{Tot}(K^\bullet \otimes_A K(A, f_1^n, \ldots, f_r^n)))$$ we obtain our claim. On the other hand, if $K^\bullet$ is an object of $D_{I^\infty\text{-torsion}}(A)$, then the complexes $K^\bullet \otimes_A A_{f_{i_0} \ldots f_{i_p}}$ have vanishing cohomology. Hence in this case the map $LHS \to K^\bullet$ is an isomorphism in $D(A)$. The construction $$R\Gamma_Z(K^\bullet) = \text{Tot}\left( K^\bullet \otimes_A (A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r})\right)$$ is functorial in $K^\bullet$ and defines an exact functor $D(A) \to D_{I^\infty\text{-torsion}}(A)$ between triangulated categories. It follows formally from the existence of the natural transformation $R\Gamma_Z \to \text{id}$ given above and the fact that this evaluates to an isomorphism on $K^\bullet$ in the subcategory, that $R\Gamma_Z$ is the desired right adjoint. \end{proof} \begin{lemma} \label{lemma-local-cohomology-and-restriction} Let $A \to B$ be a ring homomorphism and let $I \subset A$ be a finitely generated ideal. Set $J = IB$. Set $Z = V(I)$ and $Y = V(J)$. Then $$R\Gamma_Z(M_A) = R\Gamma_Y(M)_A$$ functorially in $M \in D(B)$. Here $(-)_A$ denotes the restriction functors $D(B) \to D(A)$ and ${}_A : D_{J^\infty\text{-torsion}}(B) \to D_{I^\infty\text{-torsion}}(A)$. \end{lemma} \begin{proof} This follows from uniqueness of adjoint functors as both $R\Gamma_Z((-)_A)$ and $R\Gamma_Y(-)_A$ are right adjoint to the functor $D_{I^\infty\text{-torsion}}(A) \to D(B)$, $K \mapsto K \otimes_A^\mathbf{L} B$. Alternatively, one can use the description of $R\Gamma_Z$ and $R\Gamma_Y$ in terms of alternating {\v C}ech complexes (Lemma \ref{lemma-local-cohomology-adjoint}). Namely, if $I = (f_1, \ldots, f_r)$ then $J$ is generated by the images $g_1, \ldots, g_r \in B$ of $f_1, \ldots, f_r$. Then the statement of the lemma follows from the existence of a canonical isomorphism \begin{align*} & M_A \otimes_A (A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r}) \\ & = M \otimes_B (B \to \prod\nolimits_{i_0} B_{g_{i_0}} \to \prod\nolimits_{i_0 < i_1} B_{g_{i_0}g_{i_1}} \to \ldots \to B_{g_1\ldots g_r}) \end{align*} for any $B$-module $M$. \end{proof} \begin{lemma} \label{lemma-torsion-change-rings} Let $A \to B$ be a ring homomorphism and let $I \subset A$ be a finitely generated ideal. Set $J = IB$. Let $Z = V(I)$ and $Y = V(J)$. Then $$R\Gamma_Z(K) \otimes_A^\mathbf{L} B = R\Gamma_Y(K \otimes_A^\mathbf{L} B)$$ functorially in $K \in D(A)$. \end{lemma} \begin{proof} This follows from uniqueness of adjoint functors as both $R\Gamma_Z( - ) \otimes_A^\mathbf{L} B$ and $R\Gamma_Y(- \otimes_A^\mathbf{L} B)$ are right adjoint to the functor $D_{J^\infty\text{-torsion}}(B) \to D(A)$. Alternatively, one can use the description of $R\Gamma_Z$ and $R\Gamma_Y$ in terms of alternating {\v C}ech complexes (Lemma \ref{lemma-local-cohomology-adjoint}) and use that formation of the extended {\v C}ech complex commutes with base change. \end{proof} \begin{lemma} \label{lemma-local-cohomology-vanishes} Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Let $K^\bullet$ be a complex of $A$-modules such that $f : K^\bullet \to K^\bullet$ is an isomorphism for some $f \in I$, i.e., $K^\bullet$ is a complex of $A_f$-modules. Then $R\Gamma_Z(K^\bullet) = 0$. \end{lemma} \begin{proof} Namely, in this case the cohomology modules of $R\Gamma_Z(K^\bullet)$ are both $f$-power torsion and $f$ acts by automorphisms. Hence the cohomology modules are zero and hence the object is zero. \end{proof} \begin{lemma} \label{lemma-torsion-tensor-product} Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. For $K, L \in D(A)$ we have $$R\Gamma_Z(K \otimes_A^\mathbf{L} L) = K \otimes_A^\mathbf{L} R\Gamma_Z(L) = R\Gamma_Z(K) \otimes_A^\mathbf{L} L = R\Gamma_Z(K) \otimes_A^\mathbf{L} R\Gamma_Z(L)$$ If $K$ or $L$ is in $D_{I^\infty\text{-torsion}}(A)$ then so is $K \otimes_A^\mathbf{L} L$. \end{lemma} \begin{proof} By Lemma \ref{lemma-local-cohomology-adjoint} we know that $R\Gamma_Z$ is given by $C \otimes^\mathbf{L} -$ for some $C \in D(A)$. Hence, for $K, L \in D(A)$ general we have $$R\Gamma_Z(K \otimes_A^\mathbf{L} L) = K \otimes^\mathbf{L} L \otimes_A^\mathbf{L} C = K \otimes_A^\mathbf{L} R\Gamma_Z(L)$$ The other equalities follow formally from this one. This also implies the last statement of the lemma. \end{proof} \noindent The following lemma tells us that the functor $R\Gamma_Z$ is related to cohomology with supports. \begin{lemma} \label{lemma-local-cohomology-is-local-cohomology} Let $A$ be a ring and let $I$ be a finitely generated ideal. With $Z = V(I) \subset X = \Spec(A)$ there is a functorial isomorphism $$R\Gamma_Z(K^\bullet) = R\Gamma_Z(\widetilde{K^\bullet})$$ where on the left we have (\ref{equation-local-cohomology}) and on the right we have the functor of Cohomology, Section \ref{cohomology-section-cohomology-support}. \end{lemma} \begin{proof} Denote $\mathcal{F}^\bullet = \widetilde{K^\bullet}$ be the complex of quasi-coherent $\mathcal{O}_X$-modules on $X$ associated to $K^\bullet$. By Cohomology, Section \ref{cohomology-section-cohomology-support} there exists a distinguished triangle $$R\Gamma_Z(X, \mathcal{F}^\bullet) \to R\Gamma(X, \mathcal{F}^\bullet) \to R\Gamma(U, \mathcal{F}^\bullet) \to R\Gamma_Z(X, \mathcal{F}^\bullet)[1]$$ where $U = X \setminus Z$. We know that $R\Gamma(X, \mathcal{F}^\bullet) = K^\bullet$ for example by Derived Categories of Schemes, Lemma \ref{perfect-lemma-affine-compare-bounded}. Say $I = (f_1, \ldots, f_r)$. Then we obtain a finite affine open covering $\mathcal{U} : U = D(f_1) \cup \ldots \cup D(f_r)$. By Derived Categories of Schemes, Lemma \ref{perfect-lemma-alternating-cech-complex-complex-computes-cohomology} the alternating {\v C}ech complex $$\text{Tot}(\check{\mathcal{C}}_{alt}^\bullet(\mathcal{U}, \mathcal{F}^\bullet))$$ computes $R\Gamma(U, \mathcal{F}^\bullet)$. Working through the definitions we find $$R\Gamma(U, \mathcal{F}^\bullet) = \text{Tot}\left( K^\bullet \otimes_A (\prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r})\right)$$ It is clear that $R\Gamma(X, \mathcal{F}^\bullet) \to R\Gamma(U, \mathcal{F}^\bullet)$ is given by the map from $A$ into $\prod A_{f_i}$. Hence we conclude that $$R\Gamma_Z(X, \mathcal{F}^\bullet) = \text{Tot}\left( K^\bullet \otimes_A (A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r})\right)$$ By Lemma \ref{lemma-local-cohomology-adjoint} this complex computes $R\Gamma_Z(K^\bullet)$ and we see the lemma holds. \end{proof} \begin{lemma} \label{lemma-local-cohomology-ss} Let $A$ be a ring and let $I, J \subset A$ be finitely generated ideals. Set $Z = V(I)$ and $Y = V(J)$. Then $Z \cap Y = V(I + J)$ and $R\Gamma_Y \circ R\Gamma_Z = R\Gamma_{Y \cap Z}$ as functors $D(A) \to D_{(I + J)^\infty\text{-torsion}}(A)$. For $K \in D^+(A)$ there is a spectral sequence $$E_2^{p, q} = H^p_Y(H^p_Z(K)) \Rightarrow H^{p + q}_{Y \cap Z}(K)$$ as in Derived Categories, Lemma \ref{derived-lemma-grothendieck-spectral-sequence}. \end{lemma} \begin{proof} There is a bit of abuse of notation in the lemma as strictly speaking we cannot compose $R\Gamma_Y$ and $R\Gamma_Z$. The meaning of the statement is simply that we are composing $R\Gamma_Z$ with the inclusion $D_{I^\infty\text{-torsion}}(A) \to D(A)$ and then with $R\Gamma_Y$. Then the equality $R\Gamma_Y \circ R\Gamma_Z = R\Gamma_{Y \cap Z}$ follows from the fact that $$D_{I^\infty\text{-torsion}}(A) \to D(A) \xrightarrow{R\Gamma_Y} D_{(I + J)^\infty\text{-torsion}}(A)$$ is right adjoint to the inclusion $D_{(I + J)^\infty\text{-torsion}}(A) \to D_{I^\infty\text{-torsion}}(A)$. Alternatively one can prove the formula using Lemma \ref{lemma-local-cohomology-adjoint} and the fact that the tensor product of extended {\v C}ech complexes on $f_1, \ldots, f_r$ and $g_1, \ldots, g_m$ is the extended {\v C} complex on $f_1, \ldots, f_n. g_1, \ldots, g_m$. The final assertion follows from this and the cited lemma. \end{proof} \noindent The following lemma is the analogue of More on Algebra, Lemma \ref{more-algebra-lemma-restriction-derived-complete-equivalence} for complexes with torsion cohomologies. \begin{lemma} \label{lemma-torsion-flat-change-rings} Let $A \to B$ be a flat ring map and let $I \subset A$ be a finitely generated ideal such that $A/I = B/IB$. Then base change and restriction induce quasi-inverse equivalences $D_{I^\infty\text{-torsion}}(A) = D_{(IB)^\infty\text{-torsion}}(B)$. \end{lemma} \begin{proof} More precisely the functors are $K \mapsto K \otimes_A^\mathbf{L} B$ for $K$ in $D_{I^\infty\text{-torsion}}(A)$ and $M \mapsto M_A$ for $M$ in $D_{(IB)^\infty\text{-torsion}}(B)$. The reason this works is that $H^i(K \otimes_A^\mathbf{L} B) = H^i(K) \otimes_A B = H^i(K)$. The first equality holds as $A \to B$ is flat and the second by More on Algebra, Lemma \ref{more-algebra-lemma-neighbourhood-isomorphism}. \end{proof} \noindent The following lemma was shown for $\Hom$ and $\text{Ext}^1$ of modules in More on Algebra, Lemmas \ref{more-algebra-lemma-neighbourhood-equivalence} and \ref{more-algebra-lemma-neighbourhood-extensions}. \begin{lemma} \label{lemma-neighbourhood-extensions} Let $A \to B$ be a flat ring map and let $I \subset A$ be a finitely generated ideal such that $A/I \to B/IB$ is an isomorphism. For $K \in D_{I^\infty\text{-torsion}}(A)$ and $L \in D(A)$ the map $$R\Hom_A(K, L) \longrightarrow R\Hom_B(K \otimes_A B, L \otimes_A B)$$ is a quasi-isomorphism. In particular, if $M$, $N$ are $A$-modules and $M$ is $I$-power torsion, then the canonical map $$\text{Ext}^i_A(M, N) \longrightarrow \text{Ext}^i_B(M \otimes_A B, N \otimes_A B)$$ is an isomorphism for all $i$. \end{lemma} \begin{proof} Let $Z = V(I) \subset \Spec(A)$ and $Y = V(IB) \subset \Spec(B)$. Since the cohomology modules of $K$ are $I$ power torsion, the canonical map $R\Gamma_Z(L) \to L$ induces an isomorphism $$R\Hom_A(K, R\Gamma_Z(L)) \to R\Hom_A(K, L)$$ in $D(A)$. Similarly, the cohomology modules of $K \otimes_A B$ are $IB$ power torsion and we have an isomorphism $$R\Hom_B(K \otimes_A B, R\Gamma_Y(L \otimes_A B)) \to R\Hom_B(K \otimes_A B, L \otimes_A B)$$ in $D(B)$. By Lemma \ref{lemma-torsion-change-rings} we have $R\Gamma_Z(L) \otimes_A B = R\Gamma_Y(L \otimes_A B)$. Hence it suffices to show that the map $$R\Hom_A(K, R\Gamma_Z(L)) \to R\Hom_B(K \otimes_A B, R\Gamma_Z(L) \otimes_A B)$$ is a quasi-isomorphism. This follows from Lemma \ref{lemma-torsion-flat-change-rings}. \end{proof} \section{Local cohomology for Noetherian rings} \label{section-local-cohomology-noetherian} \noindent Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Set $Z = V(I) \subset \Spec(A)$. Recall that (\ref{equation-compare-torsion}) is the functor $$D(I^\infty\text{-torsion}) \to D_{I^\infty\text{-torsion}}(A)$$ In fact, there is a natural transformation of functors \begin{equation} \label{equation-compare-torsion-functors} (\ref{equation-compare-torsion}) \circ R\Gamma_I(-) \longrightarrow R\Gamma_Z(-) \end{equation} Namely, given a complex of $A$-modules $K^\bullet$ the canonical map $R\Gamma_I(K^\bullet) \to K^\bullet$ in $D(A)$ factors (uniquely) through $R\Gamma_Z(K^\bullet)$ as $R\Gamma_I(K^\bullet)$ has $I$-power torsion cohomology modules (see Lemma \ref{lemma-adjoint}). In general this map is not an isomorphism (we've seen this in Lemma \ref{lemma-not-equal}). \begin{lemma} \label{lemma-local-cohomology-noetherian} Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. \begin{enumerate} \item the adjunction $R\Gamma_I(K) \to K$ is an isomorphism for $K \in D_{I^\infty\text{-torsion}}(A)$, \item the functor (\ref{equation-compare-torsion}) $D(I^\infty\text{-torsion}) \to D_{I^\infty\text{-torsion}}(A)$ is an equivalence, \item the transformation of functors (\ref{equation-compare-torsion-functors}) is an isomorphism, in other words $R\Gamma_I(K) = R\Gamma_Z(K)$ for $K \in D(A)$. \end{enumerate} \end{lemma} \begin{proof} A formal argument, which we omit, shows that it suffices to prove (1). \medskip\noindent Let $M$ be an $I$-power torsion $A$-module. Choose an embedding $M \to J$ into an injective $A$-module. Then $J[I^\infty]$ is an injective $A$-module, see Lemma \ref{lemma-injective-module-divide}, and we obtain an embedding $M \to J[I^\infty]$. Thus every $I$-power torsion module has an injective resolution $M \to J^\bullet$ with $J^n$ also $I$-power torsion. It follows that $R\Gamma_I(M) = M$ (this is not a triviality and this is not true in general if $A$ is not Noetherian). Next, suppose that $K \in D_{I^\infty\text{-torsion}}^+(A)$. Then the spectral sequence $$R^q\Gamma_I(H^p(K)) \Rightarrow R^{p + q}\Gamma_I(K)$$ (Derived Categories, Lemma \ref{derived-lemma-two-ss-complex-functor}) converges and above we have seen that only the terms with $q = 0$ are nonzero. Thus we see that $R\Gamma_I(K) \to K$ is an isomorphism. \medskip\noindent Suppose $K$ is an arbitrary object of $D_{I^\infty\text{-torsion}}(A)$. We have $$R^q\Gamma_I(K) = \colim \text{Ext}^q_A(A/I^n, K)$$ by Lemma \ref{lemma-local-cohomology-ext}. Choose $f_1, \ldots, f_r \in A$ generating $I$. Let $K_n^\bullet = K(A, f_1^n, \ldots, f_r^n)$ be the Koszul complex with terms in degrees $-r, \ldots, 0$. Since the pro-objects $\{A/I^n\}$ and $\{K_n^\bullet\}$ in $D(A)$ are the same by More on Algebra, Lemma \ref{more-algebra-lemma-sequence-Koszul-complexes}, we see that $$R^q\Gamma_I(K) = \colim \text{Ext}^q_A(K_n^\bullet, K)$$ Pick any complex $K^\bullet$ of $A$-modules representing $K$. Since $K_n^\bullet$ is a finite complex of finite free modules we see that $$\text{Ext}^q_A(K_n, K) = H^q(\text{Tot}((K_n^\bullet)^\vee \otimes_A K^\bullet))$$ where $(K_n^\bullet)^\vee$ is the dual of the complex $K_n^\bullet$. See More on Algebra, Lemma \ref{more-algebra-lemma-RHom-out-of-projective}. As $(K_n^\bullet)^\vee$ is a complex of finite free $A$-modules sitting in degrees $0, \ldots, r$ we see that the terms of the complex $\text{Tot}((K_n^\bullet)^\vee \otimes_A K^\bullet)$ are the same as the terms of the complex $\text{Tot}((K_n^\bullet)^\vee \otimes_A \tau_{\geq q - r - 2} K^\bullet)$ in degrees $q - 1$ and higher. Hence we see that $$\text{Ext}^q_A(K_n, K) = \text{Ext}^q_A(K_n, \tau_{\geq q - r - 2}K)$$ for all $n$. It follows that $$R^q\Gamma_I(K) = R^q\Gamma_I(\tau_{\geq q - r - 2}K) = H^q(\tau_{\geq q - r - 2}K) = H^q(K)$$ Thus we see that the map $R\Gamma_I(K) \to K$ is an isomorphism. \end{proof} \begin{lemma} \label{lemma-compute-local-cohomology-noetherian} If $A$ is a Noetherian ring and $I = (f_1, \ldots, f_r)$ an ideal. There are canonical isomorphisms $$R\Gamma_I(A) \to (A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r}) \to R\Gamma_Z(A)$$ in $D(A)$. \end{lemma} \begin{proof} This follows from Lemma \ref{lemma-local-cohomology-noetherian} and the computation of the functor $R\Gamma_Z$ in Lemma \ref{lemma-local-cohomology-adjoint}. \end{proof} \begin{lemma} \label{lemma-local-cohomology-change-rings} If $A \to B$ is a homomorphism of Noetherian rings and $I \subset A$ is an ideal, then in $D(B)$ we have $$R\Gamma_I(A) \otimes_A^\mathbf{L} B = R\Gamma_Z(A) \otimes_A^\mathbf{L} B = R\Gamma_Y(B) = R\Gamma_{IB}(B)$$ where $Y = V(IB) \subset \Spec(B)$. \end{lemma} \begin{proof} Combine Lemmas \ref{lemma-compute-local-cohomology-noetherian} and \ref{lemma-torsion-change-rings}. \end{proof} \section{Depth} \label{section-depth} \noindent In this section we revisit the notion of depth introduced in Algebra, Section \ref{algebra-section-depth}. \begin{lemma} \label{lemma-depth} Let $A$ be a Noetherian ring, let $I \subset A$ be an ideal, and let $M$ be a finite $A$-module such that $IM \not = M$. Then the following integers are equal: \begin{enumerate} \item $\text{depth}_I(M)$, \item the smallest integer $i$ such that $\text{Ext}_A^i(A/I, M)$ is nonzero, and \item the smallest integer $i$ such that $H^i_I(M)$ is nonzero. \end{enumerate} Moreover, we have $\text{Ext}^i_A(N, M) = 0$ for $i < \text{depth}_I(M)$ for any finite $A$-module $N$ annihilated by a power of $I$. \end{lemma} \begin{proof} We prove the equality of (1) and (2) by induction on $\text{depth}_I(M)$ which is allowed by Algebra, Lemma \ref{algebra-lemma-depth-finite-noetherian}. \medskip\noindent Base case. If $\text{depth}_I(M) = 0$, then $I$ is contained in the union of the associated primes of $M$ (Algebra, Lemma \ref{algebra-lemma-ass-zero-divisors}). By prime avoidance (Algebra, Lemma \ref{algebra-lemma-silly}) we see that $I \subset \mathfrak p$ for some associated prime $\mathfrak p$. Hence $\Hom_A(A/I, M)$ is nonzero. Thus equality holds in this case. \medskip\noindent Assume that $\text{depth}_I(M) > 0$. Let $f \in I$ be a nonzerodivisor on $M$ such that $\text{depth}_I(M/fM) = \text{depth}_I(M) - 1$. Consider the short exact sequence $$0 \to M \to M \to M/fM \to 0$$ and the associated long exact sequence for $\text{Ext}^*_A(A/I, -)$. Note that $\text{Ext}^i_A(A/I, M)$ is a finite $A/I$-module (Algebra, Lemmas \ref{algebra-lemma-ext-noetherian} and \ref{algebra-lemma-annihilate-ext}). Hence we obtain $$\Hom_A(A/I, M/fM) = \text{Ext}^1_A(A/I, M)$$ and short exact sequences $$0 \to \text{Ext}^i_A(A/I, M) \to \text{Ext}^i_A(A/I, M/fM) \to \text{Ext}^{i + 1}_A(A/I, M) \to 0$$ Thus the equality of (1) and (2) by induction. \medskip\noindent Observe that $\text{dept}_I(M) = \text{depth}_{I^n}(M)$ for all $n \geq 1$ for example by Algebra, Lemma \ref{algebra-lemma-regular-sequence-powers}. Hence by the equality of (1) and (2) we see that $\text{Ext}^i_A(A/I^n, M) = 0$ for all $n$ and $i < \text{depth}_I(M)$. Let $N$ be a finite $A$-module annihilated by a power of $I$. Then we can choose a short exact sequence $$0 \to N' \to (A/I^n)^{\oplus m} \to N \to 0$$ for some $n, m \geq 0$. Then $\Hom_A(N, M) \subset \Hom_A((A/I^n)^{\oplus m}, M)$ and $\text{Ext}^i_A(N, M) \subset \text{Ext}^{i - 1}_A(N', M)$ for $i < \text{depth}_I(M)$. Thus a simply induction argument shows that the final statement of the lemma holds. \medskip\noindent Finally, we prove that (3) is equal to (1) and (2). We have $H^p_I(M) = \colim \text{Ext}^p_A(A/I^n, M)$ by Lemma \ref{lemma-local-cohomology-ext}. Thus we see that $H^i_I(M) = 0$ for $i < \text{depth}_I(M)$. For $i = \text{depth}_I(M)$, using the vanishing of $\text{Ext}_A^{i - 1}(I/I^n, M)$ we see that the map $\text{Ext}_A^i(A/I, M) \to H_I^i(M)$ is injective which proves nonvanishing in the correct degree. \end{proof} \begin{lemma} \label{lemma-depth-in-ses} Let $A$ be a Noetherian ring. Let $0 \to N' \to N \to N'' \to 0$ be a short exact sequence of finite $A$-modules. Let $I \subset A$ be an ideal. \begin{enumerate} \item $\text{depth}_I(N) \geq \min\{\text{depth}_I(N'), \text{depth}_I(N'')\}$ \item $\text{depth}_I(N'') \geq \min\{\text{depth}_I(N), \text{depth}_I(N') - 1\}$ \item $\text{depth}_I(N') \geq \min\{\text{depth}_I(N), \text{depth}_I(N'') + 1\}$ \end{enumerate} \end{lemma} \begin{proof} Assume $IN \not = N$, $IN' \not = N'$, and $IN'' \not = N''$. Then we can use the characterization of depth using the Ext groups $\text{Ext}^i(A/I, N)$, see Lemma \ref{lemma-depth}, and use the long exact cohomology sequence $$\begin{matrix} 0 \to \Hom_A(A/I, N') \to \Hom_A(A/I, N) \to \Hom_A(A/I, N'') \\ \phantom{0\ } \to \text{Ext}^1_A(A/I, N') \to \text{Ext}^1_A(A/I, N) \to \text{Ext}^1_A(A/I, N'') \to \ldots \end{matrix}$$ from Algebra, Lemma \ref{algebra-lemma-long-exact-seq-ext}. This argument also works if $IN = N$ because in this case $\text{Ext}^i_A(A/I, N) = 0$ for all $i$. Similarly in case $IN' \not = N'$ and/or $IN'' \not = N''$. \end{proof} \begin{lemma} \label{lemma-depth-drops-by-one} Let $A$ be a Noetherian ring, let $I \subset A$ be an ideal, and let $M$ a finite $A$-module with $IM \not = M$. \begin{enumerate} \item If $x \in I$ is a nonzerodivisor on $M$, then $\text{depth}_I(M/xM) = \text{depth}_I(M) - 1$. \item Any $M$-regular sequence $x_1, \ldots, x_r$ in $I$ can be extended to an $M$-regular sequence in $I$ of length $\text{depth}_I(M)$. \end{enumerate} \end{lemma} \begin{proof} Part (2) is a formal consequence of part (1). Let $x \in I$ be as in (1). By the short exact sequence $0 \to M \to M \to M/xM \to 0$ and Lemma \ref{lemma-depth-in-ses} we see that $\text{depth}_I(M/xM) \geq \text{depth}_I(M) - 1$. On the other hand, if $x_1, \ldots, x_r \in I$ is a regular sequence for $M/xM$, then $x, x_1, \ldots, x_r$ is a regular sequence for $M$. Hence (1) holds. \end{proof} \begin{lemma} \label{lemma-depth-CM} Let $R$ be a Noetherian local ring. If $M$ is a finite Cohen-Macaulay $R$-module and $I \subset R$ a nontrivial ideal. Then $$\text{depth}_I(M) = \dim(\text{Supp}(M)) - \dim(\text{Supp}(M/IM)).$$ \end{lemma} \begin{proof} We will prove this by induction on $\text{depth}_I(M)$. \medskip\noindent If $\text{depth}_I(M) = 0$, then $I$ is contained in one of the associated primes $\mathfrak p$ of $M$ (Algebra, Lemma \ref{algebra-lemma-ideal-nonzerodivisor}). Then $\mathfrak p \in \text{Supp}(M/IM)$, hence $\dim(\text{Supp}(M/IM)) \geq \dim(R/\mathfrak p) = \dim(\text{Supp}(M))$ where equality holds by Algebra, Lemma \ref{algebra-lemma-CM-ass-minimal-support}. Thus the lemma holds in this case. \medskip\noindent If $\text{depth}_I(M) > 0$, we pick $x \in I$ which is a nonzerodivisor on $M$. Note that $(M/xM)/I(M/xM) = M/IM$. On the other hand we have $\text{depth}_I(M/xM) = \text{depth}_I(M) - 1$ by Lemma \ref{lemma-depth-drops-by-one} and $\dim(\text{Supp}(M/xM)) = \text{dim}(\text{Supp}(M)) - 1$ by Algebra, Lemma \ref{algebra-lemma-one-equation-module}. Thus the result by induction hypothesis. \end{proof} \begin{lemma} \label{lemma-depth-flat-CM} Let $R \to S$ be a flat local ring homomorphism of Noetherian local rings. Denote $\mathfrak m \subset R$ the maximal ideal. Let $I \subset S$ be an ideal. If $S/\mathfrak mS$ is Cohen-Macaulay, then $$\text{depth}_I(S) \geq \dim(S/\mathfrak mS) - \dim(S/\mathfrak mS + I)$$ \end{lemma} \begin{proof} By Algebra, Lemma \ref{algebra-lemma-grothendieck-regular-sequence} any sequence in $S$ which maps to a regular sequence in $S/\mathfrak mS$ is a regular sequence in $S$. Thus it suffices to prove the lemma in case $R$ is a field. This is a special case of Lemma \ref{lemma-depth-CM}. \end{proof} \begin{lemma} \label{lemma-divide-by-torsion} Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Let $M$ be an $A$-module. Let $Z = V(I)$. Then $H^0_I(M) = H^0_Z(M)$. Let $N$ be the common value and set $M' = M/N$. Then \begin{enumerate} \item $H^0_I(M') = 0$ and $H^p_I(M) = H^p_I(M')$ and $H^p_I(N) = 0$ for all $p > 0$, \item $H^0_Z(M') = 0$ and $H^p_Z(M) = H^p_Z(M')$ and $H^p_Z(N) = 0$ for all $p > 0$. \end{enumerate} \end{lemma} \begin{proof} By definition $H^0_I(M) = M[I^\infty]$ is $I$-power torsion. By Lemma \ref{lemma-local-cohomology-adjoint} we see that $$H^0_Z(M) = \Ker(M \longrightarrow M_{f_1} \times \ldots \times M_{f_r})$$ if $I = (f_1, \ldots, f_r)$. Thus $H^0_I(M) \subset H^0_Z(M)$ and conversely, if $x \in H^0_Z(M)$, then it is annihilated by a $f_i^{e_i}$ for some $e_i \geq 1$ hence annihilated by some power of $I$. This proves the first equality and moreover $N$ is $I$-power torsion. By Lemma \ref{lemma-adjoint} we see that $R\Gamma_I(N) = N$. By Lemma \ref{lemma-local-cohomology-adjoint} we see that $R\Gamma_Z(N) = N$. This proves the higher vanishing of $H^p_I(N)$ and $H^p_Z(N)$ in (1) and (2). The vanishing of $H^0_I(M')$ and $H^0_Z(M')$ follow from the preceding remarks and the fact that $M'$ is $I$-power torsion free by More on Algebra, Lemma \ref{more-algebra-lemma-divide-by-torsion}. The equality of higher cohomologies for $M$ and $M'$ follow immediately from the long exact cohomology sequence. \end{proof} \section{Torsion versus complete modules} \label{section-torsion-and-complete} \noindent Let $A$ be a ring and let $I$ be a finitely generated ideal. In this case we can consider the derived category $D_{I^\infty\text{-torsion}}(A)$ of complexes with $I$-power torsion cohomology modules (Section \ref{section-local-cohomology}) and the derived category $D_{comp}(A, I)$ of derived complete complexes (More on Algebra, Section \ref{more-algebra-section-derived-completion}). In this section we show these categories are equivalent. A more general statement can be found in \cite{Dwyer-Greenlees}. \begin{lemma} \label{lemma-complete-and-local} Let $A$ be a ring and let $I$ be a finitely generated ideal. Let $R\Gamma_Z$ be as in Lemma \ref{lemma-local-cohomology-adjoint}. Let ${\ }^\wedge$ denote derived completion as in More on Algebra, Lemma \ref{more-algebra-lemma-derived-completion}. For an object $K$ in $D(A)$ we have $$R\Gamma_Z(K^\wedge) = R\Gamma_Z(K) \quad\text{and}\quad (R\Gamma_Z(K))^\wedge = K^\wedge$$ in $D(A)$. \end{lemma} \begin{proof} Choose $f_1, \ldots, f_r \in A$ generating $I$. Recall that $$K^\wedge = R\Hom_A\left((A \to \prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_r}), K\right)$$ by More on Algebra, Lemma \ref{more-algebra-lemma-derived-completion}. Hence the cone $C = \text{Cone}(K \to K^\wedge)$ is given by $$R\Hom_A\left((\prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_r}), K\right)$$ which can be represented by a complex endowed with a finite filtration whose successive quotients are isomorphic to $$R\Hom_A(A_{f_{i_0} \ldots f_{i_p}}, K), \quad p > 0$$ These complexes vanish on applying $R\Gamma_Z$, see Lemma \ref{lemma-local-cohomology-vanishes}. Applying $R\Gamma_Z$ to the distinguished triangle $K \to K^\wedge \to C \to K[1]$ we see that the first formula of the lemma is correct. \medskip\noindent Recall that $$R\Gamma_Z(K) = K \otimes^\mathbf{L} (A \to \prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_r})$$ by Lemma \ref{lemma-local-cohomology-adjoint}. Hence the cone $C = \text{Cone}(R\Gamma_Z(K) \to K)$ can be represented by a complex endowed with a finite filtration whose successive quotients are isomorphic to $$K \otimes_A A_{f_{i_0} \ldots f_{i_p}}, \quad p > 0$$ These complexes vanish on applying ${\ }^\wedge$, see More on Algebra, Lemma \ref{more-algebra-lemma-derived-completion-vanishes}. Applying derived completion to the distinguished triangle $R\Gamma_Z(K) \to K \to C \to R\Gamma_Z(K)[1]$ we see that the second formula of the lemma is correct. \end{proof} \noindent The following result is a special case of a very general phenomenon concerning admissible subcategories of a triangulated category. \begin{proposition} \label{proposition-torsion-complete} Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. The functors $R\Gamma_Z$ and ${\ }^\wedge$ define quasi-inverse equivalences of categories $$D_{I^\infty\text{-torsion}}(A) \leftrightarrow D_{comp}(A, I)$$ \end{proposition} \begin{proof} Follows immediately from Lemma \ref{lemma-complete-and-local}. \end{proof} \noindent The following addendum of the proposition above makes the correspondence on morphisms more precise. \begin{lemma} \label{lemma-compare-RHom} With notation as in Lemma \ref{lemma-complete-and-local}. For objects $K, L$ in $D(A)$ there is a canonical isomorphism $$R\Hom_A(K^\wedge, L^\wedge) \longrightarrow R\Hom_A(R\Gamma_Z(K), R\Gamma_Z(L))$$ in $D(A)$. \end{lemma} \begin{proof} Say $I = (f_1, \ldots, f_r)$. Denote $C = (A \to \prod A_{f_i} \to \ldots \to A_{f_1 \ldots f_r})$ the alternating {\v C}ech complex. Then derived completion is given by $R\Hom_A(C, -)$ (More on Algebra, Lemma \ref{more-algebra-lemma-derived-completion}) and local cohomology by $C \otimes^\mathbf{L} -$ (Lemma \ref{lemma-local-cohomology-adjoint}). Combining the isomorphism $$R\Hom_A(K \otimes^\mathbf{L} C, L \otimes^\mathbf{L} C) = R\Hom_A(K, R\Hom(C, L \otimes^\mathbf{L} C))$$ (More on Algebra, Lemma \ref{more-algebra-lemma-internal-hom}) and the map $$L \to R\Hom_A(C, L \otimes^\mathbf{L} C)$$ (More on Algebra, Lemma \ref{more-algebra-lemma-internal-hom-diagonal}) we obtain a map $$\gamma : R\Hom_A(K, L) \longrightarrow R\Hom_A(K \otimes^\mathbf{L} C, L \otimes^\mathbf{L} C)$$ On the other hand, the right hand side is derived complete as it is equal to $$R\Hom_A(C, R\Hom_A(K, L \otimes^\mathbf{L} C)).$$ Thus $\gamma$ factors through the derived completion of $R\Hom_A(K, L)$ by the universal property of derived completion. However, the derived completion goes inside the $R\Hom_A$ by More on Algebra, Lemma \ref{more-algebra-lemma-completion-RHom} and we obtain the desired map. \medskip\noindent To show that the map of the lemma is an isomorphism we may assume that $K$ and $L$ are derived complete, i.e., $K = K^\wedge$ and $L = L^\wedge$. In this case we are looking at the map $$\gamma : R\Hom_A(K, L) \longrightarrow R\Hom_A(R\Gamma_Z(K), R\Gamma_Z(L))$$ By Proposition \ref{proposition-torsion-complete} we know that the cohomology groups of the left and the right hand side coincide. In other words, we have to check that the map $\gamma$ sends a morphism $\alpha : K \to L$ in $D(A)$ to the morphism $R\Gamma_Z(\alpha) : R\Gamma_Z(K) \to R\Gamma_Z(L)$. We omit the verification (hint: note that $R\Gamma_Z(\alpha)$ is just the map $\alpha \otimes \text{id}_C : K \otimes^\mathbf{L} C \to L \otimes^\mathbf{L} C$ which is almost the same as the construction of the map in More on Algebra, Lemma \ref{more-algebra-lemma-internal-hom-diagonal}). \end{proof} \section{Finiteness of local cohomology, I} \label{section-finiteness} \noindent We will follow Faltings approach to finiteness of local cohomology modules, see \cite{Faltings-annulators} and \cite{Faltings-finiteness}. Here is a lemma which shows that it suffices to prove local cohomology modules have an annihilator in order to prove that they are finite modules. \begin{lemma} \label{lemma-check-finiteness-local-cohomology-by-annihilator} \begin{reference} This is a special case of \cite[Lemma 3]{Faltings-annulators}. \end{reference} Let $A$ be a Noetherian ring, $I \subset A$ an ideal, $M$ a finite $A$-module, and $n \geq 0$ an integer. Let $Z = V(I)$. The following are equivalent \begin{enumerate} \item $H^i_Z(M)$ is finite for $i \leq n$, \item there exists an $e \geq 0$ such that $I^e$ annihilates $H^i_Z(M)$ for $i \leq n$, and \item there exists an ideal $J \subset A$ with $V(J) \subset Z$ such that $J$ annihilates $H^i_Z(M)$ for $i \leq n$. \end{enumerate} \end{lemma} \begin{proof} We prove the lemma by induction on $n$. For $n = 0$ we have $H^0_Z(M) \subset M$ is finite, hence (1), (2), and (3) are true. Assume that $n > 0$. \medskip\noindent If (1) is true, then, since $H^i_Z(M) = H^i_I(M)$ (Lemma \ref{lemma-local-cohomology-noetherian}) is $I$-power torsion, we see that (2) holds. It is clear that (2) implies (3). \medskip\noindent Assume (3) is true. Let $N = H^0_Z(M)$ and $M' = M/N$. By Lemma \ref{lemma-divide-by-torsion} we may replace $M$ by $M'$. Thus we may assume that $H^0_Z(M) = 0$. This means that $\text{depth}_I(M) > 0$ (Lemma \ref{lemma-depth}). Pick $f \in I$ a nonzerodivisor on $M$. After raising $f$ to a suitable power, we may assume $f \in J$ as $V(J) \subset V(I)$. Then the long exact local cohomology sequence associated to the short exact sequence $$0 \to M \to M \to M/fM \to 0$$ turns into short exact sequences $$0 \to H^i_Z(M) \to H^i_Z(M/fM) \to H^{i + 1}_Z(M) \to 0$$ for $i < n$. We conclude that $J^2$ annihilates $H^i_Z(M/fM)$ for $i < n$. By induction hypothesis we see that $H^i_Z(M/fM)$ is finite for $i < n$. Using the short exact sequence once more we see that $H^{i + 1}_Z(M)$ is finite for $i < n$ as desired. \end{proof} \noindent The following result of Faltings allows us to prove finiteness of local cohomology at the level of local rings. \begin{lemma} \label{lemma-check-finiteness-local-cohomology-locally} \begin{reference} This is a special case of \cite[Satz 1]{Faltings-finiteness}. \end{reference} Let $A$ be a Noetherian ring, $I \subset A$ an ideal, $M$ a finite $A$-module, and $n \geq 0$ an integer. Let $Z = V(I)$. The following are equivalent \begin{enumerate} \item the modules $H^i_Z(M)$ are finite for $i \leq n$, and \item for all $\mathfrak p \in \Spec(A)$ the modules $H^i_Z(M)_\mathfrak p$, $i \leq n$ are finite $A_\mathfrak p$-modules. \end{enumerate} \end{lemma} \begin{proof} The implication (1) $\Rightarrow$ (2) is immediate. We prove the converse by induction on $n$. The case $n = 0$ is clear because both (1) and (2) are always true in that case. \medskip\noindent Assume $n > 0$ and that (2) is true. Let $N = H^0_Z(M)$ and $M' = M/N$. By Lemma \ref{lemma-divide-by-torsion} we may replace $M$ by $M'$. Thus we may assume that $H^0_Z(M) = 0$. This means that $\text{depth}_I(M) > 0$ (Lemma \ref{lemma-depth}). Pick $f \in I$ a nonzerodivisor on $M$ and consider the short exact sequence $$0 \to M \to M \to M/fM \to 0$$ which produces a long exact sequence $$0 \to H^0_Z(M/fM) \to H^1_Z(M) \to H^1_Z(M) \to H^1_Z(M/fM) \to H^2_Z(M) \to \ldots$$ and similarly after localization. Thus assumption (2) implies that the modules $H^i_Z(M/fM)_\mathfrak p$ are finite for $i < n$. Hence by induction assumption $H^i_Z(M/fM)$ are finite for $i < n$. \medskip\noindent Let $\mathfrak p$ be a prime of $A$ which is associated to $H^i_Z(M)$ for some $i \leq n$. Say $\mathfrak p$ is the annihilator of the element $x \in H^i_Z(M)$. Then $\mathfrak p \in Z$, hence $f \in \mathfrak p$. Thus $fx = 0$ and hence $x$ comes from an element of $H^{i - 1}_Z(M/fM)$ by the boundary map $\delta$ in the long exact sequence above. It follows that $\mathfrak p$ is an associated prime of the finite module $\Im(\delta)$. We conclude that $\text{Ass}(H^i_Z(M))$ is finite for $i \leq n$, see Algebra, Lemma \ref{algebra-lemma-finite-ass}. \medskip\noindent Recall that $$H^i_Z(M) \subset \prod\nolimits_{\mathfrak p \in \text{Ass}(H^i_Z(M))} H^i_Z(M)_\mathfrak p$$ by Algebra, Lemma \ref{algebra-lemma-zero-at-ass-zero}. Since by assumption the modules on the right hand side are finite and $I$-power torsion, we can find integers $e_{\mathfrak p, i} \geq 0$, $i \leq n$, $\mathfrak p \in \text{Ass}(H^i_Z(M))$ such that $I^{e_{\mathfrak p, i}}$ annihilates $H^i_Z(M)_\mathfrak p$. We conclude that $I^e$ with $e = \max\{e_{\mathfrak p, i}\}$ annihilates $H^i_Z(M)$ for $i \leq n$. By Lemma \ref{lemma-check-finiteness-local-cohomology-by-annihilator} we see that $H^i_Z(M)$ is finite for $i \leq n$. \end{proof} \begin{lemma} \label{lemma-annihilate-local-cohomology} Let $A$ be a ring and let $J \subset I \subset A$ be finitely generated ideals. Let $i \geq 0$ be an integer. Set $Z = V(I)$. If $H^i_Z(A)$ is annihilated by $J^n$ for some $n$, then $H^i_Z(M)$ annihilated by $J^m$ for some $m = m(M)$ for every finitely presented $A$-module $M$ such that $M_f$ is a finite locally free $A_f$-module for all $f \in I$. \end{lemma} \begin{proof} Consider the annihilator $\mathfrak a$ of $H^i_Z(M)$. Let $\mathfrak p \subset A$ with $\mathfrak p \not \in Z$. By assumption there exists an $f \in I$, $f \not \in \mathfrak p$ and an isomorphism $\varphi : A_f^{\oplus r} \to M_f$ of $A_f$-modules. Clearing denominators (and using that $M$ is of finite presentation) we find maps $$a : A^{\oplus r} \longrightarrow M \quad\text{and}\quad b : M \longrightarrow A^{\oplus r}$$ with $a_f = f^N \varphi$ and $b_f = f^N \varphi^{-1}$ for some $N$. Moreover we may assume that $a \circ b$ and $b \circ a$ are equal to multiplication by $f^{2N}$. Thus we see that $H^i_Z(M)$ is annihilated by $f^{2N}J^n$, i.e., $f^{2N}J^n \subset \mathfrak a$. \medskip\noindent As $U = \Spec(A) \setminus Z$ is quasi-compact we can find finitely many $f_1, \ldots, f_t$ and $N_1, \ldots, N_t$ such that $U = \bigcup D(f_j)$ and $f_j^{2N_j}J^n \subset \mathfrak a$. Then $V(I) = V(f_1, \ldots, f_t)$ and since $I$ is finitely generated we conclude $I^M \subset (f_1, \ldots, f_t)$ for some $M$. All in all we see that $J^m \subset \mathfrak a$ for $m \gg 0$, for example $m = M (2N_1 + \ldots + 2N_t) n$ will do. \end{proof} \begin{lemma} \label{lemma-local-finiteness-for-finite-locally-free} Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Set $Z = V(I)$. Let $n \geq 0$ be an integer. If $H^i_Z(A)$ is finite for $0 \leq i \leq n$, then the same is true for $H^i_Z(M)$, $0 \leq i \leq n$ for any finite $A$-module $M$ such that $M_f$ is a finite locally free $A_f$-module for all $f \in I$. \end{lemma} \begin{proof} The assumption that $H^i_Z(A)$ is finite for $0 \leq i \leq n$ implies there exists an $e \geq 0$ such that $I^e$ annihilates $H^i_Z(A)$ for $0 \leq i \leq n$, see Lemma \ref{lemma-check-finiteness-local-cohomology-by-annihilator}. Then Lemma \ref{lemma-annihilate-local-cohomology} implies that $H^i_Z(M)$, $0 \leq i \leq n$ is annihilated by $I^m$ for some $m = m(M, i)$. We may take the same $m$ for all $0 \leq i \leq n$. Then Lemma \ref{lemma-check-finiteness-local-cohomology-by-annihilator} implies that $H^i_Z(M)$ is finite for $0 \leq i \leq n$ as desired. \end{proof} \section{Finiteness of pushforwards, I} \label{section-finiteness-pushforward} \noindent In this section we discuss the easiest nontrivial case of the finiteness theorem, namely, the finiteness of the first local cohomology or what is equivalent, finiteness of $j_*\mathcal{F}$ where $j : U \to X$ is an open immersion, $X$ is locally Noetherian, and $\mathcal{F}$ is a coherent sheaf on $U$. Following a method of Koll\'ar we find a necessary and sufficient condition, see Proposition \ref{proposition-kollar}. The reader who is interested in higher direct images or higher local cohomology groups should skip ahead to Section \ref{section-finiteness-pushforward-II} or Section \ref{section-finiteness-II} (which are developed independently of the rest of this section). \begin{lemma} \label{lemma-check-finiteness-pushforward-on-associated-points} Let $X$ be a locally Noetherian scheme. Let $j : U \to X$ be the inclusion of an open subscheme with complement $Z$. For $x \in U$ let $i_x : W_x \to U$ be the integral closed subscheme with generic point $x$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_U$-module. The following are equivalent \begin{enumerate} \item for all $x \in \text{Ass}(\mathcal{F})$ the $\mathcal{O}_X$-module $j_*i_{x, *}\mathcal{O}_{W_x}$ is coherent, \item $j_*\mathcal{F}$ is coherent. \end{enumerate} \end{lemma} \begin{proof} We first prove that (1) implies (2). Assume (1) holds. The statement is local on $X$, hence we may assume $X$ is affine. Then $U$ is quasi-compact, hence $\text{Ass}(\mathcal{F})$ is finite (Divisors, Lemma \ref{divisors-lemma-finite-ass}). Thus we may argue by induction on the number of associated points. Let $x \in U$ be a generic point of an irreducible component of the support of $\mathcal{F}$. By Divisors, Lemma \ref{divisors-lemma-finite-ass} we have $x \in \text{Ass}(\mathcal{F})$. By our choice of $x$ we have $\dim(\mathcal{F}_x) = 0$ as $\mathcal{O}_{X, x}$-module. Hence $\mathcal{F}_x$ has finite length as an $\mathcal{O}_{X, x}$-module (Algebra, Lemma \ref{algebra-lemma-support-point}). Thus we may use induction on this length. \medskip\noindent Set $\mathcal{G} = j_*i_{x, *}\mathcal{O}_{W_x}$. This is a coherent $\mathcal{O}_X$-module by assumption. We have $\mathcal{G}_x = \kappa(x)$. Choose a nonzero map $\varphi_x : \mathcal{F}_x \to \kappa(x) = \mathcal{G}_x$. By Cohomology of Schemes, Lemma \ref{coherent-lemma-map-stalks-local-map} there is an open $x \in V \subset U$ and a map $\varphi_V : \mathcal{F}|_V \to \mathcal{G}|_V$ whose stalk at $x$ is $\varphi_x$. Choose $f \in \Gamma(X, \mathcal{O}_X)$ which does not vanish at $x$ such that $D(f) \subset V$. By Cohomology of Schemes, Lemma \ref{coherent-lemma-homs-over-open} (for example) we see that $\varphi_V$ extends to $f^n\mathcal{F} \to \mathcal{G}|_U$ for some $n$. Precomposing with multiplication by $f^n$ we obtain a map $\mathcal{F} \to \mathcal{G}|_U$ whose stalk at $x$ is nonzero. Let $\mathcal{F}' \subset \mathcal{F}$ be the kernel. Note that $\text{Ass}(\mathcal{F}') \subset \text{Ass}(\mathcal{F})$, see Divisors, Lemma \ref{divisors-lemma-ses-ass}. Since $\text{length}_{\mathcal{O}_{X, x}}(\mathcal{F}') = \text{length}_{\mathcal{O}_{X, x}}(\mathcal{F}) - 1$ we may apply the induction hypothesis to conclude $j_*\mathcal{F}'$ is coherent. Since $\mathcal{G} = j_*(\mathcal{G}|_U) = j_*i_{x, *}\mathcal{O}_{W_x}$ is coherent, we can consider the exact sequence $$0 \to j_*\mathcal{F}' \to j_*\mathcal{F} \to \mathcal{G}$$ By Schemes, Lemma \ref{schemes-lemma-push-forward-quasi-coherent} the sheaf $j_*\mathcal{F}$ is quasi-coherent. Hence the image of $j_*\mathcal{F}$ in $j_*(\mathcal{G}|_U)$ is coherent by Cohomology of Schemes, Lemma \ref{coherent-lemma-coherent-Noetherian-quasi-coherent-sub-quotient}. Finally, $j_*\mathcal{F}$ is coherent by Cohomology of Schemes, Lemma \ref{coherent-lemma-coherent-abelian-Noetherian}. \medskip\noindent Assume (2) holds. Exactly in the same manner as above we reduce to the case $X$ affine. We pick $x \in \text{Ass}(\mathcal{F})$ and we set $\mathcal{G} = j_*i_{x, *}\mathcal{O}_{W_x}$. Then we choose a nonzero map $\varphi_x : \mathcal{G}_x = \kappa(x) \to \mathcal{F}_x$ which exists exactly because $x$ is an associated point of $\mathcal{F}$. Arguing exactly as above we may assume $\varphi_x$ extends to an $\mathcal{O}_U$-module map $\varphi : \mathcal{G}|_U \to \mathcal{F}$. Then $\varphi$ is injective (for example by Divisors, Lemma \ref{divisors-lemma-check-injective-on-ass}) and we find and injective map $\mathcal{G} = j_*(\mathcal{G}|_V \to j_*\mathcal{F}$. Thus (1) holds. \end{proof} \begin{lemma} \label{lemma-finiteness-pushforwards-and-H1-local} Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Set $X = \Spec(A)$, $Z = V(I)$, $U = X \setminus Z$, and $j : U \to X$ the inclusion morphism. Let $\mathcal{F}$ be a coherent $\mathcal{O}_U$-module. Then \begin{enumerate} \item there exists a finite $A$-module $M$ such that $\mathcal{F}$ is the restriction of $\widetilde{M}$ to $U$, \item given $M$ there is an exact sequence $$0 \to H^0_Z(M) \to M \to H^0(U, \mathcal{F}) \to H^1_Z(M) \to 0$$ and isomorphisms $H^p(U, \mathcal{F}) = H^{p + 1}_Z(M)$ for $p \geq 1$, \item given $M$ and $p \geq 0$ the following are equivalent \begin{enumerate} \item $R^pj_*\mathcal{F}$ is coherent, \item $H^p(U, \mathcal{F})$ is a finite $A$-module, \item $H^{p + 1}_Z(M)$ is a finite $A$-module, \end{enumerate} \item if the equivalent conditions in (3) hold for $p = 0$, we may take $M = \Gamma(U, \mathcal{F})$ in which case we have $H^1_Z(M) = 0$. \end{enumerate} \end{lemma} \begin{proof} By Properties, Lemma \ref{properties-lemma-extend-finite-presentation} there exists a coherent $\mathcal{O}_X$-module $\mathcal{F}'$ whose restriction to $U$ is isomorphic to $\mathcal{F}$. Say $\mathcal{F}'$ corresponds to the finite $A$-module $M$ as in (1). Note that $R^pj_*\mathcal{F}$ is quasi-coherent (Cohomology of Schemes, Lemma \ref{coherent-lemma-quasi-coherence-higher-direct-images}) and corresponds to the $A$-module $H^p(U, \mathcal{F})$. By Lemma \ref{lemma-local-cohomology-is-local-cohomology} and the general facts in Cohomology, Section \ref{cohomology-section-cohomology-support} we obtain an exact sequence $$0 \to H^0_Z(M) \to M \to H^0(U, \mathcal{F}) \to H^1_Z(M) \to 0$$ and isomorphisms $H^p(U, \mathcal{F}) = H^{p + 1}_Z(M)$ for $p \geq 1$. Here we use that $H^j(X, \mathcal{F}') = 0$ for $j > 0$ as $X$ is affine and $\mathcal{F}'$ is quasi-coherent (Cohomology of Schemes, Lemma \ref{coherent-lemma-quasi-coherent-affine-cohomology-zero}). This proves (2). Parts (3) and (4) are straightforward from (2). \end{proof} \begin{lemma} \label{lemma-finiteness-pushforward} Let $X$ be a locally Noetherian scheme. Let $j : U \to X$ be the inclusion of an open subscheme with complement $Z$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_U$-module. Assume \begin{enumerate} \item $X$ is Nagata, \item $X$ is universally catenary, and \item for $x \in \text{Ass}(\mathcal{F})$ and $z \in Z \cap \overline{\{x\}}$ we have $\dim(\mathcal{O}_{\overline{\{x\}}, z}) \geq 2$. \end{enumerate} Then $j_*\mathcal{F}$ is coherent. \end{lemma} \begin{proof} By Lemma \ref{lemma-check-finiteness-pushforward-on-associated-points} it suffices to prove $j_*i_{x, *}\mathcal{O}_{W_x}$ is coherent for $x \in \text{Ass}(\mathcal{F})$. Let $\pi : Y \to X$ be the normalization of $X$ in $\Spec(\kappa(x))$, see Morphisms, Section \ref{morphisms-section-normalization}. By Morphisms, Lemma \ref{morphisms-lemma-nagata-normalization-finite-general} the morphism $\pi$ is finite. Since $\pi$ is finite $\mathcal{G} = \pi_*\mathcal{O}_Y$ is a coherent $\mathcal{O}_X$-module by Cohomology of Schemes, Lemma \ref{coherent-lemma-finite-pushforward-coherent}. Observe that $W_x = U \cap \pi(Y)$. Thus $\pi|_{\pi^{-1}(U)} : \pi^{-1}(U) \to U$ factors through $i_x : W_x \to U$ and we obtain a canonical map $$i_{x, *}\mathcal{O}_{W_x} \longrightarrow (\pi|_{\pi^{-1}(U)})_*(\mathcal{O}_{\pi^{-1}(U)}) = (\pi_*\mathcal{O}_Y)|_U = \mathcal{G}|_U$$ This map is injective (for example by Divisors, Lemma \ref{divisors-lemma-check-injective-on-ass}). Hence $j_*i_{x, *}\mathcal{O}_{W_x} \subset j_*\mathcal{G}|_U$ and it suffices to show that $j_*\mathcal{G}|_U$ is coherent. \medskip\noindent It remains to prove that $j_*(\mathcal{G}|_U)$ is coherent. We claim Divisors, Lemma \ref{divisors-lemma-check-isomorphism-via-depth-and-ass} applies to $$\mathcal{G} \longrightarrow j_*(\mathcal{G}|_U)$$ which finishes the proof. Let $z \in X$. If $z \in U$, then the map is an isomorphism on stalks as $j_*(\mathcal{G}|_U)|_U = \mathcal{G}|_U$. If $z \in Z$, then $z \not \in \text{Ass}(j_*(\mathcal{G}|_U))$ (Divisors, Lemmas \ref{divisors-lemma-weakass-pushforward} and \ref{divisors-lemma-weakly-ass-support}). Thus it suffices to show that $\text{depth}(\mathcal{G}_z) \geq 2$. Let $y_1, \ldots, y_n \in Y$ be the points mapping to $z$. By Algebra, Lemma \ref{algebra-lemma-depth-goes-down-finite} it suffices to show that $\text{depth}(\mathcal{O}_{Y, y_i}) \geq 2$ for $i = 1, \ldots, n$. If not, then by Properties, Lemma \ref{properties-lemma-criterion-normal} we see that $\dim(\mathcal{O}_{Y, y_i}) = 1$ for some $i$. This is impossible by the dimension formula (Morphisms, Lemma \ref{morphisms-lemma-dimension-formula}) for $\pi : Y \to \overline{\{x\}}$ and assumption (3). \end{proof} \begin{lemma} \label{lemma-sharp-finiteness-pushforward} Let $X$ be an integral locally Noetherian scheme. Let $j : U \to X$ be the inclusion of a nonempty open subscheme with complement $Z$. Assume that for all $z \in Z$ and any associated prime $\mathfrak p$ of the completion $\mathcal{O}_{X, z}^\wedge$ we have $\dim(\mathcal{O}_{X, z}^\wedge/\mathfrak p) \geq 2$. Then $j_*\mathcal{O}_U$ is coherent. \end{lemma} \begin{proof} We may assume $X$ is affine. Using Lemmas \ref{lemma-check-finiteness-local-cohomology-locally} and \ref{lemma-finiteness-pushforwards-and-H1-local} we reduce to $X = \Spec(A)$ where $(A, \mathfrak m)$ is a Noetherian local domain and $\mathfrak m \in Z$. Then we can use induction on $d = \dim(A)$. (The base case is $d = 0, 1$ which do not happen by our assumption on the local rings.) Set $V = \Spec(A) \setminus \{\mathfrak m\}$. Observe that the local rings of $V$ have dimension strictly smaller than $d$. Repeating the arguments for $j' : U \to V$ we and using induction we conclude that $j'_*\mathcal{O}_U$ is a coherent $\mathcal{O}_V$-module. Pick a nonzero $f \in A$ which vanishes on $Z$. Since $D(f) \cap V \subset U$ we find an $n$ such that multiplication by $f^n$ on $U$ extends to a map $f^n : j'_*\mathcal{O}_U \to \mathcal{O}_V$ over $V$ (for example by Cohomology of Schemes, Lemma \ref{coherent-lemma-homs-over-open}). This map is injective hence there is an injective map $$j_*\mathcal{O}_U = j''_* j'_* \mathcal{O}_U \to j''_*\mathcal{O}_V$$ on $X$ where $j'' : V \to X$ is the inclusion morphism. Hence it suffices to show that $j''_*\mathcal{O}_V$ is coherent. In other words, we may assume that $X$ is the spectrum of a local Noetherian domain and that $Z$ consists of the closed point. \medskip\noindent Assume $X = \Spec(A)$ with $(A, \mathfrak m)$ local and $Z = \{\mathfrak m\}$. Let $A^\wedge$ be the completion of $A$. Set $X^\wedge = \Spec(A^\wedge)$, $Z^\wedge = \{\mathfrak m^\wedge\}$, $U^\wedge = X^\wedge \setminus Z^\wedge$, and $\mathcal{F}^\wedge = \mathcal{O}_{U^\wedge}$. The ring $A^\wedge$ is universally catenary and Nagata (Algebra, Remark \ref{algebra-remark-Noetherian-complete-local-ring-universally-catenary} and Lemma \ref{algebra-lemma-Noetherian-complete-local-Nagata}). Moreover, condition (3) of Lemma \ref{lemma-finiteness-pushforward} for $X^\wedge, Z^\wedge, U^\wedge, \mathcal{F}^\wedge$ holds by assumption! Thus we see that $(U^\wedge \to X^\wedge)_*\mathcal{O}_{U^\wedge}$ is coherent. Since the morphism $c : X^\wedge \to X$ is flat we conclude that the pullback of $j_*\mathcal{O}_U$ is $(U^\wedge \to X^\wedge)_*\mathcal{O}_{U^\wedge}$ (Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}). Finally, since $c$ is faithfully flat we conclude that $j_*\mathcal{O}_U$ is coherent by Descent, Lemma \ref{descent-lemma-finite-type-descends}. \end{proof} \begin{remark} \label{remark-closure} Let $j : U \to X$ be an open immersion of locally Noetherian schemes. Let $x \in U$. Let $i_x : W_x \to U$ be the integral closed subscheme with generic point $x$ and let $\overline{\{x\}}$ be the closure in $X$. Then we have a commutative diagram $$\xymatrix{ W_x \ar[d]_{i_x} \ar[r]_{j'} & \overline{\{x\}} \ar[d]^i \\ U \ar[r]^j & X }$$ We have $j_*i_{x, *}\mathcal{O}_{W_x} = i_*j'_*\mathcal{O}_{W_x}$. As the left vertical arrow is a closed immersion we see that $j_*i_{x, *}\mathcal{O}_{W_x}$ is coherent if and only of $j'_*\mathcal{O}_{W_x}$ is coherent. \end{remark} \begin{remark} \label{remark-no-finiteness-pushforward} Let $X$ be a locally Noetherian scheme. Let $j : U \to X$ be the inclusion of an open subscheme with complement $Z$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_U$-module. If there exists an $x \in \text{Ass}(\mathcal{F})$ and $z \in Z \cap \overline{\{x\}}$ such that $\dim(\mathcal{O}_{\overline{\{x\}}, z}) \leq 1$, then $j_*\mathcal{F}$ is not coherent. To prove this we can do a flat base change to the spectrum of $\mathcal{O}_{X, z}$. Let $X' = \overline{\{x\}}$. The assumption implies $\mathcal{O}_{X' \cap U} \subset \mathcal{F}$. Thus it suffices to see that $j_*\mathcal{O}_{X' \cap U}$ is not coherent. This is clear because $X' = \{x, z\}$, hence $j_*\mathcal{O}_{X' \cap U}$ corresponds to $\kappa(x)$ as an $\mathcal{O}_{X, z}$-module which cannot be finite as $x$ is not a closed point. \medskip\noindent In fact, the converse of Lemma \ref{lemma-sharp-finiteness-pushforward} holds true: given an open immersion $j : U \to X$ of integral Noetherian schemes and there exists a $z \in X \setminus U$ and an associated prime $\mathfrak p$ of the completion $\mathcal{O}_{X, z}^\wedge$ with $\dim(\mathcal{O}_{X, z}^\wedge/\mathfrak p) = 1$, then $j_*\mathcal{O}_U$ is not coherent. Namely, you can pass to the local ring, you can enlarge $U$ to the punctured spectrum, you can pass to the completion, and then the argument above gives the nonfiniteness. \end{remark} \begin{proposition}[Koll\'ar] \label{proposition-kollar} \begin{reference} Theorem of Koll\'ar stated in an email dated Wed, 1 Jul 2015. \end{reference} Let $j : U \to X$ be an open immersion of locally Noetherian schemes with complement $Z$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_U$-module. The following are equivalent \begin{enumerate} \item $j_*\mathcal{F}$ is coherent, \item for $x \in \text{Ass}(\mathcal{F})$ and $z \in Z \cap \overline{\{x\}}$ and any associated prime $\mathfrak p$ of the completion $\mathcal{O}_{\overline{\{x\}}, z}^\wedge$ we have $\dim(\mathcal{O}_{\overline{\{x\}}, z}^\wedge/\mathfrak p) \geq 2$. \end{enumerate} \end{proposition} \begin{proof} If (2) holds we get (1) by a combination of Lemmas \ref{lemma-check-finiteness-pushforward-on-associated-points}, Remark \ref{remark-closure}, and Lemma \ref{lemma-sharp-finiteness-pushforward}. If (2) does not hold, then $j_*i_{x, *}\mathcal{O}_{W_x}$ is not finite for some $x \in \text{Ass}(\mathcal{F})$ by the discussion in Remark \ref{remark-no-finiteness-pushforward} (and Remark \ref{remark-closure}). Thus $j_*\mathcal{F}$ is not coherent by Lemma \ref{lemma-check-finiteness-pushforward-on-associated-points}. \end{proof} \begin{lemma} \label{lemma-kollar-finiteness-H1-local} Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Set $Z = V(I)$. Let $M$ be a finite $A$-module. The following are equivalent \begin{enumerate} \item $H^1_Z(M)$ is a finite $A$-module, and \item for all $\mathfrak p \in \text{Ass}(M)$, $\mathfrak p \not \in Z$ and all $\mathfrak q \in V(\mathfrak p + I)$ the completion of $(A/\mathfrak p)_\mathfrak q$ does not have associated primes of dimension $1$. \end{enumerate} \end{lemma} \begin{proof} Follows immediately from Proposition \ref{proposition-kollar} via Lemma \ref{lemma-finiteness-pushforwards-and-H1-local}. \end{proof} \noindent The formulation in the following lemma has the advantage that conditions (1) and (2) are inherited by schemes of finite type over $X$. Moreover, this is the form of finiteness which we will generalize to higher direct images in Section \ref{section-finiteness-pushforward-II}. \begin{lemma} \label{lemma-finiteness-pushforward-general} Let $X$ be a locally Noetherian scheme. Let $j : U \to X$ be the inclusion of an open subscheme with complement $Z$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_U$-module. Assume \begin{enumerate} \item $X$ is universally catenary, \item for every $z \in Z$ the formal fibres of $\mathcal{O}_{X, z}$ are $(S_1)$. \end{enumerate} In this situation the following are equivalent \begin{enumerate} \item[(a)] for $x \in \text{Ass}(\mathcal{F})$ and $z \in Z \cap \overline{\{x\}}$ we have $\dim(\mathcal{O}_{\overline{\{x\}}, z}) \geq 2$, and \item[(b)] $j_*\mathcal{F}$ is coherent. \end{enumerate} \end{lemma} \begin{proof} Let $x \in \text{Ass}(\mathcal{F})$. By Proposition \ref{proposition-kollar} it suffices to check that $A = \mathcal{O}_{\overline{\{x\}}, z}$ satisfies the condition of the proposition on associated primes of its completion if and only if $\dim(A) \geq 2$. Observe that $A$ is universally catenary (this is clear) and that its formal fibres are $(S_1)$ as follows from More on Algebra, Lemma \ref{more-algebra-lemma-formal-fibres-normal} and Proposition \ref{more-algebra-proposition-finite-type-over-P-ring}. Let $\mathfrak p' \subset A^\wedge$ be an associated prime. As $A \to A^\wedge$ is flat, by Algebra, Lemma \ref{algebra-lemma-bourbaki}, we find that $\mathfrak p'$ lies over $(0) \subset A$. Since the formal fibre $A^\wedge \otimes_A f.f.(A)$ is $(S_1)$ we see that $\mathfrak p'$ is a minimal prime, see Algebra, Lemma \ref{algebra-lemma-criterion-no-embedded-primes}. Since $A$ is universally catenary it is formally catenary by More on Algebra, Proposition \ref{more-algebra-proposition-ratliff}. Hence $\dim(A^\wedge/\mathfrak p') = \dim(A)$ which proves the equivalence. \end{proof} \section{Trivial duality for a ring map} \label{section-trivial} \noindent Let $A \to B$ be a ring homomorphism. Consider the functor $$\Hom_A(B, -) : \text{Mod}_A \longrightarrow \text{Mod}_B,\quad M \longmapsto \Hom_A(B, M)$$ This functor is left exact and has a derived extension $R\Hom(B, -) : D(A) \to D(B)$. If $f_* : D(B) \to D(A)$ is the restriction functor, then $f_*R\Hom(B, K) = R\Hom_A(B, K)$ for every $K \in D(A)$. Since $R\Hom_A(A, K) = K$, the map $A \to B$ induces a canonical map $f_*R\Hom(B, K) \to K$ in $D(A)$ functorial in $K$. \begin{lemma} \label{lemma-right-adjoint} Let $A \to B$ be a ring homomorphism. The functor $R\Hom(B, -)$ constructed above is the right adjoint to the restriction functor $f_* : D(B) \to D(A)$. \end{lemma} \begin{proof} This is a consequence of the fact that $f_*$ and $\Hom_A(B, -)$ are adjoint functors by Algebra, Lemma \ref{algebra-lemma-adjoint-tensor-restrict}. See Derived Categories, Lemma \ref{derived-lemma-derived-adjoint-functors}. \end{proof} \begin{lemma} \label{lemma-composition-right-adjoints} Let $A \to B \to C$ be ring maps. Then $R\Hom(C, -) \circ R\Hom(B, -) : D(A) \to D(C)$ is the functor $R\Hom(C, -) : D(A) \to D(C)$. \end{lemma} \begin{proof} Follows from uniqueness of right adjoints and Lemma \ref{lemma-right-adjoint}. \end{proof} \begin{lemma} \label{lemma-RHom-ext} Let $A \to B$ be a ring homomorphism. For $K$ in $D(A)$ we have $f_*R\Hom(B, K) = R\Hom_A(B, K)$ where $f_* : D(B) \to D(A)$ is restriction. In particular $R^q\Hom(B, K) = \text{Ext}_A^q(B, K)$. \end{lemma} \begin{proof} Omitted, but see above. \end{proof} \noindent Let $A$ be a Noetherian ring. We will denote $$D_{\textit{Coh}}(A) \subset D(A)$$ the full subcategory consisting of those objects $K$ of $D(A)$ whose cohomology modules are all finite $A$-modules. This makes sense by Derived Categories, Section \ref{derived-section-triangulated-sub} because as $A$ is Noetherian, the subcategory of finite $A$-modules is a Serre subcategory of $\text{Mod}_A$. \begin{lemma} \label{lemma-exact-support-coherent} With notation as above, assume $A \to B$ is a finite ring map of Noetherian rings. Then $R\Hom(B, -)$ maps $D^+_{\textit{Coh}}(A)$ into $D^+_{\textit{Coh}}(B)$. \end{lemma} \begin{proof} 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 $\text{Ext}^i_A(B, K)$ are finite $A$-modules. Since $K$ is bounded below there is a convergent spectral sequence $$\text{Ext}^p_A(B, H^q(K)) \Rightarrow \text{Ext}^{p + q}_A(B, K)$$ This finishes the proof as the modules $\text{Ext}^p_A(B, H^q(K))$ are finite by Algebra, Lemma \ref{algebra-lemma-ext-noetherian}. \end{proof} \begin{remark} \label{remark-exact-support} Let $A$ be a ring and let $I \subset A$ be an ideal. Set $B = A/I$. In this case the functor $\Hom_A(B, -)$ is equal to the functor $$\text{Mod}_A \longrightarrow \text{Mod}_B,\quad M \longmapsto M[I]$$ which sends $M$ to the submodule of $I$-torsion. \end{remark} \begin{situation} \label{situation-resolution} Let $R \to A$ be a ring map. We will give an alternative construction of $R\Hom(A, -)$ which will stand us in good stead later in this chapter. Namely, suppose we have a differential graded algebra $(E, d)$ over $R$ and a quasi-isomorphism $E \to A$ where we view $A$ as a differential graded algebra over $R$ with zero differential. Then we have commutative diagrams $$\vcenter{ \xymatrix{ D(E, \text{d}) \ar[rd] & & D(A) \ar[ll] \ar[ld] \\ & D(R) } } \quad\text{and}\quad \vcenter{ \xymatrix{ D(E, \text{d}) \ar[rr]_{- \otimes_E^\mathbf{L} A} & & D(A) \\ & D(R) \ar[lu]^{- \otimes_R^\mathbf{L} E} \ar[ru]_{- \otimes_R^\mathbf{L} A} } }$$ where the horizontal arrows are equivalences of categories (Differential Graded Algebra, Lemma \ref{dga-lemma-qis-equivalence}). It is clear that the first diagram commutes. The second diagram commutes because the first one does and our functors are their left adjoints (Differential Graded Algebra, Example \ref{dga-example-map-hom-tensor}) or because we have $E \otimes^\mathbf{L}_E A = E \otimes_E A$ and we can use Differential Graded Algebra, Lemma \ref{dga-lemma-compose-tensor-functors-general}. \end{situation} \begin{lemma} \label{lemma-RHom-dga} In Situation \ref{situation-resolution} the functor $R\Hom(A, -)$ is equal to the composition of $R\Hom(E, -) : D(R) \to D(E, \text{d})$ and the equivalence $- \otimes^\mathbf{L}_E A : D(E, \text{d}) \to D(A)$. \end{lemma} \begin{proof} 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 $- \otimes^\mathbf{L}_E A : D(E, \text{d}) \to D(A)$. \end{proof} \begin{lemma} \label{lemma-RHom-is-tensor} In Situation \ref{situation-resolution} assume that \begin{enumerate} \item $E$ viewed as an object of $D(R)$ is compact, and \item $N = \Hom^\bullet_R(E^\bullet, R)$ computes $R\Hom(E, R)$. \end{enumerate} Then $R\Hom(E, -) : D(R) \to D(E)$ is isomorphic to $K \mapsto K \otimes_R^\mathbf{L} N$. \end{lemma} \begin{proof} Special case of Differential Graded Algebra, Lemma \ref{dga-lemma-RHom-is-tensor}. \end{proof} \begin{lemma} \label{lemma-RHom-is-tensor-special} In Situation \ref{situation-resolution} assume $A$ is a perfect $R$-module. Then $$R\Hom(A, -) : D(R) \to D(A)$$ is given by $K \mapsto K \otimes_R^\mathbf{L} M$ for $M = R\Hom(A, R) \in D(A)$. \end{lemma} \begin{proof} 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}. \end{proof} \section{Dualizing complexes} \label{section-dualizing} \noindent In this section we define dualizing complexes for Noetherian rings. \begin{definition} \label{definition-dualizing} Let $A$ be a Noetherian ring. A {\it dualizing complex} is a complex of $A$-modules $\omega_A^\bullet$ such that \begin{enumerate} \item $\omega_A^\bullet$ has finite injective dimension, \item $H^i(\omega_A^\bullet)$ is a finite $A$-module for all $i$, and \item $A \to R\Hom_A(\omega_A^\bullet, \omega_A^\bullet)$ is a quasi-isomorphism. \end{enumerate} \end{definition} \noindent 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. \begin{lemma} \label{lemma-dualizing} Let $A$ be a Noetherian ring. If $\omega_A^\bullet$ is a dualizing complex, then the functor $$D : K \longmapsto R\Hom_A(K, \omega_A^\bullet)$$ 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 equivalence $D^b_{\textit{Coh}}(A) \to D^b_{\textit{Coh}}(A)$. Moreover $D \circ D$ is isomorphic to the identity functor. \end{lemma} \begin{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 \ref{derived-remark-truncation-distinguished-triangle}. Since $\omega_A^\bullet$ has finite injective dimension we see that $R\Hom_A(\tau_{\geq n + 1}K, \omega_A^\bullet)$ has vanishing cohomology in degrees $\geq n - c$ for some constant $c$. On the other hand, we obtain a spectral sequence $$\text{Ext}_A^p(H^{-q}(\tau_{\leq n}K), \omega_A^\bullet) \Rightarrow \text{Ext}_A^{p + q}(\tau_{\leq n}K, \omega_A^\bullet) = H^{p + q}(R\Hom_A(\tau_{\leq n}K, \omega_A^\bullet))$$ which shows that these cohomology modules are finite. Since for $n > p + q + c$ this is equal to $H^{p + q}(R\Hom_A(K, \omega_A^\bullet))$ we see that $R\Hom_A(K, \omega_A^\bullet)$ is indeed 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. \end{proof} \begin{lemma} \label{lemma-detect-cohomology} 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 $H^i(K)$. \end{lemma} \begin{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 \ref{lemma-torsion-submodule-sum-injective-hulls} and that each of the modules $E = I[\mathfrak m^n]$ is of the desired type. Thus it suffices to prove that $$\Hom_{D(A)}(K, I) = \colim \Hom_{D(A)}(K, I[\mathfrak m^n])$$ 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 $\Hom$ groups above are computed by using $\Hom_{K(A)}(K^\bullet, -)$. As each $K^n$ is finite free the limit statement holds and the proof is complete. \end{proof} \noindent Let $R$ be a ring. We will say that an object $L$ of $D(R)$ is {\it invertible} if there is an open covering $\Spec(R) = \bigcup D(f_i)$ such that $L \otimes_R R_{f_i} \cong R_{f_i}[-n_i]$ for some integers $n_i$. In this case, the function $$\mathfrak p \mapsto n_\mathfrak p,\quad \text{where }n_\mathfrak p\text{ is the unique integer such that } H^{n_\mathfrak p}(L \otimes \kappa(\mathfrak p)) \not = 0$$ is locally constant on $\Spec(R)$. In particular, it follows that $L = \bigoplus H^n(L)[-n]$ which gives a well defined complex of $R$-modules (with zero differentials) representing $L$. Since each $H^n(L)$ is finite projective and nonzero for only a finite number of $n$ we also see that $L$ is a perfect object of $D(R)$. \begin{lemma} \label{lemma-equivalence-comes-from-invertible} 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 $D(A)$. \end{lemma} \begin{proof} 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 $$\text{Ext}^i_A(\kappa, \kappa) = \text{Ext}^i_A(F(\kappa), F(\kappa)) = \Hom_{D(A)}(F(\kappa), F(\kappa)[-i])$$ 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 $$\text{Ext}^i_A(A, E') = \text{Ext}^i_A(F(A), F(E')) = \Hom_{D(A)}(F(A), E[-a + i])$$ 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 $E$ (Lemma \ref{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. \end{proof} \begin{lemma} \label{lemma-dualizing-unique} 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 $D(A)$. \end{lemma} \begin{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. \end{proof} \begin{lemma} \label{lemma-dualizing-localize} 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 $B$. \end{lemma} \begin{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}. \end{proof} \begin{lemma} \label{lemma-dualizing-glue} 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 $A$. \end{lemma} \begin{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 \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. \end{proof} \begin{lemma} \label{lemma-dualizing-finite} Let $A \to B$ be a finite ring map of Noetherian rings. Let $\omega_A^\bullet$ be a dualizing complex. Then $R\Hom(B, \omega_A^\bullet)$ is a dualizing complex for $B$. \end{lemma} \begin{proof} 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}. \end{proof} \begin{lemma} \label{lemma-dualizing-quotient} Let $A \to B$ be a surjective homomorphism of Noetherian rings. Let $\omega_A^\bullet$ be a dualizing complex. Then $R\Hom(B, \omega_A^\bullet)$ is a dualizing complex for $B$. \end{lemma} \begin{proof} Special case of Lemma \ref{lemma-dualizing-finite}. \end{proof} \begin{lemma} \label{lemma-dualizing-polynomial-ring} 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 $A[x]$. \end{lemma} \begin{proof} 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}. \end{proof} \begin{proposition} \label{proposition-dualizing-essentially-finite-type} 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. \end{proposition} \begin{proof} This follows from a combination of Lemmas \ref{lemma-dualizing-localize}, \ref{lemma-dualizing-quotient}, and \ref{lemma-dualizing-polynomial-ring}. \end{proof} \begin{lemma} \label{lemma-find-function} 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 $R\Hom_A(\kappa, \omega_A^\bullet) \cong \kappa[n]$ for some $n \in \mathbf{Z}$. \end{lemma} \begin{proof} 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 $\kappa$. \end{proof} \section{Dualizing complexes over local rings} \label{section-dualizing-local} \noindent In this section $(A, \mathfrak m, \kappa)$ will be a Noetherian local ring endowed with a dualizing complex $\omega_A^\bullet$ such that the integer $n$ of Lemma \ref{lemma-find-function} is zero. More precisely, we assume that $R\Hom_A(\kappa, \omega_A^\bullet) = \kappa[0]$. In this case we will say that the dualizing complex is {\it normalized}. Observe that a normalized dualizing complex is unique up to isomorphism and that any other dualizing complex for $A$ is isomorphic to a shift of a normalized one (Lemma \ref{lemma-dualizing-unique}). \begin{lemma} \label{lemma-normalized-finite} Let