# stacks/stacks-project

Small changes to dualizing and more-algebra

 @@ -2645,13 +2645,10 @@ \section{Dualizing complexes} symmetric monoidal structure on $D(R)$ given by derived tensor product. In More on Algebra, Lemma \ref{more-algebra-lemma-invertible-derived} we have seen this means $L$ is perfect and there is an open covering $\Spec(R) = \bigcup D(f_i)$ such that we have seen this means $L$ is perfect, $L = \bigoplus H^n(L)[-n]$, this is a finite sum, each $H^n(L)$ is finite projective, and 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 particular, it follows that $L = \bigoplus H^n(L)[-n]$ which gives a well defined complex of $R$-modules (with zero differentials) representing $L$. Each $H^n(L)$ is finite projective (as it is locally free) and $H^n(L)$ nonzero for only a finite number of $n$. \begin{lemma} \label{lemma-equivalence-comes-from-invertible}
 Categories, Definition \ref{categories-definition-invertible}, and \item for every prime ideal $\mathfrak p \subset R$ there exists an $f \in R$, $f \not \in \mathfrak p$ such that $M_f \cong R_f[n]$ for some $n \in \mathbf{Z}$. $M_f \cong R_f[-n]$ for some $n \in \mathbf{Z}$. \end{enumerate} Moreover, in this case \begin{enumerate} \item[(a)] $M$ is a perfect object of $D(R)$, \item[(b)] $M = \bigoplus H^n(M)[-n]$ in $D(R)$, \item[(c)] each $H^n(M)$ is a finite projective $R$-module, \item[(d)] we can write $R = \prod_{a \leq n \leq b} R_n$ such that $H^n(M)$ corresponds to an invertible $R_n$-module. \end{enumerate} Moreover, in this case $M$ is a perfect object of $D(R)$. \end{lemma} \begin{proof} for one $i$ and equal to $1$ in that case. By Lemma \ref{lemma-lift-perfect-from-residue-field} this gives (2). \medskip\noindent In the proof above we have seen that (a) holds. Let $U_n \subset \Spec(R)$ be the union of the opens of the form $D(f)$ such that $M_f \cong R_f[-n]$. Clearly, $U_n \cap U_{n'} = \emptyset$ if $n \not = n'$. If $M$ has tor amplitude in $[a, b]$, then $U_n = \emptyset$ if $n \not \in [a, b]$. Hence we see that we have a product decomposition $R = \prod_{a \leq n \leq b} R_n$ as in (d) such that $U_n$ corresponds to $\Spec(R_n)$, see Algebra, Lemma \ref{algebra-lemma-disjoint-implies-product}. Since $D(R) = \prod_{a \leq n \leq b} D(R_n)$ and similary for the category of modules parts (b), (c), and (d) follow immediately. \end{proof}