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Small changes to dualizing and more-algebra

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aisejohan committed Nov 6, 2019
1 parent 9330540 commit bd621ecc378f672fe2526edd101c4a0dd94b17b9
Showing with 24 additions and 8 deletions.
  1. +3 −6 dualizing.tex
  2. +21 −2 more-algebra.tex
@@ -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 $n$.

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 $n \in \mathbf{Z}$.
Moreover, in this case
\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.
Moreover, in this case $M$ is a perfect object of $D(R)$.

for one $i$ and equal to $1$ in that case. By
Lemma \ref{lemma-lift-perfect-from-residue-field}
this gives (2).

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.

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