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Clarify some arguments on coherent triples

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aisejohan committed Jul 25, 2019
1 parent c53944e commit 1fcf0ae7e55c2132d8990ff69a992a7d5d222f19
Showing with 133 additions and 61 deletions.
  1. +133 −61 algebraization.tex
@@ -8193,7 +8193,7 @@ \section{Coherent triples}
\noindent
Let $(A, \mathfrak m)$ be a Noetherian local ring.
Let $f \in \mathfrak m$ be a nonzerodivisor. Set
$X = \Spec(A)$, $X_0 = \Spec(A/fA)$, $U = X \setminus V(\mathfrak a)$, and
$X = \Spec(A)$, $X_0 = \Spec(A/fA)$, $U = X \setminus V(\mathfrak m)$, and
$U_0 = U \cap X_0$.
We say $(\mathcal{F}, \mathcal{F}_0, \alpha)$ is a {\it coherent triple}
if we have
@@ -8285,30 +8285,31 @@ \section{Coherent triples}
\end{lemma}

\begin{proof}
If $\mathcal{F}', \alpha', \alpha'_0$ and
$\mathcal{F}'', \alpha'', \alpha''_0$ are two such choices, then
there exists a submodule $\mathcal{F}''' \subset \mathcal{F}''$
with $\mathcal{F}''/\mathcal{F}'''$ supported at $\{\mathfrak m\}$
and an $\mathcal{O}_X$-module map $\mathcal{F}''' \to \mathcal{F}'$
Let $\mathcal{F}', \alpha', \alpha'_0$ and
$\mathcal{F}'', \alpha'', \alpha''_0$ be two such choices.
For $n > 0$ set $\mathcal{F}'_n = \mathfrak m^n \mathcal{F}'$.
By Cohomology of Schemes, Lemma \ref{coherent-lemma-homs-over-open}
for some $n$
there exists an $\mathcal{O}_X$-module map $\mathcal{F}'_n \to \mathcal{F}''$
agreeing with the identification
$\mathcal{F}''|_U = \mathcal{F}'|_U$ determined by $\alpha'$ and $\alpha''$.
In fact, by
Then the diagram
$$
\xymatrix{
\mathcal{F}'_n/f\mathcal{F}'_n \ar[r] \ar[d] &
\mathcal{F}'/f\mathcal{F}' \ar[d]^{\alpha_0'} \\
\mathcal{F}''/f\mathcal{F}'' \ar[r]^{\alpha_0''} &
\mathcal{F}_0
}
$$
is commutative after restricting to $U_0$. Hence by
Cohomology of Schemes, Lemma \ref{coherent-lemma-homs-over-open}
we can take $\mathcal{F}''' = \mathfrak m^n \mathcal{F}''$ for
$n \gg 0$.
After increasing $n$ further we may in addition assume
the induced maps
$\mathcal{F}'''/f\mathcal{F}''' \to \mathcal{F}''/f\mathcal{F}''$
and
$\mathcal{F}'''/f\mathcal{F}''' \to \mathcal{F}'/f\mathcal{F}'$
agree with $\alpha'_0$ and $\alpha''_0$ by the same lemma
(because we have the agreement over $U_0$ and because
$\mathcal{F}'''/f\mathcal{F}''' \to \mathcal{F}'/f\mathcal{F}'$
factors through the canonical map
$\mathcal{F}'''/f\mathcal{F}''' =
\mathfrak m^n \mathcal{F}''/f\mathfrak m^n \mathcal{F}'' \to
\mathfrak m^n(\mathcal{F}''/f\mathcal{F}'')$; some details omitted).
Then we see that we have a third choice
it is commutative after restricting to
$\mathfrak m^l(\mathcal{F}'_n/f\mathcal{F}'_n)$ for some $l > 0$. Since
$\mathcal{F}'_{n + l}/f\mathcal{F}'_{n + l} \to \mathcal{F}'_n/f\mathcal{F}'_n$
factors through $\mathfrak m^l(\mathcal{F}'_n/f\mathcal{F}'_n)$
we see that after replacing $n$ by $n + l$ the diagram
is commutative. In other words, we have found a third choice
$\mathcal{F}''', \alpha''', \alpha'''_0$
such that there are maps $\mathcal{F}''' \to \mathcal{F}''$
and $\mathcal{F}''' \to \mathcal{F}'$ over $X$
@@ -8322,13 +8323,14 @@ \section{Coherent triples}
isomorphism over $U$ and since $f : \mathcal{F}'' \to \mathcal{F}''$
is injective. Clearly $\mathcal{F}'/\mathcal{F}''$ is supported on
$\{\mathfrak m\}$ hence has finite length. We have the maps
of coherent $\mathcal{O}_{X_0}$-modules
$$
\mathcal{F}''/f\mathcal{F}'' \to
\mathcal{F}'/f\mathcal{F}' \xrightarrow{\alpha'_0}
\mathcal{F}_0
$$
whose composition is $\alpha''_0$. Elementary homological algebra
gives a $6$-term exact sequence
whose composition is $\alpha''_0$ and which are isomorphisms over $U_0$.
Elementary homological algebra gives a $6$-term exact sequence
$$
\begin{matrix}
0 \to
@@ -8341,7 +8343,7 @@ \section{Coherent triples}
\end{matrix}
$$
By additivity of lengths (Algebra, Lemma \ref{algebra-lemma-length-additive})
it suffices to show that
we find that it suffices to show that
$$
\text{length}_A(
\Coker(\mathcal{F}''/f\mathcal{F}'' \to \mathcal{F}'/f\mathcal{F}')) -
@@ -8406,27 +8408,51 @@ \section{Coherent triples}
$\alpha' : \mathcal{F}'|_U \to \mathcal{F}$,
$\beta' : \mathcal{G}'|_U \to \mathcal{G}$, and
$\gamma' : \mathcal{H}'|_U \to \mathcal{H}$ by construction.
To finish the proof we'll need to construct the maps
To finish the proof we'll need to construct maps
$\alpha'_0 : \mathcal{F}'/f\mathcal{F}' \to \mathcal{F}_0$ and
$\gamma'_0 : \mathcal{H}'/f\mathcal{H}' \to \mathcal{H}_0$, except
this may not be possible with our initial choice of $\mathcal{G}'$.
Namely, the obstruction is the composition
$\gamma'_0 : \mathcal{H}'/f\mathcal{H}' \to \mathcal{H}_0$ as in
Lemma \ref{lemma-prepare-chi-triple} and fitting into
a commutative diagram
$$
\xymatrix{
0 \ar[r] &
\mathcal{F}'/f\mathcal{F}' \ar[r] \ar@{..>}[d]^{\alpha'_0} &
\mathcal{G}'/f\mathcal{G}' \ar[r] \ar[d]^{\beta'_0} &
\mathcal{H}'/f\mathcal{H}' \ar[r] \ar@{..>}[d]^{\gamma'_0} &
0 \\
0 \ar[r] &
\mathcal{F}_0 \ar[r] &
\mathcal{G}_0 \ar[r] &
\mathcal{H}_0 \ar[r] &
0
}
$$
However, this may not be possible with our initial choice of $\mathcal{G}'$.
From the displayed diagram we see the obstruction is
exactly the composition
$$
\delta :
\mathcal{F}'/f\mathcal{F}' \to
\mathcal{G}'/f\mathcal{G}' \xrightarrow{\beta'_0}
\mathcal{G}_0 \to
\mathcal{H}_0
$$
Note that the restriction of $\delta$ to $U_0$ is zero
by our choice of $\mathcal{F}'$ and $\mathcal{H}'$.
Hence by
Note that the restriction of $\delta$ to $U_0$ is zero by our choice of
$\mathcal{F}'$ and $\mathcal{H}'$. Hence by
Cohomology of Schemes, Lemma \ref{coherent-lemma-homs-over-open}
there exists an $n > 0$ such that
$\delta$ vanishes on $\mathfrak m^n \cdot (\mathcal{F}'/f\mathcal{F}')$.
Set $\mathcal{G}'_n = \mathfrak m^n \mathcal{G}'$,
there exists an $k > 0$ such that
$\delta$ vanishes on $\mathfrak m^k \cdot (\mathcal{F}'/f\mathcal{F}')$.
For $n > k$ set $\mathcal{G}'_n = \mathfrak m^n \mathcal{G}'$,
$\mathcal{F}'_n = \mathcal{G}'_n \cap \mathcal{F}'$, and
$\mathcal{H}'_n = \mathcal{G}'_n/\mathcal{F}'_n$.
Observe that $\beta'_0$ can be composed with
$\mathcal{G}'_n/f\mathcal{G}'_n \to \mathcal{G}'/f\mathcal{G}'$
to give a map
$\beta'_{n, 0} : \mathcal{G}'_n/f\mathcal{G}'_n \to \mathcal{G}_0$
as in Lemma \ref{lemma-prepare-chi-triple}.
By Artin-Rees (Algebra, Lemma \ref{algebra-lemma-Artin-Rees})
we may choose $n$ such that
$\mathcal{F}'_n \subset \mathfrak m^k \mathcal{F}'$.
As above the maps
$f : \mathcal{F}'_n \to \mathcal{F}'_n$,
$f : \mathcal{G}'_n \to \mathcal{G}'_n$, and
@@ -8442,22 +8468,32 @@ \section{Coherent triples}
$\alpha'_n : \mathcal{F}'_n|_U \to \mathcal{F}$,
$\beta'_n : \mathcal{G}'_n|_U \to \mathcal{G}$, and
$\gamma'_n : \mathcal{H}'_n|_U \to \mathcal{H}$.
Finally, we consider the composition
We consider the obstruction
$$
\delta_n :
\mathcal{F}'_n/f\mathcal{F}'_n \to
\mathcal{G}'_n/f\mathcal{G}'_n \to
\mathcal{G}'/f\mathcal{G}' \xrightarrow{\beta'_0}
\mathcal{G}'_n/f\mathcal{G}'_n
\xrightarrow{\beta'_{n, 0}}
\mathcal{G}_0 \to
\mathcal{H}_0
$$
Since the composition of the first two maps factors as
$\mathcal{F}'_n/f\mathcal{F}'_n \to
\mathfrak m^n(\mathcal{F}'/f\mathcal{F}') \to
\mathcal{G}'/f\mathcal{G}'$ we conclude $\delta_n = 0$.
as before. However, the commutative diagram
$$
\xymatrix{
\mathcal{F}'_n/f\mathcal{F}'_n \ar[r] \ar[d] &
\mathcal{G}'_n/f\mathcal{G}'_n \ar[r]_{\beta'_{n, 0}} \ar[d] &
\mathcal{G}_0 \ar[r] \ar[d] &
\mathcal{H}_0 \ar[d] \\
\mathcal{F}'/f\mathcal{F}' \ar[r] &
\mathcal{G}'/f\mathcal{G}' \ar[r]^{\beta'_0} &
\mathcal{G}_0 \ar[r] &
\mathcal{H}_0
}
$$
our choice of $n$ and our observation about $\delta$
show that $\delta_n = 0$.
This produces the desired maps
$\alpha'_{n, 0} : \mathcal{F}'_n/f\mathcal{F}'_n \to \mathcal{F}_0$,
$\beta'_{n, 0} : \mathcal{G}'_n/f\mathcal{G}'_n \to \mathcal{G}_0$, and
$\alpha'_{n, 0} : \mathcal{F}'_n/f\mathcal{F}'_n \to \mathcal{F}_0$, and
$\gamma'_{n, 0} : \mathcal{H}'_n/f\mathcal{H}'_n \to \mathcal{H}_0$.
OK, so we may use
$\mathcal{F}'_n, \alpha'_n, \alpha'_{n, 0}$,
@@ -8497,7 +8533,7 @@ \section{Coherent triples}
\chi((\mathcal{F}, \mathcal{F}_0, \alpha) \otimes
(\mathcal{L}, \mathcal{L}_0, \lambda)^{\otimes n})
$$
is a numerical polynomial of degree $\leq \dim(\text{Supp}(\mathcal{F}))$.
is a polynomial of degree $\leq \dim(\text{Supp}(\mathcal{F}))$.
\end{proposition}

\noindent
@@ -8508,9 +8544,43 @@ \section{Coherent triples}
the function is linear, etc.

\begin{proof}
There is no harm in replacing $\mathcal{F}_0$ by the quotient
of $\mathcal{F}_0$ by the functions supported on $\{\mathfrak m\}$:
this just changes the function by a constant.
We will prove this by induction on the dimension of the support of
$\mathcal{F}$.

\medskip\noindent
The base case is when $\mathcal{F} = 0$. Then either
$\mathcal{F}_0$ is zero or its support is $\{\mathfrak m\}$.
In this case we have
$$
(\mathcal{F}, \mathcal{F}_0, \alpha) \otimes
(\mathcal{L}, \mathcal{L}_0, \lambda)^{\otimes n} =
(0, \mathcal{F}_0 \otimes \mathcal{L}_0^{\otimes n}, 0) \cong
(0, \mathcal{F}_0, 0)
$$
Thus the function of the lemma is constant with value equal
to the length of $\mathcal{F}_0$.

\medskip\noindent
Induction step. Assume the support of $\mathcal{F}$ is nonempty.
Let $\mathcal{G}_0 \subset \mathcal{F}_0$ denote the submodule
of sections supported on $\{\mathfrak m\}$. Then we get a short
exact sequence
$$
0 \to (0, \mathcal{G}_0, 0) \to
(\mathcal{F}, \mathcal{F}_0, \alpha) \to
(\mathcal{F}, \mathcal{F}_0/\mathcal{G}_0, \alpha) \to 0
$$
This sequence remains exact if we tensor by the invertible
coherent triple $(\mathcal{L}, \mathcal{L}_0, \lambda)$, see
discussion above. Thus by additivity of $\chi$
(Lemma \ref{lemma-ses-chi-triple})
and the base case explained above, it suffices to prove
the induction step for
$(\mathcal{F}, \mathcal{F}_0/\mathcal{G}_0, \alpha)$.
In this way we see that we may assume $\mathfrak m$ is not
an associated point of $\mathcal{F}_0$.

\medskip\noindent
Let $T = \text{Ass}(\mathcal{F}) \cup \text{Ass}(\mathcal{F}/f\mathcal{F})$.
Since $U$ is quasi-affine, we can find $s \in \Gamma(U, \mathcal{L})$
which does not vanish at any $u \in T$, see
@@ -8556,8 +8626,9 @@ \section{Coherent triples}
\chi((\mathcal{F}, \mathcal{F}_0, \alpha) \otimes
(\mathcal{L}, \mathcal{L}_0, \lambda)^{\otimes n})
$$
is a numerical polynomial. We conclude by
Algebra, Lemma \ref{algebra-lemma-numerical-polynomial}.
is a polynomial. We conclude by a variant of
Algebra, Lemma \ref{algebra-lemma-numerical-polynomial}
for functions defined for all integers (details omitted).
\end{proof}

\begin{lemma}
@@ -8583,21 +8654,22 @@ \section{Coherent triples}
Dualizing Complexes, Lemma \ref{dualizing-lemma-depth} and
Local Cohomology, Lemma
\ref{local-cohomology-lemma-finiteness-pushforwards-and-H1-local}.
It follows that $M = \Gamma(U, \mathcal{L})$ is a finite $A$-module
for example by Local Cohomology, Lemma
\ref{local-cohomology-lemma-finiteness-for-finite-locally-free}.
Using Local Cohomology, Lemma
\ref{local-cohomology-lemma-finiteness-for-finite-locally-free}
we find that $M = \Gamma(U, \mathcal{L})$ is a finite $A$-module.
This in turn implies $\text{depth}(M) \geq 2$ for example by
part (4) of Local Cohomology, Lemma
\ref{local-cohomology-lemma-finiteness-pushforwards-and-H1-local}
or by Divisors, Lemma \ref{divisors-lemma-depth-pushforward}.
Also, we have $\mathcal{L}_0 \cong \mathcal{O}_{X_0}$
as $X_0$ is a local scheme. Hence also see that
as $X_0$ is a local scheme. Hence we also see that
$M_0 = \Gamma(X_0, \mathcal{L}_0) = \Gamma(U_0, \mathcal{L}_0|_{U_0})$
and this module is isomorphic to $A/fA$.
Thus $\mathcal{F}' = \widetilde{M}$ is a coherent $\mathcal{O}_X$-module
whose restriction to $U$ is isomorphic to $\mathcal{L}$.
The isomorphism
$\lambda : \mathcal{L}/f\mathcal{L} \to \mathcal{L}_0|_{U_0}$
and that this module is isomorphic to $A/fA$.

\medskip\noindent
By the above $\mathcal{F}' = \widetilde{M}$ is a coherent
$\mathcal{O}_X$-module whose restriction to $U$ is isomorphic to $\mathcal{L}$.
The isomorphism $\lambda : \mathcal{L}/f\mathcal{L} \to \mathcal{L}_0|_{U_0}$
determines a map $M/fM \to M_0$ on global sections
which is an isomorphism over $U_0$.
Since $\text{depth}(M) \geq 2$ we see
@@ -8669,7 +8741,7 @@ \section{Invertible modules on punctured spectra, I}
$$
n \longmapsto \chi((\mathcal{L}, \mathcal{L}_0, \lambda)^{\otimes n})
$$
is a numerical polynomial. By Lemma \ref{lemma-nonnegative-chi-triple}
is a polynomial. By Lemma \ref{lemma-nonnegative-chi-triple}
the value of this polynomial is zero if and only if
$\mathcal{L}^{\otimes n}$ is trivial.
Thus if $\mathcal{L}$ is torsion, then this
@@ -8709,7 +8781,7 @@ \section{Invertible modules on punctured spectra, I}
$$
n \longmapsto \chi((\mathcal{L}, \mathcal{L}_0, \lambda)^{\otimes n})
$$
is a numerical polynomial $P$. By Lemma \ref{lemma-nonnegative-chi-triple}
is a polynomial $P$. By Lemma \ref{lemma-nonnegative-chi-triple}
we have
$P(n) \geq 0$ for all $n \in \mathbf{Z}$ with equality if and only if
$\mathcal{L}^{\otimes n}$ is trivial. In particular $P(0) = 0$

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