completion with respect to a f.g. ideal
Thanks to Bjorn Poonen who writes
"Hi, here is an alternative proof of
http://stacks.math.columbia.edu/tag/05GG :
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Since $I$ is finitely generated,
$I^n$ is finitely generated, say by $i_1,\ldots,i_r$.
The map $M^r \stackrel{(i_1,\ldots,i_r)}\longrightarrow M$
factors as $M^r \to I^n M \to M$, a surjection followed by an injection.
Taking completions shows that
$(M^\wedge)^r \stackrel{(i_1,\ldots,i_r)}\longrightarrow M^\wedge$
factors as $(M^\wedge)^r \to (I^n M)^\wedge \to M^\wedge$,
again a surjection followed by an injection
by http://stacks.math.columbia.edu/tag/0315 part (2)
and http://stacks.math.columbia.edu/tag/0BNG , so its image
$I^n M^\wedge$ equals $(I^n M)^\wedge = \ker(M^\wedge \to M/I^n M)$.
Thus $M^\wedge / I^n M^\wedge \simeq M/I^n M$.
Taking inverse limits yields $(M^\wedge)^\wedge \simeq M^\wedge$;
that is, $M^\wedge$ is $I$-adically complete.
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If you use this, you can also delete
http://stacks.math.columbia.edu/tag/0318
which is no longer needed.
In fact, if all you want is $M^\wedge / I^n M^\wedge \simeq M/I^n M$
(and hence that $M^\wedge$ is $I$-adically complete),
i.e., if you don't care about computing $(I^n M)^\wedge$,
then you could also prove the needed case of
http://stacks.math.columbia.edu/tag/0BNG
by hand, and hence delete that lemma too, to obtain the following
proof of
http://stacks.math.columbia.edu/tag/05GG :
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Since $I$ is finitely generated,
$I^n$ is finitely generated, say by $i_1,\ldots,i_r$.
Applying http://stacks.math.columbia.edu/tag/0315 part (2)
to the surjection $M^r \stackrel{(i_1,\ldots,i_r)}\longrightarrow I^nM$
yields a surjection
\[
(M^\wedge)^r \stackrel{(i_1,\ldots,i_r)}\longrightarrow (I^n M)^\wedge
= \varprojlim_m I^n M/I^m M = \ker(M^\wedge \to M/I^n M).
\]
On the other hand, the image of
$(M^\wedge)^r \stackrel{(i_1,\ldots,i_r)}\longrightarrow M^\wedge$
is $I^n M^\wedge$. Thus $M^\wedge / I^n M^\wedge \simeq M/I^n M$.
Taking inverse limits yields $(M^\wedge)^\wedge \simeq M^\wedge$;
that is, $M^\wedge$ is $I$-adically complete.
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Best,
Bjorn"
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