20 ANDREW BAKER
in which d is the shift map with
ker d =
lim1(
n
k
Mn), coker d = lim(
n
k
Mn),
and similarly for d . The vanishing of
limsMn
n
for s = 0, 1 implies that d is an
isomorphism.
Now choose a sequence of elements an Mn with non-zero images in M0.
Define
bnk =
an if k = n,
0 if k = n,
and let b = (bn,k) be the resulting element of
n k
Mn. Defining c = (d
)−1(b),
we see that
(B.1) bnk = cnk f(cn+1 k)
for all n, k. Now fix k and consider the cnk for n k; these satisfy
cnk = f(cn+1 k),
hence they yield an element of the inverse limit of the inverse system
Mk
f
←−− Mk+1
f
←−− Mk+2
f
←− · · · ,
but
lim
n k
Mn = lim
n
Mn = 0.
Therefore cnk = 0 for n k. Using (B.1), we see that for all k n,
cnk = f
k−n(ak)
and in particular, in M0,
c0k = f
k(ak)
= 0.
This shows that c /
n k
Mn even though d (c) = b
n k
Mn. The result
follows by inspection of the diagram.
Now let R = Zp[[u]] with m = (p, u) its maximal ideal containing p and u.
Let M be the m-adic completion of
n
R which can be identified with the set of
sequences x = (xn)
n
R for which xn 0 in the m-adic topology. The group
N =
n
(pn,un) is a subgroup of M.
The category of L-complete modules is closed under products and contains all
finitely generated R-modules, therefore N and M/N are L-complete.
Lemma B.2. For each n 1, the natural map
Tor2
R
(R/mn,M/N)
−→ Tor2
R
(R/m,M/N)
is non-zero.
Proof. Since R/(pn,un) is a retract of N, it suffices to prove the result for
R/(pn,un)
in place of N. The sequence
pn,un
is regular in R, so for any R-
module K we can compute Tor∗
R
(R/(pn,un),K)
using a Koszul resolution. In
particular, if
pnK
= 0 =
unK
then we have
Tor∗
R(R/(pn,un),K)
= K
and the reduction map is the obvious epimorphism
R/(pn,un)
−→ R/(p, u). The
result follows easily from this.
20
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