in which d is the shift map with
ker d =
Mn), coker d = lim(
and similarly for d . The vanishing of
for s = 0, 1 implies that d is an
Now choose a sequence of elements an ∈ Mn with non-zero images in M0.
an if k = n,
0 if k = n,
and let b = (bn,k) be the resulting element of
Mn. Deﬁning c = (d
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
←− − · · · ,
Mn = lim
Mn = 0.
Therefore cnk = 0 for n k. Using (B.1), we see that for all k n,
cnk = f
and in particular, in M0,
c0k = f
This shows that c / ∈
Mn even though d (c) = b ∈
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
R which can be identiﬁed with the set of
sequences x = (xn) ∈
R for which xn → 0 in the m-adic topology. The group
(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
Proof. Since R/(pn,un) is a retract of N, it suﬃces to prove the result for
in place of N. The sequence
is regular in R, so for any R-
module K we can compute Tor∗
using a Koszul resolution. In
= 0 =
then we have
and the reduction map is the obvious epimorphism
−→ R/(p, u). The
result follows easily from this.