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

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