L-COMPLETE HOPF ALGEBROIDS 19
for f
G∗,
x and γ G.
A proof of the next result is sketched in [1].
Proposition A.1. The pair ( , G∗) is a Hopf algebroid.
Lemma A.2. The twisted group ring G is a simple k-algebra and every finite
dimensional G-module V is completely reducible. In particular, if V = 0 then
V
G
= 0 and there is an linear isomorphism
⊗k V
G
−→ V ; x v xv.
Proof. Since Endk A is an irreducible k-algebra, it has a unique simple module
which agrees with A as a k-module. Hence every finite dimensional module is
isomorphic to a direct sum of copies of A. Since
AG
= k, we see that V
G
= 0.
Verifying the bijectivity of the linear map is straightforward.
We can generalise this situation and still get similar results. For example, if G
is a finite group with a given epimorphism π : G −→ G, then G is semi-simple
provided that p | ker π|, see [10]. In fact the unicursal Hopf algebroid F ⊗Fp F(θ0)
of (5.2) is dual to
F (Gal(F/Fp) Fpn
×
),
where the action of the group on F is through the projection onto Gal(F/Fp).
Appendix B. Non-exactness of tensoring with a pro-free module
In [6, section 1], it was shown that in M , coproducts need not preserve left
exactness. At the suggestion of the referee, we include a more precise example
showing that tensoring with a pro-free module in M need not be a left exact functor.
This material is due to the referee to whom we are grateful for the opportunity
to include it. The main result is Theorem B.5, but we need several preparatory
technical results.
Lemma B.1. Let
M0
f
←−− M1
f
←−− M2
f
←− · · ·
be a inverse system of abelian groups for which lim
n
Mn = 0 =
lim1Mn
n
and where
each homomorphism Mn −→ M0 is non-zero. Then the induced inverse system
k∈N
M0
f
←−−
k∈N
M1
f
←−−
k∈N
M2
f
←− · · ·
satisfies
lim1
n
k∈N
Mn = 0.
Proof. For ease of notation we write
k
for
k∈N
and
n
for
n∈N0
, where
N = {0} N.
Consider the commutative square
n k
Mn
d
n k
Mn
n k
Mn
d

=
n k
Mn
19
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