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

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