1Artinian algebras
III Algebras
1.2 Artin–Wedderburn theorem
We are going to state the Artin–Wedderburn theorem for right (as opposed to
left) things, because this makes the notation easier for us.
Theorem
(Artin–Wedderburn theorem)
.
Let
A
be a semisimple right Artinian
algebra. Then
A =
r
M
i=1
M
n
i
(D
i
),
for some division algebra D
i
, and these factors are uniquely determined.
A has exactly r isomorphism classes of simple (right) modules S
i
, and
End
A
(S
i
) = {A-module homomorphisms S
i
→ S
i
}
∼
=
D
i
,
and
dim
D
i
(S
i
) = n
i
.
If A is simple, then r = 1.
If we had the left version instead, then we need to insert op’s somewhere.
Artin–Wedderburn is an easy consequence of two trivial lemma. The key idea
that leads to Artin–Wedderburn is the observation that the map
A
A
→ End
A
(
A
A
)
sending
a
to left-multiplication by
a
is an isomorphism of algebras. So we need
to first understand endomorphism algebras, starting with Schur’s lemma.
Lemma
(Schur’s lemma)
.
Let
M
1
, M
2
be simple right
A
-modules. Then either
M
1
∼
=
M
2
, or
Hom
A
(
M
1
, M
2
) = 0. If
M
is a simple
A
-module, then
End
A
(
M
)
is a division algebra.
Proof.
A non-zero
A
-module homomorphism
M
1
→ M
2
must be injective, as
the kernel is submodule. Similarly, the image has to be the whole thing since
the image is a submodule. So this must be an isomorphism, and in particular
has an inverse. So the last part follows as well.
As mentioned, we are going to exploit the isomorphism
A
A
∼
=
End
A
(
A
A
).
This is easy to see directly, but we can prove a slightly more general result, for
the sake of it:
Lemma.
(i)
If
M
is a right
A
-module and
e
is an idempotent in
A
, i.e.
e
2
=
e
, then
Me
∼
=
Hom
A
(eA, M).
(ii) We have
eAe
∼
=
End
A
(eA).
In particular, we can take e = 1, and recover End
A
(A
A
)
∼
=
A.
Proof.
(i) We define maps
me (ex 7→ mex)
Me Hom(eA, M)
α(e) α
f
1
f
2
We note that
α
(
e
) =
α
(
e
2
) =
α
(
e
)
e ∈ M e
. So this is well-defined. By
inspection, these maps are inverse to each other. So we are done.
Note that we might worry that we have to pick representatives
me
and
ex
for the map
f
1
, but in fact we can also write it as
f
(
a
)(
y
) =
ay
, since
e
is
idempotent. So we are safe.
(ii) Immediate from above by putting M = eA.
Lemma. Let M be a completely reducible right A-module. We write
M =
M
S
n
i
i
,
where
{S
i
}
are distinct simple
A
-modules. Write
D
i
=
End
A
(
S
i
), which we
already know is a division algebra. Then
End
A
(S
n
i
i
)
∼
=
M
n
i
(D
i
),
and
End
A
(M) =
M
M
n
i
(D
i
)
Proof.
The result for
End
A
(
S
n
i
i
) is just the familiar fact that a homomorphism
S
n
→ S
m
is given by an
m × n
matrix of maps
S → S
(in the case of vector
spaces over a field
k
, we have
End
(
k
)
∼
=
k
, so they are matrices with entries in
k). Then by Schur’s lemma, we have
End
A
(M) =
M
i
End
A
(M
i
)
∼
=
M
n
i
(D
i
).
We now prove Artin–Wedderburn.
Proof of Artin–Wedderburn.
If
A
is semi-simple, then it is completely reducible
as a right A-module. So we have
A
∼
=
End(A
A
)
∼
=
M
M
n
i
(D
i
).
We now decompose each
M
n
i
(
D
i
) into a sum of simple modules. We know each
M
n
i
(
D
i
) is a non-trivial
M
n
i
(
D
i
) module in the usual way, and the action of
the other summands is trivial. We can simply decompose each
M
n
i
(
D
i
) as the
sum of submodules of the form
0 0 · · · 0 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 · · · 0 0
a
1
a
2
· · · a
n
i
−1
a
n
i
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 · · · 0 0
and there are
n
i
components. We immediately see that if we write
S
i
for this
submodule, then we have
dim
D
i
(S
i
) = n
i
.
Finally, we have to show that every simple module S of A is one of the S
i
. We
simply have to note that if
S
is a simple
A
-module, then there is a non-trivial
map
f
:
A → S
(say by picking
x ∈ S
and defining
f
(
a
) =
xa
). Then in the
decomposition of
A
into a direct sum of simple submodules, there must be one
factor
S
i
such that
f|
S
i
is non-trivial. Then by Schur’s lemma, this is in fact an
isomorphism S
i
∼
=
S.
This was for semi-simple algebras. For a general right Artinian algebra, we
know that
A/J
(
A
) is semi-simple and inherits the Artinian property. Then
Artin–Wedderburn applies to A/J(A).
Some of us might be scared of division algebras. Sometimes, we can get away
with not talking about them. If
A
is not just Artinian, but finite-dimensional,
then so are the D
i
.
Now pick an arbitrary
x ∈ D
i
. Then the sub-algebra of
D
i
generated by
x
is
be commutative. So it is in fact a subfield, and finite dimensionality means it is
algebraic over
k
. Now if we assume that
k
is algebraically closed, then
x
must
live in k. So we’ve shown that these D
i
must be k itself. Thus we get
Corollary.
If
k
is algebraically closed and
A
is a finite-dimensional semi-simple
k-algebra, then
A
∼
=
M
M
n
i
(k).
This is true, for example, when k = C.
We shall end this section by applying our results to group algebras. Recall
the following definition:
Definition
(Group algebra)
.
Let
G
be a group and
k
a field. The group algebra
of G over k is
kG =
n
X
λ
g
g : g ∈ G, λ
g
∈ k
o
.
This has a bilinear multiplication given by the obvious formula
(λ
g
g)(µ
h
h) = λ
g
µ
h
(gh).
The first thing to note is that group algebras are almost always semi-simple.
Theorem
(Maschke’s theorem)
.
Let
G
be a finite group and
p - |G|
, where
p = char k, so that |G| is invertible in k, then kG is semi-simple.
Proof.
We show that any submodule
V
of a
kG
-module
U
has a complement.
Let π : U → V be any k-vector space projection, and define a new map
π
0
=
1
|G|
X
g∈G
gπg
−1
: U → V.
It is easy to see that this is a
kG
-module homomorphism
U → V
, and is a
projection. So we have
U = V ⊕ ker π
0
,
and this gives a kG-module complement.
There is a converse to Maschke’s theorem:
Theorem. Let G be finite and kG semi-simple. Then char k - |G|.
Proof.
We note that there is a simple
kG
-module
S
, given by the trivial module.
This is a one-dimensional k vector space. We have
D = End
kG
(S) = k.
Now suppose
kG
is semi-simple. Then by Artin–Wedderburn, there must be
only one summand of S in kG.
Consider the following two ideals of kG: we let
I
1
=
n
X
λ
g
g ∈ kG :
X
λ
g
= 0
o
.
This is in fact a two-sided ideal of
kG
. We also have the center of the algebra,
given by
I
2
=
n
λ
X
g ∈ kG : λ ∈ k
o
.
Now if char k | |G|, then I
2
⊆ I
1
. So we can write
kG =
kG
I
1
⊕ I
1
=
kG
I
1
⊕ I
2
⊕ · · · .
But we know
G
acts trivially on
kG
I
1
and
I
2
, and they both have dimension 1.
This gives a contradiction. So we must have char k - |G|.
We can do a bit more of representation theory. Recall that when
k
is
algebraically closed and has characteristic zero, then the number of simple
kG
-
modules is the number of conjugacy classes of
G
. There is a more general result
for a general characteristic p field:
Theorem.
Let
k
be algebraically closed of characteristic
p
, and
G
be finite.
Then the number of simple
kG
modules (up to isomorphism) is equal to the
number of conjugacy classes of elements of order not divisible by
p
. These are
known as the p-regular elements.
We immediately deduce that
Corollary.
If
|G|
=
p
r
for some
r
and
p
is prime, then the trivial module is the
only simple kG module, when char k = p.
Note that we can prove this directly rather than using the theorem, by
showing that
I
=
ker
(
kG → k
) is a nilpotent ideal, and annihilates all simple
modules.
Proof sketch of theorem.
The number of simple
kG
modules is just the number
of simple
kG/J
(
kG
) module, as
J
(
kG
) acts trivially on every simple module.
There is a useful trick to figure out the number of simple
A
-modules for a given
semi-simple A. Suppose we have a decomposition
A
∼
=
r
M
i=1
M
n
i
(k).
Then we know
r
is the number of simple
A
-modules. We now consider [
A, A
],
the k-subspace generated by elements of the form xy − yx. Then we see that
A
[A, A]
∼
=
r
M
i=1
M
n
i
(k)
[M
n
i
(k), M
n
i
(k)]
.
Now by linear algebra, we know [
M
n
i
(
k
)
, M
n
i
(
k
)] is the trace zero matrices, and
so we know
dim
k
M
n
i
(k)
[M
n
i
(k), M
n
i
(k)]
= 1.
Hence we know
dim
A
[A, A]
= r.
Thus we need to compute
dim
k
kG/J(kG)
[kG/J(kG), kG/J(kG)]
We then note the following facts:
(i) For a general algebra A, we have
A/J(A)
[A/J(A), A/J(A)]
∼
=
A
[A, A] + J(A)
.
(ii) Let g
1
, · · · , g
m
be conjugacy class representatives of G. Then
{g
i
+ [kG, kG]}
forms a k-vector space basis of kG/[kG, kG].
(iii)
If
g
1
, · · · , g
r
is a set of representatives of
p
-regular conjugacy classes, then
n
g
i
+
[kG, kG] + J(kG)
o
form a basis of kG/([kG, kG] + J(kG)).
Hence the result follows.
One may find it useful to note that [
kG, kG
] +
J
(
kG
) consists of the elements
in kG such that x
p
s
∈ [kG, kG] for some s.
In this proof, we look at
A/
[
A, A
]. However, in the usual proof of the result
in the characteristic zero looks at the center
Z
(
A
). The relation between these
two objects is that the first is the 0th Hochschild homology group of
A
, while
the second is the 0th Hochschild cohomology group of A.