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Chapter 9
Introducing Noncommutative
Algebra
We will discuss noncommutative rings and their modules, concentrating on two fundamen
tal results, the Wedderburn structure theorem and Maschke’s theorem. Further insight
into the structure of rings will be provided by the Jacobson radical.
9.1 Semisimple Modules
A vector space is the direct sum of onedimensional subspaces (each subspace consists of
scalar multiples of a basis vector). A onedimensional space is simple in the sense that
it does not have a nontrivial proper subspace. Thus any vector space is a direct sum of
simple subspaces. We examine those modules which behave in a similar manner.
9.1.1 Defnition
An
R
module
M
is
simple
if
M
6
= 0 and the only submodules of
M
are 0 and
M
.
9.1.2 Theorem
Let
M
be a nonzero
R
module. The following conditions are equivalent, and a module
satisfying them is said to be
semisimple
or
completely reducible
.
(a)
M
is a sum of simple modules;
(b)
M
is a direct sum of simple modules;
(c) If
N
is a submodule of
M
, then
N
is a direct summand of
M
, that is, there is a
submodule
N
0
of
M
such that
M
=
N
⊕
N
0
.
Proof.
(a) implies (b). Let
M
be the sum of simple modules
M
i
,
i
∈
I
.I
f
J
⊆
I
, denote
∑
j
∈
J
M
j
by
M
(
J
). By Zorn’s lemma, there is a maximal subset
J
of
I
such that the
sum de±ning
N
=
M
(
J
) is direct. We will show that
M
=
N
. First assume that
i/
∈
J
.
1
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CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA
Then
N
∩
M
i
is a submodule of the simple module
M
i
, so it must be either 0 or
M
i
.I
f
N
∩
M
i
= 0, then
M
(
J
∪{
i
}
) is direct, contradicting maximality of
J
.Thu
s
N
∩
M
i
=
M
i
,
so
M
i
⊆
N
. But if
i
∈
J
, then
M
i
⊆
N
by deFnition of
N
. Therefore
M
i
⊆
N
for all
i
,
and since
M
is the sum of all the
M
i
,wehave
M
=
N
.
(b) implies (c). This is essentially the same as (a) implies (b). Let
N
be a submodule
of
M
, where
M
is the direct sum of simple modules
M
i
,
i
∈
I
. Let
J
be a maximal
subset of
I
such that the sum
N
+
M
(
J
) is direct. If
i/
∈
J
then exactly as before,
M
i
∩
(
N
⊕
M
(
J
)) =
M
i
,so
M
i
⊆
N
⊕
M
(
J
). This holds for
i
∈
J
as well, by deFnition of
M
(
J
). It follows that
M
=
N
⊕
M
(
J
). [Notice that the complementary submodule
N
0
can be taken as a direct sum of some of the original
M
i
.]
(c) implies (a). ±irst we make several observations.
(1) If
M
satisFes (c), so does every submodule
N
. [Let
N
≤
M
, so that
M
=
N
⊕
N
0
.
If
V
is a submodule of
N
, hence of
M
, we have
M
=
V
⊕
W
f
x
∈
N
, then
x
=
v
+
w
,
v
∈
V
,
w
∈
W
w
=
x
−
v
∈
N
(using
V
≤
N
). But
v
also belongs to
N
, and
consequently
N
=(
N
∩
V
)
⊕
(
N
∩
W
)=
V
⊕
(
N
∩
W
).]
(2) If
D
=
A
⊕
B
⊕
C
, then
A
A
+
B
)
∩
(
A
+
C
). [If
a
+
b
=
a
0
+
c
, where
a,a
0
∈
A
,
b
∈
B
,
c
∈
C
, then
a
0
−
a
=
b
−
c
, and since
D
is a direct sum, we have
b
=
c
= 0 and
a
=
a
0
s
a
+
b
∈
A
.]
(3) If
N
is a nonzero submodule of
M
, then
N
contains a simple submodule.
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 Spring '11
 sdd
 Algebra

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