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But this construction forces
rx — sy = r(x — sy) = rs(x — y)
and
rx — sy = s(rx — y) = sr(x — y)
which cannot hold in general if R is noncommutative. A solution is to modify the con-
struction so that the tensor product T is only an abelian group. Later we can investigate
conditions under which T has a module structure as well.

8.8.1 De¬nitions and Comments
Let M be a right R-module and N a left R-module. (We often abbreviate this as MR and
R N .) Let f : M — N ’ P , where P is an abelian group. The map f is biadditive if it is
additive in each variable, that is, f (x+x , y) = f (x, y)+f (x , y) and f (x, y+y ) = f (x, y)+
f (x, y ) for all x, x ∈ M , y, y ∈ N . The map f is R-balanced if f (xr, y) = f (x, ry) for all
x ∈ M , y ∈ N , r ∈ R. As before, the key idea is the universal mapping property: Every
biadditive, R-balanced map can be factored through the tensor product.

8.8.2 Construction of the General Tensor Product
If MR and R N , let F be the free abelian group with basis M — N . Let G be the subgroup
of R generated by the relations
(x + x , y) ’ (x, y) ’ (x , y);
(x, y + y ) ’ (x, y) ’ (x, y );
(xr, y) ’ (x, ry)
where x, x ∈ M , y, y ∈ N , r ∈ R. De¬ne the tensor product of M and N over R as
T = M —R N = F/G
and denote the element (x, y) + G of T by x — y. Thus the general element of T is a ¬nite
sum of the form
xi — yi .
t= (1)
i

The relations force the map h : (x, y) ’ x — y of M — N into T to be biadditive and
R-balanced, so that
x — (y + y ) = x — y + x — y , (x + x ) — y = x — y + x — y, (2)
xr — y = x — ry. (3)
If f is a biadditive, R-balanced mapping from M — N to the abelian group P , then f
extends uniquely to an abelian group homomorphism from F to P , also called f . Since f
is biadditive and R-balanced, the kernel of f contains G, so by the factor theorem there
is a unique abelian group homomorphism g : T ’ P such that g(x — y) = f (x, y) for all
x ∈ M , y ∈ N . Consequently, gh = f and we have the universal mapping property:
8.8. GENERAL TENSOR PRODUCTS 27

Every biadditive, R-balanced mapping on M — N can be factored through T .
As before, any two abelian groups that satisfy the universal mapping property are
isomorphic.

8.8.3 Bimodules
Let R and S be arbitrary rings. We say that M is an S ’ R bimodule if M is both a left
S-module and a right R-module, and in addition a compatibility condition is satis¬ed:
(sx)r = s(xr) for all s ∈ S, r ∈ R. We often abbreviate this as S MR .
If f : R ’ S is a ring homomorphism, then S is a left S-module, and also a right
R-module by restriction of scalars, as in (8.7.1). The compatibility condition is satis¬ed:
(sx)r = sxf (r) = s(xr). Therefore S is an S ’ R bimodule.

8.8.4 Proposition
If S MR and R NT , then M —R N is an S ’ T bimodule.
Proof. Fix s ∈ S. The map (x, y) ’ sx — y of M — N into M —R N is biadditive and
R-balanced. The latter holds because by the compatibility condition in the bimodule
property of M , along with (3) of (8.8.2),

s(xr) — y = (sx)r — y = sx — ry.

Thus there is an abelian group endomorphism on M —R N such that x — y ’ sx — y,
and we use this to de¬ne scalar multiplication on the left by s. A symmetrical argument
yields scalar multiplication on the right by t. To check the compatibility condition,

[s(x — y)]t = (sx — y)t = sx — yt = s(x — yt) = s[(x — y)t]. ™



8.8.5 Corollary
If S MR and R N , then M —R N is a left S-module. If MR and then M —R N is a
R NT ,
right T -module.
Proof. The point is that every module is, in particular, an abelian group, hence a Z-
module. Thus for the ¬rst statement, take T = Z in (8.8.4), and for the second statement,
take S = Z. ™

8.8.6 Extensions
As in Section 8.7, we can de¬ne the tensor product of any ¬nite number of modules using
multiadditive maps (additive in each variable) that are balanced. For example, suppose
that MR , R NS and S P . If f : M —N —P ’ G, where G is an abelian group, the condition
of balance is

f (xr, y, z) = f (x, ry, z) and f (x, ys, z) = f (x, y, sz)
28 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

for all x ∈ M , y ∈ N , z ∈ P , r ∈ R, s ∈ S. An argument similar to the proof of (8.7.4)
shows that
(a) M —R N —S P ∼ (M —R N ) —S P ∼ M —R (N —S P ).
= =
If M is a right R-module, and N and P are left R-modules, then
(b) M —R (N • P ) ∼ (M —R N ) • (M —R P ).
=
This is proved as in (8.7.5), in fact the result can be extended to the direct sum of an
arbitrary (not necessarily ¬nite) number of left R-modules.
If M is a left R-module, then exactly as in (8.7.6) and (8.7.7), we have
(c) R —R M ∼ M and
=
(d) R — M ∼ M m .
m
=
Let M1 and M2 be right R-modules, and let N1 and N2 be left R-modules. If
f1 : M1 ’ N1 and f2 : M2 ’ N2 are R-module homomorphisms, the tensor product
f1 — f2 can be de¬ned exactly as in (8.7.9). As before, the key property is

(f1 — f2 )(x1 — x2 ) = f1 (x1 ) — f2 (x2 )

for all x1 ∈ M1 , x2 ∈ M2 .

8.8.7 Tensor Product of Algebras
If A and B are algebras over the commutative ring R, then the tensor product A —R B
becomes an R-algebra if we de¬ne multiplication appropriately. Consider the map of
A — B — A — B into A —R B given by

(a, b, a , b ) ’ aa — bb , a, a ∈ A, b, b ∈ B.

The map is 4-linear, so it factors through the tensor product to give an R-module homo-
morphism g : A — B — A — B ’ A — B such that

g(a — b — a — b ) = aa — bb .

Now let h : (A — B) — (A — B) ’ A — B — A — B be the bilinear map given by

h(u, v) = u — v.

If we apply h followed by g, the result is a bilinear map f : (A — B) — (A — B) ’ A — B
with

f (a — b, a — b ) = aa — bb ,

and this de¬nes our multiplication (a — b)(a — b ) on A — B. The multiplicative identity is
1A — 1B , and the distributive laws can be checked routinely. Thus A —R B is a ring that
is also an R-module. To check the compatibility condition, note that if r ∈ R, a, a ∈ A,
b, b ∈ B, then

r[(a — b)(a — b )] = [r(a — b)](a — b ) = (a — b)[r(a — b )];

all three of these expressions coincide with raa — bb = aa — rbb .
8.8. GENERAL TENSOR PRODUCTS 29

Problems For Section 8.8
We will use the tensor product to de¬ne the exterior algebra of an R-module M , where R
is a commutative ring. If p is a positive integer, we form the tensor product M —R · · ·—R M
of M with itself p times, denoted by M —p . Let N be the submodule of M —p generated
by those elements x1 — · · · — xp , with xi ∈ M for all i, such that xi = xj for some i = j.
The pth exterior power of M is de¬ned as

Λp M = M —p /N.

In most applications, M is a free R-module with a ¬nite basis x1 , . . . , xn (with 1 ¤ p ¤ n),
and we will only consider this case. To simplify the notation, we write the element
a — b — · · · — c + N of Λp M as ab · · · c. (The usual notation is a § b § · · · § c.)
1. Let y1 , . . . , yp ∈ M . Show that if yi and yj are interchanged in the product y1 · · · yp ,
then the product is multiplied by ’1.
2. Show that the products xi1 · · · xip , where i1 < · · · < ip , span Λp M .
3. Let f : M p ’ Q be a multilinear map from M p to the R-module Q, and assume that f
is alternating, that is, f (m1 , . . . , mp ) = 0 if mi = mj for some i = j. Show that f
can be factored through Λp M , in other words, there is a unique R-homomorphism
g : Λp M ’ Q such that g(y1 · · · yp ) = f (y1 , . . . , yp ).
n
Let yi = j=1 aij xj , i = 1, . . . , n. Since {x1 , . . . , xn } is a basis for M , yi can be iden-
ti¬ed with row i of A. By the basic properties of determinants, the map f (y1 , . . . , yn ) =
det A is multilinear and alternating, and f (x1 , . . . , xn ) = 1, the determinant of the identity
matrix.
4. Show that x1 · · · xn = 0 in Λn M , and that {x1 · · · xn } is a basis for Λn M .
Let I = {i1 , · · · , ip }, where i1 < · · · < ip , and write the product xi1 · · · xip as xI . Let
J be the complementary set of indices. (For example, if n = 5, p = 3, and I = {1, 2, 4},
then J = {3, 5}.) Any equation involving xI ∈ Λp M can be multiplied by xJ to produce
a valid equation in Λn M .
5. Show that the products xI of Problem 2 are linearly independent, so that Λp M is a
free R-module of rank ( n ).
p

Roughly speaking, the exterior algebra of M consists of the Λp M for all p. By con-
struction, Λ1 M = M and Λp M = 0 for p > n, since some index must repeat in any
element of Λp M . By convention, we take Λ0 M = R. Formally, the exterior powers are
assembled into a graded R-algebra

A0 • A1 • A2 • · · ·

where Ap = Λp M . Multiplication is de¬ned as in the discussion after Problem 4, that
is, if y1 · · · yp ∈ Ap and z1 · · · zq ∈ Aq , then the exterior product y1 · · · yp z1 · · · zq belongs
to Ap+q .
A ring R is said to be graded if, as an abelian group, it is the direct sum of sub-
groups Rn , n = 0, 1, 2, . . . , with Rm Rn ⊆ Rn+m for all m, n ≥ 0. [Example: R =
30 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

k[X1 , . . . , Xn ], Rn = all homogeneous polynomials of degree n.] By de¬nition, R0 is
a subring of R (because R0 R0 ⊆ R0 ), and each Rn is a module over R0 (because
R0 Rn ⊆ Rn ).
6. Suppose that the ideal I = •n≥1 Rn is generated over R by ¬nitely many elements
x1 , . . . , xr , with xi ∈ Rni . Show that Rn ⊆ S = R0 [x1 , . . . , xr ] for all n = 0, 1, . . . , so
that R = S.
7. Show that R is a Noetherian ring if and only if R0 is Noetherian and R is a ¬nitely
generated R0 -algebra.
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 one-dimensional subspaces (each subspace consists of
scalar multiples of a basis vector). A one-dimensional 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 De¬nition
An R-module M is simple if M = 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 of M such that M = N • N .

Proof. (a) implies (b). Let M be the sum of simple modules Mi , i ∈ I. If J ⊆ I, denote
j∈J Mj 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
2 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

Then N © Mi is a submodule of the simple module Mi , so it must be either 0 or Mi . If
N ©Mi = 0, then M (J ∪{i}) is direct, contradicting maximality of J. Thus N ©Mi = Mi ,
so Mi ⊆ N . But if i ∈ J, then Mi ⊆ N by de¬nition of N . Therefore Mi ⊆ N for all i,
and since M is the sum of all the Mi , we have 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 Mi , 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,
/
Mi © (N • M (J)) = Mi , so Mi ⊆ N • M (J). This holds for i ∈ J as well, by de¬nition of
M (J). It follows that M = N • M (J). [Notice that the complementary submodule N
can be taken as a direct sum of some of the original Mi .]
(c) implies (a). First we make several observations.
(1) If M satis¬es (c), so does every submodule N . [Let N ¤ M , so that M = N • N .
If V is a submodule of N , hence of M , we have M = V • W . If x ∈ N , then x = v + w,
v ∈ V , w ∈ W , so 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 + c, where a, a ∈ A,
b ∈ B, c ∈ C, then a ’ a = b ’ c, and since D is a direct sum, we have b = c = 0 and
a = a . Thus a + b ∈ A.]
(3) If N is a nonzero submodule of M , then N contains a simple submodule.
[Choose a nonzero x ∈ N . By Zorn™s lemma, there is a maximal submodule V of N
such that x ∈ V . By (1) we can write N = V • V , and V = 0 by choice of x and V .
/
If V is simple, we are ¬nished, so assume the contrary. Then V contains a nontrivial
proper submodule V1 , so by (1) we have V = V1 • V2 with the Vj nonzero. By (2),
V = (V + V1 ) © (V + V2 ). Since x ∈ V , either x ∈ V + V1 or x ∈ V + V2 , which contradicts
/ / /
the maximality of V .]
To prove that (c) implies (a), let N be the sum of all simple submodules of M . By (c)
we can write M = N • N . If N = 0, then by (3), N contains a simple submodule V .
But then V ¤ N by de¬nition of N . Thus V ¤ N © N = 0, a contradiction. Therefore
N = 0 and M = N . ™


9.1.3 Proposition
Nonzero submodules and quotient modules of a semisimple module are semisimple.


Proof. The submodule case follows from (1) of the proof of (9.1.2). Let N ¤ M , where
M = i Mi with the Mi simple. Applying the canonical map from M to M/N , we have

M/N = (Mi + N )/N.
i


This key idea has come up before; see the proofs of (1.4.4) and (4.2.3). By the second
isomorphism theorem, (Mi + N )/N is isomorphic to a quotient of the simple module Mi .
But a quotient of Mi is isomorphic to Mi or to zero, and it follows that M/N is a sum of
simple modules. By (a) of (9.1.2), M/N is semisimple. ™
9.2. TWO KEY THEOREMS 3

Problems For Section 9.1
1. Regard a ring R as an R-module. Show that R is simple if and only if R is a division
ring.
2. Let M be an R-module, with x a nonzero element of M . De¬ne the R-module homo-
morphism f : R ’ Rx by f (r) = rx. Show that the kernel I of f is a proper ideal
of R, and R/I is isomorphic to Rx.
3. If M is a nonzero R-module, show that M is simple if and only if M ∼ R/I for some
=
maximal left ideal I.
4. If M is a nonzero R-module, show that M is simple if and only if M is cyclic (that
is, M can be generated by a single element) and every nonzero element of M is a
generator.
5. What do simple Z-modules look like?
6. If F is a ¬eld, what do simple F [X]-modules look like?
7. Let V be an n-dimensional vector space over a ¬eld k. (Take n ≥ 1 so that V = 0.)
If f is an endomorphism (that is, a linear transformation) of V and x ∈ V , de¬ne
f x = f (x). This makes V into a module over the endomorphism ring Endk (V ). Show
that the module is simple.
8. Show that a nonzero module M is semisimple if and only if every short exact sequence
0 ’ N ’ M ’ P ’ 0 splits.


9.2 Two Key Theorems
If M is a simple R-module, there are strong restrictions on a homomorphism either into
or out of M . A homomorphism from one simple R-module to another is very severely
restricted, as Schur™s lemma reveals. This very useful result will be important in the proof
of Wedderburn™s structure theorem. Another result that will be needed is a theorem of
Jacobson that gives some conditions under which a module homomorphism f amounts to
multiplication by a ¬xed element of a ring, at least on part of the domain of f .


9.2.1 Schur™s Lemma
(a) If f ∈ HomR (M, N ) where M and N are simple R-modules, then f is either identically
0 or an isomorphism.
(b) If M is a simple R-module, then EndR (M ) is a division ring.

Proof. (a) The kernel of f is either 0 or M , and the image of f is either 0 or N . If f is
not the zero map, then the kernel is 0 and the image is N , so f is an isomorphism.
(b) Let f ∈ EndR (M ), f not identically 0. By (a), f is an isomorphism, and therefore
is invertible in the endomorphism ring of M . ™

The next result prepares for Jacobson™s theorem.
4 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

9.2.2 Lemma
Let M be a semisimple R-module, and let A be the endomorphism ring EndR (M ). [Note
that M is an A-module; if g ∈ A we take g • x = g(x), x ∈ M .] If m ∈ M and
f ∈ EndA (M ), then there exists r ∈ R such that f (m) = rm.
Before proving the lemma, let™s look more carefully at EndA (M ). Suppose that f ∈
EndA (M ) and x ∈ M . If g ∈ A then f (g(x)) = g(f (x)). Thus EndA (M ) consists of
those abelian group endomorphisms of M that commute with everything in EndR (M ). In
turn, by the requirement that f (rx) = rf (x), EndR (M ) consists of those abelian group
endomorphisms of M that commute with R, more precisely with multiplication by r, for
each r ∈ R. For this reason, EndA (M ) is sometimes called the double centralizer of R.
We also observe that the map taking r ∈ R to multiplication by r is a ring homomor-
phism of R into EndA (M ). [Again use rf (x) = f (rx).] Jacobson™s theorem will imply
that given any f in EndA (M ) and any ¬nite set S ⊆ M , some g in the image of this
ring homomorphism will agree with f on S. Thus in (9.2.2), we can replace the single
element m by an arbitrary ¬nite subset of M .

Proof. By (9.1.2) part (c), we can express M as a direct sum Rm • N . Now if we have a
direct sum U = V • W and u = v + w, v ∈ V , w ∈ W , there is a natural projection of U
on V , namely u ’ v. In the present case, let π be the natural projection of M on Rm.
Then π ∈ A and f (m) = f (πm) = πf (m) ∈ Rm. The result follows. ™

Before proving Jacobson™s theorem, we review some ideas that were introduced in the
exercises in Section 4.4.

9.2.3 Comments
To specify an R-module homomorphism ψ from a direct sum V — = •n Vj to a direct sum
j=1

W = •i=1 Wi , we must give, for every i and j, the i component of the image of vj ∈ Vj .
m th

Thus the homomorphism is described by a matrix [ψij ], where ψij is a homomorphism from
Vj to Wi . The ith component of ψ(vj ) is ψij (vj ), so the ith component of ψ(v1 + · · · + vn )
n
is j=1 ψij (vj ). Consequently,
® 
v1
.
ψ(v1 + · · · + vn ) = [ψij ] ° . » . (1)
.
vn

This gives an abelian group isomorphism between HomR (V — , W — ) and [HomR (Vj , Wi )],
the collection of all m by n matrices whose ij entry is an R-module homomorphism from
Vj to Wi . If we take m = n and Vi = Wj = V for all i and j, then V — = W — = V n , the
direct sum of n copies of V . Then the abelian group isomorphism given by (1) becomes

EndR (V n ) ∼ Mn (EndR (V )), (2)
=

the collection of all n by n matrices whose entries areR-endomorphisms of V . Since
composition of endomorphisms corresponds to multiplication of matrices, (2) gives a ring
isomorphism as well.
9.2. TWO KEY THEOREMS 5

9.2.4 Theorem (Jacobson)
Let M be a semisimple R-module, and let A be the endomorphism ring EndR (M ). If
f ∈ EndA (M ) and m1 , . . . , mn ∈ M , then there exists r ∈ R such that f (mi ) = rmi for
all i = 1, . . . , n.

Proof. f induces an endomorphism f (n) of M n , the direct sum of n copies of M , via

f (n) (m1 + · · · + mn ) = f (m1 ) + · · · + f (mn )

where f (mi ) belongs to the ith copy of M . Thus the matrix that represents f (n) is the
scalar matrix f I, where I is an n by n identity matrix. If B = EndR (M n ), then since
a scalar matrix commutes with everything, f (n) ∈ EndB (M n ). If m1 , . . . , mn ∈ M , then
by (9.2.2), there exists r ∈ R such that f (n) (m1 + · · · mn ) = r(m1 + · · · mn ). [Note that
since M is semisimple, so is M n .] This is equivalent to f (mi ) = rmi for all i. ™

Before giving a corollary, we must mention that the standard results that every vector
space over a ¬eld has a basis, and any two bases have the same cardinality, carry over if
the ¬eld is replaced by a division ring. Also recall that a module is said to be faithful if
its annihilator is 0.


9.2.5 Corollary
Let M be a faithful, simple R-module, and let D = EndR (M ), a division ring by (9.2.1(b)).
If M is a ¬nite-dimensional vector space over D, then EndD (M ) ∼ R, a ring isomorphism.
=

Proof. Let {x1 , . . . , xn } be a basis for M over D. By (9.2.4), if f ∈ EndD (M ), there
exists r ∈ R such that f (xi ) = rxi for all i = 1, . . . , n. Since the xi form a basis,
we have f (x) = rx for every x ∈ M . Thus the map h from R to EndD (M ) given by
r ’ gr = multiplication by r is surjective. If rx = 0 for all x ∈ M , then since M is
faithful, we have r = 0 and h is injective. Since h(rs) = gr —¦ gs = h(r)h(s), h is a ring
isomorphism. ™


Problems For Section 9.2
1. Criticize the following argument. Let M be a simple R-module, and let A = EndR (M ).
“Obviously” M is also a simple A-module. For any additive subgroup N of M that
is closed under the application of all R-endomorphisms of M is, in particular, closed
under multiplication by an element r ∈ R. Thus N is an R-submodule of M , hence
is 0 or M .
2. Let M be a nonzero cyclic module. Show that M is simple if and only if ann M , the
annihilator of M , is a maximal left ideal.
3. In Problem 2, show that the hypothesis that M is cyclic is essential.
4. Let V = F n be the n-dimensional vector space of all n-tuples with components in the
¬eld F . If T is a linear transformation on V , then V becomes an F [X]-module via
6 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

f (X)v = f (T )v. For example, if n = 2, T (a, b) = (0, a), and f (X) = a0 + a1 X + · · · +
an X n , then

f (X)(1, 0) = a0 (1, 0) + a1 T (1, 0) + a2 T 2 (1, 0) + · · · + an T n (1, 0)
= (a0 , 0) + (0, a1 )
= (a0 , a1 ).

Show that in this case, V is cyclic but not simple.
5. Suppose that M is a ¬nite-dimensional vector space over an algebraically closed ¬eld F ,
and in addition M is a module over a ring R containing F as a subring. If M is a
simple R-module and f is an R-module homomorphism, in particular an F -linear
transformation, on M , show that f is multiplication by some ¬xed scalar » ∈ F . This
result is frequently given as a third part of Schur™s lemma.
6. Let I be a left ideal of the ring R, so that R/I is an R-module but not necessarily a
ring. Criticize the following statement: “Obviously”, I annihilates R/I.



9.3 Simple and Semisimple Rings
9.3.1 De¬nitions and Comments
Since a ring is a module over itself, it is natural to call a ring R semisimple if it is
semisimple as an R-module. Our aim is to determine, if possible, how semisimple rings
are assembled from simpler components. A plausible idea is that the components are
rings that are simple as modules over themselves. But this turns out to be too restrictive,
since the components would have to be division rings (Section 9.1, Problem 1).
When we refer to a simple left ideal I of R, we will always mean that I is simple as a
left R-module. We say that the ring R is simple if R is semisimple and all simple left ideals
of R are isomorphic. [The de¬nition of simple ring varies in the literature. An advantage
of our choice (also favored by Lang and Bourbaki) is that we avoid an awkward situation
in which a ring is simple but not semisimple.] Our goal is to show that the building blocks
for semisimple rings are rings of matrices over a ¬eld, or more generally, over a division
ring.
The next two results give some properties of modules over semisimple rings.


9.3.2 Proposition
If R is a semisimple ring, then every nonzero R-module M is semisimple.


Proof. By (4.3.6), M is a quotient of a free R-module F . Since F is a direct sum of copies
of R (see (4.3.4)), and R is semisimple by hypothesis, it follows from (9.1.2) that F is
semisimple. By (9.1.3), M is semisimple. ™
9.3. SIMPLE AND SEMISIMPLE RINGS 7

9.3.3 Proposition
Let I be a simple left ideal in the semisimple ring R, and let M be a simple R-module.
Denote by IM the R-submodule of M consisting of all ¬nite linear combinations i ri xi ,
ri ∈ I, xi ∈ M . Then either IM = M and I is isomorphic to M , or IM = 0.
Proof. If IM = 0, then since M is simple, IM = M . Thus for some x ∈ M we have
Ix = 0, and again by simplicity of M , we have Ix = M . Map I onto M by r ’ rx, and
note that the kernel cannot be I because Ix = 0. Since I is simple, the kernel must be 0,
so I ∼ M . ™
=

9.3.4 Beginning the Decomposition
Let R be a semisimple ring. We regard two simple left ideals of R as equivalent if they are
isomorphic (as R-modules), and we choose a representative Ii , i ∈ T from each equivalence
class. We de¬ne the basic building blocks of R as
Bi = the sum of all left ideals of R that are isomorphic to Ii .
We have a long list of properties of the Bi to establish, and for the sake of economy we
will just number the statements and omit the words “Lemma” and “Proof” in each case.
We will also omit the end of proof symbol, except at the very end.

9.3.5
If i = j, then Bi Bj = 0. [The product of two left ideals is de¬ned exactly as in (9.3.3).]
Apply (9.3.3) with I replaced by Bi and M by Bj .

9.3.6
R = i∈T Bi
If r ∈ R, then (r) is a left ideal, which by (9.1.2) and (9.1.3) (or (9.3.2)) is a sum of
simple left ideats.

9.3.7
Each Bi is a two-sided ideal.
Using (9.3.5) and (9.3.6) we have
Bi ⊆ Bi R = Bi Bj = Bi Bi ⊆ RBi ⊆ Bi .
j

Thus RBi = Bi R = Bi .

9.3.8
R has only ¬nitely many isomorphism classes of simple left ideals I1 , . . . , It .
By (9.3.6), we can write the identity 1 of R as a ¬nite sum of elements ei ∈ Bi , i ∈ T .
t
Adjusting the notation if necessary, let 1 = i=1 ei . If r ∈ Bj where j ∈ {1, . . . , t}, then
/
by (9.3.5), rei = 0 for all i = 1, . . . , t, so r = r1 = 0. Thus Bj = 0 for j ∈ {1, . . . , t}.
/
8 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

9.3.9
t
R = •t Bi . Thus 1 has a unique representation as i=1 ei , with ei ∈ Bi .
i=1
By (9.3.6) and (9.3.8), R is the sum of the Bi . If b1 + · · · + bt = 0, with bi ∈ Bi , then

0 = ei (b1 + · · · + bt ) = ei b1 + · · · ei bt = ei bi = (e1 + · · · + et )bi = 1bi = bi .

Therefore the sum is direct.


9.3.10
If bi ∈ Bi , then ei bi = bi = bi ei . Thus ei is the identity on Bi and Bi = Rei = ei R.
The ¬rst assertion follows from the computation in (9.3.9), along with a similar com-
putation with ei multiplying on the right instead of the left. Now Bi ⊆ Rei because
bi = bi ei , and Rei ⊆ Bi by (9.3.7) and the fact that ei ∈ Bi . The proof that Bi = ei R is
similar.


9.3.11
Each Bi is a simple ring.
By the computation in (9.3.7), along with (9.3.10), Bi is a ring (with identity ei ).
Let J be a simple left ideal of Bi . By (9.3.5) and (9.3.6), RJ = Bi J = J, so J is a left
ideal of R, necessarily simple. Thus J is isomorphic to some Ij , and we must have j = i.
[Otherwise, J would appear in the sums de¬ning both Bi and Bj , contradicting (9.3.9).]
Therefore Bi has only one isomorphism class of simple left ideals. Now Bi is a sum of
simple left ideals of R, and a subset of Bi that is a left ideal of R must be a left ideal
of Bi . Consequently, Bi is semisimple and the result follows.


9.3.12
If M is a simple R-module, then M is isomorphic to some Ii . Thus there are only ¬nitely
many isomorphism classes of simple R-modules. In particular, if R is a simple ring, then
all simple R-modules are isomorphic.
By (9.3.9),
t t
{J : J ∼ Ii }
R= Bi = =
i=1 i=1

where the J are simple left ideals. Therefore
t t
{JM : J ∼ Ii }.
M = RM = Bi M = =
i=1 i=1

By (9.3.3), JM = 0 or J ∼ M . The former cannot hold for all J, since M = 0. Thus
=
∼ Ii for some i. If R is a simple ring, then there is only one i, and the result follows.
M=
9.3. SIMPLE AND SEMISIMPLE RINGS 9

9.3.13
Let M be a nonzero R-module, so that M is semisimple by (9.3.2). De¬ne Mi as the sum of
t
all simple submodules of M that are isomorphic to Ii , so that by (9.3.12), M = i=1 Mi .
Then
t
M= Bi M and Bi M = ei M = Mi , i = 1, . . . , t.
i=1

By de¬nition of Bi ,

{JMj : J ∼ Ii }
Bi Mj = =

where the J™s are simple left ideals. If N is any simple module involved in the de¬nition
of Mj , then JN is 0 or N , and by (9.3.3), JN = N implies that N ∼ J ∼ Ii . But all
==
such N are isomorphic to Ij , and therefore Bi Mj = 0, i = j. Thus

Mi = RMi = Bj Mi = Bi Mi
j

and

Bi M = Bi Mj = Bi Mi .
j

Consequently, Mi = Bi M = ei RM = ei M (using (9.3.10)), and all that remains is to
show that the sum of the Mi is direct. Let x1 + · · · + xt = 0, xi ∈ Mi . Then

0 = ei (x1 + · · · + xt ) = ei xi

since ei xj ∈ Bi Mj = 0 for i = j. Finally, by (9.3.9),

ei xi = (e1 + · · · + et )xi = xi .

9.3.14
A semisimple ring R is ring-isomorphic to a direct product of simple rings.
This follows from (9.3.9) and (9.3.5). For if ai , bi ∈ Bi , then

(a1 + · · · + at )(b1 + · · · + bt ) = a1 b1 + · · · + at bt . ™

Problems For Section 9.3
In Problems 1 and 2, let M be a semisimple module, so that M is the direct sum of simple
modules Mi , i ∈ I. We are going to show that M is a ¬nite direct sum of simple modules
if and only if M is ¬nitely generated.
1. Suppose that x1 , . . . , xn generate M . It will follow that M is the direct sum of ¬nitely
many of the Mi . How would you determine which Mi ™s are involved?
10 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

2. Conversely, assume that M is a ¬nite sum of simple modules. Show that M is ¬nitely
generated.
3. A left ideal I is said to be minimal if I = 0 and I has no proper subideal except 0.
Show that the ring R is semisimple if and only if R is a direct sum of minimal left
ideals.
4. Is Z semisimple?
5. Is Zn semisimple?
6. Suppose that R is a ring with the property that every nonzero R-module is semi-
simple. Show that every R-module M is projective, that is, every exact sequence
0 ’ A ’ B ’ M ’ 0 splits. Moreover, M is injective, that is, every exact sequence
0 ’ M ’ A ’ B ’ 0 splits. [Projective and injective modules will be studied in
Chapter 10.]
7. For any ring R, show that the following conditions are equivalent.
(a) R is semisimple;
(b) Every nonzero R-module is semisimple;
(c) Every R-module is projective;
(d) Every R-module is injective.


9.4 Further Properties of Simple Rings, Matrix Rings,
and Endomorphisms
To reach the Wedderburn structure theorem, we must look at simple rings in more detail,
and supplement what we already know about matrix rings and rings of endomorphisms.


9.4.1 Lemma
Let R be any ring, regarded as a left module over itself. If h : R ’ M is an R-module
homomorpbism, then for some x ∈ M we have h(r) = rx for every r ∈ R. Moreover, we
may choose x = h(1), and the map h ’ h(1) is an isomorphism of HomR (R, M ) and M .
This applies in particular when M = R, in which case h ∈ EndR (R).

Proof. The point is that h is determined by what it does to the identity. Thus

h(r) = h(r1) = rh(1)

so we may take x = h(1). If s ∈ R and h ∈ HomR (R, M ), we take (sh)(r) = h(rs) = rsx.
This makes HomR (R, M ) into a left R-module isomorphic to M . (For further discussion
of this idea, see the exercises in Section 10.7.) ™

Notice that although all modules are left R-modules, h is given by multiplication on
the right by x.
9.4. SIMPLE RINGS, MATRIX RINGS, AND ENDOMORPHISMS 11

9.4.2 Corollary
Let I and J be simple left ideals of the simple ring R. Then for some x ∈ R we have
J = Ix.

Proof. By the de¬nition of a simple ring (see (9.3.1)), R is semisimple, so by (9.1.2),
R = I • L for some left ideal L. Again by the de¬nition of a simple ring, I and J are
isomorphic (as R-modules). If „ : I ’ J is an isomorphism and π is the natural projection
of R on I, then „ π ∈ EndR (R), so by (9.4.1), there exists x ∈ R such that „ π(r) = rx for
every r ∈ R. Allow r to range over I to conclude that J = Ix. ™

A semisimple ring can be expressed as a direct sum of simple left ideals, by (9.1.2). If
the ring is simple, only ¬nitely many simple left ideals are needed.

9.4.3 Lemma
A simple ring R is a ¬nite direct sum of simple left ideals.

Proof. Let R = •j Ij where the Ij are simple left ideals. Changing notation if necessary,
we have 1 = y1 + · · · + ym with yj ∈ Ij , j = 1, . . . , m. If x ∈ R, then
m m
xyj ∈
x = x1 = Ij .
j=1 j=1

Therefore R is a ¬nite sum of the Ij , and the sum is direct because the original decom-
position of R is direct. ™

9.4.4 Corollary
If I is a simple left ideal of the simple ring R, then IR = R.

Proof. If J is any simple left ideal of R, then by (9.4.2), J ⊆ IR. By (9.4.3), R is a ¬nite
(direct) sum of simple left ideals, so R ⊆ IR. The reverse inclusion always holds, and the
result follows. ™

We now have some insight into the structure of simple rings.

9.4.5 Proposition
If R is a simple ring, then the only two-sided ideals of R are 0 and R.

Proof. Let J be a nonzero 2-sided ideal of R. By (9.1.3), J is a semisimple left R-module,
so by (9.1.2), J is a sum of simple left ideals of J, hence of R. In particular, J contains a
simple left ideal I. Since J is a right ideal, it follows that J = JR. Using (9.4.4), we have

J = JR ⊇ IR = R

so J = R. ™
12 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

In the literature, a simple ring is often de¬ned as a ring R whose only two-sided ideals
are 0 and R, but then extra hypotheses must be added to force R to be semisimple. See
the exercises for further discussion.


9.4.6 Corollary
Let I be a simple left ideal of the simple ring R, and let M be a simple R-module. Then
IM = M and M is faithful.

Proof. The ¬rst assertion follows from a computation that uses associativity of scalar
multiplication in a module, along with (9.4.4):

M = RM = (IR)M = I(RM ) = IM. (1)

Now let b belong to the annihilator of M , so that bM = 0. We must show that b = 0. By
a computation similar to (1) (using in addition the associativity of ring multiplication),

RbRM = RbM = R0 = 0. (2)

But RbR is a two-sided ideal of R (see (2.2.7)), so by (9.4.5), RbR = 0 or R. In the latter
case, M = RM = RbRM = 0 by (2), contradicting the assumption that M is simple.
Therefore RbR = 0, in particular, b = 1b1 = 0. ™

We are now ready to show that a simple ring is isomorphic to a ring of matrices. Let
R be a simple ring, and V a simple R-module. [V exists because R is a sum of simple
left ideals, and V is unique up to isomorphism by (9.3.12).] Let D = EndR (V ), a division
ring by Schur™s lemma (9.2.1(b)). Then (see (9.2.2)), V is a D-module, in other words,
a vector space over D. V is a faithful R-module by (9.4.6), and if we can prove that V
is ¬nite-dimensional as a vector space over D, then by (9.2.5), R is ring-isomorphic to
EndD (V ). If n is the dimension of V over D, then by (4.4.1), EndD (V ) ∼ Mn (Do ), the
=
o
ring of n by n matrices with entries in the opposite ring D .


9.4.7 Theorem
Let R be a simple ring, V a simple R-module, and D the endomorphism ring EndR (V ).
Then V is a ¬nite-dimensional vector space over D. If the dimension of this vector space
is n, then (by the above discussion),

R ∼ EndD (V ) ∼ Mn (Do ).
= =

Proof. Assume that we have in¬nitely many linearly independent elements x1 , x2 , . . . .
Let Im be the left ideal {r ∈ R : rxi = 0 for all i = 1, . . . , m}. Then the Im decrease as m
increases, in fact they decrease strictly. [Given any m, let f be a D-linear transformation
on V such that f (xi ) = 0 for 1 ¤ i ¤ m and f (xm+1 ) = 0. By Jacobson™s theorem (9.2.4),
there exists r ∈ R such that f (xi ) = rxi , i = 1, . . . , m + 1. But then rx1 = · · · = rxm = 0,
rxm+1 = 0, so r ∈ Im \ Im+1 .] Write Im = Jm • Im+1 , as in (9.1.2) part (c). [Recall
9.5. THE STRUCTURE OF SEMISIMPLE RINGS 13

from (9.1.3) that since R is semisimple, so are all left ideals.] Iterating this process, we
construct a left ideal J1 • J2 • · · · , and again by (9.1.2(c)),

R = J0 • J1 • J2 • · · · .

Therefore 1 is a ¬nite sum of elements yi ∈ Ji , i = 0, 1, . . . , t. But then

R = J0 • J1 • · · · • Jt

and it follows that Jt+1 must be 0, a contradiction. ™

Problems For Section 9.4
Problems 1“5 are the key steps in showing that a ring R is simple if and only if R is
Artinian and has no two-sided ideals except 0 and R. Thus if a simple ring is de¬ned as
one with no nontrivial two-sided ideals, then the addition of the Artinian condition gives
our de¬nition of simple ring; in particular, it forces the ring to be semisimple. The result
that an Artinian ring with no nontrivial two-sided ideals is isomorphic to a matrix ring
over a division ring (Theorem 9.4.7) is sometimes called the Wedderburn-Artin theorem.
In Problems 1“5, “simple” will always mean simple in our sense.

1. By (9.4.5), a simple ring has no nontrivial two-sided ideals. Show that a simple ring
must be Artinian.
2. If R is an Artinian ring, show that there exists a simple R-module.
3. Let R be an Artinian ring with no nontrivial two-sided ideals. Show that R has a
faithful, simple R-module.
4. Continuing Problem 3, if V is a faithful, simple R-module, and D = EndR (V ), show
that V is a ¬nite-dimensional vector space over D.
5. Continuing Problem 4, show that R is ring-isomorphic to EndD (V ), and therefore to
a matrix ring Mn (Do ) over a division ring.

In the next section, we will prove that a matrix ring over a division ring is simple; this
concludes the proof that R is simple i¬ R is Artinian with no nontrivial two-sided ideals.
(In the “if” part, semisimplicity of R follows from basic properties of matrix rings; see
Section 2.2, Problems 2, 3 and 4.)

6. If an R-module M is a direct sum •n Mi of ¬nitely many simple modules, show that
i=1
M has a composition series. (Equivalently, by (7.5.12), M is Artinian and Noetherian.)
7. Conversely, if M is semisimple and has a composition series, show that M is a ¬nite
direct sum of simple modules. (Equivalently, by Section 9.3, Problems 1 and 2, M is
¬nitely generated.)


9.5 The Structure of Semisimple Rings
We have now done all the work needed for the fundamental theorem.
14 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

9.5.1 Wedderburn Structure Theorem
Let R be a semisimple ring.

(1) R is ring-isomorphic to a direct product of simple rings B1 , . . . , Bt .
(2) There are t isomorphism classes of simple R-modules. If V1 , . . . , Vt are representatives
of these classes, let Di be the division ring EndR (Vi ). Then Vi is a ¬nite-dimensional
vector space over Di . If ni is the dimension of this vector space, then there is a ring
isomorphism

Bi ∼ EndDi (Vi ) ∼ Mni (Di ).
o
= =

Consequently, R is isomorphic to the direct product of matrix rings over division
rings. Moreover,
(3) Bi Vj = 0, i = j; Bi Vi = Vi .

Proof. Assertion (1) follows from (9.3.5), (9.3.9) and (9.3.14). By (9.3.8) and (9.3.12),
there are t isomorphism classes of simple R-modules. The remaining statements of (2)
follow from (9.4.7). The assertions of (3) follow from (9.3.13) and its proof. ™


Thus a semisimple ring can always be assembled from matrix rings over division rings.
We now show that such matrix rings can never combine to produce a ring that is not
semisimple.


9.5.2 Theorem
The ring Mn (R) of all n by n matrices with entries in the division ring R is simple.


Proof. We have done most of the work in the exercises for Section 2.2. Let Ck be the
set of matrices whose entries are 0 except perhaps in column k, k = 1 . . . , n. Then Ck
is a left ideal of Mn (R), and if any nonzero matrix in Ck belongs to a left ideal I, then
Ck ⊆ I. (Section 2.2, Problems 2, 3, 4.) Thus each Ck is a simple left ideal, and Mn (R),
the direct sum of C1 , . . . , Cn , is semisimple.
Now let I be a nonzero simple left ideal. A nonzero matrix in I must have a nonzero
entry in some column, say column k. De¬ne f : I ’ Ck by f (A) = Ak , the matrix
obtained from A by replacing every entry except those in column k by 0. Then f is an
Mn (R)-module homomorphism, since

f (BA) = (BA)k = BAk = Bf (A).

By construction, f is not identically 0, so by Schur™s lemma, f is an isomorphism. Since
the Ck are mutually isomorphic, all simple left ideals are isomorphic, proving that Mn (R)
is simple. ™
9.5. THE STRUCTURE OF SEMISIMPLE RINGS 15

9.5.3 Informal Introduction to Group Representations
A major application of semisimple rings and modules occurs in group representation
theory, and we will try to indicate the connection. Let k be any ¬eld, and let G be a ¬nite
group. We form the group algebra kG, which is a vector space over k with basis vectors
corresponding to the elements of G. In general, if G = {x1 , . . . , xm }, the elements of kG
are of the form ±1 x1 + · · · + ±m xm , where the ±i belong to k. Multiplication in kG is
de¬ned in the natural way; we set

(±xi )(βxj ) = ±βxi xj

and extend by linearity. Then kG is a ring (with identity 1k 1G ) that is also a vector space
over k, and ±(xy) = (±x)y = x(±y), ± ∈ k, x, y ∈ G, so kG is indeed an algebra over
k. [This construction can be carried out with an arbitrary ring R in place of k, and with
an arbitrary (not necessarily ¬nite) group G. The result is the group ring RG, a free
R-module with basis G.]
Now let V be an n-dimensional vector space over k. We want to describe the situation
in which “G acts linearly on V ”. We are familiar with group action (Section 5.1), but we
now add the condition that each g ∈ G determines a linear transformation ρ(g) on V . We
will write ρ(g)(v) as simply gv or g(v), so that g(±v + βw) = ±g(v) + βg(w). Thus we
can multiply vectors in V by scalars in G. Since elements of kG are linear combinations
of elements of G with coe¬cients in k, we can multiply vectors in V by scalars in kG. To
summarize very compactly,

V is a kG-module.

Now since G acts on V , (hg)v = h(gv) and 1G v = v, g, h ∈ G, v ∈ V . Thus ρ(hg) =
ρ(h)ρ(g), and each ρ(g) is invertible since ρ(g)ρ(g ’1 ) = ρ(1G ) = the identity on V .
Therefore

ρ is a homomorphism from G to GL(V ),

the group of invertible linear transformations on V . Multiplication in GL(V ) corresponds
to composition of functions.

The homomorphism ρ is called a representation of G in V ,

and n, the dimension of V , is called the degree of the representation. If we like, we can
replace GL(V ) by the group of all nonsingular n by n matrices with entries in k. In this
case, ρ is called a matrix representation.
The above process can be reversed. Given a representation ρ, we can de¬ne a linear
action of G on V by gv = ρ(g)(v), and thereby make V a kG-module. Thus representations
can be identi¬ed with kG-modules.

9.5.4 The Regular Representation
If G has order n, then kG is an n-dimensional vector space over k with basis G. We take V
to be kG itself, with gv the product of g and v in kG. As an example, let G = {e, a, a2 },
16 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

a cyclic group of order 3. V is a 3-dimensional vector space with basis e, a, a2 , and the
action of G on V is determined by

ee = e, ea = a, ea2 = a2 ;
ae = a, aa = a2 , aa2 = e;
a2 e = a2 , a2 a = e, a2 a2 = a.

Thus the matrices ρ(g) associated with the elements g ∈ G are
®  ®  ® 
1 0 0 0 0 1 0 1 0
°0 » , [a] = °1 » , [a2 ] = °0 1» .
1 0 0 0 0
[e] =
0 0 1 0 1 0 1 0 0


9.5.5 The Role of Semisimplicity
Suppose that ρ is a representation of G in V . Assume that the basis vectors of V can
be decomposed into two subsets v(A) and v(B) such that matrix of every g ∈ G has the
form

A 0
[g] = .
0 B

(The elements of A and B will depend on the particular g, but the dimensions of A and B
do not change.) The corresponding statement about V is that

V = VA • VB

where VA and VB are kG-submodules of V . We can study the representation by analyzing
its behavior on the simpler spaces VA and VB . Maschke™s theorem, to be proved in the
next section, says that under wide conditions on the ¬eld k, this decomposition process
can be continued until we reach subspaces that have no nontrivial kG-submodules. In
other words, every kG-module is semisimple. In particular, kG is a semisimple ring,
and the Wedderburn structure theorem can be applied to get basic information about
representations.
We will need some properties of projection operators, and it is convenient to take care
of this now.


9.5.6 De¬nitions and Comments
A linear transformation π on a vector space V [or more generally, a module homomor-
phism] is called a projection of V (on π(V )) if π is idempotent, that is, π 2 = π. We
have already met the natural projection of a direct sum onto a component, but there
are other possibilities. For example, let p be the projection of R2 = R • R given by
p(x, y) = x’y , ’x+y . Note that π must be the identity on π(V ), since π(π(v)) = π(v).
2 2
If we choose subspaces carefully, we can regard any projection as natural.
9.5. THE STRUCTURE OF SEMISIMPLE RINGS 17

9.5.7 Proposition
If π is a projection of V , then V is the direct sum of the image of π and the kernel of π.

Proof. Since v = π(v) + (v ’ π(v)) and π(v ’ π(v)) = 0, V = im V + ker V . To show
that the sum is direct, let v = π(w) ∈ ker π. Then 0 = π(v) = π 2 (w) = π(w) = v, so
im π © ker π = 0. ™

9.5.8 Example
For real numbers x and y, we have (x, y) = (x ’ cy)(1, 0) + y(c, 1), where c is any ¬xed
real number. Thus R2 = R(1, 0) • R(c, 1), and if we take p(x, y) = (x ’ cy, 0), then p is a
projection of R2 onto R(1, 0). By varying c we can change the complementary subspace
R(c, 1). Thus we have many distinct projections onto the same subspace R(1, 0).

Problems For Section 9.5
1. Show that the regular representation is faithful, that is, the homomorphism ρ is injec-
tive.
2. Let G be a subgroup of Sn and let V be an n-dimensional vector space over k with
basis v(1), . . . , v(n). De¬ne the action of G on V by

g(v(i)) = v(g(i)), i = 1, . . . , n.

Show that the action is legal. (V is called a permutation module.)
3. Continuing Problem 2, if n = 4, ¬nd the matrix of g = (1, 4, 3).
4. Here is an example of how a representation can arise in practice. Place√ equilateral
an
triangle in the plane V , with the vertices at v1 = (1, 0), v2 = ’ 2 , 2 3 and v3 =
11

’ 1 , ’ 1 3 ; note that v1 + v2 + v3 = 0. Let G = D6 be the group of symmetries
2 2
of the triangle, with g = counterclockwise rotation by 120 degrees and h = re¬‚ection
about the horizontal axis. Each member of D6 is of the form g i hj , i = 0, 1, 2, j = 0, 1,
and induces a linear transformation on V . Thus we have a representation of G in V
(the underlying ¬eld k can be taken as R).

With v1 and v2 taken as a basis for V , ¬nd the matrices [g] and [h] associated with g
and h.

5. Continue from Problem 4, and switch to the standard basis e1 = v1 = (1, 0), e2 =
(0, 1). Changing the basis produces an equivalent matrix representation. The matrix
representing the element a ∈ G is now of the form

[a] = P ’1 [a]P

where the similarity matrix P is the same for every a ∈ G (the key point).
Find the matrix P corresponding to the switch from {v1 , v2 } to {e1 , e2 }, and the
matrices [g] and [h] .
18 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

6. Consider the dihedral group D8 , generated by elements R (rotation) and F (re¬‚ection).
We assign to R the 2 by 2 matrix

0 1
A=
’1 0

and to F the 2 by 2 matrix

10
B= .
0 ’1

Show that the above assignment determines a matrix representation of D8 of degree 2.
7. Is the representation of Problem 6 faithful?

A very accessible basic text on group representation theory is “Representations and
Characters of Groups” by James and Liebeck.


9.6 Maschke™s Theorem
We can now prove the fundamental theorem on decomposition of representations. It is
useful to isolate the key ideas in preliminary lemmas.

9.6.1 Lemma
Let G be a ¬nite group, and k a ¬eld whose characteristic does not divide |G| (so that
division by |G| is legal). Let V be a kG-module, and ψ a linear transformation on V as
a vector space over k. De¬ne θ : V ’ V by
1
g ’1 ψg(v).
θ(v) =
|G|
g∈G

Then not only is θ a linear transformation on the vector space V , but it is also a kG-
homomorphism.

Proof. Since ψ is a linear transformation and G acts linearly on V (see (9.5.3)), θ is linear.
Now if h ∈ G, then
1
g ’1 ψg(hv).
θ(hv) =
|G|
g∈G


As g ranges over all of G, so does gh. Thus we can let x = gh, g ’1 = hx’1 , to obtain
1
hx’1 ψ(xv) = hθ(v)
θ(hv) =
|G|
x∈G

and the result follows. ™
9.6. MASCHKE™S THEOREM 19

9.6.2 Lemma
In (9.6.1), suppose that ψ is a projection of V on a subspace W that is also a kG-submodule
of V . Then θ is also a projection of V on W .

Proof. If v ∈ W , then g(v) ∈ W since W is a kG-submodule of V . Thus ψg(v) = g(v)
since ψ is a projection on W . By de¬nition of θ we have θ(v) = v. To prove that θ2 = θ,
note that since ψ maps V into the kG-submodule W , it follows from the de¬nition of θ
that θ also maps V into W . But θ is the identity on W , so

θ2 (v) = θ(θ(v)) = θ(v)

and θ is a projection. Since θ maps into W and is the identity on W , θ is a projection of
V on W . ™


9.6.3 Maschke™s Theorem
Let G be a ¬nite group, and k a ¬eld whose characteristic does not divide |G|. If V is a
kG-module, then V is semisimple.

Proof. Let W be a kG-submodule of V . Ignoring the group algebra for a moment, we
can write V = W • U as vector spaces over k. Let ψ be the natural projection of V
on W , and de¬ne θ as in (9.6.1). By (9.6.1) and (9.6.2), θ is a kG-homomorphism and
also a projection of V on W . By (9.5.7), V = im θ • ker θ = W • ker θ as kG-modules.
By (9.1.2), V is semisimple. ™

We have been examining the decomposition of a semisimple module into a direct sum
of simple modules. Suppose we start with an arbitrary module M , and ask whether M
can be expressed as M1 • M2 , where M1 and M2 are nonzero submodules. If so, we can
try to decompose M1 and M2 , and so on. This process will often terminate in a ¬nite
number of steps.


9.6.4 De¬nition
The module M is decomposable if M = M1 • M2 , where M1 and M2 are nonzero sub-
modules. Otherwise, M is indecomposable.


9.6.5 Proposition
Let M be a module with a composition series; equivalently, by (7.5.12), M is Noetherian
and Artinian. Then M can be expressed as a ¬nite direct sum •n Mi of indecomposable
i=1
submodules.

Proof. If the decomposition process does not terminate, in¬nite ascending and descending
chains are produced, contradicting the hypothesis. ™
20 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

As the above argument shows, the hypothesis can be weakened to M Noetherian or
Artinian. But (9.6.5) is usually stated along with a uniqueness assertion which uses the
stronger hypothesis:
If M has a composition series and M = •n Mi = •m Nj , where the Mi and Nj
i=1 j=1
are indecomposable submodules, then n = m and the Mi are, up to isomorphism, just a
rearrangement of the Ni .
The full result (existence plus uniqueness) is most often known as the Krull-Schmidt
Theorem. [One or more of the names Remak, Azumaya and Wedderburn are sometimes
added.] The uniqueness proof is quite long (see, for example, Jacobson™s Basic Algebra II),
and we will not need the result.
Returning to semisimple rings, there is an asymmetry in the de¬nition in that a ring is
regarded as a left module over itself, so that submodules are left ideals. We can repeat the
entire discussion using right ideals, so that we should distinguish between left-semisimple
and right-semisimple rings. However, this turns out to be unnecessary.

9.6.6 Theorem
A ring R is left-semisimple if and only if it is right-semisimple.
Proof. If R is left-semisimple, then by (9.5.1), R is isomorphic to a direct product of
matrix rings over division rings. But a matrix ring over a division ring is right-simple
by (9.5.2) with left ideals replaced by right ideals. Therefore R is right-semisimple. The
reverse implication is symmetrical. ™

Problems For Section 9.6
1. Let V be the permutation module for G = S3 (see Section 9.5, Problem 2), with basis
v1 , v2 , v3 . Give an example of a nontrivial kG-submodule of V .

In Problems 2“4, we show that Maschke™s theorem can fail if the characteristic of k
divides the order of G. Let G = {1, a, . . . , ap’1 } be a cyclic group of prime order p, and
let V be a two-dimensional vector space over the ¬eld Fp , with basis v1 , v2 . Take the
matrix of a as
1 1
[a] =
0 1

so that
1 r
[ar ] =
0 1

and [ap ] is the identity.
2. Show that W , the one-dimensional subspace spanned by v1 , is a kG-submodule of V .
3. Continuing Problem 2, show that W is the only one-dimensional kG-submodule of V .
4. Continuing Problem 3, show that V is not a semisimple kG-module.
5. Show that a semisimple module is Noetherian i¬ it is Artinian.
9.7. THE JACOBSON RADICAL 21

6. Let M be a decomposable R-module, so that M is the direct sum of nonzero submodules
M1 and M2 . Show that EndR (M ) contains a nontrivial idempotent e (that is, e2 = e
with e not the zero map and not the identity).
7. Continuing from Problem 6, suppose conversely that EndR (M ) contains a nontrivial
idempotent e. Show that M is decomposable. (Suggestion: use e to construct idem-
potents e1 and e2 that are orthogonal, that is, e1 e2 = e2 e1 = 0.)


9.7 The Jacobson Radical
There is a very useful device that will allow us to look deeper into the structure of rings.

9.7.1 De¬nitions and Comments
The Jacobson radical J(R) of a ring R is the intersection of all maximal left ideals of R.
More generally, the Jacobson radical J(M ) = JR (M ) of an R-module M is the intersection
of all maximal submodules of M . [“Maximal submodule” will always mean “maximal
proper submodule”.] If M has no maximal submodule, take J(M ) = M .
If M is ¬nitely generated, then every submodule N of M is contained in a maximal
submodule, by Zorn™s lemma. [If the union of a chain of proper submodules is M , then
the union contains all the generators, hence some member of the chain contains all the
generators, a contradiction.] Taking N = 0, we see that J(M ) is a proper submodule
of M . Since R is ¬nitely generated (by 1R ), J(R) is always a proper left ideal.
Semisimplicity of M imposes a severe constraint on J(M ).

9.7.2 Proposition
If M is semisimple, then J(M ) = 0. Thus in a sense, the Jacobson radical is an “obstruc-
tion” to semisimplicity.

Proof. Let N be any simple submodule of M . By (9.1.2), M = N • N for some sub-
module N . Now M/N ∼ N , which is simple, so by the correspondence theorem, N is
=
maximal. Thus J(M ) ⊆ N , and therefore J(M ) © N = 0. Since M is a sum of simple
modules (see (9.1.2)), J(M ) = J(M ) © M = 0 ™

Here is another description of the Jacobson radical.

9.7.3 Proposition
J(R) is the intersection of all annihilators of simple R-modules.

Proof. By Section 9.1, Problem 3, simple modules are isomorphic to R/I for maximal
left ideals I. If r annihilates all simple R-modules, then for every maximal left ideal I,
r annihilates R/I, in particular, r annihilates 1 + I. Thus r(1 + I) = I, that is, r ∈ I.
Consequently, r ∈ J(R).
Conversely, assume r ∈ J(R). If M is a simple R-module, choose any nonzero element
x ∈ M . The map fx : R ’ M given by fx (s) = sx is an epimorphism by simplicity of
22 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

M . The kernel of fx is the annihilator of x, denoted by ann(x). By the ¬rst isomorphism
theorem, M ∼ R/ ann(x). By simplicity of M , ann(x) is a maximal left ideal, so by
=
hypothesis, r ∈ ©x∈M ann(x) = ann(M ). Thus r annihilates all simple R-modules. ™


9.7.4 Corollary
J(R) is a two-sided ideal.

Proof. We noted in (4.2.6) that ann(M ) is a two-sided ideal, and the result follows
from (9.7.3). ™

In view of (9.7.4), one might suspect that the Jacobson radical is unchanged if right
rather than left ideals are used in the de¬nition. This turns out to be the case.


9.7.5 De¬nitions and Comments
The element a ∈ R is left quasi-regular (lqr) if 1 ’ a has a left inverse, right quasi-regular
(rqr) if 1 ’ a has a right inverse, and quasi-regular (qr) if 1 ’ a is invertible. Note that
if a is both lqr and rqr, it is qr, because if b(1 ’ a) = (1 ’ a)c = 1, then

b = b1 = b(1 ’ a)c = 1c = c.


9.7.6 Lemma
Let I be a left ideal of R. If every element of I is lqr, then every element of I is qr.

Proof. If a ∈ I, then we have b(1 ’ a) = 1 for some b ∈ R. Let c = 1 ’ b, so that
(1 ’ c)(1 ’ a) = 1 ’ a ’ c + ca = 1. Thus c = ca ’ a = (c ’ 1)a ∈ I. By hypothesis, c is
lqr, so 1 ’ c has a left inverse. But we know that (1 ’ c) has a right inverse (1 ’ a) [see
above], so c is rqr. By (9.7.5), c is qr and 1 ’ c is the two-sided inverse of 1 ’ a. ™


9.7.7 Proposition
The Jacobson radical J(R) is the largest two-sided ideal consisting entirely of quasi-regular
elements.

Proof. First, we show that each a ∈ J(R) is lqr, so by (9.7.6), each a ∈ J(R) is qr. If 1 ’ a
has no left inverse, then R(1 ’ a) is a proper left ideal, which is contained in a maximal
left ideal I (as in (2.4.2) or (9.7.1)). But then a ∈ I and 1 ’ a ∈ I, and therefore 1 ∈ I, a
contradiction.
Now we show that every left ideal (hence every two-sided ideal) I consisting entirely
of quasi-regular elements is contained in J(R). If a ∈ I but a ∈ J(R), then for some
/
maximal left ideal L we have a ∈ L. By maximality of L, we have I + L = R, so 1 = b + c
/
for some b ∈ I, c ∈ L. But then b is quasi-regular, so c = 1 ’ b has an inverse, and
consequently 1 ∈ L, a contradiction. ™
9.7. THE JACOBSON RADICAL 23

9.7.8 Corollary
J(R) is the intersection of all maximal right ideals of R.

Proof. We can reproduce the entire discussion beginning with (9.7.1) with left and right
ideals interchanged, and reach exactly the same conclusion, namely that the “right”
Jacobson radical is the largest two-sided ideal consisting entirely of quasi-regular ele-
ments. It follows that the “left”and “right” Jacobson radicals are identical. ™

We can now use the Jacobson radical to sharpen our understanding of semisimple
modules and rings.

9.7.9 Theorem
If M is a nonzero R-module, the following conditions are equivalent:

(1) M is semisimple and has ¬nite length, that is, has a composition series;
(2) M is Artinian and J(M ) = 0.

Proof. (1) implies (2) by (7.5.12) and (9.7.2), so assume M Artinian with J(M ) = 0.
The Artinian condition implies that the collection of all ¬nite intersections of maximal
submodules of M has a minimal element N . If S is any maximal submodule of M , then
N © S is a ¬nite intersection of maximal submodules, so by minimality of N , N © S = N ,
so N ⊆ S. Since J(M ) is the intersection of all such S, the hypothesis that J(M ) = 0
implies that N = 0. Thus for some positive integer n we have maximal submodules
M1 , . . . , Mn such that ©n Mi = 0.
i=1
Now M is isomorphic to a submodule of M = •n (M/Mi ). To see this, map x ∈ M
i=1
to (x + M1 , . . . , x + Mn ) and use the ¬rst isomorphism theorem. Since M is a ¬nite direct
sum of simple modules, it is semisimple and has a composition series. (See Section 9.4,
Problem 6) By (9.1.3) and (7.5.7), the same is true for M . ™

9.7.10 Corollary
The ring R is semisimple if and only if R is Artinian and J(R) = 0.

Proof. By (9.7.9), it su¬ces to show that if R is semisimple, then it has a composition
series. But this follows because R is ¬nitely generated, hence is a ¬nite direct sum of
simple modules (see Section 9.3, Problem 1). ™

The Jacobson radical of an Artinian ring has some special properties.

9.7.11 De¬nitions and Comments
An ideal (or left ideal or right ideal) I of the ring R is nil if each element x ∈ I is nilpotent,
that is, xm = 0 for some positive integer m; I is nilpotent if I n = 0 for some positive
integer n. Every nilpotent ideal is nil, and the converse holds if R is Artinian, as we will
prove.
24 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

9.7.12 Lemma
If I is a nil left ideal of R, then I ⊆ J(R).

Proof. If x ∈ I and xm = 0, then x is quasi-regular; the inverse of 1 ’ x is 1 + x + x2 +
· · · + xm’1 . The result follows from the proof of (9.7.7). ™


9.7.13 Proposition
If R is Artinian, then J(R) is nilpotent. Thus by (9.7.11) and (9.7.12), J(R) is the largest
nilpotent ideal of R, and every nil ideal of R is nilpotent.

Proof. Let J = J(R). The sequence J ⊇ J 2 ⊇ · · · stabilizes, so for some n we have
J n = J n+1 = · · · , in particular, J n = J 2n . We claim that J n = 0. If not, then the
collection of all left ideals Q of R such that J n Q = 0 is nonempty (it contains J n ), hence
has a minimal element N . Choose x ∈ N such that J n x = 0. By minimality of N ,
J n x = N . Thus there is an element c ∈ J n such that cx = x, that is, (1 ’ c)x = 0. But
c ∈ J n ⊆ J, so by (9.7.7), 1 ’ c is invertible, and consequently x = 0, a contradiction. ™


Problems For Section 9.7
1. Show that an R-module is M cyclic if and only if M is isomorphic to R/I for some
left ideal I, and in this case we can take I to be ann(M ), the annihilator of M .
2. Show that the Jacobson radical of an R-module M is the intersection of all kernels of
homomorphisms from M to simple R-modules.
3. If I = J(R), show that J(R/I) = 0.
4. If f is an R-module homomorphism from M to N , show that f (J(M )) ⊆ J(N ).
5. Assume R commutative, so that J(R) is the intersection of all maximal ideals of R. If
a ∈ R, show that a ∈ J(R) if and only if 1 + ab is a unit for every b ∈ R.
6. If N is a submodule of the Jacobson radical of the R-module M , show that J(M )/N =
J(M/N ).


9.8 Theorems of Hopkins-Levitzki and Nakayama
From Section 7.5, we know that a Noetherian ring need not be Artinian, and an Artinian
module need not be Noetherian. But the latter situation can never arise for rings, because
of the following result.


9.8.1 Theorem (Hopkins and Levitzki)
Let R be an Artinian ring, and M a ¬nitely generated R-module. Then M is both Artinian
and Noetherian. In particular, with M = R, an Artinian ring is Noetherian.
9.8. THEOREMS OF HOPKINS-LEVITZKI AND NAKAYAMA 25

Proof. By (7.5.9), M is Artinian. Let J be the Jacobson radical of R. By Section 9.7,
Problem 3, the Jacobson radical of R/J is zero, and since R/J is Artinian by (7.5.7), it
is semisimple by (9.7.9). Now consider the sequence

M0 = M, M1 = JM, M2 = J 2 M, . . . .

By (9.7.13), J is nilpotent, so Mn = 0 for some n. Since JMi = Mi+1 , J annihilates
Mi /Mi+1 , so by Section 4.2, Problem 6, Mi /Mi+1 is an R/J-module.
We claim that each Mi /Mi+1 has a composition series.
We can assume that Mi /Mi+1 = 0, otherwise there is nothing to prove. By (9.3.2),
Mi /Mi+1 is semisimple, and by (7.5.7), Mi /Mi+1 is Artinian. [Note that submodules of
Mi /Mi+1 are the same, whether we use scalars from R or from R/J; see Section 4.2,
Problem 6.] By Section 9.6, Problem 5, Mi /Mi+1 is Noetherian, hence has a composition
series by (7.5.12). Now intuitively, we can combine the composition series for the Mi /Mi+1
to produce a composition series for M , proving that M is Noetherian. Formally, Mn’1 ∼ =
Mn’1 /Mn has a composition series. Since Mn’2 /Mn’1 has a composition series, so does
Mn’2 , by (7.5.7). Iterate this process until we reach M . ™
We now proceed to a result that has many applications in both commutative and
noncommutative algebra.

9.8.2 Nakayama™s Lemma, Version 1
Let M be a ¬nitely generated R-module, and I a two-sided ideal of R. If I ⊆ J(R) and
IM = M , then M = 0.
Proof. Assume M = 0, and let x1 , . . . , xn generate M , where n is as small as possible.
m
(Then n ≥ 1 and the xi are nonzero.) Since xn ∈ M = IM , we can write xn = i=1 bi yi
for some bi ∈ I and yi ∈ M . But yi can be expressed in terms of the generators as
n
yi = j=1 aij xj with aij ∈ R. Thus
n
xn = bi aij xj = cj xj
i,j j=1

m
where cj = i=1 bi aij . Since I is a right ideal, cj ∈ I ⊆ J(R). (We need I to be a left
ideal to make IM a legal submodule of M .) The above equation can be written as
n’1
(1 ’ cn )xn = cj xj
j=1

and by (9.7.7), 1 ’ cn is invertible. If n > 1, then xn is a linear combination of the other
xi ™s, contradicting the minimality of n. Thus n = 1, in which case (1 ’ c1 )x1 = 0, so
x1 = 0, again a contradiction. ™
There is another version of Nakayama™s lemma, which we prove after a preliminary
result.
26 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

9.8.3 Lemma
Let N be a submodule of the R-module M , I a left ideal of R. Then M = N + IM if and
only if M/N = I(M/N ).

Proof. Assume M = N + IM , and let x + N ∈ M/N . Then x = y + z for some y ∈ N
t
and z ∈ IM . Write z = i=1 ai wi , ai ∈ I, wi ∈ M . It follows that

x + N = a1 (w1 + N ) + · · · + at (wt + N ) ∈ I(M/N ).

Conversely, assume M/N = I(M/N ), and let x ∈ M . Then
t
x+N = ai (wi + N )
i=1

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