Page 145 - Advanced Linear Algebra

Modules II: Free and Noetherian Modules 129

?
and so  ? , which is not possible since is a maximal ideal. Hence,
.
~
Thus, if is a basis for 4 over , then
8
9
(( ~ 8 ( b 8 ? 4 ( ~ dim 4 ° ?² 9°? 4 ³
and so all bases for 4 over have the same cardinality, which proves part 1 .
)
9
Finally, if spans 4 over , then b 8 ? 4 spans 4 ° ? 4 and so
8
9
dim 9°? ²4° 4³ ? ( 8 b 4  8 ? ( ( (
Thus, has cardinality at least as great as that of any basis for 4 over .
9
8
The previous theorem allows us to define the rank of a free module. The term
(
dimension is not used for modules in general.)

9
Definition Let be a commutative ring with identity. The rank rk² 4 ³ of a
nonzero free -module 4 is the cardinality of any basis for 4 . The rank of the
9

trivial module ¸ ¹ is .
Theorem 5.1 fails if the underlying ring of scalars is not commutative. The next
example describes a module over a noncommutative ring that has the
remarkable property of possessing a basis of size for any positive integer .

Example 5.1 Let be a vector space over with a countably infinite basis
-
=
8 B ~ ¸ Á Á Ã ¹. Let ²= ³ be the ring of linear operators on . Observe that
=
B²= ³ is not commutative, since composition of functions is not commutative.
The ring B is an ²= ³ -module and as such, the identity map forms a basis

B²= ³
for B . However, we can also construct a basis for ²= ³ of any desired finite
B²= ³
size . To understand the idea, consider the case ~ and define the operators

and by
² ³ ~ Á ² b ³ ~

and
² ³ ~ Á ² b ³ ~

These operators are linearly independent essentially because they are surjective
and their supports are disjoint. In particular, if

b ~
then
~ ² b ³² ³ ~ ² ³