Page 143 - Advanced Linear Algebra

Chapter 5

Modules II: Free and Noetherian Modules

The Rank of a Free Module

Since all bases for a vector space have the same cardinality, the concept of
=
vector space dimension is well-defined. A similar statement holds for free -
9
)
modules when the base ring is commutative but not otherwise .
(
Theorem 5.1 Let 4 be a free module over a commutative ring with identity.
9
)
1 Then any two bases of 4 have the same cardinality.
)
2 The cardinality of a spanning set is greater than or equal to that of a basis.
Proof. The plan is to find a vector space with the property that, for any basis
=
for 4 , there is a basis of the same cardinality for . Then we can appeal to the
=
corresponding result for vector spaces.
9
Let be a maximal ideal of , which exists by Theorem 0.23. Then ° 9 ? is a
?
field. Our first thought might be that 4 is a vector space over ° 9 ? , but that is
not the case. In fact, scalar multiplication using the field 9°? ,
?
² b ³# ~ #
is not even well-defined, since this would require that ?4 ~ ¸ ¹ . On the other
hand, we can fix precisely this problem by factoring out the submodule
? ?4~ ¸ # b Ä b # “  Á #  4¹

?
Indeed, 4° 4 is a vector space over 9° ? , with scalar multiplication defined
by
?
?
² b³²" b4³ ~ " b4
?
To see that this is well-defined, we must show that the conditions
b ? ~ b ? Z
Z
"b 4 ~ " b 4
?
?
imply