Page 160 - Advanced Linear Algebra
P. 160
144 Advanced Linear Algebra
9
is an ideal of and so 0 ~ º ² ³ » for some : . We show that
8 ~¸ Á ² ³£ ¹
is a basis for . First, suppose that
:
bÄb ~
where for . Applying this to gives
³ ~
²
9
8
and since is an integral domain, ~ . Similarly, ~ for all and so is
linearly independent.
:
8
To show that spans , since any : has finite support, there is a largest
index for which ² ~ ² ³ ³ £ . Now, if 8 , then since is well-
º º » » :
ordered, we may choose a : ± ºº »» for which ~ ~ ² ³ is as small as
8
, it follows that
possible. Then 4 . Moreover, since £ ² ³ 0
² ³ £ and ² ³ ~ ² ³ for some 9. Then
supp² c ³ ´ Á µ
and
² c ³²³ ~ ²³ c ²³ ~
8
and so ² c ³ , which implies that c ºº »» . But then
~ ² c ³ b ºº »»
8
a contradiction. Thus, is a basis for .
:
8
In a vector space of dimension , any set of linearly independent vectors is a
basis. This fails for modules. For example, is a -module of rank but the
{
{
independent set ¸ ¹ is not a basis. On the other hand, the fact that a spanning set
of size is a basis does hold for modules over a principal ideal domain, as we
now show.
Theorem 6.6 Let 4 be a free -module of finite rank , where is a principal
9
9
ideal domain. Let : ~ ¸ ÁÃÁ ¹ be a spanning set for 4 . Then is a basis
:
for 4 .
8
Proof. Let ~¸ Á Ã Á ¹ be a basis for 4 and define the map ¢ 4 ¦ 4 by
and extending to a surjective 9-homomorphism. Since 4 is free,
~
Theorem 5.6 implies that
4 ker ²³ ^ im ²³ ~ ker ²³ ^ 4
Since ker²³ is a submodule of the free module and since is a principal ideal
9
domain, we know that ker²³ is free of rank at most . It follows that
