Page 147 - Advanced Linear Algebra
P. 147
Modules II: Free and Noetherian Modules 131
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similar result holds for free -modules. We begin with the fact that 9 ² ) ³ is a
free -module. The simple proof is left to the reader.
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Theorem 5.2 Let be any set and let be a commutative ring with identity.
9
)
) of all functions from to that have finite support is a free -
The set ²9 ³ ) 9 9
8
)
module of rank (( with basis ~¸ ¹ where
% ~ if
²%³ ~ F
% £ if
)
This basis is referred to as the standard basis for ²9 ³ .
Theorem 5.3 Let 4 be an 9 -module. If ) is a basis for 4 , then 4 is
) .
isomorphic to ²9 ³
) defined by setting
Proof. Consider the map ¢4 ¦ ²9 ³
~
where is defined in Theorem 5.2 and extending to 4 by linearity. Since
maps a basis for 4 to a basis 8 ~ ¸ ¹ for ² 9 ) ³ , it follows that is an
isomorphism from 4 to 9 ² ) ³ .
Theorem 5.4 Two free -modules over a commutative ring are isomorphic if
(
)
9
and only if they have the same rank.
Proof. If 4 5 , then any isomorphism from 4 to maps a basis for 4 to
5
a basis for 5 . Since is a bijection, we have rk ² 4 ³ ~ rk ² 5 ³ . Conversely,
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8
suppose that rk²4³ ~ rk²5³ . Let be a basis for 4 and let be a basis for .
5
Since (( ~ 8 ( ( , there is a bijective map 8 ¦ 9 ¢ . This map can be extended by
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linearity to an isomorphism of 4 onto and so 4 5 .
5
We have seen that the cardinality of a minimal spanning set for a free module
(
)
4 ² is at least equal to rk 4 ³ . Let us now speak about the cardinality of maximal
linearly independent sets.
Theorem 5.5 Let be an integral domain and let 4 be a free -module. Then
9
9
all linearly independent sets have cardinality at most rk²4³ .
. Let be the
Proof. Since 4 ²9 ³ we need only prove the result for ²9 ³ 8
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field of quotients of . Then 8 ² ³ is a vector space. Now, if
8 ~¸# 0¹ ²9 ³ ²8 ³
is linearly independent over as a subset of ² 8 ³ , then is clearly linearly
8
8
9
8
independent over as a subset of 9 ² ³ . Conversely, suppose that is linearly
independent over and
9
#b Ä b # ~
