Page 140 - Advanced Linear Algebra
P. 140
124 Advanced Linear Algebra
then / is an isomorphism. Thus, a map O¢ / 4 ¢ 4 ¦ 4 is a right
c
inverse of if and only if is a range-extension of ²O ³ ¢ 4 / , the only
/
c
difference being in the ranges of the two functions. Hence, ²O ³ ¢ 4 ¦ 4
/
is the only right inverse of with image . It follows that the correspondence
/
c
/ª ² O ³ ¢ 4 ¦ 4
/
is an injection from the complements / of ker ² ³ to the right inverses of .
is a right inverse of ,
¢4 ¦ 4
Moreover, this map is a bijection, since if 9
c
then ¢4 im 9 9 ³ and is an extension of 9 ² 9 ¢ im ³ 4 , which
²
implies that
4~ im ² 9 ³ l ker ² ³
Theorem 4.16 Let 4 and 4 be -modules and let ¢ 4 ¦ 4 be an -map.
9
9
1 Let Ƣ4 4 be injective. The map
)
³ with kernel
/ª extension of ² O im²³ c /
is a bijection from the complements of im / ² ³ onto the left inverses of .
Thus, there is exactly one left inverse of for each complement of im²³
and that complement is the kernel of the left inverse.
) be surjective. The map
2 Let ¢4 ¦ 4
c
/ª ² O ³ ¢ 4 ¦ 4
/
is a bijection from the complements of ker / ² ³ to the right inverses of .
Thus, there is exactly one right inverse of for each complement / of
ker²³ and that complement is the image of the right inverse. Thus,
4~ ker ² ³ l / ker ² ³ ^ im ² ³
The last part of the previous theorem is worth further comment. Recall that if
¢= ¦ > is a linear transformation on vector spaces, then
= ker ²³ ^ im ²³
This holds for modules as well provided that ker²³ is a direct summand .
Modules Are Not as Nice as Vector Spaces
Here is a list of some of the properties of modules over commutative rings with
(
)
identity that emphasize the differences between modules and vector spaces.
)
1 A submodule of a module need not have a complement.
2 A submodule of a finitely generated module need not be finitely generated.
)
)
3 There exist modules with no linearly independent elements and hence with
no basis.
4 A minimal spanning set or maximal linearly independent set is not
)
necessarily a basis.
