Page 137 - Advanced Linear Algebra
P. 137
Modules I: Basic Properties 121
Hence, any two nonzero proper submodules of have nonzero intersection, for
{
if £ , then
{
{ { q ~
where ~ lcm ¸ Á ¹ . It follows that the only complemented submodules of {
are and ¸ ¹ .
{
In the case of vector spaces, there is an intimate connection between subspaces
and quotient spaces, as we saw in Theorem 3.6. The problem we face in
generalizing this to modules is that not all submodules are complemented.
However, this is the only problem.
Theorem 4.13 Let be a complemented submodule of 4 . All complements of
:
: 4 are isomorphic to ° : and hence to each other.
Proof. For any complement of , the first isomorphism theorem applied to
:
;
the projection ;Á: ¢4 ¦ ; gives ; 4°: .
Direct Summands and Extensions of Isomorphisms
Direct summands play a role in questions relating to whether certain module
homomorphisms ¢5 ¦ 4 can be extended from a submodule 5 4 to the
full module 4 . The discussion will be a bit simpler if we restrict attention to
epimorphisms.
If 4~ 5 l / , then a module epimorphism ¢ 5¦ 4 can be extended to an
simply by sending the elements of to zero, that is,
/
epimorphism ¢4 ¦ 4
by setting
² b ³ ~
This is easily seen to be an -map with
9
ker²³ ~ ker²³ l /
Moreover, if is another extension of with the same kernel as , then and
agree on as well as on , whence ~ 5 . Thus, there is a unique extension of
/
with kernel ker²³ l / .
is an isomorphism. If is complemented, that is,
Now suppose that ¢5 4 5
if
.~ 5 l /
then we have seen that there is a unique extension of for which ker²³ ~ / .
Thus, the correspondence
,
/ª where ker ² ³ ~ /
from complements of 5 to extensions of is an injection. To see that this
correspondence is a bijection, if is an extension of , then
¢4 ¦ 4
