Explore BrainMass

# Schur's Lemma Implies Functions

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

I have some trouble understanding the solution to the attached problem (solution included). Could you please provide some clarification of the solution. I have indicated what my points of concern are.

Show that if M1 and M2 are irreducible R modules, then any nonzero R-module homomorphism from
M1 to M2 is an isomorphism. Deduce that if M is irreducible then EndR(M) is a division ring (this result
is called Schur's Lemma). [Consider the kernel and the image.]
Let j : M1 ! M2 be a nonzero R-module homomorphism. Then ker j 6= M and Imj 6= 0. But ker j and
Imj are submodules of M1 and M2 respectively, so irreducibility implies ker j = 0 and Imj = M2, and
thus j is an isomorphism.
If M is irreducible then the above implies that any j 6= 0 in the ring EndR(M,M) has a multiplicative
inverse, so EndR(M,M) is a division ring. 
1. Why does it follow from 'Let j : M1 ! M2 be a nonzero R-module homomorphism.' that then
ker j 6= M and Imj 6= 0.
2. Why does irreducibility imply ker j = 0 and Imj = M2, and thus j is an isomorphism.
3. If M is irreducible then why does the above imply that j 6= 0 in the ring EndR(M,M) has a multiplicative
inverse, so EndR(M,M) is a division ring.

https://brainmass.com/math/linear-transformation/schurs-lemma-implies-functions-229856

## SOLUTION This solution is FREE courtesy of BrainMass!

Clarifications:
1. Since is a non-zero -homomorphism, then and . The reason is as follows.
If , then and thus is a zero -homomorphism. This is a contradiction to the condition that is a non-zero -homomorphism.
If , then and thus is a zero -homomorphism. This is a contradiction again.
Therefore, we must have and
2. Since is irreducible, then has only two submodules and . is a submodule of and from 1, then . This implies that is one-to-one.
Since is irreducible, then has only two submodules and . is a submodule of and from 1, then . This implies that is onto.
Therefore, is an isomorphism.
3. If is irreducible, and , from 1 and 2, we know that is an isomorphism. So has an inverse and its inverse is also an isomorphism. In another word, . Hence is a division ring.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!