# Schur's Lemma Implies Functions

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.

Please clarify the following:

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 Preview

Please see the attachment.

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 ...

#### Solution Summary

The expert examines schur's lemma implying functions.