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    Schur's Lemma Implies Functions

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    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.

    © BrainMass Inc. brainmass.com December 24, 2021, 7:55 pm ad1c9bdddf
    https://brainmass.com/math/linear-transformation/schurs-lemma-implies-functions-229856

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    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!

    © BrainMass Inc. brainmass.com December 24, 2021, 7:55 pm ad1c9bdddf>
    https://brainmass.com/math/linear-transformation/schurs-lemma-implies-functions-229856

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