Exercises in Classical Ring Theory, Second Edition (Problem by T.Y. Lam

By T.Y. Lam

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First assume R is semisimple, and let S1 , . . , Sr be a complete set of simple left R-modules. Then M = M1 ⊕ · · · ⊕ Mr , where Mi is the sum of all submodules of M which are isomorphic to Si . Since M is finitely generated, Mi ∼ = ni Si for suitable integers ni . 20. By Schur’s Lemma, all EndR Si are division rings, so EndR M is semisimple. If, in fact, R is simple artinian, we have r = 1 in the calculation above. Therefore, EndR M ∼ = Mn1 (EndR S1 ) is also a simple artinian ring. Comment. In the case where R M is not finitely generated, E is no longer semisimple.

After a reindexing, we may assume that I = R1 ⊕ · · · ⊕ Rs for some s ≤ r. Therefore, R/I ∼ = Rs+1 × · · · × Rr is a semisimple ring. (c) If S ∼ = Ri , we can find a surjective ring homomorphism from R to S by utilizing the ith projection of R = R1 × · · · × Rr . Conversely, suppose ϕ : R → S is a surjective ring homomorphism. After a reindexing, we may assume that ker(ϕ) = R1 × · · · × Rr−1 . ∼ Rr . ∼ R/ ker(ϕ) = Therefore, S = Comment. The solution to (c) shows that it remains true if we replace the words “simple artinian” by “indecomposable”.

Alternative Solution. Define β : R → R by β a b c d a b/n = nc . d A direct calculation shows that β is an involution on R. An involution γ on S can be defined similarly. Now consider the transpose map t : R → S, β t which is an anti-isomorphism. By composing R → R → S, we obtain an isomorphism α : R → S given by α a b c d = a nc b/n . d Comment. Here, (R, δ), (S, δ) are isomorphic as rings with involutions (with isomorphism defined by α), and (R, β), (S, γ) are isomorphic as rings with involutions (with isomorphism defined by α ).

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