A Generalization of Rickart Modules

References (28)

1. then Tr(M, r M (f n )) is a direct summand of M .
2. If M is a flat left S-module and S is a generalized right principally projective ring, then M is π-Rickart as an R-module.
3. Tr(M, r M (f n )) = eM . Since f n eM = 0, eM ≤ Tr(M, r M (f n )). Let g ∈ Hom(M, r M (f n )). so eg = g. It follows that gM ≤ egM ≤ eM or Hom(M, r M (f
4. M ≤ eM . (2) Assume that M is a flat left S-module and S is a generalized right principally projective ring. If f ∈ S, then f n S is a projective right S- module, since r S (f n ) = eS for some positive integer n and e 2 = e ∈ S. As in the proof of (1), we have Tr(M, r M (f n )) = eM . Since M is a flat left S-module and f n ∈ S, r M (f n ) is M -generated by [20, 15.9]. Again by [20, 13.5(2)], Tr(M, r M (f
5. module M , it is shown that if S is a von Neumann regular ring, then M is a Rickart module (see [12, Theorem
6. We obtain a similar result for π-Rickart modules. Lemma 3.9. Let M be a module. If S is a π-regular ring, then M is a π-Rickart module.
7. Proof. Let f ∈ S. Since S is π-regular, there exist a positive integer n and an element g in S such that f n = f n gf n . Then gf n is an idempotent of S. Now we show that r M (f n ) = (1 -gf n )M . For m ∈ M , we have f n (1 -gf n )m = (f n -f n gf n )m = (f n -f n )m = 0. Hence (1 -gf n )M ≤ r M (f n ). For the other side, if m ∈ r M (f n ), then gf n m = 0. This implies that m = (1 -gf n )m ∈ (1 -gf n )M . Therefore r M (f n ) = (1 -gf n )M . Now we recall some known facts that will be needed about π-regular rings. Lemma 3.10. Let R be a ring. Then (1) If R is π-regular, then eRe is also π-regular for any e 2 = e ∈ R. (2) If M n (R) is π-regular for any positive integer n, then so is R. References
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