Abstract
A submodule N of a module M is idempotent if N = Hom (M,N)N. The module M is fully idempotent if every submodule of M is idempotent. We prove that over a commutative ring, cyclic idempotent submodules of any module are direct summands. Counterexamples are given to show that this result is not true in general. It is shown that over commutative Noetherian rings, the fully idempotent modules are precisely the semisimple modules. We also show that the commutative rings over which every module is fully idempotent are exactly the semisimple rings. Idempotent submodules of free modules are characterized.
| Original language | English |
|---|---|
| Pages (from-to) | 2707-2722 |
| Number of pages | 16 |
| Journal | Communications in Algebra |
| Volume | 39 |
| Issue number | 8 |
| DOIs | |
| Publication status | Published - Aug 2011 |
Keywords
- Fully idempotent module
- Idempotent submodule
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