An $A-$module $M$ is called a uniform module if the intersection of any two nonzero submodules of $M$ is nonzero.
I don't know how to relate $A$ being a pid with $_AA$ being uniform. Any hint, please?
2026-04-01 21:01:45.1775077305
If $A$ is a principal ideal domain, then $_A A$ is uniform
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Any integral domain is uniform as a module over itself. Uniformity has nothing to do with being a PID.
Very late edit: I'm assuming the rings under consideration are commutative. A noncommutative domain (i.e., no zero divisors save $0$) can fail to be uniform as a module over itself. One example is a free algebra in at least two variables. In this case, $A$ being left Noetherian (for example, being a left principal ideal domain) does help and implies ${}_AA$ is uniform.