Bourbaki

Remark 4

Tag 074K, permanent. Cite it as [Bourbaki, Tag 074K].

Algebra, VII, p. 25, § 4. Finitely generated modules over a principal ideal domain

An AA-module is called simple (II, p. 37) if it is nonzero and has no submodules other than itself and 00; it is then necessarily cyclic, so finitely generated, and indecomposable; since the modules A/(pn)A/(p^n) are not simple for n≠1n \neq 1, while the modules A/(p)A/(p) are, and since AA is simple only if the ring AA is a field, we deduce that the simple AA-modules are: a) free modules of rank 1, when AA is a field; b) modules isomorphic to quotients A/(p)A/(p), where pp is an irreducible element of AA, when AA is not a field.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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Transcribed from the printed edition, 2003, Springer. Not translated.