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Georges Elencwajg
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Mnemonic: $\quad M=IM \Rightarrow m=im$

The version of Nakayama described: If $I$ is an arbitrary ideal of an arbitrary ring $A$ and if a finitely generated module $M$ satisfies $M=IM$, then there exists $i\in I$ such that for all $m\in M$ we have $m=im$.
Please notice: no noetherian nor local assumption on $A$, no assumption at all on $I$.

Georges Elencwajg
  • 47.5k
  • 14
  • 159
  • 241