Let R be a commutative finite local ring which is not a field . Let \(\displaystyle M\) be the maximal ideal of R . Can we conclude that \(\displaystyle M\neq M^2\) ?

Let R be a commutative finite local ring which is not a field . Let \(\displaystyle M\) be the maximal ideal of R . Can we conclude that \(\displaystyle M\neq M^2\) ?

yes. \(\displaystyle M\) is a finitely generated \(\displaystyle R\) module. so, by Nakayama's lemma, if \(\displaystyle IM=M\) with the ideal \(\displaystyle I \subseteq J(R)=M,\) then \(\displaystyle M=\{0\},\) i.e. \(\displaystyle R\) must be a field. contradiction!