TITLE: The maximal ideal in a local ring is finitely generated
QUESTION [6 upvotes]: Assume $m<R$ is the maximal ideal of a commutative local ring with identity, such that $m=m^2$. Is $m$ finitely generated? Is the condition $m=m^2$ redundant?
I am trying to apply Nakayama's lemma to the maximal ideal $m$, but I can't choose a finite generating system for it.

REPLY [4 votes]: Georges has fully answered the question, but I want to point out another example familiar from number theory. Let $\mathbf C_p$ denotes the $p$-adic complex numbers (i.e. the completion of the algebraic closure of $\mathbf Q_p$). Its ring of integers $\mathcal O_{\mathbf C_p}$ is a local ring with maximal ideal $\mathfrak m$ which satisfies $\mathfrak m = \mathfrak m^2$ (because every element of $\mathfrak m$ has a square root in the algebraically closed field $\mathbf C_p$, which necessarily also lies in $\mathfrak m$). On the other hand, $\mathfrak m$ is not finitely generated, because it contains elements of arbitrarily small absolute value.