Let $A$ be a local ring with maximal ideal $m$. Suppose that there exists some positive integer $k$ such that $m^k = m^{k+1}$.


Is necessarily $m^k = 0$ ?


If $m$ is finitely generated, this follows from Nakayama's lemma.



Can powers of a maximal ideal stabilize without vanishing?
The first thing that one might try seems to be a counter-example to that:


$R=k[x\_1,x\_2,\dots]$


$\mathfrak m={(x\_1,x\_2,\dots)}, I=(x\_1-x\_2^2,x\_2-x\_3^2,x\_3-x\_4^2,\dots)\subset R$.


$A=(R/I)\_{\mathfrak m}, m=\mathfrak m\_{\mathfrak m}$. 


