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.
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3$\begingroup$ If $A$ is the integers in an algebraically closed valued field (for example an algebraic closure of the p-adic numbers or any discrete valued field) then $A$ is local and $m=m^2$. $\endgroup$– ericCommented Nov 13, 2015 at 21:37
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2 Answers
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The first thing that one might try seems to be a counter-example to that:
$R=k[x_1,x_2,\dots]$
$\mathfrak m={(x_1,x_2,\dots)}, I=(x_1-x_2^2,x_2-x_3^2,x_3-x_4^2,\dots)\subset R$.
$A=(R/I)_{\mathfrak m}, m=\mathfrak m_{\mathfrak m}$.
answered Nov 13, 2015 at 20:27
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No: $A = k[[T^{(1/n)}: n \in$ ℕ$]]$
answered Nov 13, 2015 at 20:24