Let $R$ be a commutative ring and $p$ be a prime ideal which is not zero ideal. If $p^m=p^n$ for $m, n \in \mathbb{N}$, $m=n$?
Added: Sorry, what I really wanted to ask is the case $p^m=p^n\neq 0.$
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1Not if the ring is Artinian. For instance for $R = \mathbb{Z}/4\mathbb{Z}$ the prime ideal $p = (2)$ then $p^n = p^m$ for all $n,m \geq 2$. – Darth Geek May 05 '17 at 02:26
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No, not necessarily. For example, let $$R = \mathbb{Q}[x]/(x^2-x),\;p=(x),\;m = 1,\;n=2$$
quasi
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No. For example, there exists a ring with an idempotent maximal ideal:
$R=F\times F$ Where $F$ is the field of two elements. All powers of that prime ideal are equal.
It is not even true for domains: there is a domain with a nontrivial idempotent maximal ideal
It is also not true for local rings: there is a local ring with an idempotent nontrivial ideal: Take an indeterminate $x$, form the collection of powers of $x$ of the form $x^{1/2^n}$, generate an $F$ algebra with them, then take the quotient by $(x)$.
It is true for dedekind domains, I think, and possibly for a larger class, I don't know.
rschwieb
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