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Let $R$ be a local ring such that the only maximal ideal $m$ is principal and $\bigcap_{n\in\mathbb{N}}m^{n}=\lbrace 0\rbrace$. I would like to prove that any ideal $I\neq\lbrace 0\rbrace$ of $R$ is a power of $m$.

This is an exercise that had two parts. The first one was to prove that $R$ is Noetherian, which i did. Maybe its obvious to do the second part, but i couldn't. Can you help me ?

Thank you !

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Let $α$ be the generator of $m$, i.e. $m = (α)$.

Hint: First try to prove for any nonzero nonunit $x ∈ R$ that $(x) = (α^n)$ for some $n ∈ ℕ$, then use the fact that any ideal $I ≠ 0$ is finitely generated to conclude what you want to show.

I’ve already done that, but now realized you maybe wanted to do this yourself. But I will leave below what I already did, in case you want to take a peek.


Let $x ∈ R$ be nonzero nonunit, i.e. $(x) ≠ R$ and $(x) ≠ 0$. Then there’s a maximal $n ∈ ℕ$ such that $x ∈ m^n$ (because $x$ has to lie in the only maximal ideal $m$ at least and $\bigcap_{n ∈ ℕ} m^n = 0$). Write $x = rα^n$. Now $r$ cannot be in $m$, or else $n$ wouldn’t be maximal. Therefore $r ∈ R\setminus m = R^×$ and so $(x) = (α^n)$.

Since $R$ is noetherian, you can write any nontrivial ideal $I ≠ 0$ using nonzero generators $x_1, …, x_s ∈ R$ as $I = (x_1,…,x_s)$. Do the above argument for the generators: $I = (α^{n_1}, …, α^{n_s})$. Take $n = \min \{n_1, …, n_s\}$, then $I = (α^n) = m^n$.

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  • $\begingroup$ thank you for your clear answer :) $\endgroup$ – thetruth Mar 8 '14 at 22:39

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