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 !


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 $I = (α^n) = m^n$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.