I would be very grateful if someone would check my proof of the following result (this is not homework). All rings are commutative and unital.

Proposition: If $(A,\mathfrak{m})$ is a local Artinian ring and $\mathfrak{m}$ is principal, then every non-zero ideal of $A$ is a power of $\mathfrak{m}$.

Proof: I assume the following two facts:

  1. Artinian rings are Noetherian.

  2. The Jacobson radical of an Artinian ring is nilpotent.

By facts 1 and 2 combined, we see that $\mathfrak{m}$ is nilpotent. Hence, given a proper, non-zero ideal $\mathfrak{a}$ of $A$, there is some $r \geq 1$ such that $\mathfrak{a}\subseteq\mathfrak{m}^r$ (we have $\mathfrak{a}\subseteq \mathfrak{m}$ as $A$ is local) but $\mathfrak{a}\nsubseteq\mathfrak{m}^{r+1}.$ We will show that $\mathfrak{a}=\mathfrak{m}^r.$ Choose $y \in \mathfrak{a}$ such that $y \notin \mathfrak{m}^{r+1}.$ As $\mathfrak{m}$ is principal, we have $y=ax^r$ for some $a \in A$ and $x \in A$ such that $\mathfrak{m}=(x)$. But as $\mathfrak{a}\nsubseteq(x^{r+1})$ we have $a \notin (x)$ and so, as $A$ is local, we have that $a$ is a unit in $A$. Therefore $x^r=a^{-1}y \in \mathfrak{a}$ and so $\mathfrak{m}^r\subseteq\mathfrak{a}$. Q.E.D.

Many thanks!

  • $\begingroup$ It looks correct to me :) $\endgroup$ – Manos Dec 24 '14 at 2:07
  • $\begingroup$ Actually, you don't need "non zero" in the statement by the 2nd fact you use. By the way, did you try to prove it using the Cohen structure theorem? $\endgroup$ – Youngsu Dec 24 '14 at 10:37
  • $\begingroup$ Ah yes, thanks! No I didn't, as I hadn't heard of it until now - thank you again :) $\endgroup$ – user202978 Dec 24 '14 at 10:48

Alternatively, a ring in which every prime ideal is principal is a PIR, and in a PIR every proper ideal is a product of prime ideals. (But if you don't already have these facts, their proofs are more difficult than your direct proof.)


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.