Local Artinian rings with a principal maximal ideal

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!

• It looks correct to me :) – Manos Dec 24 '14 at 2:07
• 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? – Youngsu Dec 24 '14 at 10:37
• Ah yes, thanks! No I didn't, as I hadn't heard of it until now - thank you again :) – user202978 Dec 24 '14 at 10:48