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 :) Commented Dec 24, 2014 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? Commented Dec 24, 2014 at 10:37
• Ah yes, thanks! No I didn't, as I hadn't heard of it until now - thank you again :) Commented Dec 24, 2014 at 10:48