Let $(A,\mathfrak{m})$ be a local Noetherian domain of dimension one and suppose that $\mathfrak{m}$ is principal. I wish to show that every non-zero ideal of $A$ is a power of $\mathfrak{m}$. I have tried to prove it as follows and I would be most grateful if someone would check the proof for me.
Proof: I make use of three facts:
In a local Noetherian domain of dimension one, every proper, non-zero ideal contains some power of the maximal ideal.
Artinian = Noetherian of dimension zero.
In a local Artinian ring with principal maximal ideal, every non-zero ideal is a power of the maximal ideal.
Let $\mathfrak{a}$ be a non-zero ideal of $A$. If $\mathfrak{a}=A$, then $\mathfrak{a}=\mathfrak{m}^0$ so we may assume that $\mathfrak{a}$ is proper. By fact 1, we have $\mathfrak{m}^n\subset\mathfrak{a}$ for some $n\geq 2$. Let $q:A\to A/\mathfrak{m}^n$ be the (surjective) quotient homomorphism and let $\mathfrak{M}=q(\mathfrak{m})$. Observe that $\mathfrak{m}^2\neq\mathfrak{m}$, otherwise we would have $\mathfrak{m}=(0)$ by Nakayama. As prime ideals of $A$ correspond exactly to prime ideals of $A/\mathfrak{m}^n$ containing $\mathfrak{m}^n$ we see that $A/\mathfrak{m}^n$ has a unique prime ideal, namely $\mathfrak{M}\neq 0$ (non-zero as $\mathfrak{m}^2\neq\mathfrak{m}$), and so it is a local ring (with maximal ideal $\mathfrak{M}\,$). By fact 2, it follows that $A/\mathfrak{m}^n$ is Artinian. Moreover, since $\mathfrak{m}$ is principal, so is $\mathfrak{M}$. Hence, by fact 3, we have that every ideal of $A/\mathfrak{m}^n$ is a power of $\mathfrak{M}$; in particular, $q(\mathfrak{a})=\mathfrak{M}^r$ for some $r \in \mathbb{N}$ and so $\mathfrak{a}=\mathfrak{m}^r$. Q.E.D.
Many thanks!
You are right: your proof is correct and helpful, (Atiyah-MacDonald is terse).