# Discrete Valuation Rings - Atiyah & MacDonald

The following is claimed (without much proof) during the the proof of Prop 9.2 in Atiyah & MacDonald. Saurabh commented below giving the proof that was probably intended by A&M (thank you!). I have now come up with a different proof and would be very grateful if someone would tell me whether it works.

Many thanks!

Lemma. Suppose $(A,\mathfrak{m})$ is a local Noetherian domain of dimension one. Any proper, non-zero ideal $\mathfrak{a}$ of $A$ is $\mathfrak{m}$-primary.

Proof. Since $A$ is local and every non-zero prime ideal of $A$ is maximal, it follows that $\mathfrak{m}$ is the only non-zero prime ideal of $A$. Hence the quotient ring $A/\mathfrak{a}$ has only one prime ideal and, as such, it is a local Artinian ring (since the quotient $A/\mathfrak{a}$ is also Noetherian and "Artinian = Noetherian of dimension zero"). Since the maximal ideal of a local Artinian ring is nilpotent, it follows that $\mathfrak{m}^n\subseteq\mathfrak{a}$ for some $n \in \mathbb{N}$. We also have $\mathfrak{a}\subseteq\mathfrak{m}$ and so, by taking radicals, we see that $\sqrt{\mathfrak{a}}=\mathfrak{m}.$ Hence $\mathfrak{a}$ is indeed $\mathfrak{m}$-primary. //

• A radical ideal of any ideal is the intersection of prime ideals containing it. As the ring is local Noetherian of dimension 1, there is only one prime ideal containing any non-zero ideal, namely the unique maximal ideal.
– SMG
Commented Dec 2, 2014 at 16:36