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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. //

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  • $\begingroup$ 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. $\endgroup$ – SMG Dec 2 '14 at 16:36
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The unique maximal ideal is prime, and the zero ideal is prime. Every other proper ideal of the ring lies somewhere between. Since it is one dimensional, there are no prime ideals between these two.

Thus the radical of any proper nonzero ideal (the intersection of primes containing said ideal) is the maximal ideal.

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  • $\begingroup$ Ah yes! So I don't even need the Noetherian hypothesis? $\endgroup$ – user196382 Dec 2 '14 at 17:29
  • $\begingroup$ @user196382 To conclude that the radical of any nontrivial ideal is the maximal ideal, no. But to conclude that the nontrivial ideals are primary, I think you need Noetherianness. $\endgroup$ – rschwieb Dec 2 '14 at 17:43
  • $\begingroup$ I'm not sure that you do. If the radical of an ideal is maximal, then it must be primary - the proof of this doesn't require the Noetherian hypothesis, I don't think? $\endgroup$ – user196382 Dec 2 '14 at 18:04
  • $\begingroup$ @user196382 Ahh, hm, I do seem to recall something about having a maximal prime as a radical implies primaryness. I had been stuck remembering only that "prime radical does not always imply primary." Maybe it is ok without Noetherianness, in this direction. You'd better check the details $\endgroup$ – rschwieb Dec 2 '14 at 18:34

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