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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:

  1. In a local Noetherian domain of dimension one, every proper, non-zero ideal contains some power of the maximal ideal.

  2. Artinian = Noetherian of dimension zero.

  3. 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!

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    $\begingroup$ I am aware of this; however, when you are self-learning (Atiyah-MacDonald being quite terse) it is very useful to have people confirm whether you actually understand the proof (which is written above in my own words with my own extra explanations) $\endgroup$
    – user202978
    Dec 24, 2014 at 14:40
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    $\begingroup$ Thank you. (I understand that at some point one has to become self-sufficient with regard to these kind of things however at first it is very useful to have someone look over your work - I wouldn't say that I am just reproducing Atiyah-MacDonald verbatim.) $\endgroup$
    – user202978
    Dec 24, 2014 at 14:48

2 Answers 2

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You are right: your proof is correct and helpful, (Atiyah-MacDonald is terse).

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  • $\begingroup$ I am stuck trying to prove the fact 1. It would be easy if we show that the radical of every ideal under these hypothesis is prime, but how do we know it? Thank you. $\endgroup$
    – H. Jackson
    Oct 21, 2015 at 9:36
  • $\begingroup$ Let $I$ be an ideal of $A$ with $0\subset I \subset A$. The only prime ideals of $A$ are $M$ and $0$; hence $I$ is $M$-primary $\endgroup$
    – user 1
    Oct 21, 2015 at 10:29
  • $\begingroup$ A late comment, but @H.Jackson, here is how I would prove it: every ideal $I$ is contained in a maximal idea $M$. As $A$ is local, $M$ is the unique maximal ideal. Hence, noting that $\dim A=1$, (which includes $0$, for $A$ being a domain) $$nilrad(I)= \bigcap_{prime \supset I } P = \bigcap M = rad(I)=M$$ Note that $M$ is f.g. so $M^n = nilrad(I)^n \subseteq I$ (since for each generator $x_i$ exists some $n_i$ such that $x_i^{n_i} \in I$. $\endgroup$
    – Bryan Shih
    May 12, 2018 at 23:25
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In fact, you can prove it a bit simpler, notice that $A/\frak{m}^n$ is Artinian, local, and integral domain, therefore it's a field, there is a single proper ideal (0), therefore $a/\frak{m^n}$ is ideal in the field which has to be zero that is $a = \mathfrak{m}^n$

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