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I would like to know if my logic is sound.

We know that in every principal ideal domain, every ideal is multiplicatively generated.

Thus, for $a \in R$ we have: $aR = ${$ra: r \in R$}

Thus every ideal has a limit on size, right? So eventually our ascending chain of ideals will have to be the same for some ideal eventually?

Is that correct? It seems a bit too easy

user121615
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  • Well, the proof is easy: actually there are 3 equivalent ways to define noetherianity, one of which makes the solution of this question obvious. I would like to remark that $aR:{ra:r\in R}$ holds for all commutative rings with $1$, so I don't see how it makes your proof work. –  May 07 '15 at 14:01
  • The goal is to show that if We have a principal Ideal domain, it is Noetherian. By Noetherian, we mean, in a scending chain of ideals, eventually, we will reach a maximum ideal size. I don't know another way to go about doing that but by proving it using size. Can you give me a hint? – user121615 May 07 '15 at 14:06
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    Hint: Let $R$ be a PID. Consider $I_0\subseteq I_1\subseteq I_2\subseteq\cdots$

    Prove that $\bigcup_{n\in\mathbb{N}}I_n$ is an ideal that (obviously) contains all the $I_n$-s.

    But, $I=(a)$. Where does $a$ lie?

     –  May 07 '15 at 14:24
  • Or, have a look at @DietrichBurde's link. It is the same proof. ACCP and ACC are the same condition on PIDs –  May 07 '15 at 14:26
  • Oh wow, that is really simple. I feel like an idiot haha – user121615 May 07 '15 at 14:40
  • See also https://math.stackexchange.com/questions/2008433/prove-i-k-i-n-for-some-k-ge-n – Watson Nov 10 '16 at 19:51

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