I am trying to understand the proof to the following theorem:

Let $R$ be a principal ideal domain and let $I_{1}, I_{2}, ...$ be ideals in $R$ such that $I_{1} \subset I_{2} \subset \cdots$. Then there exists an integer $N$ such that $I_{n} = I_{N}$ for all $n \geq N$.

Here is the proof I typed a while ago:

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In the stablizing part of the proof, I don't understand why I let $x\in I_{N}$ then suddenly let $(x)=I_{N}$. I tried writing it again and got:

If $\bigcup I_{i}=(x)$, then $x\in I_{N}$ for some integer $N$ since $R$ is a principal ideal domain. Then for all $n\geq N$, $I_{N}\subset I_{n}\subset (x)$. Since $x\in I_{N}$, then it must also be that $I_{n}\subset I_{N}$. Then $I_{n}=I_{N}$.

However, I could also be wrong with this approach.

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    $\begingroup$ Your rewrite seems good to me. $\endgroup$ – 伽罗瓦 Nov 4 '18 at 2:31
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    $\begingroup$ I'm not sure what your question is. Yes, since $x\in I_n$ then $\mathcal{J}=(x)\subset I_n$. Is there a point in the proof that confuses you? $\endgroup$ – WSL Nov 4 '18 at 2:34

It strikes me that most if not all of the individual assertions our OP numericalorange puts forth in his efforts to construct a proof that principal ideal domains are Noetherian are in fact correct, and enter into a coherent proof at some point, but that in one or two places the exposition of the pattern of logical inference could perhaps be a little more clear. For example, I found it a bit of a challenge to follow the statement,

"Since $x \in I_{N}$, then it must also be that $I_{n}\subset I_{N}$. Then $I_{n}=I_{N}.$"

Of course this is true, for the reason that $x \in I_N$ implies $(x) \subset I_N$ and then we have

$(x) \subset I_N \subset I_n \subset J = (x), \tag 0$

which then leads directly to $I_n = I_N$ for $n \ge N$; but I think explicitly inserting the intermediate steps lends clarity to the argument.

I think it is also worth pointing out that the result binds under the somewhat weaker hypothesis that $R$ is merely a principal ideal ring; that $R$ is a domain nowhere need be invoked in what follows.

Having said these things, here is the way I present my demonstration of this fact. In the following, I take as given that the union of a nested sequence of ideals is itself and ideal; this is a very well-known and oft-used result. Now we have

$i < j \Longrightarrow I_i \subset I_j, \tag 1$

since the $I_i$ are given to be a nested sequence of ideals. Then per assumption,

$J = \displaystyle \bigcup_i I_i \; \text{is an ideal}; \tag 2$

now since every ideal in $R$ is principal,

$\exists j \in R, \; J = (j); \tag 3$

which in the light of (2) implies

$\exists N \in \Bbb N, \; j \in I_N; \tag 4$


$(j) \subset I_N \subset J = (j) \Longrightarrow I_N = (j); \tag 5$


$n \ge N \Longrightarrow (j) \subset I_N \subset I_n \subset J = (j); \tag 6$


$n \ge N \Longrightarrow I_n = I_N = (j). \tag 7$

$OE\Delta$, what we sought to prove.

In closing, I would say that the two key ideas here are ensconced in (2) and (6), which as it were "traps" $I_N$ 'twixt $(j)$ and itself.

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    $\begingroup$ I appreciate your very clear and easy to understand answer. Thanks so much for taking the time in clarifying all the rough patches! $\endgroup$ – numericalorange Nov 4 '18 at 16:23
  • $\begingroup$ @numericalorange: thank you my friend, glad to help out. And thanks for the "acceptance"! $\endgroup$ – Robert Lewis Nov 4 '18 at 17:26

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