# Proving Ascending Ideals Stabilize in PID

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:

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.

• Your rewrite seems good to me. – 伽罗瓦 Nov 4 '18 at 2:31
• 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? – 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

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

thus,

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

also,

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

therefore,

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

$$OE\Delta$$, 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.

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