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Every principal ideal domain $D$ satisfies the ACCP.

Proof. Let $(a_1) ⊆ (a_2) ⊆ (a_3) ⊆ · · ·$ be a chain of principal ideals in $D$. It can be easily verified that $I = \displaystyle{∪_{i∈N} (a_i)}$ is an ideal of $D$. Since $D$ is a PID, there exists an element $a ∈ D$ such that $ I = (a)$. Hence, $a ∈ (a_n)$ for some positive integer $n$. Then $I ⊆ (a_n) ⊆ I$. Therefore, $I = a_n$. For $t ≥ n$, $(a_t) ⊆ I = (a_n) ⊆ (a_t)$. Thus, $(a_n) = (a_t)$ for all $t ≥ n$.


I have prove $I$ is an ideal in the following way:-

Let $ x,y\in I$. Then there exist $i,j \in \mathbb{N}$ s.t. $x \in (a_i)$ & $y \in (a_j)$.
Let $k \in \mathbb{N}$ s.t $k>i,j$.
Then $x \in (a_k)$ & $y \in (a_k)$.
as $(a_k)$ is an ideal $x-y \in (a_k)\subset I$ and $rx,xr \in (a_k)\subset I$.
So $I$ is an ideal.

Is it correct?

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Your proof is right but you can let t = max(i,j) and any k > t.

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Yes. Said more simply $\ (a_1) \subseteq (a_2) \subseteq \cdots\subseteq (a_1,a_2,a_3,...)\stackrel{\rm PID} = (c_1 a_1 +\cdots + c_k a_k) \subseteq (a_k)$

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    $\begingroup$ It would not hurt to turn your decorated equal sign into an actual explanation of what you mean, as it is quite not transparent... Once you do that, I would be surprised to find anything different from what the OP wrote! $\endgroup$ – Mariano Suárez-Álvarez Jun 2 '13 at 17:36
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    $\begingroup$ The key idea of the proof is hidden in the annotation of the equal sign, rather... $\endgroup$ – Mariano Suárez-Álvarez Jun 2 '13 at 17:47
  • $\begingroup$ @Mariano The equal sign denotes an equality of ideals, deduce by the PID hypothesis. What is not transparent about that? $\endgroup$ – Key Ideas Jun 2 '13 at 17:52
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    $\begingroup$ I don't think it is constructive to continue this. As I said above, I do not find your «more simple proof» to be a complete argument: explaining what exactly the annotation means requires something equivalent to what the OP wrote (in a context where the OP is actually asking why exactly is the union of his ideals an ideal!). At best, I find what you wrote as a stenograph for a complete argument. You apparently do not agree, but that is irrelevant to my point. $\endgroup$ – Mariano Suárez-Álvarez Jun 2 '13 at 17:55
  • $\begingroup$ @Mariano That's precisely what it is intended to be, a one-line summary of the key idea of the proof (as are many of my hints). Note: I meant to write "said more simply...". I've added the "said" to avoid any further confusion. $\endgroup$ – Key Ideas Jun 2 '13 at 17:58

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