Principal Ideal Domain fulfills Ascending Chain Condition

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Theorem

Let $R$ be a principal ideal domain.

Then $R$ fulfils the ascending chain condition.


Proof

Let $I_1 \subseteq I_2 \subseteq I_3 \subseteq \dotsb$ be an ascending chain of ideals.

Build $\ds I = \bigcup_{i \mathop = 1}^\infty I_i$.

$I$ is an ideal.

Since $R$ is a principal ideal domain, $I = \ideal a$ for some $a \in R$.

Now, since $a \in I$, there is some $n$ such that $a \in I_n$.

Thus:

$\ideal a \subseteq I_n$

By definition $I_n \subset I = \ideal a$, and so $I_n = I$.

Thus:

$\forall m \ge n: I_m = I$




$\blacksquare$