Prove $I_k=I_N$ for some $k \ge N$.

Question

Let $ I_0 \subseteq I_1 \subseteq I_2 \subseteq I_3...$ be a sequence of ideals of a integral domain, $R$. Let $R$ also be a principal ideal domain. In other words for every $k \in \mathbb{N}$ assume $I_k$ is an ideal of $R$ and that $I_k \subseteq I_{k+1}$ for every $k \ge 0$. Prove there exists a natural number, $N$ such that $I_k=I_N$ for some $k \ge N$.

I can prove that the union of the sequence of ideals is also an ideal of $R$. I am having trouble formulating the proof for the rest of the statement.

I have been given the hint: Show $I_N=I$ for some $N \in \mathbb{N}$, because I is a PID, but do not know how or where to use it.

I think the easiest way to make this obvious is the proof here.. I can elaborate on that if it helps. — Bill Dubuque, Nov 10 '16 at 20:00

score 1 · Answer 1 · answered Nov 10 '16 at 19:39

As you already remarked the union of these ideals is an ideal. Since $R$ is a PID this ideal is generated by an element $x \in \bigcup_{k} I_k$, i.e there is a integer $N$ with $x \in I_N$. By definition of $x, \langle x \rangle = \bigcup_{k} I_k$ so in particular $I_N = I_{N+1} = \dots = I$.

Prove $I_k=I_N$ for some $k \ge N$.

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