Let $R$ be a principal ideal domain and suppose $I_1,I_2,....$ are ideals of $R$ with $$ I_1 \subseteq I_2 \subseteq I_3 \subseteq ....$$ The Question has two parts: 1. to show that $\cup _{i=0}^{\infty}I_i$ is an ideal. 2. to show that any ascending as above must stabilize, i.e. there is a positive integer $n$ with $I_n=I_{n+1}=...$
My problem is with the second question. I tried to assume for contradiction that for every positive integer $n$, we have $I_n \subsetneq I_{n+1}$ which mean that there is a number $x\in I_{n+1}$ which is not in $I_n$. Since we are in a PID, we can write $I_n = (d), \quad I_{n+1}=(e)$ ( where $d,e$ are the generators). I also got that $d \nmid x$, and I tried to write $\gcd(x,d)$ as a linear combination of them... The closest relationship between $x$ and $d$ that I have is $x(1-s)=r\cdot d\cdot n$, where $r,s$ came from $\gcd(x,d)=sx+rd$, and $n$ came from $\gcd(x,d) \cdot n=x$
Is it possible that it is related to the fact that we have unique factorization in PID?
I have ran out of ideas...
Any hint will be helpful!
Thank you.
As $R$ is a PID, $\cup _{i=0}^{\infty}I_i = (x)$ for some $x\in R$. Now we have that $x \in \cup _{i=0}^{\infty}I_i $, so there is some $n \in \mathbb{N}$ such that $x \in I_n$. But then $I_n \supset (x)$ and $I_n \subset \cup _{i=0}^{\infty}I_i = (x)$, so $I_n = (x) = \cup _{i=0}^{\infty}I_i$. Thus $I_n=I_{n+1}= \dots$