D.C.C condition of principal ideals on UFD

Question

The question is: Let $R$ be a UFD, and $I \neq (0)$ be an ideal of $R$. Prove that every descending chain of principal ideals containing $I$ must stabilize.

Since for an UFD, A.C.C holds for principal ideals and A.C.C implies D.C.C iff prime ideals are maximal. I have tried by saying if $a,b$ are irreducible then

1. $(a)$, $(b)$ are prime (by definition of an UFD)
2. $gcd(a,b)=1$, that is $(a,b)$ is maximal

I don't know where to proceed from here. Does anyone have a hint?

Answer 1

You're thinking about this in entirely the wrong direction. Don't think about general abstract facts about ideals; instead think very concretely in terms of the unique factorizations of elements. If $(a)$ and $(b)$ are principle ideals, then $(a)\subseteq (b)$ iff $b$ divides $a$. What does this tell you about their factorizations? Now consider what a descending chain of principal ideals means in terms of the factorizations of the generators.

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If $(a)\subset (b)$, then this means $a$ has strictly more irreducible factors (counted with multiplicity) than $b$. So if $(a_0)\supseteq (a_1)\supseteq (a_2)\supseteq\dots$ is a descending sequence of principal ideals that does not stabilize, the numbers of irreducible factors of the $a_n$ must be unbounded (since they increase infinitely many times). So no nonzero element of $R$ can be contained in $(a_n)$ for all $n$, since such an element would have to have infinitely many irreducible factors.