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Suppose $R$ is PID and $I_1 \supseteq I_2 \supseteq \cdots$ is a descending chain of ideals in $R$. I would like to prove that $\bigcap^\infty_{n=1} I_n=(0)$.

Now, since $R$ is a PID, every ideal is principal, so each $I_n=(a_n)$ for some $a_n \in R$. So I need to show that $$\bigcap^\infty_{n=1} (a_n)=(0).$$ We have $(a_1) \supseteq (a_2) \supseteq \cdots$, so $a_i \mid a_{i+1}$ for all $i$.
I am not sure what to do next.

Also, I don't think this is true if $R$ is just a UFD. What would an example of that be?

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  • $\begingroup$ An example where it doesn’t work for a UFD is $\mathbb{Z}[x]$,m with $I_n = (2,x^n)$. The intersection is $(2)\neq 0$. $\endgroup$ – Arturo Magidin Mar 13 at 16:46
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    $\begingroup$ Hint: what can a generator of the intersection of these ideals be? $\endgroup$ – Mindlack Mar 13 at 16:48
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    $\begingroup$ (You should also require proper inclusion in your chain, surely....) $\endgroup$ – Arturo Magidin Mar 13 at 16:50
  • $\begingroup$ A generator of the intersection of these ideals is the least common multiple of the generators of each ideal in the chain. $\endgroup$ – wwinters57 Mar 13 at 16:55
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Assuming strict descent (as otherwise any constant sequence is a counter-example). Let $a$ be the generator of the intersection. Then, $(a) \subset (a_i) \, \forall i$ and hence $a_i | a \, \forall i$. But since the descent is strict, none of the $a_i$'s are associates. Also $a_i | a_{i+1}$ so $\exists p_i$, a prime that divides $a_{i+1}$ but not $a_i$ for each $i$. That gives us infinitely many prime divisors for $a$, forcing $a=0$.

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You are correct that UFD does not suffice: you can have an infinite strictly decreasing chain of ideals whose intersection is not trivial. In $\mathbb{Z}[x]$, the ideals $I_n=(2,x^n)$ satisfy $I_{n+1}\subsetneq I_n$, but $\cap I_n = (2)$. But you can still leverage the UFD property to get what you want in the PID.

Note that in a UFD, if the principal ideals $(a)$ and $(b)$ satisfy $(a)\subseteq (b)$, then $b|a$; and if $(a)\subsetneq (b)$, then $b$ is a proper divisor of $a$.

Proposition. Let $R$ be a UFD, and let $(a_1)\supseteq (a_2)\supseteq\cdots\supseteq (a_n)\supseteq\cdots$ be a chain of principal ideals in $R$. If $a\neq 0$ lies in $\cap (a_k)$, then there exists $k$ such that $(a_k)=(a_{k+r})$ for all $r\geq 0$; that is, the chain stabilizes.

Proof. If $a$ is a unit, then the intersection is $R$, so each ideal is $R$ and they are all equal. So we may assume $a$ is not a unit. Since $a\neq 0$, then it has a factorization into irreducibles, $$a = p_1\cdots p_r.$$ Since $(a)\subseteq (a_n)$, then $a_n$ is a divisor of $a$. But $a$ has only finitely many proper divisors up to associates, so there are finitely many principal ideals $I$ such that $(a)\subsetneq I$. Thus, at some point, the chain $(a_i)\supseteq (a_{i+1})\supseteq\cdots$ must stabilize. $\Box$

Thus, in your PID, either your descending chain stabilizes, or else the intersection does not contain nonzero elements.

(Intuitively, the intersection is a least common multiple; but if at each step you are adding an irreducible factor, then the least common multiple would have an “infinite” factorization into irreducibles, which is impossible.)

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