# In a PID, the intersection of descending chain of ideals is trivial

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?

• An example where it doesn’t work for a UFD is $\mathbb{Z}[x]$,m with $I_n = (2,x^n)$. The intersection is $(2)\neq 0$. – Arturo Magidin Mar 13 at 16:46
• Hint: what can a generator of the intersection of these ideals be? – Mindlack Mar 13 at 16:48
• (You should also require proper inclusion in your chain, surely....) – Arturo Magidin Mar 13 at 16:50
• A generator of the intersection of these ideals is the least common multiple of the generators of each ideal in the chain. – wwinters57 Mar 13 at 16:55

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 $$a=0$$.

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.)