Exercise A0.V.2.08

Proposition

Let $R$ be a UFD and let $I$ be a non-zero ideal of $R.$ Then every descending chain of principal ideals containing $I$ stabilizes.

Proof

Consider the chain of principal ideals $(j_0) \supseteq (j_1) \supseteq (j_2) \supseteq \ldots$ such that $I \subseteq (j_n).$

In 2.01A1 we established that $(a) \subseteq (b)$ if and only if $\mathsf{F}(b) \subseteq \mathsf{F}(a),$ and that $(a) = (b)$ if and only if $\mathsf{F}(b) = \mathsf{F}(a).$

From 2.06 we know that GCD exists in UFD’s. In particular, there exists the smallest principal ideal containing $I.$ Let $d = \gcd(I).$ By minimality, any principal ideal containing $I$ must contain $(d).$ Thus, $\mathsf{F}(j_n) \subseteq \mathsf{F}(d)$ and moreover $\mathsf{F}(j_n) \subseteq \mathsf{F}(j_{n+1}).$

Suppose, for the sake of contradiction, that the chain fails to stabilize. Then we can select an infinite subchain where each inclusion is proper $(j_{\alpha_0}) \supsetneq (j_{\alpha_1}) \supsetneq (j_{\alpha_2}) \supsetneq \ldots \ .$ This gives us an infinite chain of proper inclusions of multisets of factors $\mathsf{F}(j_{\alpha_0}) \subsetneq \mathsf{F}(j_{\alpha_1}) \subsetneq \ldots \ .$ However, the sub-multiset lattice of $\mathsf{F}(d)$ is finite, since $d$ factors into a finite number of irreducibles, each with finite multiplicity, and cannot contain an infinite ascending chain. This is a contradiction.

$\blacksquare$