Exercise A0.V.2.06

Let $R$ be a domain with the property that the intersection of any family of principal ideals in $R$ is itself a principal ideal. In this case, we say that $R$ has property X.

Proposition A

Let $R$ be a domain with property X. Then $R$ has greatest common divisors.

Proof

Let $U$ be a family of elements of $R.$ Let $V$ be the family of all principal ideals of $R$ that contain the ideal $I = (u \mid u \in U).$ Clearly $(1) \in V,$ so $V$ is non-empty. By property X we know that the intersection $\bigcap V$ is itself a principal ideal. Let $(d) = \bigcap V.$ Since every member of $V$ is a superset of $I,$ $I \subseteq (d).$

Suppose $(d')$ is another principal ideal such that $I \subseteq (d').$ Then $(d') \in V$ and hence $(d) \subseteq (d').$ So $(d)$ is the smallest principal ideal that is a superset of $I.$ We conclude that $d$ is the greatest common divisor of the elements of $U.$

$\blacksquare$

Proposition B

Let $R$ be a UFD. Then $R$ has property X.

Proof

Let $V$ be a family of principal ideals of $R.$

Suppose that $(0) \in V.$ Then the intersection is the principal ideal $(0).$ We now assume that $(0) \notin V.$ ⚠️

For each $(v_\alpha) \in V$ we take the multiset of irreducible factors (modulo units) $\mathsf{F}(v_\alpha).$

Suppose that $\bigcup \mathsf{F}(v_\alpha)$ has an element $\rho$ of infinite multiplicity. In this case the intersection $\bigcap V$ is the principal ideal $(0),$ since any non-zero element $t$ of $R$ has only a finite multiplicity of the irreducible factor $\rho,$ and we can find an ideal in $V$ that does not contain $t,$ namely any ideal whose generator has a larger multiplicity of $\rho$ (which exists since $\rho$ has infinite multiplicity in $\bigcup \mathsf{F}(v_\alpha)$). We now assume that the multiplicity of any irreducible in $\bigcup \mathsf{F}(v_\alpha)$ is finite. ⚠️

Suppose that $\bigcup \mathsf{F}(v_\alpha)$ has infinite support. Then we can construct an infinite sequence of distinct irreducibles $r_0, r_1, r_2, \ldots$ such that each element of this sequence appears as a factor of the generator of some principal ideal in $V.$ A non-zero element $t$ of $R$ has a finite multiset of factors, therefore we can always find some irreducible $r_p$ such that $t$ does not have $r_p$ as a factor. In this case the intersection $\bigcap V$ is the principal ideal $(0).$ We now assume that the multiset $\bigcup \mathsf{F}(v_\alpha)$ has finite support. ⚠️

Let $\mathsf{F}(i) = \bigcup \mathsf{F}(v_\alpha)$ and let $I = (i).$ Let $J$ be the intersection of all members of $V.$

Fix $j \in J.$ If it is $0,$ then it is already an element of $I.$ Suppose it is not $0.$ Since it is an element of $J,$ it must be that $\mathsf{F}(v_\alpha) \subseteq \mathsf{F}(j)$ for all $\alpha.$ Hence, $\mathsf{F}(i) \subseteq \mathsf{F}(j)$ and hence $j \in I.$ Thus, $J \subseteq I.$

Fix $i' \in I.$ If $i'$ is $0,$ then it is already an element of $J.$ Suppose otherwise. By construction, $\mathsf{F}(v_\alpha) \subseteq \mathsf{F}(i').$ It is an element of every ideal $(v_\alpha) \in V$ and hence an element of $J.$ Thus, $I \subseteq J.$

This proves $I = J,$ which demonstrates that $R$ has property X.

$\blacksquare$