A product of ideals is an invertible module iff both factors are.

Lemma 10.120.16. Let $A$ be a ring. Let $I$ and $J$ be nonzero ideals of $A$ such that $IJ = (f)$ for some nonzerodivisor $f \in A$. Then $I$ and $J$ are finitely generated ideals and finitely locally free of rank $1$ as $A$-modules.

Proof. It suffices to show that $I$ and $J$ are finite locally free $A$-modules of rank $1$, see Lemma 10.78.2. To do this, write $f = \sum _{i = 1, \ldots , n} x_ i y_ i$ with $x_ i \in I$ and $y_ i \in J$. We can also write $x_ i y_ i = a_ i f$ for some $a_ i \in A$. Since $f$ is a nonzerodivisor we see that $\sum a_ i = 1$. Thus it suffices to show that each $I_{a_ i}$ and $J_{a_ i}$ is free of rank $1$ over $A_{a_ i}$. After replacing $A$ by $A_{a_ i}$ we conclude that $f = xy$ for some $x \in I$ and $y \in J$. Note that both $x$ and $y$ are nonzerodivisors. We claim that $I = (x)$ and $J = (y)$ which finishes the proof. Namely, if $x' \in I$, then $x'y = af = axy$ for some $a \in A$. Hence $x' = ax$ and we win. $\square$

Comment #3586 by slogan_bot on

Suggested slogan: A product of ideals is a divisor iff each factor is.

(Also: one could improve the statement according to the slogan, only assuming that $IJ$ is locally free of rank $1$.)

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