The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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

Lemma 10.119.16. 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 10.77.2. 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. $\square$


Comments (1)

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 is locally free of rank .)

There are also:

  • 7 comment(s) on Section 10.119: Factorization

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