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*}

Lemma 10.24.2. Let $R$ be a reduced ring. Then

  1. $R$ is a subring of a product of fields,

  2. $R \to \prod _{\mathfrak p\text{ minimal}} R_{\mathfrak p}$ is an embedding into a product of fields,

  3. $\bigcup _{\mathfrak p\text{ minimal}} \mathfrak p$ is the set of zerodivisors of $R$.

Proof. By Lemma 10.24.1 each of the rings $R_\mathfrak p$ is a field. In particular, the kernel of the ring map $R \to R_\mathfrak p$ is $\mathfrak p$. By Lemma 10.16.2 we have $\bigcap _{\mathfrak p} \mathfrak p = (0)$. Hence (2) and (1) are true. If $x y = 0$ and $y \not= 0$, then $y \not\in \mathfrak p$ for some minimal prime $\mathfrak p$. Hence $x \in \mathfrak p$. Thus every zerodivisor of $R$ is contained in $\bigcup _{\mathfrak p\text{ minimal}} \mathfrak p$. Conversely, suppose that $x \in \mathfrak p$ for some minimal prime $\mathfrak p$. Then $x$ maps to zero in $R_\mathfrak p$, hence there exists $y \in R$, $y \not\in \mathfrak p$ such that $xy = 0$. In other words, $x$ is a zerodivisor. This finishes the proof of (3) and the lemma. $\square$


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