Lemma 10.25.2. Let $R$ be a reduced ring. Then
$R$ is a subring of a product of fields,
$R \to \prod _{\mathfrak p\text{ minimal}} R_{\mathfrak p}$ is an embedding into a product of fields,
$\bigcup _{\mathfrak p\text{ minimal}} \mathfrak p$ is the set of zerodivisors of $R$.
Proof. By Lemma 10.25.1 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 10.17.2 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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