Lemma 10.25.4. Let $R$ be a ring. Assume that $R$ has finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$, and that $\mathfrak q_1 \cup \ldots \cup \mathfrak q_ t$ is the set of zerodivisors of $R$. Then the total ring of fractions $Q(R)$ is equal to $R_{\mathfrak q_1} \times \ldots \times R_{\mathfrak q_ t}$.

Proof. There are natural maps $Q(R) \to R_{\mathfrak q_ i}$ since any nonzerodivisor is contained in $R \setminus \mathfrak q_ i$. Hence a natural map $Q(R) \to R_{\mathfrak q_1} \times \ldots \times R_{\mathfrak q_ t}$. For any nonminimal prime $\mathfrak p \subset R$ we see that $\mathfrak p \not\subset \mathfrak q_1 \cup \ldots \cup \mathfrak q_ t$ by Lemma 10.15.2. Hence $\mathop{\mathrm{Spec}}(Q(R)) = \{ \mathfrak q_1, \ldots , \mathfrak q_ t\}$ (as subsets of $\mathop{\mathrm{Spec}}(R)$, see Lemma 10.17.5). Therefore $\mathop{\mathrm{Spec}}(Q(R))$ is a finite discrete set and it follows that $Q(R) = A_1 \times \ldots \times A_ t$ with $\mathop{\mathrm{Spec}}(A_ i) = \{ q_ i\}$, see Lemma 10.24.3. Moreover $A_ i$ is a local ring, which is a localization of $R$. Hence $A_ i \cong R_{\mathfrak q_ i}$. $\square$

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