Lemma 10.107.5. Let $R \to S$ be a ring map. The following are equivalent:

1. $R \to S$ is an epimorphism, and

2. $R_{\mathfrak p} \to S_{\mathfrak p}$ is an epimorphism for each prime $\mathfrak p$ of $R$.

Proof. Since $S_{\mathfrak p} = R_{\mathfrak p} \otimes _ R S$ (see Lemma 10.12.15) we see that (1) implies (2) by Lemma 10.107.3. Conversely, assume that (2) holds. Let $a, b : S \to A$ be two ring maps from $S$ to a ring $A$ equalizing the map $R \to S$. By assumption we see that for every prime $\mathfrak p$ of $R$ the induced maps $a_{\mathfrak p}, b_{\mathfrak p} : S_{\mathfrak p} \to A_{\mathfrak p}$ are the same. Hence $a = b$ as $A \subset \prod _{\mathfrak p} A_{\mathfrak p}$, see Lemma 10.23.1. $\square$

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