Lemma 10.107.5. Let R \to S be a ring map. The following are equivalent:
R \to S is an epimorphism, and
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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