Lemma 10.107.3. If $R \to S$ is an epimorphism of rings and $R \to R'$ is any ring map, then $R' \to R' \otimes _ R S$ is an epimorphism.

Proof. Omitted. Hint: True in any category with pushouts. $\square$

Comment #7337 by JS on

Unless I'm mistaken, this can be strengthened: if $R\to S$ is an epimorphism of rings and $R\to R'$ is any ring map, then $R'\to R'\otimes_R S$ is an isomorphism. Proof: Use either the fact that if $N$ and $M$ are $S$-modules and $R\to S$ is a ring epimorphism, then the natural map $N\otimes_R M\to N\otimes_S M$ is an isomorphism, or write $R'\otimes_R S\cong (R'\otimes_S S)\otimes_R S\cong R'\otimes_S (S\otimes_R S)\cong R'\otimes_S S\cong R'.$

Comment #7338 by on

This is not true because in particular taking $R = R'$ would give that $R = S$ and there are nontrivial epimorphisms of rings, for example $\mathbf{Z} \to \mathbf{Q}$.

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