A ring map is surjective if and only if it is a finite epimorphism.

Lemma 10.106.6. Let $R \to S$ be a ring map. The following are equivalent

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

2. $R \to S$ is surjective.

Proof. (This lemma seems to have been reproved many times in the literature, and has many different proofs.) It is clear that a surjective ring map is an epimorphism. Suppose that $R \to S$ is a finite ring map such that $S \otimes _ R S \to S$ is an isomorphism. Our goal is to show that $R \to S$ is surjective. Assume $S/R$ is not zero. The exact sequence $R \to S \to S/R \to 0$ leads to an exact sequence

$R \otimes _ R S \to S \otimes _ R S \to S/R \otimes _ R S \to 0.$

Our assumption implies that the first arrow is an isomorphism, hence we conclude that $S/R \otimes _ R S = 0$. Hence also $S/R \otimes _ R S/R = 0$. By Lemma 10.5.4 there exists a surjection of $R$-modules $S/R \to R/I$ for some proper ideal $I \subset R$. Hence there exists a surjection $S/R \otimes _ R S/R \to R/I \otimes _ R R/I = R/I \not= 0$, contradiction. $\square$

## Comments (4)

Comment #834 by on

Suggested slogan: A ring map is surjective if and only if it is a finite epimorphism.

Comment #3861 by anon on

Why does surjective imply finite? Perhaps the finiteness should be an apriori assumption?

Comment #3862 by anon on

Why does surjective imply finite? Perhaps the finiteness should be an apriori assumption?

Comment #3942 by on

A surjective ring map is a finite ring map. So everything is fine.

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