The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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 (3)

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?


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