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*}

Lemma 10.106.13. Let $R \to S$ be an epimorphism of rings. Then the cardinality of $S$ is at most the cardinality of $R$. In a formula: $|S| \leq |R|$.

Proof. The condition that $R \to S$ is an epimorphism means that each $g \in S$ satisfies $g \otimes 1 = 1 \otimes g$, see Lemma 10.106.1. We are going to use the notation introduced in Remark 10.106.12. Suppose that $g, g' \in S$ and suppose that $(P, U, V)$ is an $n$-triple which is associated to both $g$ and $g'$. Then we claim that $g = g'$. Namely, write $(P, U, V) = (X, YX, XZ)$ for a matrix factorization $(g) = YXZ$ of $g$ and write $(P, U, V) = (X', Y'X', X'Z')$ for a matrix factorization $(g') = Y'X'Z'$ of $g'$. Then we see that

\[ (g) = YXZ = UZ = Y'X'Z = Y'PZ = Y'XZ = Y'V = Y'X'Z' = (g') \]

and hence $g = g'$. This implies that the cardinality of $S$ is bounded by the number of possible triples, which has cardinality at most $\sup _{n \in \mathbf{N}} |R|^ n$. If $R$ is infinite then this is at most $|R|$, see [Ch. I, 10.13, Kunen].

If $R$ is a finite ring then the argument above only proves that $S$ is at worst countable. In fact in this case $R$ is Artinian and the map $R \to S$ is surjective. We omit the proof of this case. $\square$


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