
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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