Lemma 10.107.13. 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 10.107.1. We are going to use the notation introduced in Remark 10.107.12. 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
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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