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

$(g) = YXZ = UZ = Y'X'Z = Y'PZ = Y'XZ = Y'V = Y'X'Z' = (g')$

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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 04W0. Beware of the difference between the letter 'O' and the digit '0'.