According to this page and thence linked text, if $e : R \to S$ is an epimorphism of rings, then the cardinality of $S$ cannot exceed the cardinality of $R$. This is a non-trivial observation because epimorphisms of rings need not be surjective. Is there a "layman's" explanation of this fact, one that does not require me to learn French?
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4$\begingroup$ If I understand the link to the Stacks Project blog correctly, they rewrote this here : math.columbia.edu/algebraic_geometry/stacks-git/… $\endgroup$– Matthieu RomagnyNov 1, 2011 at 20:54
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3$\begingroup$ So...if $R$ if is an integral domain and $S$ is its field of fractions, then $e$ is automatically an isomorphism. That gives an interesting proof that a finite integral domain is a field. Or, maybe, the statement in the question should be regarded as a generalization of this fact. $\endgroup$– George LowtherNov 1, 2011 at 20:57
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2$\begingroup$ @George Lowther : I don't think you are right. Indeed, the injection of an integral domain into its field of fractions is always a ring epimorphism (direct check). $\endgroup$– Matthieu RomagnyNov 1, 2011 at 21:22
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3$\begingroup$ @Matthieu: Yes, that's what I was getting at. It is always an epimorphism and, if it is an integral domain, it is one-to-one. If $R$ is also finite then $|R|\ge|S|$ is equivalent to e being onto (hence, an isomorphism). $\endgroup$– George LowtherNov 1, 2011 at 22:33
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$\begingroup$ George, I agree if R is finite, but that assumption was neither in the original question nor in your first comment. In other words : you should really write "if R is a finite integral domain etc". Best, $\endgroup$– Matthieu RomagnyNov 2, 2011 at 13:12
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An explanation of a layman to a layman. Let $T={\rm Im}\ e$. Then embedding $T\to S$ is again an epimorphism. With respect to $T$ the ring $S$ behaves like the ring (not necessarily the field!) of fractions (compare $\mathbb{Z}$ and $\mathbb{Q}$), so it has the same cardinality. It is enough? :-) I think it is possible to give a rigorous proof by so-called "zigzag-theorem" from Theory of semigroups.
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$\begingroup$ @Boris: I don't think ring epimorphisms can be described so easily. See this question mathoverflow.net/questions/109 $\endgroup$ Nov 1, 2011 at 22:36
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$\begingroup$ @George: I tried to explain roughly. Nevertheless Anton Geraschenko in his answer said just about zig-zags. $\endgroup$ Nov 1, 2011 at 23:11
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2$\begingroup$ @Boris: I see. In fact, lemma 95.11 in the paper linked in Matthieu's comment looks like the zigzag theorem, and is used in the proof of the result asked for here (Lemma 95.13 of math.columbia.edu/algebraic_geometry/stacks-git/algebra.pdf). $\endgroup$ Nov 2, 2011 at 0:21
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$\begingroup$ What do you mean by "behaves like" a ring of fractions? That every elemente of $S$ is a quotient of elements from $T$? $\endgroup$ Nov 2, 2011 at 8:45
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$\begingroup$ @Andrej Bauer: No,I mean that the elements of $S$ "must" be algebraic expressions of elements from $T$. Don't ask me what means "must" if you want "a layman's explanation"! :-) $\endgroup$ Nov 2, 2011 at 11:44