8

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?

Andrej Bauer
  • 47,834
  • 4
    If I understand the link to the Stacks Project blog correctly, they rewrote this here : http://math.columbia.edu/algebraic_geometry/stacks-git/locate.php?tag=04W0 – Matthieu Romagny Nov 01 '11 at 20:54
  • 3
    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. – George Lowther Nov 01 '11 at 20:57
  • 2
    @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). – Matthieu Romagny Nov 01 '11 at 21:22
  • 3
    @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). – George Lowther Nov 01 '11 at 22:33
  • 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, – Matthieu Romagny Nov 02 '11 at 13:12

1 Answers1

2

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.

Boris Novikov
  • 3,080
  • @Boris: I don't think ring epimorphisms can be described so easily. See this question http://mathoverflow.net/questions/109 – George Lowther Nov 01 '11 at 22:36
  • @George: I tried to explain roughly. Nevertheless Anton Geraschenko in his answer said just about zig-zags. – Boris Novikov Nov 01 '11 at 23:11
  • 2
    @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 http://math.columbia.edu/algebraic_geometry/stacks-git/algebra.pdf). – George Lowther Nov 02 '11 at 00:21
  • What do you mean by "behaves like" a ring of fractions? That every elemente of $S$ is a quotient of elements from $T$? – Andrej Bauer Nov 02 '11 at 08:45
  • @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"! :-) – Boris Novikov Nov 02 '11 at 11:44