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Here all rings are assumed to lie in the category $\cal C$ of commutative rings with identity, and ${\cal C} (\ R\ ,\ S\ )$ is the set of all ring homomorphisms $F$ from $R$ to $S$ for which $F(1_R)=F(1_S)$. Then $F\in{\mathcal C}(\ R\ ,\ S)$ induces an $R$-module sructure on $S$. Since the functor $\otimes_R$ is right exact, it is not difficult to see that $$S\otimes_RS \cong S\cong S\otimes_SS {\mathrm{\ as \ }} R{\mathrm{-algebras}} \iff \left(S\left/{\mathrm{image}}(F)\right.\right)\otimes_RS=0.\tag{ *} $$

Intuitively, these equivalent conditions seem to show that, for $s_1,\ s_2\ \in S$, the product $s_1 \cdot s_2\in S$ is determined completely by $F(R)$, and just as for localizations, this should show that the equivalent conditions in $(*)$ should also be equivalent to other special statements.

My question if the following: please give, by a reference or a proof or a counterexample to each of the two directions of the following conjecture: $$ S\otimes_RS\cong S \iff F \mathrm{\ is\ a\ flat\ epimorphism\ in\ the \ category \ \mathcal{C}.} $$

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Please elaborate on (*). –  user26857 Feb 1 '13 at 8:19
    
@YACP: See any book on Homological Algebra for the fact that $R\otimes_RS\cong S$ and for right exactness which gives $S\rightarrow S\otimes_RS$ onto $\iff$ $S\otimes_RS$ is the zero map $\iff$ $(S/F(R))$ is the image of a zero map. –  Barbara Osofsky Feb 1 '13 at 21:01
    
I still don't get it: $S\to S\otimes_RS$ onto is the same thing with $S\otimes_RS\cong S$? If yes, why? –  user26857 Feb 1 '13 at 21:47
    
For epimorphisms of commutative rings, see mathoverflow.net/questions/109/… and the references there. –  Martin Brandenburg Feb 2 '13 at 18:41
    
@Barbara: Your statements and proofs are not correct. Be more careful, then you will see it. –  Martin Brandenburg Feb 2 '13 at 18:44

2 Answers 2

up vote 4 down vote accepted

It is a general theorem that in any category, a map $f : A \to B$ is an epimorphism if and only if the diagram

$$ \begin{matrix} A &\to& B \\ \downarrow & & \downarrow \\ B &\to& B\end{matrix} $$

is a pushout square, where the top and left arrows are $f$ and the right and bottom arrows are the identity.

In the category of commutative rings, pushouts are tensor products. That is, the following is a pushout square

$$ \begin{matrix} R &\to& S \\ \downarrow & & \downarrow \\ T &\to& S \otimes_R T\end{matrix} $$

(where the top and left arrows are the maps used to induce an $R$-module structure on $S$ and $T$)

Therefore, $R \to S$ is epimorphic if and only if the multiplication map $S \otimes_R S \to S$ is an isomorphism.

I suspect simply having $S \cong S \otimes_R S$ is not enough.

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Thanks very much. I edited $(*)$ as to read as $R$-algebras. I really appreciate the proof. –  Barbara Osofsky Feb 2 '13 at 21:30

The conjecture is true. The reference is as follows:

Daniel Lazard: "Epimorphisme plats", Seminaire Samuel, Algebre Commutative, tome 2 (1967-1968), exp. no4, pp. 1-12. "Lemme 1.0" will give you what you need.

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You're right about the link, but the link you provided was to Roby's paper, and I meant Lazard's paper from the same volume. The reason I didn't provide a proof is that it seemed what the OP really wanted was a reference, so as to know whether the result was new or not. Moreover, Lazard's 'Lemme 1.0' provides several different characterizations of when a ring map is an epimorphism. That's why I didn't provide a proof. (By the way, I can't seem to get the link to show up correctly. For anyone who wants it, replace both of the "A3"s in the link from YACP's comment with "A4"s.) –  neilme Feb 2 '13 at 18:18
    
Well, I've mixed up the links, but all is in order now. –  user26857 Feb 2 '13 at 19:19
    
Thanks for the reference. I had a copy of that paper, but I had not used it in over 40 years when I retired, and had to recycle a lot of little used reprints and preprints when I retired. Glad to get access to another copy. And you were correct that what I really wanted was confirmation that it was true, although having a proof is nice. –  Barbara Osofsky Feb 2 '13 at 21:39
    
You're welcome! –  neilme Feb 3 '13 at 0:48

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