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show/hide this revision's text 5 added 7 characters in body

(Background: In any category, an epimorphism is a morphism $f:X\to Y$ which is "surjective" in the following sense: for any two morphisms $g,h:Y\to Z$, if $g\circ f=h\circ f$, then $g=h$. Roughly, "any two functions on $Y$ that agree on the image of $X$ must agree." Even in categories where you have underlying sets, epimorphisms are not the same as surjections; for example, in the category of Hausdorff topological spaces, $f$ is an epimorphism if its image is dense.)

What do epimorphisms of (say commutative) rings look like? It's easy to verify that for any ideal $I$ in a ring $A$, the quotient map $A\to A/I$ is an epimorphism. It's also not hard to see that if $S\subset A$ is a multiplicative subset, then the localization $A\to S^{-1}A$ is an epimorphism. Here's a proof to whet your appetite.

If $g,h:S^{-1}A\to B$ are two homomorphisms that agree on $A$, then for any element $s^{-1}a\in S^{-1}A$, we have
$$g(s^{-1}a)=g(s)^{-1}g(a)=h(s)^{-1}h(a)=h(s^{-1}a)$$

Also, if $A\to B_i$ is a finite collection of epimorphisms, where the $B_i$ have disjoint support as $A$-modules, then $A\to\prod B_i$ is an epimorphism.

Is every epimorphism of rings some product of combinations of quotients and localizations? To put it another way, suppose $f: A\to B$ is an epimorphism of rings with no kernel which sends non-units to non-units and such that $B$ has no idempotents. Must $f$ be an isomorphism?
show/hide this revision's text 4 corrected typo

(Background: In any category, an epimorphism is a morphism $f:X\to Y$ which is "surjective" in the following sense: for any two morphisms $g,h:Y\to Z$, if $g\circ f=h\circ f$, then $g=h$. Roughly, "any two functions on $Y$ that agree on the image of $X$ must agree." Even in categories where you have underlying sets, epimorphisms are not the same as surjections; for example, in the category of Hausdorff topological spaces, $f$ is an epimorphism if its image is dense.)

What do epimorphisms of (say commutative) rings look like? It's easy to verify that for any ideal $I$ in a ring $A$, the quotient map $A\to A/I$ is an epimorphism. It's also not hard to see that if $S\subset A$ is a multiplicative subset, then the localization $A\to S^{-1}A$ is an epimorphism. Here's a proof to whet your appetite.

If $g,h:S^{-1}A\to B$ are two homomorphisms that agree on $A$, then for any element $s^{-1}a\in S^{-1}A$, we have
$$g(s^{-1}a)=g(s)^{-1}g(a)=f(s)^{-1}f(a)=f(s^{-1}a)$$$g(s^{-1}a)=g(s)^{-1}g(a)=h(s)^{-1}h(a)=h(s^{-1}a)$$

Also, if $A\to B_i$ is a collection of epimorphisms, where the $B_i$ have disjoint support as $A$-modules, then $A\to\prod B_i$ is an epimorphism.

Is every epimorphism of rings some product of combinations of quotients and localizations? To put it another way, suppose $f: A\to B$ is an epimorphism of rings with no kernel which sends non-units to non-units and such that $B$ has no idempotents. Must $f$ be an isomorphism?
show/hide this revision's text 3 inserted "Hausdorff" and converted to LaTeX

(Background: In any category, an epimorphism is a morphism f:X→Y $f:X\to Y$ which is "surjective" in the following sense: for any two morphisms g,h:Y→Z, $g,h:Y\to Z$, if $g\circ f=h\circ ff$, then g=h. $g=h$. Roughly, "any two functions on Y $Y$ that agree on the image of X $X$ must agree." Even in categories where you have underlying sets, epimorphisms are not the same as surjections; for example, in the category of Hausdorff topological spaces, f $f$ is an epimorphism if its image is dense.)

What do epimorphisms of (say commutative) rings look like? It's easy to verify that for any ideal I $I$ in a ring A, $A$, the quotient map A→A/I $A\to A/I$ is an epimorphism. It's also not hard to see that if S⊂A $S\subset A$ is a multiplicative subset, then the localization A→S-1A $A\to S^{-1}A$ is an epimorphism. Here's a proof to whet your appetite.

If g,h:S-1A→B $g,h:S^{-1}A\to B$ are two homomorphisms that agree on A, $A$, then for any element s-1a∈S-1A$s^{-1}a\in S^{-1}A$, we have
g(s-1a) = g(s)-1g(a) = f(s)-1f(a) = f(s-1a)
$$g(s^{-1}a)=g(s)^{-1}g(a)=f(s)^{-1}f(a)=f(s^{-1}a)$$

Also, if A→Bi$A\to B_i$ is a collection of epimorphisms, where the Bi$B_i$ have disjoint support as A-modules, $A$-modules, then A→∏Bi$A\to\prod B_i$ is an epimorphism.

Is every epimorphism of rings some product of combinations of quotients and localizations? To put it another way, suppose f:A→B $f: A\to B$ is an epimorphism of rings with no kernel which sends non-units to non-units and such that B $B$ has no idempotents. Must f $f$ be an isomorphism?
show/hide this revision's text 2 added the bit about idempotents
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