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For a commutative ring with unit $R$ let's say it has property $(*)$ if there is an epimorphism in the category of rings ${\mathbb Z}[X_1,\dots,X_n]\to R$, where the former is the polynomial ring in $n$ variables. For instance, finitely generated rings and their localisations have this property. As an example consider $\mathbb Q$, for which the map ${\mathbb Z}\to\mathbb Q$ is an epimorphism.

Does this property have a name? How is it related to other properties of a ring?

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  • $\begingroup$ It's called "quotients of $\mathbb{Z}[X_1,\dots,X_n]$"... $\endgroup$
    – Tom De Medts
    Dec 14, 2022 at 8:50
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    $\begingroup$ @Tom_de_Medts: Sorry no, that would mean that the map is surjective, which it isn't in general. You wouldn't call the field $\mathbb Q$ of rationals a quotient of $\mathbb Z$. $\endgroup$
    – user473423
    Dec 14, 2022 at 8:52
  • $\begingroup$ Wait, didn't you write "ring epimorphism"? (Perhaps you meant "epic morphism" then, which isn't the same!) $\endgroup$
    – Tom De Medts
    Dec 14, 2022 at 8:54
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    $\begingroup$ @Tom_de_Medts: The term epimorphism is defined in any catgegory and a ring epimorphism is an epimorphism in the category of rings. Otherwise I would have written surjective ring homomorphism. $\endgroup$
    – user473423
    Dec 14, 2022 at 8:56
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    $\begingroup$ I've seen the term finitely dominated used in semigroup theory which is another category where epimorphism is not the same as surjective. The origin comes from Isbell's notion of dominion. If R' is a subring of R, the dominion of R' is the subring of R consisting of all elements of R equalized by all homomorphisms from R that agree on R'. The inclusion is an epimorphism iff the dominion of R' is R. So the idea is being finitely dominated means you are the dominion of a finitely generated subring. This is the same as having an epimorphism from a finitely generated free object. $\endgroup$
    – Benjamin Steinberg
    Dec 14, 2022 at 11:52

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