Definition talk:Ring Epimorphism

This is false. The inclusion of Z into Q is a non-surjective ring epimorphism. Cbrown91 (talk) 21:09, 29 October 2016 (EDT)

An epimorphism is by definition a homomorphism which is a surjection -- please follow the links to see how it is defined on $\mathsf{Pr} \infty \mathsf{fWiki}$. If you believe that this definition is incorrect, and you can provide a link to a page (or can reference a printed work) which backs this up, please feel free to do so. In the meantime, as our own sources back up the definition given here, the page stands.

In Silver's "Noncommutative localizations and applications" (doi:10.1016/0021-8693(67)90067-1), he defines an epimorphism as a homomorphism $\phi: R \to S$ such that for any $\psi_{1}, \psi_{2}: S \to T$, if $\psi_{1} \circ \phi = \psi_{2} \circ \phi$, then $\psi_{1} = \psi_{2}$. This definition matches my experience.

Can you provide a link? I think there is a misunderstanding here. --prime mover (talk) 13:02, 30 October 2016 (EDT)

Looking more closely, I see he has defined an epimorphism as a right cancellable homomorphism. As can be seen from this link: Surjection iff Right Cancellable, a mapping is right cancellable if and only if it is a surjection. This matches our definition. Yours, I contend, is the error. --prime mover (talk) 13:29, 30 October 2016 (EDT)

An epimorphism as a homomorphism that is right cancellable with respect to other homomorphisms. Surjection iff Right Cancellable only establishes the result for sets. $\iota : \mathbb{Z} \to \mathbb{Q}$ is an epimorphism, since for any ring $R$ and ring homomorphisms $h_{1},h_{2} : \mathbb{Q} \to R$, if $h_{1} \circ \iota = h_{2} \circ \iota$, then $h_{1}(k) = h_{2}(k)$ for any integer $k$. For any $\dfrac{p}{q} \in \mathbb{Q}$, $h_{1}(\dfrac{p}{q}) = \dfrac{h_{1}(p)}{h_{1}(q)}$, since ring homomorphisms respect multiplicative inverses when they exist, and $\dfrac{h_{1}(p)}{h_{1}(q)} = \dfrac{h_{2}(p)}{h_{2}(q)} = h_{2}(\dfrac{p}{q})$, since $p,q$ are integers. The proof is similar to Inclusion of Natural Numbers in Integers is Epimorphism.

Are we talking about the same thing here? From Inclusion of Natural Numbers in Integers is Epimorphism: "The theorem statement does not assert that $\iota$ is an abstract-algebraic epimorphism. This is plainly false, as $\iota$ is not a surjection." --prime mover (talk) 12:20, 3 November 2016 (EDT)

I saw that note, and thought it was weird, since monoids are algebraic objects. Cbrown91 (talk) 17:16, 4 November 2016 (EDT)

If you're up for doing a bit of research and reading around to see who defines it as what, and which fields of mathematics call a ring epi a surjection, and which call it a right-cancellable mapping, and exactly who is "wrong" and who is "right", and at which point it begins to matter beyond personal preference for one definition over the other, then feel free to contribute. --prime mover (talk) 19:37, 4 November 2016 (EDT)

Thus, by our analysis, the inclusion of $\Z$ into $\Q$ is not a non-surjective ring epimorphism, so I believe that yours may be the false statement -- unless you can find corroborative evidence. --prime mover (talk) 04:46, 30 October 2016 (EDT)

Ring Epimorphism versus Epimorphism in the Category of Rings

I have noticed, that epimorphisms of certain algebraic structures on $\mathsf{Pr} \infty \mathsf{fWiki}$ are defined as surjective homomorphisms. This does not match the definition of epimorphism in the category of algebraic structures and explains the confusion in the discussion above. Do the given sources really define ring epimorphisms as surjective homomorphisms? I can't check at the moment. There should be some 'not to be confused with'-explanation of this including examples. --Wandynsky (talk) 00:32, 31 July 2021 (UTC)

Yes they do. The abstract algebraic concepts came first. This was all considered mainstream mathematics well before category theory was defined.

If it's different in category theory, then the latter definitions need to be written as different definitions.

My understanding is that by the time one gets to study category theory, one has already had an introduction to abstract algebra. I'm clearly wrong. --prime mover (talk) 06:07, 31 July 2021 (UTC)