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 .  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)


 * 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)