Definition talk:Ring Epimorphism

From ProofWiki
Jump to navigation Jump to search

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)