Quotient Epimorphism is Epimorphism/Ring

Theorem
Let $\left({R, +, \circ}\right)$ be a ring whose zero is $0_R$ and whose unity is $1_R$.

Let $J$ be an ideal of $R$.

Let $\left({R / J, +, \circ}\right)$ be the quotient ring defined by $J$.

Then the mapping $\phi: R \to R / J$ given by $x \in R: \phi \left({x}\right) = x + J$ is a ring epimorphism whose kernel is $J$.

This ring epimorphism is called the natural epimorphism from $\left({R, +, \circ}\right)$ onto $\left({R / J, +, \circ}\right)$.

Proof

 * The fact that $\phi$ is a homomorphism can be verified easily.


 * $\phi$ is surjective because $\forall x + J \in R / J: x + J = \phi \left({x}\right)$.

Therefore $\phi$ is an epimorphism.


 * Let $x \in \ker \left({\phi}\right)$. Then:

Thus $\ker \left({\phi}\right) = J$.