# Quotient Epimorphism is Epimorphism/Ring

## Theorem

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$.

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

Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$.

Let $\phi: R \to R / J$ be the quotient (ring) epimorphism from $R$ to $R / J$:

$x \in R: \map \phi x = x + J$

Then $\phi$ is a ring epimorphism whose kernel is $J$.

## Proof

Let $x, y \in R$.

Then:

 $\ds \map \phi {x + y}$ $=$ $\ds \paren {x + y} + J$ Definition of $\phi$ $\ds$ $=$ $\ds \paren {x + J} + \paren {y + J}$ Definition of Quotient Ring Addition $\ds$ $=$ $\ds \map \phi x + \map \phi y$

and:

 $\ds \map \phi {x \circ y}$ $=$ $\ds \paren {x \circ y} + J$ Definition of $\phi$ $\ds$ $=$ $\ds \paren {x + J} \circ \paren {y + J}$ Definition of Quotient Ring Product $\ds$ $=$ $\ds \map \phi x \, \map \phi y$

Thus $\phi$ is a homomorphism.

$\phi$ is surjective because:

$\forall x + J \in R / J: x + J = \map \phi x$

Therefore $\phi$ is an epimorphism.

Let $x \in \map \ker \phi$.

Then:

 $\ds x$ $\in$ $\ds \map \ker \phi$ $\ds \leadstoandfrom \ \$ $\ds \map \phi x$ $=$ $\ds 0_{R/J}$ Definition of Kernel of Ring Homomorphism $\ds \leadstoandfrom \ \$ $\ds x + J$ $=$ $\ds J$ $J$ is the zero of $\struct {R / J, +, \circ}$ $\ds \leadstoandfrom \ \$ $\ds x$ $\in$ $\ds J$ Left Coset Equals Subgroup iff Element in Subgroup

Thus:

$\map \ker \phi = J$

$\blacksquare$