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 $

$=$

$\ds \paren {x + J} + \paren {y + J}$

Quotient Ring Addition is Well-Defined

$\ds $

$=$

$\ds \map \phi x + \map \phi y$

Definition of $\phi$

and:

$\ds \map \phi {x \circ y}$

$=$

$\ds \paren {x \circ y} + J$

Definition of $\phi$

$\ds $

$=$

$\ds \paren {x + J} \circ \paren {y + J}$

Quotient Ring Product is Well-Defined

$\ds $

$=$

$\ds \map \phi x \, \map \phi y$

Definition of $\phi$

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:

$\map \ker \phi = J$

$\blacksquare$

Sources

1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $5$: Rings: $\S 24$. Homomorphisms: Example $46$
1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): $\S 2.2$: Homomorphisms: Lemma $2.7$
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 60.2$ Factor rings: $\text{(ii)}$