Quotient Epimorphism is Epimorphism/Ring

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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:

$\map \ker \phi = J$

$\blacksquare$


Sources

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