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
- 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: $\text{(ii)}$