Quotient Ring of Commutative Ring is Commutative
Jump to navigation
Jump to search
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 $J$.
If $\struct {R, +, \circ}$ is a commutative ring, then so is $\struct {R / J, +, \circ}$.
Proof
Let $\struct {R, +, \circ}$ be a commutative ring
That means $\circ$ is commutative on $R$.
Thus:
\(\ds \forall x, y \in R: \, \) | \(\ds \paren {x + J} \circ \paren {y + J}\) | \(=\) | \(\ds x \circ y + J\) | Definition of $\circ$ in $R / J$ | ||||||||||
\(\ds \) | \(=\) | \(\ds y \circ x + J\) | Commutativity of $\circ$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {y + J} \circ \paren {x + J}\) | Definition of $\circ$ in $R / J$ |
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 60.2$ Factor rings: $\text{(iii)}$