Quotient Ring of Commutative Ring is Commutative

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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 $J$.


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:

\(\displaystyle \forall x, y \in R: \ \ \) \(\displaystyle \paren {x + J} \circ \paren {y + J}\) \(=\) \(\displaystyle x \circ y + J\) Definition of $\circ$ in $R / J$
\(\displaystyle \) \(=\) \(\displaystyle y \circ x + J\) Commutativity of $\circ$
\(\displaystyle \) \(=\) \(\displaystyle \paren {y + J} \circ \paren {x + J}\) Definition of $\circ$ in $R / J$

$\blacksquare$


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