Congruence Relation on Ring induces Ring

Theorem

Let $\left({R, +, \circ}\right)$ be a ring.

Let $\mathcal E$ be a congruence relation on $R$ for both $+$ and $\circ$.

Let $R / \mathcal E$ be the quotient set of $R$ by $\mathcal E$.

Let $+_\mathcal E$ and $\circ_\mathcal E$ be the operations induced on $R / \mathcal E$ by $+$ and $\circ$ respectively.

Then $\left({R / \mathcal E, +_\mathcal E, \circ_\mathcal E}\right)$ is a ring.

Proof

Let $q_{\,\mathcal E}$ be the quotient mapping from $\left({R, +, \circ}\right)$ to $\left({R / \mathcal E, +_\mathcal E, \circ_\mathcal E}\right)$.

$q_{\,\mathcal E}: \left({R, +}\right) \to \left({R / \mathcal E, +_\mathcal E}\right)$ is an epimorphism
$q_{\,\mathcal E}: \left({R, \circ}\right) \to \left({R / \mathcal E, \circ _\mathcal E}\right)$ is an epimorphism.

As the morphism property holds for both $+$ and $\circ$, it follows that $q_{\,\mathcal E}: \left({R, +, \circ}\right) \to \left({R / \mathcal E, +_\mathcal E, \circ_\mathcal E}\right)$ is also an epimorphism.

From Epimorphism Preserves Rings, it follows that $\left({R / \mathcal E, +_\mathcal E, \circ_\mathcal E}\right)$ is a ring.

$\blacksquare$