Congruence Relation on Ring induces Ideal

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

Let $\mathcal E$ be a congruence relation on $R$.

Let $J = \left[\!\left[{0_R}\right]\!\right]_\mathcal E$ be the equivalence class of $0_R$ under $\mathcal E$.

Then $J$ is an ideal of $R$.

Proof
Let $J = \left[\!\left[{0_R}\right]\!\right]_\mathcal E$.

By Congruence Relation arises from Normal Subgroup, $\left({J, +}\right)$ is a normal subgroup of $\left({R, +}\right)$.

Thus the elements of $\left({R, +}\right) / \left({J, +}\right)$ are the cosets of $\left[\!\left[{0_R}\right]\!\right]_\mathcal E$ by $+$.

Now note that as $\mathcal E$ is also compatible with $\circ$, we have:


 * $\forall y \in R: \left[\!\left[{y}\right]\!\right]_\mathcal E \circ \left[\!\left[{0_R}\right]\!\right]_\mathcal E = \left[\!\left[{0_R}\right]\!\right]_\mathcal E = \left[\!\left[{0_R}\right]\!\right]_\mathcal E \circ \left[\!\left[{y}\right]\!\right]_\mathcal E$

That is:
 * $\forall x \in J, y \in R: y \circ x \in J, x \circ y \in J$

demonstrating that $J$ is an ideal of $R$.

Also see

 * Ideal Induced by Congruence Relation Defines That Congruence


 * Quotient Ring is a Ring