Congruence Relation on Ring induces Ideal

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

Let $\mathcal E$ be an equivalence relation on $R$ compatible with both $\circ$ and $+$, i.e. 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:
 * $(1): \quad J = \left[\!\left[{0_R}\right]\!\right]_\mathcal E$ is an ideal of $R$
 * $(2): \quad$ The equivalence defined by the partition $R / J$ is $\mathcal E$ itself.

Proof
This follows from the fact that, for a congruence $\mathcal E$, the quotient mapping from $R$ to $R / \mathcal E$ is an epimorphism.

As $\mathcal E$ is an equivalence relation on $R$ compatible with both $\circ$ and $+$, it is therefore a congruence on $R$ for both operations.

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 $\left({R, +}\right)$.

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

Also from Congruence Relation arises from Normal Subgroup, the equivalence defined by the partition $\left({R, +}\right) / \left({J, +}\right)$ is $\mathcal E$.

Now note that as $\mathcal E$ is also compatible with $\circ$, we also 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:
 * $\forall x \in J, y \in R: y \circ x \in R, x \circ y \in R$

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

Also see

 * Quotient Ring is a Ring