Congruence Relation and Ideal are Equivalent

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:
 * $(1a): \quad J = \left[\!\left[{0_R}\right]\!\right]_\mathcal E$ is an ideal of $R$
 * $(2a): \quad$ The equivalence defined by the quotient ring $R / J$ is $\mathcal E$ itself.

Similarly, let $J$ be an ideal of $R$.

Then:
 * $(1b): \quad J$ induces a congruence relation $\mathcal E_J$ on $R$
 * $(2b): \quad$ The ideal of $R$ defined by $\mathcal E_J$ is $J$ itself.

Proof

 * $(1a): \quad$ From Congruence Relation on Ring Induces Ideal, $J = \left[\!\left[{0_R}\right]\!\right]_\mathcal E$ is an ideal of $R$.


 * $(2a): \quad$ From Ideal Induced by Congruence Relation Defines That Congruence, the equivalence defined by the quotient ring $R / J$ is $\mathcal E$ itself.


 * $(1b): \quad$ From Ideal Induces Congruence Relation on Ring, $J$ induces a congruence relation $\mathcal E_J$ on $R$.


 * $(2b): \quad$ From Congruence Relation on Ring Induces Ideal, $J' = \left[\!\left[{0_R}\right]\!\right]_{\mathcal E_J}$ is an ideal of $R$.


 * From Ideal Induced by Congruence Relation Defines That Congruence, $J'$ is exactly the ideal of $R$ defined by $\mathcal E_J$.


 * That is, $J' = J$.