Congruence Relation and Ideal are Equivalent

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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

Part $(1a)$

This is shown on Congruence Relation on Ring induces Ideal.

$\Box$


Part $(2a)$

This is shown on Ideal induced by Congruence Relation defines that Congruence.

$\Box$


Part $(1b)$

This is shown on Ideal induces Congruence Relation on Ring.

$\Box$


Part $(2b)$