Congruence Relation and Ideal are Equivalent

Theorem

Let $\struct {R, +, \circ}$ be a ring.

Let $\EE$ be an equivalence relation on $R$ compatible with both $\circ$ and $+$, that is, a congruence relation on $R$.

Let $J = \eqclass {0_R} \EE$ be the equivalence class of $0_R$ under $\EE$.

Then:

$(1a): \quad J = \eqclass {0_R} \EE$ is an ideal of $R$

$(2a): \quad$ The equivalence defined by the quotient ring $R / J$ is $\EE$ itself.

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

Then:

$(1b): \quad J$ induces a congruence relation $\EE_J$ on $R$

$(2b): \quad$ The ideal of $R$ defined by $\EE_J$ 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 $(2b)$

This needs considerable tedious hard slog to complete it. In particular: Argument previously here didn't prove anything. This part is genuinely different from the other three To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.