# 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.

$\Box$