Ideal induced by Congruence Relation defines that Congruence

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Theorem

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

Let $\mathcal E$ be a congruence relation on $R$.

Let $J = \left[\!\left[{0_R}\right]\!\right]_\mathcal E$ be the ideal induced by $\mathcal E$.


Then the equivalence defined by the coset space $\left({R, +}\right) / \left({J, +}\right)$ is $\mathcal E$ itself.


Proof

Let $J = \left[\!\left[{0_R}\right]\!\right]_\mathcal E$.

From Congruence Relation on Ring induces Ideal, we have that $J$ is an ideal of $R$.

From Ideal is Additive Normal Subgroup, we have that $\left({J, +}\right)$ is a normal subgroup of $\left({R, +}\right)$.


From Normal Subgroup induced by Congruence Relation defines that Congruence, the equivalence defined by the partition $\left({R, +}\right) / \left({J, +}\right)$ is $\mathcal E$.

As $\mathcal E$ was the congruence relation on $R$ that was originally posited, we already know that it is compatible with $\circ$.


Thus the equivalence defined by $J$ is the same congruence relation on $R$ that gave rise to $J$ to start with.

Hence the result.

$\blacksquare$


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