Congruence Relation on Ring induces Ideal

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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 equivalence class of $0_R$ under $\mathcal E$.


Then $J$ is an ideal of $R$.


Proof

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

By Congruence Relation induces Normal Subgroup, $\left({J, +}\right)$ is a normal subgroup of $\left({R, +}\right)$.

Thus the elements of $\left({R, +}\right) / \left({J, +}\right)$ are the cosets of $\left[\!\left[{0_R}\right]\!\right]_\mathcal E$ by $+$.


We have that $\mathcal E$ is also compatible with $\circ$.

Thus from Quotient Structure is Well-Defined, and so:

$\left[\!\left[{x_1}\right]\!\right]_\mathcal E = \left[\!\left[{x_2}\right]\!\right]_\mathcal E \land \left[\!\left[{y_1}\right]\!\right]_\mathcal E = \left[\!\left[{y_2}\right]\!\right]_\mathcal E \implies \left[\!\left[{x_1 \circ y_1}\right]\!\right]_\mathcal E = \left[\!\left[{x_2 \circ y_2}\right]\!\right]_\mathcal E$

Putting $x_1 = 0_R, x_2 = x, y_1 = y_2 = y$, so that $\left[\!\left[{y_1}\right]\!\right]_\mathcal E = \left[\!\left[{y_2}\right]\!\right]_\mathcal E$ by definition:

$\left[\!\left[{0}\right]\!\right]_\mathcal E = \left[\!\left[{x}\right]\!\right]_\mathcal E \implies \left[\!\left[{0 \circ y}\right]\!\right]_\mathcal E = \left[\!\left[{x \circ y}\right]\!\right]_\mathcal E$

Hence:

$\forall y \in R: \left[\!\left[{y}\right]\!\right]_\mathcal E \circ \left[\!\left[{0_R}\right]\!\right]_\mathcal E = \left[\!\left[{0_R}\right]\!\right]_\mathcal E = \left[\!\left[{0_R}\right]\!\right]_\mathcal E \circ \left[\!\left[{y}\right]\!\right]_\mathcal E$


That is:

$\forall x \in J, y \in R: y \circ x \in J, x \circ y \in J$

demonstrating that $J$ is an ideal of $R$.

$\blacksquare$


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