Congruence Relation on Ring induces Ideal

Theorem

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

Let $\EE$ be a congruence relation on $R$.

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

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

Proof

Let $J = \eqclass {0_R} \EE$.

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

Thus the elements of $\struct {R, +} / \struct {J, +}$ are the cosets of $\eqclass {0_R} \EE$ by $+$.

We have that $\EE$ is also compatible with $\circ$.

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

$\eqclass {x_1} \EE = \eqclass {x_2} \EE \land \eqclass {y_1} \EE = \eqclass {y_2} \EE \implies \eqclass {x_1 \circ y_1} \EE = \eqclass {x_2 \circ y_2} \EE$

Putting $x_1 = 0_R, x_2 = x, y_1 = y_2 = y$, so that $\eqclass {y_1} \EE = \eqclass {y_2} \EE$ by definition:

$\eqclass 0 \EE = \eqclass x \EE \implies \eqclass {0 \circ y} \EE = \eqclass {x \circ y} \EE$

Hence:

$\forall y \in R: \eqclass y \EE \circ \eqclass {0_R} \EE = \eqclass {0_R} \EE = \eqclass {0_R} \EE \circ \eqclass y \EE$

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 $R$.

$\blacksquare$