Ideal induces Congruence Relation on Ring

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Theorem

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

Let $J$ be an ideal of $R$


Then $J$ induces a congruence relation $\mathcal E_J$ on $R$ such that $\left({R / J, +, \circ}\right)$ is a quotient ring.


Proof

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

Let $x \mathop {\mathcal E_J} y$ denote that $x$ and $y$ are in the same coset, that is:

$x \mathop {\mathcal E_J} y \iff x + N = y + N$

From Congruence Modulo Normal Subgroup is Congruence Relation, $\mathcal E_J$ is a congruence relation for $+$.


Now let $x \mathop {\mathcal E_J} x', y \mathop {\mathcal E_J} y'$.

By definition of congruence modulo $J$:

$x + \left({-x'}\right) \in J$
$y + \left({-y'}\right) \in J$


Then:

$x \circ y + \left({- x' \circ y'}\right) = \left({x + \left({-x'}\right)}\right) \circ y + x' \circ \left({y + \left({-y'}\right)}\right) \in J$

demonstrating that $\mathcal E_J$ is a congruence relation for $\circ$.

Hence the result by definition of quotient ring.

$\blacksquare$


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