Definition:Quotient Ring

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

Let $$\equiv$$ be an equivalence relation on $$R$$ compatible with both $$\circ$$ and $$+$$, i.e. a congruence relation on $$R$$.

Let $$J = \left[\left[{0_R}\right]\right]_\equiv$$ be the equivalence class of $$0_R$$ under $$\equiv$$.

Then:


 * $$J = \left[\left[{0_R}\right]\right]_\equiv$$ is an ideal of $$R$$, and the equivalence defined by the partition $$R / J$$ is $$\equiv$$ itself;


 * $$\left({R / \equiv, +_\equiv, \circ_\equiv}\right)$$ is a ring, where $$R / \equiv$$is the quotient set of $R$ by $\equiv$;

The ring $$\left({R / \equiv, +_\equiv, \circ_\equiv}\right)$$ is the quotient ring of $$R$$ and $$\equiv$$, and is the same thing as $$\left({R / J, +, \circ}\right)$$.

Similarly, if $$J$$ is an ideal of $$R$$, then $$J$$ induces a congruence relation $$\equiv_J$$ on $$R$$ such that $$\left({R / J, +, \circ}\right)$$ is itself a quotient ring.

Let addition be defined on $$\left({R / J, +, \circ}\right)$$ as here, and ring product be as defined here.

Then we are justified in our claim that $$\left({R / J, +, \circ}\right)$$ is a ring.

Proof
This follows from the fact that, for a congruence $$\equiv$$, the quotient mapping from $$R$$ to $$R / \equiv$$ is an epimorphism.

As $$\equiv$$ is an equivalence relation on $$R$$ compatible with both $$\circ$$ and $$+$$, it is therefore a congruence on $$R$$ for both operations.


 * Let $$J = \left[\left[{0_R}\right]\right]_\equiv$$.

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

The elements of $$R / J$$ are the cosets of $$\left[\left[{0_R}\right]\right]_\equiv$$, and the fact that the equivalence defined by the partition $$R / J$$ is $$\equiv$$ itself follows from the same result.


 * Now note that as $$\equiv$$ is also compatible with $$\circ$$, we also have:

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

That is:

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


 * The equivalence $$\equiv_J$$ defined on $$R$$ by $$J$$ is a congruence for both $$+$$ and $$\circ$$:

Let $$x + \left({-x'}\right), y + \left({-y'}\right) \in J$$. Then:

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

Hence the result.

Proof that $$\left({R / J, +, \circ}\right)$$ is a ring
Taking the ring axioms for $$\left({R / J, +, \circ}\right)$$ in turn:

A: Addition forms a Group
From Quotient Group of Ideal is Coset, we have that $$\left({R / J, +}\right)$$ is a Group.

The identity of $$\left({R / J, +}\right)$$ is $$J$$.

For any $$x + J \in R / J$$, the inverse of $$x + J$$ is $$\left({-x}\right) + J$$.

Thus the zero of $$\left({R/J, +, \circ}\right)$$ is $$J$$ and the negative of $$x + J$$ is $$\left({-x}\right) + J$$.

M0: Closure of Ring Product
From the definition of the Quotient Ring Product, $$\left({R / J, \circ}\right)$$ is closed.

M1: Associativity of Ring Product
Associativity can be deduced from the fact that $$\circ$$ is associative on $$R$$:

D: Distributivity of Ring Product over Addition
The distributive laws can also be checked and determined to hold for $$\left({R / J, +, \circ}\right)$$, if you were to care to do it.

Thus $$\left({R / J, +, \circ}\right)$$ is a ring.