# Quotient ring

A quotient ring is a quotient set of the elements of a ring with an induced ring structure.

## Characterization of Equivalence Relations Compatible with Ring Structure

Theorem. Let $R$ be an equivalence relation on the underlying set of a pseudo-ring $A$. Then $R(x,y)$ is compatible with addition and left (resp. right) multiplication if and only if $R(x,y)$ is equivalent to a statement of the form " $x-y \in \mathfrak{a}$", for some left (resp. right) ideal $\mathfrak{a}$ of $A$.

Proof. We prove the case for left ideals; the other case follows from passing to the opposite ring.

Suppose $R$ is an equivalence relation on $A$ compatible with addition and left multiplication. Let $\mathfrak{a}$ be the equivalence class of 0. Then $R(x,y)$ is evidently equivalent to the statement " $x-y\in \mathfrak{a}$, so it remains to show that $\mathfrak{a}$ is a left ideal of $A$.

By definition, $0\in \mathfrak{a}$, and for any $x,y\in \mathfrak{a}$, $$x+ y \equiv 0+0 \equiv 0 \pmod{R},$$ so $x+y \in \mathfrak{a}$; that is, $\mathfrak{a}$ is closed under addition. Finally, for any $x\in A$ and $y\in \mathfrak{a}$, $$xy \equiv x \cdot 0 \equiv 0 \pmod{R} ,$$ so $A\mathfrak{a} \subseteq \mathfrak{a}$. Therefore $\mathfrak{a}$ is a left ideal of $A$.

Conversely, let $\mathfrak{a}$ be any left ideal of $A$. We wish to show that " $x-y \in \mathfrak{a}$" is an equivalence relation compatible addition and left multiplication in $A$. Evidently, if $x\equiv y \pmod{\mathfrak{a}}$ and $y\equiv z \pmod{\mathfrak{a}}$, then $$x-z = (x-y)+(y-z) \in \mathfrak{a},$$ so $x\equiv z\pmod{\mathfrak{a}}$. Also, $x-x = 0$ is an element of $\mathfrak{a}$, and if $(x-y)$ is, then so is $-(x-y) = y-x$. This shows that equivalence modulo $\mathfrak{a}$ is an equivalence relation.

Now we show that equivalence modulo $\mathfrak{a}$ is compatible with addition and left multiplication. Indeed, suppose that $x-y \in \mathfrak{a}$; then for any $a\in A$, $$(a+x)- (a+y) = x-y \in \mathfrak{a},$$ so $a+x \equiv a+y \pmod{\mathfrak{a}}$. Finally, for any $a\in A$, $$a(x-y) \in \mathfrak{a},$$ since $\mathfrak{a}$ is a left ideal of $A$. $\blacksquare$

Corollary. Let $A$ be a ring, and $R(x,y)$ an equivalence relation on the elements of $A$. Then $R$ is compatible with the ring structure of $A$ if and only if it is of the form " $x-y \in \mathfrak{a}$", for some two-sided ideal $\mathfrak{a}$ of $A$.