quotient ring

Definition

Let $R$ be a ring and $I\subset R$ be an ideal. Let $R/I$ be the quotient of the abelian group $R$ by the (normal) subgroup $I$ (since $I$ is a subgroup under addition). This makes $R/I$ an abelian group with elements

$$r+Ir\in R$$

such that

$$r+I=s+I\text{in }R/I⟺r-s\in I.$$

Then we can take the product

$$(r+I)(s+I)=rs+I$$

which gives $R/I$ a ring structure. We call $R/I$ the quotient of $R$ by $I$.

Quotient map

This product defined above is well-defined and gives $R/I$ the structure of a ring with identity $1_{R/I}=1_R+I$. The canonical map

$$\begin{array}{rl}\pi :R& ⟶R/I\\ r& ⟼r+I\end{array}$$

is a surjective homomorphism with kernel $I$.

Relations to ideals and subrings

• Every ideal is the kernel of a ring homomorphism.

• Every subring is the image of a ring homomorphism (this is trivial).

Universal property

The universal property of the quotient of rings is a special case of the map above being a categorical quotient.

Let $R,S$ be rings and $I\subset R$ and ideal of $R$. For every ring homomorphism $f:R\to S$ there exists a unique ring homomorphism $\tilde{f}:R/I\to S$ which makes the following diagram commute:

Proof

Given such an $f$, define $\tilde{f}(r+I)=f(r)$.

This is well-defined because if $r+I=r^{\prime }+I$, that is $r-r^{\prime }\in I$, then

$$\tilde{f}(r+I)=f(r)=f(r^{\prime }+i)=f(r^{\prime })+f(i)=f(r^{\prime })=\tilde{f}(r^{\prime }+I).$$

Clearly $f=\tilde{f}\circ \pi$. Moreover, $\tilde{f}$ is a ring homomorphism:

$$\begin{array}{rl}\tilde{f}((r+I)+(r^{\prime }+I))& =\tilde{f}((r+r^{\prime })+I)\\ & =f(r+r^{\prime })\\ & =f(r)+f(r^{\prime })\\ & =\tilde{f}(r+I)+\tilde{f}(r^{\prime }+I)\end{array}$$

and

$$\begin{array}{rl}\tilde{f}((r+I)(r^{\prime }+I))& =\tilde{f}((r{r}^{\prime })+I)\\ & =f(r{r}^{\prime })\\ & =f(r)f(r^{\prime })\\ & =\tilde{f}(r+I)\tilde{f}(r^{\prime }+I)\end{array}$$