# Universal Property of Quotient Ring

## Theorem

Let $R, S$ be commutative rings.

Let $I \trianglelefteq R$ be an ideal of $R$.

Let $\pi : R \to R / I$ be the quotient epimorphism.

Let $f: R \to S$ be a ring homomorphism with $\map f I = \set 0$.

Then there exists a unique ring homomorphism $\overline f: R / I \to S$ such that $f = \overline f \circ \pi$.

$\xymatrix { R \ar[d]_\pi \ar[r]^{\forall f} & S \\ R/I \ar[ru]_{\exists ! \bar f} }$

## Proof

Define $\overline f: R / I \to S$ by:

$\forall r \in R: \map {\overline f} {r + I} = \map f r$

Since $f$ is a ring homomorphism, $f$ is well-defined.

Suppose for some $r_1, r_2 \in R$ that:

$r_1 + I = r_2 + I$

Since $I$ is an ideal, it contains the zero $0_R$ of $R$.

Then for some $i \in I$, we have that:

 $\ds r_1 + 0_R$ $=$ $\ds r_1$ Definition of Ring Zero $\ds$ $=$ $\ds r_2 + i$ Definition of Set Equality

Then:

 $\ds \map {\overline f} {r_1 + I}$ $=$ $\ds \map f {r_1}$ Definition of $\overline f$ $\ds$ $=$ $\ds \map f {r_2 + i}$ Since $r_1=r_2+i$ $\ds$ $=$ $\ds \map f {r_2} + \map f i$ Definition of Ring Homomorphism $\ds$ $=$ $\ds \map f {r_2} + 0_S$ Since $f(I)=\{0\}$ $\ds$ $=$ $\ds \map f {r_2}$ Definition of Ring Zero $\ds$ $=$ $\ds \map {\overline f} {r_2 + I}$ Definition of $\overline f$

so $\overline f$ is seen to be well-defined.

Next, for all $r_1, r_2 \in R$:

 $\ds \map {\overline f} {r_1 + I + r_2 + I}$ $=$ $\ds \map {\overline f} {r_1 + r_2 + I}$ Quotient Ring Addition is Well-Defined $\ds$ $=$ $\ds \map f {r_1 + r_2}$ Definition of $\overline f$ $\ds$ $=$ $\ds \map f {r_1} + \map f {r_2}$ Definition of Ring Homomorphism $\ds$ $=$ $\ds \map {\overline f} {r_1 + I} + \map {\overline f} {r_2 + I}$ Definition of $\overline f$

Therefore $\overline f$ is a homomorphism.

It remains to demonstrate uniqueness.

Suppose there were another homomorphism $g: R / I \to S$ with $f = g \circ \pi$.

Then we would require that, for all $r\in R$:

 $\ds \map g {r + I}$ $=$ $\ds \map f r$ Definition of Commutative Diagram $\ds$ $=$ $\ds \map {\overline f} {r + I}$ Definition of $\overline f$

That is:

$g = \overline f$

and so $\overline f$ is unique.

$\blacksquare$