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:

Then:

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

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

since $I$ is an ideal.

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:

That is:
 * $g = \overline f$

and so $\overline f$ is unique.

Also see

 * First Isomorphism Theorem for Rings