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
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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.
$\blacksquare$