Theorem 2

Let $A$ be a ring and $a$ a two-sided ideal of $A$. (a) The canonical mapping $\pi$ of $A$ onto $A/a$ is a ring homomorphism. (b) Let $B$ be a ring and $f$ a homomorphism of $A$ into $B$. If $f(a) = \{0\}$, there exists one and only one homomorphism $\bar{f}$ of $A/a$ into $B$ such that $f = \bar{f} \circ \pi$.

By construction, $\pi(x + y) = \pi(x) + \pi(y)$ and $\pi(xy) = \pi(x)\pi(y)$ for $x, y$ in $A$; also $\pi(1)$ is the unit $\varepsilon$ of $A/a$, whence (a).

Let $A^+$ be the additive group of $A$ and $B^+$ that of $B$; as $f$ is a homomorphism of $A^+$ into $B^+$, zero on the subgroup $a$ of $A^+$, there exists (§ 4, no. 4, Proposition 5) one and only one homomorphism $\bar{f}$ of $A^+/a$ into $B^+$ such that $f = \bar{f} \circ \pi$. Let $\xi, \eta$ be in $A/a$; choose $x, y$ in $A$ with $\pi(x) = \xi$ and $\pi(y) = \eta$; then $\xi \eta = \pi(xy)$, whence

and $\bar{f}(\varepsilon) = f(\pi(1)) = f(1)$, hence $\bar{f}$ is a ring homomorphism.