Theorem 3

Let $A$ and $B$ be rings and $f$ a homomorphism of $A$ into $B$. (a) The kernel $a$ of $f$ is a two-sided ideal of $B$. (b) The image $B' = f(B)$ of $f$ is a subring of $B$. (c) Let $\pi : A \to A/a$ and $i : B' \to B$ be the canonical morphisms. There exists one and only one morphism $\bar{f}$ of $A/a$ into $B'$ such that $f = i \circ \bar{f} \circ \pi$ and $\bar{f}$ is an isomorphism.

As $f$ is a morphism of the additive group of $A$ into that of $B$, $a$ is a subgroup of $A$. If $x \in a$ and $a \in A$, then $f(ax) = f(a)f(x) = 0$, hence $ax \in a$ and similarly $xa \in a$; hence $a$ is a two-sided ideal of $A$. Assertion (b) is obvious. As $f$ is zero on $a$, there exists a morphism $\bar{f}$ of $A/a$ into $B'$ such that $f = i \circ \bar{f} \circ \pi$ (Theorem 2). The uniqueness of $\bar{f}$ and the fact that $\bar{f}$ is an isomorphism follow from Set Theory, II, § 6, no. 4.