Proposition 5

Let $A$ and $A'$ be two rings, $f$ a homomorphism of $A$ into $A'$ and $a$ the kernel of $f$. (a) Let $B'$ be a subring of $A'$. Then $B = f^{-1}(B')$ is a subring of $A$ containing $a$. If $f$ is surjective, then $f(B) = B'$ and $f|_B$ defines when passing to the quotient an isomorphism of $B/a$ onto $B'$. (b) Let $b'$ be a left (resp. right, two-sided) ideal of $A'$. Then $b = f^{-1}(b')$ is a left (resp. right, two-sided) ideal of $A$ containing $a$.

(c) *If $b'$ is a two-sided ideal of $A'$, the composite mapping of the canonical morphism $A' \to A'/b'$ and $f : A \to A'$ defines, when passing to the quotient, an injective morphism $\bar{f}$ of $A/b$ into $A'/b'$. If $f$ is surjective, $\bar{f}$ is an isomorphism of $A/b$ onto $A'/b'$.

(d) *Suppose $f$ is surjective. Let $\Phi$ be the set of subrings (resp. left ideals, right ideals, two-sided ideals) of $A$ containing $a$. Let $\Phi'$ be the set of subrings (resp. left ideals, right ideals, two-sided ideals) of $A'$. The mappings $B \to f(B)$ and $B' \mapsto \bar{f}(B')^{-1}$ are inverse bijections of $\Phi$ onto $\Phi'$ and $\Phi'$ onto $\Phi$.

(a) and (b) are obvious, except the last assertion of (a) which follows from no. 7, Theorem 3.

The composite morphism $g : A \to A' \to A'/b'$ considered in (c) has kernel $b$ and hence $\bar{f}$ is an injective morphism of $A/b$ into $A'/b'$ (§ 8, no. 7, Theorem 3). If $f$ is surjective, $g$ is surjective and hence $\bar{f}$ is surjective.

Suppose $f$ is surjective. By the above, the mapping $\theta : B' \mapsto \bar{f}(B')^{-1}$ is a mapping of $\Phi'$ into $\Phi$. Clearly the mapping $\eta : B \mapsto f(B)$ is a mapping of $\Phi$ into $\Phi'$. Then $\theta \circ \eta = \mathrm{Id}_{\Phi}, \eta \circ \theta = \mathrm{Id}_{\Phi'},$ whence (d).