Proposition 10

Let $E$ be an a-module. For $E$ to be injective, it is necessary and sufficient that, for every ideal $a$ of $A$ and every A-homomorphism $f : a\to E$, there exist $e\in E$ such that

for every $a\in a$.

Suppose $E$ is injective; let $a$ be an ideal of $A$, $f : a\to E$ an A-homomorphism, and denote by $i : a\to A$ the canonical injection. Then the mapping

is surjective (def. 2); if $g\in\operatorname{Hom}_A(A,E)$ is such that $f=g\circ i$, one has

for every $a\in a$.

Conversely, suppose the condition of the statement is satisfied, let $M$ be an A-module, $N$ a submodule of $M$, $u : N\to E$ an A-homomorphism, and prove that there exists an A-homomorphism $\bar u : M\to E$ extending $u$ (cf. lemma 3). Let $\mathscr P$ be the set of pairs $(P,v)$ where $P$ is a submodule of $M$ containing $N$ and $v$ is a homomorphism from $P$ into $E$ extending $u$. The set $\mathscr P$ ordered by the extension relation is inductive: if $(P_j, v_j)$ is a totally ordered family of elements of $\mathscr{P}$, put $Q = \cup P_j$ and let $w : Q \to E$ be the unique mapping inducing $v_j$ on $P_j$ for every $j$; then $(Q, w) \in \mathscr{P}$ and $(Q, w)$ majorizes $(P_j, v_j)$ for every $j$. Let then $(P, v)$ be a maximal element of $\mathscr{P}$ (E, III, p. 20, th. 2); it suffices to prove that $P = M$. Let $x \in M$ and let $\alpha$ be the ideal of the $a \in A$ such that $ax \in P$; put $f(a) = v(ax)$ for $a \in \alpha$; one thus obtains an $A$-homomorphism $f : \alpha \to E$. Let then $e$ be an element of $E$ such that $f(a) = ae$ for every $a \in \alpha$. Put $P' = P + Ax$ and let $v' : P' \to E$ be the unique $A$-homomorphism such that $v'(p + ax) = v(p) + ae$ for $p \in P, a \in A$; then $(P', v')$ belongs to $\mathscr{P}$ and majorizes $(P, v)$, hence $P' = P$, that is to say $x \in P$, which completes the proof.