Corollary 1

Let M and P be A-modules and $u$ a surjective A-linear mapping from P to M. Suppose that P is projective and that there exists a nilpotent two-sided ideal $\mathfrak{a}$ of A such that the kernel N of $u$ is contained in $\mathfrak{a}P$. Let $M'$ and $M''$ be submodules of M whose direct sum is M. Then P is the direct sum of submodules $P'$ and $P''$ such that $u(P') = M'$ and $u(P'') = M''$.

Denote by B the subring of End$_A(P)$ consisting of the endomorphisms $f$ of P such that $f(N)\subset N$. Let $\overline{B}$ be the endomorphism ring of M. For $f\in B$, let $\overline{f}$ be the unique endomorphism of M such that $\overline{f}\circ u=u\circ f$. The mapping $f\mapsto \overline{f}$ is a ring homomorphism from B to B. Since the module P is projective, this homomorphism is surjective; its kernel $\mathfrak{b}$ consists of the endomorphisms $f\in B$ such that $f(P)\subset N$. Let $n$ be a natural number such that $\mathfrak{a}^n= 0$. We have

and, for every $f\in \mathfrak{b}$ and every integer $j\geqslant 0$,

because $N\subset \mathfrak{a}P$ by assumption. We therefore have $f(P)\subset \mathfrak{a}^jP$ for every natural number $j$ and every $f\in \mathfrak{b}^j$. In particular, we have $\mathfrak{b}^n= 0$.

Let $\varepsilon '$ be the projector of M with image $M'$ and kernel $M''$. By Proposition 7 applied to the ring B and the nilpotent two-sided ideal $\mathfrak{b}$, there exists an idempotent $e'$ in B such that $\overline{e'}=\varepsilon '$, that is, $\varepsilon '\circ u=u\circ e'$. Set $e''= 1-e'$, $\varepsilon ''=\overline{e''}, P'=e'$(P), and $P''=e''(P)$. Then P is the direct sum of the submodules $P'$ and $P''$, and we have

the equality $u(P'') = M''$ is proved analogously.