I'm sure I'm missing something simple. In his paper Injective modules over Noetherian rings, Pacific J. Math 8 (1958), 511-28, Matlis has lemma 1.1:
Let $R$ be a ring, $S$ and $T$ $R$-modules, and $D$ an injective submodule of $S \oplus T$. Let $E$ be an injective envelope of $D \cap S$ in $D$, and let $F$ be a complementary summand of $E$ in $D$. Thus $D=E \oplus F$; and $E$ and $F$ project monomorphically into $S$ and $T$, respectively.
For context, $R$ is a ring with $1$ (and associative). All modules are left modules and unitary. What is eluding me is the statement in the proof "It is clear that $F$ projects monomorphically into $T$." Well, only the monomophic part actually.
$F$ is a submodule of $S \oplus T$, so consists of pairs $(x,y)$ with $x \in S$ and $y \in T$. The map $g:F \rightarrow T$ is defined by $g(x,y)=y$. Then $ker(g)=\{(x,0):x \in S\}$ is isomorphic to a submodule of $S$ and $ker(g)$ is a submodule of $F \cap S$. Where to go next, or what to do instead?
Any jolt for my currently addled brain would be appreciated.
I know that $ker(g) \subseteq F \cap S$. But $F \cap S \subseteq D \cap S \subseteq E$ and $E \cap F =0$, whence $F \cap S =0$ and the result follows.