# Injective submodule of a direct sum

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.

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To anyone who is considering an answer to this question, please be advised that it is, indeed, clear. The rather trivial reason came to me in one of those "Aha" moments we hear so much about. –  Chris Leary Feb 19 at 17:13
Just for completeness's sake, would you be interesting in writing up your "aha" moment as an answer? –  Nick Feb 20 at 18:28

## 1 Answer

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.

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