TITLE: Submodule of a isotypic component
QUESTION [1 upvotes]: I'm reading Sophie Morel's notes about representation theory (see http://perso.ens-lyon.fr/sophie.morel/rep_theory_notes.pdf).
Let $R$ be a ring.
We write $S(R)$ for the set  of isomorphism classes of simple $R$-modules.
Theorem I.1.7.2.  Let $M$ be a semisimple $R$-module.
For every $S \in S(R)$, let $M_S$ be the sum of all the submodules of $M$ that are isomorphic to $S$.
Then :
(i) We have $M = \oplus_{S \in S(R)} M_S$.
(ii) There exist sets $I_S$ such that $M_S \simeq \oplus_{i \in I_S} S$ for every $S \in S(R)$.
(iii) Let $N$ be a submodule of $M$.
If we write $N = \oplus{S \in S(R)} N_S$ and $N_S \simeq \oplus_{S \in J_S} S$ as in (i) and (ii), then, for every $S \in S(R)$, $N_S = M_S \cap M$ for every $S \in S(R)$ and we can find an
injection $J_S \longrightarrow I_S$.
Lemma I.1.7.3. Let $M$ be a $R$-module, and suppose that there exists a simple $R$-module $S$ and a set $I$ and an isomorphism $M \simeq \oplus_{i \in I} S$.
Then every simple R-submodule of $M$ is isomorphic to $S$.
In the proof of (i) and (ii), it is use that every simple submodule of $M_S$ is isomorphic to $S$.
It follows from the lemma if we use (ii) and I think that (ii) is a consequence of the proof of the equivalence: a $R$-module $P$ is the direct sum of a family of simple modules $\iff$ $P$ is the sum of a family of simple modules (see Theorem I.1.3.4). Is that correct?
Can we prove that every simple submodule of $M_S$ is isomorphic to $S$ without using (ii)?

REPLY [0 votes]: Let $M$ be a semisimple $R$-module and $S \in S(R)$ and $V' \subseteq M$ a simple $R$-submodule not in $S$. As $M$ is semisimple, we have $M = V' \oplus M'$ for some submodule $M'$. Now, let $V \subseteq M$ be a simple submodule in $S$ (and thus in $M_S$). The projection $M = V' \oplus M' \to V'$ restricts to a map $V \to V'$ and since $V$ and $V'$ are non-isomorphic and simple, this map is necessarily trivial. It follows that $V \subseteq M'$ and thus $M_S \subseteq M'$. In particular, $V'$ is not contained in $M_S$ and thus every simple submodule of $M_S$ lies in $S$.
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