I want to prove that $ Rad ( \bigoplus_{i \in I} M_{i})= \bigoplus_{i \in I} Rad (M_{i})$, according to how Kasch sketched the proof in Modules and Rings in Corollary 9.1.5. I got the gist of the proof but there are a couple of steps that are a little bit confusing to me.
Suppose that $M = \bigoplus_{i \in I} M_{i}$. So, for proving that $Rad(M) \subseteq \bigoplus_{i \in I} Rad (M_{i})$ let's take $m \in Rad(M)$. Then $m = \sum_{i \in I} m_{i} \in Rad(M)$. Now taking the projection $\pi_{i}: M \to M_{i}$, where $\pi_{i}(m) = m_{i} \in Rad(M_{i})$. The part I don't get from this proof is the equality at the beginning $m = \sum_{i \in I} m_{i} \in Rad(M)$. I mean where are this $m_{i}$ at the beginning, I guess it has something to do with the fact that the radical of the module is the sum of all superfluous submodules of a module but I'm not quite sure.
To prove that $\bigoplus_{i \in I} Rad (M_{i}) \subseteq Rad(M)$. Since $M_{i} \leq M$ we got that $Rad(M_{i}) \leq Rad(M)$. Therefore $\sum_{i \in I} Rad(M_{i}) \leq Rad(M) $. Then the book states that $\sum_{i \in I} Rad(M_{i}) = \bigoplus_{i \in I} Rad (M_{i}) \leq Rad (M)$ finishing the last part of the proof. What I don't get is why $\sum_{i \in I} Rad(M_{i}) = \bigoplus_{i \in I} Rad (M_{i})$, is there any propery I'm missing here? Anyway, thanks for reading and helping me out.
(1) By definition of coproduct, every element in $M$ has the form $m = \sum_{i\in I} m_i$, where only finite $m_i$ can be nonzero. To make the proof complete, you can prove the following lemma.
Lemma 1. If $N$ and $P$ are $R$-modules, and $f: N \longrightarrow P$ is a surjective module morphism. Then $f(Rad(N)) \subset Rad(P)$.
Proof sketch: This comes from the fact that $N/ \ker f \cong P$.
(2) The sum $\oplus_{i\in I} Rad(M_i)$ is a direct sum in $M$. This is a basic property in the theory of modules as showed in the following proposition.
Proposition. If $\{N_i\}_{i\in I}$ is a family of submodules of an $R$-module $N$, and for any finite set of indices $J = \{i_1, \dots, i_j\}$, $N_{i_1} \cap (\sum_{k \in J-\{ i _1\}} N_{i_k} ) = 0$. Then $\sum_{i \in I} N_i$ is the direct sum $\oplus_{i\in I} N_i$.