semisimple ring implies semisimple module I'm attempting to do part b of the problem below.

I already did part a using the proposition and I just want to check if my attempt for b is correct.
Let $M$ be an $A$-module and as hinted by the problem, I let $\mathcal{C}$ be the collection of all submodules of $M$ that is semisimple, partial ordered by inclusion.  We know this collection is nonempty since we can pick any finitely generated submodule of $M$ and using part a.  Any chain in this collection also have an upper bound by taking the union of the chain.  Hence by Zorn's lemma, we have a submodule $N$ maximal to the condition of being semisimple, so $N = \oplus_{i \in I} S_i $ where $S_i$ is a simple submodule.
If $N = M$, then we are done.  If $N \neq M$, then we can pick $m \in M \backslash N$.  Since $ Am $ is finitely generated, by part a, we have that $Am = \oplus_{j \in J} U_j$ where $U_j$ are simple.  Since both $U_j$ and $S_i$ are simple, either $U_j \cap S_i = 0$ or the $U_j$'s might be identical to some of the $S_i$, so get rid of the repeated ones to obtain $J^* \subseteq J$.
Finally, $N \subsetneqq N + Am \subseteq N + (\oplus_{j \in J} U_j) = N \oplus (\oplus_{j \in J^*} U_j)$ which the latter is semisimple, contradict the maximality of $N$.
Is my argument above correct?  Thank you all.
Edit: Correction from Rschwieb's comment
As pointed out by Rschwieb, the sum might not intersect trivially as I carelessly thought.  So starting again from $Am  = \oplus_{j \in J} U_j$ where $U_j$ are simple.  Since $m \notin N$ and all the $U_j$'s are simple, there exist a $U_{\alpha}$ such that $U_{\alpha} \cap N = 0$.
Then $N \subsetneqq N + U_{\alpha} = N \oplus U_{\alpha}$ and the latter is semisimple, which contradict the maximality of $N$.
 A: Most is right, but the last line doesn't hold: even if the $U_j$ intersect $N$'s summands trivially, there's no reason to say that their sum intersects $N$ trivially.  And the second to the last paragraph is a little squishy about manipulating summands.
For example, $\{(x,x)\mid x\in F\}$ trivially intersects $\{(0,x)\mid x\in F\}$ and $\{(x,0)\mid x\in F\}$, but the sum of the latter two do no intersect the first trivially.
Anyhow, that extra reasoning is not even necessary if you know the basic exercise that the following are equivalent:

*

*$M$ is a direct sum of simple modules

*$M$ is a sum of simple modules

Because of this, you can just conclude "then $N+Am$ is a sum of simple modules, hence semisimple. But this contradicts the maximality of $N$. Therefore $M=N$ in the first place."
Alternatively, you can just use a single one of $Am$'s submodules to draw a contradiction, since it is true that it must be contained in $N$ or intersect $N$ trivially.  (Obviously not all of $Am$'s factors can be contained in $N$.
But your approach is correct in large part.
