Radicals and direct sums. Let A be a K-algebra and M, N be right A-submodules of a right A module L, $M \cap N =0$. How to show that $(M\oplus N) \text{rad} A = M \text{rad} A \oplus N \text{rad} A$? Let $m \in M, n\in N, x\in \text{rad} A$. Since $(m+n)x=mx+nx$, $(m+n)x \in M \text{rad} A \oplus N \text{rad} A$. Since $M, N$ are submodules, $mx \in M, nx\in N$. Therefore $ M \text{rad} A \cap N \text{rad} A = 0$. Is this true? Thank you very much.

Sure, it's obvious that for any $S \subseteq A$, we have $MS\subseteq M$ and $NS\subseteq N$. That means $MS\cap NS\subseteq M\cap N =\{0\}.$
But this has little to do with the sum being direct: the real question here is how to get the equality  
$$(M\oplus N) \text{rad} A = M \text{rad} A \oplus N \text{rad} A.$$
The definition of the $A$ action on the direct sum by "distribution" proves that the left-hand side is contained in the right-hand side.
The other containment is also true, but seeing that hinges on your proper understanding of the definition of $MI$ where $M$ is an $A$ module and $I\lhd A$. Do you see why the final containment holds?
