Semiprimitive rings and Modules I'm finding for simple proofs of the followings facts:
$\bullet$ rad$(\Lambda/rad \Lambda)=0$
$\bullet$ rad$(A/rad \Lambda \ . A)=0$
$\bullet$ rad$(A/rad \Lambda \ . A)= rad A/ (rad\Lambda. A)$
where $\Lambda$ is a left Artin ring and $A$ is a $\Lambda$-module.
The first is a consequence of the $\Lambda/rad \Lambda$ is semisimple, but this is I want to conclude. The second is like the first if we consider $rad A=rad\Lambda \ . A$, but this I want to conclude too. So I want proof these facts without using the results I cited. 
 A: In the following I denote the Jacobson radical of an $R$ module $M$ as $J(M)$.
The fact that $J(R/J(R))=\{0\}$ is an elementary fact that holds for all rings.
Consider: the maximal right ideals of $R/J(R)$ are precisely of the form $I/J(R)$ where $I$ is a maximal right ideal of $R$.  The intersection of the maximal right ideals of $R/J(R)$ is, therefore, equal to $J(R)/J(R)=\{0\}$.
Now if you additionally have that $R$ is right Artinian, then $R/J(R)$ is right and left Artinian, indeed a semisimple ring owing to the fact we just proved. All modules over such a ring are also semisimple modules.
Of course, in a semisimple module $M$, $J(M)$ has to be equal to zero. For if it were not, then suppose $S$ is a simple submodule of $J(M)$: then $S\oplus N=M$ for some (obviously maximal) submodule $N$ of $M$. But $J(M)$ cannot be contained in $N$, a contradiction.
Now since $J(R)$ annihilates $\bar M=\frac{M}{MJ(R)}$, it is a $R/J(R)$ module, which is semisimple, hence $\bar M$ is semisimple, with zero Jacobson radical.
To me, the third item seems to be proven exactly as the first point, using correspondence of submodules.
