$I$ semisimple + $R/I$ semisimple $\implies$ $R$ semisimple

Question

Let $R$ be a (not necessarily commutative) ring with unit. Let $I\subset R$ be an ideal that in turn is a ring with unit. Is there a theorem that says something like $I$ semisimple and and $R/I$ semisimple implies $R$ semisimple?

Answer 1

Let $a$ be the unit of the two-sided ideal $I$. Then the map $R\rightarrow R$, $x\rightarrow ax$ is a ring map, since $ax\cdot ay=axy$ ($a$ is the unit of $I$). The image of the map is $I$. Let the kernel be $J$. Furthermore $a$ is an idempotent, therefore $I\cap J=0$. Thus, $R=I\oplus J$.

If $A$ and $B$ are semi-simple rings, so is $A\oplus B$.

Answer 2

Here is another proof:

Let $a$ be the unit of the two-sided ideal $I\subset R$. If we proof, that $a$ is central in $R$, it is commonly known that $R=Ra\oplus R(1-a)$, where $I = Ra$ and $R(1-a)$ is isomorphic to $R/I$ by isomorphism theorems.

But it is very simple, that $a$ is central in $R$, since let $x\in R$, then $$ ax = axa = xa $$ because $ax$ and $xa$ are both in $I$ and $a$ is the unit in $I$.