Let $R$ be a nonzero ring with unity. Assume that ${R}^{0}$ is the regular $FG-module$ and $I$ and $L$ are left ideals of $R$ so that $I$ and $L$ are submodules $R^{0}$. Then show that if there exists a nonzero element $z$ in $Z(R)$ with $z^2=0$, then $R^0$ cannot be completely reducible.
Well, I know the following:
$I$ is an R-module direct summand of R iff $I= Re$ for some idempotent $e$. And if $I$ is a minimal left ideal, then $I$ is an R-module direct summand of R iff $I^2$ is not equal to $0$.
I thought about coming up with an ideal of the form $I=Rz$, where all elements are $zx=x$ for all $x$ in $I$ and showing that $I^2=0$. However, something smells fishy. Can you provide some hint to start?
Of course something's fishy. It's called a contradiction. You're looking for something like that.
Suppose $R$ is completely reducible, so that all left ideals are summands and $R$ is left and right Artinian. Since $z$ is central and $z^2=0$ you get right away that $(Rz)^2=\{0\}$. (There is nothing about "$zx=x$ for all $x\in I$" nor is it needed.)
Since $R^0$ is left Artinian, it contains a has a minimal submodule $L$ which also satisfies $L^2=\{0\}$. You already said you know $L$ is a summand iff $L^2\neq\{0\}$, so this would be a submodule that isn't a summand. That's a contradiction.