Let $R$ be an arbitrary ring and $N \vartriangleleft R$ nilpotent such that $R /N$ is semisimple. Show that N = rad(R).

Question

Let $R$ be an arbitrary ring and $N \vartriangleleft R$ nilpotent such that $R /N$ is semisimple. Show that $N = rad(R)$.

Any hints to get me started? I know that $rad(R/N)=N$ since $R /N$ is semisimple. Can I show that $rad(R)=rad(R/N)$?

Answer 1

As we know, $N\subset rad(R)$ and, for any associative ring $S$, $x\in rad(S)$ iff there exists $y\in S$ such that $x+y+xy=0$ (An element which has this property is called right-quasi-regular). So, if $x$ is r-q-r in $R$ then $\overline{x}$ is r-q-r in $\overline{R}=R/N$; hence $\overline{x}\in rad(\overline{R})=\overline{0}$. That is, $x\in N$. We have shown that $rad(R)\subset N$.

Answer 2

It seems you are using "semisimple" to mean "Jacobson semisimple", meaning it has trivial Jacobson radical. I'm going to use $J(R)$ instead of $rad$ since the latter is kind of overloaded.

In general, by definition of the Jacobson radical being the intersection of maximal right ideals, we have this fact: $R/I$ is Jacobson-semisimple only if $J(R)\subseteq I$.

On the other hand, it's easy to show that $J(R)$ contains nilpotent ideals (that's perhaps easier to see using the characterization of $J(R)$ as being the set of elements annihilating all simple modules.)

So these two things together say $J(R)\subseteq N$ and $N\subseteq J(R)$.