Inner derivations determined by idempotent elements and nilpotent elements

Question

If R is a prime ring, $e\in R$ is a non-central idempotent and $d$ is an inner derivation determined
by $e$, then $0\neq d=d^{3}=d^{5}...$
If $b\in R$ is a non-central nilpotent element of a degree $n$ and $d$ is an inner derivation determined by $b$,then $d^{2n-1}=0$,although $d \neq0$
How can we show above examples ?

Answer 1

Get a feel for the pattern:

$$bx-xb \\ b^2x-2bxb+xb^2\\ b^3x-3b^2xb+3bxb^2-xb^3$$

By this point, the formula $d^n(x)=\sum_{i=0}^{n} (-1)^i\binom{n}{i}b^{n-i}xb^{i}$ suggests itself. Prove it using induction.

You see that in order for all terms to hit zero, a power of b in every term has to hit n. $2n$ certainly works, but you can further show that $2n-1$ is sufficient.

If b is idempotent, we see the third line cancels down to the first, hence $d^3=d$, and by induction $d^{2n+1}=d$.

Finally, I hope you already see that an element is central iff its derivation is zero.