I'm trying to prove that the set $\mathrm{nilrad}(A)$ of nilpotent elements of $A$ is an ideal
Pf/ if $g\in\mathrm{nilrad}(A)$, then $g^n = 0$, for some $n>0$. Let $h$ be an element of $\mathrm{nilrad}(A)$. then $(gh)^n = (hg)^n = g^nh^n = 0\cdot0 = 0.$
also, $(-g)^n = (-1)^ng^n = (-1)^n\cdot 0 = 0.$
Now we show that $\mathrm{nilrad}(A)$ is closed under elements of $A$. Let $i$ be an element of $A$; then, $(gi)^n = g^ni^n = 0\cdot i^n = 0.$
$(g+i)^n = \sum_{i=0}^n \binom{n}{k}g^ni^{n-k}$
Also, how can we show that the set radical $\mathrm{rad}(I)$ for an ideal $I$ is an ideal? Pretty much trying to organize the proof. Thanks.