Given a commutative associative unital algebra over a field of characteristic zero.



> 
> Is it true that any derivation of it preseves its nil-radical?
> 
> 
> 


More explicitly, let $D$ be a derivation of an algebra $A$. Let $N$ denote the nil-radical of $A$.



> 
> Is it true that $D(N)\subset N?$
> 
> 
> 



Does any derivation of commutative algebra preserve its nil-radical?
Suppose $x\in N$, so that $x^n=0$ for some $n$. Then using the product rule for derivations many times, we see that
 $$ 
0=D^n(x^n)=n! D(x)^n+Y,
 $$
where $Y$ is divisible by $x$. Therefore,
 $D(x)^{n^2}=(D(x)^n)^n$ is divisible by $x^n$, and therefore vanishes. Thus, $D(x)$ is nilpotent, and therefore $D(N)\subset N$.


