TITLE: Does any derivation of commutative algebra preserve its nil-radical?
QUESTION [14 upvotes]: 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 $D(N)\subset N?$

REPLY [11 votes]: Here is another cute argument (I don't remember where I learned it, I think it is folklore). Let $P\subset A$ be an arbitrary prime ideal. We claim it contains a $D$-stable prime ideal. For this, consider the mod $P$ Taylor map 
$$ f\colon A\to (A/P)[[t]] , a \mapsto \sum_{n\geq 0} \frac{D^n(a) \textrm{ mod } P}{n!} t^n.$$
A quick computation shows that $f$ is a ring map, and that for all $a\in A$ we have $f(D(a)) = \frac{d}{dt}(f(a))$. Therefore, the kernel $Q = \mathrm{ker}(f)$ is a $D$-stable ideal of $A$. Moreover, $Q$ is prime because $(A/P)[[t]]$ is a domain, and we have $Q \subset P$ because the constant term of $f(a)$ is $a \textrm{ mod } P$. 
So every prime ideal of $A$ contains a $D$-stable prime ideal. Hence, the intersection of all prime ideals of $A$ equals the intersection of all $D$-stable prime ideals of $A$. But the former is the nilradical, and the latter is clearly $D$-stable.