# Does any derivation of commutative algebra preserve its nil-radical?

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 $$D(N)\subset N?$$

• By Vladimir Dotsenko's answer, this works over an arbitrary associative commutative ring whose underlying abelian group $(A,+)$ is torsion-free (and in particular when $(A,+)$ is torsion-free divisible, which is the context of the question). For context, it fails in finite characteristic $p$: if $A=K[x]/x^p$, then since $D(x^p)=0$ for $D$ the ordinary derivation of $K[x]$, it induces a derivation of $A$, which maps the nilpotent element $x$ to the non-nilpotent element $1$.
– YCor
Apr 8 '20 at 7:40
• Another remark is that it fails in general associative algebras, including characteristic zero: in $\mathrm{Mat}_2$, for every matrix $A$, the assignment $D_A:B\mapsto AB-BA$ is a derivation, but for the basis matrices $A=E_{21}$ and $B=E_{12}$, we have $D_A(B)=E_{22}-E_{11}$ non-nilpotent although $B^2=0$.
– YCor
Apr 8 '20 at 10:09
• @YCor In fact Dotsenko's argument shows that if $x$ is nilpotent of order $n$, where $n$ is less than the characteristic, then $Dx$ is nilpotent, possibly of higher order. So your example is sharp in this sense. Apr 14 '20 at 19:51
• @WillSawin yes, if the characteristic is prime. In general the assertion is that if $x^n=0$ and $y\mapsto pm$ is injective for every prime $p\le n$ (or equivalently if $y\mapsto n!y$ is injective) then $Dx$ is nilpotent (namely $D(x)^{n^2}=0$).
– YCor
Apr 14 '20 at 20:08

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 $$D(N)\subset N$$.
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.