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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 $D(N)\subset N?$

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    $\begingroup$ 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 $1$. $\endgroup$
    – YCor
    Commented Apr 8, 2020 at 7:40
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    $\begingroup$ 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 $B^2=0$. $\endgroup$
    – YCor
    Commented Apr 8, 2020 at 10:09
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    $\begingroup$ @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. $\endgroup$
    – Will Sawin
    Commented Apr 14, 2020 at 19:51
  • $\begingroup$ @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 $D(x)^{n^2}=0$). $\endgroup$
    – YCor
    Commented Apr 14, 2020 at 20:08

2 Answers 2

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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 $D(N)\subset N$.

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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.

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