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Let $k$ be a field. Let $A$ be a commutative $k$-algebra. A $k$-derivation is $k$-linear map $\partial:A\to A$ such that $\partial(fg)=f\partial g+g\partial f$.

Let $G$ be an affine algebraic group over $k$. If there is a dual action of $G$ on $A$, then any $\xi\in\mathrm{Lie}(G)$ defines a derivation.

It is very natural to guess that a nilpotent element is mapped by a derivation to a nilpotent element. If the derivation is from a group action, then an idea to prove this is to show that the nilradical is $G$-invariant and then $\mathfrak g$-invariant. However there may be derivations not from group actions. How about these derivations? Can they send nilpotent elements to non-nilpotent elements?

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Perhaps this is a counter-example when $k$ is of characteristic 2:
Take $A=k[x]/I$ with $I=(x^2)$, and $\partial(ax+b+I)\mapsto a+I$, which is a derivation since for all $f=ax+b+I, g=cx+d+I$ we have

$$ f\partial g+g\partial f=(ax+b+I)(c+I)+(cx+d+I)(a+I)=(\underbrace{ac+ac}_{0})x+ad+bc+I=ad+bc+I=\partial((ad+bc)x+bd+I)=\partial(fg) $$ (and $\partial$ is clearly $k$-linear)

Note that $\partial (x+I)=1+I$, so we have that a nilpotent element is sent to a non-nilpotent element.

I haven't checked it fully but I suspect a similarly defined derivation (taking the polynomial derivative of the representative) would work for $k[x]/(x^p)$ when $k$ is of characteristic $p>0$. (the only "problem part" should be the coefficient of $x^{p-1}$ in $f\partial g+ g\partial f$, which the characteristic "handles")

Mor A.
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