Timeline for Does any derivation of commutative algebra preserve its nil-radical?
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9 events
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| Apr 14, 2020 at 20:08 | comment | added | YCor | @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 $D(x)^{n^2}=0$). | |
| Apr 14, 2020 at 19:51 | comment | added | Will Sawin | @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 8, 2020 at 16:45 | answer | added | Marc | timeline score: 11 | |
| Apr 8, 2020 at 13:43 | history | became hot network question | |||
| Apr 8, 2020 at 10:09 | comment | added | YCor | 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 $B^2=0$. | |
| Apr 8, 2020 at 7:40 | comment | added | YCor | 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 $1$. | |
| Apr 8, 2020 at 6:32 | vote | accept | asv | ||
| Apr 8, 2020 at 6:24 | answer | added | Vladimir Dotsenko | timeline score: 24 | |
| Apr 8, 2020 at 5:42 | history | asked | asv | CC BY-SA 4.0 |