Timeline for Is there a really big ring of differential operators in characteristic p?
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| Mar 9, 2012 at 14:13 | comment | added | Lars |
Ah, I wonder which of Bertholot's rings he means, and what precisely is $\delta^{[p]}$. In general, if $\delta$ is an operator of order 1, then $\delta^p=0$ in $\mathcal{D}^{(n)}$ for $n>1$, but not necessarily $\mathcal{D}^{(0)}$. Modules over$\mathcal{D}^{(0)}$ are connections. Now if by $\delta^{[p]}$ you mean what I wrote as $\delta^{(p}_t$,then that′s an operator of order $p$, and cannot be considered as an elementof $\mathcal{D}^{(0)}$, but if I remember correctly, the same reasoning applies.
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| Mar 9, 2012 at 7:44 | comment | added | S. Carnahan♦ | YoungMathematician asks (in a deleted answer) whether ${\partial^{[p]}}^p=0$ holds in Berthelot's ring. | |
| Mar 8, 2012 at 18:39 | history | edited | Lars | CC BY-SA 3.0 |
Typos
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| Mar 8, 2012 at 15:28 | history | answered | Lars | CC BY-SA 3.0 |