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$k$ is a field of characteristic $p$.

$k[t]$ has canonical first-order differential operator $\partial$

As an endomorphism of $k[t]$, $\partial^p=0$.

First way to fix it: Use the divided power differential operator $\partial^{[p]}$. Shortfall: As an endomorphism of $k[t]$, ${\partial^{[p]}}^p=0$

Second way to fix it: Use crystalline differential operators. Shortfall: No higher order operators on $k[t]$.

Question:

Is there a really big ring of differential operators which contains the divided powers $\partial^[n]$ for all $n$ and which also has a natural evaluation map to $End_k(k[t])$?

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    $\begingroup$ There is the ring of diferential operators à la Grothendieck, which is infinite dimensional and has elements of all orders. It is an increasing union of matrix rings in the case of the line, though, so it is a bit strange. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Mar 8, 2012 at 0:28
  • $\begingroup$ The comment of Mariano and resulting answer by Lars are good. However, you may be interested to look here: mathoverflow.net/questions/56860/… $\endgroup$
    – B. Bischof
    Commented Mar 8, 2012 at 18:10

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Let me give some more details on Mariano's comment: The ring of differential operators a la EGA4 in this particular case will be a free $k[t]$-algebra generated by the following operators: We write $$\partial_t^{(n)}$$ for the operator which is defined by $$\partial_t^{(n)}(t^m)={m\choose n}t^{m-n}.$$ Because of this, sometimes the notation $$\partial_t^{(n)}=\frac{1}{n!}\frac{\partial^n}{\partial t^n}$$ is used.

Actually, to generate the ring, the operators $\partial_t^{(p^n)}$ suffice.

Now this ring is not noetherian, but it is an increasing union of noetherian subalgebras, lets denote them by $D^{(m)}$, which are the subalgebras generated by operators of degree $\leq p^m$.

Using partially divided powers, Berthelot abstractly defines rings $\mathcal{D}^{(m)}$ such that the full ring of differential operators $\mathcal{D}$ is the direct limit of the $\mathcal{D}^{(m)}$. The image of $\mathcal{D}^{(m)}$ in $\mathcal{D}$ is then precisely the $D^{(m)}$ that I defined ad-hoc above. The crystalline operators that you defined in the question correspond to Berthelot's $\mathcal{D}^{(0)}$.

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  • $\begingroup$ YoungMathematician asks (in a deleted answer) whether ${\partial^{[p]}}^p=0$ holds in Berthelot's ring. $\endgroup$
    – S. Carnahan ♦
    Commented Mar 9, 2012 at 7:44
  • $\begingroup$ 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. $\endgroup$
    – Lars
    Commented Mar 9, 2012 at 14:13

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