Is there a really big ring of differential operators in characteristic p? $k$ 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 $End_k(k[t])$?
 A: 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 $\mathcal{D}^{(0)}$.
