2 Typos

Let me give some more details on Mariano's comment: The ring of differential operators ala 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 write denote them by $D^{(m)}$, which are the subalgebras generated by operators of degree $\leq p^m$.

Berthelot abstractly defines, using

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 Berthelots Berthelot's $\mathcal{D}^{(0)}$.

1

Let me give some more details on Mariano's comment: The ring of differential operators ala EGA4 in this particular case will be a free $k[t]$-algebra generated by the following operators: We write $$\partial_t^{(n)}$$ 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 write them $D^{(m)}$, which are the subalgebras generated by operators of degree $\leq p^m$.

Berthelot abstractly defines, using partially divided powers, 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 Berthelots $\mathcal{D}^{(0)}$.