Is there a really big ring of differential operators in characteristic p? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T20:59:25Z http://mathoverflow.net/feeds/question/90522 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90522/is-there-a-really-big-ring-of-differential-operators-in-characteristic-p Is there a really big ring of differential operators in characteristic p? YoungMathematic 2012-03-08T00:18:41Z 2012-03-08T23:48:46Z <p>$k$ is a field of characteristic $p$.</p> <p>$k[t]$ has canonical first-order differential operator $\partial$</p> <p>As an endomorphism of $k[t]$, $\partial^p=0$.</p> <p>First way to fix it: Use the divided power differential operator $\partial^{[p]}$. Shortfall: As an endomorphism of $k[t]$, ${\partial^{[p]}}^p=0$</p> <p>Second way to fix it: Use crystalline differential operators. Shortfall: No higher order operators on $k[t]$.</p> <p>Question:</p> <p>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 $End_k(k[t])$?</p> http://mathoverflow.net/questions/90522/is-there-a-really-big-ring-of-differential-operators-in-characteristic-p/90585#90585 Answer by Lars for Is there a really big ring of differential operators in characteristic p? Lars 2012-03-08T15:28:05Z 2012-03-08T18:39:45Z <p>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.</p> <p>Actually, to generate the ring, the operators $\partial_t^{(p^n)}$ suffice. </p> <p>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 $\leq p^m$.</p> <p>Using <em>partially</em> 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 $\mathcal{D}^{(0)}$.</p>