Localizability of differential operators a la Grothendieck - MathOverflow most recent 30 from http://mathoverflow.net2013-05-23T03:01:16Zhttp://mathoverflow.net/feeds/question/81707http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/81707/localizability-of-differential-operators-a-la-grothendieckLocalizability of differential operators a la GrothendieckSasha2011-11-23T14:00:24Z2012-10-16T10:41:08Z
<p>Hello,</p>
<p>Maybe this question is trivial, so sorry</p>
<p>Let $A$ be a (comm. with 1) $k$-algebra, where $k$ is a ring (comm. with 1).</p>
<p>Then we can define the module of differential operators $D^{\leq n} (A)$, a submodule of $End_k (A,A)$ (endomorphisms of vector spaces). $D^{\leq -1} = 0$, and then inductively $D^{\leq n} = { d | [d,a]\in D^{\leq n-1}}$.</p>
<p>We have a lemma:</p>
<p>Lemma. Let $f \in A$. Then for every $d \in D^{\leq n}(A)$ we can find unique $e \in D^{\leq n}(A_f )$, such that $l\circ d = e \circ l$, where $l: A \to A_f$ is the localization map.</p>
<p>I think that I know how to prove the lemma, by induction on the order of diff. op. (just need to see how to apply operators to fractions). It gives us a map $D^{\leq n}(A) \to D^{\leq n}(A_f)$.</p>
<p>Question 1. Under which assumptions on $A/k$ this map $D^{\leq n}(A) \to D^{\leq n}(A_f)$ is a localization map (i.e. becomes an isomorphism after tensoring (say on the left, it does not matter) with $A_f$)?</p>
<p>Question 2 (my real question). If $A/k$ is finitely generated, or finitely presented, is this a localization map?</p>
<p>Somehow, I am having trouble with the "surjectivness" part. Maybe there is some reference?</p>
<p>Thank you,
Sasha</p>
http://mathoverflow.net/questions/81707/localizability-of-differential-operators-a-la-grothendieck/81718#81718Answer by Mariano Suárez-Alvarez for Localizability of differential operators a la GrothendieckMariano Suárez-Alvarez2011-11-23T16:09:18Z2011-11-23T21:03:08Z<p>Two observations (with $k$ a field of characteristic zero):</p>
<ul>
<li><p>If $A$ is a domain over a field $k$, then elements of $D(A)$ extend to elements of $D(A_S)$ for all multiplicatively closed sets $S$.</p></li>
<li><p>Your questions become easier if you ask instead about the subalgebra $\Delta(A)$ of $D(A)$ generated by $A$ and derivations: then the answer is yes to your two questions. Now, if $A$ is finitely generated and regular, then $D(A)=\Delta(A)$, so in this case the answer is yes for $D(A)$ too.</p></li>
</ul>
<p>You'll find this in the last chapter of McConnell and Robson's book on noetherian rings.</p>
<p>(Finally, $D^{\leq n}(A)$ is an $A$-bimodule which is <em>not</em> symmetric, so tensoring with $A_f$ on one side or the other is not the same thing)</p>