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Given two smooth projective schemes $X$ and $Y$ over some algebraically closed field $k$, we have $X\times Y$ with the projections $p$ to $X$ and $q$ to $Y$. Furthermore we have a "nice" sheaf of algebras $R$ on $X$, i.e. locally free and of global dimension at most dim($X$), e.g. Azumaya or something similar. Given two $p^{*}R$-modules $M$ and $N$ on $X\times Y$, which are coherent and torsion free.

Like in the commutative case, i define the i-th relative $\mathcal{E}xt$-sheaf on $Y$ to be: $\mathcal{E}xt^i_{p^{*}R,q}(M,N):=(R^i(q_{*}\mathcal{H}om_{p^{*}R}(M,-))(N)$

Can I expect them to have the same properties as in the commutative case?

For example:

(1) Do we have $\mathcal{E}xt^i_{p^{*}R,q}(M,N)=0$ for $i>dim(X)$?

(2) Given $y\in Y$ is there a map $\mathcal{E}xt^i_{p^{*}R,q}(M,N)\otimes k(y) \rightarrow Ext_R^i(M_y,N_y)$

(3) If $Ext_R^i(M_y,N_y)=0$ for all $y\in Y$ does this imply $\mathcal{E}xt^i_{p^{*}R,q}(M,N)=0$?

(4) Is there a kind of base change theorem for the $\mathcal{E}xt^i_{p^{*}R,q}(M,N)$?

Or do I need more conditions for $M$ and $N$ to have the desired properties? I'm especially interested in the case, where $M=p^{*}P$ for some $R$-module $P$ on $X$.

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  • $\begingroup$ Can we presume that R is an O_X algebra; ie, an algebra object in the category of quasi-coherent sheaves? Note that this fails for some sheaves of algebras that still have nice localization properties, like differential operators. $\endgroup$ Commented Nov 9, 2010 at 21:10
  • $\begingroup$ Yes, $R$ is supposed to be an $O_X$-algebra. I'm mainly interested in the cases where $R$ is Azumaya, i.e. etale locally just $M_n(O_X)$ or a maximal order of suitable global dimension. $\endgroup$
    – TonyS
    Commented Nov 9, 2010 at 21:39

1 Answer 1

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Questions of this type are discussed in the paper A.Kuznetsov, Hyperplane sections and derived categories, Izvestiya: Mathematics 70:3 (2006) p. 447-547, which is available at

http://www.mi.ras.ru/~akuznet/publications/HyperplaneSectionsAndDerivedCategories.pdf

See Appendix D.

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  • $\begingroup$ Thanks, that looks promising. Maybe there are some questions left after reading the appendix. $\endgroup$
    – TonyS
    Commented Nov 10, 2010 at 18:00

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