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Bernie
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Modules over an Azumaya algebra and modules over the associated Brauer-Severi variety
@Jason : Ah, I see. I wasn't thinking "etale" enough :-). Thanks a lot for your help.
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Modules over an Azumaya algebra and modules over the associated Brauer-Severi variety
@Jason. Thanks, what you wrote is what I looking for :-). Denoting your first functor by $F$ and the second one by $G$, I can show that $G\circ F=id$ using the projection formula. But if I try to compute $F\circ G$ using base change and the formula $f_{*}Hom_X(f^{*}E,N)=Hom_Y(E,f_{*}N)$ I end up with a sheaf of the form $f^{*}f_{*}(N)$ where $N$ is a bundle on $X$ with the desired properties. Why is this isomorphic to $N$? Maybe a silly question, but usually the natural morphism $f^{*}f_{*}N\rightarrow N$ is far from being an isomorphism.
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Modules over an Azumaya algebra and modules over the associated Brauer-Severi variety
@Jason : My problem is not with the sheaf $I$ (all about $I$ is in Quillen's article Higher algebriac K-theory I), but with sheaves of $End(I)$ modules. For examples, assume $M$ has rank one as an $A$-module and is locally free, then the corresponding $O_X$-module, is locally free of rank $r$ on $X$. Does every rank $r$ bundle appaer this way? Does every rank $r$ bundle on $X$ give an $A$-module of rank one after tensoring with the dual of $I$ and pushing down to $S$? I don't see how to find properties of the bundles on $X$, maybe all possible bundles can appear?
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Surjectivity locus of a morphism of families of sheaves
@abx : Can you elaborate a little bit? I don't see how this helps. My idea was to use the surjectivity of $\phi_{t_0}$ to get surjectivity of $\phi_{(x_0,t_0)}$ at some point $(x_0,t_0)$. This surjection would extend to some open subset $U$ in $X\times T$. Then my hope was that the image of $U$ under the projection could be an open subset in $T$. But then the projection had to be open. So maybe this is not a really good idea?
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How to find ideals of finite length in a power series ring with special properties?
@Mohan: If you turn your comment into a quick answer, then I can accept the answer, so the question in this form is resolved :-)
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How to find ideals of finite length in a power series ring with special properties?
@Mohan: Do you mean something like this: every morphism $f: J \rightarrow R/J$ must have image in $(x,y)/J$, because we can make things like $[y]f(x^2)=f(yx^2)=[x]f(xy)$ and so on? So no morphism can be surjective in this case.
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How to find ideals of finite length in a power series ring with special properties?
@Jason Starr: Yes, exactly. An ideal $I$ of finite length in $A$ is an ideal such that $A/I$ has finite length.
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An application of the Base Change Theorem to the moduli space of sheaves
I don't know if the two approaches are basically the same, maybe this is the case. But Mukai needs the fact that $\mathcal{I}_{\Delta}$ annihilates the sheaf, because then the relative Ext is the pushforward of a sheaf on the diagonal to the whole of $M\times M$. This sheaf is the line bundle Mukai talks about. Meanwhile I found another proof of this fact in Huybrechts and Lehn, it is in the chapter about low dimensional examples of moduli spaces on K3-surfaces, it explains what happens a little bit more.
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An application of the Base Change Theorem to the moduli space of sheaves
Yes, that is what i meant with regards to the spectral sequence, we cannot write the relative Ext in general in that form. For the second question: i don't think $\mathcal{H}om(F,F)$ is selfdual if $F$ is not locally free, that is Serre duality does not work. We only have $Ext^2(F,F)\cong Hom(F,F)^{\vee}$. I recommend to look in the article "Universal family of extensions" of H.Lange. There is a base change formula which may help you.
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