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In the first section of Altman and Kleiman's paper "Compactifying the Picard scheme", a base change map for Ext sheaves is defined. I am interested in knowing when this map is an isomorphism.

I recall the setting of the paper, so that nobody is forced to click on the link.

They start with a finitely presented morphism of schemes $f:X\to S$ and two locally finitely presented $\mathcal O_X$-modules $I,F$. Given a morphism $g:T\to S$ and a quasi-coherent $\mathcal O_T$-module $M$, they are able to define base change maps $$b^q(M):\mathscr Ext^q(I,F)\otimes_SM\to \mathscr Ext^q_{X_T}(I_T,F\otimes_SM)$$ under the additional assumption that $I$ is $S$-flat if $q\geq 1$ and $f$ is flat if $q\geq 2$.

Question. When is $b^q(k(s))$ an isomorphism for all $s\in S$?

Now, I am not sure I am interpreting Theorem 1.9 in the paper correctly, but part (i) seems to imply that if $b^q(k(s))$ is surjective at $x\in X_s$ then it is in fact an isomorphism. However, even if this was true, checking stalk surjectivity of all the maps in question does not seem to be very immediate to me. Is there some condition on the data $(I,F,f)$ which allows to conclude all $b^q(k(s))$ are isomorphisms? (We can of course assume all coherence and flatness needed to make sense of the statements.)

Thanks in advance!

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  • $\begingroup$ I think you want $M$ to be an $\mathcal{O}_S$-module, not an $\mathcal{O}_X$-module. $\endgroup$ Sep 25, 2015 at 2:10
  • $\begingroup$ @JasonStarr: Thanks! I meant $\mathcal O_T$-module. $\endgroup$
    – Brenin
    Sep 25, 2015 at 6:09
  • $\begingroup$ For the results of this form that I have seen, $F$ is $S$-flat and has proper support over $S$. For $q=0$, this is (essentially) EGA III_2, Corollaire 7.7.8, p. 70. $\endgroup$ Sep 25, 2015 at 16:54
  • $\begingroup$ @JasonStarr: thanks for your comment. Do you have a reference for the conditions you mentioned ($S$-flatness of $F$ and properness of the support)? $\endgroup$
    – Brenin
    Sep 26, 2015 at 11:07
  • $\begingroup$ That result in EGA has such a hypothesis. $\endgroup$ Sep 26, 2015 at 11:18

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