Consider the cartesian square of schemes $$ \require{AMScd} \begin{CD} X' @>{g'}>> X \\ @V{f'}VV @VV{f}V \\ S' @>>{g}> S \end{CD} $$ and the base change map $$ \eta : g^*(R^qf_*\mathcal{F}) \longrightarrow R^qf'_*(g'^*\mathcal{F}). $$ I know there exists the smooth base change theorem which assures $\eta$ is an isomorphism if $g$ is smooth, $f$ is quasi-compact, $\mathcal{F}$ is a torsion sheaf and the torsion of $\mathcal{F}$ is prime to the characteristic of the residue fields of $X$. However I need a weaker version, i.e. only in the case $q=0$. The proper base change has this weaker version where the only assumption is $g$ proper (it works for any kind of étale sheaves and this is perfect as I'm working in char 0). As in my setting all the maps are étale (they are in the small étale site of a DM stack) but not quasi-compact, I hope that the hypothesis $g$ smooth is enough to make $\eta$ an isomorphism in the case $q=0$.
2
-
$\begingroup$ Are you assuming that $X$ and $X^{\prime}$ are irreducible schemes? $\endgroup$– Armando j18eosJun 15, 2018 at 8:45
-
$\begingroup$ Not in general, no. But if it works in that case, it is useful to know. $\endgroup$– user125628Jun 15, 2018 at 20:51
Add a comment
|