I am reading EGA IV.3, Theorem 11.9.16 about universally schematically dominant morphisms, which states the following. The meaning of schematic dominance is at the end of this post.

Let $X$ be a locally Noetherian $S$-scheme, $f_{\lambda}\colon Z_{\lambda}\to X$ morphisms, $\mathscr{F}$ a coherent $O_X$ module, $\mathscr{G}_{\lambda}$ quasi-coherent $O_{Z_{\lambda}}$ modules flat over $S$, and $u_{\lambda}\colon\mathscr{F}\to (f_{\lambda})_*\mathscr{G}_{\lambda}$ morphisms. For all $s \in S$, let $i\colon X_s\to X$ and $j_{\lambda}\colon (Z_{\lambda})_s\to Z_{\lambda}$ be the inclusion of fibers, and $(u_{\lambda})_s\colon i^*\mathscr{F}\to (f_{\lambda}\times _S \operatorname{Spec} k(s))_*j^*\mathscr{G}_{\lambda}$ be the "fiber" of $u_{\lambda}$ composed with a base change map. In these settings, the index $\lambda$ runs through the same set $L$.

Then, $(u_{\lambda})_{\lambda}$ is universally schematically dominant w.r.t. $S$ iff for all $s \in S$, $((u_{\lambda})_s)_{\lambda}$ is schematically dominant.

The Only if part is clear. For the if part, first the proof tries to show that schematic dominance instead of universal one. We may assume $S = \operatorname{Spec} A$, and $X = \operatorname{Spec} B$ are affine, and $f_{\lambda}$ are identity, and when $\mathscr{F} = \tilde{M}$ and $\mathscr{G}_{\lambda} = \tilde{N_{\lambda}}$, want to show for $t\in M$ that $t = 0$ if for all $\lambda$, $u_{\lambda}(t) = 0$. (This part is EGA IV.3, 11.9.3, etc.)

Then, because $t = 0$ iff for all maximal ideal $\mathfrak{p}$ of B, $t_{\mathfrak{p}} = 0$ where $t_{\mathfrak{p}}$ is the image of $t$ in $M_{\mathfrak{p}}$, the proof says that we may assume that $B$ is local by base changes via $B\to B_{\mathfrak{p}}$. I do not understand this part as schematic dominance of $((u_{\lambda})_s)_{\lambda}$ might not be preserved. Thank you for any kinds of advices and comments in advance.

Notes on definitions

In this post, schemes mean usual ones, which are called preschemes in EGA.

In the above setting, $(u_{\lambda})_{\lambda}$ is schematically dominant if for all open subset $U\subseteq X$ and $t\in\mathscr{F}(U)$ with $u_{\lambda}(t) = 0$ for all $\lambda$, $t = 0$. $(u_{\lambda})_{\lambda}$ is universally schematically dominant w.r.t. $S$ if for all $S$-scheme $S'$, the base change of $(u_{\lambda})_{\lambda}$ via $S'\to S$, obtained in the same way as $((u_{\lambda})_s)_{\lambda}$, is schematically dominant.

EGA IV.3 is available here, for example.

  • 3
    $\begingroup$ I've improved your formatting by using \operatorname{Spec} to format $\operatorname{Spec}$, which produces better spacing. $\endgroup$ – KReiser Sep 11 '20 at 0:17
  • $\begingroup$ Thank you for your help! I will try to remember that tip. $\endgroup$ – neander Sep 11 '20 at 8:27

I think I got an easy answer that the schematic dominance for fibers is preserved when substituting $B_\mathfrak{p}$ for $B$. For all $f,f'\in B$ and the image $\mathfrak{q}\in\operatorname{Spec}$ of $\mathfrak{p}$, $$ \bigcap_{\lambda} \ker\left(M_{ff'}\otimes_A A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}}\to (N_{\lambda})_{ff'}\otimes_A A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}}\right) = 0 $$ by definition. Here, take the filtrant inductive limit of this equation over $f\in B\backslash\mathfrak{p}$ to get the desired result.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.