suppose $f:X\rightarrow Y$ is a morphism between two schemes over scheme $S.$ Do we have the morphism between their hilbert schemes, i.e. is there a natural morphism $Hilb(X/S)\rightarrow Hilb(Y/S)$ over $S$? What about the Douady spaces instead of Hilbert schemes in the analytic spaces case?
$\begingroup$
$\endgroup$
5
-
$\begingroup$ No, the Hilbert scheme is not functorial like that. For the universal closed subscheme $i:Z\hookrightarrow \text{Hilb}_{X/S}\times_S X$, the associated morphism $(\text{Id}\times f)\circ i:Z \to \text{Hilb}_{X/S} \times_S Y$ is typically not a closed immersion (often it is even ramified). $\endgroup$– Jason StarrJan 12, 2016 at 2:27
-
$\begingroup$ what about the base change case and let's assume $Y\rightarrow S$ is flat and $X\rightarrow S^\prime$ is the base change of $S^\prime\rightarrow S$? Do we have $Hilb(X/S)→Hilb(Y/S^\prime)$? $\endgroup$– user42804Jan 12, 2016 at 2:30
-
1$\begingroup$ I do not understand what you are asking. Are you asking if there is a natural $S$-morphism $\text{Hilb}_{Y\times_S S'/S'} \to \text{Hilb}_{Y/S}$? Yes there is such a morphism, and it identifies $\text{Hilb}_{Y\times_S S'/S'}$ with $\text{Hilb}_{Y/S}\times_S S'$. $\endgroup$– Jason StarrJan 12, 2016 at 2:52
-
$\begingroup$ Thanks! That's what I am asking. Is it still true for Douady spaces? $\endgroup$– user42804Jan 12, 2016 at 3:02
-
$\begingroup$ Yes, that is also true for the Douady spaces. This has nothing to do with the construction or proof of existence of the Hilbert schemes / Douady spaces. Already for the (contravariant) Hilbert functor $\underline{Hilb}_{Y/S}:\text{Sch}_S \to \text{Sets}$, there is a natural isomorphism of functors from the restriction of $\underline{Hilb}_{Y/S}$ to the faithful (usually not full) subcategory $\text{Sch}_{S'}$ and the functor $\underline{Hilb}_{Y\times_S S'/S'}$. Thus, if these functors are representable, then there is an isomorphism as above. $\endgroup$– Jason StarrJan 12, 2016 at 11:23
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
I am just posting my comments above as an answer, so that this question does not remain unanswered.
Yes, that is also true for the Douady spaces. This has nothing to do with the construction or proof of existence of the Hilbert schemes / Douady spaces. Already for the (contravariant) Hilbert functor $$ \underline{\text{Hilb}}_{Y/S}: \textbf{Schemes}_S \to \textbf{Sets},$$ fo the composition of this function with the faitful functor, $$\Phi: \textbf{Schemes}_{S'} \to \textbf{Schemes}_S,$$ there is a natural isomorphism of $\underline{\text{Hilb}}_{Y/S}\circ \Phi$ with the Hilbert functor $$ \underline{\text{Hilb}}_{Y\times_S S'/S'}: \textbf{Schemes}_{S'} \to \textbf{Sets}.$$ Thus, whenever both of these functors are representable, then there is an isomorphism as above.