I study Nitin Nitsures paper on the Construction of Hilbert and Quot Schemes and not understand the propetry (F) completely:

enter image description here

In previous chapter (Embedding Quot into Grassmanian) it was proved that there exist a morphism of functors

$\frak{Quot}$$^{\Phi, L}_{E/X/S} \to \frak{Grass}$$(W \otimes_{O_S} \operatorname{Sym}^r V, \Phi(r))$

for appropriate positive integer $r$. All used notation is explained rigorously in the linked paper on pages 7 and 24 but let me make some short remarks:

Here we deal with special case where $V$ and $W$ are vector bundles on scheme $S$, $\pi: X := \mathbb{P}(V) \to S$ and $E = \pi^*(W)$ and $L =O_X(1)$. Futhermore $\Phi$ is the Hilbert-polynomial.

Now let come to my actually question: The author intends to show that given any locally noetherian $S$-scheme $T$ and a surjective homomorphism $f: W_T \otimes_{O_T} \operatorname{Sym}^r V_T \to \mathcal{J} $ where $\mathcal{J}$ is a locally free $O_T$-module of rank $\Phi(r)$, there exists a locally closed subscheme $T' \subset T$ with universal property (F).

Question: I not understand the universal property (F): It says: Given any locally noetherian $S$-scheme $Y$ and an $S$-morphism $\phi : Y \to T$, let $f_Y$ be the pull-back of $f$, and let $\mathcal{K}_Y = \operatorname{ker}(f_Y) = \phi^* \operatorname{ker}(f)$. [...]

Why is $\operatorname{ker}(f_Y) = \phi^* \operatorname{ker}(f)$? Indeed $\phi$ wasn't assumed to be flat, therefore there is no reason with pullback functor $\phi^*$ should be left exact.

We have exact sequence

$$0 \to \mathcal{K}:= \operatorname{ker}(f) \to W_T \otimes_{O_T} \operatorname{Sym}^r V_T \xrightarrow{f} \mathcal{J} \to 0$$

applying $\phi^*$ we obtain only

$$\phi^*\operatorname{ker}(f) \to \phi^*(W_T \otimes_{O_T} \operatorname{Sym}^r V_T) \xrightarrow{f_Y} \phi^* \mathcal{J} \to 0$$

$\phi^*$ is right exact but in general not left exact (this is only true with additional assumption ie when $\phi$ is flat).

Why here nevertheless $\operatorname{ker}(f_Y) = \phi^* \operatorname{ker}(f)$ is true?

  • 2
    $\begingroup$ Hint: the exactness on the left follows from $\mathcal{J}$ being locally free $\endgroup$ – Samir Canning Aug 5 at 16:11
  • $\begingroup$ I understand, locally by freeness the sequence splits and this property is preserved after application of $\phi^*$ because $\phi^* J$ stays locally free and especially projective. And this preserves the exactness, yeah? Thank you! $\endgroup$ – Ghost in Grothendieck universe Aug 5 at 16:35

Your Answer

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

Browse other questions tagged or ask your own question.