# Use of flattening stratification (from Nitsure's construction of Hilbert and Quot schemes)

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

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?

• Hint: the exactness on the left follows from $\mathcal{J}$ being locally free Aug 5, 2020 at 16:11
• 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! Aug 5, 2020 at 16:35