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I have been recently studying the construction of the Hilbert scheme from the book 'Deformations of algebraic schemes' by Sernesi and I am stuck with one of the very first steps. Let me expose the initial setting of the construction:

As you know, the main idea is to construct the Hilbert scheme for a fixed numerical polynomial as a subscheme of a suitable grassmannian. To this end, the author fixes a numerical polynomial $P$ and an integer $m_0$ such that, for every closed subscheme $X\subset \mathbb{P}^r$ with Hilbert polynomial $P$, its sheaf of ideals $\mathcal{I}_X$ is $m_0$-regular (in the sense of Castelnuovo-Mumford regularity). Then, by general results on regularity, $N:=h^0(\mathbb{P}^r,\mathcal{I}_X(m_0))=\binom{m_0+r}{r} - P(m_0)$ (where $h^0$ stands for the dimension of $H^0(\mathbb{P}^r,\mathcal{I}_X(m_0))$.

The suitable grassmannian will be $G:=G_N(V)$, the grassmannian of $N$-dimensional subspaces of $V=H^0(\mathbb{P}^r,\mathcal{O}(m_0))$, equipped with its universal quotient bundle $V^{\vee}\otimes \mathcal{O}_G \rightarrow \mathcal{Q}$. Now, taking $p: \mathbb{P}^r\times G\rightarrow G$ the projection, the author considers the composite map of sheaves: $$(*)\;\; p^{*}\mathcal{Q}^{\vee}(-m_0) \rightarrow V\otimes \mathcal{O}_{\mathbb{P}^r\times G}(-m_0)\rightarrow \mathcal{O}_{\mathbb{P}^r\times G} $$

whose image is a sheaf of ideals, $\mathcal{J}\subset \mathcal{O}_{\mathbb{P}^r\times G}$.

And here is where I am stuck. Sernesi proceeds taking $Z$ the closed subscheme defined by $\mathcal{J}$, and then the Hilbert scheme for the polynomial $P$, denoted $H$, will be the stratum of a flattening stratification on $G$ for $Z$, whereas the universal family will be the pullback of $Z$ along the inclusion $H\hookrightarrow G$.

While I understand why the pair $(H, H\times_{G} Z)$ satisfy the universal property of Hilbert functor (basically following the rest of the proof and checking that everything works), I don't understand why we take the closed subscheme $Z$. What motivate us to look for this particular subscheme? What is the sheaf of ideals $\mathcal{J}$ describing? My intuition tells me that it should be some sort of 'incidence relation' on $\mathbb{P}^r\times G$, but I have no clue.

Moreover, what is the composition of maps $(*)$ doing in down-to-earth terms? If it helps, I guess the first map comes from the monomorphism of sheaves $\mathcal{Q}^{\vee}\rightarrow V\otimes \mathcal{O}_G$, while the second one comes from the canonical map $p^{*}p_{*}[\mathcal{O}_{\mathbb{P}^r\times G}(m_0)]\rightarrow \mathcal{O}_{\mathbb{P}^r\times G}(m_0)$ and the identification $p^{*}p_{*}[\mathcal{O}_{\mathbb{P}^r\times G}(m_0)]\otimes \mathcal{O}_{\mathbb{P}^r\times G}(-m_0) \cong V\otimes \mathcal{O}_{\mathbb{P}^r\times G}(-m_0)$.

Sorry for the length of my question, but I wanted to add some context to it. Any answer or suggestion will be very appreciated.

Thanks in advance.

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    $\begingroup$ It seems like you are looking for intuition. The point is that we can build an ideal $I \subseteq \mathcal O_{\mathbb P^r}$ by taking $\dim H^0(\mathcal O(m_0)) - N$ sections of $H^0(\mathcal O(m_0))$ which correspond to a map $\mathcal O(-m_0)^{\oplus n} \to \mathcal O$ and then taking the image--we are looking at the ideal generated by a fixed number of generators of a given degree. The moduli space parameterizing the ideal sheaves constructed in this fashion is the Grassmannian of subspaces of $H^0(\mathcal O(m_0))$, and the sheaf $\mathcal J$ is the universal ideal sheaf for this family. $\endgroup$ Commented May 16 at 14:48
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    $\begingroup$ Then CM regularity is used to show that for a fixed Hilbert polynomial, there exists an $m_0$ such that every ideal sheaf with Hilbert polynomial $P$ can be constructed in this fashion. $\endgroup$ Commented May 16 at 14:51
  • $\begingroup$ @PhilTosteson Thank you Phil for your answer. The first thing you point out is something I already understood; in fact, CM regularity tells you that every $m_0$ regular ideal sheaf is uniquely determined by its degree $m_0$ homogeneous piece, and that's why we choose that grassmannian. What I still don't get is why $\mathcal{J}$ is chosen the way I described above, using the maps of sheaves $(*)$ involving the universal bundle of the grassmannian and all that stuff. Are those maps some sort of global 'multiplication maps', in such a way that the closed subscheme defined by the image(continue) $\endgroup$ Commented May 16 at 16:41
  • $\begingroup$ @PhilTosteson has as fiber over a (closed) point of the grassmannian precisely the closed subscheme of $\mathbb{P}^r$ determined by the sheaf of ideals spanned by the subspace of $H^0(\mathbb{P}^r, \mathcal{O}(m_0))$ corresponding to the given point? $\endgroup$ Commented May 16 at 16:44
  • $\begingroup$ Yes, that's exactly how you should think ($*$). Things are slightly obscured by the fact that you're using the Grassmanian to parameterize quotients instead of subs. But if you look at the fiber of ($*$) above a given point in the Grassmanian, corresponding to a subspace $S \subseteq H^0(\mathcal O(m_0))$ (here $S = Q^{\vee}$) you will get precisely the map $S \otimes \mathcal O(-m_0) \to \mathcal O$ that generates an ideal. $\endgroup$ Commented May 16 at 19:37

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