Let $\mathrm{Grass}_{k+1,n+1}:\mathrm{Sch}^{\circ}\rightarrow\mathrm{Set}$ be the Grassmann functor, which maps a scheme $S$ to the set:
$$\left\{\mathscr{U}\subseteq\mathscr{O}_S^{n+1}:\,\,\mathscr{O}_S^{n+1}/\mathscr{U} \text{ is locally free of rank }n-k\right\}$$
We have a cover of $\mathrm{Grass}_{k+1,n+1}$ by open subfunctors $\mathrm{Grass}_{k+1,n+1}^{I}$, where $I\subseteq\{1,...,n+1\}$ ranges over all subsets with $n-k$ elements and $\mathrm{Grass}_{k+1,n+1}^{I}$ is defined as
$$\mathrm{Grass}_{k+1,n+1}^{I}(S)=\left\{\mathscr{U}\in\mathrm{Grass}_{k+1,n+1}(S):\,\, \mathscr{U}\oplus\mathscr{O}_S^{I}=\mathscr{O}_S^{n+1}\right\}$$
Let $P=\binom{m+k}{k}\in\mathbb{Q}[m]$ and $h_P:\mathrm{Sch}^{\circ}\rightarrow\mathrm{Set}$ the Hilbert functor, that associates to any scheme $S$ the set of closed subschemes $Z\subseteq S\times \mathbb{P}_{\mathbb{Z}}^n$, such that $Z$ is flat over $S$ and all fibers $Z_s$ over $S$ have Hilbert Polynomial $P$.
Is there are cover of $h_P$ by open subfunctors $h_P^{I}$, such that $\mathrm{Grass}_{k+1,n+1}^{I}\cong h_P^{I}$ ?. Since $\mathrm{Grass}_{k+1,n+1}\cong h_P$ , there has to be such a cover, but I am not really able to find one.
For an affine scheme $S=\mathrm{Spec}(R)$, I found a map $\mathrm{Grass}_{k+1,n+1}^{I}(S)\rightarrow h_P(S)$, which is defined in the following way:
$\mathscr{U}\in\mathrm{Grass}_{k+1,n+1}(S)$ is quasicoherent and therefore $\mathscr{U}=\tilde{M}$ for an $R$-module $M$. Now $\mathscr{U}\oplus\mathscr{O}_S^{I}=\mathscr{O}_S^{n+1}$ implies $M\oplus R^{I}=R^{n+1}$, so that the projection onto $R^{I}$ gives a linear map:
$$R^{n+1}\rightarrow R^{I}\cong R^{n-k}$$
Now one can try to take the components $f_1,...,f_{n-k}$ of this map and define $Z=\mathrm{Proj}\left(R[x_0,...,x_{n+1]}/(f_1,...,f_{n-k})\right)$, where the $f_i$ are viewed as linear homogeneous polynomials. Of course we could now define $h_P^{I}$ as the image of this morphism, but then it is not clear how this functor is defined for an arbitrary scheme.
I am very thankful for any thoughts on this.
