Let $S$ be a noetherian scheme, $\mathcal{E}$ a quasi-coherent sheaf on $S$ and let $d \in \mathbb{N}$. There is a Plücker embedding $\omega : \mathrm{Grass}_d(\mathcal{E}) \hookrightarrow \mathbb{P}(\wedge^d \mathcal{E})$. A very elegant functorial construction can be found in EGA I, 9.8. My question is:

How can we describe the corresponding quasi-coherent ideal $I$ on $\mathbb{P}(\wedge^d \mathcal{E})$ globally? More precisely, if $\mathcal{E}$ is coherent, then by results of EGA II there is an epimorphism $\oplus_i M_i \otimes_{\mathcal{O}_S} \mathcal{O}(n_i) \twoheadrightarrow I$ for some coherent $\mathcal{O}_S$-modules $M_i$ and integers $n_i$. I would like know if one can write this down *without* using a presentation of $\mathcal{E}$.

The answer in the special case $\mathcal{E} = \mathcal{O}_S^{\oplus I}$ for some set $I$ is well-known (at least when $S$ is a field and $I$ is finite, but the general case works the same. Does anybody know a reference where this is done?): The Plücker relations generate $I$. More precisely, let $\mathcal{O}_{\mathbb{P}}(1)$ be the universal invertible sheaf on $\mathbb{P}(\wedge^d \mathcal{E})$ together with its universal epimorphism $s : \wedge^d \mathcal{E} \otimes_{\mathcal{O}_S} \mathcal{O}_{\mathbb{P}} \twoheadrightarrow \mathcal{O}_{\mathbb{P}}(1)$ . Then define

$P : \wedge^{d-1}(\mathcal{E}) \otimes_{\mathcal{O}_S} \wedge^{d+1}(\mathcal{E}) \otimes_{\mathcal{O}_S} \mathcal{O}_{\mathbb{P}} \to \mathcal{O}_{\mathbb{P}}(2),$ $${\small f_1 \wedge \dotsc \wedge f_{d-1} \otimes e_0 \wedge \dotsc \wedge e_d \mapsto \sum_{l=0}^{d} (-1)^l s(f_1 \wedge \dotsc \wedge f_{d-1} \wedge e_k) \otimes s(e_0 \wedge \dotsc \wedge \widehat{e_k} \wedge \dotsc \wedge e_d).}$$

Then $I$ is the image of $\check{P} : \wedge^{d-1}(\mathcal{E}) \otimes_{\mathcal{O}_S} \wedge^{d+1}(\mathcal{E}) \otimes_{\mathcal{O}_S} \mathcal{O}_{\mathbb{P}}(-2) \to \mathcal{O}_{\mathbb{P}}$.

For general $\mathcal{E}$, these Plücker relations are also satisfied, but I couldn't prove the converse and meanwhile I'm convinced that we need more relations. If it helps, you may assume that $2$ is invertible on $S$.

notlooking for the equations in the classical special case $\mathrm{Grass}_d(R^n) \hookrightarrow \mathbb{P}(\wedge^d R^n)$. – Martin Brandenburg Dec 13 '12 at 9:48