# Why is $Mor_S(Spec R,SpecSym_{\mathcal{O}_S}\mathcal{Q})\cong Hom_{\mathcal{O}_S-\operatorname{mod}}(\mathcal{Q},f_{*}\mathcal{O}_{R})$?

Assume $$S$$ is a Noetherian scheme and $$f:\operatorname{Spec}(R)\longrightarrow S$$ is any morphism from an affine scheme to $$S$$. Assume $$\mathcal{Q}$$ is a coherent sheaf over $$S$$ and $$\operatorname{Sym}_{\mathcal{O}_{S}}\mathcal{Q}:=\operatorname{Tens}_{\mathcal{O}_{S}}\mathcal{Q}/(a\otimes b-b\otimes a)$$ is its associated symmetrical algebra. Nitsure (https://arxiv.org/abs/math/0504590, p.17) makes the claim that $$\begin{equation*} \operatorname{Mor}_{S}(\operatorname{Spec}(R),\operatorname{Spec}\operatorname{Sym}_{\mathcal{O}_{S}}\mathcal{Q})\cong\operatorname{Hom}_{\mathcal{O}_{S}-\operatorname{mod}}(\mathcal{Q},f_{*}\mathcal{O}_{R})\text{.} \end{equation*}$$ Why is that the case?

$$\renewcommand{\Spec}{\operatorname{Spec}}\renewcommand{\Sym}{\operatorname{Sym}}$$ Let's work locally on $$S$$, say over $$\Spec A$$ an open subscheme of $$S$$, and suppose $$\mathcal{Q}|_{\Spec A}\cong \widetilde{Q}$$ for some $$A$$-module $$Q$$. Then $$(\Spec\Sym\mathcal{Q})|_{\Spec A}\cong \Spec \Sym Q$$, and $$S$$-maps $$f^{-1}(\Spec A)\to \Spec \Sym Q$$ are in bijection with $$A$$-maps $$\Sym Q \to \Gamma(\mathcal{O}_{f^{-1}(\Spec A)},f^{-1}(\Spec A))$$ by a relative version of 01I1. But such a morphism is determined by where $$Q$$ goes, and $$\Gamma(\mathcal{O}_{f^{-1}(\Spec A)},f^{-1}(\Spec A))=(f_*\mathcal{O}_R)(\Spec A)$$, so such a morphism over $$\Spec A$$ is determined by a map $$Q\to (f_*\mathcal{O}_R)(\Spec A)$$. Assembling all of this data together over all affine open $$\Spec A\subset S$$, we have the result.