The Stacks project

Lemma 99.9.2. In Situation 99.9.1 we have $\mathrm{Hilb}_{X/B} = \mathrm{Quot}_{\mathcal{O}_ X/X/B}$.

Proof. Let $T$ be a scheme over $B$. Given an element $Z \in \mathrm{Hilb}_{X/B}(T)$ we can consider the quotient $\mathcal{O}_{X_ T} \to i_*\mathcal{O}_ Z$ where $i : Z \to X_ T$ is the inclusion morphism. Note that $i_*\mathcal{O}_ Z$ is quasi-coherent. Since $Z \to T$ and $X_ T \to T$ are of finite presentation, we see that $i$ is of finite presentation (Morphisms of Spaces, Lemma 67.28.9), hence $i_*\mathcal{O}_ Z$ is an $\mathcal{O}_{X_ T}$-module of finite presentation (Descent on Spaces, Lemma 74.6.7). Since $Z \to T$ is proper we see that $i_*\mathcal{O}_ Z$ has support proper over $T$ (as defined in Derived Categories of Spaces, Section 75.7). Since $\mathcal{O}_ Z$ is flat over $T$ and $i$ is affine, we see that $i_*\mathcal{O}_ Z$ is flat over $T$ (small argument omitted). Hence $\mathcal{O}_{X_ T} \to i_*\mathcal{O}_ Z$ is an element of $\mathrm{Quot}_{\mathcal{O}_ X/X/B}(T)$.

Conversely, given an element $\mathcal{O}_{X_ T} \to \mathcal{Q}$ of $\mathrm{Quot}_{\mathcal{O}_ X/X/B}(T)$, we can consider the closed immersion $i : Z \to X_ T$ corresponding to the quasi-coherent ideal sheaf $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_{X_ T} \to \mathcal{Q})$ (Morphisms of Spaces, Lemma 67.13.1). By construction of $Z$ we see that $\mathcal{Q} = i_*\mathcal{O}_ Z$. Then we can read the arguments given above backwards to see that $Z$ defines an element of $\mathrm{Hilb}_{X/B}(T)$. For example, $\mathcal{I}$ is quasi-coherent of finite type (Modules on Sites, Lemma 18.24.1) hence $i : Z \to X_ T$ is of finite presentation (Morphisms of Spaces, Lemma 67.28.12) hence $Z \to T$ is of finite presentation (Morphisms of Spaces, Lemma 67.28.2). Properness of $Z \to T$ follows from the discussion in Derived Categories of Spaces, Section 75.7. Flatness of $Z \to T$ follows from flatness of $\mathcal{Q}$ over $T$.

We omit the (immediate) verification that the two constructions given above are mutually inverse. $\square$


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