The Stacks project

Proof. Suppose we have a scheme $T/B$ and two quotients $\mathcal{F}_ T \to \mathcal{Q}_ i$, $i = 1, 2$ corresponding to $T$-valued points of $\mathrm{Quot}_{\mathcal{F}/X/B}$ over $B$. Denote $\mathcal{K}_1$ the kernel of the first one and set $u : \mathcal{K}_1 \to \mathcal{Q}_2$ the composition. By Flatness on Spaces, Lemma 77.8.6 there is a closed subspace of $T$ such that $T' \to T$ factors through it if and only if the pullback $u_{T'}$ is zero. This proves the diagonal is a closed immersion. Moreover, if $\mathcal{F}$ is of finite type, then $\mathcal{K}_1$ is of finite type (Modules on Sites, Lemma 18.24.1) and we see that the diagonal is of finite presentation by the same lemma. $\square$

Proof. To check $\mathrm{Quot}_{\mathcal{F}/X/B} \to B$ is separated we have to show that its diagonal is a closed immersion. This is true by Lemma 108.5.1. The second statement is part of Quot, Proposition 99.8.4. $\square$

Proof. Taking base change, this immediately reduces to the following problem: given a valuation ring $R$ with fraction field $K$, an algebraic space $X$ proper over $R$, a finite type quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$, and a coherent quotient $\mathcal{F}_ K \to \mathcal{Q}_ K$, show there exists a quotient $\mathcal{F} \to \mathcal{Q}$ where $\mathcal{Q}$ is a finitely presented $\mathcal{O}_ X$-module flat over $R$ whose generic fibre is $\mathcal{Q}_ K$. Observe that by Flatness on Spaces, Theorem 77.4.5 any finite type quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ flat over $R$ is of finite presentation. We first solve the existence of $\mathcal{Q}$ affine locally.

Affine locally we arrive at the following problem: let $R \to A$ be a finitely presented ring map, let $M$ be a finite $A$-module, let $\varphi : M_ K \to N_ K$ be an $A_ K$-quotient module. Then we may consider

The $M \to M/L$ is an $A$-module quotient which is torsion free as an $R$-module. Hence it is flat as an $R$-module (More on Algebra, Lemma 15.22.10). Since $M$ is finite as an $A$-module so is $L$ and we conclude that $L$ is of finite presentation as an $A$-module (by the reference above). Clearly $M/L$ is the unqiue such quotient with $(M/L)_ K = N_ K$.

The uniqueness in the construction of the previous paragraph guarantees these quotients glue and give the desired $\mathcal{Q}$. Here is a bit more detail. Choose a surjective étale morphism $U \to X$ where $U$ is an affine scheme. Use the above construction to construct a quotient $\mathcal{F}|_ U \to \mathcal{Q}_ U$ which is quasi-coherent, is flat over $R$, and recovers $\mathcal{Q}_ K|U$ on the generic fibre. Since $X$ is separated, we see that $U \times _ X U$ is an affine scheme étale over $X$ as well. Then $\mathcal{F}|_{U \times _ X U} \to \text{pr}_1^*\mathcal{Q}_ U$ and $\mathcal{F}|_{U \times _ X U} \to \text{pr}_2^*\mathcal{Q}_ U$ agree as quotients by the uniquess in the construction. Hence we may descend $\mathcal{F}|_ U \to \mathcal{Q}_ U$ to a surjection $\mathcal{F} \to \mathcal{Q}$ as desired (Properties of Spaces, Proposition 66.32.1). $\square$

Proof. Set $\mathcal{G} = \pi _*\mathcal{F}$. Since $\pi $ is affine we see that for any scheme $T$ over $B$ we have $\mathcal{G}_ T = \pi _{T, *}\mathcal{F}_ T$ by Cohomology of Spaces, Lemma 69.11.1. Moreover $\pi _ T$ is affine, hence $\pi _{T, *}$ is exact and transforms quotients into quotients. Observe that a quasi-coherent quotient $\mathcal{F}_ T \to \mathcal{Q}$ defines a point of $\mathrm{Quot}_{X/B}$ if and only if $\mathcal{Q}$ defines an object of $\mathcal{C}\! \mathit{oh}_{X/B}$ over $T$ (similarly for $\mathcal{G}$ and $Y$). Since we've seen in Lemma 108.4.4 that $\pi _*$ induces a morphism $\mathcal{C}\! \mathit{oh}_{X/B} \to \mathcal{C}\! \mathit{oh}_{Y/B}$ we see that if $\mathcal{F}_ T \to \mathcal{Q}$ is in $\mathrm{Quot}_{\mathcal{F}/X/B}(T)$, then $\mathcal{G}_ T \to \pi _{T, *}\mathcal{Q}$ is in $\mathrm{Quot}_{\mathcal{G}/Y/B}(T)$. $\square$

Proof. Omitted. Hint: If $(\pi _*\mathcal{F})_ T \to \mathcal{Q}$ is an element of $\mathrm{Quot}_{\pi _*\mathcal{F}/Y/B}(T)$ and for $t \in T$ we have $\text{Supp}(\mathcal{Q}_ t) \subset |X_ t|$, then the same is true for $t' \in T$ in a neighbourhood of $t$. $\square$

Proof. If $\mathcal{F}_ T \to \mathcal{Q}$ is an element of $\mathrm{Quot}_{\mathcal{F}/X/B}(T)$ then we can consider $\mathcal{G}_ T \to j_{T, *}\mathcal{F}_ T \to j_{T, *}\mathcal{Q}$. Looking at stalks one finds that this is surjective. By Lemma 108.4.4 we see that $j_{T, *}\mathcal{Q}$ is finitely presented, flat over $B$ with support proper over $B$. Thus we obtain a $T$-valued point of $\mathrm{Quot}_{\mathcal{G}/Y/B}$. This defines the morphism of the lemma. We omit the proof that this is an open immersion. Hint: If $\mathcal{G}_ T \to \mathcal{Q}$ is an element of $\mathrm{Quot}_{\mathcal{G}/Y/B}(T)$ and for $t \in T$ we have $\text{Supp}(\mathcal{Q}_ t) \subset |X_ t|$, then the same is true for $t' \in T$ in a neighbourhood of $t$. $\square$

Proof. For every scheme $T$ over $B$ the morphism $\pi _ T : X_ T \to Y_ T$ is a closed immersion. Then $\pi _{T, *}$ is an equivalence of categories between $\mathit{QCoh}(\mathcal{O}_{X_ T})$ and the full subcategory of $\mathit{QCoh}(\mathcal{O}_{Y_ T})$ whose objects are those quasi-coherent modules annihilated by the ideal sheaf of $X_ T$, see Morphisms of Spaces, Lemma 67.14.1. Since a qotient of $(\pi _*\mathcal{F})_ T$ is annihilated by this ideal we obtain the bijectivity of the map $\mathrm{Quot}_{\mathcal{F}/X/B}(T) \to \mathrm{Quot}_{\pi _*\mathcal{F}/Y/B}(T)$ for all $T$ as desired. $\square$

Proof. Let $\mathcal{K} = \mathop{\mathrm{Ker}}(\mathcal{F} \to \mathcal{G})$. By right exactness of pullbacks we find that $\mathcal{K}_ T \to \mathcal{F}_ T \to \mathcal{G}_ T \to 0$ is an exact sequecnce for all schemes $T$ over $B$. In particular, a quotient of $\mathcal{G}_ T$ determines a quotient of $\mathcal{F}_ T$ and we obtain our transformation of functors $\mathrm{Quot}_{\mathcal{G}/X/B} \to \mathrm{Quot}_{\mathcal{F}/X/B}$. This transformation is a closed immersion by Flatness on Spaces, Lemma 77.8.6. Namely, given an element $\mathcal{F}_ T \to \mathcal{Q}$ of $\mathrm{Quot}_{\mathcal{F}/X/B}(T)$, then we see that the pull back to $T'/T$ is in the image of the transformation if and only if $\mathcal{K}_{T'} \to \mathcal{Q}_{T'}$ is zero. $\square$

Proof. Set $\mathcal{G} = \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}$. Observe that $\mathcal{G}_ T = \mathcal{F}_ T \otimes _{\mathcal{O}_{X_ T}} \mathcal{L}_ T$. If $\mathcal{F}_ T \to \mathcal{Q}$ is an element of $\mathrm{Quot}_{\mathcal{F}/X/B}(T)$, then we send it to the element $\mathcal{G}_ T \to \mathcal{Q} \otimes _{\mathcal{O}_{X_ T}} \mathcal{L}_ T$ of $\mathrm{Quot}_{\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}/X/B}(T)$. This is compatible with pullbacks and hence defines a transformation of functors as desired. Since there is an obvious inverse transformation, it is an isomorphism. We omit the proof of the final statement. $\square$

Proof. Follows immediately after unwinding all the definitions. $\square$