The Stacks project

Proof. The representability of the diagonal by algebraic spaces was shown in Quot, Lemma 99.5.3. From the proof we find that we have to show $\mathit{Isom}(\mathcal{F}, \mathcal{G}) \to T$ is affine and of finite presentation for a pair of finitely presented $\mathcal{O}_{X_ T}$-modules $\mathcal{F}$, $\mathcal{G}$ flat over $T$ with support proper over $T$. This was discussed in Section 108.3. $\square$

Proof. To check $\mathcal{C}\! \mathit{oh}_{X/B} \to B$ is quasi-separated we have to show that its diagonal is quasi-compact and quasi-separated. This is immediate from Lemma 108.4.1. To prove that $\mathcal{C}\! \mathit{oh}_{X/B} \to B$ is locally of finite presentation, we have to show that $\mathcal{C}\! \mathit{oh}_{X/B} \to B$ is limit preserving, see Limits of Stacks, Proposition 102.3.8. This follows from Quot, Lemma 99.5.6 (small detail omitted). $\square$

Proof. Taking base change, this immediately reduces to the following problem: given a valuation ring $R$ with fraction field $K$ and an algebraic space $X$ proper over $R$ and a coherent $\mathcal{O}_{X_ K}$-module $\mathcal{F}_ K$, show there exists a finitely presented $\mathcal{O}_ X$-module $\mathcal{F}$ flat over $R$ whose generic fibre is $\mathcal{F}_ K$. 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. Denote $j : X_ K \to X$ the embedding of the generic fibre. As a base change of the affine morphism $\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(R)$ the morphism $j$ is affine. Thus $j_*\mathcal{F}_ K$ is quasi-coherent. Write

as a filtered colimit of its finite type quasi-coherent $\mathcal{O}_ X$-submodules, see Limits of Spaces, Lemma 70.9.2. Since $j_*\mathcal{F}_ K$ is a sheaf of $K$-vector spaces over $X$, it is flat over $\mathop{\mathrm{Spec}}(R)$. Thus each $\mathcal{F}_ i$ is flat over $R$ as flatness over a valuation ring is the same as being torsion free (More on Algebra, Lemma 15.22.10) and torsion freeness is inherited by submodules. Finally, we have to show that the map $j^*\mathcal{F}_ i \to \mathcal{F}_ K$ is an isomorphism for some $i$. Since $j^*j_*\mathcal{F}_ K = \mathcal{F}_ K$ (small detail omitted) and since $j^*$ is exact, we see that $j^*\mathcal{F}_ i \to \mathcal{F}_ K$ is injective for all $i$. Since $j^*$ commutes with colimits, we have $\mathcal{F}_ K = j^*j_*\mathcal{F}_ K = \mathop{\mathrm{colim}}\nolimits j^*\mathcal{F}_ i$. Since $\mathcal{F}_ K$ is coherent (i.e., finitely presented), there is an $i$ such that $j^*\mathcal{F}_ i$ contains all the (finitely many) generators over an affine étale cover of $X$. Thus we get surjectivity of $j^*\mathcal{F}_ i \to \mathcal{F}_ K$ for $i$ large enough. $\square$

Proof. Let $(T \to B, \mathcal{F})$ be an object of $\mathcal{C}\! \mathit{oh}_{X/B}$. We claim

1. $(T \to B, \pi _{T, *}\mathcal{F})$ is an object of $\mathcal{C}\! \mathit{oh}_{Y/B}$ and

2. for $T' \to T$ we have $\pi _{T', *}(X_{T'} \to X_ T)^*\mathcal{F} = (Y_{T'} \to Y_ T)^*\pi _{T, *}\mathcal{F}$.

Part (b) guarantees that this construction defines a functor $\mathcal{C}\! \mathit{oh}_{X/B} \to \mathcal{C}\! \mathit{oh}_{Y/B}$ as desired.

Let $i : Z \to X_ T$ be the closed subspace cut out by the zeroth fitting ideal of $\mathcal{F}$ (Divisors on Spaces, Section 71.5). Then $Z \to B$ is proper by assumption (see Derived Categories of Spaces, Section 75.7). On the other hand $i$ is of finite presentation (Divisors on Spaces, Lemma 71.5.2 and Morphisms of Spaces, Lemma 67.28.12). There exists a quasi-coherent $\mathcal{O}_ Z$-module $\mathcal{G}$ of finite type with $i_*\mathcal{G} = \mathcal{F}$ (Divisors on Spaces, Lemma 71.5.3). In fact $\mathcal{G}$ is of finite presentation as an $\mathcal{O}_ Z$-module by Descent on Spaces, Lemma 74.6.7. Observe that $\mathcal{G}$ is flat over $B$, for example because the stalks of $\mathcal{G}$ and $\mathcal{F}$ agree (Morphisms of Spaces, Lemma 67.13.6). Observe that $\pi _ T \circ i : Z \to Y_ T$ is quasi-finite as a composition of quasi-finite morphisms and that $\pi _{T, *}\mathcal{F} = (\pi _ T \circ i)_*\mathcal{G})$. Since $i$ is affine, formation of $i_*$ commutes with base change (Cohomology of Spaces, Lemma 69.11.1). Therefore we may replace $B$ by $T$, $X$ by $Z$, $\mathcal{F}$ by $\mathcal{G}$, and $Y$ by $Y_ T$ to reduce to the case discussed in the next paragraph.

Assume that $X \to B$ is proper. Then $\pi $ is proper by Morphisms of Spaces, Lemma 67.40.6 and hence finite by More on Morphisms of Spaces, Lemma 76.35.1. Since a finite morphism is affine we see that (b) holds by Cohomology of Spaces, Lemma 69.11.1. On the other hand, $\pi $ is of finite presentation by Morphisms of Spaces, Lemma 67.28.9. Thus $\pi _{T, *}\mathcal{F}$ is of finite presentation by Descent on Spaces, Lemma 74.6.7. Finally, $\pi _{T, *}\mathcal{F} $ is flat over $B$ for example by looking at stalks using Cohomology of Spaces, Lemma 69.4.2. $\square$

Proof. Omitted. Hint: If $\mathcal{F}$ is an object of $\mathcal{C}\! \mathit{oh}_{Y/B}$ over $T$ and for $t \in T$ we have $\text{Supp}(\mathcal{F}_ t) \subset |X_ t|$, then the same is true for $t' \in T$ in a neighbourhood of $t$. $\square$

Proof. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be the sheaf of ideals cutting out $X$ as a closed subspace of $Y$. Recall that $\pi _*$ induces an equivalence between the category of quasi-coherent $\mathcal{O}_ X$-modules and the category of quasi-coherent $\mathcal{O}_ Y$-modules annihilated by $\mathcal{I}$, see Morphisms of Spaces, Lemma 67.14.1. The same, mutatis mutandis, is true after base by $T \to B$ with $\mathcal{I}$ replaced by the ideal sheaf $\mathcal{I}_ T = \mathop{\mathrm{Im}}((Y_ T \to Y)^*\mathcal{I} \to \mathcal{O}_{Y_ T})$. Analyzing the proof of Lemma 108.4.4 we find that the essential image of $\mathcal{C}\! \mathit{oh}_{X/B} \to \mathcal{C}\! \mathit{oh}_{Y/B}$ is exactly the objects $\xi = (T \to B, \mathcal{F})$ where $\mathcal{F}$ is annihilated by $\mathcal{I}_ T$. In other words, $\xi $ is in the essential image if and only if the multiplication map

is zero and similarly after any further base change $T' \to T$. Note that

Hence the vanishing of the multiplication map on $T'$ is representable by a closed subspace of $T$ by Flatness on Spaces, Lemma 77.8.6. $\square$

Proof. This is immediately clear if $I$ is finite, because the functions $t \mapsto \chi _ i(t)$ are locally constant. If $I$ is infinite, then we write

and we denote $P' = P|_{I'}$. Then we have

Therefore, $\mathcal{C}\! \mathit{oh}^ P_{X/B}$ is always an algebraic stack and the morphism $\mathcal{C}\! \mathit{oh}^ P_{X/B} \subset \mathcal{C}\! \mathit{oh}_{X/B}$ is always a flat closed immersion, but it may no longer be an open substack. (We leave it to the reader to make examples). However, if $B$ is locally Noetherian, then so is $\mathcal{C}\! \mathit{oh}_{X/B}$ by Lemma 108.4.2 and Morphisms of Stacks, Lemma 101.17.5. Hence if $U \to \mathcal{C}\! \mathit{oh}_{X/B}$ is a smooth surjective morphism where $U$ is a locally Noetherian scheme, then the inverse images of the open and closed substacks $\mathcal{C}\! \mathit{oh}^{P'}_{X/B}$ have an open intersection in $U$ (because connected components of locally Noetherian topological spaces are open). Thus the result in this case. $\square$

Proof. We may work étale locally on $B$, hence we may assume that $B$ is affine. In this case we may perform absolute Noetherian reduction; we suggest the reader skip the proof. Namely, say $B = \mathop{\mathrm{Spec}}(\Lambda )$. Write $\Lambda = \mathop{\mathrm{colim}}\nolimits \Lambda _ i$ as a filtered colimit with each $\Lambda _ i$ of finite type over $\mathbf{Z}$. For some $i$ we can find a morphism of algebraic spaces $X_ i \to \mathop{\mathrm{Spec}}(\Lambda _ i)$ which is separated and of finite presentation and whose base change to $\Lambda $ is $X$. See Limits of Spaces, Lemmas 70.7.1 and 70.6.9. Then after increasing $i$ we may assume there exist perfect objects $E_{1, i}, \ldots , E_{r, i}$ in $D(\mathcal{O}_{X_ i})$ whose derived pullback to $X$ are isomorphic to $E_1, \ldots , E_ r$, see Derived Categories of Spaces, Lemma 75.24.3. Clearly we have a cartesian square

and hence we may appeal to Lemma 108.4.8 to finish the proof. $\square$