Lemma 99.10.1. Let $S$ be a scheme. Let $f : X \to B$ be a morphism of algebraic spaces over $S$ which is flat, of finite presentation, and proper. The natural map

is representable by open immersions.

The Picard stack for a morphism of algebraic spaces was introduced in Examples of Stacks, Section 95.16. We will deduce it is an open substack of the stack of coherent sheaves (in good cases) from the following lemma.

Lemma 99.10.1. Let $S$ be a scheme. Let $f : X \to B$ be a morphism of algebraic spaces over $S$ which is flat, of finite presentation, and proper. The natural map

\[ \mathcal{P}\! \mathit{ic}_{X/B} \longrightarrow \mathcal{C}\! \mathit{oh}_{X/B} \]

is representable by open immersions.

**Proof.**
Observe that the map simply sends a triple $(T, g, \mathcal{L})$ as in Examples of Stacks, Section 95.16 to the same triple $(T, g, \mathcal{L})$ but where now we view this as a triple of the kind described in Situation 99.5.1. This works because the invertible $\mathcal{O}_{X_ T}$-module $\mathcal{L}$ is certainly a finitely presented $\mathcal{O}_{X_ T}$-module, it is flat over $T$ because $X_ T \to T$ is flat, and the support is proper over $T$ as $X_ T \to T$ is proper (Morphisms of Spaces, Lemmas 67.30.4 and 67.40.3). Thus the statement makes sense.

Having said this, it is clear that the content of the lemma is the following: given an object $(T, g, \mathcal{F})$ of $\mathcal{C}\! \mathit{oh}_{X/B}$ there is an open subscheme $U \subset T$ such that for a morphism of schemes $T' \to T$ the following are equivalent

$T' \to T$ factors through $U$,

the pullback $\mathcal{F}_{T'}$ of $\mathcal{F}$ by $X_{T'} \to X_ T$ is invertible.

Let $W \subset |X_ T|$ be the set of points $x \in |X_ T|$ such that $\mathcal{F}$ is locally free in a neighbourhood of $x$. By More on Morphisms of Spaces, Lemma 76.23.8. $W$ is open and formation of $W$ commutes with arbitrary base change. Clearly, if $T' \to T$ satisfies (b), then $|X_{T'}| \to |X_ T|$ maps into $W$. Hence we may replace $T$ by the open $T \setminus f_ T(|X_ T| \setminus W)$ in order to construct $U$. After doing so we reach the situation where $\mathcal{F}$ is finite locally free. In this case we get a disjoint union decomposition $X_ T = X_0 \amalg X_1 \amalg X_2 \amalg \ldots $ into open and closed subspaces such that the restriction of $\mathcal{F}$ is locally free of rank $i$ on $X_ i$. Then clearly

\[ U = T \setminus f_ T(|X_0| \cup |X_2| \cup |X_3| \cup \ldots ) \]

works. (Note that if we assume that $T$ is quasi-compact, then $X_ T$ is quasi-compact hence only a finite number of $X_ i$ are nonempty and so $U$ is indeed open.) $\square$

Proposition 99.10.2. Let $S$ be a scheme. Let $f : X \to B$ be a morphism of algebraic spaces over $S$. If $f$ is flat, of finite presentation, and proper, then $\mathcal{P}\! \mathit{ic}_{X/B}$ is an algebraic stack.

**Proof.**
Immediate consequence of Lemma 99.10.1, Algebraic Stacks, Lemma 94.15.4 and either Theorem 99.5.12 or Theorem 99.6.1
$\square$

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