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76.9 Flattening a map

This section is the analogue of More on Flatness, Section 38.23. In particular the following result is a variant of More on Flatness, Theorem 38.23.3.

Theorem 76.9.1. In Situation 76.7.1 assume

  1. $f$ is of finite presentation,

  2. $\mathcal{F}$ is of finite presentation, flat over $B$, and pure relative to $B$, and

  3. $u$ is surjective.

Then $F_{iso}$ is representable by a closed immersion $Z \to B$. Moreover $Z \to S$ is of finite presentation if $\mathcal{G}$ is of finite presentation.

Proof. Let $\mathcal{K} = \mathop{\mathrm{Ker}}(u)$ and denote $v : \mathcal{K} \to \mathcal{F}$ the inclusion. By Lemma 76.7.5 we see that $F_{u, iso} = F_{v, zero}$. By Lemma 76.8.5 applied to $v$ we see that $F_{u, iso} = F_{v, zero}$ is representable by a closed subspace of $B$. Note that $\mathcal{K}$ is of finite type if $\mathcal{G}$ is of finite presentation, see Modules on Sites, Lemma 18.24.1. Hence we also obtain the final statement of the lemma. $\square$

Lemma 76.9.2. In Situation 76.7.1. Assume

  1. $f$ is locally of finite presentation,

  2. $\mathcal{F}$ is locally of finite presentation and flat over $B$,

  3. the support of $\mathcal{F}$ is proper over $B$, and

  4. $u$ is surjective.

Then the functor $F_{iso}$ is an algebraic space and $F_{iso} \to B$ is a closed immersion. If $\mathcal{G}$ is of finite presentation, then $F_{iso} \to B$ is of finite presentation.

Proof. Let $\mathcal{K} = \mathop{\mathrm{Ker}}(u)$ and denote $v : \mathcal{K} \to \mathcal{F}$ the inclusion. By Lemma 76.7.5 we see that $F_{u, iso} = F_{v, zero}$. By Lemma 76.8.6 applied to $v$ we see that $F_{u, iso} = F_{v, zero}$ is representable by a closed subspace of $B$. Note that $\mathcal{K}$ is of finite type if $\mathcal{G}$ is of finite presentation, see Modules on Sites, Lemma 18.24.1. Hence we also obtain the final statement of the lemma. $\square$

We will use the following (easy) result when discussing the Quot functor.

Lemma 76.9.3. In Situation 76.7.1. Assume

  1. $f$ is locally of finite presentation,

  2. $\mathcal{G}$ is of finite type,

  3. the support of $\mathcal{G}$ is proper over $B$.

Then $F_{surj}$ is an algebraic space and $F_{surj} \to B$ is an open immersion.

Proof. Consider $\mathop{\mathrm{Coker}}(u)$. Observe that $\mathop{\mathrm{Coker}}(u_ T) = \mathop{\mathrm{Coker}}(u)_ T$ for any $T/B$. Note that formation of the support of a finite type quasi-coherent module commutes with pullback (Morphisms of Spaces, Lemma 66.15.1). Hence $F_{surj}$ is representable by the open subspace of $B$ corresponding to the open set

\[ |B| \setminus |f|(\text{Supp}(\mathop{\mathrm{Coker}}(u))) \]

see Properties of Spaces, Lemma 65.4.8. This is an open because $|f|$ is closed on $\text{Supp}(\mathcal{G})$ and $\text{Supp}(\mathop{\mathrm{Coker}}(u))$ is a closed subset of $\text{Supp}(\mathcal{G})$. $\square$


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