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77.9 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 77.9.1. In Situation 77.7.1 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 77.7.5 we see that F_{u, iso} = F_{v, zero}. By Lemma 77.8.5 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 77.9.2. In Situation 77.7.1. 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 77.7.5 we see that F_{u, iso} = F_{v, zero}. By Lemma 77.8.6 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 77.9.3. In Situation 77.7.1. 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 67.15.1). 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 66.4.8. 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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