Lemma 77.8.2. In Situation 77.8.1. Let T \to S be a quasi-compact morphism of schemes such that the base change u_ T is zero. Then exists a closed subscheme Z \subset S such that (a) T \to S factors through Z and (b) the base change u_ Z is zero. If \mathcal{F} is a finite type \mathcal{O}_ X-module and the scheme theoretic support of \mathcal{F} is quasi-compact, then we can take Z \to S of finite presentation.
Proof. Let U \to X be a surjective étale morphism of algebraic spaces where U = \coprod U_ i is a disjoint union of affine schemes (see Properties of Spaces, Lemma 66.6.1). By Lemma 77.7.3 we see that we may replace X by U. In other words, we may assume that X = \coprod X_ i is a disjoint union of affine schemes X_ i. Suppose that we can prove the lemma for u_ i = u|_{X_ i}. Then we find a closed subscheme Z_ i \subset S such that T \to S factors through Z_ i and u_{i, Z_ i} is zero. If Z_ i = \mathop{\mathrm{Spec}}(R/I_ i) \subset \mathop{\mathrm{Spec}}(R) = S, then taking Z = \mathop{\mathrm{Spec}}(R/\sum I_ i) works. Thus we may assume that X = \mathop{\mathrm{Spec}}(A) is affine.
Choose a finite affine open covering T = T_1 \cup \ldots \cup T_ m. It is clear that we may replace T by \coprod _{j = 1, \ldots , m} T_ j. Hence we may assume T is affine. Say T = \mathop{\mathrm{Spec}}(R'). Let u : M \to N be the homomorphisms of A-modules corresponding to u : \mathcal{F} \to \mathcal{G}. Then N is a flat R-module as \mathcal{G} is flat over S. The assumption of the lemma means that the composition
is zero. Let z \in M. By Lazard's theorem (Algebra, Theorem 10.81.4) and the fact that \otimes commutes with colimits we can find free R-module F_ z, an element \tilde z \in F_ z, and a map F_ z \to N such that u(z) is the image of \tilde z and \tilde z maps to zero in F_ z \otimes _ R R'. Choose a basis \{ e_{z, \alpha }\} of F_ z and write \tilde z = \sum f_{z, \alpha } e_{z, \alpha } with f_{z, \alpha } \in R. Let I \subset R be the ideal generated by the elements f_{z, \alpha } with z ranging over all elements of M. By construction I maps to zero in R' and the elements \tilde z map to zero in F_ z/IF_ z whence in N/IN. Thus Z = \mathop{\mathrm{Spec}}(R/I) is a solution to the problem in this case.
Assume \mathcal{F} is of finite type with quasi-compact scheme theoretic support. Write Z = \mathop{\mathrm{Spec}}(R/I). Write I = \bigcup I_\lambda as a filtered union of finitely generated ideals. Set Z_\lambda = \mathop{\mathrm{Spec}}(R/I_\lambda ), so Z = \mathop{\mathrm{colim}}\nolimits Z_\lambda . Since u_ Z is zero, we see that u_{Z_\lambda } is zero for some \lambda by Lemma 77.7.4. This finishes the proof of the lemma. \square
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