Lemma 76.8.2. In Situation 76.8.1. 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 65.6.1). By Lemma 76.7.3 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 76.7.4. This finishes the proof of the lemma. $\square$

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