Situation 76.8.1. Let $S = \mathop{\mathrm{Spec}}(R)$ be an affine scheme. Let $X$ be an algebraic space over $S$. Let $u : \mathcal{F} \to \mathcal{G}$ be a map of quasi-coherent $\mathcal{O}_ X$-modules. Assume $\mathcal{G}$ flat over $S$.

## 76.8 Making a map zero

This section has no analogue in the corresponding chapter on schemes.

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. $\square$

Lemma 76.8.3. Let $A$ be a ring. Let $u : M \to N$ be a map of $A$-modules. If $N$ is projective as an $A$-module, then there exists an ideal $I \subset A$ such that for any ring map $\varphi : A \to B$ the following are equivalent

$u \otimes 1 : M \otimes _ A B \to N \otimes _ A B$ is zero, and

$\varphi (I) = 0$.

**Proof.**
As $N$ is projective we can find a projective $A$-module $C$ such that $F = N \oplus C$ is a free $R$-module. By replacing $u$ by $u \oplus 1 : F = M \oplus C \to N \oplus C$ we see that we may assume $N$ is free. In this case let $I$ be the ideal of $A$ generated by coefficients of all the elements of $\mathop{\mathrm{Im}}(u)$ with respect to some (fixed) basis of $N$.
$\square$

Lemma 76.8.4. In Situation 76.8.1. Let $T \subset S$ be a subset. Let $s \in S$ be in the closure of $T$. For $t \in T$, let $u_ t$ be the pullback of $u$ to $X_ t$ and let $u_ s$ be the pullback of $u$ to $X_ s$. If $X$ is locally of finite presentation over $S$, $\mathcal{G}$ is of finite presentation^{1}, and $u_ t = 0$ for all $t \in T$, then $u_ s = 0$.

**Proof.**
To check whether $u_ s$ is zero, is étale local on the fibre $X_ s$. Hence we may pick a point $x \in |X_ s| \subset |X|$ and check in an étale neighbourhood. Choose

as in Proposition 76.5.1. Let $T' \subset S'$ be the inverse image of $T$. Observe that $s'$ is in the closure of $T'$ because $S' \to S$ is open. Hence we reduce to the algebra problem described in the next paragraph.

We have an $R$-module map $u : M \to N$ such that $N$ is projective as an $R$-module and such that $u_ t : M \otimes _ R \kappa (t) \to N \otimes _ R \kappa (t)$ is zero for each $t \in T$. Problem: show that $u_ s = 0$. Let $I \subset R$ be the ideal defined in Lemma 76.8.3. Then $I$ maps to zero in $\kappa (t)$ for all $t \in T$. Hence $T \subset V(I)$. Since $s$ is in the closure of $T$ we see that $s \in V(I)$. Hence $u_ s = 0$. $\square$

It would be interesting to find a “simple” direct proof of either Lemma 76.8.5 or Lemma 76.8.6 using arguments like those used in Lemmas 76.8.2 and 76.8.4. A “classical” proof of this lemma when $f : X \to B$ is a projective morphism and $B$ a Noetherian scheme would be: (a) choose a relatively ample invertible sheaf $\mathcal{O}_ X(1)$, (b) set $u_ n : f_*\mathcal{F}(n) \to f_*\mathcal{G}(n)$, (c) observe that $f_*\mathcal{G}(n)$ is a finite locally free sheaf for all $n \gg 0$, and (d) $F_{zero}$ is represented by the vanishing locus of $u_ n$ for some $n \gg 0$.

Lemma 76.8.5. In Situation 76.7.1. Assume

$f$ is of finite presentation, and

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

Then $F_{zero}$ is an algebraic space and $F_{zero} \to B$ is a closed immersion. If $\mathcal{F}$ is of finite type, then $F_{zero} \to B$ is of finite presentation.

**Proof.**
By Lemma 76.6.5 the module $\mathcal{G}$ is universally pure relative to $B$. In order to prove that $F_{zero}$ is an algebraic space, it suffices to show that $F_{zero} \to B$ is representable, see Spaces, Lemma 64.11.3. Let $B' \to B$ be a morphism where $B'$ is a scheme and let $u' : \mathcal{F}' \to \mathcal{G}'$ be the pullback of $u$ to $X' = X_{B'}$. Then the associated functor $F'_{zero}$ equals $F_{zero} \times _ B B'$. This reduces us to the case that $B$ is a scheme.

Assume $B$ is a scheme. We will show that $F_{zero}$ is representable by a closed subscheme of $B$. By Lemma 76.7.2 and Descent, Lemmas 35.37.2 and 35.39.1 the question is local for the étale topology on $B$. Let $b \in B$. We first replace $B$ by an affine neighbourhood of $b$. Choose a diagram

and $b' \in B'$ mapping to $b \in B$ as in Lemma 76.5.2. As we are working étale locally, we may replace $B$ by $B'$ and assume that we have a diagram

with $B$ and $X'$ affine such that $\Gamma (X', g^*\mathcal{G})$ is a projective $\Gamma (B, \mathcal{O}_ B)$-module and $g(|X'|) \supset |X_ b|$. Let $U \subset X$ be the open subspace with $|U| = g(|X'|)$. By Divisors on Spaces, Lemma 70.4.10 the set

is constructible in $B$. By Lemma 76.6.3 part (2) we see that $E$ contains $\mathop{\mathrm{Spec}}(\mathcal{O}_{B, b})$. By Morphisms, Lemma 29.22.4 we see that $E$ contains an open neighbourhood of $b$. Hence after replacing $B$ by a smaller affine neighbourhood of $b$ we may assume that $\text{Ass}_{X/B}(\mathcal{G}) \subset g(|X'|)$.

From Lemma 76.6.6 it follows that $u : \mathcal{F} \to \mathcal{G}$ is injective if and only if $g^*u : g^*\mathcal{F} \to g^*\mathcal{G}$ is injective, and the same remains true after any base change. Hence we have reduced to the case where, in addition to the assumptions in the theorem, $X \to B$ is a morphism of affine schemes and $\Gamma (X, \mathcal{G})$ is a projective $\Gamma (B, \mathcal{O}_ B)$-module. This case follows immediately from Lemma 76.8.3.

We still have to show that $F_{zero} \to B$ is of finite presentation if $\mathcal{F}$ is of finite type. This follows from Lemma 76.7.4 combined with Limits of Spaces, Proposition 69.3.10. $\square$

Lemma 76.8.6. In Situation 76.7.1. Assume

$f$ is locally of finite presentation,

$\mathcal{G}$ is an $\mathcal{O}_ X$-module of finite presentation flat over $B$,

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

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

**Proof.**
If $f$ is of finite presentation, then this follows immediately from Lemmas 76.8.5 and 76.3.6. This is the only case of interest and we urge the reader to skip the rest of the proof, which deals with the possibility (allowed by the assumptions in this lemma) that $f$ is not quasi-separated or quasi-compact.

Let $i : Z \to X$ be the closed subspace cut out by the zeroth fitting ideal of $\mathcal{G}$ (Divisors on Spaces, Section 70.5). Then $Z \to B$ is proper by assumption (see Derived Categories of Spaces, Section 74.7). On the other hand $i$ is of finite presentation (Divisors on Spaces, Lemma 70.5.2 and Morphisms of Spaces, Lemma 66.28.12). There exists a quasi-coherent $\mathcal{O}_ Z$-module $\mathcal{H}$ of finite type with $i_*\mathcal{H} = \mathcal{G}$ (Divisors on Spaces, Lemma 70.5.3). In fact $\mathcal{H}$ is of finite presentation as an $\mathcal{O}_ Z$-module by Algebra, Lemma 10.6.4 (details omitted). Then $F_{zero}$ is the same as the functor $F_{zero}$ for the map $i^*\mathcal{F} \to \mathcal{H}$ adjoint to $u$, see Lemma 76.7.6. The sheaf $\mathcal{H}$ is flat relative to $B$ because the same is true for $\mathcal{G}$ (check on stalks; details omitted). Moreover, note that if $\mathcal{F}$ is of finite type, then $i^*\mathcal{F}$ is of finite type. Hence we have reduced the lemma to the case discussed in the first paragraph of the proof. $\square$

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