The Stacks project

Lemma 37.30.6. Consider $n$ commutative diagrams of schemes

\[ \xymatrix{ X_ i \ar[rr] \ar[dr]_{f_ i} & & X \ar[dl]^ f \\ & S } \]

with $f_ i : X_ i \to S$ and $f : X \to S$ satisfying the hypotheses of Lemma 37.30.1. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module and let $\mathcal{L}_ i$ be the pullback to $X_ i$. Let $Z \subset S$, resp. $Z_ i \subset S$ be the locally closed subscheme constructed in Lemma 37.30.1 for $(f, \mathcal{L})$, resp. $(f_ i, \mathcal{L}_ i)$ so that $Z \subset \bigcap _{i = 1, \ldots , n} Z_ i$. If $s \in Z$ and

\[ H^1(X_ s, \mathcal{O}) \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} H^1(X_{i, s}, \mathcal{O}) \]

is injective, then $Z \cap U = (\bigcap _{i = 1, \ldots , n} Z_ i) \cap U$ (scheme theoretic intersection) for some open neighbourhood $U$ of $s$.

Proof. This lemma is a variant of Lemma 37.30.5 and we strongly urge the reader to read that proof first; this proof is basically a copy of that proof with minor modifications. It follows from the description of (scheme valued) points of $Z$ and the $Z_ i$ that $Z \subset \bigcap _{i = 1, \ldots , n} Z_ i$ where we take the scheme theoretic intersection. Thus we may replace $S$ by the scheme theoretic intersection $\bigcap _{i = 1, \ldots , n} Z_ i$. After shrinking $S$ to an affine open neighbourhood of $s$ we may assume that $\mathcal{L}_ i = \mathcal{O}_{X_ i}$ for $i = 1, \ldots , n$. Let $E = Rf_*\mathcal{L}$ and $E_ i = Rf_{i, *}\mathcal{L}_ i = Rf_{i, *}\mathcal{O}_{X_ i}$. These are perfect complexes whose formation commutes with arbitrary change of base (Derived Categories of Schemes, Lemma 36.30.4). In particular we see that

\[ E \otimes _{\mathcal{O}_ S}^\mathbf {L} \kappa (s) = R\Gamma (X_ s, \mathcal{L}_ s) = R\Gamma (X_ s, \mathcal{O}_{X_ s}) \]

The second equality because $s \in Z$. Set $h_ j = \dim _{\kappa (s)} H^ j(X_ s, \mathcal{O}_{X_ s})$. After shrinking $S$ we can represent $E$ by a complex

\[ \mathcal{O}_ S \to \mathcal{O}_ S^{\oplus h_1} \to \mathcal{O}_ S^{\oplus h_2} \to \ldots \]

see More on Algebra, Lemma 15.74.6 (strictly speaking this also uses Derived Categories of Schemes, Lemmas 36.3.5 and 36.10.7). Similarly, we may assume $E_ i$ is represented by a complex

\[ \mathcal{O}_ S \to \mathcal{O}_ S^{\oplus h_{i, 1}} \to \mathcal{O}_ S^{\oplus h_{i, 2}} \to \ldots \]

where $h_{i, j} = \dim _{\kappa (s)} H^ j(X_{i, s}, \mathcal{O}_{X_{i, s}})$. By functoriality of cohomology we have a map

\[ E \longrightarrow E_ i \]

in $D(\mathcal{O}_ S)$ whose formation commutes with change of base. Since the complex representing $E$ is a finite complex of finite free modules and since $S$ is affine, we can choose a map of complexes

\[ \xymatrix{ \mathcal{O}_ S \ar[r]_ d \ar[d]_{a_ i} & \mathcal{O}_ S^{\oplus h_1} \ar[r] \ar[d]_{b_ i} & \mathcal{O}_ S^{\oplus h_2} \ar[r] \ar[d]_{c_ i} & \ldots \\ \mathcal{O}_ S \ar[r]^{d_ i} & \mathcal{O}_ S^{\oplus h_{i, 1}} \ar[r] & \mathcal{O}_ S^{\oplus h_{i, 2}} \ar[r] & \ldots } \]

representing the given map $E \to E_ i$. Since $s \in Z$ we see that the trivializing section of $\mathcal{L}_ s$ pulls back to a trivializing section of $\mathcal{L}_{i, s} = \mathcal{O}_{X_{i, s}}$. Thus $a_ i \otimes \kappa (s)$ is an isomorphism, hence after shrinking $S$ we see that $a_ i$ is an isomorphism. Finally, we use the hypothesis that $H^1(X_ s, \mathcal{O}) \to \bigoplus _{i = 1, \ldots , n} H^1(X_{i, s}, \mathcal{O})$ is injective, to see that there exists a $h_1 \times h_1$ minor of the matrix defining $\oplus b_ i$ which maps to a nonzero element in $\kappa (s)$. Hence after shrinking $S$ we may assume that $(b_1, \ldots , b_ n) : \mathcal{O}_ S^{h_1} \to \bigoplus _{i = 1, \ldots , n} \mathcal{O}_ S^{h_{i, 1}}$ is injective. However, since $\mathcal{L}_ i = \mathcal{O}_{X_ i}$ we see that $d_ i = 0$ for $i = 1, \ldots n$. It follows that $d = 0$ because $(b_1, \ldots , b_ n) \circ d = (\oplus d_ i) \circ (a_1, \ldots , a_ n)$. In this way we see that the trivializing section of $\mathcal{L}_ s$ lifts to a section of $\mathcal{L}$ over $X$. A straightforward topological argument (omitted) shows that this means that $\mathcal{L}$ is trivial after possibly shrinking $S$ a bit further. $\square$


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