Lemma 37.33.5. Consider a commutative diagram of schemes

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

with $f' : X' \to S$ and $f : X \to S$ satisfying the hypotheses of Lemma 37.33.1. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module and let $\mathcal{L}'$ be the pullback to $X'$. Let $Z \subset S$, resp. $Z' \subset S$ be the locally closed subscheme constructed in Lemma 37.33.1 for $(f, \mathcal{L})$, resp. $(f', \mathcal{L}')$ so that $Z \subset Z'$. If $s \in Z$ and

\[ H^1(X_ s, \mathcal{O}) \longrightarrow H^1(X'_ s, \mathcal{O}) \]

is injective, then $Z \cap U = Z' \cap U$ for some open neighbourhood $U$ of $s$.

**Proof.**
We may replace $S$ by $Z'$. After shrinking $S$ to an affine open neighbourhood of $s$ we may assume that $\mathcal{L}' = \mathcal{O}_{X'}$. Let $E = Rf_*\mathcal{L}$ and $E' = Rf'_*\mathcal{L}' = Rf'_*\mathcal{O}_{X'}$. 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 $h_ i = \dim _{\kappa (s)} H^ i(X_ s, \mathcal{O}_{X_ s})$. 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 15.75.6 (strictly speaking this also uses Derived Categories of Schemes, Lemmas 36.3.5 and 36.10.7). Similarly, we may assume $E'$ is represented by a complex

\[ \mathcal{O}_ S \to \mathcal{O}_ S^{\oplus h'_1} \to \mathcal{O}_ S^{\oplus h'_2} \to \ldots \]

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

\[ E \longrightarrow E' \]

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

\[ \xymatrix{ \mathcal{O}_ S \ar[r]_ d \ar[d]_ a & \mathcal{O}_ S^{\oplus h_1} \ar[r] \ar[d]_ b & \mathcal{O}_ S^{\oplus h_2} \ar[r] \ar[d]_ c & \ldots \\ \mathcal{O}_ S \ar[r]^{d'} & \mathcal{O}_ S^{\oplus h'_1} \ar[r] & \mathcal{O}_ S^{\oplus h'_2} \ar[r] & \ldots } \]

representing the given map $E \to E'$. Since $s \in Z$ we see that the trivializing section of $\mathcal{L}_ s$ pulls back to a trivializing section of $\mathcal{L}'_ s = \mathcal{O}_{X'_ s}$. Thus $a \otimes \kappa (s)$ is an isomorphism, hence after shrinking $S$ we see that $a$ is an isomorphism. Finally, we use the hypothesis that $H^1(X_ s, \mathcal{O}) \to H^1(X'_ s, \mathcal{O})$ is injective, to see that there exists a $h_1 \times h_1$ minor of the matrix defining $b$ which maps to a nonzero element in $\kappa (s)$. Hence after shrinking $S$ we may assume that $b$ is injective. However, since $\mathcal{L}' = \mathcal{O}_{X'}$ we see that $d' = 0$. It follows that $d = 0$. 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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