Lemma 33.30.1. Let $A \to B$ be a faithfully flat ring map. Let $X$ be a quasi-compact and quasi-separated scheme over $A$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module whose pullback to $X_ B$ is trivial. Then $H^0(X, \mathcal{L})$ and $H^0(X, \mathcal{L}^{\otimes -1})$ are invertible $H^0(X, \mathcal{O}_ X)$-modules and the multiplication map induces an isomorphism

$H^0(X, \mathcal{L}) \otimes _{H^0(X, \mathcal{O}_ X)} H^0(X, \mathcal{L}^{\otimes -1}) \longrightarrow H^0(X, \mathcal{O}_ X)$

Proof. Denote $\mathcal{L}_ B$ the pullback of $\mathcal{L}$ to $X_ B$. Choose an isomorphism $\mathcal{L}_ B \to \mathcal{O}_{X_ B}$. Set $R = H^0(X, \mathcal{O}_ X)$, $M = H^0(X, \mathcal{L})$ and think of $M$ as an $R$-module. For every quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ with pullback $\mathcal{F}_ B$ on $X_ B$ there is a canonical isomorphism $H^0(X_ B, \mathcal{F}_ B) = H^0(X, \mathcal{F}) \otimes _ A B$, see Cohomology of Schemes, Lemma 30.5.2. Thus we have

$M \otimes _ R (R \otimes _ A B) = M \otimes _ A B = H^0(X_ B, \mathcal{L}_ B) \cong H^0(X_ B, \mathcal{O}_{X_ B}) = R \otimes _ A B$

Since $R \to R \otimes _ A B$ is faithfully flat (as the base change of the faithfully flat map $A \to B$), we conclude that $M$ is an invertible $R$-module by Algebra, Proposition 10.83.3. Similarly $N = H^0(X, \mathcal{L}^{\otimes -1})$ is an invertible $R$-module. To see that the statement on tensor products is true, use that it is true after pulling back to $X_ B$ and faithful flatness of $R \to R \otimes _ A B$. Some details omitted. $\square$

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