Lemma 68.16.2. In Situation 68.16.1 for an $A$-module $M$ we have $p_*(M \otimes _ A \mathcal{O}_ X) = \widetilde{M}$ and $\Gamma (X, M \otimes _ A \mathcal{O}_ X) = M$.
Proof. The equality $p_*(M \otimes _ A \mathcal{O}_ X) = \widetilde{M}$ follows from the equality $\Gamma (X, M \otimes _ A \mathcal{O}_ X) = M$ as $p_*(M \otimes _ A \mathcal{O}_ X)$ is a quasi-coherent module on $\mathop{\mathrm{Spec}}(A)$ by Morphisms of Spaces, Lemma 66.11.2. Observe that $\Gamma (X, \bigoplus _{i \in I} \mathcal{O}_ X) = \bigoplus _{i \in I} A$ by Lemma 68.5.1. Hence the lemma holds for free modules. Choose a short exact sequence $F_1 \to F_0 \to M$ where $F_0, F_1$ are free $A$-modules. Since $H^1(X, -)$ is zero the global sections functor is right exact. Moreover the pullback $p^*$ is right exact as well. Hence we see that
\[ \Gamma (X, F_1 \otimes _ A \mathcal{O}_ X) \to \Gamma (X, F_0 \otimes _ A \mathcal{O}_ X) \to \Gamma (X, M \otimes _ A \mathcal{O}_ X) \to 0 \]
is exact. The result follows. $\square$
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