**Proof.**
To prove this we consider the fibre product diagram

\[ \xymatrix{ X' = \mathop{\mathrm{Spec}}(B) \times _{\mathop{\mathrm{Spec}}(A)} X \ar[r]_-\pi \ar[d]_{f'} & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(B) \ar[r] & \mathop{\mathrm{Spec}}(A) } \]

Note that $f'$ is a proper morphism, see Morphisms, Lemma 29.41.5. Also, $B$ is a finitely generated $A$-algebra, and hence Noetherian (Algebra, Lemma 10.31.1). This implies that $X'$ is a Noetherian scheme (Morphisms, Lemma 29.15.6). Note that $X'$ is the relative spectrum of the quasi-coherent $\mathcal{O}_ X$-algebra $\mathcal{B}$ by Constructions, Lemma 27.4.6. Since $\mathcal{F}$ is a quasi-coherent $\mathcal{B}$-module we see that there is a unique quasi-coherent $\mathcal{O}_{X'}$-module $\mathcal{F}'$ such that $\pi _*\mathcal{F}' = \mathcal{F}$, see Morphisms, Lemma 29.11.6 Since $\mathcal{F}$ is finite type as a $\mathcal{B}$-module we conclude that $\mathcal{F}'$ is a finite type $\mathcal{O}_{X'}$-module (details omitted). In other words, $\mathcal{F}'$ is a coherent $\mathcal{O}_{X'}$-module (Lemma 30.9.1). Since the morphism $\pi : X' \to X$ is affine we have

\[ H^ p(X, \mathcal{F}) = H^ p(X', \mathcal{F}') \]

by Lemma 30.2.4. Thus (1) follows from Lemma 30.19.2. Given $\mathcal{L}$ as in (2) we set $\mathcal{L}' = \pi ^*\mathcal{L}$. Note that $\mathcal{L}'$ is ample on $X'$ by Morphisms, Lemma 29.37.7. By the projection formula (Cohomology, Lemma 20.54.2) we have $\pi _*(\mathcal{F}' \otimes \mathcal{L}') = \mathcal{F} \otimes \mathcal{L}$. Thus part (2) follows by the same reasoning as above from Lemma 30.16.2.
$\square$

## Comments (2)

Comment #8106 by Laurent Moret-Bailly on

Comment #8217 by Stacks Project on

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