Higher direct images of quasi-coherent modules vanish along affine morphisms

Let $V\subseteq S$ be affine. By [F1] the preimage $f^{-1}V$ is an affine scheme, and $\mathcal{F}{\mathrm{\mid }}_{f^{-1}V}$ is quasi-coherent by [F3]. Applying [F2] to the affine scheme $f^{-1}V$ and this restriction gives $H^q(f^{-1}V,\mathcal{F})=0$ for every $q>0$.

Fix $q>0$ and let ${\mathcal{P}}_q$ be the presheaf $V\mapsto H^q(f^{-1}V,\mathcal{F})$ on the open subsets of $S$. By [F4] the sheaf $R^q{f}_{\ast }\mathcal{F}$ is ${\mathcal{P}}_q$'s sheafification, so by [F5] its stalk at a point $s\in S$ is the stalk of ${\mathcal{P}}_q$ at $s$.

Compute that stalk as a colimit. By [F6] the stalk of ${\mathcal{P}}_q$ at $s$ is the filtered colimit of the groups ${\mathcal{P}}_q(V)=H^q(f^{-1}V,\mathcal{F})$ over the open neighbourhoods $V\ni s$, and by the same fact the affine open neighbourhoods of $s$ — which form a basis — are cofinal in this system, so the colimit may be computed over them alone. Every value occurring there vanishes by step 1.1, and a filtered colimit of zero groups with zero transition maps is zero, so $(R^q{f}_{\ast }\mathcal{F}{)}_s=0$ for every $s\in S$.

The sheaf $R^q{f}_{\ast }\mathcal{F}$ has zero stalk at every point of $S$ by step 2.1, so by [F7] it is the zero sheaf; this holds for every $q>0$, which is the assertion. Boundary cases: if $S=\mathrm{\varnothing }$ there is nothing to check and the claim is vacuous; if $X=\mathrm{\varnothing }$ or $\mathcal{F}=0$, step 1.1 applies at every affine $V$ with value $0$; if $f$ is the identity or $S$ is affine, step 1.1 is the same computation. The Axiom of Choice is inherited from [F2] and [F4], and no further selection is made beyond the choice of the pointwise affine neighbourhoods implicit in [F6].