Let $f :X \to Y$ be an affine morphism. Prove that the direct image sheaf $f_*\mathscr{O}_X$ is a quasi-coherent $\mathscr{O}_Y$-module.
One of the equivalent definitions of a quasi-coherent $\mathscr{O}_X$-module $\mathscr{F}$ is that for all affine opens $V\subset U$, the natural map $\mathscr{F}(V)\otimes_{\mathscr{O}_X(U)}\mathscr{O}_X(V)\to \mathscr{F}(V)$ is an isomorphism of $\mathscr{O}_X(V)$-modules.
My try was as follows:
Let $V\subset U$ be affine opens in $Y$. Since $f$ is an affine morphism, $f^{-1}V\subset f^{-1}U$ is an inclusion of affine opens. Now, $\mathscr{O}_X$ itself is a quasi-coherent $\mathscr{O}_X$-module, so the natural map $$ \mathscr{O}_X(f^{-1}U)\otimes_{\mathscr{O}_X(f^{-1}U)} \mathscr{O}_X(f^{-1}V) \longrightarrow \mathscr{O}_X(f^{-1}V) $$ is an isomorphism. This is of course just $$ f_*\mathscr{O}_X(U) \otimes_{\mathscr{O}_X(f^{-1}U)} \mathscr{O}_X(f^{-1}V) \longrightarrow f_*\mathscr{O}_X(V) $$ What we want to show is that $$ f_*\mathscr{O}_X(U) \otimes_{\mathscr{O}_Y(U)} \mathscr{O}_Y(V) \longrightarrow f_*\mathscr{O}_X(V) $$ is an isomorphism.
I believe that there should not be much to it, but I can not see it. Perhaps that we can use the fact that $A\otimes_B B\cong A$ ? Is there any slick way to do this exercise ? Any help is appreciated
Challenge: "Let f:X→Y be an affine morphism. Prove that the direct image sheaf f∗OX is a quasi-coherent OY-module."
Reply: A sheaf of $\mathcal{O}_Y$-modules $F$ is quasi coherent iff there is an open affine cover $U_i:=Spec(A_i)$ of $Y$ with $F_{U_i}\cong \tilde{F}_i$ where $F_i$ is a left $A_i$-module. If $f$ is affine it follows for every affine open subscheme $U:=Spec(A) \subseteq Y$ the inverse image scheme satisfies $f^{-1}(U)=Spec(B)$. Let $U_i:=Spec(A_i)$ be an affine open cover of $Y$ with $V_i:=f^{-1}(U_i)=Spec(B_i)$. It follows $f_*(\mathcal{O}_X)(U_i):=\mathcal{O}_X(V_i)=B_i$ and $B_i$ is a left $A_i$-module by definition.
If $X:=Spec(B), Y:=Spec(A)$ with $\phi: A \rightarrow B$ inducing $f$ we get
$$I1.\text{ }f_*\mathcal{O}_X(D(a)):=\mathcal{O}_X(f^{-1}(D(a))=\mathcal{O}_X(D(\phi(a))=B_{\phi(a)} \cong B\otimes_A A_a \cong \tilde{B}(D(a)).$$
Hence $f_*\mathcal{O}_X$ and $\tilde{B}$ agree on the basic open sets $D(a)$. You should verify that this implies they are the same sheaf. What you must do is to prove there is a morphism of sheaves
$$f_*\mathcal{O}_X \rightarrow \tilde{B}$$
restricting to $I1$ on an open cover consisting of basic open sets $D(a)$.
From this it follows that $f_*(\mathcal{O}_X)\cong \tilde{B}$ and is a quasi coherent $\mathcal{O}_Y$-module. Hence $f_*(\mathcal{O}_X)$ is a quasi coherent $\mathcal{O}_Y$-module. More generally if $E$ is a quasi coherent $\mathcal{O}_X$-module it follows
$$f_*E(U_i):=E(V_i):=E_i$$
where $E_i$ is a $B_i$-module and $A_i$-module, hence $f_*E$ is a quasi coherent $\mathcal{O}_Y$-module. See Proposition II.5.2 in Hartshorne.