Quasi coherent module over affine schemes I am stuck by this question from Liu's algebraic geometry textbook on quasi-coherent modules.
Let X be an affine scheme $\mbox {Spec} A $. Let $\mathcal {F} $ be a quasi-coherent $\mathcal{O}_{X} $ module. Show that for any affine open subset U of X we have a canonical isomorphism
$\mathcal {F}(X)\otimes_{A}\mathcal {O}_{X}(U)\cong\mathcal {F}(U) $. 
I have tried to reduce to the problem to the really simple case where $ U=D (f) $ for $ f\in A $, which is really easy but I can't prove it for general case where U is just any affine open subset. Can somebody give me some hints? Thanks!
EDIT: I am still working on it when I realise that I might need this: Can someone tell me if this is true?
If $ u\in U $ where U is an affine subset as before, then is
$\mathcal {F} _{u}=\tilde {\mathcal{F}(X)}_{u}=\tilde {\mathcal {F}(U)}_{u}\cong \mathcal {F}(U)\otimes_{\mathcal{O}_{X}(U)}\mathcal{O}_{X}(U)_{u} $ 
the above expression true because it is quasi coherent over affine schemes?
 A: When $U$ is basic-open, it holds by definition. In general, there is a canonical homomorphism $F(X) \otimes_A \mathcal{O}_X(U) \to F(U)$ adjoint to the restriction map $F(X) \to F(U)$. If $U$ is quasi-compact, it is a finite union of basic-open subsets $U_i$ and their intersections $U_{ij} := U_i \cap U_j$ are also basic-open. Now look at the following commutative diagram:
$$\begin{array}{c} 0 & \rightarrow & F(X) \otimes_A \mathcal{O}_X(U) & \rightarrow & \bigoplus_i F(X) \otimes_A \mathcal{O}_X(U_i) & \rightrightarrows & \bigoplus_{i,j} F(X) \otimes_A \mathcal{O}_X(U_{ij}) \\ && \downarrow && \downarrow && \downarrow \\\\ 0 & \rightarrow & F(U) & \rightarrow & \bigoplus_i F(U_i) & \rightrightarrows & \bigoplus_{i,j} F(U_{ij})\end{array}$$
The bottom row is exact. The top row is exact when $F$ is flat i.e. when $F(X)$ is flat over $A$. The two vertical homomorphisms on the right are isomorphisms. Hence, the vertical homomorphism on the left is an isomorphism, too. If $F$ is not flat, we choose a free resolution $F'' \to F' \to F \to 0$. Then $F''(X) \to F'(X) \to F(X) \to 0$ is exact (because we have quasi-coherent modules on an affine schemes and $F \mapsto F(X)$ is actually an equivalence of categories) and likewise for $U$ when $U$ is affine. Then the commutative diagram with exact rows
$$\begin{array}{c} F''(X) \otimes_A \mathcal{O}_X(U) & \rightarrow & F'(X) \otimes_A \mathcal{O}_X(U) & \rightarrow & F(X) \otimes_A \mathcal{O}_X(U) & \rightarrow & 0 \\ \downarrow && \downarrow && \downarrow & \\ F''(U) & \rightarrow & F'(U) & \rightarrow & F(U) & \rightarrow & 0 \end{array}$$
finishes the proof.
Actually this offers another proof, which even works when $U$ is just quasi-compact: Clearly the claim holds when $F=\mathcal{O}_X$, but then also for $F=\mathcal{O}_X^{\oplus I}$ for finite sets $I$. Since $U$ is quasi-compact, $F \mapsto F(U)$ (and likewise for $X$) commutes with filtered colimits, so that the claim even holds for $F=\mathcal{O}_X^{\oplus I}$ with any set $I$. Now choose free resolutions to obtain the result for arbitrary $F$.
A: I attempted the question but it will be too long as a comment in the preceding answer, hope someone can help me verify. (And because I am typing this on the phone :( )
I need to show that $\mathcal {F}(X)\otimes_{A}\mathcal {O}_{X}(U)\cong \mathcal {F}(U) $.
The $\mathcal{O}_{X} $-module $ \mathcal {O}_{X}|_{U}$ is quasi-coherent because $ i: U\hookrightarrow X $ corresponds to morphisms of affine schemes.
Then localising at $ u\in U $, we have
$ (\mathcal {F}\otimes_{\mathcal {O}_{X}}i_{*}\mathcal {O}_{X}|_{U})_{u}= \tilde {i_{*}\mathcal {F}|_{U}(X)}_{u} ,$
Which is the one that needs checking. 
Then the two schemes are isomorphic and so we get the conclusion.
