Quasi-coherent sheaves on varieties I am reading Kempf's book "Algebraic varieties".
On page 55 the author considers a sheaf of rings ${\mathcal{A}}$ (commutative, unital) on a topological space $X$. 
An ${\mathcal{A}}$-module ${\mathcal{M}}$ is said to be quasi-coherent if there is an open cover $X=\bigcup X_i$ such that we have exact sequences of ${\mathcal{A}}|_{X_i}$-modules
$$
 {\mathcal{A}}|_{X_i}^{\oplus J}\to {\mathcal{A}}|_{X_i}^{\oplus I}\to {\mathcal{M}}|_{X_i}\to 0
$$
(where the sets $I$ and $J$ can be infinite).
Let $M$ be an ${\mathcal{A}}(X)$-module. Then we can form an ${\mathcal{A}}$-module $M\otimes_{{\mathcal{A}}(X)} {\mathcal{A}}$ by taking it to be the sheaf associated to the presheaf
$$
U\mapsto M\otimes_{{\mathcal{A}}(X)}{\mathcal{A}}(U).
$$
The author notices that if $M$ is the cokernel of a homomorphism
$$
\psi\colon {\mathcal{A}}(X)^{\oplus J}\to {\mathcal{A}}(X)^{\oplus I},
$$ 
then we have an exact sequence
$$
{\mathcal{A}}(X)^{\oplus J}\to {\mathcal{A}}(X)^{\oplus I}\to M\otimes_{{\mathcal{A}}(X)} {\mathcal{A}} \to 0.
$$
Then the author remarks that thus on a Noetherian space $X$, an ${\mathcal{A}}$-module ${\mathcal{M}}$ is quasi-coherent if and only if it locally has the form
$M_i\otimes_{{\mathcal{A}}(X_i)} ({\mathcal{A}}|_{X_i})$.
Question. How can one prove the assertion of this last remark? Where does one use the assumption that $X$ is Noetherian? 
 A: You can always express an $A$ module $M$ as a cokernel of $A^{J}\to A^I$ (take a set of generators indicized by $I$, define $A^I\to M$ and then do the same with the kernel and $J$). So, if $\mathcal{M}$ is locally of the form $M_i\otimes_{\mathcal{A}(X_i)}(\mathcal{A}|_{X_i})$, then $\mathcal{M}$ is quasi-coherent because you can express it locally as
$$\mathcal{A}|_{X_i}^{J}\to\mathcal{A}|_{X_i}^{I}\to M_i\otimes_{\mathcal{A}(X_i)}(\mathcal{A}|_{X_i})=\mathcal{M}|_{X_i}\to 0$$
Here you don't have used the assumption that $X$ is noetherian yet. Now, if $\mathcal{M}$ is quasi-coherent and $X$ is noetherian, you have 
$${\mathcal{A}}|_{X_i}^{J}\to {\mathcal{A}}|_{X_i}^{I}\to {\mathcal{M}}|_{X_i}\to 0$$
Now take $\{a_j\}_{j\in J}$, $\{a_i'\}_{i\in I},$ the canonical basis of $\mathcal{A}(X_i)^{J}$, $\mathcal{A}(X_i)^{I}$ and for all $j$ take the image $b_j\in\mathcal{A}|_{X_i}^{I}$ of $a_j$. Locally, $b_j$ is expressed as a linear combination of only a finite number of $a_i'$, but this is true on the whole $X_i$ because $X$ is noetherian (you have to do this for all $j$'s at the same time, so it's not enough to say that this is true locally without $X$ notherian, this is the crucial point).
So, $\mathcal M|_{X_i}$ is the sheafification of
$$U\mapsto\mathcal{A}(U)^I/\mathcal{A}(U)^J$$
that is $\mathcal{A}(X_i)^I/\mathcal{A}(X_i)^J\otimes_{\mathcal{A}(X_i)}(\mathcal{A}|_{X_i})$.
Edit: as requested, adding more details.
Call $P$ the presheaf $U\mapsto \mathcal{A}(U)^I/\mathcal{A}(U)^J$. You have an obvious map $\phi:P\to \mathcal{M}$, induced by $\mathcal{A}^I_{X_i}\to\mathcal{M}$. $\phi$ is an isomorfism everytime you localize to a point $x\in X_i$: this is precisely the fact that ${\mathcal{A}}|_{X_i}^{J}\to {\mathcal{A}}|_{X_i}^{I}\to {\mathcal{M}}|_{X_i}\to 0$ is exact. Hence, when you sheafify $P$, you get an isomorphism.
You need the fact about the $b_j$'s because $\mathcal{A}^J(X_i)$ is different from $\mathcal{A}(X_i)^J$ (you have to sheafify) hence it's not always true that ${\mathcal{A}}|_{X_i}^{J}\to {\mathcal{A}}|_{X_i}^{I}$ is induced by ${\mathcal{A}}(X_i)^{J}\to {\mathcal{A}}(X_i)^{I}$.
