Let $X$ be a scheme with structure sheaf $\mathcal O_X$ and let $\mathcal B$ be a quasi-coherent $\mathcal O_X$-algebra. Let $\mathcal F$ be any sheaf of $\mathcal B$-modules.

In EGA I, § 9, Prop. 9.6.1 it is asserted that then the following holds:

Then $\mathcal F$ is quasi-coherent over $\mathcal B$ if and only if $\mathcal F$ is quasi-coherent over $\mathcal O_ X$.

The proof of the implication $\Rightarrow$ is easy, but the other direction is hard to digest for me: EGA argues as follows: we may work locally on $X$, hence assume $X=Spec(A)$, whence $\mathcal B$ corresponds to a $A$-algebra $B$ and $\mathcal F$ corresponds to a $B$-module $M$ (namely, because $\mathcal B$ and $\mathcal F$ are quasi-coherent over $\mathcal O_X$).

Then it reads without explanation that $M$ is isomorphic to the cokernel of a map

$(*) \quad B^{(I)} \rightarrow B^{(J)}$,

and hence by taking the tilde-construction we see that $\mathcal F$ is quasi-coherent over $\mathcal B$.

Apriori, one should only be able to write $M$ as a cokernel of a map $(*)$ with $B$ replaced by $A$, but tensoring with $B$ doesn't seem to give $M$ as cokernel, but $M\otimes_A B$.

My question is why one may write $M$ as a cokernel of a map $(*)$, as claimed in EGA.

Or any other proof of the desired implication

"$\mathcal F$ quasi-coherent over $\mathcal O_X \Rightarrow \mathcal F$ quasi-coherent over $\mathcal B$"

would be very appreciated.

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    $\begingroup$ $M$ is already a $B$-module, so you can definitely write $M$ as the cokernel of $(*)$. But then again quasi-coherence is somewhat tautological for modules over rings. $\endgroup$ – Zhen Lin Jan 7 '13 at 16:07

Basically Zhen already answered the question, but let me add this as an answer so that it won't bump up again (and to see if this really answered the question):

By assumption $M$ is a $B$-module, hence any presentation with generators and relations gives an exact sequence $B^{(I)} \to B^{(J)} \to M \to 0$ of $B$-modules. The rest of the EGA proof is quite self-explanatory; if not you can ask further questions.

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