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> Let $f : X \to Y$ be an [fpqc morphism](http://stacks.math.columbia.edu/tag/022B) of schemes, and let $\mathcal{G}$ be an $\mathcal{O}\_{Y}$-module (on the [small Zariski](http://stacks.math.columbia.edu/tag/020T) site) such that $f^{\ast}\mathcal{G}$ is quasi-coherent. Is $\mathcal{G}$ necessarily quasi-coherent?
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I'd be happy to see any answers to the above question with "fpqc" replaced by "fppf" or "etale" as well.



Is "quasi-coherent" an fpqc-local property of modules?
Yes, this is an application of descent theory (with a bit of care). Let $F = f^{\ast}(G)$ on $X$, a quasi-coherent $O\_X$-module by hypothesis. Then for the maps $p\_1, p\_2: X \times\_Y X \rightrightarrows X$ we have an evident composite isomorphism $$\theta: p\_1^{\ast}(F) \simeq (f \circ p\_1)^{\ast}(G) = (f \circ p\_2)^{\ast}(G) \simeq p\_2^{\ast}(F)$$ that satisfies the usual cocycle condition; i.e., $\theta$ is a descent datum. Hence, by fpqc descent for quasi-coherent sheaves, we obtain a quasi-coherent $O\_Y$-module $G'$ and an $O\_X$-linear isomorphism $\alpha: f^{\ast}(G') \simeq F := f^{\ast}(G)$ respecting the descent data on both sides.


Let $f':X \times\_Y X \rightarrow Y$ be the natural map, and $F' = {f'}^{\ast}(G)$, so $F'$ is naturally identified with each of $p\_1^{\ast}(F)$ and $p\_2^{\ast}(F)$ compatibly with $\theta$ (via how $\theta$ is defined). The equality of $f'$ with $f \circ p\_1$ and $f \circ p\_2$ thereby defines two $O\_Y$-linear maps
$$f\_{\ast}(F) \rightrightarrows f'\_{\ast}(F')$$
whose equality is the quasi-coherent $G'$ by design. But $G$ is visibly an $O\_Y$-submodule of $f\_{\ast}(F)$ (since $f$ is surjective and faithfully flat between local rings) and as such is contained inside the equalizer $G'$, so we have $G \subset G'$. The problem of checking equality thereby reduces to comparing stalks at each $y \in Y$. 


If $x \in f^{-1}(y)$ is a point then $O\_y \rightarrow O\_x$ is faithfully flat, so by fpqc descent for modules we see that
$$G\_y = \ker(F\_x \rightrightarrows F\_x \otimes\_{O\_y} O\_x).$$
But the right side coincides with $G'\_y$ since $G'$ is a *quasi-coherent* descent of $F$, so $G\_y = G'\_y$ via the natural map; i.e., the inclusion $G \subset G'$ is an equality on $y$-stalks.


