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Let $f : X \to Y$ be an fpqc morphism of schemes, and let $\mathcal{G}$ be an $\mathcal{O}_{Y}$-module (on the small Zariski site) such that $f^{\ast}\mathcal{G}$ is quasi-coherent. Is $\mathcal{G}$ necessarily quasi-coherent?

I'd be happy to see any answers to the above question with "fpqc" replaced by "fppf" or "etale" as well.

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  • $\begingroup$ G is an $O_Y$-module in what topology? $\endgroup$
    – Yosemite Sam
    Commented Nov 29, 2016 at 3:54
  • $\begingroup$ Sorry, I mean the small Zariski site of $Y$. $\endgroup$
    – Minseon Shin
    Commented Nov 29, 2016 at 4:15

1 Answer 1

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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.

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  • $\begingroup$ Thanks for your answer. I am confused by some steps and am hoping you would clarify. (1) Why is the canonical map $G \to f_{\ast}(F)$ injective? If $G$ were quasi-coherent, then this is the statement that $N \to N \otimes_{B} A$ is injective for any $B$-module $N$. (2) Why is $G_y = \ker(F_x \rightrightarrows F_x \otimes_{O_y} O_x)$? If $G$ were quasi-coherent, then the map $\widetilde{G_{y}} \to i_{y}^{\ast}G$ is an isomorphism (here $i_{y} : {\rm Spec}(O_{y}) \to Y$ is the natural inclusion) and we can use fpqc descent for modules. $\endgroup$
    – Minseon Shin
    Commented Nov 29, 2016 at 6:56
  • $\begingroup$ I just edited away some unnecessary passage to the affine case which wasn't used in the end. Anyway, injectivity of $G\rightarrow f_*(F)$ is seen on stalks by composing with $f_*(F)_y \rightarrow F_x$ for $x\in f^{-1}(y)$ and noting that $F_x = G_y \otimes_{O_y} O_x$ by the definition of $F$ as $f^*(G)$ (and clearly the composite map $G_y\rightarrow F_x$ is the natural map that is injective since $O_x$ is faithfully flat over $O_y$). This also underlies the equalizer description of $G_y$ that you ask about. $\endgroup$
    – nfdc23
    Commented Nov 29, 2016 at 6:58

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