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I'm currently interested in the following result:

Let $f: X \to Y$ be a fpqc morphism of schemes. Then there is an equivalence of categories between quasi-coherent sheaves on $Y$ and "descent data" on $X$. Namely, the second category consists of quasi-coherent sheaves $\mathcal{F}$ on $X$ with an isomorphism $p_{1}^*(\mathcal{F}) \simeq p_2^*(\mathcal{F})$, where $p_1, p_2: X \times_Y X \to X$ are the two projections. Edit: There is a further condition; a diagram involving an iterated fibered product is required to commute as well.

In Demazure-Gabriel's Introduction to Algebraic Geometry and Algebraic Groups, it is proved (under the name ffqc (sic) descent theorem) that the sequence $$ X \times_Y X \to^{p_1, p_2} X \to Y$$ is a coequalizer in the category of locally ringed spaces under the above hypotheses. If I am not mistaken, this is the same as the theorem that says that representable functors are sheaves in the fpqc topology. On the other hand, D-G give a fairly explicit description of the quotient space.

Question: For a coequalizer diagram of (locally) ringed spaces, $$A \to^{f,g} B \to C,$$ is there a descent diagram for quasi-coherent sheaves on $A,B,C$? In particular, does the D-G form of the descent theorem directly, by itself, imply the more general one for quasi-coherent sheaves?

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  • $\begingroup$ After asking this question, I've heard that that the coequalizer condition (which I believe is also referred to in the literature as "$X \to Y$ is an effective epimorphism), is rather weak and does not suffice for many purposes, e.g. because it is not preserved under non-flat base-change (well, though this particular instance is, because fpqc is preserved under base-change). $\endgroup$ – Akhil Mathew Sep 5 '10 at 15:09
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    $\begingroup$ Perhaps mathoverflow is the right place for your question. By the way, I'm impressed (actually, I'm totally amazed) by your mathematical outputs on the internet since you're only 18 ... $\endgroup$ – Martin Brandenburg Sep 6 '10 at 11:21
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    $\begingroup$ Ok, it's posted (mathoverflow.net/questions/37970/…) on MO now; thank you for the kind words. $\endgroup$ – Akhil Mathew Sep 7 '10 at 12:05

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