Friday, 11 July 2014

ag.algebraic geometry - Do coequalizers in RingSpc automatically lead to descent?

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. Also, a diagram involving an iterated fibered product is required to commute as well (the cocycle condition).



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 (or very closely equivalent to) 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 schemes,
$$A rightrightarrows^{f,g} B to C,$$
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?



My guess is the answer is no. First, I've heard that the coequalizer condition above is actually very weak. So, suppose instead we have that $A rightrightarrows^{f,g} B to C,$ is a coequalizer and any base-change of it is a coequalizer. Does that imply that there is a descent diagram for quasi-coherent sheaves?



My guess is that the answer to the modified question is still no, for the meta-reason that Vistoli in FGA Explained spends much more time on proving descent for quasi-coherent sheaves than proving that the fpqc topology is subcanonical. On the other hand, I'd like to see a counterexample.



(N.B. I initially asked this question on Math.SE. I was advised to re-ask it here.)

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