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.


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.


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?

  • $\begingroup$ 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). $\endgroup$ Sep 5, 2010 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 ... $\endgroup$ Sep 6, 2010 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. $\endgroup$ Sep 7, 2010 at 12:05

1 Answer 1


This question was subsequently crossposted to MO and answered there by BCnrd. Here is the accepted answer in a community-wiki post:

Initial question has a negative answer even for affine schemes. Let $B$ = Spec($R$) equipped with an action by a finite group $G$, and define $R' = \prod_{g \in G} R$ and $A$ = Spec($R'$). Let $A \rightrightarrows B$ be the natural maps. Then $C$ := Spec($R^G$) is easily checked to be the coequalizer in the category of schemes (even in the category of locally ringed spaces), yet the descent diagram (if true) would say that for any $G$-equivariant $R$-module $M$ the natural map $R \otimes_{R^G} M^G \rightarrow M$ is an isomorphism (since $M^G$ corresponds to the equalizer sheaf for $M$ under $A \rightrightarrows B$). Or even if you don't want to be so abstract, merely positing that $M = R \otimes_{R^G} N$ for some $R^G$-module $N$ forces $N = M^G$ when $R$ is faithfully flat over $R^G$ and $N$ is a finite free $R^G$-module.

For example, let $R$ be the ring of integers of a quadratic field $K$, $G$ the order-2 Galois group, so $R^G = \mathbf{Z}$ is a PID. Take any finite torsion-free (= locally free) $R$-module $M$. Then automatically if $M$ were to have the form $R \otimes_{R^G} N$ we see that $N$ must be torsion-free and hence free over $\mathbf{Z} = R^G$. In particular, if $M$ is an invertible $R$-module then $N$ must have rank 1 and hence it would be forced that $M$ is free of rank 1 over $R$. So to make a counterexample, we just need a nontrivial line bundle $M$ over $R$ to admit a $G$-action over the one on $R$. Take $M$ to be a non-principal ramified prime ideal (which always exists when the discriminant is not prime, or 4 times a prime, or whatever the condition to deal with 2...you know what I mean).

[It is no coincidence that such an example is related to the non-etale locus for the finite flat map $\mathbf{Z} \rightarrow O_K$, since away from that the map is exactly the "Galois" instance of finite etale descent, for which everything does work exactly as in the familiar case of Galois descent for vector spaces over fields.]


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