Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.)

share|improve this question
3  
Concept of "quasi-coh." and "base change" are not good notions (in a useful way) for abstract loc. ringed spaces. Can make def'n of quasi-coh. in cases with coh. str. sheaf (related to local injlim of coherent sheaves), such as for complex-analytic spaces (see "Coherent Analytic Sheaves"). But surprises happen, such as lost under direct limits; see Example 2.1.10 in my paper "Relative ampleness in rigid geometry", which applies to the unit disc in the complex plane. Likewise, base change for C-analytic spaces is specific to that category. Rewrite question to stick to schemes. –  BCnrd Sep 7 '10 at 13:01
2  
Dear Akhil: Did you try to find any example of a coequalizer diagram of schemes whose formation as such is compatible with arbitrary base change on $C$ and yet which is not arising from an fpqc cover (recall fpqc topology mixes in Zar. topology too, by the way)? The difficulty of making examples where that happens would be a good reason that nobody tries to prove theorems with that hypothesis away from the fpqc case (where more tools are available). There is a kind of "non-flat descent" (I think due to Oort) worked out in appendix to one of Kleiman's exposes in SGA6; very hard to use. –  BCnrd Sep 7 '10 at 13:18
    
Thanks! I hadn't tried to find such an example (I don't really have any intuition for coequalizers and how to construct them outside from fpqc maps, only equalizers). I need to think a bit to understand your other comment, will respond shortly. –  Akhil Mathew Sep 7 '10 at 13:54
    
I've now read it carefully: Great--I'd accept it if you posted it as an answer. –  Akhil Mathew Sep 7 '10 at 18:17
add comment

1 Answer

up vote 4 down vote accepted

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.]

share|improve this answer
    
Thanks! . –  Akhil Mathew Sep 7 '10 at 20:39
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.