Suppose $A \rightarrow B$ is a faithfully flat map of rings. Then the Amitsur complex is exact:

$0 \rightarrow A \rightarrow B \rightarrow B \otimes_A B \rightarrow \dots$

(the second map is $id \otimes 1 - 1 \otimes id$, and the subsequent maps are alternating sums of the different ways of putting in a 1.)

That this is exact makes sense! Its saying that element of $b\in B$ such that $1\otimes b=b\otimes 1$ comes from $A$. The way to prove it is exact is roughly the following (the details can be found in Milne's online notes on étale cohomology.)

First, one can check that the sequence is exact if the first map has a section. Second, it is exact iff it is exact after a faithfully flat base change. Finally, if we base change by $B$, we have a section to the first map: the map we want to find a section to is $B \rightarrow B\otimes_A B, b \mapsto b\otimes 1$, and this has a section given by multiplication.

Why do we care? This fact is important in showing schemes are sheaves on the étale site. (Again, this can be found in Milne's notes.)

Okay, so my question is: what is really going on behind the scenes? I know a section is supposed to somehow "solve an equation", but I have no idea what is really being solved here. The proof feels like magic.

(One note: Its probably instructive to look at the special case of $A \rightarrow \prod A_f$ , a cover by basic open affines. I believe this proof is showing that the sequence is exact simultaneously on each member of the cover, and because exactness is a local property we have that the sequence is exact.)


2 Answers 2


I think it feels like magic because there's something tautological going on. The story should really culminate with the definition of "faithfully flat" rather than begin with it.

As you suggested, let's consider the case where you cover an affine scheme Spec(A) by finitely many basic open affines Spec(Af). The cover Spec(B)=Spec(∏Af) has the following special property:

(∗) If you want to check exactness of any sequence of A-modules, it's enough to check exactness after restricting to Spec(B) (i.e. tensoring with B over A).

Now if you'd like to show that the Amitsur complex is exact, you know it's enough to find a section B→A. By (∗), we know what it's actually enough to find a section locally, and in the case Spec(B)=∐Spec(Af)→Spec(A), it's obvious that there's a section locally. This is all very tautological, but it allows you to prove that schemes are sheaves in the Zariski topology, which maybe doesn't impress you so much.

But we have extra assumptions in this argument. We didn't really need B to be of the form ∏Af, we just needed it to satisfy (∗). So let's make a new definition, saying that B is "faithfully flat" over A if (∗) holds. Now we get the result: "If B is faithfully flat over A, then the Amitsur complex is exact." As a consequence, we can show that schemes are sheaves in the faithfully flat topology on affine schemes, but all we've really done is shown that "schemes are sheaves in the strongest topology for which this proof works," and just defined "faithfully flat" to be that topology. Delightfully, we can prove (with commutative algebra) that lots of different properties of affine schemes and morphisms of affine schemes are local in this topology.

Okay, but we'd really like to prove things about schemes and morphisms of schemes, not about affine schemes and morphisms of affine schemes. I think part of your confusion arises from the desire to understand the etale topology (a good desire), rather than continuing the strategy of making definitions which make results trivial (or at least straight-forward). If you indulge your generality tooth, you'll ask the question,

What is the strongest topology on the category of schemes so that every cover can be understood as a combination of (a) Zariski covers and (b) faithfully flat covers of affine schemes by affine schemes?

The answer is the fpqc topology. Basically by construction, if a property of schemes (resp. morphisms of schemes) is local in the Zariski topology and is local in the faithfully flat topology when you restrict to the category of affine schemes, then it is local in the fpqc topology. Similarly, if a functor is a sheaf in the Zariski topology, and its restriction to the category of affine schemes is a sheaf in the faithfully flat topology, then it's a sheaf in the fpqc topology. In particular, we get that

This sounds very impressive given that nobody understands what general fpqc morphisms look like (I think), but that's just because we defined the fpqc topology to be whatever it has to be to make the straight-forward proofs work. One thing we do know is that fppf morphisms are fpqc (by EGA IV, Corollary 1.10.4). In particular, etale morphisms are fpqc.

  • $\begingroup$ Except for the fact that étale morphisms are not required to be quasi-compact. Also, I think that there are several conventions in the literature about the definition of fppf; sometimes fppf morphismes are not required to be quasi-compact, and in this case the most cautious authors write fplpf for faithfully flat locally of finite presentation. $\endgroup$ May 29, 2010 at 12:47
  • $\begingroup$ @Matthieu: fpqc morphisms are not required to be quasi-compact either, despite the name. Just as fppf morphisms are faithfully flat and locally of finite presentation, fpqc morphisms are faithfully flat and "locally quasi-compact" (meaning that for every point $x$ in the source and any affine neighborhood $V$ of the image of $x$, there is a quasi-compact neighborhood $U$ of $x$ which is is faithfully flat over $V$). $\endgroup$ May 29, 2010 at 21:47

The more general name this sort of argument goes by is descent theory, and chapter six of Bosch, Lutkebohmert, and Raynaud's "Neron Models" is a good introduction. The general idea is that you have some object (say, a quasi-coherent sheaf) defined on the "open sets" in a cover (it could be a Zariski cover, or an etale cover, or an fpqc cover, or...) of $X$, and you want to know whether you can glue it in some unique way to get something on $X$ itself.

That was vague, so let's consider the case when we have a quasi-coherent sheaf of modules $M_f$ on $\text{Spec}A_f$, where $A\rightarrow \prod A_f$ is a cover by open affines. Let's set $B=\oplus A_f$ and $M'=\oplus M_f$. We want a quasi-coherent sheaf $M$ on $\text{Spec} A$. What's the most obvious requirement on the $M_f$? Well, they have to agree on overlaps. That is, if $U=\text{Spec}B=\coprod\text{Spec}A_f$, and $p_1,p_2$ are the projections $U\times_A U\rightarrow U$, we need an isomorphism $p_1^* M'\rightarrow p_2^* M'$, because $U\times_A U$ is the disjoint union of all the overlaps. Another way of saying this is that if $M'=M\otimes_A B$, then $M\rightarrow M\otimes_A B\rightrightarrows M\otimes_A B\otimes_A B$ is exact. Note that this is not a sufficient condition for $M$ to exist! Line bundles are locally trivial, so of course there are isomorphisms on the overlaps, but not every choice of such isomorphisms is compatible with triple overlaps ($U\times_A U\times_A U$).

As for what the assumption of a section is doing there, it's saying that locally, your map of schemes $\text{Spec}B\rightarrow \text{Spec}A$ has a section. Not Zariski-locally, fpqc locally, and that's important. One reason the Zariski topology isn't so great to work with is that a lot of morphisms that "look like" they should be fiber bundles (that is, locally trivial in some sense) really aren't. But they tend to be in finer topologies, like the etale topology and the fpcq topology. The "base change to get a section" trick you mention above is saying that fpqc-locally, $\text{Spec}B\rightarrow \text{Spec}A$ behaves like a fiber bundle, in that it has lots of sections to work with. If the map happens to be etale, that's just saying that locally on the base, the total space is a bunch of disjoint copies of the base, which is definitely a topological property we want.

In the specific application you mention, showing that schemes are sheaves in the etale topology, what you're actually going to do is show that they're sheaves in the fpqc topology, and deduce that they're sheaves in any coarser topology, like the etale topology.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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