We continue in the quest towards descent theory. Today, we discuss the fpqc topology and prove the fundamental fact that representable functors are sheaves.

We now describe another topology on the category of schemes. First, we need the notion of an fpqc morphism.

Definition 1 A morphism of schemes ${f: X \rightarrow Y}$ is called fpqc if the following conditions are satisfied:

1. ${f}$ is faithfully flat (i.e., flat and surjective)
2. ${f}$ is quasi-compact.

Indeed, “fpqc” is an abbreviation for “fidelement plat et quasi-compact.” It is possible to carry out faithfully flat descent with a weaker notion of fpqc morphism, for which I refer you to Vistoli’s part of FGA explained.

As with many interesting classes of morphisms of schemes, we have a standard list of properties.

Proposition 2

1. Fpqc morphisms are closed under base-change and composition.
2. If ${f: X \rightarrow Y, g: X' \rightarrow Y'}$ are fpqc morphisms of ${S}$-schemes, then ${f \times_S f': X \times_S X' \rightarrow Y \times_S Y'}$ is fpqc.

Proof: We shall omit the proof, since the properties of flatness, quasi-compactness, and surjectivity are all (as is well-known) preserved under base-change, composition, and products. This can be looked up in EGA 1 (except for flatness, for which you need to go to EGA 4 or Hartshorne III). $\Box$

So we have the notion of fpqc morphism. Next, we use this to define a topology.

Definition 3 Consider the category ${\mathfrak{C}}$ of ${S}$-schemes, for ${S}$ a fixed base-scheme. The fpqc topology on ${\mathfrak{C}}$ is defined as follows: A collection of arrows ${\left\{U_i \rightarrow U\right\}}$ is said to be a cover of ${U}$ if the map ${\coprod U_i \rightarrow U}$ is an fpqc morphism.

This implies in particular that each ${U_i \rightarrow U}$ is a flat morphism. We need now to check that this is indeed a topology.

1. An isomorphism is obviously an fpqc morphism, so an isomorphism is indeed a cover.
2. If ${\left\{U_i \rightarrow U\right\}}$ is a fpqc cover and ${V \rightarrow U}$, then the morphism ${\coprod( U_i \times_U V )\rightarrow V }$ is equal to the base-change ${(\coprod U_i) \times_U V \rightarrow V}$, hence is fpqc.
3. Suppose ${\left\{U^i_j \rightarrow U_i\right\}}$ is a cover for each ${i}$ and ${\left\{U_i \rightarrow U\right\}}$ is a cover, I claim that ${\left\{U_j^i \rightarrow U\right\}}$ is a cover. Indeed, we have that$\displaystyle \coprod_{i,j} U^{j}_i \rightarrow U$factors through$\displaystyle \coprod_{i,j} U^{i}_j \rightarrow \coprod_i {U_i} \rightarrow U$and we know that each morphism in the composition is flat (since the coproduct of flat morphisms is flat) and quasi-compact (since the coproduct of quasi-compact morphisms is quasi-compact). Similarly for surjectivity. It follows that ${\left\{U^i_j \rightarrow U \right\}}$ is an fpqc cover.

So we have another topology on the category of schemes, which is very fine in that it is finer than many other topologies of interest (e.g. the fppf and etale topologies, which I will discuss at some other point).

Note that, strictly speaking, the topology thus defined is not finer than the usual Zariski topology! The reason is that if we have an open set ${U \subset X}$, the inclusion ${U \rightarrow X}$ need not be quasi-compact. It is, however, finer than the “finite Zariski topology,” when we only allow finite open covers, on the category of noetherian schemes. But this is a cop-out. For the more general definition of fpqc morphism that gives a better theory, I refer the reader to Vistoli’s article in FGA Explained (also openly available on the arXiv). For simplicity, however, I will restrict myself to a special case which is still rather interesting.

This topology is really awesome because, first of all:

Theorem 4 Any representable functor on the category ${\mathfrak{C}}$ of ${S}$-schemes is a sheaf in the fpqc topology.

In other words, if we have an fpqc cover ${\left\{U_i \rightarrow U\right\}}$ of a scheme ${U}$ and morphsims ${U_i \rightarrow X}$ that glue on the fibered products, then there is a unique morphism ${U \rightarrow X}$ that pulls back to each of these.

Write ${\overline{U} = \coprod U_i}$. There is a map ${\overline{U} \rightarrow X}$, by assumption, which is a fpqc morphism, and the two pull-backs to ${\overline{U} \times_U \overline{U} = \coprod_{i,j} U_i \times_U U_j}$ are equal. We need to show that ${\overline{U} \rightarrow X}$ factors through ${U \rightarrow X}$.

In particular, we are reduced to proving:

Lemma 5 Let ${\overline{U} \rightarrow U}$ be a fpqc morphism of ${S}$=schemes. Then the diagram$\displaystyle \overline{U}\times_U \overline{U} \rightrightarrows^{p_1, p_2} \overline{U} \rightarrow U,$

is a coequalizer in the category of ${S}$-schemes.

Recall what this means translated into informal English. To hom out of ${U}$ is the same thing as homming out of ${\overline{U}}$ such that the two pull-backs to ${\overline{U} \times_U \overline{U}}$ are the same.

It is in fact true (cf. Demazure-Gabriel or EGA IV-2 2.3.12) that in the lemma, the topology of ${U}$ is the quotient topology of ${U}$ under the equivalence relation generated by ${p_1(z) \sim p_2(z)}$ (furthermore, the diagram is a coequalizer in the category of locally ringed spaces, which implies this). This lets you deduce all sorts of dandy descent-ish results to the effect that if a base-change of a given morphism by a fpqc morphism has a given property (e.g. is closed, open, etc.), then so is the initial un-base-changed morphism.

Anyway, back to the lemma. To prove this, we will show that for any scheme ${X}$, the diagram

$\displaystyle \hom_S(U, X) \rightarrow \hom_S(\overline{U}, X) \rightrightarrows \hom_S(\overline{U} \times_U \overline{U},X) \ \ \ \ \ (1)$

is an equalizer diagram, which is the meaning of the lemma. First, note that ${\overline{U } \rightarrow U}$ is an epimorphism in the category of ${S}$-schemes. This implies that the first map in the above diagram of sets is an injection. This is because ${\overline{U} \rightarrow U}$ is surjective on the underlying topological spaces and the maps on the local rings are injective, by:

Lemma 6 Let ${A \rightarrow B}$ be a flat local homorphism of local rings. Then it is injective.

Proof: One approach is to argue that ${A \rightarrow B}$ is faithfully flat because it is flat, and for each (unique!) maximal ideal ${\mathfrak{m} \subset A}$, we have ${\mathfrak{m}B \subsetneq B}$ because ${ A \rightarrow B}$ is local. Any faithfully flat extension of rings is injective, because if ${A \rightarrow B}$ is faithfully flat, we can tensor with the faithfully flat module ${B}$ to get a morphism of ${A}$-algebras,

$\displaystyle B \rightarrow B \otimes_A B,$

which is injective (it has a section ${b\otimes b' \rightarrow bb'}$!), and consequently (by faithful flatness), the original map ${A \rightarrow B}$ is itself injective.

This completes the proof. $\Box$

Remark: In an arbitrary category with fibered products, an epimorphism ${X \rightarrow Y}$ does not generally induce a coequalizer diagram:

$\displaystyle X \times_Y X \rightarrow X \rightarrow Y;$

if it does, it is called an effective epimorphism. The conclusion of this result is that an fpqc morphism is an effective epimorphism. (Note that any base-change of it is an effective epimorphism, since a base-change of a fpqc morphism is still fpqc.)

Reduction to the case of ${X}$ affine: But we still need to show that if a morphism ${\overline{U} \rightarrow X}$ glues (i.e. pulls back to the same thing in ${\overline{U} \times_U \overline{U}}$), then it comes from something in ${\hom_S(U,X)}$. To see this, we can assume first of all that ${X}$ is affine. The reason is that we can cover ${X}$ by open affines ${\left\{W_i\right\}}$. We can replicate the situation of the lemma with ${\overline{U} \times_{X} W_i \rightarrow W_i}$ and ${\overline{U} \times_X W_i \rightarrow W_i}$. If we can prove that the lemma is true for open affines, then given a map ${\overline{U} \rightarrow X}$, we can show that there are maps ${W_i \rightarrow X}$ that come from ${\overline{U} \times_X W_i \rightarrow W_i}$. These necessarily restrict to the same thing on ${U \times_X (W_i \cap W_j)}$, though, since they restrict to the same thing on ${\overline{U} \times_X (W_i \cap W_j)}$, and

$\displaystyle \overline{U} \times_X (W_i \cap W_j) \rightarrow U \times_X (W_i \cap W_j)$

is an epimorphism in the category of ${S}$-schemes by the above. So we can glue them (in the Zariski topology!) to get ${U \rightarrow X}$.

So in showing that (1) is a coequalizer, we need only restrict to the case of ${X}$ affine. And now we are going to reduce to the case of ${U}$ affine.

Reduction to the case of ${U}$ affine: If ${W \subset U}$ is open and affine, then the map ${\overline{U} \times_U W \rightarrow W}$ satisfies the same condition of the lemma: it is a fpqc morphism. Suppose we know that (1) is exact when ${U}$ is an affine scheme (such as ${W}$). Throughout, ${X}$ is fixed (say affine by the above paragraph).

If we have the data ${\overline{U} \rightarrow X}$ pulling back equally by ${p_1, p_2}$, then we get data ${\overline{U} \times_U W \rightarrow X}$ for each ${W}$ open affine pulling back equally by the canonical projections corresponding to ${\overline{U} \times_U W \rightarrow W}$. It thus follows that if (1) is exact for ${U}$ affine and we have the data ${\overline{U} \rightarrow X}$, then we can (by descent for the case of one affine) define a bunch of maps ${W \rightarrow X}$ for each ${W \subset U}$ open affine that pull back to ${\overline{U} \times_U W \rightarrow X}$. These must glue on ${W \cap W' \subset U}$ because they pull back to the same thing in ${\overline{U}\times_U (W \cap W')}$.

Thus, we are reduced to the case ${U}$ affine (and ${X}$ affine).

Reduction to the case ${\overline{U}}$ affine: We are now reduced to proving that (1) is exact for ${U,X}$ affine and ${\overline{U} \rightarrow U}$ a fpqc morphism. Next, we will reduce even further, to the case of ${\overline{U}}$ affine.

Now ${U}$ is affine, so ${\overline{U}}$ is quasi-compact. There is a cover of ${\overline{U}}$ by a finite number of open affines ${V_i \subset \overline{U}}$. The coproduct ${V = \coprod V_i = \mathrm{Spec} \prod \Gamma(V_i, \mathcal{O}_{V_i})}$ is affine and surjects onto ${\overline{U}}$ by, in fact, a fpqc morphism.

There is a diagram:

The vertical maps are injections because ${V \rightarrow \overline{U}}$ (and all its base-changes, products) are fpqc, hence epimorphisms by what has already been proved. But the first leftmost vertical map is an isomorphism. Now if the bottom map is an equalizer (which would be the case if we had proved the result for affine schemes), an easy diagram chase then shows that the top one is an equalizer too. The details of the diagram chase are best checked for oneself.

The final case: Whew! So, finally, our lemma, and thus the theorem on representable functors being sheaves, will be proved if we show that:

Lemma 7 Suppose ${\overline{U}, U, X}$ are affine schemes and ${\overline{U} \rightarrow U}$ a fpqc morphism. Then (1) is exact.

Proof: Suppose ${\overline{U} = \mathrm{Spec} B, U = \mathrm{Spec} A, X = \mathrm{Spec} R}$. Then there is a homomorphism ${ A\rightarrow B}$ which makes ${B}$ into a faithfully flat ${A}$-algebra. For now, assume that ${S = \mathbb{Z}}$ so we don’t have to worry about “${S}$-morphisms,” only regular morphisms.

We will show that the sequence of sets

$\displaystyle A \rightarrow B \rightrightarrows B \otimes_A B$

is exact (i.e. an equalizer of sets) for ${B}$ a faithfully flat ${A}$-algebra. It follows that if we have a homomorphism ${R \rightarrow B}$ whose two images in ${B \otimes_A B}$ are the same, the image actually lies in ${A}$.

Now we already know that ${A \rightarrow B}$ is injective. We need to show that the map ${f: b \rightarrow (1 \otimes b - b \otimes 1)}$ has kernel equal to the image of ${A}$. In particular, the associated sequence of ${A}$-modules ${A \rightarrow B \rightarrow B \otimes_A B}$ needs to be shown to be exact.

Suppose there is a section of ${s: A \rightarrow B}$, i.e. a ring homomorphism ${B \rightarrow A}$ such that ${A \rightarrow B \rightarrow A}$ is the identity (equivalently, a morphism of ${A}$-algebras ${B \rightarrow A}$). Then if ${x \in \ker f \subset B}$, we have ${1 \otimes x = x \otimes 1}$. There is a homomorphism ${s \otimes 1: B \otimes_A B \rightarrow A \otimes_A B = B}$. We have

$\displaystyle x = (s \otimes 1)(1 \otimes x) = (s \otimes 1)(x \otimes 1) = sx,$

so that ${x \in A}$.

In general, we don’t have a section. But we do have one if we make a faithfully flat base-change. If ${A \rightarrow B}$ is faithfully flat, then ${B \rightarrow B_B = B \otimes_A B}$ is faithfully flat too. If we have the sequence ${A \rightarrow B \rightarrow B \otimes_A B}$, we can base-change by ${\otimes_A B}$ to get the sequence

$\displaystyle B \rightarrow B_B \rightarrow B_{B}\otimes_B B_B.$

But this is exact because ${B_B \rightarrow B}$ admits an obvious section, namely multiplication. So the original sequence is itself exact because ${B}$ is faithfully flat.

The proof of the lemma, and thus of the full theorem on representable functors, is thus complete for ${S = \mathbb{Z}}$. In particular, the fpqc topology on the category of plain schemes is such that every representable functor is a sheaf.

The general case can be deduced as follows. Suppose ${\overline{U} \rightarrow U}$ is a map of ${S}$-schemes and ${\overline{U} \rightarrow X}$ is an ${S}$-morphism that pulls back appropriately. Then the above reasoning says that we can get a morphism ${U \rightarrow X}$ that pulls back to this, but it needn’t be an ${S}$-morphism a priori.

So we need to show that ${U \rightarrow X \rightarrow S}$ equals ${U \rightarrow S}$. To see this, pull-back both to ${\overline{U}}$. On the one hand, we get ${\overline{U} \rightarrow U \rightarrow X \rightarrow S}$, which is equal to the structure ${\overline{U} \rightarrow S}$, because ${\overline{U} \rightarrow X}$ is an ${S}$-morphism. On the other hand, we get ${\overline{U} \rightarrow S}$ because ${\overline{U} \rightarrow U}$ is an ${S}$-morphism. Since ${\overline{U} \rightarrow U}$ is an epimorphism of schemes, it follows that ${U \rightarrow X \rightarrow S}$ equals ${U \rightarrow S}$. In particular, ${\hom_S(-, X)}$ is also a sheaf. This completes the proof of the theorem. $\Box$

This method of proof can in fact be used to obtain the following general fact:

Proposition 8 Suppose ${F}$ is a contravariant functor from the category of ${S}$-schemes to ${\mathbf{Sets}}$. Suppose that

1. ${F}$ is a sheaf in the Zariski topology
2. If ${U \rightarrow V}$ is a faithfully flat morphism of affine schemes, then $\displaystyle F(V) \rightarrow F(U) \rightrightarrows F(U \times_V U)$ is exact.

Then ${F}$ is a sheaf in the fpqc topology.

For a proof, see FGA Explained. I believe it is also in FGA.