## 34.9 The fpqc topology

Definition 34.9.1. Let $T$ be a scheme. An fpqc covering of $T$ is a family of morphisms $\{ f_ i : T_ i \to T\} _{i \in I}$ of schemes such that each $f_ i$ is flat and such that for every affine open $U \subset T$ there exists $n \geq 0$, a map $a : \{ 1, \ldots , n\} \to I$ and affine opens $V_ j \subset T_{a(j)}$, $j = 1, \ldots , n$ with $\bigcup _{j = 1}^ n f_{a(j)}(V_ j) = U$.

To be sure this condition implies that $T = \bigcup f_ i(T_ i)$. It is slightly harder to recognize an fpqc covering, hence we provide some lemmas to do so.

Lemma 34.9.2. Let $T$ be a scheme. Let $\{ f_ i : T_ i \to T\} _{i \in I}$ be a family of morphisms of schemes with target $T$. The following are equivalent

1. $\{ f_ i : T_ i \to T\} _{i \in I}$ is an fpqc covering,

2. each $f_ i$ is flat and for every affine open $U \subset T$ there exist quasi-compact opens $U_ i \subset T_ i$ which are almost all empty, such that $U = \bigcup f_ i(U_ i)$,

3. each $f_ i$ is flat and there exists an affine open covering $T = \bigcup _{\alpha \in A} U_\alpha$ and for each $\alpha \in A$ there exist $i_{\alpha , 1}, \ldots , i_{\alpha , n(\alpha )} \in I$ and quasi-compact opens $U_{\alpha , j} \subset T_{i_{\alpha , j}}$ such that $U_\alpha = \bigcup _{j = 1, \ldots , n(\alpha )} f_{i_{\alpha , j}}(U_{\alpha , j})$.

If $T$ is quasi-separated, these are also equivalent to

1. each $f_ i$ is flat, and for every $t \in T$ there exist $i_1, \ldots , i_ n \in I$ and quasi-compact opens $U_ j \subset T_{i_ j}$ such that $\bigcup _{j = 1, \ldots , n} f_{i_ j}(U_ j)$ is a (not necessarily open) neighbourhood of $t$ in $T$.

Proof. We omit the proof of the equivalence of (1), (2), and (3). From now on assume $T$ is quasi-separated. We prove (4) implies (2). Let $U \subset T$ be an affine open. To prove (2) it suffices to show that for every $t \in U$ there exist finitely many quasi-compact opens $U_ j \subset T_{i_ j}$ such that $f_{i_ j}(U_ j) \subset U$ and such that $\bigcup f_{i_ j}(U_ j)$ is a neighbourhood of $t$ in $U$. By assumption there do exist finitely many quasi-compact opens $U'_ j \subset T_{i_ j}$ such that such that $\bigcup f_{i_ j}(U'_ j)$ is a neighbourhood of $t$ in $T$. Since $T$ is quasi-separated we see that $U_ j = U'_ j \cap f_ j^{-1}(U)$ is quasi-compact open as desired. Since it is clear that (2) implies (4) the proof is finished. $\square$

Lemma 34.9.3. Let $T$ be a scheme. Let $\{ f_ i : T_ i \to T\} _{i \in I}$ be a family of morphisms of schemes with target $T$. The following are equivalent

1. $\{ f_ i : T_ i \to T\} _{i \in I}$ is an fpqc covering, and

2. setting $T' = \coprod _{i \in I} T_ i$, and $f = \coprod _{i \in I} f_ i$ the family $\{ f : T' \to T\}$ is an fpqc covering.

Proof. Suppose that $U \subset T$ is an affine open. If (1) holds, then we find $i_1, \ldots , i_ n \in I$ and affine opens $U_ j \subset T_{i_ j}$ such that $U = \bigcup _{j = 1, \ldots , n} f_{i_ j}(U_ j)$. Then $U_1 \amalg \ldots \amalg U_ n \subset T'$ is a quasi-compact open surjecting onto $U$. Thus $\{ f : T' \to T\}$ is an fpqc covering by Lemma 34.9.2. Conversely, if (2) holds then there exists a quasi-compact open $U' \subset T'$ with $U = f(U')$. Then $U_ j = U' \cap T_ j$ is quasi-compact open in $T_ j$ and empty for almost all $j$. By Lemma 34.9.2 we see that (1) holds. $\square$

Lemma 34.9.4. Let $T$ be a scheme. Let $\{ f_ i : T_ i \to T\} _{i \in I}$ be a family of morphisms of schemes with target $T$. Assume that

1. each $f_ i$ is flat, and

2. the family $\{ f_ i : T_ i \to T\} _{i \in I}$ can be refined by an fpqc covering of $T$.

Then $\{ f_ i : T_ i \to T\} _{i \in I}$ is an fpqc covering of $T$.

Proof. Let $\{ g_ j : X_ j \to T\} _{j \in J}$ be an fpqc covering refining $\{ f_ i : T_ i \to T\}$. Suppose that $U \subset T$ is affine open. Choose $j_1, \ldots , j_ m \in J$ and $V_ k \subset X_{j_ k}$ affine open such that $U = \bigcup g_{j_ k}(V_ k)$. For each $j$ pick $i_ j \in I$ and a morphism $h_ j : X_ j \to T_{i_ j}$ such that $g_ j = f_{i_ j} \circ h_ j$. Since $h_{j_ k}(V_ k)$ is quasi-compact we can find a quasi-compact open $h_{j_ k}(V_ k) \subset U_ k \subset f_{i_{j_ k}}^{-1}(U)$. Then $U = \bigcup f_{i_{j_ k}}(U_ k)$. We conclude that $\{ f_ i : T_ i \to T\} _{i \in I}$ is an fpqc covering by Lemma 34.9.2. $\square$

Lemma 34.9.5. Let $T$ be a scheme. Let $\{ f_ i : T_ i \to T\} _{i \in I}$ be a family of morphisms of schemes with target $T$. Assume that

1. each $f_ i$ is flat, and

2. there exists an fpqc covering $\{ g_ j : S_ j \to T\} _{j \in J}$ such that each $\{ S_ j \times _ T T_ i \to S_ j\} _{i \in I}$ is an fpqc covering.

Then $\{ f_ i : T_ i \to T\} _{i \in I}$ is an fpqc covering of $T$.

Proof. We will use Lemma 34.9.2 without further mention. Let $U \subset T$ be an affine open. By (2) we can find quasi-compact opens $V_ j \subset S_ j$ for $j \in J$, almost all empty, such that $U = \bigcup g_ j(V_ j)$. Then for each $j$ we can choose quasi-compact opens $W_{ij} \subset S_ j \times _ T T_ i$ for $i \in I$, almost all empty, with $V_ j = \bigcup _ i \text{pr}_1(W_{ij})$. Thus $\{ S_ j \times _ T T_ i \to T\}$ is an fpqc covering. Since this covering refines $\{ f_ i : T_ i \to T\}$ we conclude by Lemma 34.9.4. $\square$

Lemma 34.9.6. Any fppf covering is an fpqc covering, and a fortiori, any syntomic, smooth, étale or Zariski covering is an fpqc covering.

Proof. We will show that an fppf covering is an fpqc covering, and then the rest follows from Lemma 34.7.2. Let $\{ f_ i : U_ i \to U\} _{i \in I}$ be an fppf covering. By definition this means that the $f_ i$ are flat which checks the first condition of Definition 34.9.1. To check the second, let $V \subset U$ be an affine open subset. Write $f_ i^{-1}(V) = \bigcup _{j \in J_ i} V_{ij}$ for some affine opens $V_{ij} \subset U_ i$. Since each $f_ i$ is open (Morphisms, Lemma 29.25.10), we see that $V = \bigcup _{i\in I} \bigcup _{j \in J_ i} f_ i(V_{ij})$ is an open covering of $V$. Since $V$ is quasi-compact, this covering has a finite refinement. This finishes the proof. $\square$

The fpqc1 topology cannot be treated in the same way as the fppf topology2. Namely, suppose that $R$ is a nonzero ring. We will see in Lemma 34.9.14 that there does not exist a set $A$ of fpqc-coverings of $\mathop{\mathrm{Spec}}(R)$ such that every fpqc-covering can be refined by an element of $A$. If $R = k$ is a field, then the reason for this unboundedness is that there does not exist a field extension of $k$ such that every field extension of $k$ is contained in it.

If you ignore set theoretic difficulties, then you run into presheaves which do not have a sheafification, see [Theorem 5.5, Waterhouse-fpqc-sheafification]. A mildly interesting option is to consider only those faithfully flat ring extensions $R \to R'$ where the cardinality of $R'$ is suitably bounded. (And if you consider all schemes in a fixed universe as in SGA4 then you are bounding the cardinality by a strongly inaccessible cardinal.) However, it is not so clear what happens if you change the cardinal to a bigger one.

For these reasons we do not introduce fpqc sites and we will not consider cohomology with respect to the fpqc-topology.

On the other hand, given a contravariant functor $F : \mathit{Sch}^{opp} \to \textit{Sets}$ it does make sense to ask whether $F$ satisfies the sheaf property for the fpqc topology, see below. Moreover, we can wonder about descent of object in the fpqc topology, etc. Simply put, for certain results the correct generality is to work with fpqc coverings.

Lemma 34.9.7. Let $T$ be a scheme.

1. If $T' \to T$ is an isomorphism then $\{ T' \to T\}$ is an fpqc covering of $T$.

2. If $\{ T_ i \to T\} _{i\in I}$ is an fpqc covering and for each $i$ we have an fpqc covering $\{ T_{ij} \to T_ i\} _{j\in J_ i}$, then $\{ T_{ij} \to T\} _{i \in I, j\in J_ i}$ is an fpqc covering.

3. If $\{ T_ i \to T\} _{i\in I}$ is an fpqc covering and $T' \to T$ is a morphism of schemes then $\{ T' \times _ T T_ i \to T'\} _{i\in I}$ is an fpqc covering.

Proof. Part (1) is immediate. Recall that the composition of flat morphisms is flat and that the base change of a flat morphism is flat (Morphisms, Lemmas 29.25.8 and 29.25.6). Thus we can apply Lemma 34.9.2 in each case to check that our families of morphisms are fpqc coverings.

Proof of (2). Assume $\{ T_ i \to T\} _{i\in I}$ is an fpqc covering and for each $i$ we have an fpqc covering $\{ f_{ij} : T_{ij} \to T_ i\} _{j\in J_ i}$. Let $U \subset T$ be an affine open. We can find quasi-compact opens $U_ i \subset T_ i$ for $i \in I$, almost all empty, such that $U = \bigcup f_ i(U_ i)$. Then for each $i$ we can choose quasi-compact opens $W_{ij} \subset T_{ij}$ for $j \in J_ i$, almost all empty, with $U_ i = \bigcup _ j f_{ij}(U_{ij})$. Thus $\{ T_{ij} \to T\}$ is an fpqc covering.

Proof of (3). Assume $\{ T_ i \to T\} _{i\in I}$ is an fpqc covering and $T' \to T$ is a morphism of schemes. Let $U' \subset T'$ be an affine open which maps into the affine open $U \subset T$. Choose quasi-compact opens $U_ i \subset T_ i$, almost all empty, such that $U = \bigcup f_ i(U_ i)$. Then $U' \times _ U U_ i$ is a quasi-compact open of $T' \times _ T T_ i$ and $U' = \bigcup \text{pr}_1(U' \times _ U U_ i)$. Since $T'$ can be covered by such affine opens $U' \subset T'$ we see that $\{ T' \times _ T T_ i \to T'\} _{i\in I}$ is an fpqc covering by Lemma 34.9.2. $\square$

Lemma 34.9.8. Let $T$ be an affine scheme. Let $\{ T_ i \to T\} _{i \in I}$ be an fpqc covering of $T$. Then there exists an fpqc covering $\{ U_ j \to T\} _{j = 1, \ldots , n}$ which is a refinement of $\{ T_ i \to T\} _{i \in I}$ such that each $U_ j$ is an affine scheme. Moreover, we may choose each $U_ j$ to be open affine in one of the $T_ i$.

Proof. This follows directly from the definition. $\square$

Definition 34.9.9. Let $T$ be an affine scheme. A standard fpqc covering of $T$ is a family $\{ f_ j : U_ j \to T\} _{j = 1, \ldots , n}$ with each $U_ j$ is affine, flat over $T$ and $T = \bigcup f_ j(U_ j)$.

Since we do not introduce the affine site we have to show directly that the collection of all standard fpqc coverings satisfies the axioms.

Lemma 34.9.10. Let $T$ be an affine scheme.

1. If $T' \to T$ is an isomorphism then $\{ T' \to T\}$ is a standard fpqc covering of $T$.

2. If $\{ T_ i \to T\} _{i\in I}$ is a standard fpqc covering and for each $i$ we have a standard fpqc covering $\{ T_{ij} \to T_ i\} _{j\in J_ i}$, then $\{ T_{ij} \to T\} _{i \in I, j\in J_ i}$ is a standard fpqc covering.

3. If $\{ T_ i \to T\} _{i\in I}$ is a standard fpqc covering and $T' \to T$ is a morphism of affine schemes then $\{ T' \times _ T T_ i \to T'\} _{i\in I}$ is a standard fpqc covering.

Proof. This follows formally from the fact that compositions and base changes of flat morphisms are flat (Morphisms, Lemmas 29.25.8 and 29.25.6) and that fibre products of affine schemes are affine (Schemes, Lemma 26.17.2). $\square$

Lemma 34.9.11. Let $T$ be a scheme. Let $\{ f_ i : T_ i \to T\} _{i \in I}$ be a family of morphisms of schemes with target $T$. Assume that

1. each $f_ i$ is flat, and

2. every affine scheme $Z$ and morphism $h : Z \to T$ there exists a standard fpqc covering $\{ Z_ j \to Z\} _{j = 1, \ldots , n}$ which refines the family $\{ T_ i \times _ T Z \to Z\} _{i \in I}$.

Then $\{ f_ i : T_ i \to T\} _{i \in I}$ is an fpqc covering of $T$.

Proof. Let $T = \bigcup U_\alpha$ be an affine open covering. For each $\alpha$ the pullback family $\{ T_ i \times _ T U_\alpha \to U_\alpha \}$ can be refined by a standard fpqc covering, hence is an fpqc covering by Lemma 34.9.4. As $\{ U_\alpha \to T\}$ is an fpqc covering we conclude that $\{ T_ i \to T\}$ is an fpqc covering by Lemma 34.9.5. $\square$

Definition 34.9.12. Let $F$ be a contravariant functor on the category of schemes with values in sets.

1. Let $\{ U_ i \to T\} _{i \in I}$ be a family of morphisms of schemes with fixed target. We say that $F$ satisfies the sheaf property for the given family if for any collection of elements $\xi _ i \in F(U_ i)$ such that $\xi _ i|_{U_ i \times _ T U_ j} = \xi _ j|_{U_ i \times _ T U_ j}$ there exists a unique element $\xi \in F(T)$ such that $\xi _ i = \xi |_{U_ i}$ in $F(U_ i)$.

2. We say that $F$ satisfies the sheaf property for the fpqc topology if it satisfies the sheaf property for any fpqc covering.

We try to avoid using the terminology “$F$ is a sheaf” in this situation since we are not defining a category of fpqc sheaves as we explained above.

Lemma 34.9.13. Let $F$ be a contravariant functor on the category of schemes with values in sets. Then $F$ satisfies the sheaf property for the fpqc topology if and only if it satisfies

1. the sheaf property for every Zariski covering, and

2. the sheaf property for any standard fpqc covering.

Moreover, in the presence of (1) property (2) is equivalent to property

1. the sheaf property for $\{ V \to U\}$ with $V$, $U$ affine and $V \to U$ faithfully flat.

Proof. Assume (1) and (2) hold. Let $\{ f_ i : T_ i \to T\} _{i \in I}$ be an fpqc covering. Let $s_ i \in F(T_ i)$ be a family of elements such that $s_ i$ and $s_ j$ map to the same element of $F(T_ i \times _ T T_ j)$. Let $W \subset T$ be the maximal open subset such that there exists a unique $s \in F(W)$ with $s|_{f_ i^{-1}(W)} = s_ i|_{f_ i^{-1}(W)}$ for all $i$. Such a maximal open exists because $F$ satisfies the sheaf property for Zariski coverings; in fact $W$ is the union of all opens with this property. Let $t \in T$. We will show $t \in W$. To do this we pick an affine open $t \in U \subset T$ and we will show there is a unique $s \in F(U)$ with $s|_{f_ i^{-1}(U)} = s_ i|_{f_ i^{-1}(U)}$ for all $i$.

By Lemma 34.9.8 we can find a standard fpqc covering $\{ U_ j \to U\} _{j = 1, \ldots , n}$ refining $\{ U \times _ T T_ i \to U\}$, say by morphisms $h_ j : U_ j \to T_{i_ j}$. By (2) we obtain a unique element $s \in F(U)$ such that $s|_{U_ j} = F(h_ j)(s_{i_ j})$. Note that for any scheme $V \to U$ over $U$ there is a unique section $s_ V \in F(V)$ which restricts to $F(h_ j \circ \text{pr}_2)(s_{i_ j})$ on $V \times _ U U_ j$ for $j = 1, \ldots , n$. Namely, this is true if $V$ is affine by (2) as $\{ V \times _ U U_ j \to V\}$ is a standard fpqc covering and in general this follows from (1) and the affine case by choosing an affine open covering of $V$. In particular, $s_ V = s|_ V$. Now, taking $V = U \times _ T T_ i$ and using that $s_{i_ j}|_{T_{i_ j} \times _ T T_ i} = s_ i|_{T_{i_ j} \times _ T T_ i}$ we conclude that $s|_{U \times _ T T_ i} = s_ V = s_ i|_{U \times _ T T_ i}$ which is what we had to show.

Proof of the equivalence of (2) and (2') in the presence of (1). Suppose $\{ T_ i \to T\}$ is a standard fpqc covering, then $\coprod T_ i \to T$ is a faithfully flat morphism of affine schemes. In the presence of (1) we have $F(\coprod T_ i) = \prod F(T_ i)$ and similarly $F((\coprod T_ i) \times _ T (\coprod T_ i)) = \prod F(T_ i \times _ T T_{i'})$. Thus the sheaf condition for $\{ T_ i \to T\}$ and $\{ \coprod T_ i \to T\}$ is the same. $\square$

The following lemma is here just to point out set theoretical difficulties do indeed arise and should be ignored by most readers.

Lemma 34.9.14. Let $R$ be a nonzero ring. There does not exist a set $A$ of fpqc-coverings of $\mathop{\mathrm{Spec}}(R)$ such that every fpqc-covering can be refined by an element of $A$.

Proof. Let us first explain this when $R = k$ is a field. For any set $I$ consider the purely transcendental field extension $k_ I = k(\{ t_ i\} _{i \in I})/k$. Since $k \to k_ I$ is faithfully flat we see that $\{ \mathop{\mathrm{Spec}}(k_ I) \to \mathop{\mathrm{Spec}}(k)\}$ is an fpqc covering. Let $A$ be a set and for each $\alpha \in A$ let $\mathcal{U}_\alpha = \{ S_{\alpha , j} \to \mathop{\mathrm{Spec}}(k)\} _{j \in J_\alpha }$ be an fpqc covering. If $\mathcal{U}_\alpha$ refines $\{ \mathop{\mathrm{Spec}}(k_ I) \to \mathop{\mathrm{Spec}}(k)\}$ then the morphisms $S_{\alpha , j} \to \mathop{\mathrm{Spec}}(k)$ factor through $\mathop{\mathrm{Spec}}(k_ I)$. Since $\mathcal{U}_\alpha$ is a covering, at least some $S_{\alpha , j}$ is nonempty. Pick a point $s \in S_{\alpha , j}$. Since we have the factorization $S_{\alpha , j} \to \mathop{\mathrm{Spec}}(k_ I) \to \mathop{\mathrm{Spec}}(k)$ we obtain a homomorphism of fields $k_ I \to \kappa (s)$. In particular, we see that the cardinality of $\kappa (s)$ is at least the cardinality of $I$. Thus if we take $I$ to be a set of cardinality bigger than the cardinalities of the residue fields of all the schemes $S_{\alpha , j}$, then such a factorization does not exist and the lemma holds for $R = k$.

General case. Since $R$ is nonzero it has a maximal prime ideal $\mathfrak m$ with residue field $\kappa$. Let $I$ be a set and consider $R_ I = S_ I^{-1} R[\{ t_ i\} _{i \in I}]$ where $S_ I \subset R[\{ t_ i\} _{i \in I}]$ is the multiplicative subset of $f \in R[\{ t_ i\} _{i \in I}]$ such that $f$ maps to a nonzero element of $R/\mathfrak p[\{ t_ i\} _{i \in I}]$ for all primes $\mathfrak p$ of $R$. Then $R_ I$ is a faithfully flat $R$-algebra and $\{ \mathop{\mathrm{Spec}}(R_ I) \to \mathop{\mathrm{Spec}}(R)\}$ is an fpqc covering. We leave it as an exercise to the reader to show that $R_ I \otimes _ R \kappa \cong \kappa (\{ t_ i\} _{i \in I}) = \kappa _ I$ with notation as above (hint: use that $R \to \kappa$ is surjective and that any $f \in R[\{ t_ i\} _{i \in I}]$ one of whose monomials occurs with coefficient $1$ is an element of $S_ I$). Let $A$ be a set and for each $\alpha \in A$ let $\mathcal{U}_\alpha = \{ S_{\alpha , j} \to \mathop{\mathrm{Spec}}(R)\} _{j \in J_\alpha }$ be an fpqc covering. If $\mathcal{U}_\alpha$ refines $\{ \mathop{\mathrm{Spec}}(R_ I) \to \mathop{\mathrm{Spec}}(R)\}$, then by base change we conclude that $\{ S_{\alpha , j} \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(\kappa ) \to \mathop{\mathrm{Spec}}(\kappa )\}$ refines $\{ \mathop{\mathrm{Spec}}(\kappa _ I) \to \mathop{\mathrm{Spec}}(\kappa )\}$. Hence by the result of the previous paragraph, there exists an $I$ such that this is not the case and the lemma is proved. $\square$

 The letters fpqc stand for “fidèlement plat quasi-compacte”.
 A more precise statement would be that the analogue of Lemma 34.7.7 for the fpqc topology does not hold.

Comment #1450 by on

Would it be possible to make the comment "We claim that there does not exist a set $A$ of fpqc-coverings of $Spec(R)$ such that every fpqc-covering can be refined by an element of $A$." a remark, with a tag, so it can be cited? Or, is there a reference where this was first observed?

Comment #1459 by on

Dear David Roberts, OK I made it into an actual lemma (later today it will get a tag). I do think one can find this statement somewhere in the literature (somewhere in SGA?), but I do not know where. I would love to add a reference if you find one. Also: I added a reference to the paper by Waterhouse on fpqc sheafification, see commit, which shows clearly that the problem is there.

Comment #3805 by Fernando on

Maybe I'm being silly and after sleeping I will regret my comment, but the proof in 022H (Lemma 33.9.13) is not using the finitude of the standard fpqc covering $\{ U_j \longrightarrow U \}_{j}$. More precisely, by the same proof one can conclude the analogous assertion with 2 modified to fpqc coverings ${T_i \longrightarrow T}_{i \in I}$ such that each $T_i$ is affine (where $I$ is not necessarily finite). From this, one can conclude that a sheaf satisfying descent for the flat topology (ie, coverings are just jointly surjective collections of faithfully flat morphisms) would also satisfy descent for the fpqc topology. This cannot be true as the flat topology is not subcanonical by Vistoli's example (gluing the local rings of an integral smooth curve along the generic point).

Comment #3807 by on

Dear Fernando, yes you are being silly.

Comment #3808 by on

To clarify, the proof only shows the implication in one direction (because the proof of the other direction is trivial exactly because Zariski coverings and standard fpqc coverings are fpqc coverings and your families of morphisms aren't fpqc coverings so this direction would fail right off the bat).

Comment #3809 by Fernando on

Dear Johan, thanks for replying. By the way, in my comment I inverted the order (it should be fpqc descent implying flat descent). In any case, ouch! That was indeed a stupid mistake as I suspected.

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