Famously, the fpqc topology is so big that no one set of fpqc covers of an affine scheme is cofinal in all fpqc covers. The Stacks Project gives a construction using field extensions with arbitrarily large infinite transcendence basis, which are all fpcq maps.

This means that it is difficult to see whether the collection of isomorphism classes of fpqc-locally trivial principal $G$-bundles, for a fixed group scheme $G$, on a given scheme form a set, rather than a proper class. If $G$ is nice (e.g. finite type/smooth/insert your favourite condition here), then standard reductions show that any fpqc bundle is in fact locally trivial in a far smaller topology, such as the étale topology. I think (from reading this thesis) that even if we are working with a group scheme over a field $k$ such that $G\to Spec(k)$ is fppf, then we should probably only get set-many isomorphism classes, as the stack of $G$-bundles is then algebraic. So any group scheme that would give a proper class of non-isomorphic bundles would have to be rather nasty. (Also, I believe we need to be in positive characteristic, to get maximum nastiness...)

My vague plan for constructing an example would be to consider fpqc $G$-bundles on $Spec(k)$ ($G$ a group scheme over $k$) via cocycles over the covers that the Stacks project describes at the above link. If a purely transcendental extension $k\to k\{I\}$ (where $I$ is a set of arbitrary cardinality) made $Spec(k\{I\}) \to Spec(k)$ a torsor (so the extension is 'Galois' for a suitable interpretation of that term), then one could consider homomorphisms $\alpha\colon Gal(k\{I\}) \to G$ and the associated $G$-bundles. It seems possible that one could figure out when two such bundles are isomorphic, possibly using a cardinality bound coming from data attached to $G$. Or not, I really don't know.

Can we make a call in either direction? Does the above sketch give proper class-many non-isomorphic bundles, or can we prove that for every fpqc group scheme $G$ and base scheme $X$ there is only a set of isomorphism classes of fpqc principal $G$-bundles on $X$?

  • $\begingroup$ See this blog post: math.columbia.edu/~dejong/wordpress/?p=591 $\endgroup$
    – darx
    Aug 23, 2016 at 13:16
  • $\begingroup$ @darx ah, excellent. I would like to comment and tell Johan he should probably assume κ regular in his definition but comments are closed. $\endgroup$
    – David Roberts
    Aug 23, 2016 at 22:15
  • $\begingroup$ There is an example that Johan gives where the torsors (up to isomorphism) form a proper class, but I don't grasp yet what's going on: stacks.math.columbia.edu/tag/04AF $\endgroup$
    – David Roberts
    Aug 23, 2016 at 22:40
  • $\begingroup$ @DavidRoberts: That example is not saying what you think it does. It has no set-theoretic significance. He is just showing by an example that over any field there is an affine group scheme that is not fppf which has torsors that are not split by an fppf cover of the base (or equivalently, by the Nullstellensatz, by a finite extension of the base field). Thinking about how descent theory works for properties of morphisms, it isn't surprising that this happens (and one would expect almost any non-fppf group scheme over a field to admit a torsor not split by a finite extension of the field). $\endgroup$
    – nfdc23
    Aug 24, 2016 at 1:09
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    $\begingroup$ @DavidRoberts: Does "property under discussion" mean "not a set" (rather, a proper class)? I do not see any such comment with such an assertion at that link. What statement of his do you think is conveying that assertion? I think you are misunderstanding something in what he has written in his comments. $\endgroup$
    – nfdc23
    Aug 24, 2016 at 1:15

1 Answer 1


In more down-to-earth terms, you are asking whether there is always a set of $G$-torsors for the fpqc topology over a scheme $S$ such any every $G$-torsor for the fpqc topology over $S$ is isomorphic to one of those in the set. The answer is "yes" when $G$ is flat over $S$ (as is automatic over a field).

The question asks about something being a "proper class", which I am told in the comment by Q. Yuan below means "class that is not a set", so the answer to the question in the title is then "no" under that flatness hypothesis on $G$. I don't see a reason for interest in torsors for for the fpqc topology on $S$ relative to a group scheme that isn't flat over the base, so I will regard the $S$-flatness of $G$ as a basic hypothesis one should impose.

The real content is in the following simple lemma:

Lemma: Let $A$ be a ring, $B$ an $A$-algebra, and $A'$ a faithfully flat $A$-algebra. For an infinite set $I$, if the $A'$-algebra $B' = A' \otimes_A B$ admits a set of at most $|I|$ generators $($equivalently, $B'$ is a quotient of a polynomial algebra $A'[X_i]_{i \in I}$ over $A'$$)$ then so does $B$ over $A$.

This lemma and its proof are motivated by the more well-known fact (and its proof!) that if $B'$ is finitely generated (resp. finitely presented) over $A'$ then the same holds for $B$ over $A$ (EGA IV$_2$,, but the proof is easier for this controlled "infinitely generated" property.

Proof: Pick a set of generators of $B'$ over $A'$ indexed by a subset $J \subset I$. Write each as a finite sum of elementary tensors, and consider the set $S$ of elements of $B$ that arise in those tensors. Since $I$ is infinite, it is clear that $|S| \le |I|$. Hence, it suffices to show that $B$ coincides with its $A$-subalgebra $B_0$ generated by the elements in $S$.

By flatness of $A'$ over $A$, the natural map $A' \otimes_A B_0 \rightarrow A' \otimes_A B$ is injective. But by design of $B_0$ in terms of $S$, this latter injection is also surjective. Hence, by faithful flatness the inclusion $B_0 \hookrightarrow B$ is an equality too.

QED Lemma

Now we consider $G$-torsors for the fpqc topology on $S$, where $G$ is $S$-flat (hence faithfully flat, due to the identity section, but whatever). It suffices to show that for each open affine $U = {\rm{Spec}}(A) \subset S$ (or really just the members of whatever fixed affine open cover you wish) there is a set's worth of isomorphism classes of $G$-torsors for the fpqc topology over $U$. That is, we can assume $S$ is affine, say with coordinate ring $A$. By choosing finitely many affine opens in $G$ which cover the quasi-compact identity section, we can shrink $S$ a bit more so that there exists an open affine $V \subset G$ containing the identity section (so $V \rightarrow S$ is faithfully flat between affines). Let $I$ be an infinite cardinal so that each affine open subscheme of $G$ (e.g., $V$) admits a set of at most $|I|$ generators for its coordinate ring as an $A$-algebra (such an $I$ clearly exists); for what follows it would be enough to pick $I$ that "works" for the members of a single affine open covering of $G$.

For any given faithfully flat $A$-algebra $R$, there is a set's worth of isomorphism classes of $G$-torsors for the fpqc topology over $S$ which admit an $R$-point (since by fpqc descent these are controlled by the descent data encoded in ${\rm{H}}^1(R/A, G)$; note that not all such descent may be effective when $G$ isn't relatively affine).

Hence, it suffices to find a set's worth of such $R$ so that every $G$-torsor for the fpqc topology on $S$ splits over one of those $A$-algebras $R$ (i.e. admits an $R$-point over $A$). We will show that the set (!) of isomorphism classes of faithfully flat $A$-algebras $R$ that admit a set of $A$-algebra generators of size at most $|I|$ does the job.

Let $E$ be a $G$-torsor for the fpqc topology over $S$. Thus, there exists a faithfully flat $A$-algebra $A'$ then $E(A')$ is non-empty, so $E_{A'} \simeq G_{A'}$ as $A'$-schemes. In particular, $V_{A'}$ is an affine open subscheme of $E_{A'}$ admitting at most $|I|$ generators as an $A'$-algebra. Its image in $E$ is quasi-compact, so is contained in a quasi-compact open $\Omega \subset E$. The map $\Omega \rightarrow S = {\rm{Spec}}(A)$ is faithfully flat because after fpqc base change to $A'$ it becomes $\Omega_{A'} \rightarrow {\rm{Spec}}(A')$ that is faithfully flat (as $\Omega_{A'}$ is identified with an open subscheme of $G_{A'}$ containing $V_{A'}$).

By quasi-compactness, $\Omega$ is covered by a finite set of affine open subschemes ${\rm{Spec}}(B_j)$ with $1 \le j \le n$. Thus, $B = \prod_{j=1}^n B_j$ is a flat $A$-algebra that is faithfully flat since ${\rm{Spec}}(B) = \coprod {\rm{Spec}}(B_j) \rightarrow \Omega$ is surjective. Hence, $E(B)$ is non-empty with $B$ faithfully flat over $A$. We claim that $B$ admits a set of at most $|I|$ generators as an $A$-algebra (which will do the job). It suffices to do the same for each $B_j$ separately.

Now finally we do something that isn't formal nonsense, namely we invoke the Lemma: it suffices to show that $B_j \otimes_A A'$ admits a set of at most $|I|$ generators as an $A'$-algebra. But the spectrum of this $A'$-algebra is an affine open subscheme of $E_{A'} = G_{A'}$, so it suffices to show that every affine open $W$ in $G_{A'}$ admits a set of at most $|I|$ generators for its coordinate ring as an $A'$-algebra. By design, $G$ is covered by affine opens each of whose coordinate rings admits a set of at most $|I|$ generators as an $A$-algebra. Thus, the same holds for $G_{A'}$ relative to $A'$ for the base-change covering, and consequently for any basic affine open of one of those.

Recall the fact that for any two affine opens in a scheme, their overlap is covered by affine opens which are simulatenously affine open in each. Hence, the affine $W$ is covered by basic affine opens (relative to $W$!) whose coordinate rings each admit a set of at most $|I|$ generators as an $A'$-algebra. By quasi-compactness, $W$ is covered by finitely many of those. Since $I$ is infinite, the numerators in such fractional expressions give a set of size at most $|I|$ that generate the coordinate ring of $W$ as an $A'$-algebra.

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    $\begingroup$ Sets are classes, but proper classes are by definition classes that are not sets. $\endgroup$ Aug 23, 2016 at 17:31
  • $\begingroup$ I'm not an algebraic geometer, rather a category theorist, so I was hoping for a really nasty counterexample, to test the boundaries of this concept. Just to confirm, by saying G is flat rather than fpqc, you are making a stronger statement than I was considering? $\endgroup$
    – David Roberts
    Aug 23, 2016 at 22:22
  • $\begingroup$ You were imposing fpqc-local triviality on the torsors but not making any hypothesis on the structure morphism $G \rightarrow S$ (e.g., that it is flat, or even flat and quasi-compact). Of course, over a field the flatness is automatic, so over a field I am addressing the same generality as the question. Anyway, so yes: for any scheme $S$ and an $S$-group $G$ that is flat (not assumed quasi-compact over $S$) there exists a set of $G$-torsors for the fpqc topology on $S$ such that any $G$-torsor over $S$ for the fpqc topology is isomorphic to one of the $G$-torsors in this set. $\endgroup$
    – nfdc23
    Aug 24, 2016 at 1:12
  • $\begingroup$ Ok, I got it now. Of course, I'd love to see something non-flat that breaks this result, but that's enough for this question for today. $\endgroup$
    – David Roberts
    Aug 24, 2016 at 1:35

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