Famously, the [fpqc topology](http://stacks.math.columbia.edu/tag/022A) 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](http://stacks.math.columbia.edu/tag/0BBK) 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](http://www.math.harvard.edu/theses/senior/wang/wang.pdf)) 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 $X$?



Is there a group scheme G such that there is a proper class of non-isomorphic fpqc G-bundles over some fixed base?
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$\_2$, 2.7.1.1), 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.


