Let $V$ be an algebraic variety over a field $K$. Is there a constant $d = d(V) \in \mathbb{N}$ such that for any variety $W$ defined over $K$ and isomorphic to $V$ over the algebraic closure of $K$, there is a field extension $F/K$ with $[F:K] \leq d$ such that $V$ and $W$ are isomorphic over $F$?
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2 Answers
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Yes if $K=\mathbb{R}$ for example, but no in general.
Namely this fails for curves of genus $1$, over $\mathbb{Q}$, say. Given an elliptic curve $E$ over $\mathbb{Q}$ and a positive integer $d$, a general $d$-covering of $E$ will contain a divisor of degree $d^2$, but no divisors of smaller degree. In particular, it will only become isomorphic to $E$ over a field extension of degree $d^2$.
answered Jul 7, 2014 at 17:37
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$\begingroup$ Please explain the procedure you are using to produce from $E$ other Galois-twisted forms over the ground field. Your answer is a bit unclear on the precise construction you have in mind and how you know it doesn't have divisors of degree below $d^2$. $\endgroup$ Commented Jul 7, 2014 at 20:49
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1$\begingroup$ Pull-back along what then? The covering map has degree $d^2$. $\endgroup$– R.P.Commented Jul 7, 2014 at 21:44
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1$\begingroup$ @user52824: it is well-known that $H^1(K,E)$ has elements of arbitrarily high order $d$. Such an element $\alpha$ can be lifted to $H^1(K,E[d])$, which parametrizes $d$-coverings up to isomorphism (this follows from the fact that the automorphism group over $\overline{K}$ of the diagram $E \stackrel{[d]}{\rightarrow} E$ is $E[d]$ as Galois modules). If $X \rightarrow E$ is the associated $d$-covering then by "period divides index" any divisor on $X$ has degree divisible by $d$. I think Daniel claims that the index is "generically" equal to $d^2$, but I can't make this statement more rigorous. $\endgroup$– R.P.Commented Jul 7, 2014 at 22:02
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1$\begingroup$ @René and Daniel: Actually, already the fact that $H^1(K,E)$ has elements of arbitrarily large order is enough for the OP's purposes. Identifying $E$ with a group of automorphisms of $E$ as an algebraic variety, given by translation by points, identifies $H^1(K,E)$ with a set of forms of $E$, i.e. curves that are isomorphic to $E$ over $\bar{K}$ as algebraic curves. Now, if a class in this cohomology group is trivialised by a finite Galois extension $F/K$ of degree $n$, then it has order dividing $n$. No need to talk about coverings and "period divides index", I think. $\endgroup$– Alex B.Commented Jul 7, 2014 at 22:47
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1$\begingroup$ I completely agree with that. Note however that "period divides index" is proved exactly by your observation, namely that multiplication by $[F:K]$ factors through restriction from $K$ to $F$. $\endgroup$– R.P.Commented Jul 7, 2014 at 23:19
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This question is most "reasonable" when $V$ is projective, since at least then the automorphism functor is represented by a (locally finite type) $K$-scheme $G_V := {\rm{Aut}}_{V/K}$ and so one can try to apply to $G_V^0$ the structure theory of algebraic groups (and then separately grapple with the component group, if nontrivial). In particular, if the automorphism scheme is smooth then the question amounts to asking if all elements in the Galois cohomology set ${\rm{H}}^1(K_s/K,G_V(K_s))$ are split by finite extensions of $K$ of bounded degree, a question one could hope to analyze by using the structure of $G_V^0$ as an algebraic group (and of $\pi_0(G_V)(\overline{K})$ as discrete group with continuous Galois action). In general it isn't known (not even for smooth projective surfaces in characteristic 0) if $\pi_0(G_V)({\overline{K}})$ is even finitely generated
Every abelian variety $A$ of dimension $g > 0$ over a number field $K$ provides extreme counterexamples when $A(K)$ has rank $r < g[K:\mathbf{Q}]$ (e.g., $r=0$), in the sense that if $p$ is a prime unramified in $K$ and $S$ is the set of $p$-adic places of $K$ then the $S$-relaxed "Tate-Shafarevich" group of isomorphism classes of $A$-torsors that are locally trivial away from $S$ has infinite $p$-part yet has finite $p^n$-torsion for every $n \ge 1$.
So already over number fields you're going to expect to have problems when $G_V^0$ has nontrivial "abelian part" (in the sense of Chevalley's structure theorem) if you don't demand that $W$ becomes isomorphic to $V$ at all places of $K$. If you do impose such a condition then you're OK over number fields provided that you can somehow control the influence of $\pi_0(G_V)$ (which is probably hopeless in general).
On the other hand, if $G_V$ is affine and you're over a number field then the answer is positive since the influence of the $K$-finite $\pi_0(G_V)$ is controlled by bounded degree, the unipotent radical of $G_V^0$ is invisible to the problem, the influence of degree-1 cohomology of the maximal torus quotient of the maximal reductive quotient $(G_V^0)^{\rm{red}}$ is controlled by the degree of the splitting field of the torus, and finally the influence of the degree-1 cohomology of the semisimple derived group of $(G_V^0)^{\rm{red}}$ is controlled by the degree of its simply connected central cover (as the Hasse Principle identifies the degree-1 cohomology of simply connected semisimple groups with cohomology at real places).
Over more general fields $K$, where one may know very little about the degree-1 cohomology of simply connected semisimple groups, probably nothing much can be said in general. For example, consider smooth projective quadrics of dimension at least 2. For a non-degenerate quadratic space $(M,q)$ over $K$ of dimension at least 3 and the associated smooth quadric $V = (q=0) \subset \mathbf{P}(M^{\ast})$ we have $G_V = {\rm{PGO}}(q)$, so one basically runs into two problems: can one control the degree of splitting fields for Brauer classes with controlled torsion-order (e.g., 2-torsion, which might not come from quaternion algebras in general), and can one get a handle on ${\rm{H}}^1(K, {\rm{Spin}}(q))$? For general $K$ these seem out of reach, so one should focus on some class of interesting $K$ in order to get nice answers.
