Let $K$ and $L$ be two subfields of some field. If a variety is defined over both $K$ and $L$, does it follow that the variety can be defined over their intersection?

$\begingroup$ This follows very easily for the very special case of elliptic curves, using the $j$invariant. $\endgroup$ – Regenbogen Mar 27 '10 at 0:25

15$\begingroup$ Someone has voted to close this question as "offtopic". Could you please explain how a nontrivial arithmetic geometry question is offtopic for Math Overflow? $\endgroup$ – Pete L. Clark Mar 27 '10 at 1:35

1$\begingroup$ I had this question and another question open and accidentally clicked to close this one rather than the other one. $\endgroup$ – Harry Gindi Mar 27 '10 at 2:43

1$\begingroup$ ...and, as I pointed out on meta yesterday, this is an "unrecoverable" mistake! You can't currently undo a vote to close; you have to wait until the question is closed and then vote to reopen :) $\endgroup$ – Kevin Buzzard Mar 27 '10 at 7:19

1$\begingroup$ @Kevin: Agreed, this is certainly an undesirable feature which we are hoping will be changed in the future. But the moral is clear: voting to close is a serious thing. Don't do it on a whim, and pay attention while you're doing it! $\endgroup$ – Pete L. Clark Mar 27 '10 at 16:18
Yes, if varieties are interpreted as subvarieties closed subschemes of base extensions of a fixed ambient variety scheme (e.g., affine space or projective space).
More precisely, suppose that $k \subseteq F$ are fields and the variety $X$ is an $F$subvariety a closed subscheme of $\mathbf{P}^n_F$. Say for a field $K$ with $k \subseteq K \subseteq F$ that "$X$ is defined over $K$" if $X$ is the base extension of some subvariety of $\mathbf{P}^n_K$. Then $X$ has a minimal field of definition $E$ with $k \subseteq E \subseteq F$, characterized by the property that for any field $K$ with $k \subseteq K \subseteq F$, we have that $X$ is defined over $K$ if and only if $K$ contains $E$.
The same statement holds if $\mathbf{P}^n$ is replaced by any fixed $k$variety $k$scheme $Y$.
(Note: this answer does not contradict Pete's. This is just a different interpretation of the question.)
EDIT: As Brian points out, I was indeed assuming that my varieties were closed in the ambient space. The statement about minimal field of definition is not even true for open subschemes in characteristic $p$. For example, if $k=\mathbb{F}_p$ and $F=k(t)$ and $Y=\operatorname{Spec} k[x]$ and $X=\operatorname{Spec} F[x,1/(xt)]$, then $X$ is the base extension of $\operatorname{Spec} F^{p^n}[x,1/(x^{p^n}t^{p^n})]$, and hence is definable over $F^{p^n}$ for all $n$, but not over the intersection of all these fields, which is just $k$.
On the other hand, the intersection of any finite number of fields of definition is still a field of definition.
I have generalized to schemes as suggested by Brian.

4$\begingroup$ This is true with "variety" replaced by "scheme", "subvariety" replaced by "closed subscheme", and projective space over F replaced with base change to $F$ of arbitrary scheme over $k$ (used in place of projective space over $k$). In general seems delicate to relax "closed subscheme" to "locally closed subscheme" when ground field not perfect, since reduced structure on closed complement $C$ inside schemetheoretic closure may not commute with extn on ground field, so no reason field of definition over $k$ for $C$ matches that of $X$. Bjorn, are you sure no problem even in projective space? $\endgroup$ – BCnrd Mar 27 '10 at 14:13

2$\begingroup$ By the way, a reference for existence of minimal field of definition over $k$ is EGA IV$_2$, section 4.8 (especially 4.8.11). $\endgroup$ – BCnrd Mar 27 '10 at 14:17

$\begingroup$ Thank you, Brian. Indeed, I was thinking of closed subvarieties only. For open subschemes, the strong form of the statement, about a minimal field of definition, can fail even if k and F are both algebraically closed. $\endgroup$ – Bjorn Poonen Mar 27 '10 at 15:38


1$\begingroup$ As long as we are speaking of closed subschemes, or we are in characteristic 0, the same holds for infinite intersections, since in this case there is a unique minimal field of definition in the sense of this answer. $\endgroup$ – Bjorn Poonen Apr 3 '10 at 13:32
No, not in general. Assuming we start with a variety over an algebraically closed field, this would be true if the variety could be defined over its field of moduli, which would then be the unique minimal field of definition. (In particular it is true if the variety has no nontrivial automorphisms.) Note also that it holds trivially for genus zero curves  there are no moduli!  and for genus one curves, by the theory of the $j$invariant.
Therefore the simplest counterexample seems to be a genus $2$ curve over $\mathbb{Q}$. Mestre has shown that the obstruction to such a curve being defined over its field of moduli is a quaternion algebra over $\mathbb{Q}$  i.e., the curve can be defined over its field of moduli iff this quaternion algebra is split, i.e., isomorphic to a $2 \times 2$ matrix algebra.
Let $C$ be a genus $2$ curve with field of moduli $\mathbb{Q}$ and nontrivial Mestre obstruction. (Such curves certainly exist. Let me know if you want an explicit example.) Then $C$ may be defined over a quadratic field $K$ iff $K$ splits the quaternion algebra. By the structure theory of quaternion algebras, $C$ can therefore be defined over infinitely many quadratic fields, but not over $\mathbb{Q}$.
Addendum: Here is a nice paper by S. Baba and H. Granath:
http://www.math.kau.se/granath/research/qm.pdf
It deals with the Mestre obstruction on certain genus 2 curves and includes specific examples.

1$\begingroup$ So, as is often the case, the obstruction lies in an H^2, right? H^2(Q,G_m)[2] 'is' the quaternion algebras over Q, the split one being the trivial element. $\endgroup$ – Kevin Buzzard Mar 27 '10 at 7:23

1$\begingroup$ @Kevin: Yes, that's right. It's even better to think of the obstruction as lying in H^2(Q,\mu_2) (the same group!), because $\mu_2$ is the subgroup of automorphisms of the genus 2 curve generated by the hyperelliptic involution. [It has since been shown that if the automorphism group is strictly larger, the genus 2 curve can be defined over its field of moduli.] And in general, the obstruction to V being defined over its field of moduli is an element of H^2(K,Aut(V)). This needs to be interpreted gerbeically when Aut(V) is nonabelian, which is fortunately not the case here. $\endgroup$ – Pete L. Clark Mar 27 '10 at 16:16

$\begingroup$ Note by the way that Mestre's obstruction is defined for genus 2 curves over an arbitrary field of characteristic different from 2. $\endgroup$ – Pete L. Clark Mar 27 '10 at 16:50
No, not even for genus $0$ curves, if "$X$ is defined over $K$" means that the variety $X$, initially defined over an extension $F$ of $K$, is to be isomorphic to the base extension of some variety over $K$. (Pete in his answer was implicitly assuming that he was allowed to base extend to an algebraic closure of $F$ before going down to $K$.)
For finite extensions $F \supseteq E$ of $\mathbb{Q}_p$, if $X$ is the genus $0$ curve over $F$ corresponding to the nonsplit quaternion algebra, then $X$ is a base extension of a genus $0$ curve over $E$ if and only if $[F:E]$ is odd (because $\operatorname{Br}(E)[2] \to \operatorname{Br}(F)[2]$ is multiplication by $[F:E]$ from $\frac{1}{2} \mathbb{Z}/\mathbb{Z}$ to itself).
So if $F$ is an $A_4$extension of $k:=\mathbb{Q}_2$ (such an extension exists), and $K$ and $L$ are the subfields of $F$ fixed by two $3$cycles in $A_4$, and $X$ is the genus $0$ curve over $F$ corresponding to the nonsplit quaternion algebra, then $X$ is a base extension of a genus $0$ curve over $K$, and similarly over $L$, but not over $K \cap L = \mathbb{Q}_2$.

$\begingroup$ +1: This is another reasonable interpretation of the question. I believe that I construed the question in the sense that "defined over K" is most commonly used, but it is valuable to have these multiple answers for multiple interpretations. $\endgroup$ – Pete L. Clark Mar 27 '10 at 16:48

4$\begingroup$ Pete, personally I'd never use the terminology "defined over $K$" for a subfield $K$ of $F$ to have a meaning not implying that scalar extension back to $F$ recovers what we originally began with over $F$. Seems too confusing. I'm surprised you suggest your interpretation is the most commonly used. $\endgroup$ – BCnrd Mar 27 '10 at 20:06

1$\begingroup$ @Brian: I agree with your first sentence: if we start with a variety over a named field $F$, then yes, it makes no sense to change the $F$model. But, since classical algebraic geometry predates arithmetic geometry, in practice I have found that in this context most people start out with a variety defined over an algebraically closed field (or even over C). Or they start out with a point on a coarse moduli space, like $\mathcal{M}_g$, in which it is natural to go up to the algebraic closure before going down. Anyway, in any given situation, you should of course say what you mean! $\endgroup$ – Pete L. Clark Mar 27 '10 at 22:31