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Is there a standard example of two abelian varieties $A$, $B$ over some number field $k$ which are $k_v$-isomorphic for every place $v$ of $k$ but not $k$-isomorphic ?

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4 Answers 4

up vote 24 down vote accepted

(If you upvote this answer, please consider upvoting the answers by Felipe Voloch and David Speyer too, since this answer builds on their ideas.)

The smallest examples are in dimension $2$. Let $E$ be any elliptic curve over $\mathbf{Q}$ without complex multiplication, e.g., $X_0(11)$. We will construct two twists of $E^2$ that are isomorphic over $\mathbf{Q}_p$ for all $p \le \infty$ but not isomorphic over $\mathbf{Q}$.

Let $K:=\mathbf{Q}(\sqrt{-1},\sqrt{17})$. Let $G:=\operatorname{Gal}(K/\mathbf{Q}) = (\mathbf{Z}/2\mathbf{Z})^2$. Let $\alpha \colon G \to \operatorname{GL}_2(\mathbf{Z}) = \operatorname{Aut}(E^2)$ be a homomorphism sending the two generators to the reflections in the coordinate axes of $\mathbf{Z}^2$, and let $A$ be the $K/\mathbf{Q}$-twist of $E^2$ given by $\alpha$. Define $\beta$ and $B$ similarly, but with the lines $y=x$ and $y=-x$ in place of the coordinate axes. The representations $\alpha$ and $\beta$ of $G$ on $\mathbf{Z}^2$ are not conjugate: only the former is such that the lattice vectors fixed by nontrivial elements of $G$ generate all of $\mathbf{Z}^2$. Thus $A$ and $B$ are not isomorphic over $\mathbf{Q}$.

On the other hand, every decomposition group $D_p$ in $G$ is smaller than $G$ since $-1$ is a square in $\mathbf{Q}_{17}$ and $17$ is a square in $\mathbf{Q}_2$. Also, the restrictions of $\alpha$ and $\beta$ to any proper subgroup of $G$ are conjugate: any single line spanned by a primitive vector in $\mathbf{Z}^2$ can be mapped to any other by an element of $\operatorname{GL}_2(\mathbf{Z})$. Thus $A$ and $B$ become isomorphic after base extension to $\mathbf{Q}_p$ for any $p \le \infty$. $\square$

Remark: The abelian surfaces $A$ and $B$ constructed above are isogenous even over $\mathbf{Q}$, because the $\mathbf{Z}^2$ with one Galois action can be embedded into the $\mathbf{Z}^2$ with the other Galois action: rotate $45^\circ$ and dilate.

Remark: The nonexistence of examples in dimension $1$ follows from these two well-known facts:

1) Twists of an elliptic curve over a field $k$ of characteristic $0$ are classified by $H^1(k,\mu_n)=k^\times/k^{\times n}$ where $n$ is 2, 4, or 6.

2) If $n<8$, the map $k^\times/k^{\times n} \to \prod_v k_v^\times/k_v^{\times n}$ is injective.

[Edit: This answer was edited to simplify the construction and to add those remarks at the end.]

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I wanted to see such an explicit example. Many thanks. –  Chandan Singh Dalawat Dec 30 '09 at 3:19
    
Now there is a standard example. Many thanks for the simplification. –  Chandan Singh Dalawat Dec 31 '09 at 4:36
    
Could you please change Q_17 to {\mathbf Q}_17 ? –  Chandan Singh Dalawat Dec 31 '09 at 4:37
    
The reason I wrote it as Q_17 was that using mathbf there gave me TeX errors. Maybe someone who knows why this is happening can fix it. –  Bjorn Poonen Dec 31 '09 at 19:19

Here's a slight variant of Felipe Voloch's answer, for those who don't have a favorite group cohomology class. Let $C$ be an abelian variety over $\mathbb{Q}$. Suppose that all the $\overline{\mathbb{Q}}$ automorphisms of $C$ are defined over $\mathbb{Q}$ and let $P$ be this automorphism group.

Take two classes in $H^1(\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}), P)$ which are distinct, but become equal in $H^1(\mathrm{Gal}(\overline{\mathbb{Q}_v}/\mathbb{Q}_v), P)$ for every $v$. The corresponding twists of $C$ should give you the examples you want.

How have I made things easier? Because I made the action of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on $P$ trivial, I can describe the group cohmology explicitly as $$H^1(\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}), P) \cong \mathrm{Hom}(\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}), P)/P.$$ Here $P$ acts by conjugation on the target.

Since $P$ is finite, any of these Hom's factor through $\mathrm{Gal}(K/\mathbb{Q})$ for some finite extension $K/\mathbb{Q}$.

So we are now reduced to the following: We must find finite groups $G$ and $P$, an extension $K/\mathbb{Q}$ with Galois group $G$, an abelian variety with automorphism group $P$ and two maps $\alpha$, $\beta: G \to P$ such that

  • $\alpha$ and $\beta$ are not conjugated to each other by any element of $P$ but
  • when we restrict to any decomposition subgroup group, $\alpha$ and $\beta$ become conjugate.

Take $G=(\mathbb{Z}/2)^2$ and $P=S_6 \times (\mathbb{Z}/2)$. We will not use the $(\mathbb{Z}/2)$ factor at all in the following; the reason it is there is that the automorphism group of an abeliabn variety always contains a central involution, namely $-1$. Feel free to think of $P$ as $S_6$.

Take $K/\mathbb{Q}$ to be any biquadratic extension in which no prime is completely ramified. This condition assures that no decomposition group is the whole of $G$. Let $\alpha$ send the generators of $G$ to the elements $(12)(56)$ and $(34)(56)$ of $S_6$. Let $\beta$ send the generators of $G$ to $(12)(34)$ and $(13)(24)$. Then $\alpha$ and $\beta$ are not conjugate in $S_6$, but they become conjugate when restricted to any of the three cyclic subgroups.

The one missing step is to construct an abelian variety with automorphism group $S_6 \times (\mathbb{Z}/2)$, and all automorphisms defined over $\mathbb{Q}$. Dror Spieser, in the comments, points out that we can just take the restriction of scalars of an elliptic curve (without CM) defined over an $S_6$ extension of $\mathbb{Q}$. I still don't have a good construction of this but, thanks to Bjorn's answer, I don't need one.

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How about taking the Weil restriction to the rationals of an elliptic curve over a galois field with galois group S_6? –  Dror Speiser Dec 29 '09 at 19:38
    
@Dror: That doesn't seem to give the right automorphism group: the Weil restriction is geometrically a product of 6 elliptic curves, so its automorphism group contains (Z/2Z)^6. –  Bjorn Poonen Dec 30 '09 at 2:17
    
@David: Saying that no prime is totally ramified in K/Q ensures only that the inertia groups are not all of G. –  Bjorn Poonen Dec 30 '09 at 2:18

If $A,B$ are as stated, then $B$ must be a twist of $A$ which is everywhere locally trivial, so $B$ gives a class in $H^1(k,G)$ (where $G$ is the automorphism group of $A$), which is everywhere locally trivial. So, pick a group $G$ that you know has everywhere locally trivial but globally non-trivial class in $H^1(k,G)$ and make it act on an abelian variety. For instance you can make the group act on a curve and therefore on its jacobian.

As for your actual question, if there is a "standard" such example, I guess the answer is no.

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I was going to post pretty much the same answer, but then got scared by the fact that I couldn't rig things so that the automorphism group was equal to G, only that it contained G, so I wasn't sure if the construction would definitely work. –  Kevin Buzzard Dec 29 '09 at 17:52

Selmer's curve $3X^3+4y^3+5z^3=0$ is a non-example (see the comment below) but somwhat relevant. See Theorem 1 in Mazur's article titled ON THE PASSAGE FROM LOCAL TO GLOBAL IN NUMBER THEORY.

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The pair (Selmer's curve, Jacobian of Selmer's curve) are everywhere locally isomorphic, but not globally so. However, Selmer's curve is not an abelian variety, because it has no rational points. So this is very close to, but not the same as, what was asked for. –  David Speyer Dec 29 '09 at 16:25
    
Yes, that's right. –  Idoneal Dec 29 '09 at 16:39

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