Everywhere locally isomorphic abelian varieties 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 ?
 A: 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.
A: (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.]
A: 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.
A: 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.
