[**Edit:** Question 1 has been moved [elsewhere](https://mathoverflow.net/questions/11316/smooth-proper-schemes-over-z-with-points-everywhere-locally) so that an answer to Question 2 can be *accept*ed.]


**Question 2.** *Is there a number field* $K$, *and a smooth proper scheme* $X\to\operatorname{Spec}(\mathfrak{o})$ *over its ring of integers, such that* $X(K\_v)\neq\emptyset$ *for every place* $v$ *of* $K$, *and yet* $X(K)=\emptyset$ ?


I believe the answer is Yes.


**Remark.** Let $K$ be a real quadratic field, $\mathfrak{o}$ the ring of integers of $K$, and $A$ the quaternion algebra over $K$ which is ramified exactly at the two real places. Then the conic $C$ corresponding to $A$ is a smooth projective $\mathfrak{o}$-scheme such that $C(\mathfrak{o})=\emptyset$ (because $C(K\_v)=\emptyset$ for each of the real places $v$). But if we insist that $C(K\_v)\neq\emptyset$ at these two real places $v$, then $A$ would have to split at these $v$ (in addition to all the finite places), and we would have $C=\mathbb{P}\_{1,\mathfrak{o}}$.


More generally, let $K$ be a number field, $\mathfrak{o}$ its ring of integers, and let $C$ be a smooth proper $\mathfrak{o}$-scheme whose generic fibre $C\_{K}$ is a twisted $K$-form of the projective space of some dimension $n>0$. If $C$ has points everywhere locally, then $C=\mathbb{P}\_{n,\mathfrak{o}}$. This remark shows that $X$ cannot be a twisted form of a projective space.



Smooth proper schemes over rings of integers with points everywhere locally
Chandan asked Vladimir and me for an example of an elliptic curve over a real quadratic field that has everywhere good reduction and non-trivial sha, with an explicit genus $1$ curve representing some element of sha. Here's one we found:


The elliptic curve $y^2+xy+y = x^3+x^2-23x-44$ over $\mathbb Q$ (Cremona's reference 4225m1) has reduction type III at 5 and 13. These become I0\* over $K=\mathbb Q(\sqrt{65})$, and I0\* can be killed by a quadratic twist. Specifically, the original curve can also be written as $y^2 = x^3+5x^2-360x-2800$ over $\mathbb Q$, and its quadratic twist over $K$.


$E: \sqrt{65}Uy^2 = x^3+5x^2-360x-2800$


has everywhere good reduction over $K$; here $U = 8+\sqrt{65}$ is the fundamental unit of $K$ of norm $-1$. 


Now 2-descent in Magma says that the 2-Selmer group of $E/K$ is $(\mathbb Z/2\mathbb Z)^4$, of which $(\mathbb Z/2\mathbb Z)^2$ is accounted by torsion. So it has either has rank over K or non-trivial Sha[2], and according to BSD its rank is 0 as L(E/K,1)<>0 (again in Magma). Actually, because $K$ is totally real, I think results like those of Bertolini and Darmon might prove that E has Mordell-Weil rank $0$ over $K$ unconditionally. So it has non-trivial Sha[2]. After some slightly painful minimisation, one of its non-trivial elements corresponds to a homogeneous space 


$C: y^2 = (23562U+1462)x^4 + (4960U+240)x^3 + (1124U-291)x^2 + (141U-833)x + (50U-733)$


with $U$ as above. So here is a curve such that $J(C)$ has everywhere good reduction and the Hasse principle fails for C.


Hope this helps!


Tim


