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

There is no doubt that such examples as in David Speyer's response exist: indeed, they exist in great abundance in the following sense:

Let $k_1$ be any number field, and let $E_{/k_1}$ be any elliptic curve with integral $j$-invariant. Then it has potentially good reduction, meaning that there is a finite extension $k_2/k_1$ such that $E_{/k_2}$ is the generic fiber of an abelian scheme over $\mathbb{Z}_{k_2}$. Furthermore, let $N$ be your favorite integer which is greater than $1$. Then there exists a degree $N$ field extension $k_3/k_2$ such that the Shafarevich-Tate group of $E_{/k_3}$ has an element of order $N$ (in fact, one can arrange to have at least $M$ elements of order $N$ for your favorite positive integer $M$): see Theorem 3 of

http://math.uga.edu/~pete/ClarkSharif2009.pdf

Since good reduction is preserved by base extension, the genus one curve $C_{/k_3}$ corresponding to the locally trivial principal homogeneous space of $E_{/k_3}$ of period $N$ gives an affirmative answer to Question 2.

Specific examples of elliptic curves over quadratic fields with everywhere good reduction are known: see e.g. the survey paper

http://mathnet.kaist.ac.kr/pub/trend/shkwon.pdf

where the following example appears and is attributed to Tate:

$E: y^2 + xy + \epsilon^2 y = x^3, \ \epsilon = \frac{5+\sqrt{29}}{2}$,

has everywhere good reduction over $k = \mathbb{Q}(\sqrt{29})$. Indeed, the given equation is smooth over $\mathbb{Z}_k$, since the discriminant is $-\epsilon^{10}$ and $\epsilon$ is a unit in $\mathbb{Z}_k$.

If this elliptic curve happens itself to have nontrivial Sha, great. If not, the theoretical results above imply that a quadratic extension of it will have a nontrivial $2$-torsion element of Sha, i.e., there will exist some hyperelliptic quartic equation

$y^2 + p(x)y + q(x) = 0$

with $p(x), q(x)$ in the ring of integers of some quadratic extension $K$ of $\mathbb{Q}(\sqrt{29})$, which is smooth over $\mathbb{Z}_K$ and violates the local-global principle.

If someone is interested in actually computing the equation, I would say a better strategy is searching for elliptic curves defined over quadratic fields with everywhere good reduction until you find one which already has a 2-torsion element in its Shafarevich-Tate group. (I don't see how to guarantee this theoretically, but I would be surprised if it were not possible.) Then it is easy to write down the defining equation.

Regarding question 2, does the following work? Let $E$ be a rational elliptic curve with integer $j$-invariant. Then there is a number field $K$ so that $E \times_{\mathbb{Q}} K$ has a smooth model over $\mathcal{O}_K$. Roughly, $\mathbb{Q}(j^{1/6})$ should work, but there might be some subtleties at 2 and 3. If Sha of $E \times_{\mathbb{Q}} K$ is nontrivial, then I think an element of Sha should correspond to a torsor for $E \times_{\mathbb{Q}} K$ with the required property.

I don't understand the elliptic curve tables well enough to know how to search them for an example like this, but presumably one of our readers does.