# Smooth proper schemes over rings of integers with points everywhere locally

[Edit: Question 1 has been moved elsewhere 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.

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I cannot for the life of me find a definition of a proper scheme online. Just so I understand why an ordinary variety violating the Hasse principle isn't an answer to this question, could someone provide such a definition? –  Qiaochu Yuan Jan 6 '10 at 0:00
"Proper" is a property of a map, not a scheme. It's the map to Spec(Z) that's proper, not the scheme. It morally means "all fibres compact". The reason an ordinary (projective) variety violating the Hasse principle may not be a counterexample is because the OP wants the map to be smooth too, and smooth over Spec(Z) means good reduction at all primes. –  Kevin Buzzard Jan 6 '10 at 15:56
Part of what is so annoying about this question is that I know very few examples of schemes which are proper and smooth over Spec Z. So I basically keep trying to modify Kevin Buzzard's example, and failing. –  David Speyer Jan 7 '10 at 12:48

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 Q (Cremona's reference 4225m1) has reduction type III at 5 and 13. These become I0* over K=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 Q, and its quadratic twist over K

E: sqrt(65)*U*y^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 (Z/2Z)^4, of which (Z/2Z)^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

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That's very nice. Thank you. –  Chandan Singh Dalawat Jan 11 '10 at 4:06

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.

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My choice of Tate's example was made independently of Chandan's comment above: it's just a coincidence. Anyway, there are plenty of other examples, probably including some which have nontrivial Sha over the ground field. –  Pete L. Clark Jan 8 '10 at 9:03
That's very nice, Pete; the question seems to have been tailor-made for you! I'm tempted to "accept" your answer, but perhaps we should wait for someone to write down an explicit equation of a genus-$1$ curve $C$ over a quadratic field $K$ whose jacobian $J$ has good reduction everywhere and such that $[C]$ is an order-$2$ element of $\operatorname{Sha}(J,K)$. It would have the same appeal as Selmer's example ($3x^3+4y^3+5z^3=0$) or Tate's example ($y^2+xy+\varepsilon^2y=x^3$). –  Chandan Singh Dalawat Jan 8 '10 at 10:42

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.

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You are right; this is the reason why I said I believe the answer is Yes. There are examples in the literature of abelian varieties $A$ over number fields $K$ which have everywhere good reduction, and let's hope someone can find an explicit example where moreover $\operatorname{Sha}(A,K)\neq0$, as you suggest, and write down all the details. –  Chandan Singh Dalawat Jan 8 '10 at 4:46
Tate's example : over $K=\mathbb{Q}(\sqrt{29})$ , the elliptic curve $E:y^2+xy+\varepsilon^2y=x^3$ , where $\varepsilon=(5+\sqrt{29})/2$, has good reduction everywhere because its discriminant is a unit. I don't know whether there is a finite extension $L|K$ with $\operatorname{Sha}(E,L)\neq0$. Another possibility is to start with a genus-$1$ curve $C$ over some number field $K$ which has points everywhere locally but no $K$-points, and look for a finite extension $L|K$ over which the jacobian of $C$ acquires good reduction but $C$ does not acquire an $L$-point. –  Chandan Singh Dalawat Jan 8 '10 at 8:22
@David: By the way, I don't think $\mathbb{Q}(j^{\frac{1}{6}})$ is even roughly correct -- for instance, nothing stops $j$ from being a perfect $6$th power! I know that it is sufficient to trivialize the $3$ and $4$ torsion, but this leads to an extension of quite large degree. Frustratingly, I think I used to know more about this but I am having trouble remembering it now. –  Pete L. Clark Jan 8 '10 at 9:01

If the fibres of the morphism $f: X\rightarrow\mathrm{Spec}(\mathbb{Z})$ have dimension $\leq 1$ the following facts are interesting regarding question 1:

1. By a theorem of Minkowski the field $\mathbb{Q}$ has no unramified extensions. If the fibres of $f$ have dimension $0$ smoothness is the same as being etale thus leading to an unramified extension of $\mathbb{Q}$.

2. A theorem of Fontaine proved in 1985 says that there exist no proper smooth curves over $\mathbb{Q}$ of genus $g\geq 1$ with good reduction everywhere.

Thus the case of a non-rational curve of genus $0$ remains, that is a curve with $g=0$ and without a rational point.

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That doesn't work. If f:C -> Z is a curve of genus zero, then omega_{C/Z}^{-1} has degree 2 and f^*(omega_{C/Z}^{-1}) is a free Z module of rank 3. (Locally free over a general base, but Z is a PID.) So C can be written as a conic in P^2_{Z}. The Hasse principle applies to conics. –  David Speyer Jan 8 '10 at 11:44
Put another way, the only smooth proper curve over $\mathbb{Q}$ which has good reduction everywhere is $\mathbb{P}_1$. –  Chandan Singh Dalawat Jan 8 '10 at 12:18
In fact this was already explained by Poonen in a "motivational comment" to his question linked to above. –  Pete L. Clark Jan 8 '10 at 13:54