[Edit: Question 1 has been moved elsewhere so that an answer to Question 2 can be accepted.]

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.

  • $\begingroup$ 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? $\endgroup$ Jan 6, 2010 at 0:00
  • 1
    $\begingroup$ "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. $\endgroup$ Jan 6, 2010 at 15:56
  • $\begingroup$ 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. $\endgroup$ Jan 7, 2010 at 12:48

4 Answers 4


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!



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


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


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.

  • $\begingroup$ 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. $\endgroup$ Jan 8, 2010 at 9:03
  • $\begingroup$ 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 ($y^2+xy+\varepsilon^2y=x^3$). $\endgroup$ Jan 8, 2010 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.

  • $\begingroup$ 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. $\endgroup$ Jan 8, 2010 at 4:46
  • $\begingroup$ 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. $\endgroup$ Jan 8, 2010 at 8:22
  • $\begingroup$ @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. $\endgroup$ Jan 8, 2010 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.

  • $\begingroup$ 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. $\endgroup$ Jan 8, 2010 at 11:44
  • $\begingroup$ Put another way, the only smooth proper curve over $\mathbb{Q}$ which has good reduction everywhere is $\mathbb{P}_1$. $\endgroup$ Jan 8, 2010 at 12:18
  • 1
    $\begingroup$ In fact this was already explained by Poonen in a "motivational comment" to his question linked to above. $\endgroup$ Jan 8, 2010 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.