7
$\begingroup$

It's well known that there are no elliptic curves over Spec $\mathbb{Z}$, but it's unclear (to me at least) if the proof generalizes.

My question is: If $S$ is a connected scheme such that has every prime as a residue characteristic, then can there exist an elliptic curve over $S$?

Improve this question
$\endgroup$
2
  • $\begingroup$ @KestutisCesnavicius Of course! Thank you for pointing that out! Well that was resolved quickly. $\endgroup$
    – Will Chen
    Sep 23, 2015 at 22:34
  • 1
    $\begingroup$ @KestutisCesnavicius If you'd like to post that as an answer I can accept it and nobody will have to read my stupid question anymore :-D $\endgroup$
    – Will Chen
    Sep 23, 2015 at 22:52

3 Answers 3

Reset to default
15
$\begingroup$

Yes, there can. Choose any elliptic curve $E$ over $\mathbb{Q}$ with potential good reduction (for instance, a curve with potential CM) and pass to a number field $K$ over which the reduction is everywhere good. Then $E_K$ extends to an elliptic curve over the ring of integers $\mathcal{O}_K$, and the latter has all primes as residue characteristics.

Improve this answer
$\endgroup$
19
$\begingroup$

More interesting is the question of which fields admit what sorts of elliptic curves having everywhere good reduction. For example, how about quadratic fields? Here are a couple of results.

Theorem (Rohrlich [1]) Let $K$ be an imaginary quadratic field and $j$ the invariant of some fixed isomorphism class of elliptic curves with complex multiplication by the ring of integers of $K$. Put $F=\mathbb Q(j)$ and $H=K(j)$. Then there is an elliptic curve with invariant $j$, defined over $F$, which has good reduction at every place of $F$ if and only if the discriminant $−d$ of $K$ is divisible by at least two primes congruent to 3 modulo 4.

Theorem (Setzer [2]) There are no elliptic curves having good reduction everywhere over $\mathbb Q(\sqrt{-m})$ for $m=1,2,3,5,6,7,10,13,14,15,17,21,22,\ldots$ .

  1. Rohrlich, David E., Elliptic curves with good reduction everywhere. J. London Math. Soc. (2) 25 (1982), no. 2, 216–222.
  2. Setzer, Bennett, Elliptic curves over complex quadratic fields. Pacific J. Math. 74 (1978), no. 1, 235–250.
Improve this answer
$\endgroup$
7
$\begingroup$

Another perspective: Let $K$ be any number field and let $E$ be any elliptic curve over $K$ with $j$-invariant in $\mathcal{O}_K$. Then there is an extension $L$ of $K$ such that $E \times_K L$ extends to an elliptic curve over $\mathcal{O}_L$.

Proof: By properness of the stack $\overline{M}_{1,1}$, there is some $L$ so that we get a stable genus one curve with one marked point over $\mathrm{Spec}\ \mathcal{O}_L$. We just need to rule out the possibility that some fibers are nodal. If the fiber over a prime $\pi$ of $\mathcal{O}_L$ is nodal, than a computation with Tate curves shows that $v_{\pi}(j)<0$, contradicting that $j \in \mathcal{O}_L$. $\square$

I think it is worth bringing this up, as well as the CM solutions above, because there are many more integral $j$-invariants than there are $j$-invariants with CM.

Improve this answer
$\endgroup$
1
  • 7
    $\begingroup$ That's a fancy way to do it. :) The fact that elliptic curves with integral $j$-invariant have everywhere good reduction over an extension field (which is iff) follows easily if char $\mathfrak p\ge5$ using the Weierstrass eqn, and can even be done for 2 and 3 with more work. There's also the Serre-Tate (Neron-Ogg-Shafarevich) criterion which says good reduction iff the action of Galois on appropriate torsion is unramified. So adjoining some torsion gives a field of everywhere good reduction. In particular, I think you can take your $L$ to be something like $K(E[15])$. $\endgroup$
    – Joe Silverman
    Sep 24, 2015 at 13:49

Your Answer

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

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