MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a follow-up question after this

The set-up is almost the same as before,

Let $k$ be a number field, $p$ be a rational prime. Let $A$ be an abelian variety over $k$ which has a good reduction at all primes $\mathfrak{p}\subset k$ lying over $p$.

Suppose also that $k(A[p])\neq k$. Then

Does there exist a prime $\mathfrak{p}\subset k$ lying over $p$ such that $\mathfrak{p}$ is ramified in $k(A[p])$?

An easy case is when $\zeta_p=\exp(\frac{2\pi i}{p})\notin k$, since in this case, $p$ is totally ramified in $\mathbb{Q}(\zeta_p)$.

So, I am interested in the case when $\zeta_p\in k\neq k(A[p])$.

share|cite|improve this question
up vote 2 down vote accepted

F.Voloch's $p=2$ counterexample $y^2 = x(x^2-d)$ with $d \equiv 1 \bmod 4$ still works. Yes, the curve has bad reduction at $2$, but it has potential good reduction, so it will work over some number field $k$.

An explicit $p=2$ example over ${\bf Q}$ is the curve $[1,1,1,0,0]$, a.k.a. $X_1(15): y^2+xy+y=x^3+x^2$, whose $2$-torsion field is ${\bf Q}(\sqrt{-15})$. This didn't take very long to find because it's the first candidate in the Antwerp tables (given that it must have odd conductor and nontrivial $2$-torsion).

share|cite|improve this answer
In F.Voloch's example, what if the finite extension $k$ contains $\mathbb{Q}(\sqrt d)$? – i707107 Jul 4 '13 at 17:10
Does there exist a finite extension $k$ which allows good reduction everywhere, and $k$ doesn't contain $\sqrt d$? – i707107 Jul 4 '13 at 17:11
By the way, the example you gave is great! – i707107 Jul 4 '13 at 17:14

I think not, at least if you look at a single prime $\mathfrak{p}$. The point is that, unless all the p-torsion points of the reduction of $A$ are rational over the residue field, the extension $k(A[p])/k$ will not be totally ramified at the prime. So by replacing $k$ with a suitable ramified extension, you should be able to make $k(A[p])/k$ unramified and nontrivial (at least sometimes).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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