MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 might be a very simple question, and that might be the reason that I could not find any reference on this.

My question is

Let $A$ be an abelian variety defined over a number field $k$, and $N$ the conductor. Let $m\geq 2$. Consider the division field $k(A[m])$. Let $\mathfrak{p}$ be a prime ideal in $k$ that divides $m$. Also suppose that $k(A[m])\neq k$. Then is it true that $$\mathfrak{p} \textrm{ is ramified in } k(A[m]) ? $$

If this is not true, then can anyone provide a counterexample?

I know that if $\mathfrak{p}$ does not divide $mN$, then it should be unramified.

share|cite|improve this question
Counterexample: take an elliptic curve $y^2=x(x^2-d)$ with $d \equiv 1 \pmod 4$ over the rationals and $m=2$. – Felipe Voloch May 16 '13 at 20:31
Thank you. This is also very clear. So $\mathbb{Q}(E[2])=\mathbb{Q}(\sqrt{d})$, and primes ramify in $\mathbb{Q}$ are precisely the prime divisors of $d$. – i707107 May 16 '13 at 20:50
Maybe I should have put, $m$ avoids all primes of bad reduction. – i707107 May 16 '13 at 21:09
up vote 2 down vote accepted

What if $m=pq$ with $\mathfrak p \mid p$ and $p\ne q$ and $k(A[p])=k$? Then the $p$-torsion doesn't cause ramification since its defined over $k$, and the $q$-torsion won't cause $\mathfrak p$ ramification (assuming $A$ has good reduction at the primes lying over $p$ and $q$).

It gets more interesting if you assume that $m$ is a power of $p$.

share|cite|improve this answer
Thank you for clear counterexample. I'd like to know what happens when $m$ is a power of $p$. – i707107 May 16 '13 at 20:43

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.