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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.

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    $\begingroup$ Counterexample: take an elliptic curve $y^2=x(x^2-d)$ with $d \equiv 1 \pmod 4$ over the rationals and $m=2$. $\endgroup$
    – Felipe Voloch
    Commented May 16, 2013 at 20:31
  • $\begingroup$ 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 $d$. $\endgroup$
    – Sungjin Kim
    Commented May 16, 2013 at 20:50
  • $\begingroup$ Maybe I should have put, $m$ avoids all primes of bad reduction. $\endgroup$
    – Sungjin Kim
    Commented May 16, 2013 at 21:09

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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 $p$.

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  • $\begingroup$ Thank you for clear counterexample. I'd like to know what happens when $m$ is a power of $p$. $\endgroup$
    – Sungjin Kim
    Commented May 16, 2013 at 20:43

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