This is a follow-up question after [this](https://mathoverflow.net/questions/130868/ramification-in-division-field-of-abelian-varieties)


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 $\zeta\_p\in k\neq k(A[p])$.



Ramification in Division field of Abelian Varieties II
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](http://modular.math.washington.edu/Tables/antwerp/table1/small_82.jpg) (given that it must have odd conductor and nontrivial $2$-torsion).


