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. 



Ramification in Division field of Abelian Varieties 
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 $p$.


