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

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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).

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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
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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).

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