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Let $E$ be an elliptic curve, $K$ a number field so that $\operatorname{rank}_{K}(E)\geq 1,$ $p$ a prime of $\mathbb{Q}$ at which $E$ reduces well.

We know (see for instance Silverman, The Arithmetic of Elliptic Curves, Proposition VIII.1.5(b)) that $K(p^{-1}E(K))/K$ is unramified outside the set consisting of the primes of bad reduction, the primes dividing $p$ and the archmedean places.

My question is towards the reverse: Is there always a place $v$ that is either infinite or of bad reduction and ramifies in $K(p^{-1}E(K))/K$?

It would suffice for my purposes if such a place $v$ existed for all but finitely many $p.$

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    $\begingroup$ If $\nu$ is a place of (bad) multiplicative reduction, then adjoining the $p$-torsion (which comes with adjoining all $p$-division) is already ramified for $p$ large. In the terms of the Tate parametrization, you are adjoining the $p$-th root of $q$ and $v(q) > 0$. $\endgroup$
    – Felipe Voloch
    Commented Aug 8, 2016 at 18:24
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    $\begingroup$ Even when there is no place of multiplicative reduction: Infinitely many $p$ would give an infinite unramified abelian extension of $K$ which is impossible by class field theory. $\endgroup$
    – Chris Wuthrich
    Commented Aug 9, 2016 at 8:29
  • $\begingroup$ @ChrisWuthrich: I am looking for bad or infinite places; if $p$ is a good prime then $K(p^{-1}E(K))/K$ would still be ramified at $p.$ $\endgroup$
    – The Thin Whistler
    Commented Aug 9, 2016 at 12:25
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    $\begingroup$ If $p>2$ and $K$ has a real place $\mathfrak r$, then since the $p$-division points are rational over the extension, the $p$-th roots of unity will be in the extension too (from the $e_p$-pairing), so $\mathfrak r$ must ramify. $\endgroup$
    – Lubin
    Commented Aug 27, 2016 at 14:51

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