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I expect answers to these questions are known, or at least partial answers are known:

  1. Let $E$ be a rank 0 elliptic curve defined over $\mathbb{Q}$ and let $p$ be an odd prime. Is it possible that the rank of $E$ does not go up in any degree $p$ extension?
  2. Are there rank 0 elliptic curves, for which the rank always goes up in a degree $p$ extension ($p\neq 2$)?
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    $\begingroup$ 1. No for $p=3$. Write $E$ as smooth cubic. Any generic line defined over $\mathbb{Q}$ meets $E$ in three points defined over a cubic extension. They cannot all be torsion points so there is a cubic field for which $E$ has positive rank. $\endgroup$ Feb 8 '19 at 21:13
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    $\begingroup$ 2. Look at papers by Hershy Kisilevsky and co-authors. For $p>7$ we expect only finitely many cyclic extensions of degree $p$ where the rank grows. $\endgroup$ Feb 8 '19 at 21:15
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Expanding on Chris's answer, let $E/K$ be an elliptic curve defined over a number field. If you embed $E$ using the linear system $|n(O)|$ with $n\ge3$, you'll get $E$ as a smooth curve of degree $n$ in $\mathbb P^{n-1}_K$. Taking the intersection of $E$ with a generic hyperplane $H$ defined over $K$, for most choices of $H$ (in a Hilbert irreducibility sense), you should get $E\cap H=\{P_1,\ldots,P_n\}$ with $K(P_1)/K$ an extension of degree $n$, with $P_1$ non-torsion, and indeed, with $P_1$ independent from the points in $E(K)$. Filling in the details would prove:

Theorem Let $E/K$ be an elliptic curve defined over a number field. Then for every $n\ge1$ there exists an extension $L/K$ with $[L:K]=n$ and $\operatorname{rank}E(L)\ge\operatorname{rank}E(K)+1$.

Of course, this doesn't contradict Chris's second comment, since that refers to extensions $L/K$ that are Galois with cyclic Galois group.

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