# Increase in rank of elliptic curves

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 ($$p\neq 2$$)?
• 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. Feb 8 '19 at 21:13
• 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. Feb 8 '19 at 21:15

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.