There are infinitely many elliptic curves over any number field with rank 1, and for any quadratic extension L/K, infinitely many curves satisfy rank E(K) = rank E(L) = 1.
An elliptic surface with infinitely many fibers for which the rank does not jump
1 Pith paper cite this work. Polarity classification is still indexing.
abstract
Let $E$ be a nonisotrivial elliptic curve over $\mathbb{Q}(T)$ and denote the rank of the abelian group $E(\mathbb{Q}(T))$ by $r$. For all but finitely many $t\in \mathbb{Q}$, specialization will give an elliptic curve $E_t$ over $\mathbb{Q}$ for which the abelian group $E_t(\mathbb{Q})$ has rank at least $r$. Conjecturally, the set of $t\in\mathbb{Q}$ for which $E_t(\mathbb{Q})$ has rank exactly $r$ has positive density. We produce the first known example for which $E_t(\mathbb{Q})$ has rank $r$ for infinitely many $t\in\mathbb{Q}$. For our particular $E/\mathbb{Q}(T)$ which has rank $0$, we will make use of a theorem of Green on $3$-term arithmetic progressions in the primes to produce $t\in\mathbb{Q}$ for which $E_t$ has only a few bad primes that we understand well enough to perform a $2$-descent.
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Rank one elliptic curves and rank stability
There are infinitely many elliptic curves over any number field with rank 1, and for any quadratic extension L/K, infinitely many curves satisfy rank E(K) = rank E(L) = 1.