Existence of Elements of ${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}$ of all Prime Orders

Proposition 7.1   Let $p$ be a prime number. Then Conjecture 4.1 implies that there exists infinitely many twisted powers $A$ of some elliptic curve such that ${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})[p]\neq \{0\}$.

Proof. Most of the proposition can be proved using a single elliptic curve. When ordered by conductor, the first elliptic curve $E$ over $\mathbb{Q}$ with positive rank has prime conductor $37$ and is defined by the Weierstrass equation $y^2 + y = x^3 - x$. Table 1 of [5] shows that $E$ is isolated in its isogeny class, so [9, Exercise 4] implies that representations $\rho_{E,p}$ are irreducible. Since $\ord_{37}(j(E))=-1$, $\overline{c}_{37}=1$. Thus all odd primes $p\neq 37$ are rigid for $E$. The proposition then follows for all odd primes $p\neq 37$ by Theorem 6.1.

We complete the proof as follows. Exactly the same argument applied to the unique elliptic curve of conductor $43$ proves the proposition for all odd primes $p\neq 43$. Finally, Bölling proved in [2] that for every $j\in\mathbb{Q}$ there is an elliptic curve $E$ with $j$-invariant $j$ such that infinitely many twists $E'$ of $E$ have ${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(E'/\mathbb{Q})[2]\neq\{0\}$. $\qedsymbol$