We procede by contradiction, {\it i.e.} we assume that for1all cyclic subgroup $C$ of order $p$ of $E(\bar\B Q)$,2the pair $(E,C)$ can be defined over ${\B Q(\sqrt{p})}$.3We choose such a pair $(E_0,C_0)$ over ${\B Q(\sqrt{p})}$.45We show first that such a situation prevents that $j(E)=0$ or6$j(E)=1728$. Indeed in those cases7$E$ has complex multiplication by an order $R_K$ of $K=\B Q[\sqrt{-1}]$ or8$\B Q[\sqrt{-3}]$.9Consider the map $\rho$ :10$\Gal(\bar \B Q/\B Q(\sqrt p))\longrightarrow{\rm Aut}\,E_0(\bar\B Q)[p]$.11If its image had an element of order $p$, so would have12$\rho(\Gal(\bar \B Q/K(\sqrt p)))$. Let $L_p$ be the class field of ray $p$13of $K$. It contains $\B Q(\sqrt p)$ since $p\equiv 1\pmod 4$.14By the theory15of complex multiplication,16$\rho(\Gal(\bar \B Q/L_p))$ is trivial. By class field theory, $\Gal(L_p/K)$17has no element of order $p$, since $K$ has class number $1$. Therefore18$\rho(\Gal(\bar \B Q/\B Q(\sqrt p)))$ has no element of order $p$.19Since it is contained in the Borel subgroup of ${\rm Aut}\,E_0(\bar\B Q)[p]$20which stabilizes $C_0$, it is an abelian group. By the theory of complex21multiplication, it is the semi-direct product of $\Gal(L_p/K(\sqrt p))$ and22$\Gal(K(\sqrt p)/\B Q(\sqrt p))\simeq\Gal(\B C/\B R)$. Such a group23is not abelian since24$\Gal(L_p/K(\sqrt p))$ is not a $2$-group, hence the contradiction.2526We can now assume that all twists of $E$ are quadratic.27We now show that the group28$\Gal(\bar\B Q/{\B Q(\sqrt p)})$ acts by scalars29on the $\B F_p$-vector space $E_0(\bar\B Q)[p]$. For this it30suffices to show that all subgroups of order $p$ of31$E_0(\bar\B Q)[p]$ are stable by $\Gal(\bar\B Q/{\B Q(\sqrt p)})$.3233Let $C_1$ be a cyclic subgroup of order $p$ of $E_0(\bar\B Q)[p]$.34By assumption, there exists a quadratic twist $E_1$ of $E_0$ and35a cyclic subgroup $C_1'$ of $E_1(\bar\B Q)[p]$ of order $p$, such that36the image of $C_1$ by the isomorphism $E_0\simeq E_1$ is $C'_1$.37Since the Galois action on $E_1(\bar\B Q)[p]$ is the one obtained38by twisting by a character the action on $E_0(\bar\B Q)[p]$.39The group $\Gal(\bar\B Q/{\B Q(\sqrt p)})$ leaves $C_1$ stable.40That implies that $\Gal(\bar\B Q/\B Q(\sqrt p))$ acts41by scalars on $E_0(\bar\B Q)[p]$. Denote by ~$\alpha$ the corresponding42character of $\Gal(\bar\B Q/\B Q(\sqrt p))$.4344Because of the Weil pairing, $\alpha^2$ coincides45with the cyclotomic character modulo~$p$, and it factors through46$\Gal(\B Q(\mu_p)/\B Q(\sqrt p))$. But, when47$p\equiv 1\!\!\pmod 4$, the group $\Gal(\B Q(\mu_p)/{\B Q(\sqrt p)})$ is of48even order, and the characters modulo~$p$ form a group generated by the49reduction modulo~$p$ of the cyclotomic character, which, therefore,50can not be a square.5152