CoCalc Public Fileswww / papers / merel2 / notes_v5
Author: William A. Stein
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1We procede by contradiction, {\it i.e.} we assume that for
2all cyclic subgroup $C$ of order $p$ of $E(\bar\B Q)$,
3the pair $(E,C)$ can be defined over ${\B Q(\sqrt{p})}$.
4We choose such a pair $(E_0,C_0)$ over ${\B Q(\sqrt{p})}$.
5
6We show first that such a situation prevents that $j(E)=0$ or
7$j(E)=1728$. Indeed in those cases
8$E$ has complex multiplication by an order $R_K$ of $K=\B Q[\sqrt{-1}]$ or
9$\B Q[\sqrt{-3}]$.
10Consider the map $\rho$ :
11$\Gal(\bar \B Q/\B Q(\sqrt p))\longrightarrow{\rm Aut}\,E_0(\bar\B Q)[p]$.
12If its image had an element of order $p$, so would have
13$\rho(\Gal(\bar \B Q/K(\sqrt p)))$. Let $L_p$ be the class field of ray $p$
14of $K$. It contains $\B Q(\sqrt p)$ since $p\equiv 1\pmod 4$.
15By the theory
16of complex multiplication,
17$\rho(\Gal(\bar \B Q/L_p))$ is trivial. By class field theory, $\Gal(L_p/K)$
18has no element of order $p$, since $K$ has class number $1$. Therefore
19$\rho(\Gal(\bar \B Q/\B Q(\sqrt p)))$ has no element of order $p$.
20Since it is contained in the Borel subgroup of ${\rm Aut}\,E_0(\bar\B Q)[p]$
21which stabilizes $C_0$, it is an abelian group. By the theory of complex
22multiplication, it is the semi-direct product of $\Gal(L_p/K(\sqrt p))$ and
23$\Gal(K(\sqrt p)/\B Q(\sqrt p))\simeq\Gal(\B C/\B R)$. Such a group
24is not abelian since
25$\Gal(L_p/K(\sqrt p))$ is not a $2$-group, hence the contradiction.
26
27We can now assume that all twists of $E$ are quadratic.
28We now show that the group
29$\Gal(\bar\B Q/{\B Q(\sqrt p)})$ acts by scalars
30on the $\B F_p$-vector space $E_0(\bar\B Q)[p]$. For this it
31suffices to show that all subgroups of order $p$ of
32$E_0(\bar\B Q)[p]$ are stable by $\Gal(\bar\B Q/{\B Q(\sqrt p)})$.
33
34Let $C_1$ be a cyclic subgroup of order $p$ of $E_0(\bar\B Q)[p]$.
35By assumption, there exists a quadratic twist $E_1$ of $E_0$ and
36a cyclic subgroup $C_1'$ of $E_1(\bar\B Q)[p]$ of order $p$, such that
37the image of $C_1$ by the isomorphism $E_0\simeq E_1$ is $C'_1$.
38Since the Galois action on $E_1(\bar\B Q)[p]$ is the one obtained
39by twisting by a character the action on $E_0(\bar\B Q)[p]$.
40The group $\Gal(\bar\B Q/{\B Q(\sqrt p)})$ leaves $C_1$ stable.
41That implies that $\Gal(\bar\B Q/\B Q(\sqrt p))$ acts
42by scalars on $E_0(\bar\B Q)[p]$. Denote by ~$\alpha$ the corresponding
43character of $\Gal(\bar\B Q/\B Q(\sqrt p))$.
44
45Because of the Weil pairing, $\alpha^2$ coincides
46with the cyclotomic character modulo~$p$, and it factors through
47$\Gal(\B Q(\mu_p)/\B Q(\sqrt p))$. But, when
48$p\equiv 1\!\!\pmod 4$, the group $\Gal(\B Q(\mu_p)/{\B Q(\sqrt p)})$ is of
49even order, and the characters modulo~$p$ form a group generated by the
50reduction modulo~$p$ of the cyclotomic character, which, therefore,
51can not be a square.
52