# Elliptic curve E and Galois representation

Assume that an elliptic curve $E$ over $\Bbb Q$ has a reducible mod $p$ representation. Then

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## Q: How can we find a curve $E'$ isogenous to $E$ such that $E'$ has a point of order $p$ as well as two independent rational point of order $2$.

This occurs in checking that for prime $p > 5$, the mod $p$ representation $$\rho_{E_{abc},p} \colon {\mathrm{Gal}}(\overline{{\Bbb Q}}/{\Bbb Q}) \to {\mathrm{GL}}_2({\Bbb F}_p)$$ is irreducible, where $E_{abc}$ is Frey-Hellegouarch curve $E_{abc} \colon y^2 = x(x - a^p)(x + b^p)$ associated to triple $a,b,c \in {\Bbb Z}$ such that $a \equiv 3 (4), b \equiv 0 (2)$, and $abc \not=0$.

To show this, assume to the contrary that $\rho_{E_{abc},p}$ is reducible, then the semi-simplification of $E_{abc}[p]$ should be the direct sum of ${\Bbb Z}/p{\Bbb Z}$ and $\mu_p$. Then, we have $E'$ isogeneous to $E_{abc}$ having the property that $E'$ has a rational point of order $p$ as well as two independent rational point of order $2$, which will be shattered by Mazur's theorem. Thus we see $\rho_{E_{abc},p}$ is irreducible as desired.

I cannot understand this reasoning, which caused my questions. Pierre

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Could you provide more context for your questions? For example, does "look at connected components" suffice for first question? If $p>3$, and the point of order $p$ in the second question is rational, then Mazur's classification of torsion subgroups says that $E'$ doesn't exist. –  S. Carnahan Jul 3 '14 at 12:56
Yes, this is because I would like to show that the Serre's mod $l$ representation $$– Pierre Jul 3 '14 at 15:30 ## 2 Answers For your first question, if your elliptic curve is semistable, then what you write is true (and fortunately for your application Frey curves are semistable). To see this, if E[p] is reducible, we have an exact sequence:$$ 0 \to A \to E[p] \to B \to 0  where $A$ and $B$ are just isomorphic to ${\bf F}_p$, but stable under the action of $G_{\bf Q}$. Let $\chi_A$ and $\chi_B$ be the corresponding ${\bf F}_p^\times$-valued characters giving the action on $A$ and $B$ respectively. Since the determinant of the Galois representation on $E[p]$ is $\omega$, the mod $p$ cyclotomic character, we have $\chi_A \cdot \chi_B = \omega$. We can then write $\chi_A = \omega \chi$ and $\chi_B = \chi^{-1}$.

Now I claim that the character $\chi$ is unramified away from $p$ in the semistable case. Indeed, for $q \neq p$, if $q$ doesn't divide the conductor of $E$, then the representation is unramified at $q$ and we are fine. If $q$ divides the conductor of $E$, since $E$ is semistable, $E$ is a Tate curve at $q$. In particular, the Galois representation $E[p]$ locally at $q$ has an unramified quotient. Thus, either $\omega \chi$ or $\chi$ is unramified at $q$ which implies $\chi$ is unramified at $q$ (since $\omega$ is unramified at $q$).

But we can also apply this argument at $q = p$ as $E$ is still a Tate curve at $p$. In this case, we see that either $\chi$ or $\omega \chi$ is unramified at $p$. (This argument is essentially Lemma 6 in Serre's article: Propriétés galoisiennes des points d'ordre fini des courbes elliptiques.)

In the first case, $\chi$ is unramified everywhere and thus trivial. In the second case, $\omega \chi$ is unramified everywhere and $\chi = \omega^{-1}$. Thus $( \chi_A, \chi_B)$ equals either $(1, \omega)$ or $(\omega,1)$ and $E[p]$ is an extension of $\mu_p$ by ${\bf Z}/p{\bf Z}$ or vice-versa.

On to your second question: the 2-torsion part is easy to handle in the case of Frey curves. Their explicit Weierstrass equation visibly gives two independent points of order 2. For the point of order $p$, if either $E[p]$ is a split extension or if ${\bf Z}/p{\bf Z}$ is a sub of $E[p]$ then we are immediately done without applying an isogeny. If $E[p]$ is a non-split extension with $\mu_p$ as the sub, then simply consider the isogenous curve $E' = E/\mu_p$. Then $E'[p]$ contains ${\bf Z}/p{\bf Z}$ as sub and we are done.

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Thanks Robert!! Pierre –  Pierre Jul 13 '14 at 10:27

Both questions have incorrect expectations. This has already been noted for the second question by S. Carnahan in the comments. For the first question, take any elliptic curve for which the semisimplification of $E[p]$ is $\mathbb{Z}/p\mathbb{Z} \oplus \mu_p$ and twist $E$ by a quadratic character $\chi\colon \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \rightarrow \{\pm 1\}$ to arrive at a curve $A$ over $\mathbb{Q}$ with $A[p]$ reducible, yet $A[p]^{ss} \cong \chi \oplus \mu_p \chi$.

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But $\mu_p$ (connected) and $\mathbb{Z}/p$ (etale) are the only group schemes of order $p$ over $\mathbb{Q}$, so $\mu_p$ has to be the connected component of the identity of $E[p]$, and the connected-etale sequence for $E[p]$ has to read: $0 \to \mu_p \to E[p] \to \mathbb{Z}/p \to 0$. (This picture continues to hold over $\mathbb{Z}_p$ when $p$ is a prime of good reduction for $E$.) –  Vesselin Dimitrov Jul 3 '14 at 13:19
The only finite connected group scheme over $\mathbb{Q}$ is $0$. In particular, $\mu_p$ over $\mathbb{Q}$ (as over any field of characteristic different from $p$) is not connected, but instead is etale (e.g., base change to $\mathbb{Q}(\zeta_p)$ to see this). –  Kestutis Cesnavicius Jul 3 '14 at 13:44
My apology, and thanks for correcting me: indeed $\mu_p$ and $\mathbb{Z}/p$ become isomorphic over $\mathbb{Q}(\zeta_p)$. What I had in mind was the connectedness of $\mu_p$ over $\mathbb{Z}_p$. Given this, am I right in saying that when $E$ has good reduction at $p$, the connected-etale sequence for $E[p]_{/\mathbb{Z}_p}$ is $0 \to \mu_p \to E[p] \to \mathbb{Z} / p \to 0$? –  Vesselin Dimitrov Jul 3 '14 at 13:56
I think there is still room for twisting by characters unramified at $p$: over $\mathbb{Z}_p^{nr}$, the connected-etale sequence will be as you claim, but not necessarily so over $\mathbb{Z}_p$ (for $p \neq 2$). E.g., the etale part over $\mathbb{Z}_p$ is $\mathbb{Z}/p\mathbb{Z}$ iff the Frobenius action on the $p$-torsion of the reduction is trivial, and you can twist by an unramified quadratic character to make it nontrivial. For $p = 2$ what you claim is true because $\mathbb{Z}/2\mathbb{Z}$ has no nontrivial forms. –  Kestutis Cesnavicius Jul 3 '14 at 14:11