Property of a Galois representation associated to an elliptic curve

In the paper "Fermat's Last Theorem" by H. Darmon, F. Diamond, R. Taylor (2007), theorem 2.15 says that for $$l>3$$ the Galois representation $$\rho_{E.l}$$ is irreducible.

My question in general: Is the Galois representation always not irreducible at $$l =2$$ or $$3$$, for all elliptic curves?

If it is not, let $$E :y^2=x^3+ax+b$$ where $$ab \in \mathbb Q$$ (and we know the value of $$ab$$, not all elliptic curve). What are the steps to prove that the Galois representation for this specific curve is irreducible at $$l=2$$?

For an elliptic curve over $$\Bbb{Q}$$ or any field of characteristic $$0$$

$$x^3+ax+b = (x-e_1)(x-e_2)(x-e_3)$$ then since $$-(x,y) = (x,-y)$$ we see $$(\infty,\infty),(e_1,0),(e_2,0),(e_3,0)$$ is in the 2-torsion and it is the whole of it so $$\rho_{E,2}$$ (the $$GL_2(\Bbb{F}_2)$$ action of $$G_\Bbb{Q}$$ on $$E \cong \Bbb{Z/2Z}\times \Bbb{Z/2Z}$$) is the Galois group of $$x^3+ax+b$$'s splitting field. There are 4 cases : $$x^3+ax+b$$ splits completely over $$\Bbb{Q}$$, or it is a quadratic times a linear, or it is irreducible with $$-4a^3-27b^2$$ not a square, or irreducible with $$-4a^3-27b^2$$ a square.

• In the two latter cases the Galois group acts transitively on the points of order $$2$$ , thus on the 3 subspaces of $$\Bbb{F}_2^2$$, so $$\rho_{E,2}$$ is irreducible over $$\Bbb{F}_2$$. When the discriminant is a square the splitting field is a degree 3 cyclic extension $$Gal(K/\Bbb{Q}) = \{1,\sigma,\sigma^2\}$$ and (identifying $$e_j$$ with the point $$(e_j,0) \in E$$) $$E = e_1\Bbb{F}_2+e_2\Bbb{F}_2,e_1+e_2=e_3,\rho_{E,2}(\sigma)= \pmatrix{0 & 1 \\ 1 & 1}$$ and the characteristic polynomial of this matrix is $$x(x+1)+1=(x-\zeta_3)(x-\zeta_3^2)$$ which is separable thus it is conjugate to a diagonal matrix over $$\Bbb{F}_2(\zeta_3)$$ and $$\rho_{E,2}$$ is reducible over $$\Bbb{F}_2(\zeta_3)$$.

• When $$-4a^3-27b^2$$ is not a square then $$Gal(K/\Bbb{Q})\cong S_3$$, it contains an element of order 2 sending $$e_1$$ to itself and permuting $$e_2$$ and $$e_3=e_1+e_2$$ so its matrix is $$\pmatrix{1 & 0 \\ 1 & 1}$$ and the representation is irreducible over $$\overline{\Bbb{F}}_2$$.

• When $$x^3+ax+b= (x+cx+d)(x-e_1)$$ then the representation factorizes through $$Gal(\Bbb{Q}(\sqrt{c^2-4d}/\Bbb{Q})= \{1,\sigma\}$$ $$\rho_{E,2}(\sigma) = \pmatrix{1 & 0 \\ 1 & 1}$$ this matrix is not conjugate to any diagonal matrix thus $$\rho_{E,2}$$ is irreducible over $$\overline{\Bbb{F}}_2$$.

• When $$x^3+ax+b$$ splits completely then the representation is trivial and reducible.

• so if any elliptic curve have point of order 2 and it is speliting field is irreducible with $-4a^3-27b^3$ than $\rho_{E.2}$ is irreducible , i mean this work for all elliptic curve or for weistrass form just – Abdallahchaibeddrra Oct 9 at 18:48
• hmm ? We are dealing with elliptic curve over $\Bbb{Q}$ (or any field of characteristic 0) the 2 torsion always exists – reuns Oct 9 at 18:57
• "irreducible over $Q$" ?? $E$ is a $\Bbb{F}_2$ vector space, the representation is to send elements of $Gal(\overline{\Bbb{Q}}/\Bbb{Q})$ to matrices in $GL_2(\Bbb{F}_2)$. – reuns Oct 9 at 19:21
• can i get your email i want to talk with you more about this subject – Abdallahchaibeddrra Oct 9 at 19:30
• no feel free to add your thoughts in comments, or in your question or in an answer to it – reuns Oct 9 at 19:32