Currently I am reading the book, Fermat’s Last Theorem written by Darmon, Diamond and Taylor. (You can find this pdf online http://www.math.mcgill.ca/darmon/pub/Articles/Expository/05.DDT/paper.pdf) In proposition 2.8 part (b), it says if one has an elliptic curve defined over $\mathbf{Q}$, for a prime number $\ell$, consider the $\ell$-adic representation attached to the tate module of this elliptic curve: $$\rho_{E,l}:G_{\mathbf{Q}}\rightarrow GL_{2}(\mathbf{Z}_{\ell})$$ Then $\rho_{E,\ell}$ is absolutely irreducible for all $\ell$. The proof of this statement refers to Serre's book 'Abelian l-adic Representations and Elliptic Curves'. However, in that book, it only shows the irreducibility for an elliptic curve without complex multiplication. I am wondering the idea of proving the irreducibility in complex multiplication case; any references would be appreciated.


An elliptic curve over $\mathbf Q$ cannot have complex multiplication (defined over $\mathbf Q$). It's possible for a rational elliptic curve to have extra endomorphisms, but these will only be defined over a finite extension.

But let's instead take an elliptic curve $E$ over a number field $K$ with complex multiplication. Then the associated Galois representation is reducible*!

Indeed, if $$\rho_{E,\ell}:G_K\to \mathrm{GL}_2(\overline{\mathbf Q}_\ell)$$ is the associated $\ell$-adic representation, then its not too hard to check that $$\mathrm{End}(E)\otimes_\mathbf Z\overline{\mathbf Q}_\ell\hookrightarrow\mathrm{End}(\rho_{E,\ell}).$$

In particular, if $\mathrm{End}(E)\ne \mathbf Z$, then $\mathrm{End}(\rho_{E,\ell})$ is not a field, so $\rho_{E,\ell}$ is reducible. Its subrepresentations are one dimensional Galois representations, which by class field theory, correspond to the Grossencharacters of $E$.

In fact the above map is an isomorphism (by Faltings' theorem). So if $\mathrm{End}(E) = \mathbf Z$, then $\mathrm{End}(\rho_{E,\ell})$ is a field, so $\rho_{E,\ell}$ is irreducible.

If $E$ does not have complex multiplcation over $K$, but obtains extra endomorphisms over a finite extension, then the above argument shows that $\rho_{E, \ell}$ is irreducible. However, $\rho_{E, \ell}$ will not be surjective. By Mackey theory, since $\rho_{E, \ell}$ is irreducible, but $\rho_{E, \ell}|_{G_L}$ is reducible for some $L$, we find that $\rho_{E, \ell}$ is induced from a character of a quadratic extension. In particular, its image cannot be $\mathrm{GL}_2(\mathbf Z_\ell)$.

*By reducible, I mean that it becomes reducible over the algebraic closure $\overline{\mathbf Q}_\ell$. It may still be irreducible over $\mathbf Z_\ell$.

  • $\begingroup$ Thanks for the answer. Maybe there is a terminology confusion here. By 'defined over $\mathbf{Q}$', I mean the elliptic curve has a Weierstrass model whose coefficients are in $\mathbf{Q}$. So in this case, I think an elliptic curve defined over $\mathbf{Q}$ could have complex multiplication, e.g $y^2=x^3-x$ whose $\mathrm{End}(E)=\mathbf{Z}[i]$. Also, could you explain why $End(\rho_{E,l})$ is not a field. i.e why $\mathrm{End}(E)\otimes\overline{\mathbf Q}_\ell$ could not be $\overline{\mathbf Q}_\ell$. $\endgroup$ – Yudong Qiu Feb 14 at 15:43
  • $\begingroup$ In the case of $y^2 = x^3-x$, the isogeny which corresponds to $i$ is given by $(x,y) \mapsto (-x, iy)$. As a map of algebraic curves, this map is not defined over $\mathbf Q$. In that sense, $E/\mathbf Q$ does not have CM, but $E/\mathbf Q(i)$ does (one often says that $E$ has potential CM). $\endgroup$ – Mathmo123 Feb 14 at 17:12
  • $\begingroup$ For your second question, you can think dimensionally: if $\mathrm{End}(E)\otimes\mathbf Q$ is an $n$-dimensional $\mathbf Q$ vector space, then $\mathrm{End}(E)\otimes\overline{\mathbf Q}_\ell$ will be an $n$-dimensional $\overline{\mathbf Q}_\ell$ vector space. $\endgroup$ – Mathmo123 Feb 14 at 17:16
  • $\begingroup$ The potential CM vs CM distinction is somewhat contentious, and many authors define a curve to be CM if it has extra endomorphisms defined over $\overline{\mathbf Q}$. However, from the perspective of Galois representations, the distinction is important: CM reps are reducible, potential CM reps are irreducible. $\endgroup$ – Mathmo123 Feb 14 at 17:18
  • $\begingroup$ Maybe I see your confusion? The tensor product is taken over $\mathbf Z$. So $\mathbf Z[i]\otimes_\mathbf Z \overline{\mathbf Q}_\ell$ is $\overline{\mathbf Q}_\ell ^2$. $\endgroup$ – Mathmo123 Feb 14 at 17:20

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