2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $\mathbf Z_\ell$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.