3
$\begingroup$

For a CM elliptic curve $E$ and its Grössencharakter, their conductors are both supported on bad primes of $E$. Moreover, by comparing their functional equation, there should be some obvious relations. I think this is how to deduce the functional equation for a CM elliptic curve from the one for its Grössencharakter.

So what's the exact relation between their conductors?

P.S. I guessed this should be easy to find, but it is not so.

P.P.S. By checking functional equations, I guess there should be a square relation. But this is not a proof.

P.P.P.S. Hey guys, I find a proof in the famous Serre-Tate Good reduction of abelian varieties, theorem 12, Page 514. Indeed it is Theorem 6. And the relation should be exactly square

$\endgroup$
2
  • 8
    $\begingroup$ You need to be a little careful about the field. Let $E/F$ have CM, where $F=\mathbb Q(j(E))$, let $K$ be the CM field, and let $L=FK$. Then $L(E/F,s) = L(s,\psi_{E/F})$, while $L(E/L,s)=L(s,\psi_{E/L})L(s,\overline{\psi_{E/L}})$. $\endgroup$ Oct 30, 2015 at 16:09
  • 1
    $\begingroup$ Why the downvote? $\endgroup$
    – Fan Zheng
    Nov 27, 2015 at 0:22

1 Answer 1

7
$\begingroup$

Here is a completely overkill explanation that nevertheless answers the question.

The question of the relation between the conductor of a CM elliptic curve and its associated Grössencharakter is related to something much stronger, namely automorphic induction.

Given a quadratic extension of number fields $E/F$ and a Hecke character $\chi$ of $E^{\times} \backslash \mathbb{A}_E^{\times}$ such that $\chi$ does not factor through the global norm map $N_{\mathbb{A}_E / \mathbb{A}_F} \colon \mathbb{A}_E^{\times} \to \mathbb{A}_F^{\times}$, there exists a cuspidal automorphic representation $\pi$ of $\mathrm{GL}_2(\mathbb{A}_F)$ that is automorphically induced from $\chi$. In particular, their $L$-functions are the same: \[L(s,\pi) = L(s,\chi).\] The central character of $\pi$ is given by $\omega_{\pi} = \chi \vert_{\mathbb{A}_F} \omega_{E/F}$, where $\omega_{E/F}$ is the quadratic Hecke character of $F^{\times} \backslash \mathbb{A}_F^{\times}$ of conductor $\mathfrak{d}_{E/F}$, the discriminant of $E/F$, associated to the quadratic extension $E/F$ via global class field theory. In particular, if $\pi$ has trivial central character $\omega_{\pi}$, then $\chi \vert_{\mathbb{A}_F} = \omega_{E/F}$.

Via a purely local argument, one can describe the local components of $\pi$ in terms of the local components of $\chi$. In particular, if $\mathfrak{F}$ is the conductor of $\chi$ (viewed as an integral ideal of the ring of integers $\mathcal{O}_E$ of $E$), then the conductor $\mathfrak{q}$ of $\pi$ (viewed as an integral ideal of $\mathcal{O}_F$) is given by \[\mathfrak{q} = \mathfrak{d}_{E/F} N _{E/F}(\mathfrak{F}),\] where $N_{E/F} \colon E^{\times} \to F^{\times}$ is the norm map. So if $\pi$ has trivial central character, then $\mathfrak{d}_{E/F}$ divides $N _{E/F}(\mathfrak{F})$, and consequently $\mathfrak{d}_{E/F}^2$ divides $\mathfrak{q}$.

In the case of a CM elliptic curve, by the modularity theorem this corresponds to a cuspidal automorphic representation of $\mathrm{GL}_2(\mathbb{A}_{\mathbb{Q}})$ with the same conductor, so the above discussion shows the relation between the conductor of the associated Grössencharakter.

$\endgroup$
3
  • $\begingroup$ This is Congling, :D $\endgroup$
    – user42690
    Oct 30, 2015 at 17:04
  • $\begingroup$ I thought it was you! $\endgroup$ Oct 30, 2015 at 17:04
  • $\begingroup$ Cool answer, but I think we can solve it without modularity theorem $\endgroup$
    – user42690
    Oct 30, 2015 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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