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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

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    $\begingroup$ 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
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    $\begingroup$ Why the downvote? $\endgroup$
    – Fan Zheng
    Nov 27, 2015 at 0:22

1 Answer 1

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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.

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  • $\begingroup$ 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

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