0
$\begingroup$

Let $K$ be a imaginary quadratic field, and $E/K$ be elliptic curve which has CM over $K$. Let $ψ_E$ be Hecke(Grossencharacter) character of $E/K$. Let fix prime ideal $I$ of $K$. Then, why $ψ_E(I)$ is prime element of $K_I$(K's completion at $I$) ?

Back ground : I heard formal group of $E$, $ \hat{E}$ is Lubin Tate for prime element $ψ(I)$, but it seems not clear to me why it is prime element of local field $K_I$.

Improve this question
$\endgroup$
12
  • 1
    $\begingroup$ Hit: if $h$ is the class number of $K$, then what can you say about $\psi(I)^h$? $\endgroup$
    – David Loeffler
    Commented Aug 25, 2022 at 6:13
  • 1
    $\begingroup$ In this situation, automatically $h＝1$, so it is just $ψ(I)$. $\endgroup$
    – Duality
    Commented Aug 25, 2022 at 15:25
  • $\begingroup$ Could you give me one more hint ? $\endgroup$
    – Duality
    Commented Aug 26, 2022 at 3:06
  • $\begingroup$ Sorry, I meant to let $h$ be the order of the ray class group modulo the conductor $f$ of $\psi$. Then $I^h$ is trivial in the ray class group, so $I^h = (\lambda)$ for some $\lambda \in K^\times$ with $\lambda = 1 \bmod f$. Hence $\psi(I)^h = \lambda$. So what can you say about the valuation of $\psi(I)$ at the prime $I$? Or at any other prime? $\endgroup$
    – David Loeffler
    Commented Aug 26, 2022 at 6:08
  • $\begingroup$ I want to prove is $ord_I(ψ(I))＝0$. $ord_I(ψ(I)^h)＝ord_I(λ)$ so $h・ord_I(ψ(I))＝ord_I(λ)$. On the other hand, $h＝ord_I(λ)ord_I(a)$ for some ideal $a$ thus $ord_I(ψ(I))＝1/ord_I(a)$ , but this cannot be $0$ so I'm confused. $\endgroup$
    – Duality
    Commented Aug 29, 2022 at 8:11

0

Reset to default

You must log in to answer this question.