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Let $K$ be a imaginary quadratic field, $R_K$ be ring of integers of $K$, and $E/K$ be elliptic curve which has CM over $K$. Let $\psi_E$ be Hecke (Grössencharakter) character of $E/K$. Let fix prime ideal $I＝(\pi)$ of $K$. Then, why does $[I](P)＝0$ ($P\in E$) imply $[\psi(I)](P)＝0$?

If I could write $\psi(I)$ like $\psi(I)＝a\pi$ for some $a\in R_K$, $[\psi(I)](P)＝[a][\pi](P)＝0$, but I'm not confident.

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This is essentially the same as your other recent CM-theory question, in a mild disguise; for both questions the point is that $\psi(I)$ is a generator of $I$. This follows easily from the fact that $\psi$ takes values in $K$, and for principal ideals $(\lambda)$ with generators sufficiently congruent to 1 we have $\psi(\ (\lambda)\ ) = \lambda$.

Perhaps an example would be helpful. Consider the elliptic curve of conductor 49, with Weierstrass equation $y^2 + xy = x^3 - x^2 - 2x - 1$. (This is 49a1 in Cremona's tables, and 49a4 in the LMFDB; it also happens to be the modular curve $X_0(49)$, but that isn't important here.)

This curve has CM by the ring of integers $R_K = \mathbb{Z}[(1 + \sqrt{-7})/2]$ of $\mathbb{Q}(\sqrt{-7})$, and one can write down the corresponding Groessencharacter explicitly:

For $\mathfrak{a}$ any integral ideal of $R_K$ coprime to the ideal $\mathfrak{f} = \sqrt{-7} R_K$, the value $\psi(\mathfrak{a})$ is the unique generator of $\mathfrak{a}$ whose image in $(R_K / \mathfrak{f})^\times \cong (\mathbb{Z} / 7\mathbb{Z})^\times$ is a square.

Note this makes sense, because $R_K$ has class number 1 and its only units are $\pm 1$; so every $\mathfrak{a}$ has exactly two generators, say $x$ and $-x$, and as $-1$ is not a square mod 7, precisely one of these is a square.

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  • $\begingroup$ In the proof of $ψ(I)$ is a generator of $I$, could you tell me why you chose ray class group mod conductor $f$ of $ψ$ ? Another mod ray class group does not work ? $\endgroup$
    – Duality
    Commented Sep 5, 2022 at 16:46
  • $\begingroup$ Well, obviously it wouldn't work, because it wouldn't give the information we need. We know what $\psi$ looks like on ideals with generators of a particular kind; and those are precisely the ideals having trivial image in the ray class group mod the conductor of $\psi$. $\endgroup$
    – David Loeffler
    Commented Sep 5, 2022 at 18:02
  • $\begingroup$ Let $h$ be a order of ray class group mod conducor of $ψ$. In the process $p^h=(λ),ψ(p)^h=ψ(p^h)=ψ((λ))=λ$, where did you used $λ≡１mod f$? $\endgroup$
    – Duality
    Commented Sep 6, 2022 at 5:19
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    $\begingroup$ Otherwise how would one justify the step $\psi( (\lambda) ) = \lambda$? $\endgroup$
    – David Loeffler
    Commented Sep 6, 2022 at 8:31
  • $\begingroup$ What about when $E$ is defined over $L⊃K$ ? $ψ(p)$ is not in general prime element of $R_K$ ? $\endgroup$
    – Duality
    Commented Dec 30, 2022 at 1:02

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