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.



Why does $[I](P)＝0$ ($P\in E$) imply $[\psi(I)](P)＝0$ ? ($\psi$ is Hecke character of elliptic curve)
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](https://beta.lmfdb.org/EllipticCurve/Q/49/a/4); 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.


