Hecke Characters

My question today concerns Hecke characters or Größencharaktere. I'm doing a project on complex multiplication of elliptic curves and need some help understanding Hecke characters. My main references are Silverman's advanced topics on elliptic curves and Ireland-Rosen Classical Intro to Number Theory.

In Silverman, [II.9.2](see below for print of the theorem) gives us a Hecke character for the base field [see (a)]. (b) mentions how the character is related to the curve. But the rest of the way, Silverman did not make use of the property that the character is unramified at $\mathfrak{P}$ [see (b)]. So it seems that characters are defined only on the base field and has nothing to do with the curve, which would be silly, wouldn't it? Although $\alpha_{E/L}$ may yet have a hand in bringing the curve in.
So these are my questions:

1. How are Hecke characters defined for each curve wrt to a certain ground field?
2. How does (b) contribute to how we find the character?
3. What are the invariants on the Hecke character that makes it unique to a certain $E/K$?

Hope these questions make sense and a BIG thank you!

Theorem 9.2. Let $E/L$ be an elliptic curve with complex multiplication by the ring of integers $R_K$ of $K$, assume that $K\subset L$, and let $\alpha_{E/L}: \mathbf{A}^{*}_{L}\rightarrow K^*$ be the map such that

(i) $\alpha R_K=(s)$, where $(s)\subset K$ is the ideal of $s$.
(ii) For any fractional ideal $\mathfrak{a}\subset K$ and any analytic isomorphism $$f:\mathbb{C}/\mathfrak{a}\rightarrow E(\mathbb{C})$$ the following diagram commutes:

$$\begin{array}[c]{ccc} K/\mathfrak{a}&\stackrel{\alpha s^{-1}}{\longrightarrow}&K/\mathfrak{a}\\ \downarrow\scriptstyle{f}&&\downarrow\scriptstyle{f}\\ E(L^{\text{ab}})&\stackrel{[x,L]}{\longrightarrow}&E(L^{\text{ab}}) \end{array}.$$

For any idele $s\in\mathbf{A}^{*}_{L}$, let $s_{\infty}\in \mathbb{C}^*$ be the component of $s$ correspoinding to the unique archimedean absolute value on $K$. Define the map $$\psi_{E/L}: \mathbf{A}^{*}_{L}\rightarrow \mathbb{C}^*,\quad\psi_{E/L}(x)=\alpha_{E/L}(x)N^L_K(x^{-1})_{\infty}$$ (a) $\psi_{E/L}$ is a Hecke character of L.
(b) Let $\mathfrak{P}$ be a prime of $L$. Then $\psi_{E/L}$ is unramified at $\mathfrak{P}$ iff $E$ has good reduction at $\mathfrak{P}$.

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You may find this resource helpful: www2.imperial.ac.uk/~tsg/Index_files/… In particular, see Theorem 2.41 on pg 12. – Jeff H Jul 3 '13 at 15:33
Hi Jeff, I don't immediately see the link(not the URL). That resource is pitched a little too beyond me I'm afraid. Do you mind explaining it if you have the time? Thank you for your comment though. – BlackAdder Jul 3 '13 at 16:16
Ah, yes, those notes are dealing in a much more general setting (automorphic representations) than the one you care about (elliptic curves). Sorry about that. I don't have time to write something myself at the moment, and in any case, there are people here much more qualified than me to do so. I'm sorry that link wasn't more helpful! – Jeff H Jul 3 '13 at 16:21
No worries Jeff, I'm just grateful to have any sort of help. Thanks again! One last thing, pg169, Proposition 4.1 of Cassels-Frohlich's Algebraic Number Theory seems like it has something to do with my question, but I'm not too sure. – BlackAdder Jul 3 '13 at 16:31

Let $\upsilon$ be a prime ideal of $L$. We denote the completion of $L$ at $\upsilon$ with $L_\upsilon$ and the integer ring with $R_\upsilon$. Recall $A^\ast_L = \{\alpha = (\alpha_\upsilon) \in \Pi_\upsilon L^\ast_\upsilon \shortmid \alpha_\upsilon \in R^\ast_\upsilon,$ almost for all $\upsilon \}$. We define as follows;

$\psi_{E/L}$ is unramified at $\upsilon$ $=_{def}$ $\psi_{E/L}(R^\ast_\upsilon) = 1$.

Thus, (b) means

$\psi_{E/L}(R^\ast_\upsilon) = 1$ iff $E$ has good reduction at $\upsilon$.

I dare to state that (b) doesn't contribute to find the character.

I guess you confuse the unramification of a character with the unramification of a prime ideal. I recommend you to read Tate’s paper Fourier analysis in number fields and Hecke's zeta-functions. You can find it on the net.

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