First question on MathOverflow, I hope it is appropriate for this site. There are two related questions.


Let $K$ be a number field, $G\_K = Gal(\overline{K}/K)$, $p$ a prime, and
$$\chi\_1,\chi\_2:G\_K\rightarrow\overline{\mathbb{Q}}\_p^\times$$
be continuous characters such that $\ker(\chi\_1) = \ker(\chi\_2)$. 


My first question is, are the two fields generated by the values of $\chi\_1$ and $\chi\_2$ the same? That is, does 
$$\mathbb{Q}\_p(\chi\_1(\sigma), \sigma\in G\_K) = \mathbb{Q}\_p(\chi\_2(\sigma),\sigma\in G\_K)$$
As far as I can tell, this is true if $\chi\_1,\chi\_2$ have finite image, but I'm not sure if it's true if they have infinite image.


My second question is, does there exists $\sigma\in G\_{\mathbb{Q}\_p} = Gal(\overline{\mathbb{Q}}\_p/\mathbb{Q}\_p)$ such that for all $\tau\in G\_K$, $\chi\_1(\tau) = \sigma(\chi\_2(\tau))$?


If $E = \mathbb{Q}\_p(\chi\_1(\sigma), \sigma\in G\_K) = \mathbb{Q}\_p(\chi\_2(\sigma),\sigma\in G\_K)$, then one could define $\sigma$ by the rule $\sigma(\chi\_2(\tau)) = \chi\_1(\tau)$, show $\sigma\in Gal(E/\mathbb{Q}\_p)$, and take an extension of $\sigma$ to an element of $Gal(\overline{\mathbb{Q}}\_p/\mathbb{Q}\_p)$. Then the answer to the second question would be yes if the answer to the first question is yes.


I don't know how to show that $\sigma$ defined as above is a well defined element of $Gal(E/\mathbb{Q}\_p)$.



If two Hecke characters cut out the same field, are they Galois conjugates?
The answer to both your questions is "no".


Take $K = \mathbf{Q}$. Then there is a unique $\mathbf{Z}\_p$-extension of $K$ (contained in $\mathbf{Q}(\zeta\_{p^\infty})$) which gives us a surjection $G\_K \to \Gamma$ where $\Gamma$ is isomorphic to $\mathbf{Z}\_p$.


Now, what are the continuous characters $\mathbf{Z}\_p \to \overline{\mathbf{Q}}\_p^\times$? It turns out that for every $a \in \overline{\mathbf{Q}}\_p^\times$ with $|a - 1| < 1$, there's a unique character sending $1 \in \mathbf{Z}\_p$ to $a$. If $a$ is not a root of unity, then this character is an injection.


So pick two values $a, a'$ which aren't roots of unity, don't generate the same extension of $\mathbf{Q}\_p$, and aren't Galois-conjugate. Then this gives you two continuous characters $\chi, \chi': G\_K \to \overline{\mathbf{Q}}\_p^\times$ with the same kernel which are the required counterexamples.


With a little more work you can even find counterexamples with characters that are locally algebraic (and thus arise from algebraic Groessencharacters).


