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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 $Gal(E/\mathbb{Q}_p)$.

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  • $\begingroup$ what sort of characters do you have in mind? What is the topology on ${\overline {\mathbb Q}}^*$? Is it discrete? With respect to this topology, do you want the characters to be continuous? $\endgroup$
    – Venkataramana
    Commented Apr 20, 2016 at 23:17
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    $\begingroup$ Note that your notion of "Hecke character" is non-standard. The standard notion is a continuous complex-valued character of the idele class group of $K$. A subset of these true Hecke characters can be identified, by class field theory, with the continuous complex-valued characters of $G_K$, which have finite image automatically. $\endgroup$
    – GH from MO
    Commented Apr 20, 2016 at 23:30
  • $\begingroup$ @GHfromMO thanks for the comment, I realize the mistake I made. I'm rewriting the question in order to address this. I'm using the definition in terms of ideals instead the idele class group. $\endgroup$
    – Will Dukeminier
    Commented Apr 21, 2016 at 1:28

1 Answer 1

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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).

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  • $\begingroup$ Beautiful, this clears up a lot for me, thanks so much! $\endgroup$
    – Will Dukeminier
    Commented Apr 21, 2016 at 20:50

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