Let $F$ be a number field and $\chi:\mathrm{Gal}(\overline{\mathbb{Q}}/F)\to\overline{\mathbb{Q}_\ell}^{\times}$ ($\ell$ a prime) a Galois character. My question is: Can we find a finite extension $K/F$ such that $\chi_{|\mathrm{Gal}(\overline{\mathbb{Q}}/K)}=1$

  • $\begingroup$ this might be what you're looking for $\endgroup$ – Ryan Keleti Jan 16 '19 at 23:08

By Galois theory, your question is equivalent to asking whether all $\ell$-adic Galois characters have finite image. Unlike with complex-valued characters, there are plenty of infinite image $\ell$-adic Galois characters.

The most important example is the $\ell$-adic cyclotomic character. Take $F=\mathbb Q$ and define $\chi$ as follows:

$$ \begin{align} \mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q) &\to \mathrm{Gal}(\mathbb Q(\zeta_{\ell^\infty})\ /\ \mathbb Q)\\ &= \varprojlim_{n}\ \mathrm{Gal}(\mathbb Q(\zeta_{\ell^n})/\mathbb Q)\\ &=\varprojlim_{n}\ (\mathbb Z/\ell^n\mathbb Z)^\times\\ &= \mathbb Z_\ell^\times\subset \mathbb Q_\ell^\times. \end{align} $$

Here $\zeta_{\ell^n}$ is a primitive $\ell^n$-th root of unity, and $\mathbb Q(\zeta_{\ell^\infty})$ is the field obtained by adjoining all $\ell$-power roots of unity. This map is surjective (onto $\mathbb Z_\ell^\times)$, so has infinite image. It only becomes trivial after restriction to $\mathbb Q(\zeta_{\ell^\infty})$, which is an infinite extension.

  • $\begingroup$ Thank you very much for your answer. $\endgroup$ – AZMEH Jan 21 '19 at 23:02

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