# A question about Galois characters

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 $$\chi_{|\mathrm{Gal}(\overline{\mathbb{Q}}/K)}=1$$

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

## 1 Answer

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.

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