1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ this might be what you're looking for $\endgroup$ – Ryan Keleti Jan 16 '19 at 23:08
2
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.