3
$\begingroup$

Let $\mathbb{A}_{\mathbb{Q}}^{\times}$ be the ring of ideles over the rational $\mathbb{Q}$. Let $\{ \chi : \mathbb{Q}^\times \backslash \mathbb{A}_{\mathbb{Q}}^{\times} : \chi_\infty = 1, \chi^n =1 \} $ be the set of Hecke characters of order $n$ with trivial infinity part.

I want to whether they are finite in number corresponding bijectively to the Dirichlect characters $(\mathbb{Z}/n\mathbb{Z})^\times \to \mathbb{C}^\times$.

Improve this question
$\endgroup$
1
  • $\begingroup$ You must have misunderstood something. The claim as stated isn't correct. $\endgroup$
    – Fan Zheng
    Commented Aug 1, 2017 at 8:24

1 Answer 1

Reset to default
7
$\begingroup$

The answer is obviously no, for cardinality reasons.

Just take $n=2$. Then there is only one homomorphism $(\mathbb{Z}/2\mathbb{Z})^\times \to \mathbb{C}^\times$, however there are many Hecke characters of order $2$. (e.g. take $$\widehat{\mathbb{Z}}^\times \to \mathbb{Z}_p^\times \to \{\pm 1\}, \quad x_p \mapsto \left(\frac{x_p \bmod p}{p}\right) $$ given by the Legendre symbol.)

The correct relationship between Dirichlet characters and Hecke characters is given by matching up the ramification of the Hecke characters with the conductor of the Dirichet character. See here for further info: How to associate a Dirichlet character to a Tate character?

Improve this answer
$\endgroup$
1
  • 3
    $\begingroup$ @user49908: Note also that $\chi_\infty$ is only trivial if the associated primitive Dirichlet character is even. If the associated primitive Dirichlet character is odd, then $\chi_\infty$ is the sign character $\mathbb{R}^\times\to\{\pm 1\}$. You can find a concise account over number fields here (see pages 7-8): arxiv.org/abs/1402.1332 $\endgroup$
    – GH from MO
    Commented Aug 1, 2017 at 17:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .