How to associate a Dirichlet character to a Tate character?

Question

A Dirichlet character is a multiplicative map from (Z/N)* to $S^1$. A Tate character is a continuous map from I/Q to $S^1$, where I is the Idele group of Q. It is always claimed that they are equivalent but I never see a convincing source.

There is something involved called $F^N/P^N$ where $F^N$ is the group of fraction ideals not involving primes dividing $N$ and $P^N$ is the group of fraction ideals generated by $a$ such that a=1 mod N.

I cannot see why a map on $F^N/P^N$ can be associated to a Dirichlet character, either.

Answer 1

I'll tell you how to move between Dirichlet characters and finite-order Hecke characters (which is what your question seems to be).

Let $\chi:(\mathbf{Z}/N)^\times\rightarrow\mathbf{C}^\times$ be a Dirichlet character. It induces a Dirichlet character $\chi_M:(\mathbf{Z}/M)^\times\rightarrow\mathbf{C}^\times$ for $M$ divisible by $N$ by composing with the canonical projection. You can take the inverse limit $$\widehat{\chi}:=\underset{\longleftarrow}{\lim}\chi_M:\widehat{\mathbf{Z}}^\times=\underset{\longleftarrow}{\lim}(\mathbf{Z}/M)^\times\rightarrow\mathbf{C}^\times.$$ One has the following isomorphism $$I\cong\mathbf{Q}^\times\times\mathbf{R}^\times_{>0}\times\widehat{\mathbf{Z}}^\times$$ given by $$x\mapsto \left(\frac{x_\infty}{|x|},|x|,\frac{x^{(\infty)}}{x_\infty}|x|\right).$$ Note that $\mathbf{Q}^\times\subseteq I$ maps to the factor $\mathbf{Q}^\times$ on the right. Then you can obtain a character $\omega_\chi:I\rightarrow\mathbf{C}^\times$ by composing $\widehat{\chi}$ with the projection to $\widehat{\mathbf{Z}}^\times$ in the above isomorphism. This will give you a character of $I$ trivial on $\mathbf{Q}^\times$ and hence a Hecke character which is finite order by construction.

Conversely, given a finite order Hecke character $\omega:I/\mathbf{Q}^\times\rightarrow\mathbf{C}^\times$ use the above isomorphism to view it as a finite order character of $\mathbf{R}^\times_{>0}\times\widehat{\mathbf{Z}}$. The only finite order character on $\mathbf{R}^\times$ is the trivial character so $\omega$ induces a finite order character $\widehat{\mathbf{Z}}^\times\rightarrow\mathbf{C}^\times$. Since it's finite order it factors through some finite quotient and gives you a Dirichlet character $\chi_\omega:(\mathbf{Z}/N)^\times\rightarrow\mathbf{C}^\times$.

The first section of Hida's book Modular forms and Galois cohomology covers this material.

Added: I just noticed you were also asking for the relation with characters on the group of fractional ideals. The relation is because $(\mathbf{Z}/N)^\times$ can be viewed as the ray class group mod $N(\infty)$ over $\mathbf{Q}$ which is a quotient of the group of fractional ideals prime to $N$. I'd suggest looking at Childress' book Class field theory section 3.2 for a nice exposition of this.

Answer 2

Taking the question in the title literally this is quite easy. Here's how to take a (finite order) Tate character and derive the corresponding Dirichlet character (omitting most details).

Let $\chi:I/\mathbb{Q}\to S^1$ be a finite order character (see Boyarsky's comment). For each prime $p$, $\chi$ restricts to a character $\chi_p$ of $\mathbb{Q}_p^*$. Then $p$ is unramified if $\chi$ is trivial on $\mathbb{Z}_p^*$. All but finitely many $p$ are unramified (why?). Define $\tilde\chi(p)=\chi_p(p)$ for unramfied $p$ and $\tilde\chi(p)=0$ for ramified $p$. Extend $\tilde\chi$ by strict multiplicativity to all positive integers.

The harder part is going the other way :-) For more details see for example Heilbronn's article in Algebraic Number Theory (ed. Cassels and Frohlich).

Answer 3

I think it easier to see this from an adelic point of view. I prefer to use $I =\mathbb{A}^\times$ for the ideles.

Using strong approximation:

$$ \mathbb{A}^\times = \mathbb{Q}^\times \times \prod\limits_p \mathbb{Z}_p^\times \times \mathbb{R}^\times $$

Accordingly, we can factor a Hecke character $\chi$ on $\mathbb{Q}^\times \backslash \mathbb{A}^\times $ to character $\chi_p$ for all $p$ (all but finitely many trivial by Tychonoff's theorem) and $\chi_\infty$.

Now a continuous character $\chi_p$ on $\mathbb{Z}_p^\times$ must factor through $\mathbb{Z}_p^\times /(1 + p^k\mathbb{Z}_p) \cong ( \mathbb{Z} / p^k)^\times$.

In the end, we use the Chinese remainder theorem (this is essentially the approximation property above)

$$ (\mathbb{Z}/N)^\times = \prod\limits_{p^k || N }(\mathbb{Z}/p^k)^\times.$$