TITLE: How to associate a Dirichlet character to a Tate character?
QUESTION [8 upvotes]: 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.

REPLY [3 votes]: 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)
$$ (\mathbb{Z}/N)^\times = \prod\limits_{p^k || N }(\mathbb{Z}/p^k)^\times.$$