Let $$U(N)=\{x \in \widehat{\mathbb{Z}}^\times \mid x\equiv 1 \bmod N\mathbb{Z} \}$$
In the isomorphism in the Lemma below, why does $p_p \in \mathbb{Q}_p^\times$ go to $p \bmod N$ if $p$ is not a factor of $N$? This is from Hida's book Modular Forms and Galois cohomology.
I think $p_p$ means the adele $(x_q)_q$ where $x_q=1$ if $q \neq p$ and $x_q=p$ if $q=p$. Then $rat(x)=p$ and so $u=(u_q)_q$ where $u_q=p^{-1}$ if $q\neq p$ and $u_q=1$ if $q=p$. But I don't know what element of $(\mathbb{Z}/N\mathbb{Z})^\times$ this gets sent to.
There is an issue of convention here.
As you have seen, $p_p \in \Bbb A^\times$ corresponds to the element $p \times (p^{-1}, \dots, 1_p) \times p^{-1}$ in $\Bbb Q^\times \times U(1) \times \Bbb R_+^\times$.
Therefore, if we define the isomorphism $\Bbb A^\times / (\Bbb Q^\times \times U(N) \times \Bbb R_+^\times) \rightarrow U(1) /U(N)$ simply as the projection to the $U(1)$ component in $\Bbb A^\times$, then the image of $p_p$ goes to $p^{-1} \mod N$.
Alternatively you may define the isomorphism as the inverse of the above isomorphism. Then the image of $p_p$ is just $p\mod N$. This seems to be what Hida uses, but he definitely wrote it in a very confusing way.
These two conventions correspond to the two different conventions of class field theory, i.e. by sending a uniformizer to an arithmetic or a geometric Frobenius.