Let K be a global function field, L an extension of K and $N_{L|K}$ the divisor norm given by $N(P)=p^{f(P|p)}$ for prime divisors and extended linearly. Denote by $\mathcal{D}_{K}$ the divisor group of $K$. For an effective divisor $m\in \mathcal{D}_{K}$ let $D^{m}:=\{D\in\mathcal{D}_{K}: supp(D)\cap suppp(m)=\emptyset\}$ be the group of divisors with support disjoint to $m$ and $\mathcal{P}_{m}:=\{div_{K}(f)\in\mathcal{P}_{K}| f\in K^{\times}, f\equiv1 \text{ mod } p^{ord_{p}(m)} \text{ in } \mathcal{O}\text{ for all places $p$ of $K$} \}$ where $\mathcal{P}_{K}$ denotes the group of principle divisors of $K$. The ray class group is then defined by $Cl_{m}:=\mathcal{D}^{m}/\mathcal{P}_{m}$. Let $\hat m$ be the conorm of $m$, where the conorm is defined by $conorm(p):=\sum_{P|p}e(P|p)*P$ and is extended linearly. It is stated that the divisor norm is a well defined map from $Cl_{\hat m}$ to $Cl_{m}$.
It is not clear to me why $N_{L|K}$ takes $\mathcal{P}_{\hat m}$ to $\mathcal{P}_{m}$. I know that $N_{L|K}(div_{L}(f))=div_{K}(N_{L|K}(f))$ where the second norm symbol is the usual field norm of $L|K$. So it remains to show that $div_{K}(N_{L|K}(f))\in\mathcal{P}_{m}$. If someone could help me out? A proof for the number field case would also be enough since its probably the same argument.