Equivalent Definitions of Ideal Norm Let $A \subseteq B$ be Dedekind domains, with $K \subseteq L$ their quotient fields with $L/K$ finite and separable.  If $J$ is a fractional ideal of $B$, then $$ J = \prod\limits_{\mathfrak B} \mathfrak{B}^{v_{\mathfrak B} J} $$ where the product runs through the prime ideals of $B$, $v_{\mathfrak B}$ is the valuation obtained from the discrete valuation ring $B_{\mathfrak B}$ (normalized by setting $v_{\mathfrak B}(\varpi) = 1$ for a generator $\varpi$ of $\mathfrak B B_{\mathfrak B}$), and $v_{\mathfrak B}(J) = Min \{ v_{\mathfrak B}(j) : j \in J\}$.  
Let $P = \mathfrak B \cap A$ for each $\mathfrak B$.  Serge Lang defines the norm $N_{L/K}(J)$ to be $\prod\limits_{\mathfrak B} P^{v_{\mathfrak B}(J^{f(\mathfrak B/P)})}$, where $f(\mathfrak B/P)$ is the dimension $[B/\mathfrak B : A / P]$.  
On the other hand, Frohlich in Algebraic Number Theory gives a different definition.  For each prime ideal $P$ of $A$, $J_P = J(A \setminus P)^{-1}$ and $B_P$ are free $A_P$ modules and contain a basis for $L/K$, so there is a $K$-module isomorphism $\phi_P: L \rightarrow L$ for which $\phi_P(B_P) = J_P$.  The determinant $det(\phi_P) \in K$ is unique up to a unit in $A_P$, so the principal ideal $det(\phi_P)A_P$ is the same regardless of the choice of $\phi_P$.  The norm $N_{L/K}(J)$ is then defined as $$ \prod\limits_{P} P^{v_P(det(\phi_P))} $$ How does one prove these definitions are equivalent?
 A: I thought I dare to share some thoughts on the problem:
Define $\displaystyle N_{L/K}(J):=\prod_{\mathfrak{B}}\mathfrak{B}^{v_p(J)f(\mathfrak{B}/P)}$ and $\displaystyle N_{L/K}'(J):=\prod_{P}P^{v_P(det(\phi_p))}$.


*

*Both $N_{L/K}$ and $N_{L/K}'$ are homomorphisms $Id(B) \rightarrow Id(A)$ where $Id$ denotes the ideal groups (This requires some work in the case of $N_{L/K}'$). As $Id(B)$ is the free abelian group over the prime ideals, it suffices to see that $N_{L/K}$ and $N_{L/K}'$ coincide on a prime ideal $\mathfrak{P} \in Id(B)$ over $P \in Id(A)$.

*Consider the exponent of $N_{L/K}'$. The $K$-vector space isomorphism $\phi_P:L \rightarrow L$ restricts to an $A_P$-module homomorphism $\phi_P:B_P \rightarrow B_P$ with image $\mathfrak{P}(A\setminus P)^{-1}= \mathfrak{P}B_P$. 

*As $A_P$ is a PID, we can apply the Smith Normal Form to $\phi_P$ to get $e_1 \mid \dots \mid e_n \in A_P$ such that $\phi$ is represented by a diagonal matrix with diagonal entries $e_1,\dots,e_n$. Hence, $det(\phi_P)A_p = (\prod_ie_i)A_p$. Also by the structure theorem $$B_P/\mathfrak{P}B_P =coker \,\phi\cong \prod_{i=1}^nA_P/e_iA_P.$$

*As the left side is already an $A_P/PA_P$-module, so is the right side (compare annihilators of $A_P$-modules) and the above isomorphism is an isomorphism of $A_P/PA_P$-vector spaces. Thus for every $i$ either $e_iA_P=A_P$  or $e_iA_P=PA_P$. By comparing $A_P/PA_P$-dimensions we get  $$v_P((det \phi_P)) = v_P(\prod e_iA_P) = \dim(\prod_{i=1}^nA_P/e_iA_P) = \dim(B_P/\mathfrak{P}B_P)=f(\mathfrak{P}B_P\mid PA_P).$$
I suspect that $f(\mathfrak{P}B_P\mid PA_P)=f(\mathfrak{P} \mid P)$ which would give the desired result $N_{L/K}'(\mathfrak{P}) = P^{f(\mathfrak{P}\mid P)}$ but I am not sure. Of course it is perfectly possible that I made plenty of mistakes on the way.
