Showing $B_P$ is a finitely generated module over $A_P$ where $P$ is a prime ideal in a Dedekind domain. Suppose we have $A$ a Dedekind domain, $K$ its field of fractions. Let $L$ be a field extension of $K$ of degree $n$ and also that it is separable. 
Let $B$ be the integral closure of $A$ in $L$. 

Let $P$ be a prime ideal in $A$. I am having trouble seeing why $B_P := (A \backslash P)^{-1} B$ is a finitely generated $A_P$-module. 

This was skipped in a lecture and I was trying to figure out, but I am not quite seeing it at the moment. I would greatly appreciate any hints/answers! Thanks! 
 A: Hint:
This is because separability implies $B$ itself is a finitely generated $A$-module. Furthermore, it is unrelated to $A$ being a Dedekind domain: it is enough that $A$ be noetherian and integrally closed.
A: Proof in several steps.  You are probably familiar with many of the steps, but for the sake of a complete argument here is some overkill:

1 .  Let $v_1, ... , v_n$ be a basis for $L/K$.  Show that there exists a basis $w_1, ... , w_n$ for $L/K$ such that $Tr(v_iw_j) = \delta_{ij}$, where $Tr$ is the trace map $L \rightarrow K$.

Hint: consider $L$ as a vector space over $K$.  Define a function $B: L \times L \rightarrow K$ by $B(x,y) = Tr(xy)$.  Let $L^{\ast}$ be the dual of $L$ as a vector space over $K$, i.e. $L = \textrm{Hom}_K(L,K)$.  Show that there is an isomorphism $\Phi: L \rightarrow L^{\ast}$ defined by basis elements by $\Phi(v_i) = B(v_i,-)$.  
Recall from linear algebra that there is a basis $v_1^{\ast}, ... , v_n^{\ast}$ of $L^{\ast}$, where $v_i^{\ast}$ is the function from $L$ to $K$ defined by $v_i^{\ast}(c_1v_1 + \cdots + c_nv_n) = c_i$.  You can then define $w_i$ to be $\Phi^{-1}(v_i^{\ast})$, and show it has the required property.

2 .  To show that $B$ is finitely generated as an $A$-module (from which it follows that $B_P$ is finitely generated over $A_P$), show that this is done if we have shown that $B$ is contained in a finitely generated $A$-module.

Hint: a module $M$ over a given ring (Noetherian or not) is called Noetherian if every ascending chain of submodules terminates, or equivalently every submodule of $M$ is finitely generated.  Quotients and submodules of Noetherian modules are Noetherian; conversely if $N$ is a submodule of $M$ and $N$ and $M/N$ are both Noetherian, then so is $M$.  If $A$ is a Noetherian ring, show that $A^k = \bigoplus\limits_{i=1}^k A$ is a Noetherian $A$-module. A finitely generated module over $A$ is the same thing as a quotient $A^k/N$ for some $k$ and some submodule $N$ of $A^k$.  Hence a finitely generated module over $A$ is Noetherian.  Thus any submodule of a finitely generated $A$-module is itself finitely generated.  

3 .  Show that there exists a basis $v_1, ... , v_n$ for $L/K$ such that the $v_i$ are integral over $A$, i.e. $v_i \in B$.

Hint: $x \in L$ is integral over $A$ if and only if its monic minimal polynomial has coefficients in $A$.

4 .  Assume $v_i$ is a basis for $L/K$ with $v_i \in B$.  Let $w_1, ... , w_n$ be the dual basis from (1).  Show that there exists a $c \in A$ such that $cw_i$ are integral over $A$.

(same as (3)).

5 .  If $x \in B$, show that $Tr(x) \in A$.

The coefficients of the minimal polynomial of $x$ over $K$ lie in $A$.

6 .  Show that $B \subseteq Ac^{-1}v_1 + \cdots + Ac^{-1}v_n$.

If $z \in B$, write $z = c_1v_1 + \cdots + c_nv_n$ for $c_i \in K$.  If you can show that $cc_i \in A$ for all $i$, then you are done.  To do this, multiply both sides by $cw_i$ and take the trace.  Use the fact that $Tr(B) \subseteq A$.
