I want to prove the theorem below, I came at to some point but then confused and can not go further:
Theorem. Let $A$ be a commutative ring with unity, $K=Frac(A)$, $L/K$ a finite extension and $B$ be the integral closure of $A$ in $L$. I want to show that if $B$ is a finitely generated free $A-$module with basis $S =\{b_1,\dots, b_n\}$ then $S$ is also a basis for $L$ over $K$.
Now, we also know that $Frac(B)=L$.
So, let $\alpha=\cfrac{c}{d} \in L$ with $c,d \in B$. Then $g(\alpha) =0$ for some $g(X) \in B[X]$. Let us say that $g(X) = X^m + b_{m-1}X^{m-1} + \dots +b_0$. Since $B = \{ \alpha \in L : \exists f \in A[X] | f(\alpha) = 0\}$, then for each $b_i$, we have some $g_i(X) \in A[X]$ with $g_i(b_i)=0$.
Using this fact, I want to manipulate $g(X)$ to get some polynomial in $A[X]$ and conclude the result. Is my idea correct? Either case, how can I finish the proof?
Every element $x$ of $L$ can we written as $\frac{b}{a}$ with $b \in B$ and $a \in A \setminus \{0\}$ (see for example p. 8 of Algebraic number theory by Jürgen Neukirch).
Now let $x = \frac{b}{a} \in L$, with $b \in B$ and $a \in A \setminus \{0\}$. Since $(b_1,\cdots,b_n)$ is a basis of the $A$-module $B$, we can write $b = \sum_{i=1}^n c_i b_i$ for some $c_1,\dots,c_n \in A$. Thus $x = \sum_{i=1}^n \frac{c_i}{a} b_i$ where $\frac{c_i}{a} \in K$ for all $i$. Hence $(b_1,\cdots,b_n)$ spans $L$ as a vector space over $K$.
Moreover, if a linear combination $\sum_{i=1}^n c_i b_i$ with $c_i \in K$ for all $i$ is zero, then clearing the denominators of the $c_i$'s and using that $(b_1,\cdots,b_n)$ is a basis of the $A$-module $B$, it is easily seen that $c_i = 0$ for all $i$.
Therefore $(b_1,\cdots,b_n)$ is a basis of the vector space $L$ over $K$.