Suppose that $A\subset B$ is an integral extension of domains and $A$ is integrally closed. Also assume that, if $F_A$ and $F_B$ the quotient fields of $A$ and $B$ then $F_B$ is a simple extension of $F_A$, i.e., $F_B=F_A(b)$ with $b\in B$. Let $q(X)\in A[X]$ be the minimal polynomial of $b$ and $D$ its discriminant.
I want to prove that any $z\in B$ can be written as $\frac{T(b)}{D}$ where $T(X)\in A[X]$ has degree $<\deg(q)$.
We can certainly write $z=T(b)/\lambda $ for some $\lambda\in A$, but I am struggling to see where the discriminant comes into play.
What I 've tried so far is use the expression $\lambda z=T(b)$ and the fact that $\lambda z \in B$ integral over A, so I can find monic polynomial $p(X)$ s.t $p(\lambda z)=p(T(b))=0$ and notice that $p\circ T (X)$ must be divided by q, but it doesn't get me anywhere.
Any hints would be welcome