Representation of elements in a field Let $A$ and $B$ be integrals domains, such that $A$ is integral over $B$. Writing $K(A)$ for the field of fractions, suppose that $K(A)$ is generated over $K(B)$ by a single element in $A$, say $\eta$. If $q\in B[X]$ is the minimal polynomial of $\eta$, how do I prove that every element in $A$ can be written as
$$
y = \frac{T(\eta)}{D},
$$
where $T\in B[X]$ is a polynomial of $\deg T< \deg q$ and $D$ is the discriminant of $q$.
In my case, $A=O_n/I$, the germs of holomorphic functions in $\mathbb{C}^n$ and $I$ is a prime ideal. $B=O_k$ and $I\cap O_k=\{0\}$, so $B$ can be seen as a subring of $A$. Finally $\eta = z_{k+1}$ and I wish to represent the elements $z_j$ for $j=k+2,\ldots, n$. This is a claim in the book of Gunning and Rossi, "Analytic Functions of Several Complex Variables", p. 95, and they refer to Zariski and Samuel, but where? Can someone provide a proof?
Thanks!
 A: The statement only makes sense under the additional hypothesis that the field extension $K(A)/K(B)$ is separable. Indeed, if it were not separable then the discriminant $D$ would be zero.
The statement is true if $B$ is integrally closed. I'm not sure whether it is true otherwise.
In any case, in your particular situation of germs of holomorphic functions, $K(A)/K(B)$ is separable and $B$ is integrally closed, so that the following applies.
Suppose that $K(A)/K(B)$ is separable and that $B$ is integrally closed. Let $\langle\cdot,\cdot\rangle$ be the trace form on the $K(B)$-vector space $K(A)$ defined by
$$
\langle x,y\rangle=\mathrm{tr}(xy)
$$
for all $x,y\in K(A)$, where $\mathrm{tr}$ denotes the trace with respect to the field extension $K(A)/K(B)$. The trace form is a symmetric $K(B)$-bilinear form. 
Let $n$ be the degree of the extension $K(A)/K(B)$, i.e., $n=\deg(q)$.
Since $K(A)/K(B)$ is separable, there are $n$ distinct $K(B)$-linear embeddings
$$
\sigma_i\colon K(A)\rightarrow \overline{K(B)}
$$
of the field $K(A)$ into the algebraic closure $\overline{K(B)}$ of $K(B)$.
Then, it is well known that
$$
\mathrm{tr}(x)=\sum\sigma_i(x)
$$
for all $x\in K(A)$.
The preceding formula allows us to express the discriminant of the trace form with respect to the $K(B)$-basis $1,\eta,\ldots,\eta^{n-1}$ in terms of the discriminant $D$ of the minimal polynomial $q$ of $\eta$ over $K(B)$:
$$
\det(\mathrm{tr}(\eta^i\eta^j))=\det(\sigma_i(\eta^j))^2=D^2.
$$
It follows, in particular, that the trace form is nondegenerate, but that will be of no use to us.
Let $A_0$ be the free $B$-module generated by $1,\eta,\ldots,\eta^{n-1}$. Since $\eta\in A$, one has $A_0\subseteq A$. We need to show that $A\subseteq \frac1DA_0$.
If $A$ is a free $B$-module then this can be proved as follows.
Let $v_1,\ldots,v_n$ be a $B$-basis of $A$.
Then $v_1,\ldots,v_n$ is a $K(B)$-basis of $K(A)$. Let $L$ be the $K(B)$-linear automorphism of $K(A)$ determined by that $L(v_i)=
\eta^{i-1}$ for $i=1,\ldots,n$. One has $L(A)=A_0$. Since $A_0\subseteq A$, one has $\det(L)\in B$.
The discriminant of the trace form with respect to the basis $v_1,\ldots,v_n$ satisfies
$$
\det(\mathrm{tr}(v_iv_j))=\det(L^{-1})^2\det(\mathrm{tr}(\eta^i\eta^j))^2=\det(L)^{-2}D^2,
$$
and belongs to $B$. Since $B$ is integrally closed, one deduces that $\det(L)^{-1}D\in B$, i.e., $\det(L)^{-1}\in\frac1DB$.
It follows that
$$
A=L^{-1}(A_0)\subseteq\tfrac1DA_0.
$$
The general case of $A$ not necessarily free over $B$ follows from localization. Indeed, $A$ is a locally free $B$-module since $B$ is integrally closed, and one can apply the free case locally in order to obtain $A\subseteq \frac1DA_0$ globally.
