TITLE: Euler's formula for trace of $x^i/g'(x)$
QUESTION [5 upvotes]: Let $L/K$ be a finite separable extension of fields generated by an element $x$ with minimal polynomial $g(X)=0$. Assume that $K$ is the fraction field of a Dedekind domain $R$, and $S$ is its integral closure in $L$. Let $t_{L/K}$ be the trace.
Then, according to Froehlich ANT, page 17, it follows from Euler's formulae that
$$t_{L/K} (x^i/g'(x)) \in R$$
I have no idea which are these Euler's formulae, can someone help? The only thing that came to my mind here was to write
$$x^i/g'(x) = \frac {x^i} {\prod_{\alpha \neq x \text{ a root of g}} (x-\alpha)}$$

REPLY [5 votes]: Write $g(X)/(X-x) = X^{n-1} +\beta_{n-2}X^{n-2}...+\beta_0$.
Let $\{\alpha_i\}_{i=1}^n$ be the roots of $g(X)$, where $\alpha_1 =x$. Observe that by what I mentioned before $$\sum_i \frac {g(X)}{(X-\alpha_i)} \frac {\alpha_i^r}{g'(\alpha_i)}= X^r, \quad\forall\, 0\le r\le n-1.$$
Define the trace of a polynomial $P(X)=\sum_i^n a_iX^i$ to be the polynomial $\tau( P(X)):=\sum_i^n t_{L/K}(a_i)X^i$. Then
$$\tau\left( \frac {g(X)}{(X-x)} \frac {x^r}{g'(x)}\right) = X^r$$
So that, by comparing coefficients, $t_{L/K} (\beta_j \frac {x^i}{g'(x)})= \delta_{ij}$.
It follows that the $\{ \frac{x^i}{g'(x)}\}$ has the dual basis $\{\beta_i\}$ (in $L$). As Shankman pointed out, this finishes the proof.