In A. Fröhlich's article *Local Fields* in *Algebraic Number Theory*, the following claim is made: if $R$ is a Dedekind domain with field of fractions $K$, $L$ is a finite separable extension of $K$ and $S$ is the integral closure of $R$ in $L$, and $x$ is an element of $S$ with minimal polynomial $g$, then, "by Euler's formulae",


$$\text{tr}\_{L/K}(x^i/g'(x)) \in R$$
for each $0 \leq i \leq n-1$, where $n=\text{deg } g$.


Which formulae of Euler are being referred to? The claim can be proven by the Lagrange interpolation formula; in fact the given quantity is $1$ if $i=n-1$, and $0$ for $0 \leq i < n-1$. However, I have no idea what proof Fröhlich has in mind. I also cannot resist pointing out the humor in appealing to Euler's "formulae" without further precision. Perhaps the formulae in question are well-known, and I am the only one who has not been invited to the party? In any case, more details would be greatly appreciated!


Thank you.



Which formulae of Euler is Fröhlich referring to?
I believe the reference is to this formula of Euler (see [here](http://tinyurl.com/bw2a9gm)): If $P(x)/Q(x)$ is a rational function and $ax+b$ is a simple factor of $Q(x)$, then the coefficient of $1/(ax+b)$ in the partial fraction decomposition of $P/Q$ is given by
$$ \lim\_{x\to \frac{-b}{a}} \frac{a P(x)}{Q'(x)}. $$




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To see how this applies here, proceed as Serre does in *Local Fields*. That is, write (in a suitable extension of $L$)
$$ \frac{1}{g(X)} = \sum\_{k=1}^n \frac{a\_k}{X-x\_k} \qquad (\*) $$
where the $x\_k$ are the conjugates of $x$. Then a formal application of Euler's formula gives
$$ a\_k = \lim\_{X \to x\_k} \frac{1}{g'(X)} = \frac{1}{g'(x\_k)}. $$
Now expand both sides of (\*) as power series in $1/X$ and compare coefficients.


