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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.

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See Lemma 2 (Euler) on p.56 of Serre's Local Fields. – Chandan Singh Dalawat May 7 '12 at 5:56
Serre makes a similar comment in "Writing mathematics badly" (it's a lecture, google it to find a video), that if you want to make sure your article or lecture's title contains no information at all then call it "On a theorem of Euler". – Gunnar Þór Magnússon May 7 '12 at 8:31
This is also in the proof of Prop. 2 on p. 58 of Lang's Algebraic Number Theory. – KConrad May 7 '12 at 14:06
up vote 5 down vote accepted

I believe the reference is to this formula of Euler (see here): 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 $$ \lim_{x\to \frac{-b}{a}} \frac{a P(x)}{Q'(x)}. $$

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.

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Wonderful! Thanks for writing that out, Faisal. – Bruno Joyal May 7 '12 at 18:09
No problem! $ $ – Faisal May 8 '12 at 16:08

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