## Which formulae of Euler is Fröhlich referring to?

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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.

-

3
See Lemma 2 (Euler) on p.56 of Serre's Local Fields. – Chandan Singh Dalawat May 7 2012 at 5:56

7
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 Magnusson May 7 2012 at 8:31

1
This is also in the proof of Prop. 2 on p. 58 of Lang's Algebraic Number Theory. – KConrad May 7 2012 at 14:06

## 1 Answer

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.

-

Wonderful! Thanks for writing that out, Faisal. – Bruno May 7 2012 at 18:09
No problem!  – Faisal May 8 2012 at 16:08
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 26, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9229533076286316, "perplexity_flag": "head"}