Thank you for this nice lemma, I am fond of this kind of results.

It seems to me (if I am not wrong) that the argument of the second proof holds

also whenever the problem is considered in any finite abelian group of odd cardinality,

working one coset after the other (and may be even in non abelian groups ?!).

I was wondering if the first proof does also hold ?

Best regards from Paris,

Eric

]]>The difficult step is to prove that, for some $x$, $\sum_{i=0}^{n-1} \xi^{-i} \sigma^i(x)$ is nonzero. Your argument does that by finding an $x$ such that the $\sigma^i(x)$ are linearly independent – the normal basis theorem. Artin’s Lemma is just the statement that the maps themselves are linearly independent, so the map $x \to \sum_{i=0}^{n-1} \xi^{-i} \sigma^i(x)$ is nonzero and hence has a nonzero element in the image.

]]>http://www.numdam.org/numdam-bin/browse?id=CM_1935__1_

or directly.

http://archive.numdam.org/ARCHIVE/CM/CM_1935__1_/CM_1935__1__106_0/CM_1935__1__106_0.pdf

Most sources do give 1935 for the paper.

]]>http://www.numdam.org/item?id=CM_1935__1__106_0 ]]>

Let me also mention that the proper French version can be so much worse… A friend of mine, with last name Рудченко, thought that it will be made into Rudchenko (just one more letter) and this is how he signs his papers. But he became Roudtchenko in his `foreign’ passport issued 20 years ago (this is at least how it used to be thanks to the huge French historic or cultural influence)… Some French names becomes readable in Russian, Houellebecq shortens to 7 letters (one of which is the same obscure ь which makes makes `l’ in his last name sounds soft).

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