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The proof of Hilbert's Theorem 90 about cyclic extensions goes like this: Let $\sigma$ be the generator of the Galois group of order $n$ and let $b$ have norm $1$, i.e. $b \sigma(b) \cdots \sigma^{n-1}(b) = 1$. For an element $c$ consider

$a := c + b \sigma(c) + \dotsc + b \sigma(b) \cdots \sigma^{n-2}(b) \sigma^{n-1}(c)$.

Then one verifies $b \sigma(a) = a$. We may choose $c$ such that $a \neq 0$ since characters are linearly independent. QED

Question 1. Is there a motivation for the choice of $a$? It is clear that it works, but why did Hilbert or rather Kummer (feel free to add historical details) came up with this sum?

Was he influenced by Lagrange's resolvents? But if this is the source, I would have to ask the same question which idea underlies the definition these resolvents; again here I only know that they solve the cubic. After a quick glance at the original papers I could not find any explanation.

Feel free to forget about history and give another, perhaps more modern motivation of the choice of $a$. I already tried to "get" $a$ starting with

$a = b \sigma(a) = b \sigma(b \sigma(a)) = b \sigma(a) \sigma^2(a) = \dotsc = b \sigma(b) \cdots \sigma^k(b) \sigma^{k+1}(a)$.

Perhaps also the following interpretation helps: We look for a fixed point of the function $b \sigma(-)$. Now perhaps there is a connection with others fixed point theorems, in case I already asked about the motiviation of their proofs.

Question 2. Linear independence of characters is proven indirectly, so the choice of $c$ is above is not canonical. Is there any chance to get an explicit choice of $c$? This would be great because then for every solvable separable polynomial with known Galois group one would get explicit generators of the Galois extension. This is because Hilbert's Theorem 90 makes it possible to classify cyclic extensions, which are the intermediate steps; actually there $b$ is in the ground field, perhaps this simplifies both questions.

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For the sake of keeping the question self-contained I think you should state the theorem: it won't exactly take long. –  Qiaochu Yuan Aug 17 '11 at 17:36
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Isn't this rather close to mathoverflow.net/questions/21110/… ? –  Franz Lemmermeyer Aug 17 '11 at 17:38
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I agree with Qiaochu Yuan. Many people don't know theorems by names. –  André Henriques Aug 17 '11 at 19:01
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The link to Wikipedia is broken. Also, notice that there are several versions of Hilbert's 90, even in the Wikipedia page. It is clear which one you refer to, but I agree with Qiaochu that it would be nice to see the statement here, for this page to be self-contained. –  Álvaro Lozano-Robledo Aug 17 '11 at 19:39
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@A. Henriques: Moreover, many people who do know a theorem called "Hilbert 90" mean a result about group cohomology that Hilbert didn't know and whose equivalence with Hilbert's result takes some work to establish... –  Noam D. Elkies Aug 17 '11 at 19:39

1 Answer 1

up vote 8 down vote accepted

The map $T : a \mapsto b \sigma(a)$ is linear and has order $n$. It follows straightforwardly that $c + T c + ... + T^{n-1} c$ is a fixed point of $T$.

More generally, let $V$ be a representation of a finite group $G$ over a field of characteristic not dividing $|G|$ containing the values of every character of $G$ over the algebraic closure. Let $\chi$ be the character of an irreducible representation of $G$. Then

$$v \mapsto \frac{1}{|G|} \sum_{g \in G} \overline{\chi(g)} gv$$

is the projection from $V$ to the isotypic component $V_{\chi}$ of $V$. When $G$ is a cyclic group we recover Lagrange resolvents. In particular when $\chi$ is the trivial representation, the above is the projection from $V$ to its $G$-invariant subspace.

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Of course, it's just averaging! Thank you Qiaochu. –  Martin Brandenburg Aug 17 '11 at 18:41
    
(Although I accept this answer, Question 2 remains open. So answers to this are still wellcome) –  Martin Brandenburg Aug 17 '11 at 21:56

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