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.

doknow 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