Let $L/K$ be a finite cyclic Galois extension of degree $n$ with Galois group $G$ generated by $\sigma$. Hilbert's Theorem 90 tells us that if $c \in L^\times$ satisfies $N_{L/K}(c) = c\sigma(c)\cdots\sigma^{n-1}(c) = 1$, then $c$ can be written as $b/\sigma(b)$, for some $b \in L^\times$.

My question is: how many such $b$ are there? Clearly if we have $b/\sigma(b) = c$ then we can multiply $b$ by any element $x$ of $K^\times$ to get $b' = bx$ still satisfying the equation. Are there any other solutions?


1 Answer 1


It turns out there really is only a $K^\times$'s worth of solutions: We can rewrite the key equation as $b=\sigma(b)c$. Assume that also $b' = \sigma(b')c$. Then $b' = bx$, for some $x \in L^\times$. Then $\sigma(b') = \sigma(b)\sigma(x)$, so $b' = \sigma(b')c = \sigma(b)c\sigma(x) = b\sigma(x)$, and thus $x = \sigma(x)$.

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    $\begingroup$ So for $K=\Bbb F_2$ there is only one choice for $b$. $\endgroup$ Nov 28, 2023 at 15:34

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