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Assume that $F\subseteq K$ is a Galois field extension. Clearly, for every $\sigma \in \mbox{Gal}(K/F)$ and $a\in K$, $$tr_{K/F}(a-\sigma(a) )=0$$ Is it true that, whenever $x\in K$ is an element such that $tr_{K/F}(x)=0$, there are $a\in K$ and $\sigma\in \mbox{Gal}(K/F)$ such that $x=a-\sigma(a)$?

user333618
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  • https://math.stackexchange.com/q/639310/96384 – Torsten Schoeneberg Aug 04 '19 at 00:15
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    Your converse is true if the Galois group G is cyclic, say generated by $\sigma$. It also holds true when replacing the additive group $K^+$ by the multiplicative group $K^$. Namely, for $x\in K^$ such that N$(x)=1$ (where N is the norm map of $K/F$), there exist $a\in K^$ s.t. $x=a/\sigma (a)$. This is the celebrated Hilbert's thm.90. A more general result, when G is not necessarily cyclic, asserts the nullity of the 1-st cohomology group of G acting on $K^$ (resp. $K^+$) : $H^1(G,K^*)$ (resp. $H^1(G, K^+)) = 0$. See e.g. Lang's "Algebra", chap.VIII, §§6 and 10. – nguyen quang do Aug 04 '19 at 19:52

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No. Take a normal basis $K = \sum_{g \in G} g(\beta) F$ then the equation $$\sum_{g \in G} c(g)g(\beta)=\alpha- \sigma(\alpha), \qquad \alpha = \sum_g a(g) g(\beta)$$ yields $$c(g) = a(g)-a(\sigma^{-1} g)$$ having a solution iff for each $g \in \langle\sigma\rangle\setminus G$, $\sum_{n=1}^{ord(\sigma)} c(\sigma^n g)=0$ which is stronger (for $G$ non cyclic) than just $\sum_{g \in G} c(g)=0$.

For example $K = \Bbb{Q}(\beta),\beta=\sqrt{2}+\sqrt{3}+1$, $\scriptstyle\sum_{g \in G} c(g)g(\beta)=1 (\sqrt{2}+\sqrt{3}+1)+2(\sqrt{2}-\sqrt{3}+1)+3(-\sqrt{2}+\sqrt{3}+1)-6(-\sqrt{2}-\sqrt{3}+1)$

reuns
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