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Let $L/K$ be a Galois extensions. It seems to be true to me, after some tinkering, that some $x\in L$ is a primitive element of the extension $L/K$ iff $\sigma(x)\neq x$ for all $\sigma\in \text{Gal}(L/K)$, where $\sigma$ isn't the identity (though I'm not sure about the "iff" part). How could that be proven ?

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For any $x\in L$, $\sigma\in\mathrm{Gal}(L/K)$, $\sigma(x)=x$ if and only if $\sigma$ fixes $K(x)$ pointwise. So, $K(x)=L$ if and only if the only automorphism fixing $K(x)$ is 1, if and only if the only automorphism fixing $x$ is 1.

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$x\in L$ is a primitive element of $L/K$ if and only if $K(x) = L$.

By the fundamental theorem of Galois theory, $K(x) = L$ if and only if $Gal(L/K(x)) = 1$.

$Gal(L/K(x)) = 1$ if and only if $\sigma(x) \neq x$ for all $\sigma \in Gal(L/K) - \{1\}$.