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Question. For a finite Galois cyclic extension $E/F$, prove that there is a primitive element $\alpha$ such that $$\sum_{\sigma\in\operatorname{Gal}(E/F)}u_{\sigma}\sigma(\alpha) = 0$$ for $u_{\sigma} \in\{0,\pm 1\}$ implies that $u_{\sigma} = 0$ for each $\sigma$.

The Context below illustrated the motivation for the above question.

Motivating example. Take $E/F = \Bbb Q[\zeta]/\Bbb Q$ where $\zeta$ is a primitive $n^{th}$ root of unity. If the set $\{\zeta,...,\zeta^{n-1}\}$ is linearly independent, then the trace is primitive for the intermediate extension over $\Bbb Q$:$$\operatorname{Fix}\left<\sigma_k\right> = \Bbb Q\left[\zeta+\zeta^k+\zeta^{k^2}+...\right]=\Bbb Q\left[T_{\left.\Bbb Q[\zeta]\middle/\operatorname{Fix}\left<\sigma_k\right>\right.}(\zeta)\right].$$

The first equality follows from linearly independency, and the second equality follows from the fact that for a Galois extension $E/F$ and $a\in E$, $$T_{E/F}(a) = \sum_{\sigma\in \operatorname{Gal}(E/F)}\sigma(a).$$

If you are not familiar with this example, I illustrated a more general context below in detail.

In general let $E/F$ be a finite (Galois) cyclic extension of degree $n$, pick $\alpha\in E$ to be a primitive element and let $\operatorname{Gal}(E/F) = \left<\sigma\right>.$ We are interested in if the following is still true for some intermediate extension $M$: $$M = F\left[T_{E/M}(\alpha)\right].$$

Let $F[\alpha+\sigma^k(\alpha)+\sigma^{k^2}(\alpha)+...]=M$. To show $M = \operatorname{Fix}\left<\sigma^k\right>$, by the Fundamental Theorem of Galois Theory, it's sufficient to show $$\left| \operatorname{Gal}\left(E/M\right)\right| \leq \left|\left<\sigma^k\right>\right|.$$ Let $\tau\in \operatorname{Gal}(E/M)$ and write $\tau = \sigma^t$, then $$\sigma^t(\alpha)+\sigma^t\sigma^k(\alpha)+\sigma^t\sigma^{k^2}(\alpha)+... = \alpha+\sigma^k(\alpha)+\sigma^{k^2}(\alpha)+...$$ If $t$ is not a power of $k$, the above will produce a linearly dependent set of distinct powers of $\sigma$. In conclusion:

If $E/F$ is a finite cyclic Galois extension and $\alpha\in E$ is a primitive element such that $\{\sigma(\alpha)\}_{\sigma\in\operatorname{Gal}(E/F)}$ is linearly independent over $F$, then for any intermediate extension $E/M/F$,$$M = F\left[T_{E/M}(\alpha)\right].$$

It's a consequence of normal basis theorem(see here) that such element $\alpha$ always exists for a finite field extension $E/F$. I am wondering if things are simpler for just this case above. So below is my question, which is a special case of the Normal Basis Theorem.

Question. For a finite Galois cyclic extension $E/F$, prove that there is a primitive element $\alpha$ such that $$\sum_{\sigma\in\operatorname{Gal}(E/F)}u_{\sigma}\sigma(\alpha) = 0$$ for $u_{\sigma} \in\{0,\pm 1\}$ implies that $u_{\sigma} = 0$ for each $\sigma$.

William Sun
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  • Are you sure about the motivating example? What about the case $n=8$, $k=5$, so $k^2\equiv1\pmod n$. If $\zeta=e^{\pi i/4}$ is a primitive eighth root of unity, then $\zeta^k=\zeta^5=-\zeta$, and therefore the trace $\zeta+\zeta^5=0$ cannot generate the fixed field $\Bbb{Q}(i)$ of $\sigma_5$. – Jyrki Lahtonen Aug 27 '23 at 06:26
  • I recall a discussion here with the outcome that there is a related problem with the $n$th roots of unity if and only if $n$ is not square free. – Jyrki Lahtonen Aug 27 '23 at 06:31
  • I just found this old question. See the discussion there. It is specific to the case of roots of unity. I add the link simply to give further reading related to my objection. – Jyrki Lahtonen Aug 27 '23 at 06:37
  • True, the extension $\Bbb{Q}(\zeta_8)/\Bbb{Q}$ is not cyclic. Instead, let's consider $\Bbb{Q}(\zeta_9)/\Bbb{Q}$, a cyclic extension of degree six. The automorphism $\sigma_4$ defined by $\sigma_4(\zeta_9)=\zeta_9^4$ generates a subgroup of order three. Here also the trace $$\zeta_9+\zeta_9^4+\zeta_9^7=\zeta(1+\zeta_9^3+\zeta_9^6)=0$$ as the sum in parens is the sum of all the third roots of unity (including $1$). – Jyrki Lahtonen Aug 27 '23 at 06:42
  • @JyrkiLahtonen My apologies. I should have specified that the motivating example works when $\Bbb Q[\zeta]/\Bbb Q$ is cyclic, or else $1,\zeta^1,...,\zeta^{n-1}$ might not be linearly independent. I will add it to the question body. – William Sun Aug 27 '23 at 12:24
  • It's still not right. Look at the case $n=9$ in me preceding comment. $1,\zeta^3,\zeta^6$ are not linearly independent. Hence neither are $\zeta,\zeta^4,\zeta^7$. But their sum is one of the traces! – Jyrki Lahtonen Aug 27 '23 at 12:29
  • @JyrkiLahtonen You are right. I edited the question again. I need to change my faulty belief that $1,\zeta^1,...$ should often be linearly independent... – William Sun Aug 27 '23 at 12:37
  • Ok. Now it is correct in a way. but $1,\zeta,\zeta^2,\ldots,\zeta^{n-1}$ is NEVER linearly independent because their sum is zero. What suffices for the example is the case where $n$ is a prime. That is equivalent to the assumption that $\zeta,\zeta^2,\ldots,\zeta^{n-1}$ are linearly independent. Which is just what the doctor ordered! In other words, you need $n$ to be a prime and you also need to ditch $\zeta^0=1$ from the basis. – Jyrki Lahtonen Aug 27 '23 at 14:29
  • Thank you for pointing them out. I’m glad that fortunately, these mistakes doesn’t affect the validity of the argument for the cyclic extension part… – William Sun Aug 27 '23 at 18:18
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    Closely Related: https://math.stackexchange.com/questions/131757/a-proof-of-the-normal-basis-theorem-of-a-cyclic-extension-field – William Sun Aug 29 '23 at 17:33

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