Let p be an odd prime integer and let $\zeta$ be a primitive p-th root of unity. Let $\alpha$ be a non-trivial cyclotomic unit of $\mathbb Q(\zeta)$, i.e. an element of the form $1+\zeta+\dots+\zeta^{r-1}$, with $2\le r < p$. Does $\alpha$ generate a normal basis of $\mathbb{Q}(\zeta)$? In other words does the Galois conjugates of $\alpha$ form a basis of $\mathbb Q(\zeta)$?
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$\begingroup$ Clearly I asked the wrong question. I posted another question with the real question I am interested in. $\endgroup$– Angel del RioCommented Apr 6, 2015 at 15:35
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$\begingroup$ The answer is positive. I got a solution by René Schoof $\endgroup$– Angel del RioCommented Jul 29, 2015 at 12:02
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No, in general $\alpha$ doesnt generate a normal basis. The smallest counterexample is $p=5$, $r=3$. In fact for all $p\geq 5$, $r=(p+1)/2$ gives a counterexample: For $1\leq i\leq p-1$, let $\varphi_i\in \text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ be the automorphism given by $\zeta\mapsto \zeta^i$. A small computation shows that $\varphi_i(\alpha)+\varphi_{p-i}(\alpha)=1$ for all $i$. In particular we have $\alpha-\varphi_2(\alpha)-\varphi_{p-2}(\alpha)+\varphi_{p-1}(\alpha)=0$ and hence the conjugates of $\alpha$ are not linearly independent for $p\geq 5$.
There are other counterexamples as well: For $p=11$ the counterexamples are given by $r=3$, $r=6$ and $r=9$.
