0
$\begingroup$

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 $\mathbb Q(\zeta)$?

$\endgroup$
  • $\begingroup$ Clearly I asked the wrong question. I posted another question with the real question I am interested in. $\endgroup$ – Angel del Rio Apr 6 '15 at 15:35
  • $\begingroup$ The answer is positive. I got a solution by René Schoof $\endgroup$ – Angel del Rio Jul 29 '15 at 12:02
3
$\begingroup$

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 $r=9$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.