Let $p$ be a prime number and let $\zeta_{p}$ be the pth roots of unity and let $\mathbb{Q}(\zeta_{p})/\mathbb{Q}$ be a Galois extension and $G=\operatorname{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})$ be its Galois group we know that $|G|=\phi(p)=p-1.$
And let $S=\{1,\zeta,\zeta^{2},.....,\zeta^{p-2}\}$ be a basis of the Galois extension $\mathbb{Q}(\zeta_{p})/\mathbb{Q}$ and let $m_{\alpha}(x)=\alpha x $ be a linear operator, and $[m_{\alpha}]$ be its matrix representation.
Let $\alpha=1+\zeta +\zeta^{2}+....+\zeta^{m}. $
And $\det[m_{\alpha}]$ be its determinant. I want to show that $|\det[m_{\alpha}]| \neq 1$ for some value of $m>c$ where $c$ is a number. In other words, is there any good lower bound which helps as to show that $|\det[m_{\alpha}]| \neq 1$ ?
If $\beta=1-\zeta$, we have $$[m_\beta]=\begin{pmatrix}1&0&0&\cdots&0&1\\-1&1&0&\cdots&0&1\\0&-1&1&\cdots&0&1\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&1&1\\0&0&0&\cdots&-1&2\end{pmatrix} =:A_{p-1}$$ and by developing along the top row, say, we find $$\det A_n=\det A_{n-1}+(-1)^{n-1}\det(-I_{n-1})=\det A_{n-1}+1$$ and with $\det A_1=2$ ultimately $$\det [m_\beta]=p.$$ The determinant of a linear map does not depend on the choice of basis. Therefore, we more generally have $\det [m_{1-\xi}]=p$ for any primitive $p$th root $\xi\in\Bbb Q(\zeta)$, in particular for $\xi=\zeta^{m+1}$ with $p\nmid m+1$. As $\alpha=\frac{1-\zeta^{m+1}}{1-\zeta}$, we conclude that $$ \det [m_\alpha]=\frac{\det [m_{1-\zeta^{m+1}}]}{\det[m_\beta]}=1$$ for all $m\not\equiv -1\pmod p$. On the other hand, for $m\equiv -1\pmod p$, we clearly have $\alpha=0$ and so $\det [m_\alpha]=0$.