Upper bound for determinant of matrix representation of linear operator Let $p$ be a prime number and let $\zeta_{p}$ be the $p$-th 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} $ where $m|p-1\,.$
I am looking for an upper bound for $\det[m_{\alpha}]$.
The only upper bounds which I know are the following
$$\det(A) \leq \bigg(\frac{\operatorname{trace}(A)}{n}\bigg)^n\\[3ex]
\det(A) \leq \prod_{i=1}^n A_{ii}$$
The first one needs $[m_{\alpha}]$ to be Hermitian or symmetric, the second one needs $[m_{\alpha}] $ to be orthogonal.
But I can't prove that $[m_{\alpha}] $ has any property of the above listed ones.
What is a good upper bound for $\,\det[m_{\alpha}] $?
 A: Let $\psi:\operatorname{mult}(\Bbb Q[\zeta]) \to \Bbb Q^{p \times p}$ denote the map corresponding to the circulant matrix representation of $\Bbb Q[\zeta]$. That is: for polynomials $f$, we define $\psi(m_{f(\zeta)}) = f(P)$, where $P$ denotes the permutation matrix
$$
P = \pmatrix{&&&1\\ 1\\ &\ddots \\ &&1}.
$$
We note that this map can be written in the form $\psi = \operatorname{id}_{\Bbb Q} \oplus \psi'$, where $\psi'$ is a ring isomorphism between $\operatorname{mult}(\Bbb Q[\zeta])$ and $\Bbb Q^{(p-1) \times (p-1)}$. I claim (without proof) that this allows us to deduce that the map $\psi$ is determinant preserving, i.e. $\det(m_{f(\zeta)}) = \det(f(P))$.
As is explained on the wiki page, the determinant of $f(P)$ may be computed with the formula
$$
\det(f(P)) = \prod_{j=0}^{p-1} f(\zeta^j).
$$
Your $\alpha$ can be written as $\alpha = f(\zeta)$, where $f(x) = 1 + x^2 + \cdots + x^{m} = \frac{x^{m-1} - 1}{x - 1}$. Thus, your determinant can be computed as
$$
\det(m_{\alpha}) = \prod_{j=0}^{p-1} f(\zeta^j) = \prod_{j=0}^{p-1} \frac{\zeta^{m+1}  - 1}{\zeta - 1}.
$$
From there, we have the inequality
$$
|\det(m_{\alpha})| = \prod_{j=0}^{p-1} \frac{|\zeta^{m+1}  - 1|}{|\zeta - 1|} \leq
|\det(m_{\alpha})| = \prod_{j=0}^{p-1} \frac{|\zeta^{m+1}|  + 1}{|\zeta - 1|} = \frac{2^p}{2 \sin(\pi /p)} = \frac{2^{p-1}}{\sin(\pi /p)}.
$$
Perhaps you will find this bound to be useful.
