Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th root of unity in $\mathbb{C}$. Let $K = \mathbb{Q}(\zeta)$. Let $A$ be the ring of algebraic integers in $K$. Let $G$ be the Galois group of $\mathbb{Q}(\zeta)/\mathbb{Q}$. $G$ is isomorphic to $(\mathbb{Z}/l\mathbb{Z})^*$. Hence $G$ is a cyclic group of order $l - 1$. Let $\sigma$ be one of its generators. Then $\sigma(\zeta) = \zeta^r$, where $r$ is a primitive root mod $l$. Let $f$ be a positive divisor of $l - 1$. Let $e = (l - 1)/f$. There exists a unique subgroup $G_f$ of $G$ whose order is $f$. Let $K_f$ be the fixed subfield of $K$ by $G_f$. Then $e = [K_f : \mathbb{Q}]$. Let $A_f$ be the ring of algebraic integers in $K_f$.
We introduce the following algebraic integers.
\begin{align*} \eta_0 &= \zeta + \sigma^e(\zeta) + \sigma^{2e}(\zeta) + \dots + \sigma^{(f-1)e}(\zeta) \\ \eta_1 &= \sigma(\zeta) + \sigma^{e+1}(\zeta) + \dots + \sigma^{(f-1)e + 1}(\zeta) \\ &\vdots \\ \eta_{e-1} &= \sigma^{e-1}(\zeta) + \sigma^{2e-1}(\zeta) + \dots + \sigma^{fe - 1}(\zeta) \end{align*}
$\eta_0, \ldots, \eta_{e-1}$ are called periods of length $f$.
Applying $\sigma$ to $\eta_0$, we get $\eta_1$. Similarly,
$\eta_0 \to \eta_1 \to \dots \to \eta_{e-1} \to \eta_0$
My question: Is the following proposition true? If yes, how would you prove this?
Proposition The set $\{\eta_0, \ldots, \eta_{e-1}\}$ is an integral base of $A_f$.