$$\begin{bmatrix} 0 &-1 \\ 1 & 0\end{bmatrix}\to\begin{bmatrix} 1 &-1 \\ 1 & 0\end{bmatrix}\to\begin{bmatrix} 1 &-1 \\ 0 & 1\end{bmatrix}\to\begin{bmatrix} 1 &0 \\ 0 & 1\end{bmatrix}$$
and
$$\begin{bmatrix} 2 &0 \\ 0 & 1/2\end{bmatrix}\to\begin{bmatrix} 2 &0 \\ 2 & 1/2\end{bmatrix}\to \begin{bmatrix} 1 &-1/4 \\ 2 & 1/2\end{bmatrix}\to \begin{bmatrix} 1 &-1/4 \\ 0 & 1\end{bmatrix}\to \begin{bmatrix} 1 &0 \\ 0 & 1\end{bmatrix}$$

We have the chain of intermediate fields $$F=K_0\leq K_1 \leq \ldots \leq K_r=K$$

Let $\omega$ be the $n$th primitive root of $1$.

We consider the chain $$F\leq F(\omega)\leq F_1\leq \ldots \leq F_r=K(\omega)$$ with $F_i=K_i(\omega)$.

We have that $F_i=F_{i-1}K_i$ and $\text{Gal}(F_i/F_{i-1})\hookrightarrow \text{Gal}(K_i/K_{i-1})$.

Why does it hold that the exponent of $\text{Gal}(F_i/F_{i-1})$ divides the exponent of $\text{Gal}(K_i/K_{i-1})$ ?

a1*v1+a2*v2+...+an*vn=0 => a1=a2=...=an=0
that's the definition for linear independent vectors, so when i try to solve a1*v1+a2*v2+a3*v3+a4*v4=0 i get a polynomial with a comination of a1,a2,a3,a4 for coefficients and those combinations have to be equal to 0 because the RHS is a zero polynomial, and if by solving i get a1=a2=a3=a4=0 that means the vectors are linearly independent?

@user379685 I think you can look at the coordinate matrix $\begin{bmatrix}
-1 &-2 &0 &1 \\
1&-1 &1 &1 \\
2&4 &0 &2 \\
1& 2& 0& 1
\end{matrix}$. I think this is not invertible.