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Let $mu=e^{2pi i/9}$ . Let $u_j:=(mu^j,mu^{2j}, mu ^{3j},...,mu^{9j})^T in mathbb C^9$, for $j=1,...,9$.

Let $V$ be the vector subspace of $mathbb C^9$ spanned by ${u_2,u_3,u_4,u_5,u_6,u_7}$ .

I can show that there exists $A in M_{3 times 6} (mathbb C)$ such that

$V={(x_1,...,x_9)in mathbb C^9 : (x_3,x_6,x_9)^T = A (x_1,x_2,x_4,x_5,x_7,x_8)^T }$ .

If On a special kind of $6$-dimensional vector subspace of $mathbb C^9$ is true, then this $A$ is unique and it is of the form $A=begin{pmatrix} a_1 &a_2 &a_3&a_4&a_5&a_6 \ a_5&a_6&a_1&a_2&a_3&a_4 \ a_3&a_4&a_5&a_6&a_1&a_2 end{pmatrix}$.

Let $L$ be the smalles t subsfield of $mathbb C$ generated by the entries of $A$.

I can show that $L subseteq mathbb Q(mu) $ , and hence $L/mathbb Q$ is a Galois-extension.

My question is : What is the extension degree $[L : mathbb Q]$ ?

Since $[mathbb Q(mu):mathbb Q]=phi(9)=6$, so $[L:mathbb Q]=1,2,3, $ or $6$, but exactly which one is it ?

Let's re-write slightly using the facts you have established.

Let $K=mathbb{Q}[mu]$ where $mu$ is a primitive $9$-th root of unity. For $j=0,dots, 8$ let $u_j$ be the column vector whose $i$-th component (for $i=0,dots, 8$) is $mu^{ji}$.

Let $M$ be the matrix whose columns are $u_0,dots, u_8$. Note that we have the orthogonality conditions $bar{M}^{T}M=9I$. Amongst other things this shows that the nine columns are linearly independent.

It is convenient to re-order the rows and columns in this order: $0,3,6,1,2,4,8,7,5$.

Let $V$ be the $K$-span of $u_2,u_3,u_4,u_4,u_6,u_7$.

We are given that $V$ is the null space of some $B:=[I_3 A]$, and that the $A$ in question is unique. Again, for convenience we've replaced $A$ of the question by its negative. Let us then find $A$ explicitly.

As the columns $u_2,u_3,u_4,u_4,u_6,u_7$ are in the null space of $B$ the rows of $B$ are orthogonal to these six vectors. However, by the orthogonality relations we have that the three rows $bar{u_{0}}^T,bar{u_{1}}^T,bar{u_{8}}^T$ are orthogonal to the six vectors. These three rows form the matrix
$$
B_1:=begin{bmatrix}
1 & 1 & 1 &1 & 1 & 1 &1 & 1 & 1\
1 & omega^2 & omega &mu^{-1} & mu^{-2} & mu^{-4}&mu & mu^2 & mu^4\
1 & omega & omega^2 &mu & mu^2 & mu^4 &mu^{-1} & mu^{-2} & mu^{-4}\
end{bmatrix}=[Omega Delta bar{Delta}], text{ say,}
$$
where $omega:=mu^3$.

Now $bar{Omega}^{T}B_1$ also annihilates the six vectors $u_2,u_3,u_4,u_4,u_6,u_7$, and as $Omega$ also satisfies the orthogonality conditions we have
$$
bar{Omega}^{T}B_1=[ 3I (bar{Omega}^{T}Delta) (bar{Omega}^{T}bar{Delta})].
$$

By the uniqueness of $A$, then, we have
$$
A=frac{1}{3}[(bar{Omega}^{T}Delta) (bar{Omega}^{T}bar{Delta})].
$$

It is now routine to calculate the entries of $A$. They all lie in the degree $3$ extension $mathbb{Q} [mu+mu^{-1}, mu^2+mu^{-2},mu^4+mu^{-4}]$ and they are not all rational.