On a special kind of six dimensional vector subspace of $\mathbb C^9$ related to the primitive $9$-th root of unity

Question

Let $\mu=e^{2\pi i/9}$ . 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 ?

Answer 1

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.