Let $F_4$ be the field ${\{\overline{00}, \overline{01}, \overline{10}, \overline{11}}\}$, where the elements are in $F_2^2$, which has the $+$ operation of bitwise addition, $00 = \mathbf{0}$, the zero element of $F_4$, $01 = \mathbf{1}$, the one in $F_4$ and the following ( sorry I do not know how to create a multiplication table on latex); $ \overline{10}*\overline10 = \overline{11}, \overline{10} * \overline{11} = \overline{01}, \overline{11} * \overline{11} = \overline{10}$.
Let $\mathbf{\alpha} = \overline{10}$
Let $Z$ be the 4-ary linear code with generator matrix $\ G= \left[ {\begin{array}{cccccc} 1 & 0 & 0 & 1 & \alpha & \alpha\\ 0 & 1 & 0 & \alpha & 1 & \alpha\\ 0 & 0 & 1 & \alpha & \alpha & 1\\ \end{array} } \right] $
Now, I want to show that $W_Z(x,y) = x^6 + 3f(x,y)$, where f(x,y) is a polynomial with integer coefficients.
Now, I know $W_Z(x,y) = \sum_{\mathbf{z} \in Z} x^{(n - w(\mathbf{z}))}y^{w(\mathbf{z})}$, where $w(\mathbf{z})$ is the weight of a codevector $\mathbf{z} \in Z$.
Now,I know tht the $x^6$ term comes from there being one codevector with weight $\mathbf{0}$, so the problem is reduced to showing that for all $ i = 1,...,6$ , the number of codevectors $ \in Z$ of weight $i$ is divisible by 3. This is where I'm stuck; any help would be appreciated!
Hint: If $\bf{z}$ is a codeword, then $\alpha\bf{z}$ and $\alpha^2\bf{z}$ are codewords of the same Hamming weight. They differ from each other unless _____ (you fill in the blank).