I'm looking through the lecture notes by Alexandar Barg ENEE626:Error-Correcting Codes Lecture 13.
We are given an example which says:
"Consider the Hamming Code $H_{4}[15,11,3]$.
Let $\alpha$ be a primitive $14^{th}$ root of unity. We can construct a BCH code $C$ with zeroes $\alpha$ and $\alpha_{3}$. The parity check matrix is:
$$H = \begin{bmatrix} 1 & \alpha & \alpha^{2} & \alpha^{3} & \alpha^{4} & \dots & \alpha^{14} \\ 1 & \alpha^{3} & \alpha^{6} & \alpha^{9} & \alpha^{12} & \dots & \alpha^{12} \end{bmatrix} $$
This is a $8$ by $15$ parity check matrix with all eight rows linearly independent. So, we have to a $[15,7]$ code. Note that the dimension is $n-2m$, which will be true for any BCH code. It remains to prove that $d=5$."
My question is, how do we know that $\alpha$ and $\alpha^{3}$ are the zeroes? And how is the above parity check matrix an $8$ by $15$ matrix? It looks like a $2$ by $15$ matrix?
"Let $\alpha$ be a $14^{th}$ root of unity"
This statement is false, as the $n^{th}$ roots of unity are the roots of the polynomial $x^{n}-1$. As we are working over $\mathbb{F}_{15}$, $n=15$ and as such $\alpha$ must be a $15^{th}$ root of unity.
"This is a $8$ by $15$ parity check matrix with all eight rows linearly independent. So, we have a [$15,7$] code. "
The parity check matrix shown is incorrect. $H$ should only have one row of $\alpha$, as the {$15, 11, 3$] Hamming code is a SINGLE error correcting code. If it were a DOUBLE error-correcting code, then we would have two rows of $\alpha$ and $\alpha_{3}$ respectively. Recall, the $\alpha_{i}$'s are the primitive elements of $\mathbb{F}_{15}$ and as the length of each of the primitive elements is $4$, then each of the elements of $\alpha$ consists of column vectors of length $4$ (the binary expression of the primitive elements). As such, the parity check matrix is a $4 \times 15$ matrix rather than an $8 \times 15$ matrix as the lecture notes had suggested.