We need a binary code of length \(5\text{,}\) so we need to choose \(m\) such that \(2^m-1\) is a multiple of \(5\text{,}\) which first happens when \(m=4\text{.}\) Then if \(\beta\) is the primitive generator of \(\F_{2^4}=\F_2[\beta]/(\beta^4+\beta+1)\) we can choose \(\alpha=\beta^3\) to get our element of order \(5\text{.}\) Since the code is narrow-sense we take \(b=1\text{.}\) To get a designed distance of \(3\) we need a dimension of \(5-3+1=3\) so a parity-check matrix for the underlying RS code is
\begin{equation*}
H_{\RS} = \begin{bmatrix}
1 \amp \beta^3 \amp \beta^6 \amp \beta^9 \amp \beta^{12} \\
1 \amp \beta^6 \amp \beta^{12} \amp \beta^3 \amp \beta^9
\end{bmatrix}\text{.}
\end{equation*}
As in the example in class, the second row corresponding to the root condition \(c(\alpha^2)=0\) is implied by the first rowβs root condition of \(c(\alpha)=0\text{.}\) So we construct a check matrix for the BCH code by replacing the first row of \(H_{\RS}\) with the column vector coordinate representations of these elements of \(\F_{2^4}\) in terms of the basis \(\{1,\beta,\beta^2,\beta^3\}\) of \(\F_{2^4}\) over \(\F_{2}\) to get
\begin{equation*}
H_{\mathrm{BCH}}=\begin{bmatrix}
1 \amp 0 \amp 0 \amp 0 \amp 1\\
0 \amp 0 \amp 0 \amp 1 \amp 1\\
0 \amp 0 \amp 1 \amp 0 \amp 1\\
0 \amp 1 \amp 1 \amp 1 \amp 1
\end{bmatrix}\text{.}
\end{equation*}
This matrix has full rank (seen quickly by swapping row 2 and row 4), so this is the parity-check matrix for the code and this BCH code is a \([5,1,5]_2\) code, since a quick check shows that \((1,1,1,1,1)\) is orthogonal to \(H_{\mathrm{BCH}}\text{.}\)