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Show that \(w(t)\) is a code polynomial if and only if \(s_i = 0\) for all \(i\).

BCH codes have very attractive error correction algorithms. Let \(C\) be a BCH code in \(R_n\), and suppose that a code polynomial \(c(t) = c_0 + c_1 t + \cdots + c_{n-1} t^{n-1}\) is transmitted. Let \(w(t) = w_0 + w_1 t + \cdots w_{n-1} t^{n-1}\) be the polynomial in \(R_n\) that is received. If errors have occurred in bits \(a_1, \ldots, a_k\), then \(w(t) = c(t) + e(t)\), where \(e(t) = t^{a_1} + t^{a_2} + \cdots + t^{a_k}\) is the *error polynomial*. The decoder must determine the integers \(a_i\) and then recover \(c(t)\) from \(w(t)\) by flipping the \(a_i\)th bit. From \(w(t)\) we can compute \(w( \omega^i ) = s_i\) for \(i = 1, \ldots, 2r\), where \(\omega\) is a primitive \(n\)th root of unity over \({\mathbb Z}_2\). We say the *syndrome* of \(w(t)\) is \(s_1, \ldots, s_{2r}\).

Show that \(w(t)\) is a code polynomial if and only if \(s_i = 0\) for all \(i\).

Show that
\begin{equation*}s_i = w( \omega^i) = e( \omega^i) = \omega^{i a_1} + \omega^{i a_2} + \cdots + \omega^{i a_k}\end{equation*}
for \(i = 1, \ldots, 2r\). The *error-locator polynomial* is defined to be
\begin{equation*}s(x) = (x + \omega^{a_1})(x + \omega^{a_2}) \cdots (x + \omega^{a_k}).\end{equation*}

Recall the \((15,7)\)-block BCH code in Example 22.19. By Theorem 8.13, this code is capable of correcting two errors. Suppose that these errors occur in bits \(a_1\) and \(a_2\). The error-locator polynomial is \(s(x) = (x + \omega^{a_1})(x + \omega^{a_2})\). Show that \begin{equation*}s(x) = x^2 + s_1 x + \left( s_1^2 + \frac{s_3}{s_1} \right).\end{equation*}

Let \(w(t) = 1 + t^2 +t^4 + t^5 + t^7 + t^{12} + t^{13}\). Determine what the originally transmitted code polynomial was.