## Section22.2Polynomial Codes

With knowledge of polynomial rings and finite fields, it is now possible to derive more sophisticated codes than those of Chapter 8. First let us recall that an $$(n, k)$$-block code consists of a one-to-one encoding function $$E:{\mathbb Z}^{k}_{2} \rightarrow {\mathbb Z}^{n}_{2}$$ and a decoding function $$D:{\mathbb Z}^{n}_{2} \rightarrow {\mathbb Z}^{k}_{2}\text{.}$$ The code is error-correcting if $$D$$ is onto. A code is a linear code if it is the null space of a matrix $$H \in {\mathbb M}_{k \times n}({\mathbb Z}_2)\text{.}$$

We are interested in a class of codes known as cyclic codes. Let $$\phi : {\mathbb Z}_2^k \rightarrow {\mathbb Z}_2^n$$ be a binary $$(n,k)$$-block code. Then $$\phi$$ is a cyclic code if for every codeword $$(a_1, a_2, \ldots, a_n )\text{,}$$ the cyclically shifted $$n$$-tuple $$(a_n, a_1, a_2, \ldots, a_{n - 1} )$$ is also a codeword. Cyclic codes are particularly easy to implement on a computer using shift registers [2, 3].

###### Example22.14.

Consider the $$(6,3)$$-linear codes generated by the two matrices

\begin{equation*} G_1 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \quad \text{and} \quad G_2 = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\text{.} \end{equation*}

Messages in the first code are encoded as follows:

\begin{equation*} \begin{array}{rclccrcl} (000) & \mapsto & (000000) & & & (100) & \mapsto & (100100) \\ (001) & \mapsto & (001001) & & & (101) & \mapsto & (101101) \\ (010) & \mapsto & (010010) & & & (110) & \mapsto & (110110) \\ (011) & \mapsto & (011011) & & & (111) & \mapsto & (111111). \end{array} \end{equation*}

It is easy to see that the codewords form a cyclic code. In the second code, 3-tuples are encoded in the following manner:

\begin{equation*} \begin{array}{rclccrcl} (000) & \mapsto & (000000) & & & (100) & \mapsto & (111100) \\ (001) & \mapsto & (001111) & & & (101) & \mapsto & (110011) \\ (010) & \mapsto & (011110) & & & (110) & \mapsto & (100010) \\ (011) & \mapsto & (010001) & & & (111) & \mapsto & (101101). \end{array} \end{equation*}

This code cannot be cyclic, since $$(101101)$$ is a codeword but $$(011011)$$ is not a codeword.

### SubsectionPolynomial Codes

We would like to find an easy method of obtaining cyclic linear codes. To accomplish this, we can use our knowledge of finite fields and polynomial rings over $${\mathbb Z}_2\text{.}$$ Any binary $$n$$-tuple can be interpreted as a polynomial in $${\mathbb Z}_2[x]\text{.}$$ Stated another way, the $$n$$-tuple $$(a_0, a_1, \ldots, a_{n - 1} )$$ corresponds to the polynomial

\begin{equation*} f(x) = a_0 + a_1 x + \cdots + a_{n-1} x^{n - 1}\text{,} \end{equation*}

where the degree of $$f(x)$$ is at most $$n - 1\text{.}$$ For example, the polynomial corresponding to the $$5$$-tuple $$(10011)$$ is

\begin{equation*} 1 + 0 x + 0 x^2 + 1 x^3 + 1 x^4 = 1 + x^3 + x^4\text{.} \end{equation*}

Conversely, with any polynomial $$f(x) \in {\mathbb Z}_2[x]$$ with $$\deg f(x) \lt n$$ we can associate a binary $$n$$-tuple. The polynomial $$x + x^2 + x^4$$ corresponds to the $$5$$-tuple $$(01101)\text{.}$$

Let us fix a nonconstant polynomial $$g(x)$$ in $${\mathbb Z}_2[x]$$ of degree $$n - k\text{.}$$ We can define an $$(n,k)$$-code $$C$$ in the following manner. If $$(a_0, \ldots, a_{k - 1})$$ is a $$k$$-tuple to be encoded, then $$f(x) = a_0 + a_1 x + \cdots + a_{k - 1} x^{k - 1}$$ is the corresponding polynomial in $${\mathbb Z}_2[x]\text{.}$$ To encode $$f(x)\text{,}$$ we multiply by $$g(x)\text{.}$$ The codewords in $$C$$ are all those polynomials in $${\mathbb Z}_2[x]$$ of degree less than $$n$$ that are divisible by $$g(x)\text{.}$$ Codes obtained in this manner are called polynomial codes.

###### Example22.15.

If we let $$g(x)= 1 + x^3\text{,}$$ we can define a $$(6,3)$$-code $$C$$ as follows. To encode a $$3$$-tuple $$( a_0, a_1, a_2 )\text{,}$$ we multiply the corresponding polynomial $$f(x) = a_0 + a_1 x + a_2 x^2$$ by $$1 + x^3\text{.}$$ We are defining a map $$\phi : {\mathbb Z}_2^3 \rightarrow {\mathbb Z}_2^6$$ by $$\phi : f(x) \mapsto g(x) f(x)\text{.}$$ It is easy to check that this map is a group homomorphism. In fact, if we regard $${\mathbb Z}_2^n$$ as a vector space over $${\mathbb Z}_2\text{,}$$ $$\phi$$ is a linear transformation of vector spaces (see Exercise 20.5.15, Chapter 20). Let us compute the kernel of $$\phi\text{.}$$ Observe that $$\phi ( a_0, a_1, a_2 ) = (000000)$$ exactly when

\begin{align*} 0 + 0x + 0x^2 + 0x^3 + 0x^4 + 0 x^5 & = (1 + x^3) ( a_0 + a_1 x + a_2 x^2 )\\ & = a_0 + a_1 x + a_2 x^2 + a_0 x^3 + a_1 x^4 + a_2 x^5\text{.} \end{align*}

Since the polynomials over a field form an integral domain, $$a_0 + a_1 x + a_2 x^2$$ must be the zero polynomial. Therefore, $$\ker \phi = \{ (000) \}$$ and $$\phi$$ is one-to-one.

To calculate a generator matrix for $$C\text{,}$$ we merely need to examine the way the polynomials $$1\text{,}$$ $$x\text{,}$$ and $$x^2$$ are encoded:

\begin{align*} (1 + x^3) \cdot 1 & = 1 + x^3\\ (1 + x^3)x & = x + x^4\\ (1 + x^3)x^2 & = x^2 + x^5\text{.} \end{align*}

We obtain the code corresponding to the generator matrix $$G_1$$ in Example 22.14. The parity-check matrix for this code is

\begin{equation*} H = \begin{pmatrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{pmatrix}\text{.} \end{equation*}

Since the smallest weight of any nonzero codeword is $$2\text{,}$$ this code has the ability to detect all single errors.

Rings of polynomials have a great deal of structure; therefore, our immediate goal is to establish a link between polynomial codes and ring theory. Recall that $$x^n - 1 = (x - 1)( x^{n - 1} + \cdots + x + 1)\text{.}$$ The factor ring

\begin{equation*} R_n = {\mathbb Z}_2[x]/ \langle x^n - 1 \rangle \end{equation*}

can be considered to be the ring of polynomials of the form

\begin{equation*} f(t) = a_0 + a_1 t + \cdots + a_{n - 1} t^{n - 1} \end{equation*}

that satisfy the condition $$t^n = 1\text{.}$$ It is an easy exercise to show that $${\mathbb Z}_2^n$$ and $$R_n$$ are isomorphic as vector spaces. We will often identify elements in $${\mathbb Z}_2^n$$ with elements in $${\mathbb Z}[x] / \langle x^n - 1 \rangle\text{.}$$ In this manner we can interpret a linear code as a subset of $${\mathbb Z}[x] / \langle x^n - 1 \rangle\text{.}$$

The additional ring structure on polynomial codes is very powerful in describing cyclic codes. A cyclic shift of an $$n$$-tuple can be described by polynomial multiplication. If $$f(t) = a_0 + a_1 t + \cdots + a_{n - 1} t^{n - 1}$$ is a code polynomial in $$R_n\text{,}$$ then

\begin{equation*} tf(t) = a_{n - 1} + a_0 t + \cdots + a_{n - 2} t^{n - 1} \end{equation*}

is the cyclically shifted word obtained from multiplying $$f(t)$$ by $$t\text{.}$$ The following theorem gives a beautiful classification of cyclic codes in terms of the ideals of $$R_n\text{.}$$

Let $$C$$ be a linear cyclic code and suppose that $$f(t)$$ is in $$C\text{.}$$ Then $$t f(t)$$ must also be in $$C\text{.}$$ Consequently, $$t^k f(t)$$ is in $$C$$ for all $$k \in {\mathbb N}\text{.}$$ Since $$C$$ is a linear code, any linear combination of the codewords $$f(t), tf(t), t^2f(t), \ldots, t^{n-1}f(t)$$ is also a codeword; therefore, for every polynomial $$p(t)\text{,}$$ $$p(t)f(t)$$ is in $$C\text{.}$$ Hence, $$C$$ is an ideal.

Conversely, let $$C$$ be an ideal in $${\mathbb Z}_2[x]/\langle x^n + 1\rangle\text{.}$$ Suppose that $$f(t) = a_0 + a_1 t + \cdots + a_{n - 1} t^{n - 1}$$ is a codeword in $$C\text{.}$$ Then $$t f(t)$$ is a codeword in $$C\text{;}$$ that is, $$(a_1, \ldots, a_{n-1}, a_0)$$ is in $$C\text{.}$$

Theorem 22.16 tells us that knowing the ideals of $$R_n$$ is equivalent to knowing the linear cyclic codes in $${\mathbb Z}_2^n\text{.}$$ Fortunately, the ideals in $$R_n$$ are easy to describe. The natural ring homomorphism $$\phi : {\mathbb Z}_2[x] \rightarrow R_n$$ defined by $$\phi[f(x)] = f(t)$$ is a surjective homomorphism. The kernel of $$\phi$$ is the ideal generated by $$x^n - 1\text{.}$$ By Theorem 16.34, every ideal $$C$$ in $$R_n$$ is of the form $$\phi(I)\text{,}$$ where $$I$$ is an ideal in $${\mathbb Z}_2[x]$$ that contains $$\langle x^n - 1 \rangle\text{.}$$ By Theorem 17.20, we know that every ideal $$I$$ in $${\mathbb Z}_2[x]$$ is a principal ideal, since $${\mathbb Z}_2$$ is a field. Therefore, $$I = \langle g(x) \rangle$$ for some unique monic polynomial in $${\mathbb Z}_2[x]\text{.}$$ Since $$\langle x^n - 1 \rangle$$ is contained in $$I\text{,}$$ it must be the case that $$g(x)$$ divides $$x^n - 1\text{.}$$ Consequently, every ideal $$C$$ in $$R_n$$ is of the form

\begin{equation*} C = \langle g(t) \rangle = \{ f(t)g(t) : f(t) \in R_n \text{ and } g(x) \mid (x^n - 1) \text{ in } {\mathbb Z}_2[x] \}\text{.} \end{equation*}

The unique monic polynomial of the smallest degree that generates $$C$$ is called the minimal generator polynomial of $$C\text{.}$$

###### Example22.17.

If we factor $$x^7 - 1$$ into irreducible components, we have

\begin{equation*} x^7 - 1 = (1 + x)(1 + x + x^3)(1+ x^2 + x^3)\text{.} \end{equation*}

We see that $$g(t) = (1 + t + t^3)$$ generates an ideal $$C$$ in $$R_7\text{.}$$ This code is a $$(7, 4)$$-block code. As in Example 22.15, it is easy to calculate a generator matrix by examining what $$g(t)$$ does to the polynomials 1, $$t\text{,}$$ $$t^2\text{,}$$ and $$t^3\text{.}$$ A generator matrix for $$C$$ is

\begin{equation*} G = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\text{.} \end{equation*}

In general, we can determine a generator matrix for an $$(n, k)$$-code $$C$$ by the manner in which the elements $$t^k$$ are encoded. Let $$x^n - 1 = g(x) h(x)$$ in $${\mathbb Z}_2[x]\text{.}$$ If $$g(x) = g_0 + g_1 x + \cdots + g_{n-k} x^{n-k}$$ and $$h(x) = h_0 + h_1 x + \cdots + h_k x^k\text{,}$$ then the $$n \times k$$ matrix

\begin{equation*} G = \begin{pmatrix} g_0 & 0 & \cdots & 0 \\ g_1 & g_0 & \cdots & 0 \\ \vdots & \vdots &\ddots & \vdots \\ g_{n-k} & g_{n-k-1} & \cdots & g_0 \\ 0 & g_{n-k} & \cdots & g_{1} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & g_{n-k} \end{pmatrix} \end{equation*}

is a generator matrix for the code $$C$$ with generator polynomial $$g(t)\text{.}$$ The parity-check matrix for $$C$$ is the $$(n - k) \times n$$ matrix

\begin{equation*} H = \begin{pmatrix} 0 & \cdots & 0 & 0 & h_k & \cdots & h_0 \\ 0 & \cdots & 0 & h_k & \cdots & h_0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ h_k & \cdots & h_0 & 0 & 0 & \cdots & 0 \end{pmatrix}\text{.} \end{equation*}

We will leave the details of the proof of the following proposition as an exercise.

###### Example22.19.
\begin{equation*} x^7 - 1 = g(x) h(x) = (1 + x + x^3)(1 + x + x^2 + x^4)\text{.} \end{equation*}

Therefore, a parity-check matrix for this code is

\begin{equation*} H = \begin{pmatrix} 0 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 & 1 & 0 & 0 \end{pmatrix}\text{.} \end{equation*}

To determine the error-detecting and error-correcting capabilities of a cyclic code, we need to know something about determinants. If $$\alpha_1, \ldots, \alpha_n$$ are elements in a field $$F\text{,}$$ then the $$n \times n$$ matrix

\begin{equation*} \begin{pmatrix} 1 & 1 & \cdots & 1 \\ \alpha_1 & \alpha_2 & \cdots & \alpha_n \\ \alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ \alpha_1^{n-1} & \alpha_2^{n-1} & \cdots & \alpha_n^{n-1} \end{pmatrix} \end{equation*}

is called the Vandermonde matrix. The determinant of this matrix is called the Vandermonde determinant. We will need the following lemma in our investigation of cyclic codes.

We will induct on $$n\text{.}$$ If $$n = 2\text{,}$$ then the determinant is $$\alpha_2 - \alpha_1\text{.}$$ Let us assume the result for $$n - 1$$ and consider the polynomial $$p(x)$$ defined by

\begin{equation*} p(x) = \det \begin{pmatrix} 1 & 1 & \cdots & 1 & 1 \\ \alpha_1 & \alpha_2 & \cdots & \alpha_{n-1} & x \\ \alpha_1^2 & \alpha_2^2 & \cdots & \alpha_{n-1}^2 & x^2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ \alpha_1^{n-1} & \alpha_2^{n-1} & \cdots & \alpha_{n-1}^{n-1} & x^{n-1} \end{pmatrix}\text{.} \end{equation*}

Expanding this determinant by cofactors on the last column, we see that $$p(x)$$ is a polynomial of at most degree $$n-1\text{.}$$ Moreover, the roots of $$p(x)$$ are $$\alpha_1, \ldots, \alpha_{n-1}\text{,}$$ since the substitution of any one of these elements in the last column will produce a column identical to the last column in the matrix. Remember that the determinant of a matrix is zero if it has two identical columns. Therefore,

\begin{equation*} p(x) = (x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_{n-1}) \beta\text{,} \end{equation*}

where

\begin{equation*} \beta = (-1)^{n + n} \det \begin{pmatrix} 1 & 1 & \cdots & 1 \\ \alpha_1 & \alpha_2 & \cdots & \alpha_{n-1} \\ \alpha_1^2 & \alpha_2^2 & \cdots & \alpha_{n-1}^2 \\ \vdots & \vdots & \ddots & \vdots \\ \alpha_1^{n-2} & \alpha_2^{n-2} & \cdots & \alpha_{n-1}^{n-2} \end{pmatrix}\text{.} \end{equation*}

By our induction hypothesis,

\begin{equation*} \beta = (-1)^{n+n} \prod_{1 \leq j \lt i \leq n - 1} (\alpha_i - \alpha_j)\text{.} \end{equation*}

If we let $$x = \alpha_n\text{,}$$ the result now follows immediately.

The following theorem gives us an estimate on the error detection and correction capabilities for a particular generator polynomial.

Suppose that

\begin{equation*} g( \omega^r) = g(\omega^{r + 1}) = \cdots = g( \omega^{r + s - 1}) = 0\text{.} \end{equation*}

Let $$f(x)$$ be some polynomial in $$C$$ with $$s$$ or fewer nonzero coefficients. We can assume that

\begin{equation*} f(x) = a_{i_0} x^{i_0} + a_{i_1} x^{i_1} + \cdots + a_{i_{s - 1}} x^{i_{s - 1}} \end{equation*}

be some polynomial in $$C\text{.}$$ It will suffice to show that all of the $$a_i$$'s must be 0. Since

\begin{equation*} g( \omega^r) = g(\omega^{r + 1}) = \cdots = g( \omega^{r + s - 1}) = 0 \end{equation*}

and $$g(x)$$ divides $$f(x)\text{,}$$

\begin{equation*} f( \omega^r) = f(\omega^{r + 1}) = \cdots = f( \omega^{r + s - 1}) = 0\text{.} \end{equation*}

Equivalently, we have the following system of equations:

\begin{align*} a_{i_0} (\omega^r)^{i_0} + a_{i_1} (\omega^r)^{i_1} + \cdots + a_{i_{s - 1}} (\omega^r)^{i_{s - 1}} & = 0\\ a_{i_0} (\omega^{r + 1})^{i_0} + a_{i_1} (\omega^{r + 1})^{i_2} + \cdots + a_{i_{s-1}} (\omega^{r+1})^{i_{s-1}} & = 0\\ & \vdots\\ a_{i_0} (\omega^{r + s - 1})^{i_0} + a_{i_1} (\omega^{r + s - 1})^{i_1} + \cdots + a_{i_{s - 1}} (\omega^{r + s - 1})^{i_{s - 1}} & = 0\text{.} \end{align*}

Therefore, $$(a_{i_0}, a_{i_1}, \ldots, a_{i_{s - 1}})$$ is a solution to the homogeneous system of linear equations

\begin{align*} (\omega^{i_0})^r x_0 + (\omega^{i_1})^r x_1 + \cdots + (\omega^{i_{s - 1}})^r x_{n - 1} & = 0\\ (\omega^{i_0})^{r + 1} x_0 + (\omega^{i_1})^{r + 1} x_1 + \cdots + (\omega^{i_{s - 1}})^{r + 1} x_{n - 1} & = 0\\ & \vdots\\ (\omega^{i_0})^{r + s - 1} x_0 + (\omega^{i_1})^{r + s - 1} x_1 + \cdots + (\omega^{i_{s - 1}})^{r + s - 1} x_{n - 1} & = 0\text{.} \end{align*}

However, this system has a unique solution, since the determinant of the matrix

\begin{equation*} \begin{pmatrix} (\omega^{i_0})^r & (\omega^{i_1})^r & \cdots & (\omega^{i_{s-1}})^r \\ (\omega^{i_0})^{r+1} & (\omega^{i_1})^{r+1} & \cdots & (\omega^{i_{s-1}})^{r+1} \\ \vdots & \vdots & \ddots & \vdots \\ (\omega^{i_0})^{r+s-1} & (\omega^{i_1})^{r+s-1} & \cdots & (\omega^{i_{s-1}})^{r+s-1} \end{pmatrix} \end{equation*}

can be shown to be nonzero using Lemma 22.20 and the basic properties of determinants (Exercise). Therefore, this solution must be $$a_{i_0} = a_{i_1} = \cdots = a_{i_{s - 1}} = 0\text{.}$$

### SubsectionBCH Codes

Some of the most important codes, discovered independently by A. Hocquenghem in 1959 and by R. C. Bose and D. V. Ray-Chaudhuri in 1960, are BCH codes. The European and transatlantic communication systems both use BCH codes. Information words to be encoded are of length $$231\text{,}$$ and a polynomial of degree $$24$$ is used to generate the code. Since $$231 + 24 = 255 = 2^8-1\text{,}$$ we are dealing with a $$(255, 231)$$-block code. This BCH code will detect six errors and has a failure rate of $$1$$ in $$16$$ million. One advantage of BCH codes is that efficient error correction algorithms exist for them.

The idea behind BCH codes is to choose a generator polynomial of smallest degree that has the largest error detection and error correction capabilities. Let $$d = 2r + 1$$ for some $$r \geq 0\text{.}$$ Suppose that $$\omega$$ is a primitive $$n$$th root of unity over $${\mathbb Z}_2\text{,}$$ and let $$m_i(x)$$ be the minimal polynomial over $${\mathbb Z}_2$$ of $$\omega^i\text{.}$$ If

\begin{equation*} g(x) = \lcm[ m_1(x), m_{2}(x), \ldots, m_{2r}(x)]\text{,} \end{equation*}

then the cyclic code $$\langle g(t) \rangle$$ in $$R_n$$ is called the BCH code of length $$n$$ and distance $$d\text{.}$$ By Theorem 22.21, the minimum distance of $$C$$ is at least $$d\text{.}$$

(1) $$\Rightarrow$$ (2). If $$f(t)$$ is in $$C\text{,}$$ then $$g(x) \mid f(x)$$ in $${\mathbb Z}_2[x]\text{.}$$ Hence, for $$i = 1, \ldots, 2r\text{,}$$ $$f( \omega^i) = 0$$ since $$g( \omega^i ) = 0\text{.}$$ Conversely, suppose that $$f( \omega^i) = 0$$ for $$1 \leq i \leq d\text{.}$$ Then $$f(x)$$ is divisible by each $$m_i(x)\text{,}$$ since $$m_i(x)$$ is the minimal polynomial of $$\omega^i\text{.}$$ Therefore, $$g(x) \mid f(x)$$ by the definition of $$g(x)\text{.}$$ Consequently, $$f(x)$$ is a codeword.

(2) $$\Rightarrow$$ (3). Let $$f(t) = a_0 + a_1 t + \cdots + a_{n - 1}v t^{n - 1}$$ be in $$R_n\text{.}$$ The corresponding $$n$$-tuple in $${\mathbb Z}_2^n$$ is $${\mathbf x} = (a_0 a_1 \cdots a_{n - 1})^\transpose\text{.}$$ By (2),

\begin{equation*} H {\mathbf x} = \begin{pmatrix} a_0 + a_1 \omega + \cdots + a_{n-1} \omega^{n-1} \\ a_0 + a_1 \omega^2 + \cdots + a_{n-1} (\omega^2)^{n-1} \\ \vdots \\ a_0 + a_1 \omega^{2r} + \cdots + a_{n-1} (\omega^{2r})^{n-1} \end{pmatrix} = \begin{pmatrix} f(\omega) \\ f(\omega^2) \\ \vdots \\ f(\omega^{2r}) \end{pmatrix} = 0 \end{equation*}

exactly when $$f(t)$$ is in $$C\text{.}$$ Thus, $$H$$ is a parity-check matrix for $$C\text{.}$$

(3) $$\Rightarrow$$ (1). By (3), a code polynomial $$f(t) = a_0 + a_1 t + \cdots + a_{n - 1} t^{n - 1}$$ is in $$C$$ exactly when $$f(\omega^i) = 0$$ for $$i = 1, \ldots, 2r\text{.}$$ The smallest such polynomial is $$g(t) = \lcm[ m_1(t),\ldots, m_{2r}(t)]\text{.}$$ Therefore, $$C = \langle g(t) \rangle\text{.}$$

###### Example22.23.

It is easy to verify that $$x^{15} - 1 \in {\mathbb Z}_2[x]$$ has a factorization

\begin{equation*} x^{15} - 1 = (x + 1)(x^2 + x + 1)(x^4 + x + 1)(x^4 + x^3 + 1)(x^4 + x^3 + x^2 + x + 1)\text{,} \end{equation*}

where each of the factors is an irreducible polynomial. Let $$\omega$$ be a root of $$1 + x + x^4\text{.}$$ The Galois field $$\gf(2^4)$$ is

\begin{equation*} \{ a_0 + a_1 \omega + a_2 \omega^2 + a_3 \omega^3 : a_i \in {\mathbb Z}_2 \text{ and } 1 + \omega + \omega^4 = 0 \}\text{.} \end{equation*}

By Example 22.8, $$\omega$$ is a primitive $$15$$th root of unity. The minimal polynomial of $$\omega$$ is $$m_1(x) = 1 + x + x^4\text{.}$$ It is easy to see that $$\omega^2$$ and $$\omega^4$$ are also roots of $$m_1(x)\text{.}$$ The minimal polynomial of $$\omega^3$$ is $$m_2(x) = 1 + x + x^2 + x^3 + x^4\text{.}$$ Therefore,

\begin{equation*} g(x) = m_1(x) m_2(x) = 1 + x^4 + x^6 + x^7 + x^8 \end{equation*}

has roots $$\omega\text{,}$$ $$\omega^2\text{,}$$ $$\omega^3\text{,}$$ $$\omega^4\text{.}$$ Since both $$m_1(x)$$ and $$m_2(x)$$ divide $$x^{15} - 1\text{,}$$ the BCH code is a $$(15, 7)$$-code. If $$x^{15} -1 = g(x)h(x)\text{,}$$ then $$h(x) = 1 + x^4 + x^6 + x^7\text{;}$$ therefore, a parity-check matrix for this code is

\begin{equation*} \left( \begin{array}{ccccccccccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right)\text{.} \end{equation*}