section-poly-codes.ipynb

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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 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

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

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

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

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

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

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 .

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 20.4.15, Chapter 20). Let us compute the kernel of $\phi\text{.}$ Observe that $\phi ( a_0, a_1, a_2 ) = (000000)$ exactly when

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

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:

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

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

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

Since the smallest weight of any nonzero codeword is 2, 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 $R_n\text{.}$

Theorem22.16

A linear code $C$ in ${\mathbb Z}_2^n$ is cyclic if and only if it is an ideal in $R_n = {\mathbb Z}[x] / \langle x^n - 1 \rangle\text{.}$

Proof

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

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

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

Example22.17

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

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

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

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

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

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

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

Proposition22.18

Let $C = \langle g(t) \rangle$ be a cyclic code in $R_n$ and suppose that $x^n - 1 = g(x) h(x)\text{.}$ Then $G$ and $H$ are generator and parity-check matrices for $C\text{,}$ respectively. Furthermore, $HG = 0\text{.}$

Example22.19

In Example 22.17,

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

Therefore, a parity-check matrix for this code is

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

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

Lemma22.20

Let $\alpha_1, \ldots, \alpha_n$ be elements in a field $F$ with $n \geq 2\text{.}$ Then

$$\begin{equation*} \det \begin{pmatrix} 1 & 1 & \cdots & 1 \\ \alpha_1 & \alpha_2 & \cdots & \alpha_n \\ \alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ \alpha_1^{n-1} & \alpha_2^{n-1} & \cdots & \alpha_n^{n-1} \end{pmatrix} = \prod_{1 \leq j \lt i \leq n} (\alpha_i - \alpha_j). \end{equation*}$$

In particular, if the $\alpha_i$'s are distinct, then the determinant is nonzero.

Proof

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

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

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,

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

where

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

By our induction hypothesis,

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

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.

Theorem22.21

Let $C = \langle g(t) \rangle$ be a cyclic code in $R_n$ and suppose that $\omega$ is a primitive $n$th root of unity over ${\mathbb Z}_2\text{.}$ If $s$ consecutive powers of $\omega$ are roots of $g(x)\text{,}$ then the minimum distance of $C$ is at least $s + 1\text{.}$

Proof

Suppose that

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

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 $f(x)\text{,}$

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

Equivalently, we have the following system of equations:

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

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

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

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, 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

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

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

Theorem22.22

Let $C = \langle g(t) \rangle$ be a cyclic code in $R_n\text{.}$ The following statements are equivalent.

1. The code $C$ is a BCH code whose minimum distance is at least $d\text{.}$

2. A code polynomial $f(t)$ is in $C$ if and only if $f( \omega^i) = 0$ for $1 \leq i \lt d\text{.}$

3. The matrix

   $$\begin{equation*} H = \begin{pmatrix} 1 & \omega & \omega^2 & \cdots & \omega^{n-1}\\ 1 & \omega^2 & \omega^{4} & \cdots & \omega^{(n-1)(2)} \\ 1 & \omega^3 & \omega^{6} & \cdots & \omega^{(n-1)(3)} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & \omega^{2r} & \omega^{4r} & \cdots & \omega^{(n-1)(2r)} \end{pmatrix} \end{equation*}$$

   is a parity-check matrix for $C\text{.}$

Proof

(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 ${\mathbf x} = (a_0 a_1 \cdots a_{n - 1})^{\rm t}\text{.}$ 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

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

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

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

By Example 22.8, $\omega$ is a primitive 15th 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

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