Extension Field

GF(2²) – Example of an extension field

In "Example 2" of the chapter "Some Basics of Algebra" it has already been shown that the finite set of numbers $\{0,\ 1,\ 2,\ 3\}$ ⇒ $q = 4$ does not satisfy the properties of a Galois field $\rm GF(4)$. Rather, the following tables result for the "addition modulo 4" and the "multiplication modulo 4":

$$ \begin{array}{c} {\rm modulo}\hspace{0.15cm}{\it q} = 4\\ \end{array}\hspace{0.25cm} \Rightarrow\hspace{0.25cm}\text{Addition: } \left[ \begin{array}{c|cccccc} + & 0 & 1 &2 & 3 \\ \hline 0 & 0 & 1 &2 & 3 \\ 1 & 1 & 2 &3 & 0 \\ 2 & 2 & 3 &0 & 1 \\ 3 & 3 & 0 &1 & 2 \end{array} \right] \hspace{-0.1cm} ,\hspace{0.25cm}\text{Multiplication: } \left[ \begin{array}{c|cccccc} \cdot & 0 & 1 &2 & 3 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 3 \\ 2 & 0 & 2 & 0 & 2 \\ 3 & 0 & 3 & 2 & 1 \\ \end{array} \right] . $$

For $z_i = 2$ there is no multiplicative inverse ${\rm Inv_M}(z_i)$. This can be seen from the fact that no single element $z_i ∈ \{0,\ 1,\ 2,\ 3\}$ satisfies the condition $2 · z_i = 1$.

On the other hand, if we start from the binary Galois field ${\rm GF}(2) = \{0,\ 1\}$ and extend it according to the equation

\[{\rm GF}(2^2)= \big\{k_0+k_1\cdot \alpha \ \big | \ k_0, k_1\in{\rm GF}(2) = \{ 0, 1\} \big \}\hspace{0.05cm}, \]

then the likewise finite set $\{0,\ 1,\ \alpha,\ 1 + \alpha\}$ ⇒ order is further $q=4$.

Performing the arithmetic operations modulo $p(\alpha) = \alpha^2 + \alpha + 1$ we get the following result:

$$ \begin{array}{c} {\rm modulo}\hspace{0.15cm}{\it p}(\alpha)= \alpha^2 + \alpha + 1\\ \end{array}\hspace{0.25cm} \Rightarrow\hspace{0.25cm} \left[ \begin{array}{c|cccccc} + & 0 & 1 & \alpha & 1\!+\!\alpha \\ \hline 0 & 0 & 1 & \alpha & 1\!+\!\alpha \\ 1 & 1 & 0 & 1\!+\!\alpha & \alpha \\ \alpha & \alpha & 1\!+\!\alpha & 0 & 1 \\ 1\!+\!\alpha & 1\!+\!\alpha & \alpha & 1 & 0 \end{array} \right] \hspace{-0.1cm} ,\hspace{0.5cm} \left[ \begin{array}{c|cccccc} \cdot & 0 & 1 & \alpha & 1\!+\!\alpha \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & \alpha & 1\!+\!\alpha \\ \alpha & 0 & \alpha & 1\!+\!\alpha & 1 \\ 1\!+\!\alpha & 0 & 1\!+\!\alpha & 1 & \alpha \end{array} \right] .$$

In this regard, it should be noted:

The neutral elements of addition or multiplication are still $N_{\rm A} = 0$ and $N_{\rm M} = 1$.

Since there is no difference between addition and subtraction in modulo arithmetic $\alpha + \alpha = \alpha - \alpha = 0$.
For all $z_i$ thus holds: The additive inverse of $z_i$ is the element $z_i$ itself.

The entries in the multiplication table are obtained according to the following calculations:

\[\big [ \alpha \cdot (1+\alpha) \big ] \hspace{0.15cm}{\rm mod} \hspace{0.15cm} p(\alpha) = (\alpha^2 + \alpha) \hspace{0.15cm}{\rm mod} \hspace{0.15cm} (\alpha^2 + \alpha + 1)= 1\hspace{0.05cm},\]

\[\big [ \alpha \cdot \alpha \big ] \hspace{0.15cm}{\rm mod} \hspace{0.15cm} p(\alpha) = (\alpha^2 ) \hspace{0.15cm}{\rm mod} \hspace{0.15cm} (\alpha^2 + \alpha + 1)= 1+\alpha\hspace{0.05cm},\]

\[\big [ (1+\alpha) \cdot (1+\alpha) \big ] \hspace{0.15cm}{\rm mod} \hspace{0.15cm} p(\alpha) = (\alpha^2 + 1) \hspace{0.15cm}{\rm mod} \hspace{0.15cm} (\alpha^2 + \alpha + 1)= \alpha\hspace{0.05cm}.\]

Thus, the multiplicative inverses exist for all elements except the zero element:

\[{\rm Inv_M}( 1) = 1 \hspace{0.05cm},\hspace{0.2cm}{\rm Inv_M}(\alpha) = 1+\alpha \hspace{0.05cm},\hspace{0.2cm}{\rm Inv_M}(1+\alpha) = \alpha \hspace{0.05cm}.\]

$\text{Intermediate result:}$

The set $\{0, \ 1, \ \alpha, \ 1 + \alpha\}$ together with the two operations addition and multiplication modulo $p(\alpha)= \alpha^2 + \alpha + 1$ represents a Galois field. The order is $q = 4$.
This Galois field, denoted by $\rm GF(2^2) = GF(4)$ satisfies all the requirements mentioned in thr "previous chapter" .
In contrast to the Galois field $\rm GF(3) = \{0, \ 1, \ 2\}$ with the property that $q = 3$ is a prime number, $\rm GF(2^2)$ is called an extension field.

Reducible and irreducible polynomials

The polynomial $p(\alpha)$ and thus the equation of determination $p(\alpha) = 0$ must not be given arbitrarily. The polynomial used on the last page

$$p(\alpha)= \alpha^2 + \alpha + 1$$

is suitable. Now we try another polynomial, namely $p(\alpha)= \alpha^2 + 1$.

$$ \begin{array}{c} {\rm modulo}\hspace{0.15cm}{\it p}(\alpha)= \alpha^2 + 1\\ \end{array}\hspace{0.25cm} \Rightarrow\hspace{0.25cm} \left[ \begin{array}{c|cccccc} + & 0 & 1 & \alpha & 1\!+\!\alpha \\ \hline 0 & 0 & 1 & \alpha & 1\!+\!\alpha \\ 1 & 1 & 0 & 1\!+\!\alpha & \alpha \\ \alpha & \alpha & 1\!+\!\alpha & 0 & 1 \\ 1\!+\!\alpha & 1\!+\!\alpha & \alpha & 1 & 0 \end{array} \right] \hspace{-0.1cm} ,\hspace{0.5cm} \left[ \begin{array}{c|cccccc} \cdot & 0 & 1 & \alpha & 1\!+\!\alpha \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & \alpha & 1\!+\!\alpha \\ \alpha & 0 & \alpha & 1 &1\!+\!\alpha \\ 1\!+\!\alpha & 0 & 1\!+\!\alpha & 1\!+\!\alpha & 0 \end{array} \right] .$$

The addition table is identical in both cases and also the multiplication tables differ only by the four entries in the two bottom rows and the two last columns:

From $p(\alpha) = 0$ now follows for the product $\alpha \cdot \alpha = 1$ and the product $(1 +\alpha) \cdot (1 +\alpha) $ gives the zero element. The mixed product is $\alpha \cdot (1 +\alpha) = 1 +\alpha $.

In the last row of the multiplication table and also in the last column there is now no "$1$" ⇒ Concerning the condition $p(\alpha)= \alpha^2 + 1= 0$ consequently the multiplicative inverse to $1 +\alpha$ does not exist.

But thus the finite set $\{0, \ 1, \ \alpha, \ 1 + \alpha\}$ together with arithmetic operations modulo $p(\alpha)= \alpha^2 + 1$ does not satisfy the conditions of an extension fields either $\rm GF(2^2) $.

$\text{Let us summarize:}$

From the binary Galois field $\rm GF(2) = \{0, \ 1\}$ an extension field $\rm GF(2^2)$ can be formulated with the aid of a polynomial of degree $m = 2$ with binary coefficients:

\[p(x) = x^2 + k_1 \cdot x + k_0 \hspace{0.05cm}, \hspace{0.45cm}k_0\hspace{0.05cm},\hspace{0.1cm}k_1 \in \{0, 1\} \hspace{0.05cm}.\]

Note: The renaming of the variable $\alpha$ to $x$ has only formal meaning with regard to later pages.

In the present case there is only one suitable polynomial $p_1(x)= x^2 + x + 1$. All other possible polynomials of degree $m = 2$, namely,

\[p_2(x) = x^2 + 1 \hspace{0.06cm} = (x+1) \cdot (x+1)\hspace{0.05cm},\]

\[p_3(x) =x^2 \hspace{0.76cm} = x \cdot x \hspace{0.05cm},\]

\[p_4(x) = x^2 + x = (x+1) \cdot x\hspace{0.05cm}, \]

can be factorized and do not yield extension fields.

The polynomials $p_2(x)$, $p_3(x)$ and $p_4(x)$ are called "reducible".
The conclusion is obvious that only "irreducible polynomials" such as $p_1(x)$ are suitable for an extension fields

.

Interpretation of the new element "alpha"

We further consider the field ${\rm GF}(2^2) = \{0, \ 1,\ \alpha,\ 1 + \alpha\}$ corresponding to the following two operational tables, based on the constraint $p(\alpha)= \alpha^2 + \alpha + 1 = 0$ (irreducible ploynomial):

$$ \begin{array}{c} {\rm modulo}\hspace{0.15cm} p(\alpha)= \alpha^2 + \alpha + 1\\ \end{array}\hspace{0.25cm} \Rightarrow\hspace{0.25cm} \left[ \begin{array}{c|cccccc} + & 0 & 1 & \alpha & 1\!+\!\alpha \\ \hline 0 & 0 & 1 & \alpha & 1\!+\!\alpha \\ 1 & 1 & 0 & 1\!+\!\alpha & \alpha \\ \alpha & \alpha & 1\!+\!\alpha & 0 & 1 \\ 1\!+\!\alpha & 1\!+\!\alpha & \alpha & 1 & 0 \end{array} \right] \hspace{-0.1cm} ,\hspace{0.5cm} \left[ \begin{array}{c|cccccc} \cdot & 0 & 1 & \alpha & 1\!+\!\alpha \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & \alpha & 1\!+\!\alpha \\ \alpha & 0 & \alpha & 1\!+\!\alpha & 1 \\ 1\!+\!\alpha & 0 & 1\!+\!\alpha & 1 & \alpha \end{array} \right] .$$

Transition from ${\rm GF}(2)$ to ${\rm GF}(2^2)$

But what is the meaning of the new element $\alpha$?

The polynomial $p(\alpha)= \alpha^2 + \alpha + 1 $ has no zero in ${\rm GF}(2) = \{0, \ 1\}$ . This further implies that $\alpha$ can be neither $0$ nor $1$ .

If $\alpha= 0$ resp. $\alpha= 1$, then moreover two of the four set elements $\{0,\ 1,\ \alpha,\ 1 + \alpha\}$ would be identical respectively: Either "$0$" and "$\alpha$" as well as "$1$" and "$1+\alpha$" or "$1$" and "$\alpha$" as well as "$0$" and "$1+\alpha$".

Much more the one-dimensional field ${\rm GF}(2)$ gets a second dimension by the introduction of the element $\alpha$ . It is thus extended to the Galois field ${\rm GF}(2^2)$ as shown in the accompanying diagram.

The element $\alpha$ has similar meaning as the imaginary unit ${\rm j}$, by which one extends the set of real numbers under the constraint ${\rm j}^2 + 1 = 0$ to the set of complex numbers.

$\text{Common representation of the binary extension field}\ \ {\rm GF}(2^2)\text{:}$

Due to the identity $\alpha^2 = 1 + \alpha$, which follows from the constraint $p(\alpha) = 0$ , one can write in the same way ${\rm GF}(2^2) = \{0,\ 1,\ \alpha,\ \alpha^2\}$ where now the following operation tables hold:

$$ \begin{array}{c} {\rm modulo}\hspace{0.15cm} p(\alpha)= \alpha^2 + \alpha + 1\\ \end{array}\hspace{0.25cm} \Rightarrow\hspace{0.25cm} \left[ \begin{array}{c | cccccc} + & 0 & 1 & \alpha & \alpha^2 \\ \hline 0 & 0 & 1 & \alpha & \alpha^2 \\ 1 & 1 & 0 & \alpha^2 & \alpha \\ \alpha & \alpha & \alpha^2 & 0 & 1 \\ \alpha^2 & \alpha^2 & \alpha & 1 & 0 \end{array} \right] \hspace{-0.1cm} ,\hspace{0.5cm} \left[ \begin{array}{c | cccccc} \cdot & 0 & 1 & \alpha & \alpha^2 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & \alpha & \alpha^2 \\ \alpha & 0 & \alpha &\alpha^2 & 1 \\ \alpha^2 & 0 & \alpha^2 & 1 & \alpha \end{array} \right] .$$

Polynomials over a finite field

$\text{Definition:}$ A polynomial in a finite field ${\rm GF}(P)$, where $P$ denotes a prime number, has the following form:

\[a(x) = \sum_{i = 0}^{m} a_i \cdot x^{i} = a_0 + a_1 \cdot x + a_2 \cdot x^2 + \hspace{0.1cm}\text{...} \hspace{0.1cm} + a_m \cdot x^{m} \hspace{0.05cm}.\]

To note:

All coefficients $a_i $ are elements of the field: $a_i \in {\rm GF}(P)$.

If the leading coefficient $a_m ≠ 0$, then $m$ indicates the degree of the polynomial.

Let us consider a second polynomial with degree $M$,

\[b(x) = \sum_{i = 0}^{M} b_i \cdot x^{i} = b_0 + b_1 \cdot x + b_2 \cdot x^2 + \hspace{0.1cm}\text{...} \hspace{0.1cm} + b_M \cdot x^{M} \hspace{0.05cm},\]

then we get for the sum (resp. difference) and the product respectively in ${\rm GF}(P)$:

\[a(x) \pm b(x) = \sum_{i = 0}^{{\rm max}\hspace{0.05cm}(m, \hspace{0.05cm}M)} \hspace{0.15cm}(a_i \pm b_i) \cdot x^{i} \hspace{0.05cm},\]

\[a(x) \cdot b(x) = \sum_{i = 0}^{m + M} \hspace{0.15cm}c_i \cdot x^{i}\hspace{0.05cm},\hspace{0.5cm} c_i = \sum_{j = 0}^{i}\hspace{0.15cm}a_j \cdot b_{i-j} \hspace{0.05cm}.\]

$\text{Example 1:}$ $a(x) = x^3 + x + 1$ and $b(x) = x^2 + x + 1$ are valid.

In the binary Galois field ⇒ ${\rm GF}(2)$ results according to the above equations for the sum, difference and product of the two polynomials:

\[s(x) = a(x) + b(x) = x^3 + x^2 \hspace{0.05cm}, \]

\[d(x) = a(x) - b(x) = x^3 + x^2 = s(x)\hspace{0.05cm},\]

\[c(x) = a(x) \cdot b(x) =\sum_{i = 0}^{3 + 2} \hspace{0.15cm}c_i \cdot x^{i}\hspace{0.05cm},\hspace{0.5cm} c_i = \sum_{j = 0}^{i}\hspace{0.15cm}a_j \cdot b_{i-j} \hspace{0.05cm}.\]

With $a_0 = a_1 = a_3 = b_0 = b_1 =b_2 = 1$ und $a_2 = a_4 = a_5 = b_3 = b_4 =b_5 = 0$ we obtain:

\[c_0 = a_0 \cdot b_0 = 1 \cdot 1 = 1 \hspace{0.05cm},\]

\[c_1 = a_0 \cdot b_1 + a_1 \cdot b_0 = 1 \cdot 1 + 1 \cdot 1 = 0 \hspace{0.05cm},\]

\[c_2 =a_0 \cdot b_2 + a_1 \cdot b_1 + a_2 \cdot b_0 = 1 \cdot 1 + 1 \cdot 1 + 0 \cdot 1 = 0 \hspace{0.05cm},\]

\[c_3 = a_0 \cdot b_3 + a_1 \cdot b_2 + a_2 \cdot b_1 + a_3 \cdot b_0 = 1 \cdot 0 + 1 \cdot 1 + 0 \cdot 1 + 1 \cdot 1 = 0 \hspace{0.05cm},\]

\[c_4=a_0 \cdot b_4 + a_1 \cdot b_3 + \hspace{0.05cm}\text{...}\hspace{0.05cm}+ \hspace{0.05cm}a_4 \cdot b_0 =1 \cdot 0 + 1 \cdot 0 + 0 \cdot 1 + 1 \cdot 1 + 0 \cdot 1 = 1 \hspace{0.05cm},\]

\[c_5 = a_0 \cdot b_5 + a_1 \cdot b_4 + \hspace{0.05cm}\text{...}\hspace{0.05cm}+ \hspace{0.05cm} a_5 \cdot b_0 =1 \cdot 0 + 1 \cdot 0 + 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 1 + 0 \cdot 1= 1 \]

\[\Rightarrow \hspace{0.3cm} c(x) = x^5 + x^4 +1 \hspace{0.05cm}.\]

In the Galois field ${\rm GF}(3)$ other results are obtained due to the modulo 3 operations:

\[s(x) = (x^3 + x + 1) + (x^2 + x + 1) = x^3 + x^2 + 2x + 2\hspace{0.05cm},\]

\[d(x) = (x^3 + x + 1) - (x^2 + x + 1) = x^3 + 2x^2 \hspace{0.05cm},\]

\[c(x) = (x^3 + x + 1) \cdot (x^2 + x + 1) = x^5 + x^4 + 2x^3 + 2x^2 + 2x +1\hspace{0.05cm}.\]

$\text{Definition:}$ A polynomial $a(x)$ is called reducible if it can be represented as the product of two polynomials $p(x)$ and $q(x)$ each of lower degree:

\[a(x) = p(x) \cdot q(x) \hspace{0.05cm}.\]

If this factorization is not possible, that is

\[a(x) = p(x) \cdot q(x) + r(x)\hspace{0.05cm},\hspace{0.5cm} r(x) \ne 0\]

holds, then the polynomial is called an irreducible or prime.

$\text{Example 2:}$ It holds $b(x) = x^3 + x + 1$, then $p_1(x) = x^2 + x + 1$ and $p_2(x) = x^2 + 1$ are valid.

The graph on the left illustrates the modulo 2 multiplication $a(x)= b(x) \cdot p_1(x)$. The result is $a(x) = x^5 + x^4 + 1$.

Example of polynomial multiplication and division

In the right part of the above graph, the modulo 2 division $q(x)= a(x)/ p_2(x)$ is shown with the result $q(x) = x^3 + x^2 + x + 1$.

This leaves the remainder $r(x) = x$.

According to this calculation alone $a(x) = x^5 + x^4 + 1$ could well be an irreducible polynomial.

However, the proof that the polynomial $a(x) = x^5 + x^4 + 1$ is indeed irreducible would only be given if $a(x)/p(x)$ yields a remainder for all

\[p(x) = \sum_{i = 0}^{m} a_i \cdot x^{i} = a_m \cdot x^{m} + a_{m-1} \cdot x^{m-1} + \hspace{0.1cm}\text{...} \hspace{0.1cm}+ a_2 \cdot x^2 + a_1 \cdot x + a_0 \hspace{0.05cm}.\]

This proof would require (almost) $2^5 = 32$ divisions in the present example.

Based on our left-hand calculation, we can immediately see here that $a(x)$ is certainly not an irreducible polynomial,
since e.g. $a(x) = x^5 + x^4 + 1$ divided by $p_1(x) = x^2 + x + 1$ yields the polynomial $b(x) = x^3 + x + 1$ with no remainder.

Generalized definition of an extension field

We assume the following:

A Galois field ${\rm GF}(P)$, where $P$ denotes a prime number,

an irreducible polynomial $p(x)$ over ${\rm GF}(P)$ of degree $m$:

\[p(x) = a_m \cdot x^{m} + a_{m-1} \cdot x^{m-1} + \hspace{0.1cm}\text{...} \hspace{0.1cm}+ a_2 \cdot x^2 + a_1 \cdot x + a_0 \hspace{0.05cm}, \hspace{0.3cm} a_i \in {\rm G}(P)\hspace{0.05cm}, \hspace{0.15cm}a_m \ne 0\hspace{0.05cm}. \]

With the above conditions generally applies:

$\text{Definition:}$ Let

$P$ be a prime number,
$m$ be an integer,
$p(x)$ be an irreducible polynomial of degree $m$ and
$p(\alpha) = 0$ hold.

An extension field can then be described as follows:

\[{\rm GF}(P^m)= \Big\{ k_{m-1} \hspace{0.01cm}\cdot \hspace{0.02cm}\alpha^{m-1} \hspace{0.05cm}+ \hspace{0.05cm}\text{...} \hspace{0.05cm}+ \hspace{0.05cm}k_1 \hspace{0.01cm}\cdot \hspace{0.02cm} \alpha \hspace{0.05cm}+ \hspace{0.05cm} k_0\hspace{0.05cm} \Big{\vert}\hspace{0.02cm} \ k_i\in{\rm GF}(P) = \{ 0, 1, \hspace{0.05cm}\text{...} \hspace{0.05cm}, P-1\}\Big \}.\]

The addition and multiplication in this extension field then corresponds to polynomial addition and polynomial multiplication modulo $p(\alpha)$.

So: A Galois field ${\rm GF}(q)$ with $q$ elements can be specified whenever the element number can be written in the form $q = P^m$
$(P$ denotes a prime number, $m$ be integer$)$.

Possible Galois fields ${\rm GF}(q)$ for $q ≤ 64$

The diagram shows for which $q$–values a Galois field can be constructed. For the shaded values no finite field can be given.

Further it is to be noted:

The yellow highlighted positions $q=P$ ⇒ $m = 1$ mark sets of numbers $\{0,\ 1,\hspace{0.05cm}\text{ ...} \hspace{0.05cm},\ q- 1\}$ with Galois properties, see page "Definition of a Galois Field".
The other background colors mark extension fields with $q=P^m$, $m ≥ 2$. For $q ≤ 64$ these are based on the primes $2$, $3$, $5$ and $7$.
Highlighted in red are binary fields ⇒ $q=2^m$, $m ≥ 1$, which will be considered in more detail on the next page.
All other extension fields are labeled in blue.

Binary extension fields – Primitive polynomials

Irreducible and primitive polynomials

In the following we consider binary extension fields with $q$ elements:

\[q = 2^m \hspace{0.15cm}(m \ge 2) \hspace{0.3cm} \Rightarrow\hspace{0.3cm} q = 4,\ 8,\ 16, 32,\ 64,\ \text{...}\]

In the table all irreducible polynomials of the Galois field ${\rm GF}(2)$ are given for $2 ≤ m ≤ 6$.

The polynomials in columns 2 and 3 are not only irreducible, but additionally also primitive.

Primitive polynomials also provide the basis for the "realization of pseudo–noise generators".

Before we turn to the definition of a primitive polynomial, we shall first mention the peculiarities of "primitive elements" using the example of

\[{\rm GF}(q) = \{\hspace{0.05cm}z_0 = 0,\hspace{0.1cm} z_1 = 1,\hspace{0.05cm} \text{...}\hspace{0.05cm} , \hspace{0.05cm}z_{q-1}\}.\]

The element $z_i = \beta$ is then called primitive ,

if the power $\beta^{\hspace{0.05cm}i}$ modulo $q$ for the first time for $i = q-1$ leads to the result "$1$" so that

$\beta^{\hspace{0.05cm}i}$ for $1 ≤ i ≤ q- 1$ yields exactly the elements $z_1$, ... , $z_{q-1}$ i.e. all elements of ${\rm GF}(q)$ except the zero element $z_0 = 0$.

$\text{Example 3:}$ From the number set $Z_5 = \{0,\ 1,\ 2,\ 3,\ 4\}$, the numberss "$2$" and "$3$" are primitive elements because of

\[2^1 \hspace{-0.1cm} = \hspace{-0.1cm} 2\hspace{0.05cm},\hspace{0.2cm} 2^2 = 4\hspace{0.05cm},\hspace{0.2cm} 2^3 = 8 \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 5 = 3\hspace{0.05cm},\hspace{0.2cm} 2^4 = 16 \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 5 = 1\hspace{0.05cm},\]

\[3^1 \hspace{-0.1cm} = \hspace{-0.1cm} 3\hspace{0.05cm},\hspace{0.2cm} 3^2 = 9\hspace{0.1cm}{\rm mod} \hspace{0.1cm} 5 = 4\hspace{0.05cm},\hspace{0.2cm} 3^3 = 27 \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 5 = 2\hspace{0.05cm},\hspace{0.2cm} 3^4 = 81 \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 5 = 1\]

On the other hand, "$4$" is not a primitive element, because already $4^2 = 1$:

\[4^1 = 4\hspace{0.05cm},\hspace{0.2cm} 4^2 = 16 \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 5 = 1\hspace{0.05cm},\hspace{0.2cm} 4^3 = 64 \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 5 = 4\hspace{0.05cm},\hspace{0.2cm} 4^4 = 256 \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 5 = 1\hspace{0.05cm}.\]

$\text{Definition:}$ An irreducible polynomial is called at the same time a "primitive polynomial"
if the root $\alpha$ with respect to the polynomial $p(x)$ is a primitive element of ${\rm GF}(q)$. Then holds

\[{\rm GF}(q) = \{\hspace{0.1cm}\alpha^{-\infty} = 0\hspace{0.05cm},\hspace{0.1cm} \alpha^{0} = 1,\hspace{0.05cm}\hspace{0.2cm} \alpha\hspace{0.05cm},\hspace{0.2cm} \alpha^{2},\hspace{0.2cm} \text{...} \hspace{0.1cm} , \hspace{0.2cm}\alpha^{q-2}\hspace{0.1cm}\}\hspace{0.05cm}. \]

All polynomials given in the second column of the above table are both irreducible and primitive.

If $p_1(x)$ is a primitive polynomial, then the polynomial reciprocal $p_2 (x) = x^m \cdot p_1(x^{-1})$ to it is also primitive.

All polynomials in the third column are reciprocal to the polynomial in the second column. For example, for $m = 3$:

\[p_1(x) = x^3 + x + 1 \hspace{0.3cm} \Rightarrow\hspace{0.3cm}p_2(x) = x^3 \cdot \big[x^{-3} + x^{-1} + 1 \big]= x^3 + x^2 + 1 \hspace{0.05cm}.\]

The irreducible polynomials of column 4, on the other hand, are not primitive; they play only a minor role in describing error correction procedures.

$\text{Example 4:}$ For clarification of these statements we consider exemplarily

the Galois field $\rm GF(2^3) = GF(8)$, as well as

the polynomial $p(x) = x^3 + x + 1$.

From the condition $p(\alpha) = 0$ we obtain in $\rm GF(2^3)$ further:

\[\alpha^3 + \alpha + 1 = 0 \hspace{0.3cm} \Rightarrow\hspace{0.3cm}\alpha^3 = \alpha + 1 \hspace{0.05cm},\]

and thus for the powers $\alpha^{i}$ of the root for $i ≥ 4$:

\[\alpha^4 = \alpha \cdot \alpha^3 = \alpha \cdot (\alpha + 1) = \alpha^2 + \alpha \hspace{0.05cm},\]

\[\alpha^5 = \alpha^2 \cdot \alpha^3 = \alpha^2 \cdot (\alpha + 1) = \alpha^3 + \alpha^2 = \alpha^2 + \alpha + 1 \hspace{0.05cm},\]

\[\alpha^6 = \alpha^3 \cdot \alpha^3 = (\alpha + 1) \cdot (\alpha + 1) = \alpha^2 + \alpha + \alpha + 1= \alpha^2 + 1 \hspace{0.05cm},\]

\[\alpha^7 = \alpha^4 \cdot \alpha^3 = (\alpha^2 + \alpha) \cdot (\alpha + 1) = \alpha^3 + \alpha^2 + \alpha^2 + \alpha = \alpha + 1 + \alpha = 1 = \alpha^0 \hspace{0.05cm}.\]

$\text{Example 5:}$ The elements $z_0$, $z_1$, ... , $z_7$ of the Galois field $\rm GF(2^3)$ can be represented according to the adjacent table as follows:

Elements of $\rm GF(2^3)$ in three different representations

as powers of $\alpha$ ⇒ "exponent representation",

as polynomials of the form $k_2 \cdot \alpha^2 + k_1 \cdot \alpha + k_0$ with binary coefficients $k_2$, $k_1$, $k_0$ ⇒ "polynomial representation",

as vectors of coefficients $(k_2, \ k_1, \ k_0)$ ⇒ "coefficient representation".

⇒ For addition (or subtraction) of two elements polynomial and vector representation are equally suitable,
where the components are to be added $\text{modulo 2}$, for example:

\[z_5 + z_7 =(\alpha^2 + \alpha) + (\alpha^2 + 1) = \alpha + 1 = \alpha^3 = z_4 \hspace{0.05cm},\]

\[{\rm oder}\hspace{0.15cm} z_5 + z_7 =(110) + (101) = (011) = z_4 \hspace{0.05cm},\]

\[\hspace{0.15cm} z_1 + z_2 + z_3 =(001) + (010) + (100)= (111) = z_6 \hspace{0.05cm}.\]

⇒ For multiplications, the exponent representation is well suited, as the following examples show:

$\rm GF(2^3)$ in 3D representation

\[z_3 \cdot z_4 =\alpha^2 \cdot \alpha^3 = \alpha^{2+3}= \alpha^{5} = z_6 \hspace{0.05cm},\]

\[z_0 \cdot z_5 =\alpha^{-\infty} \cdot \alpha^4 = \alpha^{-\infty} = z_0 \hspace{0.05cm},\]

\[z_5 \cdot z_7 = \alpha^4 \cdot \alpha^6 = \alpha^{10}= \alpha^{7} \cdot \alpha^{3} = 1 \cdot \alpha^{3}= z_4 \hspace{0.05cm}.\]

It can be seen that the exponents result modulo $q-1$ $($in this example modulo $7)$.

⇒ The bottom graph shows the finite extension field $\rm GF(2^3)$ in a three-dimensional representation:

The axes are labeled $\alpha^0 =1$, $\alpha^1$ and $\alpha^2$.

The $2^3 = 8$ points in the three-dimensional space are labeled with the coefficient vectors.

The assignment of the coefficients $k_2$, $k_1$, $k_0$ to the axes is made clear by color.

Exercises for the chapter

Exercise 2.3: Reducible and Irreducible Polynomials

Exercise 2.3Z: Polynomial Division

Exercise 2.4: GF(2 to the Power of 2) Representation Forms

Exercise 2.4Z: Finite and Infinite Fields

Exercise 2.5: Three Variants of GF(2 power 4)

Exercise 2.5Z: Some Calculations about GF(2 power 3)

Exercise 2.6: GF(P power m). Which P, which m?