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

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Three representations for  ${\rm GF}(2^2)$

Opposite you can see for the extension field  $\rm GF(2^2)$  the addition– as well as the multiplication table in three different variants:

• the polynomial representation,
• the coefficient vector representation,
• the exponent representation.

Hints:

• The exercise refers to the chapter  extension fields.
• All necessary information about  ${\rm GF}(2^2)$  can be found on the  first page  of this chapter.
• In the subtask (4) the following expressions are considered:

$$A = z_2 \cdot z_2 + z_2 \cdot z_3 + z_3 \cdot z_3,$$

$$B = (z_0 + z_1 + z_2) \cdot (z_0 + z_1 + z_3).$$

Questions

1

What characteristics can be recognized from the polynomial representation?

	The elements  $\alpha$  and  $1 + \alpha$  are neither  $0$  nor  $1$.
	The arithmetic operations are modulo  $2$.
	The arithmetic operations are performed modulo  $4$.
	One recognizes the result  $\alpha^2 + \alpha + 1 = 0$  from the addition table.
	One recognizes the result  $\alpha^2 + \alpha + 1 = 0$  from the multiplication table.

2

What is the relationship between the coefficient vector and the polynomial representation? Let  $k_0 ∈ \{0, \, 1\}$  and  $k_1 ∈ \{0, \, 1\}$ hold.

	$(k_0 \ k_1)$  refers to the element  $k_1 \cdot \alpha + k_0$.
	$(k_1 \ k_0)$  refers to the element  $k_1 \cdot \alpha + k_0$.
	There is no relationship between the two representations.

3

How are polynomial and exponent representation related?

	No connections can be seen.
	The elements  $0, \ 1$  and  $\alpha$  are the same in both representations.
	The element  $1 + \alpha$  is $\alpha^2$ in the exponent representation .
	The element  $\alpha^2$  of the exponent representation stands for  $\alpha \cdot (1 + \alpha)$.

4

Calculate the expressions  $A$  and  $B$  according to these three forms of representation. Which statements are true?

	It is true  $A = z_0$,
	It holds  $A = z_2$,
	It holds  $B = z_1$,
	It holds  $B = z_3$.

Solution

Solution

(1)  The proposed solutions 1, 2 and 5 are applicable.

Justification:

• If $\alpha = 0$ or $\alpha = 1$, the pseudo element $\alpha$ would be indistinguishable from the other two ${\rm GF}(2)$–elements $0$ and $1$.
• The modulo $2$ calculation can be recognized from the addition table. For example, $1 + 1 = 0, \ \alpha + \alpha = 0, \ (1 + \alpha) + (1 + \alpha) = 0$, etc.
• From the multiplication table we see that $\alpha^2 = \alpha \cdot \alpha = 1 + \alpha$ holds (3rd row, 3rd column). Thus also

$$\alpha^2 + \alpha + 1 = 0.$$

(2)  Correct is solution suggestion 2. Thus

• "$01$" for the element "$1$" and
• "$10$" for the element "$\alpha$".

(3)  Correct are the solutions 2 and 3:

• It is true that $\alpha^0 = 1$ and $\alpha^1 = \alpha$.
• For the underlying polynomial $p(x) = x^2 + x + 1$, it further follows from $p(\alpha) = 0$:

$$\alpha^2 +\alpha + 1 = 0 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} \alpha^2 =\alpha + 1 \hspace{0.05cm}.$$

(4)  According to the tables of polynomial representation holds:

$$A \hspace{-0.15cm} \ = \ \hspace{-0.15cm} z_2 \cdot z_2 + z_2 \cdot z_3 + z_3 \cdot z_3 = \alpha \cdot \alpha + \alpha \cdot (1+\alpha) + (1+\alpha) \cdot (1+\alpha) = (1+\alpha) + (1) + (\alpha) = 0 = z_0 \hspace{0.05cm},$$

$$ B \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (z_0 + z_1 + z_2) \cdot (z_0 + z_1 + z_3) = (0 + 1 + \alpha) \cdot (0 + 1 + 1+ \alpha) = (1+\alpha) \cdot \alpha = 1 = z_1 \hspace{0.05cm}.$$

Therefore, the proposed solutions 1 and 2 are correct.

The same results are obtained with the coefficient vector representation:

$$A \hspace{-0.15cm} \ = \ \hspace{-0.15cm} z_2 \cdot z_2 + z_2 \cdot z_3 + z_3 \cdot z_3 = (10) \cdot (10) + (10) \cdot (11) + (11) \cdot (11) = (11) + (01) + (10) = (00) = 0 = z_0 \hspace{0.05cm},$$

$$B \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (z_0 + z_1 + z_2) \cdot (z_0 + z_1 + z_3) = [(00) + (01) + (10)] \cdot [(00) + (01) + (11)] =(11) \cdot (10) = (01) = z_1 \hspace{0.05cm}.$$

And finally with the exponent representation:

$$A \hspace{-0.15cm} \ = \ \hspace{-0.15cm} z_2 \cdot z_2 + z_2 \cdot z_3 + z_3 \cdot z_3 = \alpha^1 \cdot \alpha^1 + \alpha^1 \cdot \alpha^2 + \alpha^2 \cdot \alpha^2 = \alpha^2 + \alpha^3 + \alpha^4 = \alpha^2 + \alpha^0 + \alpha^1 = 0 = z_0 \hspace{0.05cm},$$

$$B \hspace{-0.15cm} \ = \ \hspace{-0.15cm}(z_0 + z_1 + z_2) \cdot (z_0 + z_1 + z_3) = [0 + \alpha^0 + \alpha^1] \cdot [0 + \alpha^0 + \alpha^2] = \alpha^2 \cdot \alpha^1 = \alpha^3 = \alpha^0 = z_1 \hspace{0.05cm}.$$