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

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$\rm GF(2^3)$  elements;  polynomial  $p(x) = x^3 + x + 1$

We now consider the extension field with eight elements   ⇒   $\rm GF(2^3)$  according to the adjacent table. Since the underlying polynomial

$$p(x) = x^3 + x +1 $$

EN_KC_Z_2_5.png

is both irreducible and primitive, the Galois field at hand can be stated in the following form:

$${\rm GF}(2^3) = \{\hspace{0.1cm}0\hspace{0.05cm},\hspace{0.1cm} 1,\hspace{0.05cm}\hspace{0.1cm} \alpha\hspace{0.05cm},\hspace{0.1cm} \alpha^{2}\hspace{0.05cm},\hspace{0.1cm} \alpha^{3}\hspace{0.05cm},\hspace{0.1cm} \alpha^{4}\hspace{0.05cm},\hspace{0.1cm} \alpha^{5}\hspace{0.05cm},\hspace{0.1cm} \alpha^{6}\hspace{0.1cm}\}\hspace{0.05cm}. $$

The element  $\alpha$  results thereby as solution of the equation  $p(\alpha) = 0$  in the Galois field  $\rm GF(2)$.

• This gives the following constraint:

$$\alpha^3 + \alpha +1 = 0\hspace{0.3cm} \Rightarrow\hspace{0.3cm} \alpha^3 = \alpha +1\hspace{0.05cm}.$$

• The following calculations apply to the other elements:

$$\alpha^4 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot \alpha^3 = \alpha \cdot (\alpha + 1) = \alpha^2 + \alpha \hspace{0.05cm},$$

$$\alpha^5 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot \alpha^4 = \alpha \cdot (\alpha^2 +\alpha) = \alpha^3 + \alpha^2 = \alpha^2 + \alpha + 1\hspace{0.05cm},$$

$$\alpha^6 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot \alpha^5 = \alpha \cdot (\alpha^2 +\alpha + 1)= \alpha^3 + \alpha^2 + \alpha= \alpha + 1 + \alpha^2 + \alpha = \alpha^2+ 1\hspace{0.05cm}.$$

In this exercise you are to do some algebraic transformations in the  Galois field $\rm GF(2^3)$ . Among other things you are asked for the multiplicative inverse of the element  $\alpha^4$ . Then it must hold:

$$\alpha^4 \cdot {\rm Inv_M}( \alpha^4) = 1 \hspace{0.05cm}.$$

Hints:

• This exercise belongs to the chapter  "extension field".
• This exercise is intended as a supplement to the slightly more difficult  "Exercise 2.5" .

Questions

1

Which of the statements are true for the higher powers of  $\alpha^{i} \ (i ≥ 7)$  true?

	$\alpha^7 = 1$,
	$\alpha^8 = \alpha$,
	$\alpha^{13} = \alpha^2 + 1$,
	$\alpha^i = \alpha^{i \ \rm mod \, 7}$.

2

Which transformation is allowed for  $A = \alpha^8 + \alpha^6 - \alpha^2 + 1$ ?

	$A = 1$,
	$A = \alpha$,
	$A = \alpha^2$,
	$A = \alpha^3$,
	$A = \alpha^4$.

3

Which transformation is allowed for  $B = \alpha^{16} - \alpha^{12} \cdot \alpha^3$  permissible?

	$B = 1$,
	$B = \alpha$,
	$B = \alpha^2$,
	$B = \alpha^3$,
	$B = \alpha^4$.

4

What transformation is allowed for  $C = \alpha^3 + \alpha$ ?

	$C = 1$,
	$C = \alpha$,
	$C = \alpha^2$,
	$C = \alpha^3$,
	$C = \alpha^4$.

5

What transformation is allowed for  $D = \alpha^4 + \alpha$ ?

	$D = 1$,
	$D = \alpha$,
	$D = \alpha^2$,
	$D = \alpha^3$,
	$D = \alpha^4$.

6

Which transformation is allowed for  $E = A \cdot B \cdot C/D$ ?

	$E = 1$,
	$E = \alpha$,
	$E = \alpha^2$,
	$E = \alpha^3$,
	$E = \alpha^4$.

7

What statements hold for the multiplicative inverse to  $\alpha^2 + \alpha$?

	${\rm Inv_M}(\alpha^2 + \alpha) = 1$,
	${\rm Inv_M}(\alpha^2 + \alpha) = \alpha + 1$,
	${\rm Inv_M}(\alpha^2 + \alpha) = \alpha^3$,
	${\rm Inv_M}(\alpha^2 + \alpha) = \alpha^4$.

Solution

Solution

(1)  For example, using the table given in the front, you can find:

$$\alpha^7 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot \alpha^6 = \alpha \cdot (\alpha^2 + 1) = \alpha^3 + \alpha = (\alpha + 1) + \alpha = 1 \hspace{0.05cm},$$

$$\alpha^8 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot \alpha^7 = \alpha \cdot 1 = \alpha\hspace{0.05cm},$$

$$\alpha^{13} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha^7 \cdot \alpha^6 = 1 \cdot \alpha^6 = \alpha^2 + 1\hspace{0.05cm}.$$

The table can therefore be continued modulo $7$. This means: All proposed solutions are correct.

(2)  Correct is the proposed solution 2 because of.

• $\alpha^8 = \alpha$ according to subtask (1),
• $\alpha^6 = \alpha^2 + 1$ (according to table), and
• $-\alpha^2 = \alpha^2$ (operations in the binary Galois field).

So applies:

$$A = \alpha^8 + \alpha^6 - \alpha^2 + 1 = \alpha + (\alpha^2 + 1) + \alpha^2 + 1 = \alpha \hspace{0.05cm}.$$

(3)  With $\alpha^{16} = \alpha^{16-14} = \alpha^2$ sowie $\alpha^{12} \cdot \alpha^3 = \alpha^{15} = \alpha^{15-14} = \alpha$ we obtain the proposed solution 5:

$$B = \alpha^2 + \alpha= \alpha^4 \hspace{0.05cm}.$$

(4)  It holds $\alpha^3 = \alpha + 1$ und damit $C = \alpha^3 + \alpha = \alpha + 1 + \alpha = 1$   ⇒   Proposed solution 1.

(5)  With $\alpha^4 = \alpha^2 + \alpha$ we obtain $D = \alpha^4 + \alpha = \alpha^2$   ⇒   Proposed solution 3.

(6)  Correct is the proposed solution 4:

$$E = A \cdot B \cdot C/D = \alpha \cdot \alpha^4 \cdot 1/\alpha^2 = \alpha^3 \hspace{0.05cm}.$$

(7)  According to the table, $\alpha^2 + \alpha = \alpha^4$ holds. Therefore must be valid:

$$\alpha^4 \cdot {\rm Inv_M}( \alpha^4) = 1 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} {\rm Inv_M}( \alpha^2 + \alpha) = {\rm Inv_M}( \alpha^4) = \alpha^{-4} = \alpha^3 \hspace{0.05cm}.$$

Because of $\alpha^3 = \alpha + 1$ the proposed solutions 2 and 3 are correct.