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

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Underlying tables for
addition and multiplication

A Galois field  ${\rm GF}(q)$  with  $q = P^m$  elements defined by the adjacent tables is to be analyzed

• for addition  $($marked with "$+$"$)$,  and

• for multiplication  $($marked with "$\hspace{0.05cm}\cdot\hspace{0.05cm}$)".

This Galois field   ${\rm GF}(q) = \{\hspace{0.1cm}z_0,\hspace{0.1cm} z_1,\hspace{0.05cm} \text{...} , \hspace{0.1cm}z_{q-1}\}$   satisfies all the requirements for a finite field listed in the chapter  "Some Basics of Algebra" . Thus,  commutative, associative and distributive laws are also satisfied.

Furthermore there is

• a neutral element with respect to addition   ⇒   $N_{\rm A}$:

$$\exists \hspace{0.15cm} z_j \in {\rm GF}(q)\text{:} \hspace{0.25cm}z_i + z_j = z_i \hspace{0.3cm} \Rightarrow \hspace{0.3cm} z_j = N_{\rm A} \hspace{0.25cm}{\rm (zero\hspace{0.15cm}element)} \hspace{0.05cm},$$

• a neutral element with respect to multiplication   ⇒   $N_{\rm M}$:

$$\exists \hspace{0.15cm} z_j \in {\rm GF}(q)\text{:} \hspace{0.25cm}z_i \cdot z_j = z_i \hspace{0.3cm} \Rightarrow \hspace{0.3cm} z_j = N_{\rm M} \hspace{0.25cm}{\rm (identity\hspace{0.15cm}element)} \hspace{0.05cm},$$

• for all elements  $z_i$  an additive inverse   ⇒   ${\rm Inv_A}(z_i)$:

$$\forall \hspace{0.15cm} z_i \in {\rm GF}(q)\hspace{0.15cm} \exists \hspace{0.15cm} {\rm Inv_A}(z_i) \in {\rm GF}(q)\text{:}$$

$$z_i + {\rm Inv_A}(z_i) = N_{\rm A} = {\rm "\hspace{-0.15cm}0\hspace{-0.1cm}"}\hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm short}\text{:}\hspace{0.15cm} {\rm Inv_A}(z_i) = - z_i \hspace{0.05cm}, $$

• for all elements  $z_i$  except the zero element a multiplicative inverse  ⇒  ${\rm Inv_M}(z_i)$:

$$\forall \hspace{0.15cm} z_i \in {\rm GF}(q),\hspace{0.15cm} z_i \ne N_{\rm A} \hspace{0.15cm} \exists \hspace{0.15cm} {\rm Inv_M}(z_i) \in {\rm GF}(q)\text{:}$$

$$z_i \cdot {\rm Inv_M}(z_i) = N_{\rm M} = {\rm "\hspace{-0.15cm}1\hspace{-0.1cm}"} \hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm short}\text{:}\hspace{0.15cm} {\rm Inv_M}(z_i) = z_i^{-1} \hspace{0.05cm}. $$

Hints:

• This exercise belongs to the chapter  "Extension Field".

• In the tables,  the elements   $z_0, \hspace{0.05cm} \text{...} \hspace{0.1cm} , \ z_8$   are called  "coefficient vectors".

• For example,  "$2 \hspace{0.03cm}1$"  stands for  "$2 \cdot \alpha + 1$".

Questions

1

Specify the parameters of the Galois field considered here.

$P \ = \ $

$m \ = \ $

$q \ = \ $

2

What is the neutral element of addition?

	The neutral element of addition is  $N_{\rm A} = \,$ "$0\hspace{0.03cm}0$",
	The neutral element of addition is  $N_{\rm A} = \,$ "$0\hspace{0.03cm}1$".

3

What is the neutral element of multiplication?

	The neutral element of multiplication is  $N_{\rm M} = \,$ "$0\hspace{0.03cm}0$",
	The neutral element of multiplication is  $N_{\rm M} = \,$ "$0\hspace{0.03cm}1$".

4

What statements are true regarding additive inverses?

	It holds  ${\rm Inv_A} ($"$0\hspace{0.03cm}2$") $\, = \, $ "$0\hspace{0.03cm}1$",
	It holds  ${\rm Inv_A} ($"$1\hspace{0.03cm}1$") $\, = \, $ "$2\hspace{0.03cm}2$",
	It holds  ${\rm Inv_A} ($"$2\hspace{0.03cm}2$") $\, = \, $ "$0\hspace{0.03cm}0$".

5

Which of the following statements are true about the multiplication?

	The multiplication is defined here modulo  $p(\alpha) = \alpha^2 + 2$.
	The multiplication is defined here modulo  $p(\alpha) = \alpha^2 + 2\alpha + 2$.

6

What statements are true regarding multiplicative inverses?

	There is a multiplicative inverse for all elements  $z_i ∈ {\rm GF}(P^m)$  .
	It holds  ${\rm Inv_M} ($"$1\hspace{0.03cm}2$") $\, = \, $"$1\hspace{0.03cm}0$".
	It holds  ${\rm Inv_M} ($"$2\hspace{0.03cm}1$") $\, = \, $ "$1\hspace{0.03cm}2$".

7

Does  $($"$2\hspace{0.03cm}0$" $\, + \,$ "$1\hspace{0.03cm}2$"$)$ $\, \cdot\, $ "$1\hspace{0.03cm}2$" $\, = \, $"$2\hspace{0.03cm}0$" $\, \cdot\, $ "$1\hspace{0.03cm}2$" $\, + \, $"$1\hspace{0.03cm}2$" $\, \cdot\, $ "$1\hspace{0.03cm}2$" hold?

	Yes.
	No.

Solution

Solution

(1)  Each element consists of two ternaries   ⇒   $\underline{P = 3}, \ \underline{m = 2}$. There are $q = P^m = 3^8 = \underline{9 \rm \hspace{0.15cm}elements}$.

(2)  Correct is the proposed solution 1:

• The neutral element of the addition $(N_{\rm A})$ satisfies for all $z_i ∈ {\rm GF}(P^m)$ the condition $z_i + N_{\rm A} = z_i$.
• From the addition table it can be read that "$0\hspace{0.03cm}0$" satisfies this condition.

(3)  Correct is the proposed solution 2:

• The neutral element of the multiplication $(N_{\rm M})$ must always satisfy the condition $z_i \cdot N_{\rm M} = z_i$.
• From the multiplication table, $N_{\rm M} = \, "0\hspace{0.03cm}1"$.
• In polynomial notation, this corresponds to $k_1 = 0$ and $k_0 = 1$:

$$k_1 \cdot \alpha + k_0 = 1 \hspace{0.05cm}.$$

(4)  With the polynomial representation, the following calculations result:

$${\rm Inv_A}("\hspace{-0.05cm}0\hspace{0.03cm}2") \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(2) = (-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = 1 \hspace{0.25cm}\Rightarrow \hspace{0.25cm}{\rm Vector}\hspace{0.15cm}"\hspace{-0.05cm}0\hspace{0.03cm}1"\hspace{0.05cm},$$

$${\rm Inv_A}("\hspace{-0.05cm}1\hspace{0.03cm}1")\hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(\alpha + 1) = \big[(-\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] + \big[(-1) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] =2\alpha + 2 \hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm Vector}\hspace{0.15cm}"\hspace{-0.05cm}2\hspace{0.03cm}2"\hspace{0.05cm},$$

$${\rm Inv_A}("\hspace{-0.05cm}2\hspace{0.03cm}2")\hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(2\alpha + 2) = \big[(-2\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] + \big[(-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] =\alpha + 1 \hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm Vector}\hspace{0.15cm}"\hspace{-0.05cm}1\hspace{0.03cm}1"\hspace{0.05cm}.$$

Consequently, only the first two proposed solutions are correct.

However, the exercise can also be solved without calculation using the addition table alone.

• For example, you can find the inverse of "$2\hspace{0.03cm}2$" by looking for the column with the entry "$0\hspace{0.03cm}0$" in the last row.
• You find the column labeled "$1\hspace{0.03cm}1$" and thus ${\rm Inv_A}("2\hspace{0.03cm}2") = \, "1\hspace{0.03cm}1"$.

(5)  Multiplying $\alpha$ (vector "$1\hspace{0.03cm}0$") by itself gives $\alpha^2$.

• If the first proposed solution were valid, the condition $\alpha^2 + 2 = 0$ and thus $\alpha^2 = (-2) \, {\rm mod} \, 3 = 1$, thus yielding the vector "$0\hspace{0.03cm}1$."
• Assuming the second proposed solution, it follows from the condition $\alpha^2 + 2\alpha + 2 = 0$ in polynomial notation.

$$\alpha^2 = [(-2\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3] + [(-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3] = \alpha + 1 $$

and thus the coefficient vector "$1\hspace{0.03cm}1$".

In the multiplication table, in row 4, column 4, we find exactly the entry "$1\hspace{0.03cm}1$" → So, the correct one is proposed solution 2.

(6)  The multiplicative inverse to "$1\hspace{0.03cm}2$" can be found in row 6 of the multiplication table as the column with the entry "$0\hspace{0.03cm}1$"  
⇒  So the proposed solution 2 is correct in contrast to proposal 3. Namely, ${\rm Inv_M}("21") = \, "2\hspace{0.03cm}0"$ holds.

We check these results considering $\alpha^2 + 2\alpha + 2 = 0$ by multiplications:

$$"1\hspace{0.03cm}2" \hspace{0.05cm}\cdot \hspace{0.05cm}"1\hspace{0.03cm}0" \hspace{0.15cm} \Rightarrow \hspace{0.15cm} (\alpha + 2) \cdot \alpha = \alpha^2 + 2\alpha = (-2\alpha-2) + 2\alpha = -2 \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = 1 \hspace{0.15cm} \Rightarrow \hspace{0.15cm} {\rm Vector}\hspace{0.15cm}"0\hspace{0.03cm}1" \hspace{0.15cm} \Rightarrow \hspace{0.15cm}{\rm multiplicative \hspace{0.15cm}inverse}\hspace{0.05cm}.$$

$$"2\hspace{0.03cm}1" \hspace{0.05cm}\cdot \hspace{0.05cm}"1\hspace{0.03cm}2" "21" \hspace{0.05cm}\cdot \hspace{0.05cm} "12" \hspace{0.15cm} \Rightarrow \hspace{0.15cm} (2\alpha + 1) \cdot (\alpha + 2) = 2 \alpha^2 + \alpha + 4\alpha + 2 = 2 \alpha^2 + 5\alpha + 2 = 2 \cdot (-2\alpha - 2) + 5\alpha + 2 = (\alpha - 2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = \alpha +1 $$

$$\hspace{2.725cm} \Rightarrow \ {\rm Vector}\hspace{0.15cm}"1\hspace{0.03cm}1" \hspace{0.15cm} \Rightarrow \hspace{0.15cm}{\rm no\hspace{0.15cm}multiplicative \hspace{0.15cm}inverse}\hspace{0.05cm}.$$

The solution suggestion 1 is therefore not correct, because there is no multiplicative inverse for "$00$".

(7)  The two expressions agree   ⇒   YES, as the following calculations show:

$$("20" + "12") \ \cdot "12" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "02"\cdot "12" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "21"\hspace{0.05cm},$$

$$"20" \cdot "12" + "12" \cdot "12" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "02" + "22" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "21"\hspace{0.05cm}.$$

This means:   The distributive law has been proved at least on a single example.