Channel Coding/Extension Field - Revision history

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← Older revision		Revision as of 14:27, 16 March 2026
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	{{Display}}		{{Display}}
			[[de:Kanalcodierung/Erweiterungskörper]]

Hwang at 16:10, 23 November 2022

2022-11-23T16:10:48Z

← Older revision		Revision as of 18:10, 23 November 2022
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	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 3:}$  From the number set  $Z_5 = \{0,\ 1,\ 2,\ 3,\ 4\}$,  the ~~numberss~~  "$2$"  and  "$3$"  are primitive elements because of		$\text{Example 3:}$  From the number set  $Z_5 = \{0,\ 1,\ 2,\ 3,\ 4\}$,  the numbers  "$2$"  and  "$3$"  are primitive elements because of

	::<math>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},</math>		::<math>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},</math>

Hwang at 16:08, 23 November 2022

2022-11-23T16:08:14Z

← Older revision		Revision as of 18:08, 23 November 2022
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	\hspace{0.05cm}.</math>		\hspace{0.05cm}.</math>

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

	::<math>c_0 = a_0 \cdot b_0 = 1 \cdot 1 = 1 \hspace{0.05cm},</math>		::<math>c_0 = a_0 \cdot b_0 = 1 \cdot 1 = 1 \hspace{0.05cm},</math>

Hwang at 16:03, 23 November 2022

2022-11-23T16:03:27Z

← Older revision		Revision as of 18:03, 23 November 2022
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	*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.<br>		*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.<br>

	*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) $.<br><br>		*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 field either  $\rm GF(2^2) $.<br><br>

	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=

Guenter at 10:45, 22 November 2022

2022-11-22T10:45:11Z

← Older revision		Revision as of 12:45, 22 November 2022
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	Further it is to be noted:		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 section  [[Channel_Coding/Some_Basics_of_Algebra#Definition_of_a_Galois_field\|"Definition of a Galois Field"]].<br>		#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 section  [[Channel_Coding/Some_Basics_of_Algebra#Definition_of_a_Galois_field\|"Definition of a Galois Field"]].<br>
	#~~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 ~~ $7$.<br>		#Other background colors mark extension fields with  $q=P^m$,   $m ≥ 2$.  For  $q ≤ 64$  these are based on the primes  $2$,  $3$,  $5$,  $7$.<br>
	#Highlighted in red are binary fields   ⇒   $q=2^m$,   $m ≥ 1$, which will be considered in more detail in the next section.		#Highlighted in red are binary fields   ⇒   $q=2^m$,   $m ≥ 1$, which will be considered in more detail in the next section.
	# All other extension fields are labeled in blue.<br>		# All other extension fields are labeled in blue.<br>

Guenter at 10:43, 22 November 2022

2022-11-22T10:43:22Z

← Older revision		Revision as of 12:43, 22 November 2022
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	::<math>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\}		::<math>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}.</math>		\hspace{0.05cm}.</math>
	<i>Note:</i>   The renaming of the variable  $\alpha$  to  $x$  has only formal meaning with regard to later sections.<br>		<u>Note:</u>   The renaming of the variable  $\alpha$  to  $x$  has only formal meaning with regard to later sections.<br>
	*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,		*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,

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	::<math>p_4(x) = x^2 + x = (x+1) \cdot x\hspace{0.05cm}, </math>		::<math>p_4(x) = x^2 + x = (x+1) \cdot x\hspace{0.05cm}, </math>
	:can be factorized and do not yield extension fields.		: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 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}}.<br>		*The conclusion is obvious that only  "irreducible polynomials"  such as  $p_1(x)$  are suitable for an extension fields}}.<br>

Guenter at 10:33, 22 November 2022

2022-11-22T10:33:20Z

← Older revision		Revision as of 12:33, 22 November 2022
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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Intermediate result:}$		$\text{Intermediate result:}$
	*The set  $\{0, \ 1, \ \alpha, \ 1 + \alpha\}$  together with ~~the two~~ operations  "addition"  and  "multiplication"  modulo  $p(\alpha)= \alpha^2 + \alpha + 1$  represents a ~~<i>~~Galois field~~</i>. The~~ order is  $q = 4$.		*The set  $\{0, \ 1, \ \alpha, \ 1 + \alpha\}$  together with operations  "addition"  and  "multiplication"  modulo  $p(\alpha)= \alpha^2 + \alpha + 1$  represents a  "Galois field"  $($order  $q = 4)$.

	*This Galois field, denoted by  $\rm GF(2^2) = GF(4)$  satisfies all the requirements mentioned in the  [[Channel_Coding/Some_Basics_of_Algebra#Definition_of_a_Galois_field\| "previous chapter"]] .<br>		*This Galois field, denoted by  $\rm GF(2^2) = GF(4)$  satisfies all the requirements mentioned in the  [[Channel_Coding/Some_Basics_of_Algebra#Definition_of_a_Galois_field\| "previous chapter"]] .<br>

	*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.}}<br>		*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.}}<br>

Guenter at 10:28, 22 November 2022

2022-11-22T10:28:20Z

← Older revision		Revision as of 12:28, 22 November 2022
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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Intermediate result:}$		$\text{Intermediate result:}$
	*The set  $\{0, \ 1, \ \alpha, \ 1 + \alpha\}$  together with the two operations ~~<i>~~addition~~</i>~~  and ~~<i>~~multiplication~~</i>~~  modulo  $p(\alpha)= \alpha^2 + \alpha + 1$  represents a <i>Galois field</i>. The order is  $q = 4$.		*The set  $\{0, \ 1, \ \alpha, \ 1 + \alpha\}$  together with the two operations  "addition"  and  "multiplication"  modulo  $p(\alpha)= \alpha^2 + \alpha + 1$  represents a <i>Galois field</i>. The order is  $q = 4$.
	*This Galois field, denoted by  $\rm GF(2^2) = GF(4)$  satisfies all the requirements mentioned in the  [[Channel_Coding/Some_Basics_of_Algebra#Definition_of_a_Galois_field\| "previous chapter"]] .<br>		*This Galois field, denoted by  $\rm GF(2^2) = GF(4)$  satisfies all the requirements mentioned in the  [[Channel_Coding/Some_Basics_of_Algebra#Definition_of_a_Galois_field\| "previous chapter"]] .<br>
	*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.}}<br>		*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.}}<br>

Hwang at 14:46, 19 November 2022

2022-11-19T14:46:45Z

← Older revision		Revision as of 16:46, 19 November 2022
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	$\text{Intermediate result:}$		$\text{Intermediate result:}$
	*The set  $\{0, \ 1, \ \alpha, \ 1 + \alpha\}$  together with the two operations <i>addition</i>  and <i>multiplication</i>  modulo  $p(\alpha)= \alpha^2 + \alpha + 1$  represents a <i>Galois field</i>. The order is  $q = 4$.		*The set  $\{0, \ 1, \ \alpha, \ 1 + \alpha\}$  together with the two operations <i>addition</i>  and <i>multiplication</i>  modulo  $p(\alpha)= \alpha^2 + \alpha + 1$  represents a <i>Galois field</i>. The order is  $q = 4$.
	*This Galois field, denoted by  $\rm GF(2^2) = GF(4)$  satisfies all the requirements mentioned in ~~thr~~  [[Channel_Coding/Some_Basics_of_Algebra#Definition_of_a_Galois_field\| "previous chapter"]] .<br>		*This Galois field, denoted by  $\rm GF(2^2) = GF(4)$  satisfies all the requirements mentioned in the  [[Channel_Coding/Some_Basics_of_Algebra#Definition_of_a_Galois_field\| "previous chapter"]] .<br>
	*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.}}<br>		*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.}}<br>