Aufgaben:Exercise 2.4Z: Finite and Infinite Fields - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:57:46Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:57, 16 March 2026
Line 150:		Line 150:

	[[Category:Channel Coding: Exercises\|^2.2 Extension Field^]]		[[Category:Channel Coding: Exercises\|^2.2 Extension Field^]]
	[[de:Aufgaben:2.~~4Z_Endliche_und_unendliche_Körper~~]]		[[de:Aufgaben:Aufgabe 2.4Z: Endliche und unendliche Körper]]

Maintenance script: Add German interlanguage link

2026-03-16T12:31:19Z

Add German interlanguage link

← Older revision		Revision as of 14:31, 16 March 2026
Line 150:		Line 150:

	[[Category:Channel Coding: Exercises\|^2.2 Extension Field^]]		[[Category:Channel Coding: Exercises\|^2.2 Extension Field^]]
			[[de:Aufgaben:2.4Z_Endliche_und_unendliche_Körper]]

Guenter at 16:25, 3 October 2022

2022-10-03T16:25:04Z

← Older revision		Revision as of 18:25, 3 October 2022
Line 88:		Line 88:
	'''(1)'''  In the data section it says accordingly:		'''(1)'''  In the data section it says accordingly:

	A polynomial of degree $m$ is called ''reducible'' in the field $\mathcal{F}$ if it is in the form		A polynomial of degree  $m$  is called  "reducible"  in the field  $\mathcal{F}$  if it is can be represented the form
	:$$p(x)= \prod_{i = 1}^m (x-x_i) = (x - x_1) \cdot (x - x_2) \cdot \hspace{0.05cm}\text{ ...}\hspace{0.05cm} \cdot (x - x_m) $$		:$$p(x)= \prod_{i = 1}^m (x-x_i) = (x - x_1) \cdot (x - x_2) \cdot \hspace{0.05cm}\text{ ...}\hspace{0.05cm} \cdot (x - x_m) $$

	~~can be represented~~ and is valid for all zeros $x_i ∈ \mathcal{F}$. If this is not possible, one speaks of an ''irreducible'' polynomial.		and this is valid for all zeros  $x_i ∈ \mathcal{F}$.   If this is not possible,  one speaks of an  "irreducible"  polynomial.

	In real number space, for ~~each~~  $m = 2$  zeros  $x_1$  and  $x_2$ ~~hold~~:		In real number space,  it hold for the  $m = 2$  zeros  $x_1$  and  $x_2$:
	:$$p_1(x) \text{:}\hspace{0.3cm}x_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} +{\rm j}\hspace{0.05cm},\hspace{0.2cm}x_2 = -{\rm j}\hspace{0.05cm},$$		:$$p_1(x) \text{:}\hspace{0.3cm}x_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} +{\rm j}\hspace{0.05cm},\hspace{0.2cm}x_2 = -{\rm j}\hspace{0.05cm},$$
	:$$p_2(x) \text{:}\hspace{0.3cm}x_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} +1\hspace{0.05cm},\hspace{0.2cm}x_2 = -1\hspace{0.05cm},$$		:$$p_2(x) \text{:}\hspace{0.3cm}x_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} +1\hspace{0.05cm},\hspace{0.2cm}x_2 = -1\hspace{0.05cm},$$
Line 99:		Line 99:
	:$$p_4(x) \text{:}\hspace{0.3cm}x_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} +1\hspace{0.05cm},\hspace{0.2cm}x_2 = -2\hspace{0.05cm}.$$		:$$p_4(x) \text{:}\hspace{0.3cm}x_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} +1\hspace{0.05cm},\hspace{0.2cm}x_2 = -2\hspace{0.05cm}.$$

	Thus the <u>proposed solutions 1 and 3</u> are correct:		Thus the  <u>proposed solutions 1 and 3</u>  are correct:
	*The two zeros of $p_2(x)$ and $p_4(x)$ are both real. Thus, these are certainly reducible polynomials.		*The two zeros of  $p_2(x)$  and  $p_4(x)$  are both real.  Thus,  these are certainly reducible polynomials.
	*The other two polynomials, on the other hand, have no real zeros (rather imaginary or complex ones) and are irreducible in the real field according to the above definition.
			*The other two polynomials,  on the other hand,  have no real zeros  (rather imaginary or complex ones)  and they are irreducible in the real field according to the above definition.




	'''(2)'''  Correct is the <u>proposed solution 3</u>:		'''(2)'''  Correct is the  <u>proposed solution 3</u>:
	$p_3 = x^2 + x + 1$  is the only irreducible polynomial in the Galois field  $\rm GF(2^2)$. In the [[Channel_Coding/Extension_Field#GF.2822.29_.E2.80.93_Example_of_an_extension_field\|"theory section"]] the additions and the multiplication table were given for this.
			$p_3 = x^2 + x + 1$  is the only irreducible polynomial in the Galois field  $\rm GF(2^2)$.  In the [[Channel_Coding/Extension_Field#GF.2822.29_.E2.80.93_Example_of_an_extension_field\|"theory section"]]  the additions and the multiplication table were given for this.

	For the other polynomials holds:		For the other polynomials holds:
	* The polynomial  $p_1(x)$  is reducible in  ${\rm GF}(2) = \{0, \, 1\}$  since this polynomial can be factorized:		* The polynomial  $p_1(x)$  is reducible in  ${\rm GF}(2) = \{0, \, 1\}$  since this polynomial can be factorized:
	:$$p_1(x) = x^2 + 1 = (x+1)^2\hspace{0.05cm}.$$		:$$p_1(x) = x^2 + 1 = (x+1)^2\hspace{0.05cm}.$$
	* Since in  $\rm GF(2)$  there is no difference between sum and difference, the polynomial  $p_2(x) = x^2 - 1$  is also reducible.		* Since in  $\rm GF(2)$  there is no difference between sum and difference,  the polynomial  $p_2(x) = x^2 - 1$  is also reducible.
	*The polynomial  $p_4(x) = x^2 + x - 2$ is  unsuitable for  $\rm GF(2)$  alone, since not all polynomial coefficients are $0$ or $1$.
	*The "$2$" would only be possible in the Galois field $\rm GF(3)$.

			*The polynomial  $p_4(x) = x^2 + x - 2$  is  unsuitable for  $\rm GF(2)$  alone,  since not all polynomial coefficients are  $0$  or  $1$.  The  "$2$"  would only be possible in the Galois field  $\rm GF(3)$.



	~~'''(3)'''  Correct are the <u>last three proposed solutions</u>:~~
	* The set  $\mathcal{N}$  is not a field, since already the subtraction is not allowed for all elements, for example  $2 - 3 = -1 ∉ \mathcal{N}$.
	* Also the set  $\mathcal{Z}$  of integers is not a field, since for example the equation  $2 \cdot z = 1$  is not to be satisfied for any  $z ∈ \mathcal{N}$ .

			'''(3)'''  Correct are the  <u>last three proposed solutions</u>:
			* The set  $\mathcal{N}$  is not a field,  since already the subtraction is not allowed for all elements,  for example  $2 - 3 = -1 ∉ \mathcal{N}$.

			* Also the set  $\mathcal{Z}$  of integers is not a field,  since for example the equation  $2 \cdot z = 1$  is not to be satisfied for any  $z ∈ \mathcal{N}$.


	~~'''(4)'''  Correct are <u>answers 1 and 3</u>:~~
	* It is $\mathcal{Q} ⊂ \mathcal{R}$ (rational numbers are a subset of real numbers) and $\mathcal{R} ⊂ \mathcal{C}$ (real numbers are a subset of the complex numbers).
	*Thus also $\mathcal{Q} ⊂ \mathcal{C}$.
	*For finite fields, ${\rm GF}(2^m) ⊂ {\rm GF}(2^M)$ means that $m < M$ must hold.


			'''(4)'''  Correct are  <u>the answers 1 and 3</u>:
			* It is  $\mathcal{Q} ⊂ \mathcal{R}$  (rational numbers are a subset of real numbers)  and  $\mathcal{R} ⊂ \mathcal{C}$  (real numbers are a subset of the complex numbers).

			*Thus also  $\mathcal{Q} ⊂ \mathcal{C}$.
			*For finite fields,  ${\rm GF}(2^m) ⊂ {\rm GF}(2^M)$  means that  $m < M$  must hold.

	'''(5)'''  Correct is the <u>proposed solution 2</u>:
	* The set  $\mathcal{C}$  of complex numbers is an extension of the real numbers  $(\mathcal{R})$  into a second dimension. For this can be written:

			'''(5)'''  Correct is only the  <u>proposed solution 2</u>:
			* The set  $\mathcal{C}$  of complex numbers is an extension of the real numbers  $(\mathcal{R})$  into a second dimension.  For this can be written:
	:$$\mathcal{C} = \{k_0 + {\rm j} \cdot k_1\hspace{0.05cm}\| \hspace{0.05cm}k_0 \in {\mathcal{R}}, k_1 \in \mathcal{R}\}\hspace{0.05cm}.$$		:$$\mathcal{C} = \{k_0 + {\rm j} \cdot k_1\hspace{0.05cm}\| \hspace{0.05cm}k_0 \in {\mathcal{R}}, k_1 \in \mathcal{R}\}\hspace{0.05cm}.$$
	* $\rm GF(2^2)$  is an extension of the finite field  $\rm GF(2) = \{0, \, 1\}$  into a second dimension:		* $\rm GF(2^2)$  is an extension of the finite field  $\rm GF(2) = \{0, \, 1\}$  into a second dimension:
	:$${\rm GF}(2^2) = \{k_0 + \alpha \cdot k_1\hspace{0.05cm}\| \hspace{0.05cm}k_0 \in {\rm GF}(2), k_1 \in {\rm GF}(2)\}\hspace{0.05cm}.$$		:$${\rm GF}(2^2) = \{k_0 + \alpha \cdot k_1\hspace{0.05cm}\| \hspace{0.05cm}k_0 \in {\rm GF}(2), k_1 \in {\rm GF}(2)\}\hspace{0.05cm}.$$

	*The imaginary unit  ${\rm j} ∉ \mathcal{C}$  results as the solution of the equation  ${\rm j}^2 + 1 = 0$, while the new element of  $\rm GF(2^2)$  is denoted by  $\alpha ∉ \rm GF(2)$  and follows from the equation  $\alpha^2 + \alpha + 1 = 0$~~ ~~.		*The imaginary unit  ${\rm j} ∉ \mathcal{C}$  results as the solution of the equation   ${\rm j}^2 + 1 = 0$,  while the new element of  $\rm GF(2^2)$  is denoted by  $\alpha ∉ \rm GF(2)$  and follows from the equation   $\alpha^2 + \alpha + 1 = 0$.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 15:51, 3 October 2022

2022-10-03T15:51:06Z

← Older revision		Revision as of 17:51, 3 October 2022
Line 40:		Line 40:

	Hints:		Hints:
	* This exercise belongs to the chapter  [[Channel_Coding/Extension_Field\|"Extension ~~fields~~"]].		* This exercise belongs to the chapter  [[Channel_Coding/Extension_Field\|"Extension Field"]].

	* Above you can see illustrations of the Italian mathematicians  [https://en.wikipedia.org/wiki/Gerolamo_Cardano Gerolamo Cardano]  and  [https://en.wikipedia.org/wiki/Rafael_Bombelli Rafael Bombelli], who first introduced imaginary numbers to solve algebraic equations, and of  [https://en.wikipedia.org/wiki/%C3%89variste_Galois Évariste Galois], who established the foundations of finite fields at a very young age.		* Above you can see illustrations of the Italian mathematicians  [https://en.wikipedia.org/wiki/Gerolamo_Cardano Gerolamo Cardano]  and  [https://en.wikipedia.org/wiki/Rafael_Bombelli Rafael Bombelli], who first introduced imaginary numbers to solve algebraic equations, and of  [https://en.wikipedia.org/wiki/%C3%89variste_Galois Évariste Galois], who established the foundations of finite fields at a very young age.

Guenter at 15:50, 3 October 2022

2022-10-03T15:50:18Z

← Older revision		Revision as of 17:50, 3 October 2022
Line 28:		Line 28:
	The first two subtasks are related to the classification of polynomials.		The first two subtasks are related to the classification of polynomials.

	A degree-m polynomial is called  "reducible"  in the field  $\mathcal{F}$  if it is in the form		*A degree-m polynomial is called  "reducible"  in the field  $\mathcal{F}$  if it is representable in the form
	:$$p(x)= \prod_{i = 1}^m (x-x_i) = (x - x_1) \cdot (x - x_2) \cdot \hspace{0.05cm}\text{ ...}\hspace{0.05cm} \cdot (x - x_m) $$		:$$p(x)= \prod_{i = 1}^m (x-x_i) = (x - x_1) \cdot (x - x_2) \cdot \hspace{0.05cm}\text{ ...}\hspace{0.05cm} \cdot (x - x_m) $$

	~~is representable~~ and holds for all zeros $x_i ∈ \mathcal{F}$~~. If this is not possible, one speaks of an ''irreducible polynomial''~~.		and holds for all zeros  $x_i ∈ \mathcal{F}$.

			*If this is not possible,  one speaks of an  "irreducible polynomial".


Line 38:		Line 39:


			Hints:
			* This exercise belongs to the chapter  [[Channel_Coding/Extension_Field\|"Extension fields"]].

			* Above you can see illustrations of the Italian mathematicians  [https://en.wikipedia.org/wiki/Gerolamo_Cardano Gerolamo Cardano]  and  [https://en.wikipedia.org/wiki/Rafael_Bombelli Rafael Bombelli], who first introduced imaginary numbers to solve algebraic equations, and of  [https://en.wikipedia.org/wiki/%C3%89variste_Galois Évariste Galois], who established the foundations of finite fields at a very young age.
	~~Hints:~~
	* This exercise belongs to the chapter  [[Channel_Coding/Extension_Field\|"extension fields"]].
	* Above you can see illustrations of the Italian mathematicians  [https://en.wikipedia.org/wiki/Gerolamo_Cardano "Gerolamo Cardano"]  and  [https://en.wikipedia.org/wiki/Rafael_Bombelli "Rafael Bombelli"], who first introduced imaginary numbers to solve algebraic equations, and of  [https://en.wikipedia.org/wiki/%C3%89variste_Galois "Évariste Galois"], who established the foundations of finite fields at a very young age.

Guenter at 14:37, 2 October 2022

2022-10-02T14:37:22Z

← Older revision		Revision as of 16:37, 2 October 2022
Line 2:		Line 2:

	[[File:P_ID2511__KC_Z_2_4.png\|right\|frame\|Some pioneers of mathematics]]		[[File:P_ID2511__KC_Z_2_4.png\|right\|frame\|Some pioneers of mathematics]]
	In mathematics, different sets of numbers are distinguished:		In mathematics,  different sets of numbers are distinguished:
	* the set of natural numbers:  $\mathcal{N} = \{0, \, 1, \, 2, \hspace{0.05cm}\text{ ...} \}$,		* the set of natural numbers:  $\mathcal{N} = \{0, \, 1, \, 2, \hspace{0.05cm}\text{ ...} \}$,

	* the set of integers:  $\mathcal{Z} = \{\text{ ...}, \, -1, \, 0, \, +1, \hspace{0.05cm}\text{ ...} \}$,		* the set of integers:  $\mathcal{Z} = \{\text{ ...}, \, -1, \, 0, \, +1, \hspace{0.05cm}\text{ ...} \}$,
	* the set of rational numbers:  $\mathcal{Q} = \{m/n\}$ with $m ∈ \mathcal{Z}, \ n ∈ \mathcal{Z} \, \backslash \, \{0\}$,
			* the set of rational numbers:  $\mathcal{Q} = \{m/n\}$   with   $m ∈ \mathcal{Z}, \ n ∈ \mathcal{Z} \, \backslash \, \{0\}$,

	* the set  $\mathcal{R}$  of real numbers,		* the set  $\mathcal{R}$  of real numbers,
	* the set of complex numbers:  $\mathcal{C}= \{a + {\rm j} \cdot b\}$  with  $a ∈ \mathcal{R}, \ b ∈ \mathcal{R}$  and the imaginary unit  $\rm j$.

			* the set of complex numbers:  $\mathcal{C}= \{a + {\rm j} \cdot b\}$  with   $a ∈ \mathcal{R}, \ b ∈ \mathcal{R}$  and the imaginary unit  "$\rm j$".

	Such a set is called a  '''field''' in the algebraic sense if (and only if) the four arithmetic operations addition, subtraction, multiplication and division are allowed in it and the results are representable in the same field. Some related definitions can be found in the  [[Channel_Coding/Some_Basics_of_Algebra#Group.2C_ring.2C_field_-_basic_algebraic_concepts\|"theory section"]]. So much up front:  Not all of the sets listed above are fields.

	Besides, there are also  [[Channel_Coding/Some_Basics_of_Algebra#Definition_of_a_Galois_field\|"finite fields"]] , which in our learning tutorial are also called  Galois field  ${\rm GF}(P^m)$  where		Such a set is called a  "'''field'''"  in the algebraic sense if  (and only if)  the four arithmetic operations addition, subtraction, multiplication and division are allowed in it and the results are representable in the same field.

			Some related definitions can be found in the  [[Channel_Coding/Some_Basics_of_Algebra#Group.2C_ring.2C_field_-_basic_algebraic_concepts\|"theory section"]].  So much up front:  Not all of the sets listed here are fields.

			Besides,  there are also  [[Channel_Coding/Some_Basics_of_Algebra#Definition_of_a_Galois_field\|"finite fields"]],  which in our learning tutorial are also called  "Galois field"  ${\rm GF}(P^m)$  where
	* $P ∈ \mathcal{P}$  denotes a prime number, and		* $P ∈ \mathcal{P}$  denotes a prime number, and

	* $m ∈ \mathcal{N}$  denotes a natural number.		* $m ∈ \mathcal{N}$  denotes a natural number.


	If the exponent  $m ~~≥ is~~ 2$, it is called an  [[Channel_Coding/Extension_Field\|"extension field"]]. In this exercise we restrict ourselves to extension fields to the base  $P = 2$.		If the exponent  $m \ge 2$,  it is called an  [[Channel_Coding/Extension_Field\|"extension field"]].  In this exercise we restrict ourselves to extension fields to the base  $P = 2$.

			The first two subtasks are related to the classification of polynomials.

	~~The first two subtasks are related to the classification of polynomials.~~ A degree $m$ polynomial is called ''reducible''  in the field  $\mathcal{F}$ if it is in the form		A degree-m polynomial is called  "reducible"  in the field  $\mathcal{F}$  if it is in the form
	:$$p(x)= \prod_{i = 1}^m (x-x_i) = (x - x_1) \cdot (x - x_2) \cdot \hspace{0.05cm}\text{ ...}\hspace{0.05cm} \cdot (x - x_m) $$		:$$p(x)= \prod_{i = 1}^m (x-x_i) = (x - x_1) \cdot (x - x_2) \cdot \hspace{0.05cm}\text{ ...}\hspace{0.05cm} \cdot (x - x_m) $$

Noah at 15:02, 31 August 2022

2022-08-31T15:02:36Z

← Older revision		Revision as of 17:02, 31 August 2022
Line 17:		Line 17:


	If the exponent  $m ≥ is 2$, it is called an  [[Channel_Coding/Extension_Field\|extension field]]. In this exercise we restrict ourselves to extension fields to the base  $P = 2$.		If the exponent  $m ≥ is 2$, it is called an  [[Channel_Coding/Extension_Field\|"extension field"]]. In this exercise we restrict ourselves to extension fields to the base  $P = 2$.

	The first two subtasks are related to the classification of polynomials. A degree $m$ polynomial is called ''reducible''  in the field  $\mathcal{F}$ if it is in the form		The first two subtasks are related to the classification of polynomials. A degree $m$ polynomial is called ''reducible''  in the field  $\mathcal{F}$ if it is in the form

Noah at 15:02, 31 August 2022

2022-08-31T15:02:19Z

Show changes

Guenter at 16:57, 27 August 2022

2022-08-27T16:57:50Z

← Older revision		Revision as of 18:57, 27 August 2022
Line 136:		Line 136:


	[[Category:Channel Coding: Exercises\|^2.2 ~~Erweiterungskörper~~^]]		[[Category:Channel Coding: Exercises\|^2.2 Extension Field^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 2.4Z: Endliche und unendliche Körper to Aufgaben:Exercise 2.4Z: Finite and Infinite Fields

2022-08-27T15:59:53Z

Guenter moved page Aufgaben:Aufgabe 2.4Z: Endliche und unendliche Körper to Aufgaben:Exercise 2.4Z: Finite and Infinite Fields

← Older revision	Revision as of 17:59, 27 August 2022
(No difference)