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

Guenter at 16:25, 3 October 2022

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Guenter at 15:51, 3 October 2022

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Guenter at 15:50, 3 October 2022

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Guenter at 14:37, 2 October 2022

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Noah at 15:02, 31 August 2022

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Noah at 15:02, 31 August 2022

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Guenter at 16:57, 27 August 2022

2022-08-27T16:57:50Z

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 Aufgabe 2.4Z: Endliche und unendliche Körper to Exercise 2.4Z: Finite and Infinite Fields

Javier: Text replacement - "”" to """

2021-05-28T14:25:45Z

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 '''(1)'''&nbsp; 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&nbsp; $m$&nbsp; is called&nbsp; "reducible"&nbsp; in the field&nbsp; $\mathcal{F}$&nbsp; if it is can be represented the form
 :$$p(x)= \prod_{i = 1}^m (x-x_i) = (x - x_1)  \cdot  (x - x_2) \cdot \hspace{0.05cm}\text{ ...}\hspace{0.05cm}  \cdot (x - x_m) $$
-can be represented and is valid for all zeros $x_i &#8712; \mathcal{F}$. If this is not possible, one speaks of an ''irreducible'' polynomial.
+and this is valid for all zeros&nbsp; $x_i &#8712; \mathcal{F}$. &nbsp; If this is not possible,&nbsp; one speaks of an&nbsp; "irreducible"&nbsp; polynomial.
-In real number space, for each&nbsp; $m = 2$&nbsp; zeros&nbsp; $x_1$&nbsp; and&nbsp; $x_2$ hold:
+In real number space,&nbsp; it hold for the&nbsp; $m = 2$&nbsp; zeros&nbsp; $x_1$&nbsp; and&nbsp; $x_2$:
 :$$p_1(x) \text{:}\hspace{0.3cm}x_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} +{\rm j}\hspace{0.05cm},\hspace{0.2cm}x_2 = -{\rm j}\hspace{0.05cm},$$
 :$$p_2(x) \text{:}\hspace{0.3cm}x_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} +1\hspace{0.05cm},\hspace{0.2cm}x_2 = -1\hspace{0.05cm},$$
@@ Line 99: / Line 99: @@
 :$$p_4(x) \text{:}\hspace{0.3cm}x_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} +1\hspace{0.05cm},\hspace{0.2cm}x_2 = -2\hspace{0.05cm}.$$
-Thus the <u>proposed solutions 1 and 3</u> are correct:
+Thus the&nbsp; <u>proposed solutions 1 and 3</u>&nbsp; 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&nbsp; $p_2(x)$&nbsp; and&nbsp; $p_4(x)$&nbsp; are both real.&nbsp; Thus,&nbsp; 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.
-'''(2)'''&nbsp; Correct is the <u>proposed solution 3</u>:
+'''(2)'''&nbsp; Correct is the&nbsp; <u>proposed solution 3</u>:
 $p_3 = x^2 + x + 1$&nbsp; is the only irreducible polynomial in the Galois field &nbsp;$\rm GF(2^2)$. 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:
 * The polynomial&nbsp; $p_1(x)$&nbsp; is reducible in&nbsp; ${\rm GF}(2) = \{0, \, 1\}$&nbsp; since this polynomial can be factorized:
 :$$p_1(x) = x^2 + 1 = (x+1)^2\hspace{0.05cm}.$$
-* Since in&nbsp; $\rm GF(2)$&nbsp; there is no difference between sum and difference, the polynomial&nbsp; $p_2(x) = x^2 - 1$&nbsp; is also reducible.
+* Since in&nbsp; $\rm GF(2)$&nbsp; there is no difference between sum and difference,&nbsp; the polynomial&nbsp; $p_2(x) = x^2 - 1$&nbsp; is also reducible.
 '''(5)'''&nbsp; Correct is the <u>proposed solution 2</u>:
 * The set&nbsp; $\mathcal{C}$&nbsp; of complex numbers is an extension of the real numbers&nbsp; $(\mathcal{R})$&nbsp; into a second dimension. For this can be written:
 :$$\mathcal{C} = \{k_0 + {\rm j} \cdot k_1\hspace{0.05cm}| \hspace{0.05cm}k_0 \in {\mathcal{R}}, k_1 \in \mathcal{R}\}\hspace{0.05cm}.$$
 * $\rm GF(2^2)$&nbsp; is an extension of the finite field&nbsp; $\rm GF(2) = \{0, \, 1\}$&nbsp; into a second dimension:
 :$${\rm GF}(2^2) = \{k_0 + \alpha \cdot k_1\hspace{0.05cm}| \hspace{0.05cm}k_0 \in {\rm GF}(2), k_1 \in {\rm GF}(2)\}\hspace{0.05cm}.$$
-*The imaginary unit&nbsp; ${\rm j} &#8713; \mathcal{C}$&nbsp; results as the solution of the equation&nbsp; ${\rm j}^2 + 1 = 0$, while the new element of&nbsp; $\rm GF(2^2)$&nbsp; is denoted by&nbsp; $\alpha &#8713; \rm GF(2)$&nbsp; and follows from the equation&nbsp; $\alpha^2 + \alpha + 1 = 0$&nbsp;.
+*The imaginary unit&nbsp; ${\rm j} &#8713; \mathcal{C}$&nbsp; results as the solution of the equation &nbsp; ${\rm j}^2 + 1 = 0$,&nbsp; while the new element of&nbsp; $\rm GF(2^2)$&nbsp; is denoted by&nbsp; $\alpha &#8713; \rm GF(2)$&nbsp; and follows from the equation &nbsp; $\alpha^2 + \alpha + 1 = 0$.
 {{ML-Fuß}}

@@ Line 40: / Line 40: @@
 Hints:
-* This exercise belongs to the chapter&nbsp; [[Channel_Coding/Extension_Field|"Extension fields"]].
+* This exercise belongs to the chapter&nbsp; [[Channel_Coding/Extension_Field|"Extension Field"]].
 * Above you can see illustrations of the Italian mathematicians&nbsp; [https://en.wikipedia.org/wiki/Gerolamo_Cardano Gerolamo Cardano]&nbsp; and&nbsp; [https://en.wikipedia.org/wiki/Rafael_Bombelli Rafael Bombelli], who first introduced imaginary numbers to solve algebraic equations, and of&nbsp; [https://en.wikipedia.org/wiki/%C3%89variste_Galois Évariste Galois], who established the foundations of finite fields at a very young age.

@@ Line 28: / Line 28: @@
 The first two subtasks are related to the classification of polynomials.
-A degree-m polynomial is called&nbsp; "reducible"&nbsp; in the field&nbsp; $\mathcal{F}$&nbsp; if it is in the form
+*A degree-m polynomial is called&nbsp; "reducible"&nbsp; in the field&nbsp; $\mathcal{F}$&nbsp; if it is representable in the form
 :$$p(x)= \prod_{i = 1}^m (x-x_i) = (x - x_1)  \cdot  (x - x_2) \cdot  \hspace{0.05cm}\text{ ...}\hspace{0.05cm} \cdot (x - x_m) $$
-is representable and holds for all zeros $x_i &#8712; \mathcal{F}$. If this is not possible, one speaks of an ''irreducible polynomial''.
+and holds for all zeros&nbsp; $x_i &#8712; \mathcal{F}$.
@@ Line 38: / Line 39: @@
 * Above you can see illustrations of the Italian mathematicians&nbsp; [https://en.wikipedia.org/wiki/Gerolamo_Cardano Gerolamo Cardano]&nbsp; and&nbsp; [https://en.wikipedia.org/wiki/Rafael_Bombelli Rafael Bombelli], who first introduced imaginary numbers to solve algebraic equations, and of&nbsp; [https://en.wikipedia.org/wiki/%C3%89variste_Galois Évariste Galois], who established the foundations of finite fields at a very young age.

@@ Line 2: / Line 2: @@
 [[File:P_ID2511__KC_Z_2_4.png|right|frame|Some pioneers of mathematics]]
-In mathematics, different sets of numbers are distinguished:
+In mathematics,&nbsp; different sets of numbers are distinguished:
 * the set of natural numbers:&nbsp; $\mathcal{N} = \{0, \, 1, \, 2, \hspace{0.05cm}\text{ ...} \}$,
 * the set of integers:&nbsp; $\mathcal{Z} = \{\text{ ...}, \, -1, \, 0, \, +1, \hspace{0.05cm}\text{ ...}  \}$,
 * the set of rational numbers:&nbsp; $\mathcal{Q} = \{m/n\}$ with $m &#8712; \mathcal{Z}, \ n &#8712; \mathcal{Z} \, \backslash \, \{0\}$,
 * the set&nbsp; $\mathcal{R}$&nbsp; of real numbers,
-Besides, there are also&nbsp; [[Channel_Coding/Some_Basics_of_Algebra#Definition_of_a_Galois_field|"finite fields"]] , which in our learning tutorial are also called&nbsp; Galois field&nbsp; ${\rm GF}(P^m)$&nbsp; where
+Such a set is called a&nbsp; "'''field'''"&nbsp; in the algebraic sense if&nbsp; (and only if)&nbsp; the four arithmetic operations addition, subtraction, multiplication and division are allowed in it and the results are representable in the same field.
 * $P &#8712; \mathcal{P}$&nbsp; denotes a prime number, and
 * $m &#8712; \mathcal{N}$&nbsp; denotes a natural number.
-If the exponent&nbsp; $m &#8805; is 2$, it is called an&nbsp; [[Channel_Coding/Extension_Field|"extension field"]]. In this exercise we restrict ourselves to extension fields to the base&nbsp; $P = 2$.
+If the exponent&nbsp; $m \ge 2$,&nbsp; it is called an&nbsp; [[Channel_Coding/Extension_Field|"extension field"]].&nbsp; In this exercise we restrict ourselves to extension fields to the base&nbsp; $P = 2$.
-The first two subtasks are related to the classification of polynomials. A degree $m$ polynomial is called ''reducible''&nbsp; in the field&nbsp; $\mathcal{F}$ if it is in the form
+A degree-m polynomial is called&nbsp; "reducible"&nbsp; in the field&nbsp; $\mathcal{F}$&nbsp; if it is in the form
 :$$p(x)= \prod_{i = 1}^m (x-x_i) = (x - x_1)  \cdot  (x - x_2) \cdot  \hspace{0.05cm}\text{ ...}\hspace{0.05cm} \cdot (x - x_m) $$

← Older revision		Revision as of 15:02, 31 August 2022
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−	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

← Older revision		Revision as of 16:57, 27 August 2022
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−	[[Category:Channel Coding: Exercises\|^2.2 ~~Erweiterungskörper~~^]]	+	[[Category:Channel Coding: Exercises\|^2.2 Extension Field^]]

@@ Line 105: / Line 105: @@
 * Da in&nbsp; $\rm GF(2)$&nbsp; kein Unterschied zwischen Summe und Differenz besteht, ist auch das Polynom&nbsp; $p_2(x) = x^2 - 1$&nbsp; reduzibel.
 * Das Polynom&nbsp; $p_4(x) = x^2 + x - 2$ ist&nbsp; schon allein deshalb für&nbsp; $\rm GF(2)$&nbsp; ungeeignet, da nicht alle Polynomkoeffizienten $0$ oder $1$ sind.
-*Die "$2$&rdquo; wäre nur im Galoisfeld $\rm GF(3)$ möglich.
+*Die "$2$" wäre nur im Galoisfeld $\rm GF(3)$ möglich.