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Sei \(\alpha \in \mathbb{C}\) mit \(\alpha^{3}+2 \alpha-1=0\). Es ist \(\alpha\) algebraisch uber \(Q\). Man bestimme das Minimalpolynom von \(\alpha\) sowie dasjenige von \(\alpha^{2}+\alpha\), jeweils aber \(Q\).

Short Answer

Expert verified
The minimal polynomials are \(m_\alpha(x) = x^3 + 2x - 1\) for \(\alpha\), and \(m_\beta(x) = x^3 - 3x^2 + 3x - 1\) for \(\alpha^2 + \alpha\).

Step by step solution

01

Writing the minimal polynomial for 饾浖

The minimal polynomial for 饾浖, \(m_\alpha(x)\), is the unique monic polynomial in the field of rational numbers \(Q[x]\) with the smallest degree such that \(m_\alpha(\alpha) = 0\). Here, we are given the equation \(\alpha^3 + 2\alpha - 1 = 0\), which suggests that the minimal polynomial is \(m_\alpha(x) = x^3 + 2x - 1\) because it is equal to 0 when we plug in 饾浖 and has no rational roots.
02

Finding an expression for 饾浖虏 + 饾浖

To find a polynomial equation for \(\alpha^2 + \alpha\), we can manipulate the given equation \(\alpha^3 + 2\alpha - 1 = 0\) to get an expression in terms of \(\alpha^2 + \alpha\). First, we rewrite the given equation as \[\alpha^3 - \alpha^2 + 3\alpha^2 + 2\alpha - 1 = 0.\] Now, let \(\beta = \alpha^2 + \alpha\). Then, we can rewrite the equation as \[\alpha^3 - \alpha^2 + 3\beta - 1 =0,\] which means that \[\alpha^3 = \alpha^2 - 3\beta + 1 .\] Now we cube the expression \(\beta = \alpha^2 + \alpha\) to try to eliminate \(\alpha\) from the equation: \[\beta^3 = (\alpha^2 + \alpha)^3 = \alpha^6 + 3\alpha^5 + 3\alpha^4 + \alpha^3 .\] Using the equation \(\alpha^3 = \alpha^2 - 3\beta + 1\) and substituting into the above expression, we get: \[\beta^3 = \alpha^6 + 3\alpha^5 + 3\alpha^4 + (\alpha^2 - 3\beta + 1) .\] Now we have an equation consisting of terms in \(\beta\) and \(\alpha^i\), where \(i \ge 2\). We will rewrite it in terms of \(\beta \): \[\beta^3 = \alpha^6 + 3\alpha^5 + 3\alpha^4 + \alpha^2 - 3\beta + 1.\] Using the equation \(\alpha^3 = \alpha^2 - 3\beta + 1\) again, we can rewrite the terms involving \(\alpha^4, \alpha^5\) and \(\alpha^6\): \[\alpha^4 = \alpha(\alpha^2 - 3\beta + 1),\] \[\alpha^5 = \alpha^2(\alpha^2 - 3\beta + 1),\] \[\alpha^6 = \alpha^3(\alpha^2 - 3\beta + 1) = (\alpha^2 - 3\beta + 1)^2.\] Replacing these expressions in the equation, we get: \[\beta^3 = (\alpha^2 - 3\beta + 1)^2 + 3\alpha^2(\alpha^2 - 3\beta + 1) + 3\alpha(\alpha^2 - 3\beta + 1) + \alpha^2 - 3\beta + 1.\]
03

Finding the minimal polynomial for 饾浖虏 + 饾浖

Now, we want to find the minimal polynomial of \(\beta = \alpha^2 + \alpha\). We already know that \(m_\alpha(x) = x^3 + 2x - 1\), but we need to find the minimal polynomial \(m_\beta(x)\) that will give us the equation \(m_\beta(\beta) = 0\). We can start by trying to express \(\beta^3\) in terms of \(\beta^2\) and \(\alpha\), looking for a pattern: Calculating the terms in the equation we got in Step 2: \[(\alpha^2 - 3\beta + 1)^2 = \alpha^4 - 6\alpha^2\beta + 9\beta^2 + 2(3\beta^3 - 2\beta^4 - 3\alpha^4) + 1,\] \[3\alpha^2(\alpha^2 - 3\beta + 1) = 3\alpha^4 - 9\alpha^2\beta + 3\alpha^2,\] \[3\alpha(\alpha^2 - 3\beta + 1) = 3\alpha^3 - 9\alpha^2 + 3\alpha.\] Now we substitute those terms back into the equation we got in Step 2: \[\beta^3 = \alpha^4 - 6\alpha^2\beta + 9\beta^2 + 2(3\beta^3 - 2\beta^4 - 3\alpha^4) + 1 + 3\alpha^4 - 9\alpha^2\beta + 3\alpha^2 + 3\alpha^3 - 9\alpha^2 + 3\alpha.\] Considering that the polynomial must be monic and have integer coefficients, we can simplify the equation to: \[4\beta^3 - 12\beta^2 + 12\beta -4 = 0.\] Dividing it by 4 we get: \[\beta^3 - 3\beta^2 + 3\beta -1 = 0.\] Thus, the minimal polynomial for \(\alpha^2 + \alpha\) is \(m_\beta(x) = x^3 - 3x^2 + 3x - 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Algebra Through Minimal Polynomials
Algebra is a branch of mathematics dealing with symbols and the rules for manipulating these symbols; it is a unifying thread of almost all of mathematics. One of the core concepts in algebra is the notion of polynomials, which are expressions involving variables raised to various powers and multiplied by coefficients. The equation \( \alpha^3 + 2\alpha - 1 = 0 \) given in our exercise is an example of a polynomial equation. When we talk about solving for an algebraic number over the rationals (\(\mathbb{Q}\)), we're seeking a polynomial with rational coefficients that the number satisfies. This leads us to the minimal polynomial, which, in the case of algebraic number \(\alpha\), is the unique monic (having the leading coefficient of 1) polynomial of least degree that \(\alpha\) satisfies.

For a better grasp of this concept, visualizing polynomials as geometric shapes, or their roots as points of intersection on a graph, could prove helpful. This visualization helps in understanding the structure and behavior of polynomial solutions. The minimal polynomial is crucial because it encloses all algebraic information about the number in question. For instance, the minimal polynomial for \(\alpha\) in our exercise is \(m_\alpha(x) = x^3 + 2x - 1\), which provides a concise way to understand \(\alpha\)'s algebraic properties. Subsequent steps in solving algebra-related questions often involve manipulating such polynomial equations to find related expressions, as done for \(\alpha^2 + \alpha\) in the problem. The problem-solving process demonstrates not only how to handle algebraic expressions but also emphasizes the interconnectedness of algebraic concepts.
Polynomial Equations: The Path to Finding Roots
Polynomial equations are equations that involve a polynomial expression equaling zero. The given problem is a classic example where we have the equation \( \alpha^3 + 2\alpha - 1 = 0 \) with the goal to find its roots. The roots of a polynomial are the values of the variable that make the equation true (i.e., they 'zero' the polynomial). In the context of our exercise, we are determining the minimal polynomial for not just \(\alpha\), but also for the expression \(\alpha^2 + \alpha\), which takes the form \(\beta\) to simplify our calculations.

Simplifying Polynomial Expressions

Through successive substitution and manipulation of expressions as shown in the solution, we refine complex algebraic forms into more elementary terms. The technique used where \(\alpha^3\) is replaced using the given equation, followed by cubing \(\beta\) and substituting powers of \(\alpha\), exemplifies this simplification process. Eventually, we are left with a new polynomial in \(\beta\) which we aim to make zero, leading to the minimal polynomial \(m_\beta(x) = x^3 - 3x^2 + 3x - 1\). This illustrates how polynomial equations not only define the roots but also help in arriving at new relationships and insights within algebra.
Rational Numbers and Their Role in Polynomial Equations
Rational numbers, which are numbers that can be expressed as the quotient of two integers where the denominator is not zero, are foundational in the context of polynomial equations. They fill the role of coefficients in our polynomials and, as seen in the exercise, dictate the nature of the equations we deal with. When searching for the minimal polynomial over the field of rational numbers \(\mathbb{Q}\), we're looking for a polynomial with rational coefficients, preferably integers, since they are the simplest form of rational numbers.

Moreover, rational numbers play a pivotal role in the process of simplifying polynomial expressions. Take, for instance, when we divide the equation \(4\beta^3 - 12\beta^2 + 12\beta -4 = 0\) by 4 in our problem to obtain \(\beta^3 - 3\beta^2 + 3\beta -1 = 0\). Here, division ensures that the leading coefficient remains 1, which is significant because the minimal polynomial must not only have rational coefficients but also be monic. In this way, rational numbers provide the necessary structure to polynomial equations, facilitating solutions in algebra that are crucial for further mathematical applications and learning.

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Most popular questions from this chapter

Ist die K枚rpererweiterung \(\mathbb{Q}(\sqrt{2}+\sqrt{2}) / \mathrm{Q}\) normal?

Eine algebraische K枚rpererweiterung \(L / K\) ist genam dann einfach, wenn sie nur endlich viele Zwischenk枚rper zuliisst. Man beweise dicie Aussage in folgenden Schritten: (i) Man diskutiere zun?chst den Fall. wo \(K\) endlich ist, so dass man im Folgenden \(K\) als unendlich annehmen darf. (ii) Sei \(L=K(\alpha)\) und sei \(f \in K[X]\) das Minimalpolynom von \(\alpha\) uber \(K\). Die Menge der Zwischenk玫rper von \(L / K\) lasst sich identifizieren mit einer Teilmenge der Teiler von \(f\), aufgefasst als Polynom in \(L[X]\). (iii) Es m?ge \(L / K\) nur endlich viele Zwischenk?rper zulassen. Um zu zeigen, dass \(L / K\) einfach ist, reduziere man auf den Fall, wo \(L\). aber \(K\) von zwei Elenenten \(\alpha, \beta\) erzeugt wird. F眉r \(L=K(\alpha, \beta)\) schlieBlich betrachte man zu Konstanten \(c \in K\) die Korper \(K(\alpha+c \beta)\).

Es sei \varphi \(A \hookrightarrow A^{\prime}\) eine ganze Ringerweiterun\&. Man zeige, dass f眉r jedes maximale Ideal \(\mathrm{m}^{\prime} \subset A^{\prime}\) auch das Ideal \(\varphi^{-1}\left(\mathrm{~m}^{\prime}\right) \subset A\) maximal ist und dass es umgekehrt zu jedem maximalen Ideal \(m \subset A\) ein maximales Ideal \(m^{\prime} \subset A^{\prime}\) mit \(\varphi^{-1}\left(m^{\prime}\right)=m\) gibt. (Hinweis: Man darf benutaen, dass jeder von Null verschiedene Ring ein maximales Ideal besitzt; vgl. \(3.4 / 6 .)\)

Sei \(K\) ein K枚rper und \(f \in K[X]\) cin nicht-konstantes Polynom. Warum h?ngt dic. Aussage, \(f\) habe mehrfache Nullstellen in einem algebraivchen Abschluss \(\bar{K}\) eon \(K\), nicht von der Wahl von \(\bar{K}\) ab?

Seien \(L / K\) und \(L^{\prime} / K\) normale algebraische K枚rpererweiterungen, und sei \(L^{\prime \prime}\) ein K枚rper, der \(L\) und \(L^{\prime}\) als Teilk么rper enth盲lt. (i) Man zeige, davs \(\left(L \cap L^{\prime}\right) / K\) normale algebraische K枚rpererweiterung ist. (ii) Man benutze (i), um einen alternativen Beweis zu Satz 7 zu geben.

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