Chapter 20: Problem 1
Es sei \(a \in \mathbb{C}\) eine Wurzel des Polynoms \(P=X^{3}+3 X-2 \in \mathbb{Q}[X]\) (a) Zeigen Sie, dass \(P\) irreduzibel ist. (b) Stellen Sie die Elemente \(a^{-1},(1+a)^{-1}\) und \(\left(1-a+a^{2}\right)\left(5+3 a-2 a^{2}\right)\) von \(Q(a)\) als Q-Linearkombinationen von \(\left\\{1, a, a^{2}\right\\}\) dar.
Short Answer
Step by step solution
Show that P is irreducible over \( \mathbb{Q} \)
Find the inverse of \( a \)
Find the inverse of \( 1 + a \)
Express \((1-a+a^{2})(5+3a-2a^{2})\)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Root Theorem
For instance, consider our polynomial \( P(X) = X^3 + 3X - 2 \). Here, the constant term is -2, and the leading coefficient (the coefficient of \( X^3 \)) is 1. Therefore, the potential rational roots are the divisors of -2 divided by 1, which are \( \pm 1, \pm 2 \).
By evaluating these potential roots in the given polynomial, none of them results in zero, indicating that \( P \) has no rational roots, supporting the idea that \( P \) is irreducible over \( \mathbb{Q} \). This is significant because if a polynomial is irreducible over the rationals, it cannot be factored into lower-degree polynomials with rational coefficients.
Complex Numbers
In the context of polynomial equations, when the Rational Root Theorem fails, as seen with polynomial \( P(X) = X^3 + 3X - 2 \), we can turn to complex numbers. Each polynomial of degree \( n \) has exactly \( n \) roots in the complex number system due to the Fundamental Theorem of Algebra.
When \( a \) is a root in the complex numbers, it provides a bridge to solving polynomial equations that seem unsolvable in the realm of just real numbers. This concept aids in expressing elements like \( a^{-1} \) or \((1+a)^{-1}\) that relate back to the polynomial and its powers.
Polynomial Degree
For example, our polynomial \( P(X) = X^3 + 3X - 2 \) is of degree 3. This implies that there are 3 possible roots, considering both real and complex solutions.
Knowing the degree helps in understanding the solutions: For a degree 3 polynomial, if the roots are not all rational or do not factorize as expected, we explore root extensions within the complex plane, guided by complex algebra to determine a solution set that complements our degree's expectations.
- Degree dictates polynomial behavior.
- It guides solutions in complex system exploration.
- Provides root expectations via the polynomial's degree.
Field Extensions
In our exercise, we started from \( \mathbb{Q} \), the field of rational numbers, and extended it to \( \mathbb{Q}(a) \), where \( a \) is a root of the polynomial \( P(X) = X^3 + 3X - 2 \).
Fields like \( \mathbb{Q}(a) \) allow us to work algebraically with roots of polynomials and form combinations in simpler terms. When you express elements like \( a^{-1} \) or \( (1+a)^{-1} \) as linear combinations \( c_0 + c_1a + c_2a^2 \), it demonstrates the power of field extensions to make seemingly complex algebraic entities manageable. This process illustrates how by introducing a construct like \( a \), we augment our field to encompass all linear combinations of its elements over \( \mathbb{Q} \).
- Field extensions allow solving polynomials outside original fields.
- They transform and simplify algebraic expression handling.
- Provide a structured way to include polynomial roots into algebraic systems.