Problem 4
Question
Suppose that \(\varepsilon\) is a primitive \(m\) th root of unity over \(\mathbb{Q}\), where \(m>2\). Let \(\eta=\varepsilon+\varepsilon^{-1}\). Show that \([Q(\varepsilon): \mathbb{Q}(\eta)]=2\), find the minimal polynomial for \(\varepsilon\) over \(\mathbb{Q}(\eta)\) and identify the Galois group \(\Gamma[\mathbb{Q}(\varepsilon): \mathbb{Q}(\eta)]\)
Step-by-Step Solution
Verified Answer
The degree \([\mathbb{Q}(\varepsilon): \mathbb{Q}(\eta)]\) is 2. The minimal polynomial is \(x^2 - \eta x + 1\). The Galois group is \(\mathbb{Z}_2\).
1Step 1: Understand Primitive Root of Unity
A primitive \( m \)th root of unity, denoted \( \varepsilon \), is a complex number such that \( \varepsilon^m = 1 \) and \( \varepsilon^k eq 1 \) for any smaller positive integer \( k \). This ensures that \( \varepsilon\) is not just any root of unity but specifically one with minimal order \( m \).
2Step 2: Define \( \eta \)
Given \( \varepsilon \) as a primitive \( m \)th root of unity, define \( \eta = \varepsilon + \varepsilon^{-1} \). Since \( \varepsilon^{-1} = \overline{\varepsilon} \) in this context, \( \eta \) is a real number. It's a trace element, representing the sum of \( \varepsilon \) and its complex conjugate.
3Step 3: Determine a Minimal Polynomial for \( \varepsilon \) over \( \mathbb{Q}(\eta) \)
Consider the expression \( (x - \varepsilon)(x - \varepsilon^{-1}) = x^2 - \eta x + 1 \). Replace \( \varepsilon \) and \( \varepsilon^{-1} \) with any complex conjugate pair that satisfies this relation, confirming that the minimal polynomial for \( \varepsilon \) over \( \mathbb{Q}(\eta) \) is \( x^2 - \eta x + 1 \).
4Step 4: Show the Degree is 2
The minimal polynomial \( x^2 - \eta x + 1 \) is of degree 2. This implies that the field extension degree \( [\mathbb{Q}(\varepsilon): \mathbb{Q}(\eta)] \) is indeed 2, as the degree of the polynomial is the minimal degree required to express \( \varepsilon \) over \( \mathbb{Q}(\eta) \).
5Step 5: Identify the Galois Group
The Galois group \( \Gamma[\mathbb{Q}(\varepsilon): \mathbb{Q}(\eta)] \) corresponds to the symmetries in the roots of the polynomial \( x^2 - \eta x + 1 \). With two roots, \( \varepsilon \) and \( \varepsilon^{-1} \), the Galois group is isomorphic to \( \mathbb{Z}_2 \), the cyclic group of order 2, as it only swaps \( \varepsilon \) and \( \varepsilon^{-1} \).
Key Concepts
Primitive Roots of UnityMinimal PolynomialField ExtensionsGalois Groups
Primitive Roots of Unity
When we talk about primitive roots of unity, we are diving into the heart of complex numbers and their fascinating symmetries. A primitive root of unity is often denoted as \( \varepsilon \) and satisfies the condition \( \varepsilon^m = 1 \), where \( m \) is a positive integer. This means \( \varepsilon \) is a solution to the equation \( x^m - 1 = 0 \). But there's more! To be considered 'primitive', \( \varepsilon \) must not satisfy \( \varepsilon^k = 1 \) for any positive integer \( k < m \). Essentially, \( \varepsilon \) is the first root to complete the cycle after \( m \) applications.
- Primitive roots of unity are crucial for understanding symmetries in numbers.
- They are used extensively in fields like algebra and number theory.
Minimal Polynomial
The concept of a minimal polynomial is essential in field theory and is closely tied to the idea of algebraic numbers. When you have a complex number like \( \varepsilon \), which is a primitive \( m \)-th root of unity, its behavior can be expressed through a polynomial with the smallest possible degree.Such a polynomial for \( \varepsilon \) becomes vital as it provides the simplest way to describe \( \varepsilon \) over a given field, here \( \mathbb{Q}(\eta) \). Notably, for the expression \( \varepsilon + \varepsilon^{-1} \) as \( \eta \), the minimal polynomial turns out to be \( x^2 - \eta x + 1 \).
- The minimal polynomial is always monic, meaning the leading coefficient is 1.
- It provides insights into the degree of extension when moving from one field to another.
Field Extensions
Field extensions are central in algebra, particularly in Galois Theory. When we seek to understand \( \mathbb{Q}(\varepsilon) \), a field extension of \( \mathbb{Q}(\eta) \), what we're really looking at is how one field can build upon another.In this specific case, our base field \( \mathbb{Q}(\eta) \) is extended to \( \mathbb{Q}(\varepsilon) \) using \( \varepsilon \), the primitive \( m \)-th root of unity. This is characterized by the expression \( [\mathbb{Q}(\varepsilon) : \mathbb{Q}(\eta)] = 2 \), indicating that the degree of the extension is 2.
- Field extensions illuminate how different field structures relate and interact.
- The degree of the extension indicates the 'dimensions' one field adds upon another.
Galois Groups
Galois groups are intriguing entities that provide a bridge between algebra and geometry by describing how roots of polynomials can be transformed. In the context of our primitive root of unity \( \varepsilon \), the Galois group \( \Gamma[\mathbb{Q}(\varepsilon) : \mathbb{Q}(\eta)] \) captures the symmetries in the polynomial \( x^2 - \eta x + 1 \).This polynomial has roots \( \varepsilon \) and \( \varepsilon^{-1} \), and the Galois group explains how these roots can 'swap' places. Structurally, it is isomorphic to the cyclic group \( \mathbb{Z}_2 \), indicating that there are only two elements, effectively swapping the roots.
- Galois groups tell us how fields can map onto each other.
- They offer a rich framework for understanding 'reversible' processes in algebra.
Other exercises in this chapter
Problem 2
Show that if \(n\) is odd then \(\Phi_{2 n}(x)=\Phi_{n}(-x)\).
View solution Problem 3
Show that if \(p\) is a prime then $$ \Phi_{p^{n}}(x)=1+x^{p^{n-1}}+x^{2 p^{n-1}}+\cdots+x^{(p-1) p^{n-1}} $$
View solution Problem 5
Find the Galois groups of \(x^{4}+1\) and \(x^{5}+1\) over \(\mathbb{Q}\)
View solution Problem 6
Suppose that \(p\) is a prime which does not divide \(m\), and let \(\varepsilon\) be a primitive \(m\) th root of unity over \(\mathbb{Z}_{p}\). Show that \(\l
View solution