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Chapter 12: Problem 8

Let \(E=Q(\sqrt{2}, \sqrt{3}, i),\) and consider the intermediate fields \(\mathbb{Q}\) \subseteq \(K \subseteq E\). $$ \text { Describe } \operatorname{Gal}(E / Q) $$

Short Answer

Expert verified
\(\operatorname{Gal}(E/\mathbb{Q}) \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2\).

Step by step solution

01

Understand the Field Extension

The field extension \(E = \mathbb{Q}(\sqrt{2}, \sqrt{3}, i)\) is a splitting field of the polynomials: \(x^2 - 2\), \(x^2 - 3\), and \(x^2 + 1\) over \(\mathbb{Q}\). This means \(E\) contains all roots of these polynomials.
02

Determine the Degree of the Extension

We calculate the degree of the extension \([E: \mathbb{Q}]\). Each square root introduces a factor of 2, and \(i = \sqrt{-1}\) introduces another factor of 2. So, \([E: \mathbb{Q}] = 2 \times 2 \times 2 = 8\).
03

Identify Automorphisms

The Galois group \(\operatorname{Gal}(E/\mathbb{Q})\) consists of automorphisms of \(E\) that leave \(\mathbb{Q}\) fixed. There are eight elements since \([E: \mathbb{Q}] = 8\). Each automorphism is determined by how it permutes \(\sqrt{2}, \sqrt{3},\) and \(i\).
04

Describe the Galois Group

Since \(\operatorname{Gal}(E/\mathbb{Q})\) is comprised of all combinations of sign changes for \(\sqrt{2}, \sqrt{3},\) and \(i\), it is isomorphic to \(\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2\). This is a direct product of three cyclic groups of order 2, with each generator corresponding to conjugations over each quadratic factor.

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Key Concepts

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

Field Extension
In Galois Theory, a field extension is the addition of elements from a larger field to a smaller one, creating a bigger field. For instance, consider the field extension as \( E = \mathbb{Q}(\sqrt{2}, \sqrt{3}, i) \). Here, \( E \) is an extension field of \( \mathbb{Q} \), which means \( E \) contains \( \mathbb{Q} \) and possibly additional elements.
Field extensions often help in understanding the structure of polynomial roots. The field \( E \) is formed by adjoining square roots of 2 and 3, and the imaginary unit \( i \), which is \( \sqrt{-1} \). Hence, \( E \) is so-called because it includes all possible roots of the relevant polynomials over \( \mathbb{Q} \).
  • The core purpose is to evaluate how one field can be extended from another by adding roots of certain polynomials.
  • This concept serves as a bridge for defining automorphisms and understanding Galois groups.
Galois Group
The Galois group, denoted as \( \operatorname{Gal}(E/\mathbb{Q}) \), encapsulates the symmetries of a field extension. Essentially, it's a set of automorphisms鈥攆unctions that map a field onto itself, preserving the field's operations. Here, the specific field extension is \( E = \mathbb{Q}(\sqrt{2}, \sqrt{3}, i) \).
The elements within the Galois group \( \operatorname{Gal}(E/\mathbb{Q}) \) are those automorphisms which keep the base field \( \mathbb{Q} \) intact. Each element represents a potential "reshuffling" of the roots in \( E \) while maintaining the structure of \( \mathbb{Q} \).
  • In more simple terms, think of it like a set of symmetry operations you can perform on the field without breaking its rules.
  • This group typically helps determine how the elements of \( E \) relate to the entire field system.
Automorphisms
Automorphisms in the context of a Galois extension are crucial transformations. They map the field onto itself while preserving both addition and multiplication. For the field \( E = \mathbb{Q}(\sqrt{2}, \sqrt{3}, i) \), automorphisms help us see how different expressions of the same field can exist and interrelate.
In this specific example, there are eight automorphisms because \([E: \mathbb{Q}] = 8\), corresponding to each combination of sign changes of \( \sqrt{2} \), \( \sqrt{3} \), and \( i \).
  • Each automorphism in \( \operatorname{Gal}(E/\mathbb{Q}) \) represents a different element in this set of possible transformations.
  • Understanding these transformations allows mathematicians to analyze how field elements are organized and manipulated.
Splitting Field
A splitting field is a minimal field in which a polynomial breaks down into linear factors, also known as roots. For \( x^2 - 2 \), \( x^2 - 3 \), and \( x^2 + 1 \), the field \( E = \mathbb{Q}(\sqrt{2}, \sqrt{3}, i) \) acts as a splitting field. This is because it contains all the roots for these specific polynomials.
The role of a splitting field is central in Galois Theory because it provides a complete environment where the behavior of polynomial roots can be fully explored.
  • In a splitting field, complex roots such as those of \( x^2 + 1 \) (notably \( i \) and \(-i\)) are explicitly represented.
  • It helps illustrate the process of decomposing more complex field constructs into simpler, fully understandable parts.

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Most popular questions from this chapter

In Exercises 1 through 3 express the indicated symmetric functions of \(n\) indeterminates in terms of the elementary symmetric functions in \(n\) indeterminates. $$ n=2 \quad\left(x_{1}-x_{2}\right)^{2} $$

In Exercises 8 through 13 determine for the indicated \(n\) whether or not the regular \(n\) -gon is constructible by marked ruler and compass. $$ n=14 $$

In Exercises 1 through 8 express the splitting field of the indicated polynomial \(f(x) \in \mathbb{Q}[x]\) as a radical extension of \(\mathrm{Q}\). $$ x^{2}+2 x+2 $$

Let \(\phi: \mathrm{C} \rightarrow \mathrm{C}\) be defined by \(\phi(a+b i)=a-b i\) (complex conjugation). Show that \(\phi\) is an automorphism of \(\mathrm{C}\).

For the indicated quartic polynomial find (a) the resolvant cubic (b) the discriminant \(D\) (c) the Galois group over \(Q\) of the resolvant cubic (d) the Galois group over \({Q}\) of the quartic $$ x^{4}+2 $$
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