Problem 5 - A Course in Galois Theory

Question

Find the Galois groups of \(x^{4}+1\) and \(x^{5}+1\) over \(\mathbb{Q}\)

Step-by-Step Solution

Answer

The Galois group of \(x^4 + 1\) is \(C_2 \times C_2\) and for \(x^5 + 1\) it is \(C_4\).

Step 1: Understanding the Problem

We need to find the Galois group of the polynomials \(x^4 + 1\) and \(x^5 + 1\) over \(\mathbb{Q}\). The Galois group of a polynomial is the group of all field automorphisms of its splitting field that fix the base field, in this case \(\mathbb{Q}\). Understanding the roots of these polynomials is crucial, as the roots will determine the structure of the Galois group.

Step 1: Roots of \(x^4 + 1\)

The equation \(x^4 + 1 = 0\) can be rewritten as \(x^4 = -1\). The roots are the fourth roots of \(-1\), which are complex numbers. This can be simplified using Euler's formula: \(x = e^{i\pi/4}, e^{3i\pi/4}, e^{5i\pi/4}, e^{7i\pi/4}\). These correspond to \( e^{\pi i/4}, e^{3\pi i/4}, e^{5\pi i/4}, e^{7\pi i/4} \).

Step 2: Splitting Field for \(x^4 + 1\)

The splitting field of \(x^4 + 1\) is \(\mathbb{Q}(i, \sqrt{2})\). This is because \(e^{i\pi/4} = \cos(\pi/4) + i\sin(\pi/4)\) which equals \(\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\), components of which require \(i\) and \(\sqrt{2}\) to be expressed. Therefore the degree of the splitting field is 4.

Step 3: Galois Group of \(x^4 + 1\)

The Galois group of \(x^4 + 1\) over \(\mathbb{Q}\) is isomorphic to \(C_2 \times C_2\), the Klein four-group. This is because the roots are symmetric, and the Galois group of the extension \(\mathbb{Q}(i, \sqrt{2})/\mathbb{Q}\) corresponds to the automorphisms fixing \(\mathbb{Q}\) but freely assigning the signs of \(i\) and \(\sqrt{2}\).

Step 4: Roots of \(x^5 + 1\)

The polynomial \(x^5 + 1 = 0\) is rewritten as \(x^5 = -1\). This can be rewritten in exponential form as \(x = e^{i\pi (2k + 1)/5}\) for \(k=0,1,2,3,4\). These are the primitive 10th roots of unity \(\zeta_{10}\) excluding the principal root \(-1\).

Step 5: Splitting Field for \(x^5 + 1\)

The splitting field for \(x^5 + 1\) is \(\mathbb{Q}(\zeta_{10})\). The degree of this extension over \(\mathbb{Q}\) is \(\varphi(10) = 4\), where \(\varphi\) is the Euler totient function. So the Galois group has 4 elements.

Step 6: Galois Group of \(x^5 + 1\)

The Galois group of \(x^5 + 1\) over \(\mathbb{Q}\) is isomorphic to \(C_4\), the cyclic group of order 4. This arises because \(\mathbb{Q}(\zeta_{10})\) corresponds to a cyclic extension of degree 4 over \(\mathbb{Q}\).

Key Concepts

Splitting Fields

Understanding the concept of splitting fields is a critical step when working with Galois groups. In simple terms, the splitting field of a polynomial is the smallest field extension in which the polynomial's roots can be found. Imagine you have a polynomial, like \(x^4 + 1\). Its roots are not neatly contained within the set of rational numbers, \(ackslashmathbb{Q}\). Instead, you have to incorporate additional numbers, in this case, complex numbers, to fully "capture" all the roots.

• For \(x^4 + 1\), the splitting field is \(ackslashmathbb{Q}(i, ackslashsqrt{2})\). This field contains all roots of \(x^4 + 1 =0\), which cannot be found in \(ackslashmathbb{Q}\).
• Similarly, for \(x^5 + 1\), we require \(ackslashmathbb{Q}(\zeta_{10})\), where \(\zeta_{10}\) are the primitive 10th roots of unity, ensuring we include all possible solutions.

By constructing splitting fields, we step into the world where the original polynomial neatly breaks down into well-behaved factors, allowing us to leverage field theory to study symmetries among its roots.

Field Automorphisms

Field automorphisms are like special transformations of fields. They keep the "structure" of the field intact while possibly moving elements around. Consider them as invisible hands that shift elements within a field without breaking the field's rules.

• For a polynomial's splitting field over \(\mathbb{Q}\), a field automorphism must keep \(\mathbb{Q}\) unchanged. That means rational numbers stay put, while other numbers from the splitting field might switch places.
• For instance, in the extension \(ackslashmathbb{Q}(i, ackslashsqrt{2})\), one automorphism might switch \(i\) with \(-i\) but leaves \(ackslashsqrt{2}\) untouched. Another might do the opposite, or even swap both.

These reshufflings matter because they form the Galois group. The automorphisms must satisfy all field properties and may reflect deeper symmetries among the polynomial's roots. They help us understand how these roots interrelate, leading us closer to revealing the structure of the Galois group.

Complex Roots

Complex roots are often the key to deciphering the behavior of polynomials, especially when dealing with Galois theory. Polynomials like \(x^4 + 1\) or \(x^5 + 1\), which cannot be decomposed into rational factors, often have complex numbers as roots.
The roots of \(x^4 + 1 = 0\) are found using Euler's formula and are expressed as complex numbers, specifically those involving powers of \(e^{i\pi/4}\). These represent angles in the complex plane. When graphed, they form symmetric arrangements which are often indicative of their algebraic properties.

• Each complex root can be seen as a point on a circle of radius 1 in the complex plane, providing both magnitude and direction (angle).
• For \(x^5 = -1\), the roots are primitive 10th roots of unity, excluding \(-1\). These are spaced evenly on the unit circle, showcasing the aesthetic symmetry that occurs in complex roots.

Understanding complex roots allows us to see beyond real numbers into an extended universe of numbers, where algebraic properties tied to symmetries and transformations become vividly clear. This insight is invaluable for constructing Galois groups and exploring the harmony within algebraic equations.