Problem 3 - A Course in Galois Theory

Question

Suppose that \(K(\alpha): K\) is a simple extension and that \(\alpha\) is transcendental over \(K\). Let \(\sigma\) be the automorphism of \(K(\alpha)\) which fixes \(K\) and sends \(\alpha\) to \(1 /(1-\alpha)\). Verify that \(\sigma^{3}\) is the identity, and determine the fixed field of \(\sigma\).

Step-by-Step Solution

Answer

\( \sigma^3 \) is the identity and the fixed field is \( K \).

Step 1: Understand Transcendental and Automorphism

Since \( \alpha \) is transcendental over \( K \), there are no polynomial equations with coefficients in \( K \) that \( \alpha \) satisfies. The automorphism \( \sigma \) is an isomorphism from \( K(\alpha) \) to itself that fixes \( K \) and maps \( \alpha \) to \( \frac{1}{1-\alpha} \).

Step 2: Apply \(\sigma\) Repeatedly

First, apply \( \sigma \) to \( \alpha \): \( \sigma(\alpha) = \frac{1}{1-\alpha} \).Next, find \( \sigma^2(\alpha) = \sigma(\sigma(\alpha)) \):\[ \sigma^2(\alpha) = \sigma(\frac{1}{1-\alpha}) = 1 - \frac{1}{1-\alpha} = \frac{\alpha}{\alpha-1.} \]Finally, find \( \sigma^3(\alpha) = \sigma(\sigma^2(\alpha)) \):\[ \sigma^3(\alpha) = \sigma\left(\frac{\alpha}{\alpha - 1}\right) = \frac{1}{1 - \frac{\alpha}{\alpha - 1}}. \]

Step 3: Simplify \(\sigma^3(\alpha)\)

Simplify \( \sigma^3(\alpha) \):\[ \sigma^3(\alpha) = \frac{1}{1 - \frac{\alpha}{\alpha - 1}} = \frac{1}{\frac{\alpha-1-\alpha}{\alpha-1}} = \alpha. \]Thus, \( \sigma^3(\alpha) = \alpha \), showing that \( \sigma^3 \) is indeed the identity on \( \alpha \).

Step 4: Determine the Fixed Field of \(\sigma\)

The fixed field of \( \sigma \) are those elements \( \beta \) in \( K(\alpha) \) such that \( \sigma(\beta) = \beta \). Since \( \sigma^3 \) is the identity, \( \sigma^3(\beta) = \beta \) for any element in the field. Elements fixed by \( \sigma \) must satisfy the condition \( \sigma(\beta) = \beta \), meaning \( \beta \) must be invariant under all transformations by \( \sigma \). For this specific transformation, \( K \) itself, being common to all automorphisms of this form, forms the fixed field since no nontrivial elements of \( K(\alpha) \) remain invariant. Thus, the fixed field is \( K \).

Key Concepts

Field Extensions

In the realm of Galois theory, understanding field extensions is pivotal. A field extension involves a base field, let's refer to it as \( K \), and a larger field containing \( K \), which we can denote as \( K(\alpha) \).
In this context, \( K(\alpha) : K \) is a simple extension. This means we're extending the field \( K \) by adding an element \( \alpha \) that is not originally in \( K \).

• A finite field extension is one where the larger field has a finite number of dimensions over the base field when viewed as a vector space.
• An infinite field extension has an infinite number of such dimensions.

In our exercise, the extension involves adding a transcendental element, \( \alpha \), to \( K \). This means that \( \alpha \) does not satisfy any polynomial equation with coefficients from \( K \). Therefore, the extension is infinite. Field extensions with transcendental elements are particularly intriguing as they pave the way for more complex structures in higher mathematics.

Automorphisms

An automorphism is essentially a symmetry of a field extension. In simpler terms, it’s a function mapping the field onto itself, preserving the operations of addition and multiplication. In our exercise, we consider the automorphism \( \sigma \) of the field \( K(\alpha) \) that fixes \( K \). In practice, this means \( \sigma \) leaves every element of \( K \) unchanged and applies a specific transformation to \( \alpha \).

• The key property of an automorphism is that it must be bijective – every element in the field should map to exactly one element in the field, and vice versa.
• A significant characteristic is that it respects operations – it fulfills \( \sigma(x+y) = \sigma(x) + \sigma(y) \) and \( \sigma(x \cdot y) = \sigma(x) \cdot \sigma(y) \).

In our scenario, \( \sigma \) transforms \( \alpha \) into \( \frac{1}{1-\alpha} \). Repetitions of this process demonstrate a cyclic behavior, and ultimately, applying \( \sigma \) three times returns \( \alpha \) to its original value, emphasizing that \( \sigma^3 \) is the identity automorphism.

Transcendental Elements

Transcendental elements like \( \alpha \) in our exercise, play an essential role in field theory. These are elements that are not roots of any non-zero polynomial equation with coefficients from a given field, \( K \) in this instance. Hence, they are not algebraic over \( K \).

• If an element is algebraic, it satisfies some polynomial equation with coefficients in the base field, providing a finite relationship between the elements of the extension and the base field.
• A transcendental element implies the extension is infinite – no such bounding relationship exists.

Understanding transcendental elements helps in exploring the concept of infinite extensions and how transformations affect these elements. In the exercise, \( \alpha \) is transformed by an automorphism that cyclically alters it, underscoring the non-restrictive nature of transcendental elements in complex field extensions.

Fixed Fields

The concept of fixed fields is crucial in analyzing the effect of automorphisms within a field extension. A fixed field under an automorphism consists of all those elements that appear unchanged when the automorphism is applied. In this exercise, the fixed field of \( \sigma \), which acts on \( K(\alpha) \), is determined by identifying elements that are invariant under this transformation.

• The fixed field is always a subfield – it must contain \( K \) and all elements fixed by the automorphism.
• It offers a reflection of the invariants of the automorphism – those characteristics of the field that remain constant under transformation.

For the automorphism \( \sigma \) in our exercise, which maps \( \alpha \) to different states, the fixed field simply is \( K \) itself. This occurs because no elements of the extended field \( K(\alpha) \) other than \( K \) remain unchanged after \( \sigma \) is applied. This simplistic conclusion shows how some automorphisms can reveal simplicity amidst complexity by noting invariance in seemingly vast field extensions.