Problem 4 - A Book of Abstract Algebra

Question

Find a basis of \(\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})\) over \(\mathbb{Q}\), and describe the elements of \(\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})\).

Step-by-Step Solution

Answer

\( \mathbb{Q}(2+\u0074[3]{4})\) has a basis \( \{1, \sqrt{2}, \sqrt{2}\sqrt[3]{4}, \sqrt[3]{4}, (\sqrt[3]{4})^2, \sqrt{2}(\sqrt[3]{4})^2\} \) over \( \mathbb{Q} \).

Step 1: Overview of the Problem

We need to find a basis for the field extension \( \mathbb{Q}(\sqrt{2} + \sqrt[3]{4}) \) over \( \mathbb{Q} \). This involves determining the dimension of the extension field \( \mathbb{Q}(\sqrt{2}, \sqrt[3]{4}) \) and constructing a set of algebraic numbers that can be used to express any element in the field as a linear combination with rational coefficients.

Step 1: Understand the Elements

The elements in \( \mathbb{Q}(\sqrt{2} + \sqrt[3]{4}) \) are rational combinations of \( \sqrt{2} \) and \( \sqrt[3]{4} \). Define the elements \( \alpha = \sqrt{2} \) and \( \beta = \sqrt[3]{4} \), so the field is \( \mathbb{Q}(\alpha, \beta) \). We need to find the minimal polynomials for \( \alpha \) and \( \beta \) over \( \mathbb{Q} \).

Step 2: Minimal Polynomial of \( \alpha \)

The number \( \alpha = \sqrt{2} \) is a root of the polynomial \( x^2 - 2 = 0 \). This polynomial is irreducible over \( \mathbb{Q} \), so it is the minimal polynomial for \( \sqrt{2} \).

Step 3: Minimal Polynomial of \( \beta \)

The number \( \beta = \sqrt[3]{4} \) is a root of the polynomial \( x^3 - 4 = 0 \). This polynomial is irreducible over \( \mathbb{Q} \), thus serving as the minimal polynomial for \( \sqrt[3]{4} \).

Step 4: Determine the Degree of the Extension

The degree of \( \mathbb{Q}(\alpha, \beta) \) over \( \mathbb{Q} \) is the product of the degrees of the minimal polynomials of \( \alpha \) and \( \beta \). Therefore, \( [\mathbb{Q}(\alpha, \beta) : \mathbb{Q}] = 2 \times 3 = 6 \).

Step 5: Identify a Basis

A basis for the extension \( \mathbb{Q}(\alpha, \beta) \) over \( \mathbb{Q} \) is given by the set \( \{1, \alpha, \alpha\beta, \beta, \beta^2, \alpha\beta^2\} \). This is because any element of \( \mathbb{Q}(\alpha, \beta) \) can be expressed as a linear combination of these elements with coefficients in \( \mathbb{Q} \).

Step 6: Describe the Elements of the Field

Each element of the extension can be written as \( a + b\alpha + c\beta + d\alpha\beta + e\beta^2 + f\alpha\beta^2 \) where \( a, b, c, d, e, f \in \mathbb{Q} \). This shows the set \( \{1, \alpha, \alpha\beta, \beta, \beta^2, \alpha\beta^2\} \) spans the field.

Key Concepts

Minimal Polynomial

A minimal polynomial is a key concept when dealing with algebraic numbers. It is the smallest degree monic polynomial (leading coefficient equal to 1) over a given field that has the algebraic number as a root. The minimal polynomial provides vital information to understand the properties and degree of field extensions.

For example, let's consider the minimal polynomial of \( \alpha = \sqrt{2} \). The function \( x^2 - 2 \) is the polynomial for which \( \alpha \) is a root, and no lower degree polynomial with rational coefficients can have \( \alpha \) as a root. Similarly, the minimal polynomial of \( \beta = \sqrt[3]{4} \) is the polynomial \( x^3 - 4 \), which means this is the simplest form such polynomial over \( \mathbb{Q} \).

Understanding minimal polynomials helps determine other properties of the field extension, such as its degree.

Basis of a Vector Space

In the context of field extensions, the idea of a "basis" is similar to that for vector spaces. A basis is a set of elements from which any element of the vector space (in this case, the field extension) can be expressed as a linear combination.

For the field extension \( \mathbb{Q}(\alpha, \beta) \), where \( \alpha = \sqrt{2} \) and \( \beta = \sqrt[3]{4} \), a basis is comprised of the set \( \{1, \alpha, \alpha\beta, \beta, \beta^2, \alpha\beta^2\} \). This means any element of this extended field can be written in the form \( a + b\alpha + c\beta + d\alpha\beta + e\beta^2 + f\alpha\beta^2 \), where \( a, b, c, d, e, f \) belong to \( \mathbb{Q} \).

Thus, a basis is essential for understanding the structure and representation of elements in a field extension, similar to how a basis is important for understanding vector spaces.

Degree of a Field Extension

The degree of a field extension represents how many copies of the base field fit into the extended field, analogous to how vectors are expressed in a vector space. It also indicates the dimension of the vector space when the extension field is viewed as a vector space over the base field.

In our example, the degree of the field extension \( \mathbb{Q}(\alpha, \beta) \) over \( \mathbb{Q} \) is found by multiplying the degrees of the minimal polynomials of \( \alpha \) and \( \beta \). Since \( x^2 - 2 = 0 \) and \( x^3 - 4 = 0 \) both serve as minimal polynomials, their degrees are 2 and 3, respectively.

Therefore, the degree of the extension \( \mathbb{Q} (\alpha, \beta) \) is \( 2 \times 3 = 6 \). This tells us that \( \mathbb{Q}(\alpha, \beta) \) has six dimensions as a vector space over \( \mathbb{Q} \).

Algebraic Numbers

Algebraic numbers are numbers that are roots of non-zero polynomial equations with rational coefficients. In simpler terms, if you have an equation like \( x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 = 0 \), where the "a" coefficients are rational, and the equation equals zero with some value of \( x \), then that value of \( x \) is an algebraic number.

Both \( \sqrt{2} \) and \( \sqrt[3]{4} \) are examples of algebraic numbers since they solve polynomial equations over \( \mathbb{Q} \). The key distinction between algebraic and transcendental numbers is that the latter cannot be roots of any such polynomial equations with rational coefficients.

Algebraic numbers form a vast and important class of numbers in mathematics, facilitating the study of field extensions and allowing for the construction of fields that have more complex elements than the rationals alone.