Problem 2 - A Book of Abstract Algebra

Question

Show that the degree of \(Q(\sqrt{2}, \sqrt{3}, \sqrt{5})\) over \(\mathbb{Q}\) is 8 .

Step-by-Step Solution

Answer

The degree of \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})\) over \(\mathbb{Q}\) is 8.

Step 1: Identify Field Extensions

Let's consider the field extensions involved in forming \(Q(\sqrt{2}, \sqrt{3}, \sqrt{5})\) over \(\mathbb{Q}\). We construct this field through a series of extensions starting from \(\mathbb{Q}\) to \(\mathbb{Q}(\sqrt{2})\), then \(\mathbb{Q}(\sqrt{2}, \sqrt{3})\), and finally reaching \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})\).

Step 2: Determine Degree of First Extension

The degree of the extension \(\mathbb{Q}(\sqrt{2}) / \mathbb{Q}\) is 2 since \(\sqrt{2}\) is not in \(\mathbb{Q}\) and satisfies the minimal polynomial \(x^2 - 2\).

Step 3: Determine Degree of Second Extension

After extending to \(\mathbb{Q}(\sqrt{2})\), consider \(\mathbb{Q}(\sqrt{3})\). The degree of the extension \(\mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q}(\sqrt{2})\) is also 2, as \(\sqrt{3}\) is not in \(\mathbb{Q}(\sqrt{2})\) and satisfies \(x^2 - 3\), which is irreducible over \(\mathbb{Q}\).

Step 4: Determine Degree of Third Extension

Now, we have \(\mathbb{Q}(\sqrt{2}, \sqrt{3})\), and we extend by \(\sqrt{5}\). The degree of extension \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}) / \mathbb{Q}(\sqrt{2}, \sqrt{3})\) is 2 because \(\sqrt{5}\) is not in \(\mathbb{Q}(\sqrt{2}, \sqrt{3})\) and satisfies \(x^2 - 5\), also irreducible.

Step 5: Calculate Overall Degree of the Extension

The total degree of the field extension \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}) / \mathbb{Q}\) is calculated as the product of the degrees of each individual extension: \[2 \times 2 \times 2 = 8\].

Key Concepts

Degree of Extension

The degree of a field extension is a fundamental concept in field theory. It refers to the "size" of an extension field relative to its base field, often measured by the dimension of the extension field as a vector space over the base field.
Let's consider an extension field denoted as \(F\) over a base field \(K\). The degree of the extension, \([F:K]\), is the number of vector space bases of \(F\) as a vector space over \(K\).
In simple terms, it counts how many copies of the base field are needed to span the extension field.

• For example, if \( [F:K] = 2 \), this implies that two dimensions are needed over \(K\) to form \(F\).


• In the context of the problem, the degree of \( \mathbb{Q} (\sqrt{2}, \sqrt{3}, \sqrt{5}) \) over \( \mathbb{Q} \) was shown to be \(8\). By consecutively evaluating the degree of smaller intermediate extensions, we determined the combined degree by multiplication.

Minimal Polynomial

A minimal polynomial is the polynomial of the smallest degree for which a given algebraic element is a root.
In simpler terms, for any algebraic number \( \alpha \), the minimal polynomial is the characteristic polynomial with the smallest degree and leading coefficient of 1 (monic polynomial) having \( \alpha \) as a root.
This concept is crucial because:

• It helps in determining whether certain roots are within a given field.
• It provides the degree for the simple extensions like \(\mathbb{Q}(\sqrt{2})\).

For example, the minimal polynomial for \(\sqrt{2}\) is \(x^2 - 2\). This means that \( \sqrt{2} \) satisfies this polynomial, which is irreducible over \( \mathbb{Q} \).
Hence, the degree of the extension \( \mathbb{Q}(\sqrt{2}) / \mathbb{Q} \) is 2, as the smallest polynomial containing this root has a degree of 2.

Irreducible Polynomial

An irreducible polynomial is a non-constant polynomial that cannot be expressed as the product of two non-constant polynomials over a given field.
Understanding irreducible polynomials is crucial in field theory because they are the building blocks of polynomial factorization within the field.

• A polynomial is labeled irreducible if it cannot be factored into smaller degree polynomials with coefficients in the same field.


• For instance, polynomials \(x^2 - 2\), \(x^2 - 3\), and \(x^2 - 5\) utilized in constructing the field \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})\) are irreducible over \(\mathbb{Q}\).

These irreducible polynomials confirm that each additional extension (\( \sqrt{2}, \sqrt{3}, \sqrt{5}\)) increases the degree of the extension by 2 for each step, respecting the structure of the field and validating the non-reducibility.

Algebraic Number Fields

Algebraic number fields are advanced structures in number theory and abstract algebra. These fields are finite extensions of the rational numbers \(\mathbb{Q}\).
Algebraic number fields encapsulate roots of polynomials within them, thus enriching \(\mathbb{Q}\) with new numbers like radicals and roots of unity.

• For example, \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})\) is an algebraic number field, as it involves finite extensions of \(\mathbb{Q}\) using radicals, specifically \(\sqrt{2}, \sqrt{3},\) and \(\sqrt{5}\).


• These fields retain many properties of \(\mathbb{Q}\) while extending it, facilitating a greater range of operations.

Algebraic number fields are pivotal in generalizing properties of numbers, factoring polynomials, and solving equations that are irreducible over base fields like \(\mathbb{Q}\). They offer a broader perspective to analyze the behavior of numbers beyond rational solutions.