Problem 3 - Applied Discrete Structures

Question

Determine the splitting field of \(x^{4}-5 x^{2}+6\) over \(\mathbb{Q}\).

Step-by-Step Solution

Answer

The splitting field is \(\mathbb{Q}(\sqrt{2}, \sqrt{3})\).

Step 1: Factor the Polynomial

First, we need to factor the polynomial \(x^{4}-5x^{2}+6\). Notice that this can be treated as a quadratic in terms of \(x^2\). Let's set \(y = x^2\), then our equation becomes \(y^2 - 5y + 6\). This is a simple quadratic equation which can be factored as \((y - 2)(y - 3)\). Hence, the factorization in terms of \(x\) is \((x^2 - 2)(x^2 - 3)\).

Step 2: Find the Roots of Each Quadratic

The polynomial splits into two quadratic equations: \(x^2 - 2 = 0\) and \(x^2 - 3 = 0\). Finding the roots of these equations gives us: \(x = \pm \sqrt{2}\) and \(x = \pm \sqrt{3}\), respectively. Therefore, the roots of the original polynomial are \(\pm \sqrt{2}\) and \(\pm \sqrt{3}\).

Step 3: Determine the Splitting Field

The splitting field of a polynomial is the smallest field extension of \(\mathbb{Q}\) that contains all the roots of the polynomial. In this case, the roots are \(\sqrt{2}\), \(-\sqrt{2}\), \(\sqrt{3}\), and \(-\sqrt{3}\). The splitting field will be \(\mathbb{Q}(\sqrt{2}, \sqrt{3})\), which is the field obtained by adjoining \(\sqrt{2}\) and \(\sqrt{3}\) to \(\mathbb{Q}\).

Key Concepts

Field Extension

In mathematics, particularly in the field of algebra, a **field extension** is a way of expanding a given field to include additional elements that are not initially present. It’s like upgrading a system to a more complex version without altering the original framework. A field extension of a field \( \mathbb{Q} \) (the field of rational numbers) means building a larger field that retains the properties of \( \mathbb{Q} \) but includes new elements, such as roots of a polynomial that didn’t exist within \( \mathbb{Q} \).

The concept of field extension is crucial for understanding the solution to the problem at hand, where we need to determine the splitting field of the polynomial \( x^4 - 5x^2 + 6 \). Here, the field extension is required to incorporate the roots of the polynomial that aren’t rational numbers. For our polynomial, the roots are \( \pm \sqrt{2} \) and \( \pm \sqrt{3} \). To include these roots, the field \( \mathbb{Q} \) needs to be extended to \( \mathbb{Q}(\sqrt{2}, \sqrt{3}) \).

This new field allows calculations involving these roots, making it possible to solve equations or conduct operations that weren't solvable in \( \mathbb{Q} \) alone. Field extensions are a key concept in many areas of algebra and number theory, helping to bridge the gap between simple rational fields and more intricate algebraic structures.

Polynomial Roots

**Polynomial roots** are the solutions to the equation formed when a polynomial is set equal to zero. Roots provide valuable information about the behavior and characteristics of a polynomial function. When a polynomial is expressed in the form of an equation like \( x^4 - 5x^2 + 6 = 0 \), finding its roots involves factoring the polynomial and solving for the values of \( x \) that make it zero.

In the exercise provided, the polynomial \( x^4 - 5x^2 + 6 \) is factored using a substitution! Let \( y = x^2 \), transforming the original polynomial into \( y^2 - 5y + 6 = 0 \). This quadratic equation was straightforward to factor into \( (y - 2)(y - 3) \). Re-substituting back for \( x \), we obtained the factored expression \( (x^2 - 2)(x^2 - 3) \).

Finding the roots of these factors, \( x = \pm \sqrt{2} \) and \( x = \pm \sqrt{3} \), illustrates a common scenario in algebra where new kinds of numbers (like these roots) must be considered to fully understand the polynomial's behavior. Recognizing and calculating polynomial roots is essential for not only solving equations but also for comprehending the nature of polynomial functions in algebra and calculus.

Quadratic Factorization

**Quadratic factorization** deals with expressing a quadratic expression as a product of its linear factors. This procedure simplifies solving equations, as it transforms the equation into a form where the principle of zero product can be applied. If you have a quadratic equation like \( ax^2 + bx + c = 0 \), the goal is to rewrite it as \((px + q)(rx + s) = 0\).

In our polynomial, \( x^4 - 5x^2 + 6 \), because the highest degree term is expressed as \((x^2)^2\), it can be viewed as a quadratic in disguise. By setting \( y = x^2 \), the polynomial morphs into \( y^2 - 5y + 6 \), a form suitable for traditional quadratic factoring techniques. Factoring this yields \((y - 2)(y - 3)\), which upon reverting the substitution, provides \((x^2 - 2)(x^2 - 3)\).

Breaking down polynomials using quadratic factorization is a fundamental technique in algebra. It simplifies complex expressions and reveals the roots or solutions to equations. Understanding this process is pivotal for anyone studying polynomial functions, as it allows for deeper insight into how polynomials behave and how their solutions interact.