Problem 5 - A Book of Abstract Algebra

Question

Find the root field of \(a(x)=\left(x^{2}-3\right)\left(x^{3}-1\right)\) over \(\mathbf{Q} .\) ANSWER The complex roots of \(a(x)\) are \(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm \sqrt{3} i)\), so the root field is \(\mathbb{Q}\left(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm \sqrt{3} i)\right)\). The same field can be written more simply as \(\mathbb{Q}(\sqrt{3}, i)\). Find irreducible polynomials \(a(x)\) over \(\mathbb{Q}\), and \(b(x)\) over \(\mathbb{( i )}\), such that \(Q(i, \sqrt{3})\) is the root field of \(a(x)\) over \(\mathbb{Q}\), and is the root field of \(b(x)\) over \(Q(i)\). Then do the same for \(Q(\sqrt{-2}, \sqrt{-3})\).

Step-by-Step Solution

Answer

Use \(a(x) = (x^2 - 3)(x^2 + 1)\) over \(\mathbb{Q}\) and \(b(x) = x^2 - 3\) over \(\mathbb{Q}(i)\), similarly use \((x^2 + 2)(x^2 + 3)\) over \(\mathbb{Q}\) for \(\mathbb{Q}(\sqrt{-2}, \sqrt{-3})\).

Step 1: Identify Roots of Each Factor

Let's start by finding the roots of each factor of the polynomial \(a(x) = (x^2 - 3)(x^3 - 1)\). The first factor \(x^2 - 3 = 0\) gives us roots \(x = \pm \sqrt{3}\). The second factor \(x^3 - 1 = 0\) gives us roots \(1, \frac{-1 + \sqrt{3}i}{2}, \frac{-1 - \sqrt{3}i}{2}\) which are the cube roots of unity.

Step 2: Construct the Root Field

The complex roots from step 1 are \(\pm \sqrt{3}, 1, \frac{-1 + \sqrt{3}i}{2}, \frac{-1 - \sqrt{3}i}{2}\). Thus, the root field, including all these roots over \(\mathbb{Q}\), can be described as \(\mathbb{Q}(\sqrt{3}, i)\) since both \(\sqrt{3}\) and \(i\) (imaginary unit) can generate all given roots.

Step 3: Find Irreducible Polynomial over \(\mathbb{Q}\)

For \(\mathbb{Q}(i, \sqrt{3})\), we need a polynomial with \(\pm \sqrt{3}\) and \(i\) as roots. The polynomial \(x^2 + 1\) is irreducible over \(\mathbb{Q}\) with \(i\) as a root, and \(x^2 - 3\) is irreducible over \(\mathbb{Q}\) with \(\sqrt{3}\) as a root. Therefore, \(a(x) = (x^2 - 3)(x^2 + 1)\) is a polynomial over \(\mathbb{Q}\) having \(\mathbb{Q}(i, \sqrt{3})\) as its root field.

Step 4: Find Irreducible Polynomial over \(\mathbb{Q}(i)\)

Next, we consider \(b(x)\) over \(\mathbb{Q}(i)\). \(\sqrt{3}\) is not in \(\mathbb{Q}(i)\), thus is a root of the irreducible polynomial \(x^2 - 3\) over \(\mathbb{Q}(i)\). Therefore, \(b(x) = x^2 - 3\) is irreducible over \(\mathbb{Q}(i)\).

Step 5: Similar Procedure for \(\mathbb{Q}(\sqrt{-2}, \sqrt{-3})\)

For the field \(\mathbb{Q}(\sqrt{-2}, \sqrt{-3})\), the irreducible polynomial over \(\mathbb{Q}\) is \((x^2 + 2)(x^2 + 3)\), since both \(x^2 + 2\) and \(x^2 + 3\) have roots \(\sqrt{-2}\) and \(\sqrt{-3}\), respectively, and are irreducible over \(\mathbb{Q}\). Similarly, the irreducible polynomial over \(\mathbb{Q}(i)\) would be \(x^2 + 3\) for \(\sqrt{-3}\).

Key Concepts

Root Field

A root field is an advanced concept in field theory essential for understanding polynomials and their roots. In simple terms, a root field is the smallest possible field containing all the roots of a given polynomial. Let's break this down:

• Polynomials can have one or more roots, which are solutions to the equation formed by setting the polynomial equal to zero.
• Plotting these roots on any number line or complex plane shows where the polynomial evaluates to zero.
• The root field is effectively constructed by including these roots in the least used field, ensuring all operations on these elements remain closed within that field.

In our case, the polynomial consisted of complex roots such as \(\sqrt{3}\) and cube roots of unity that required introducing the elements \(z = rac{-1 + \sqrt{3}i}{2}\) and its conjugate. This is why the root field here is \(\mathbb{Q}(i, \sqrt{3})\), ensuring we capture all behaviors and operations on these roots.

Irreducible Polynomials

Irreducible polynomials play a key role in understanding the structure of fields. They are the building blocks in constructing fields because they cannot be factored further over the base field. An irreducible polynomial is one that cannot be expressed as a product of two smaller degree polynomials with coefficients in the given field. Let's illustrate this using our case:

• Over field \(\mathbb{Q}\), we have \(\pm \sqrt{3}\) and \(\pm \u001bi\) as roots.
• The polynomial \(\left(x^2 + 1\right)\) is irreducible over \(\mathbb{Q}\) as it cannot be simplified further into factors without leaving \(\mathbb{Q}\) (since \(\u001bi\) is not part of \(\mathbb{Q}\)).
• Similarly, \(\left(x^2 - 3\right)\) remains irreducible over \(\mathbb{Q}\) because no two rational numbers can multiply to -3 and sum up to 0 within \(\mathbb{Q}\).

Knowing how to find irreducible polynomials helps us identify the minimal polynomials that define root fields over a base field.

Complex Roots

Complex roots emerge when dealing with polynomials that do not yield real solutions. They occur in conjugate pairs when derived from a polynomial with real coefficients. Understanding these complex numbers is crucial because:

• They allow us to complete the field in terms of solutions, enabling full comprehension of polynomial behavior.
• The imaginary unit \(\u001bi\) satisfies the equation \(\i^2 = -1\), which extends the reals to complex numbers.

In our problem, after solving \(\left(x^2 - 3\right)(x^3 - 1)\), we identified the complex roots as \(\pm \sqrt{3}, 1, rac{1}{2}(-1 \pm \sqrt{3}i)\).
These roots underline the necessity for a field comprising complex numbers to explain all polynomial solutions, thus introducing the complex numbers within \(\mathbb{Q}(i, \sqrt{3})\).

Construction of Fields

The construction of fields is foundational for constructing new mathematical territories where specific operations are defined and closed. Fields can be built from smaller fields by adding roots of irreducible polynomials, which help generate elements not initially present. This process is pertinent for understanding algebraic structures and is achieved through steps such as:

• Identifying the necessary elements (roots) missing from the original field.
• Establishing minimal polynomials that extend these elements into the field, ensuring the field remains closed under operations.

In constructing \(\mathbb{Q}(i, \sqrt{3})\), we've used irreducible polynomials \(\left(x^2 + 1\right)\) and \(\left(x^2 - 3\right)\) to introduce \(\u001bi\) and \(\sqrt{3}\), thus covering all roots needed from \(\left(x^2 - 3\right)(x^3 - 1)\).
This expansion allows us to define a more comprehensive root field, which includes all roots of the polynomial over a field initially containing only rational numbers.