Problem 2 - 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 \(\mathbb{Q}(\sqrt{3}, i)\). Prove that \(x^{2}-3\) and \(x^{2}-2 x-2\) are both irreducible over \(\mathbb{Q}\). Then find their root fields over \(\mathbf{Q}\) and show they are the same.

Step-by-Step Solution

Answer

The root field of \(a(x)\) over \(\mathbb{Q}\) is \(\mathbb{Q}(\sqrt{3}, i)\); both polynomials are irreducible over \(\mathbb{Q}\) with root fields \(\mathbb{Q}(\sqrt{3})\).

Step 1: Understand the Problem

We need to find the field generated by the roots of the polynomial \(a(x)=(x^2 - 3)(x^3 - 1)\) over \(\mathbb{Q}\), and prove some irreducibility properties. Additionally, we have to show that both polynomials \(x^2-3\) and \(x^2-2x-2\) are irreducible over \(\mathbb{Q}\), find their root fields, and demonstrate these fields coincide.

Step 2: Find the Roots of the Polynomials

The roots of \(x^2 - 3\) are \(\pm \sqrt{3}\). The roots of \(x^3 - 1\) are found using the equation \(x^3 = 1\), yielding roots \(1, \frac{1}{2}(-1 + i\sqrt{3}), \frac{1}{2}(-1 - i\sqrt{3})\). Therefore, the roots of \(a(x)\) are \(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm i\sqrt{3})\).

Step 3: Identify the Root Field

The root field of \(a(x)\) over \(\mathbb{Q}\) is \(\mathbb{Q}(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm i\sqrt{3}))\). Simplifying, we note that this field is generated by extending \(\mathbb{Q}\) by \(\sqrt{3}\) and \(i\), as these generate the rest of the roots, resulting in the field \(\mathbb{Q}(\sqrt{3}, i)\).

Step 4: Prove Irreducibility of \(x^2-3\) Over \(\mathbb{Q}\)

A polynomial of degree 2 is irreducible over \(\mathbb{Q}\) if it has no rational roots. Using the Rational Root Theorem, the potential rational roots of \(x^2 - 3\) are \(\pm 1, \pm 3\). Evaluating the polynomial at these values shows they do not lead to 0, hence \(x^2 - 3\) is irreducible over \(\mathbb{Q}\).

Step 5: Prove Irreducibility of \(x^2-2x-2\) Over \(\mathbb{Q}\)

Similarly, for the polynomial \(x^2 - 2x - 2\), using the discriminant \(\Delta = b^2 - 4ac\), we find \(\Delta = 4 + 8 = 12\). The square root of the discriminant \(\sqrt{12} = 2\sqrt{3}\) is irrational, indicating the polynomial is irreducible over \(\mathbb{Q}\) because it does not split into rational roots.

Step 6: Find and Compare the Root Fields of Both Polynomials

The root field of \(x^2 - 3\) is \(\mathbb{Q}(\sqrt{3})\). The roots of \(x^2-2x-2\) are \(1 \pm \sqrt{3}\), implying its root field is also \(\mathbb{Q}(\sqrt{3})\). These root fields are identical, affirming the intended conclusion.

Key Concepts

Irreducible Polynomial

An irreducible polynomial is a polynomial that cannot be factored into the product of two non-constant polynomials with coefficients in a given field. In simple terms, it's as simple as a polynomial can get, assuming we are only allowed to use coefficients from a particular field, like the rational numbers \( \mathbb{Q} \).

An example of an irreducible polynomial can be illustrated with \( x^2 - 3 \). This polynomial has no rational roots, which means there are no numbers in \( \mathbb{Q} \) that when squared and subtracted by 3 result in zero. Therefore, \( x^2 - 3 \) cannot be factored further using rational coefficients, thus it's considered irreducible over the rationals.

Similarly, another example is \( x^2 - 2x - 2 \). By calculating its roots and checking that they are not rational, we establish that this polynomial too is irreducible over \( \mathbb{Q} \). Both \( x^2 - 3 \) and \( x^2 - 2x - 2 \) persist as the simplest form within the rational numbers, characterizing the notion of irreducibility.

Rational Root Theorem

The Rational Root Theorem is a powerful tool used to determine whether a polynomial has any rational roots and, if so, what they might be. It states that if a polynomial \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 \) has a rational root \( \frac{p}{q} \), then \( p \) is a factor of the constant term \( a_0 \) and \( q \) is a factor of the leading coefficient \( a_n \).

For example, consider \( x^2 - 3 \). Here, the constant term is \(-3\) and the leading coefficient is \(1\). Therefore, the possible rational roots could be \( \pm 1, \pm 3 \). But testing these values, you find that none satisfy the equation \( x^2 - 3 = 0 \), confirming that the polynomial has no rational roots and thus is irreducible over \( \mathbb{Q} \).

This theorem helps in understanding and proving the irreducibility of polynomials by systematically eliminating the possibility of rational roots.

Discriminant

The discriminant is an expression derived from the coefficients of a polynomial and provides crucial insights into the nature of the roots of the polynomial. For a quadratic polynomial \( ax^2 + bx + c \), the discriminant \( \Delta \) is given by \( b^2 - 4ac \).

The discriminant helps determine the type of roots without finding the roots explicitly:

• If \( \Delta > 0 \), the polynomial has two distinct real roots.
• If \( \Delta = 0 \), it has exactly one real root (a repeated root).
• If \( \Delta < 0 \), it has two complex roots which are conjugates.

In the case of \( x^2 - 2x - 2 \), the discriminant is 12, computed as \( 4 - (-8) \). Since \( 12 > 0 \), this indicates irrational roots, specifically making \( \sqrt{12} = 2\sqrt{3} \), confirming the lack of rational roots and hence irreducibility over \( \mathbb{Q} \).

Complex Roots

Complex roots are solutions to polynomial equations that appear when a polynomial has a negative discriminant. Specifically, they arise as pairs known as complex conjugates, which are of the form \( a + bi \) and \( a - bi \), ensuring that polynomial coefficients remain real if they initially were.

When dealing with polynomials like \( x^3 - 1 \), the roots include complex numbers since the equation doesn't resolve neatly into solely real factors. For instance, this polynomial gives us roots such as \( \frac{1}{2}(-1 \pm \sqrt{3}i) \), complementing the real root 1, introducing complexity into the solution framework.

Complex roots help extend the foundational understanding of root fields, showcasing that not all roots visible over the rationals \( \mathbb{Q} \) need to be real. By solving for such roots, polynomials enrich the number field with imaginary numbers, an essential component when describing the full set of solutions to polynomial equations.