Problem 1 - A Book of Abstract Algebra

Question

Computing a Galois Group 1 Show that \(Q(i, \sqrt{2})\) is the root field of \(\left(x^{2}+1\right)\left(x^{2}-2\right)\) over \(\mathbb{Q}\).

Step-by-Step Solution

Answer

The field \(\mathbb{Q}(i, \sqrt{2})\) is the root field containing all roots of \((x^2+1)(x^2-2)\).

Step 1: Identify the roots of the polynomials

First, consider the polynomial \((x^2 + 1)(x^2 - 2)\). The roots of the polynomial \(x^2 + 1 = 0\) are \(i\) and \(-i\). For the polynomial \(x^2 - 2 = 0\), the roots are \(\sqrt{2}\) and \(-\sqrt{2}\). Therefore, the roots of \((x^2 + 1)(x^2 - 2)\) are \(i, -i, \sqrt{2}, \) and \(-\sqrt{2}\).

Step 2: Understand the concept of a root field

A field is a root field of a polynomial if it is the smallest field that contains all the roots of the polynomial. The field \(\mathbb{Q}(i, \sqrt{2})\) contains the numbers \(i, -i, \sqrt{2},\) and \(-\sqrt{2}\) because it is generated by \(i\) and \(\sqrt{2}\) over \(\mathbb{Q}\). Hence, it contains all roots of the polynomial \((x^2 + 1)(x^2 - 2)\).

Step 3: Confirm minimal field requirement

To confirm that \(\mathbb{Q}(i, \sqrt{2})\) is indeed the root field, check that any smaller field over which the polynomial factors must contain both \(i\) and \(\sqrt{2}\). Neither \(i\) (root of \(x^2 + 1\)) nor \(\sqrt{2}\) (root of \(x^2 - 2\)) are in \(\mathbb{Q}\), requiring \(\mathbb{Q}(i, \sqrt{2})\) to be the minimal field containing all roots. Thus, it fits the definition of a root field.

Key Concepts

Root Field

A root field is a fascinating and essential concept in abstract algebra. It is essentially the smallest field that encompasses all the roots of a given polynomial. For a polynomial such as \((x^2 + 1)(x^2 - 2)\), the root field is the smallest field containing all roots of the two quadratics involved in the polynomial.
The significance of a root field lies in its minimal size and structure that still includes all possible roots. In our exercise, \(\mathbb{Q}(i, \sqrt{2})\) is identified as the root field for the polynomial. This is because it includes both imaginary numbers like \(i\) and \(-i\), as well as real roots like \(\sqrt{2}\) and \(-\sqrt{2}\).

• Ensures all polynomial roots are present
• Is minimal, containing no extraneous elements beyond those needed for the polynomial’s roots

Understanding this concept is key to comprehending larger algebraic structures and their extensions.

Polynomial Roots

Understanding polynomial roots is crucial as they form the basis for constructing field extensions. The roots of a polynomial are the values that satisfy the equation when set to zero.
In our context, for the polynomial \((x^2 + 1)(x^2 - 2)\), the roots are derived by solving the individual components \(x^2 + 1 = 0\) and \(x^2 - 2 = 0\).

• For \(x^2 + 1 = 0\), the roots are \(i\) and \(-i\)
• For \(x^2 - 2 = 0\), the roots are \(\sqrt{2}\) and \(-\sqrt{2}\)

These roots are pivotal as they guide the construction of the root field \(\mathbb{Q}(i, \sqrt{2})\). Recognizing these numbers as roots helps in forming field extensions that are vital for understanding Galois groups and their properties.

Minimal Field

A minimal field is a smaller, yet sufficient field that contains all necessary elements required for the roots of a given polynomial. To identify the minimal field, one must verify that no smaller field encompassing the polynomial’s roots exists.
In our example, \(\mathbb{Q}(i, \sqrt{2})\) serves as the minimal field for the polynomial \((x^2 + 1)(x^2 - 2)\).
This minimality ensures that not a single element is unneeded for the polynomial's complete factorization:

• It contains \(i\) and \(-i\) from \(x^2 + 1\)
• It includes \(\sqrt{2}\) and \(-\sqrt{2}\) from \(x^2 - 2\)

Establishing \(\mathbb{Q}(i, \sqrt{2})\) as the minimal field is essential because it ensures efficiency in computations and further investigations within the field, particularly when evaluating field extensions and Galois groups.

Field Extension

Field extension is a critical concept that involves expanding a smaller field by adjoining new elements, typically the roots of a polynomial, hence creating a larger field. In our context, we start with the field \(\mathbb{Q}\), known as the rationals.
By adjoining \(i\) and \(\sqrt{2}\), we extend \(\mathbb{Q}\) to \(\mathbb{Q}(i, \sqrt{2})\), thereby encompassing all roots from \((x^2 + 1)(x^2 - 2)\):

• Adding \(i\) provides solutions for \(x^2 + 1 = 0\)
• Adding \(\sqrt{2}\) resolves \(x^2 - 2 = 0\)

Creating such a field extension is fundamental to applying concepts within the Galois theory. It allows mathematicians to understand the symmetry and structure of polynomials through their roots and gain deeper insights into how fields relate to each other.