Chapter 32: Problem 1
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 \(\mathbb{Q}\).
Short Answer
Expert verified
The field \(\mathbb{Q}(i, \sqrt{2})\) is the root field containing all roots of \((x^2+1)(x^2-2)\).
Step by step solution
01
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 \(-\sqrt{2}\).
02
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 \((x^2 + 1)(x^2 - 2)\).
03
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.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \(-\sqrt{2}\).
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
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 \(x^2 - 2 = 0\).
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 \(-\sqrt{2}\)
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:
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 \(x^2 - 2\)
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 \((x^2 + 1)(x^2 - 2)\):
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 \(x^2 - 2 = 0\)