Chapter 10: Problem 1
Find the splitting field \(K\) in \(C\) of the indicated polynomial \(f(x)\) over \(Q\), and determine \([K: \mathbb{Q}]\). $$ f(x)=x^{4}-1 $$
Short Answer
Expert verified
The splitting field is \(\mathbb{Q}(i)\), with \([\mathbb{Q}(i) : \mathbb{Q}] = 2\).
Step by step solution
01
Factor the polynomial
Factor the polynomial \(f(x) = x^4 - 1\) as much as possible over \(\mathbb{Q}\). We have \(x^4 - 1 = (x^2 - 1)(x^2 + 1)\). These can be further factored as \(x^2 - 1 = (x - 1)(x + 1)\) and \(x^2 + 1 = (x - i)(x + i)\) where \(i\) is the imaginary unit (\(i^2 = -1\)).
02
Determine the roots
The factored form \((x - 1)(x + 1)(x - i)(x + i)\) reveals the roots of \(x^4 - 1\) as \(1, -1, i,\) and \(-i\).
03
Identify generators of the splitting field
The splitting field \(K\) over \(\mathbb{Q}\) is the smallest field containing all the roots \(1, -1, i,\) and \(-i\). Since \(1\) and \(-1\) are already in \(\mathbb{Q}\), the field generated is \(\mathbb{Q}(i)\).
04
Determine the degree of the extension
The degree \([K: \mathbb{Q}]\) is equal to the degree of the minimal polynomial of \(i\) over \(\mathbb{Q}\). The minimal polynomial for \(i\) is \(x^2 + 1\), which has a degree of 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Polynomial Roots
A polynomial root is a solution to the equation where the polynomial equals zero. These roots are crucial because they reveal the values at which a polynomial changes direction, crosses the x-axis, or touches it without crossing. For the polynomial \(x^4 - 1\), the roots are the values that make the equation zero. By factoring, we find the roots to be \(1, -1, i,\) and \(-i\).
- \(x - 1 = 0\) gives the root \(1\).
- \(x + 1 = 0\) leads to the root \(-1\).
- \(x - i = 0\) gives the complex root \(i\).
- \(x + i = 0\) provides the complex root \(-i\).
Exploring Field Extensions
A field extension is a way to expand our number system to include new numbers not present in the original field. In simple terms, if our original field is \( \mathbb{Q} \), which includes rational numbers, a field extension like \( \mathbb{Q}(i) \) includes all rational combinations of \(i\).
- The original field is \( \mathbb{Q} \) (the field of rational numbers).
- The extension field is \( \mathbb{Q}(i) \), encompassing all elements of the form \(a + bi\), where \(a, b \in \mathbb{Q}\).
Deciphering Minimal Polynomials
The minimal polynomial of a number is the smallest degree monic polynomial over the original field, for which the number is a root. In the context of our problem, the minimal polynomial for \(i\) over \( \mathbb{Q} \) is essential.
- The minimal polynomial of \(i\) over \( \mathbb{Q} \) is \(x^2 + 1\).
- This polynomial is monic (leading coefficient is 1) and cannot be factored further over \( \mathbb{Q} \).
- It's the smallest degree polynomial having \(i\) as a root, with coefficients in \( \mathbb{Q} \).
Determining the Degree of Extension
The degree of an extension, denoted \([K: \mathbb{Q}]\), measures how much larger our extended field is compared to the original. It is a crucial aspect for understanding the complexity added by including new elements.
- The degree is the number of dimensions the extension field has as a vector space over the original field.
- In our problem, \([\mathbb{Q}(i) : \mathbb{Q}] = 2\).
- This degree is equivalent to the degree of the minimal polynomial \(x^2 + 1\) for \(i\) over \( \mathbb{Q} \).