Chapter 16: Problem 1
(a) Use the definition of a field to show that \(\mathbb{Q}(\sqrt{2})\) is a field. (b) Use the definition of vector space to show that \(\mathbb{Q}(\sqrt{2})\) is a vector space over \(\mathbb{Q}\). (c) Prove that \(\\{1, \sqrt{2}\\}\) is a basis for the vector space \(\mathbb{Q}(\sqrt{2})\) over \(\mathbb{Q}\), and, therefore, the dimension of \(\mathbb{Q}(\sqrt{2})\) over \(\mathbb{Q}\) is 2 .
Short Answer
Step by step solution
Definition of a Field
Demonstrating Field Properties in \(\mathbb{Q}(\sqrt{2})\)
Definition of a Vector Space
Demonstrating Vector Space Properties in \(\mathbb{Q}(\sqrt{2})\) over \(\mathbb{Q}\)
Showing \(\{1, \sqrt{2}\}\) is a Basis
Determine the Dimension of \(\mathbb{Q}(\sqrt{2})\) over \(\mathbb{Q}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Field Theory
Here are the critical properties of a field:
- Closure: The result of adding or multiplying any elements of the field remains in the field.
- Associativity: Addition and multiplication are associative; meaning, for any elements \(a\), \(b\), and \(c\) in the field, \((a+b)+c = a+(b+c)\) and \((a\cdot b)\cdot c = a\cdot(b\cdot c)\).
- Commutativity: Addition and multiplication are commutative; thus, \(a+b = b+a\) and \(a\cdot b = b\cdot a\).
- Identities: There exist distinct elements 0 and 1, acting as the additive and multiplicative identities, respectively.
- Inverses: Every element has an additive inverse, and every non-zero element has a multiplicative inverse.
Vector Spaces
A vector space must satisfy certain axioms:
- Closure: Adding vectors, or multiplying them by scalars from the field, results in new vectors that are still within the vector space.
- Associativity and Commutativity: Vector addition is associative and commutative.
- Identity and Inverses: There's a zero vector (the additive identity), and every vector has an additive inverse.
- Scalar Multiplication: Vectors can be scaled by elements of the field \(F\), showcasing distributive properties.
Linear Independence
To determine if a set of vectors \(\{v_1, v_2, \ldots, v_n\}\) is linearly independent:
- Form a linear combination: Check if \(c_1v_1 + c_2v_2 + \cdots + c_nv_n = 0\) only when all \(c_i = 0\).
- Unique solutions: For linear independence, the only solution to expressing one vector as a combination of others is the trivial one, where all coefficients \(c_i\) are zero.
Basis and Dimension
A basis can be thought of as the "building blocks" for creating every vector in the space through linear combinations. For instance, to find a basis:
- Span: The set must span the vector space, meaning any vector in the space can be expressed as a linear combination of the basis vectors.
- Linear Independence: The basis vectors must be linearly independent.