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Chapter 30: Problem 4

Give a basis for the indicated vector space over the field. $$ Q(\sqrt{2}) \text { over } Q $$

Short Answer

Expert verified
The basis for the vector space \\( Q(\sqrt{2}) \\\) over \\( Q \\\) is \\( \{1, \sqrt{2}\} \\\).

Step by step solution

01

Understanding the Problem

We need to find a basis for the extension field \( Q(\sqrt{2}) \) over the base field \( Q \). This involves finding the smallest set of elements in \( Q(\sqrt{2}) \) such that any element in \( Q(\sqrt{2}) \) can be expressed as a linear combination of these elements with coefficients from \( Q \).
02

Identifying Elements of the Extension Field

Recall that if \( \alpha = \sqrt{2} \), then \( Q(\sqrt{2}) = \{ a + b\sqrt{2} \mid a, b \in Q \} \). Elements of this field have the form \( a + b\sqrt{2} \), where \( a \) and \( b \) are rational numbers.
03

Choosing the Basis

A basis for \( Q(\sqrt{2}) \) over \( Q \) must allow for every element \( a + b\sqrt{2} \) to be expressed as a linear combination. The simplest basis is \( \{1, \sqrt{2}\} \), since any element \( a + b\sqrt{2} \) can indeed be written as \( a\cdot1 + b\cdot\sqrt{2} \).
04

Verifying Linear Independence

To verify that \( \{1, \sqrt{2}\} \) is a basis, we check for linear independence. Suppose \( c_1\cdot1 + c_2\cdot\sqrt{2} = 0 \) for rational \( c_1 \) and \( c_2 \). This implies \( c_1 + c_2\sqrt{2} = 0 \). Equating the rational and irrational parts gives \( c_1 = 0 \) and \( c_2 = 0 \), confirming that the set is linearly independent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Field Extension
A field extension is essentially a bigger field that contains a smaller field. In this exercise, the field extension is denoted as \( Q(\sqrt{2}) \). Here, \( Q \) represents the set of all rational numbers, and by adding \( \sqrt{2} \), we create a new field that includes everything from \( Q \) as well as additional elements formed using \( \sqrt{2} \).
  • The idea is to extend our existing field \( Q \) by adjoining \( \sqrt{2} \), an element not originally in \( Q \).
  • This expanded field consists of all numbers that can be expressed in the form \( a + b\sqrt{2} \), where \( a \) and \( b \) are rational numbers.
  • Field extensions like \( Q(\sqrt{2}) \) help in solving equations that couldn't be solved within the original field \( Q \).
Understanding field extensions is key to working with different algebraic structures efficiently, allowing for the expression and solution of more complex equations. This concept is very useful in various areas of mathematics, including algebra and number theory.
Linear Combination
In the context of field extensions and vector spaces, a linear combination is a way to build new vectors from a set of known vectors using certain coefficients. For any given set of elements such as \( \{1, \sqrt{2}\} \), elements of the field \( Q(\sqrt{2}) \) can be written as linear combinations of these basis elements.
  • A linear combination of \( \{1, \sqrt{2}\} \) involves multiplying each by some rational coefficients and then adding the results.
  • In this case, an arbitrary element from \( Q(\sqrt{2}) \) is \( a + b\sqrt{2} \) for rational \( a \) and \( b \).
  • Thus, it can be expressed as \( a \cdot 1 + b \cdot \sqrt{2} \), showcasing the idea of linear combinations.
The ability to express elements as linear combinations is fundamental in defining the structure of vector spaces. This allows the manipulation of components in a systematic way, which is essential for solving equations or determining vector dependencies.
Rational Numbers
Rational numbers, denoted by \( Q \), are those numbers that can be expressed as a fraction where both the numerator and denominator are integers, and the denominator is not zero. This includes integers, fractions, and finite decimals.
  • Rational numbers form the base field over which we are extending in this exercise.
  • They are crucial because they serve as the coefficients in linear combinations of our basis \( \{1, \sqrt{2}\} \).
  • Any element \( a + b\sqrt{2} \) in our field extension involves rational numbers \( a \) and \( b \).
The properties of rational numbers, such as closure under addition, subtraction, and multiplication, make them a reliable field for constructing more advanced number systems through extensions. This foundational body plays a significant role in algebra and arithmetic.
Linear Independence
Linear independence is a property that determines whether a set of vectors (or elements) can be uniquely expressed without relying on the others. In simpler terms, none of the vectors in a linearly independent set can be written as a linear combination of the others.
  • For our exercise, we check the linear independence of the set \( \{1, \sqrt{2}\} \).
  • This means we want to ensure that if \( c_1 \cdot 1 + c_2 \cdot \sqrt{2} = 0 \), then both rational coefficients \( c_1 \) and \( c_2 \) must be zero.
  • The rational and irrational parts lead to equations that confirm \( c_1 = 0 \) and \( c_2 = 0 \), proving that no element is a combination of others.
Understanding linear independence helps in establishing the minimal set of elements needed to span a space. It's a crucial test to ascertain the basis of vectors, ensuring that they are the most efficient representation possible, which is invaluable in mathematical problem-solving.

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Most popular questions from this chapter

The dimension over \(F\) of a finite-dimensional vector space \(V\) over a field \(F\) is the minimum number of vectors required to span \(V\).

Mark each of the following true or false. a. The sum of two vectors is a vector. b. The sum of two scalars is a vector. c. The product of two scalars is a scalar. d. The product of a scalar and a vector is a vector. e. Every vector space has a finite basis. f. The vectors in a basis are linearly dependent. g. The 0 -vector may be part of a basis. h. If \(F \leq E\) and \(\alpha \in E\) is algebraic over the field \(F\), then \(\alpha^{2}\) is algebraic over \(F\). 1\. If \(F \leq E\) and \(\alpha \in E\) is algebraic over the field \(F\), then \(\alpha+\alpha^{2}\) is algebraic over \(F\). j. Every vector space has a basis.

Let \(F\) be any field. Consider the "system of \(m\) simultaneous linear equations in \(n\) unknowns" $$ \begin{gathered} a_{11} X_{1}+a_{12} X_{2}+\cdots+a_{1 n} X_{n}=b_{1} \\ a_{21} X_{1}+a_{22} x_{2}+\cdots+a_{21} X_{n}=b_{2} \\ \vdots \\ a_{m 1} X_{1}+a_{m 2} X_{2}+\cdots+a_{m n} X_{n}=b_{m} \end{gathered} $$ where \(a_{i j}, b_{i} \in F\). a. Show that the "system has a solution," that is, there exist \(X_{1}, \ldots, X_{n} \in F\) that satisfy all \(m\) equations, if and only if the vector \(\beta=\left(b_{1}, \cdots, b_{m}\right)\) of \(F^{m}\) is a linear combination of the vectors \(\alpha_{j}=\left(a_{1} j, \cdots, a_{m j}\right)\). (This resuht is straightforwand to prove, being practically the definition of a solution, but should really be regarded as the fundamental existence theorem for a simultaneous solution of a system of linear equations.) b. From part (a), show that if \(n=m\) and \(\left\\{\alpha_{j} \mid j=1, \ldots, n\right\\}\) is a basis for \(F^{n}\), then the system always has a unique solution.

Correct the definition of the italicized term without reference to the text, if correction is needed, so that it is in a form acceptable for publication. A basis for a vector space \(V\) over a field \(F\) is a set of vectors in \(V\) that span \(V\) and are linearly dependent.

Determine whether the given set of vectors is a basis for \(\mathbb{R}^{3}\) over \(\mathbb{R}\). $$ \\{(-1,1,2),(2,-3,1),(10,-14,0)\\} $$
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