Problem 2 - Applied Discrete Structures

Question

(a) Determine the splitting field of \(f(x)=x^{2}+1\) over \(\mathbb{R}\). This means consider the polynomial \(f(x)=x^{2}+1 \in \mathbb{R}[x]\) and find the smallest field that contains \(\mathbb{R}\) and all the zeros of \(f(x) .\) Denote this field by \(\mathbb{R}(i)\) (b) \(\mathbb{R}(i)\) is more commonly referred to by a different name. What is it? (c) Show that \(\\{1, i\\}\) is a basis for the vector space \(\mathbb{R}(i)\) over \(\mathbb{R}\). What is the dimension of this vector space (over \(\mathbb{R}) ?\)

Step-by-Step Solution

Answer

(a) \(\mathbb{R}(i)\) is the splitting field, known as \(\mathbb{C}\); (b) It's the complex numbers; (c) \({1, i}\) is a basis, making the dimension 2 over \(\mathbb{R}\).

Step 1: Identify the roots of the polynomial

The polynomial given is \(f(x) = x^2 +1\). We need to find the roots of this polynomial. Setting \(f(x) = 0\) gives us the equation \(x^2 + 1 = 0\). Solving for \(x\) gives us \(x^2 = -1\). The solutions to this are \(x = i\) and \(x = -i\), where \(i\) is the imaginary unit satisfying \(i^2 = -1\). These roots are not real numbers.

Step 2: Define the splitting field over \(\mathbb{R}\)

The splitting field of a polynomial over a field is the smallest field extension of the original field that contains all the roots of the polynomial. For \(f(x) = x^2 + 1\), the smallest field that contains \(\mathbb{R}\) and the roots \(i\) and \(-i\) is \(\mathbb{R}(i)\). This is because any field containing \(i\) must also contain \(-i\), since the reals already contain all real numbers.

Step 3: Recognize \(\mathbb{R}(i)\) by its common name

The field \(\mathbb{R}(i)\) is commonly referred to as the complex numbers, denoted as \(\mathbb{C}\). Thus, the splitting field of \(f(x) = x^2 + 1\) over \(\mathbb{R}\) is the complex number field \(\mathbb{C}\).

Step 4: Show that \({1, i}\) is a basis

A basis for a vector space over a field is a set of vectors that is linearly independent and spans the space. To show \({1, i}\) is a basis for \(\mathbb{R}(i)\) over \(\mathbb{R}\), we must show that each element in \(\mathbb{R}(i)\) can be written as a linear combination \(a + bi\) where \(a, b \in \mathbb{R}\), and that \(1\) and \(i\) are linearly independent. Since neither element is a scalar multiple of the other, they are linearly independent.

Step 5: Determine the dimension of the vector space

The dimension of a vector space is the number of vectors in its basis. Since \({1, i}\) is a basis for \(\mathbb{R}(i)\) over \(\mathbb{R}\), and there are two vectors, the dimension of \(\mathbb{R}(i)\) over \(\mathbb{R}\) is 2.

Key Concepts

Splitting Field

A splitting field of a polynomial is often the smallest extension field in which the polynomial can be expressed as a product of linear factors. Let's look at the polynomial \( f(x) = x^2 + 1 \). To find the splitting field over \( \mathbb{R} \), we need to find its roots. Solving \( x^2 + 1 = 0 \) reveals the roots are \( i \) and \( -i \), where \( i \) is the imaginary unit satisfying \( i^2 = -1 \). Now, these roots aren't part of \( \mathbb{R} \) because they are not real numbers. Therefore, we need a field extension that includes these roots. This extension is denoted as \( \mathbb{R}(i) \). Simply put, \( \mathbb{R}(i) \) is the smallest field that contains both the real numbers and \( i \), making it the splitting field of \( f(x) = x^2 + 1 \).

Complex Numbers

The field \( \mathbb{R}(i) \) you arrive at when determining the splitting field of \( x^2 + 1 \) over the reals is more commonly known as the set of complex numbers, \( \mathbb{C} \).Complex numbers are numbers that have both a real part and an imaginary part, usually represented as \( a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit. - The real part is \( a \).- The imaginary part is \( b \cdot i \).The complex field \( \mathbb{C} \) encompasses a much richer range of numbers than \( \mathbb{R} \). It allows solutions for any polynomial equation, making it extremely useful in both mathematics and applied fields like engineering and physics.

Vector Space Basis

In linear algebra, a basis for a vector space is a set of vectors that are linearly independent and span the space. For the complex field \( \mathbb{R}(i) \) over the reals, the set \( \{1, i\} \) acts as a basis.- Any complex number can be written as \( a + bi \), where both \( a \) and \( b \) are real numbers.To prove \( \{1, i\} \) is a basis:

• **Linear Independence**: Neither 1 nor \( i \) is a scalar multiple of the other, meaning they are independent.
• **Spanning**: Any element of \( \mathbb{R}(i) \), or \( \mathbb{C} \), can be represented as a linear combination of \( 1 \) and \( i \).

Thus, \( \{1, i\} \) forms a complete basis for the vector space.

Dimension of Vector Space

The dimension of a vector space is the number of vectors in any of its bases. In the case of the complex numbers over the reals, we have determined the basis to be \( \{1, i\} \).- This indicates that every complex number \( a + bi \) can be derived from \( 1 \) and \( i \).With exactly two vectors in \( \{1, i\} \), the dimension of the vector space \( \mathbb{R}(i) \) over \( \mathbb{R} \) is 2. This means you need two numbers (real coefficients in this case) to express any element in this space. In comparison, the dimension of \( \mathbb{R} \) over itself is 1, showing how \( \mathbb{C} \) provides a more complex, multidimensional structure for expressing numbers.