Problem 3
Question
In the following, let \(F\) be a subfield of \(\mathbb{C}\). An injective homomorphism \(h: F \rightarrow \mathrm{C}\) is called a monomorphism; it is obviously an isomorphism \(F \rightarrow h(F)\). Let \(F \subseteq K\), with \([K: F]=n\). If \(h: F \rightarrow \mathbb{C}\) is a monomorphism, prove that there are exactly \(n\) monomorphisms \(K \rightarrow \mathbb{C}\) which are extensions of \(h\)
Step-by-Step Solution
Verified Answer
There are exactly \(n\) monomorphisms from \(K\) to \(\mathbb{C}\) extending \(h\), as they correspond to the \(n\) roots of the minimal polynomial of \(a\) over \(F\).
1Step 1: Understanding the Setting
We are given that \( F \) is a subfield of \( \mathbb{C} \) and an injective homomorphism \( h: F \rightarrow \mathbb{C} \), which is a monomorphism. We also have an extension field \( K \) such that \( [K: F] = n \). Our task is to show that there are exactly \( n \) extensions of this monomorphism to \( K \rightarrow \mathbb{C} \).
2Step 2: Applying the Primitive Element Theorem
Since \( [K:F] = n \), by the Primitive Element Theorem, \( K \) can be expressed as \( K = F(a) \) for some element \( a \in K \). Thus, any element in \( K \) can be written as a polynomial in \( a \) with coefficients in \( F \).
3Step 3: Analyzing the Minimal Polynomial
The element \( a \) has a minimal polynomial over \( F \) of degree \( n \) since \( [K:F] = n \). This minimal polynomial has \( n \) roots in \( \mathbb{C} \), as it can be factored completely over \( \mathbb{C} \).
4Step 4: Constructing Extensions
Each root \( b_i \in \mathbb{C} \) of the minimal polynomial defines a homomorphism \( h_i: K \rightarrow \mathbb{C} \) that extends \( h \). Specifically, map \( a \) to one of its roots \( b_i \), and every element in \( K = F(a) \) can be extended by applying \( h \) to the coefficients from \( F \).
5Step 5: Counting the Extensions
Since there are \( n \) distinct roots for the minimal polynomial, there are exactly \( n \) such extensions of \( h \) to \( K \rightarrow \mathbb{C} \). Hence, there are exactly \( n \) monomorphisms extending \( h \).
Key Concepts
MonomorphismInjective HomomorphismPrimitive Element TheoremMinimal PolynomialComplex Numbers
Monomorphism
In the context of fields and homomorphisms, a **monomorphism** refers to an injective homomorphism between two algebraic structures. In simpler terms, it is a one-to-one function that preserves structure from one field to another. This is important because it ensures there are no two different elements in the subfield that map to the same element in the larger field, maintaining the distinctness of elements.
It's crucial to understand that a monomorphism is a special type of function: it both respects operations (addition and multiplication) and is injective.
It's crucial to understand that a monomorphism is a special type of function: it both respects operations (addition and multiplication) and is injective.
- Preserves field operations: For any elements \( x, y \) in field \( F \), the monomorphism \( h \) satisfies \( h(x+y) = h(x) + h(y) \) and \( h(xy) = h(x) \cdot h(y) \).
- Injectivity: If \( h(x) = h(y) \), then \( x = y \). This means no two different elements from \( F \) can have the same image in \( C \).
Injective Homomorphism
An **injective homomorphism** is essentially a monomorphism. It is a function or mapping that connects elements from one field to another while preserving their structure, and ensuring that distinct elements remain distinct.
An injective homomorphism is characterized by its properties:
An injective homomorphism is characterized by its properties:
- It maintains the operation of addition and multiplication from the source field \( F \) to the target, such as \( h(x+y) = h(x) + h(y) \) and \( h(xy) = h(x) \cdot h(y) \).
- This homomorphism is injective, meaning \( h(x) = h(y) \) implies \( x = y \) within the field \( F \). Thus, it ensures there are no overlapping images.
Primitive Element Theorem
The **Primitive Element Theorem** is a cornerstone of field theory, providing a simple characterization of finite field extensions. It states that any finite separable extension of fields can be generated by a single element. This means if \( [K:F] = n \), the field \( K \) can be expressed as \( F(a) \) for some element \( a \) in \( K \).
Here's the breakdown:
Here's the breakdown:
- The extension field \( K \) can be seen as a collection of elements that relate to \( F \) through \( a \), essentially forming \( K = F(a) \).
- All elements in \( K \) are expressible as polynomials of \( a \) with coefficients from \( F \).
Minimal Polynomial
A **minimal polynomial** is a unique monic polynomial of lowest degree that captures an element's algebraic properties over a field. If \( a \in K \) is an element whose codescendant field is \( F \), and \( [K:F] = n \), then the degree of this minimal polynomial is precisely \( n \).
Characteristics include:
Characteristics include:
- It is a polynomial of least degree with coefficients from \( F \) for which \( p(a) = 0 \).
- If \( p(x) \) is the minimal polynomial for \( a \), then \( p(x) \) has exactly \( n \) roots in \( \mathbb{C} \).
Complex Numbers
**Complex numbers** are an extension of the real numbers through the introduction of the imaginary unit \( i \) where \( i^2 = -1 \). They form a complete field under addition, subtraction, multiplication, and division.
- Every complex number is in the form \( a + bi \), where \( a \) and \( b \) are real numbers.
- They are essential in field theory because every polynomial over the complex numbers can be completely factored into linear factors.
Other exercises in this chapter
Problem 2
Find the root field of \(\mathrm{x}^{2}+1\) over \(\mathbb{Z}_{3} .\) ANSWER By the basic theorem of field extensions, $$ \mathbb{Z}_{3}[x] /\left\langle x^{2}+
View solution Problem 2
Find the root field of \(a(x)=\left(x^{2}-3\right)\left(x^{3}-1\right)\) over \(\mathbf{Q} .\) ANSWER The complex roots of \(a(x)\) are \(\pm \sqrt{3}, 1, \frac
View solution Problem 3
Find \(c\) such that \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{-5})=Q(c)\).
View solution Problem 3
Find the root field of \(a(x)=\left(x^{2}-3\right)\left(x^{3}-1\right)\) over \(\mathbf{Q} .\) ANSWER The complex roots of \(a(x)\) are \(\pm \sqrt{3}, 1, \frac
View solution