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Chapter 31: Problem 3

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 \(h\)

Short Answer

Expert verified
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 \(F\).

Step by step solution

01

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 \( K \rightarrow \mathbb{C} \).
02

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 \( F \).
03

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 \( \mathbb{C} \).
04

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 \( F \).
05

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 \( h \).

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Key Concepts

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

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.
  • 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 \( C \).
Monomorphisms give us insight into how fields relate and how structures grow from simpler to more complex.
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:
  • 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.
Such functions are vital in field theory as they allow us to extend the properties of smaller fields to larger complex structures, while preserving important characteristics.
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:
  • 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 \( F \).
This theorem simplifies understanding extensions, as it allows intricate fields to be expressed in terms of simpler ones by adding just one element. It is a powerful tool in field theory as it provides a clear method of building complex fields from basic constituents.
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:
  • 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 \( \mathbb{C} \).
Understanding the minimal polynomial helps in finding all possible extensions or mappings from one field to another, especially in the presence of complex numbers.
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.
Complex numbers play a crucial role in field extensions, as they ensure that all roots of polynomial equations are accounted for. The completeness of the complex number field makes it an ideal setting for exploring algebraic structures and relationships.

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Most popular questions from this chapter

If \(F \subseteq I \subseteq K\) and \(K\) is a root field of \(a(x)\) over \(F\), then \(K\) is a root field of \(a(x)\) over \(I .\)

Find the root field of \(\mathrm{x}^{2}+1\) over \(\mathbb{Z}_{3} .\) ANSWER By the basic theorem of field extensions, $$ \mathbb{Z}_{3}[x] /\left\langle x^{2}+1\right\rangle \cong \mathbb{Z}_{3}(u) $$ where \(u\) is a root of \(x^{2}+1 .\) In \(\mathbb{Z}_{3}(u), x^{2}+1=(x+u)(x-u)\), because \(u^{2}+1=0\). Since \(\mathbb{Z}_{3}(u)\) contains \(\pm u\), it is the root field of \(x^{2}+1\) over \(\mathbb{Z}_{3} .\) Note that \(\mathbb{Z}_{3}(u)\) has nine elements, and its addition and multiplication tables are easy to construct. (See Chapter 27, Exercise C.) Show that, in any extension of \(\mathbb{Z}_{3}\) which contains a root \(u\) of $$ a(x)=x^{3}+2 x+1 \in \mathbb{Z}_{3}[x] $$ it happens that \(u+1\) and \(u+2\) are the remaining two roots of \(a(x)\). Use this fact to find the root field of \(x^{3}+2 x+1\) over \(\mathbb{Z}_{3}\). Write its addition and multiplication tables

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 \(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm \sqrt{3} i)\), so the root field is \(\mathbb{Q}\left(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm \sqrt{3} i)\right)\). The same field can be written more simply as \(\mathbb{Q}(\sqrt{3}, i)\). Prove that \(x^{2}-3\) and \(x^{2}-2 x-2\) are both irreducible over \(\mathbb{Q}\). Then find their root fields over \(\mathbf{Q}\) and show they are the same.

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 \(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm \sqrt{3} i)\), so the root field is \(\mathbb{Q}\left(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm \sqrt{3} i)\right)\). The same field can be written more simply as \(\mathbb{Q}(\sqrt{3}, i)\). Find the root field of \(x^{4}-2\), first over \(Q\), then over \(\mathbb{R}\).

Let \(F\) be a field. An irreducible polynomial \(p(x)\) in \(F[x]\) is said to be separable over \(F\) if it has no multiple roots in any extension of \(F\). If \(p(x)\) does have a multiple root in some extension, it is inseparable over \(F\). Prove the following : If \(F\) has characteristic 0 , every irreducible polynomial in \(F[x]\) is separable. Thus, for characteristic 0, there is no question whether an irreducible polynomial is separable or not. However, for characteristic \(p \neq 0\), it is different. This case is treated next. In the following problems, let \(F\) be a field of characteristic \(p \neq 0\)
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