Problem 4 - A Book of Abstract Algebra

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 \(F \rightarrow h(F)\). Prove: The only possible monomorphism \(h: \mathbf{Q} \rightarrow \mathrm{C}\) is \(h(x)=x\). Thus, any monomorphism \(h: \mathbb{Q}(a) \rightarrow C\) necessarily fixes \(\mathbb{Q}\).

Step-by-Step Solution

Answer

The only monomorphism \( h: \mathbb{Q} \rightarrow \mathbb{C} \) is the identity \( h(x) = x \). This ensures any monomorphism fixes \( \mathbb{Q} \).

Step 1: Introduction

To demonstrate that the only monomorphism \( h: \mathbb{Q} \rightarrow \mathbb{C} \) is \( h(x) = x \), we'll utilize properties of field homomorphisms and fundamental properties of rational numbers \( \mathbb{Q} \).

Step 1: Understand Field Homomorphisms

A field homomorphism \( h: F \rightarrow \mathbb{C} \) is a function that preserves the operations of addition and multiplication, and maps the multiplicative identity from \( F \) to the multiplicative identity in \( \mathbb{C} \).

Step 2: Apply Properties of Rational Numbers

Since \( \mathbb{Q} \) is the smallest field containing \( 1 \), any field homomorphism \( h \) must map \( 1 \) in \( \mathbb{Q} \) to \( 1 \) in \( \mathbb{C} \). Thus, for any rational number \( x=n/m \) with \( n, m \in \mathbb{Z}, \ h(x) = h(n/m) \).

Step 3: Compute for Integers and Rational Numbers

Because homomorphisms preserve addition and multiplication, \( h(n) = n \cdot h(1) = n \) for any integer \( n \). For rational \( x = n/m \), \( h(x) = h(n/m) = h(n)/h(m) = n/m = x \). This implies \( h \) fixes all rational numbers.

Step 5: Conclusion: Monomorphism Fixation

Since every rational number is fixed by \( h \), the only possible monomorphism \( h \) from \( \mathbb{Q} \) to \( \mathbb{C} \) is the identity function. Hence, any monomorphism must fix \( \mathbb{Q} \).

Key Concepts

Monomorphism

A monomorphism is essentially a fancy term for an injective (or one-to-one) homomorphism between fields. In our context, this means there's a function, let's call it \( h \), that maps elements from one field to another, while respecting the operations of addition and multiplication. For \( h \) to be a monomorphism, it must meet two key criteria. First, \( h \) must be injective, meaning if you take two different elements from the starting field, their images under \( h \) have to be different as well. Secondly, it has to respect the structure of a field.

• For any two elements \( a \) and \( b \) in the field, \( h(a + b) = h(a) + h(b) \)
• Similarly, \( h(a\cdot b) = h(a)\cdot h(b) \)

Moreover, if the monomorphism is from the rational numbers \( \mathbf{Q} \) to the complex numbers \( \mathbb{C} \), as in the original exercise, it highlights a specific uniqueness. This implies that for our function \( h \), mapping a number \( x \) should only be possible as \( h(x) = x \) for all rational numbers \( x \). Hence, no complex trickery or changes in value occur in this process when \( \mathbf{Q} \) is involved.

Rational Numbers

Rational numbers are the set of numbers that can be expressed as the fraction of two integers. Typically denoted as \( \mathbb{Q} \), these numbers include all integers, as well as fractions like \( \frac{1}{2} \) or \( \frac{-4}{3} \). What's crucial about rational numbers in mathematical operations and transformations is their completeness in terms of field properties:

• They are closed under addition, subtraction, multiplication, and division (except by zero).
• They encompass both the positive fractions and negative fractions.

When a homomorphism involves these numbers, especially lending to monomorphisms, special rules apply. An injective homomorphism involving \( \mathbf{Q} \) must map each number to an equivalent one in the target field without affecting their basic trait, which is why the identity \( h(x) = x \) is crucial. In essence, the machine of monomorphism must keep rational numbers completely intact when mapping them to complex numbers.

Complex Numbers

Complex numbers, denoted by \( \mathbb{C} \), are even broader than rational numbers. A complex number is expressed in the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit with the property \( i^2 = -1 \).

• They can represent solutions to equations that have no solutions in the rational numbers, such as \( x^2 + 1 = 0 \) whose solution is \( x = \pm i \).
• They cover a more expansive type of field, including both real and imaginary parts.

Monomorphisms involving complex numbers typically hint at a kind of structural maintenance; they transform a subfield (like the rational numbers) into a segment of this larger, more comprehensive field while holding onto original properties. In relations like those illustrated in the original exercise, the inherent properties of complex numbers offer a vast ground. However, the conditions of monomorphisms like \( h: \mathbb{Q} \rightarrow \mathbb{C} \) enforce simple identity mappings, owing largely to the steadfastness of rational numbers.

Injective Function

An injective function, also called a one-to-one function, is a type of mapping where every distinct element of a set is associated with a unique element in another set. This means that different inputs always lead to different outputs; there's no overlapping in the target set. To put it simply, if \( f(a) = f(b) \), it must follow that \( a = b \).

• Injective functions help ensure that each element remains distinct when mapped from one set to another.
• They preserve individuality in the context of function mappings.

In terms of field homomorphisms, injective functions become monomorphisms. They serve as the crucial component that guaratees mappings from initial fields, like \( \mathbb{Q} \), remain unique when transitioning into another field, such as \( \mathbb{C} \). Thus, these functions not only preserve structure but ensure that no information is lost, establishing a clear path between two mathematical landscapes. This is pivotal when demonstrating proofs, like the one in the exercise, where an injective nature helps confirm mappings are just the identity function, \( h(x) = x \).