Chapter 31: Problem 4
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 \(\mathbb{Q}\).
Short Answer
Step by step solution
Introduction
Understand Field Homomorphisms
Apply Properties of Rational Numbers
Compute for Integers and Rational Numbers
Conclusion: Monomorphism Fixation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Monomorphism
- For any two elements \( a \) and \( b \) in the field, \( h(a + b) = h(a) + h(b) \)
- Similarly, \( h(a\cdot b) = h(a)\cdot h(b) \)
Rational Numbers
- They are closed under addition, subtraction, multiplication, and division (except by zero).
- They encompass both the positive fractions and negative fractions.
Complex Numbers
- 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.
Injective Function
- 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.