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Chapter 12: Problem 22

Let \(F \subset E\) be algebraic number fields. Show that any isomorphism of \(F\) into \(\mathbb{C}\) extends in exactly \([E: F]\) ways to an isomorphism of \(E\) into \(\mathbb{C}\).

Short Answer

Expert verified
Any isomorphism of F into \( \backslash mathbb{C} \) extends to \( [E : F] \) isomorphisms of E into \( \backslash mathbb{C} \).

Step by step solution

01

- Understand the Problem

You need to show that any isomorphism of a given algebraic number field F into the complex numbers \(\backslash mathbb{C}\) extends to exactly \([E : F]\) isomorphisms of a larger field E into \(\backslash mathbb{C}\).
02

- Consider the Degree of Extension

Recognize that \([E : F]\) represents the degree of the field extension from F to E. This is the number of distinct embeddings of E into \( \backslash mathbb{C} \) that fix F pointwise.
03

- Identify the Generators

Assume that E is generated by elements over F. Let \( \backslash alpha_1, \backslash alpha_2, ..., \backslash alpha_n \) be the generators.
04

- Extend the Isomorphism

Take an isomorphism \( \varphi : F \rightarrow \backslash mathbb{C} \). Extend \( \varphi \) to E by defining it on each generator \( \alpha_i \) of E. This extension needs to preserve the additive and multiplicative structures of E.
05

- Construct the Embeddings

Since \( \alpha_i \) are algebraic over F, each \( \alpha_i \) satisfies a minimal polynomial in F. The roots of this polynomial can be used to define possible extensions. Therefore, each such root corresponds to a possible embedding.
06

- Count the Embeddings

Using the degrees of the minimal polynomials, conclude that there are exactly \( [E :F] \) such embeddings extending the original isomorphism \( \varphi \). Each embedding corresponds to a different choice of roots for the minimal polynomials.

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Key Concepts

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

Algebraic Number Fields
An algebraic number field is a finite extension of the rational numbers \(\backslash mathbb{Q} \). These fields contain numbers satisfying polynomial equations with coefficients in \(\backslash mathbb{Q} \). For example, the field \(\backslash mathbb{Q}( \backslash sqrt{2} )\) is an algebraic number field, because every number in it can be expressed as a \(\backslash mathbb{Q} \)-linear combination of 1 and \(\backslash sqrt{2} \). Algebraic number fields have a nice structure that allows extension and embedding properties to be studied using tools from both field theory and algebraic number theory.
Field Extensions
A field extension is a pair of fields E and F where F is a subfield of E. In other words, every element of F is also an element of E. The degree of the extension \([E : F]\) measures the 'size' of E relative to F, specifically the dimension of E as a vector space over F. For example, \(\backslash mathbb{C} \) (the field of complex numbers) can be considered as a field extension of \(\backslash mathbb{R} \) (the field of real numbers) with degree 2.
Minimal Polynomials
A minimal polynomial of an algebraic element \(\backslash alpha \) over a field F is the monic polynomial of lowest degree having \(\backslash alpha \) as a root with coefficients in F. For example, the minimal polynomial of \(\backslash sqrt{2} \) over \(\backslash mathbb{Q} \) is \(x^2 - 2\). Roots of minimal polynomials are critical in constructing field extensions and determining how elements from F can be related to elements in a larger field E.
Isomorphism Extension
An isomorphism extension is about extending a field isomorphism (a bijective homomorphism that preserves the field structure) from a smaller field F to a larger field E. This process involves understanding how generators of E over F can be expressed in \(\backslash mathbb{C} \) consistent with the given isomorphism on F. The key result discussed here is that any isomorphism of F into \(\backslash mathbb{C} \) can be extended to exactly \[E : F\] isomorphisms of E into \(\backslash mathbb{C} \).
Complex Numbers
The complex numbers \( \backslash mathbb{C} \) form an algebraically closed field, meaning that every non-constant polynomial with complex coefficients has a complex root. Because \( \backslash mathbb{C} \) includes solutions to all polynomials, it's a perfect setting for extending fields coming from algebraic number theory. This characteristic facilitates the multiple embeddings of field extensions into \( \backslash mathbb{C} \) and provides a robust framework for understanding algebraic structures.

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Most popular questions from this chapter

If \(R \subset S\) are integral domains \(\alpha \in S\) is said to be integral over \(R\) if \(\alpha^{m}+b_{1} \alpha^{m-1}+\cdots\) \(+b_{m}=0\) for suitable \(m ; b_{1}, \ldots, b_{m} \in R \cdot S\) is called integral over \(R\) if every element of \(S\) is integral over \(R\). Prove that if \(S\) is integral over \(R\) then \(S\) is a field iff \(R\) is a field.

Find the minimal polynomial for \(\sqrt{3}+\sqrt{7}\).

Let \(K=F_{2}(x)\) and \(L=K(\sqrt{x})\). Show that the trace map is identically zero. (Recall, \(F_{2}\) is the finite field with two elements.)

Let \(F\) be an algebraic number field with ring of integers \(D\). Show that if \(P\) and \(Q\) are distinct prime ideals then \(\left(P^{a}, Q^{b}\right)=D\), where \(a\) and \(b\) are positive integers.

If \(D\) is the ring of integers in an algebraic number field and \(\boldsymbol{P}\) is a prime ideal such that \(\mathfrak{P}=(\alpha)\) then show that \(\alpha\) is irreducible.
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