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Chapter 30: Problem 1

Prove each of the following: If the number \(a\) is a root of an irreducible polynomial \(p(x) \in \mathbb{Q}[x]\) whose degree is not a power of 2, then \(a\) is not a constructible number.

Short Answer

Expert verified
The root of an irreducible polynomial with a degree not a power of 2 is not constructible.

Step by step solution

01

Understand Constructible Numbers

A constructible number is a real or complex number that can be constructed using a finite number of steps involving addition, subtraction, multiplication, division, and square root extractions starting from integers. These numbers correspond to lengths that can be constructed with a straightedge and compass.
02

Define Constructible Numbers in Terms of Field Extensions

A number is constructible if it can be obtained from the rational numbers \ \( \mathbb{Q} \ \) through a sequence of field extensions, each of which has degree equal to 2. This means the degree of the extension field over \ \( \mathbb{Q} \ \) in which the number resides must be a power of 2.
03

Understand the Implication of Irreducible Polynomial

If \ \( a \ \) is a root of an irreducible polynomial \ \( p(x) \ \) in \ \( \mathbb{Q}[x] \), the irreducibility implies that the minimal polynomial of \ \( a \ \) over \ \( \mathbb{Q} \) is \ \( p(x) \). Thus, the degree of \ \( p(x) \) is the degree of the field extension \ \( \mathbb{Q}(a)/\mathbb{Q} \).
04

Check Degree Condition

Given that \ \( p(x) \) has a degree that is not a power of 2, we deduce that the field extension degree \ \( [\mathbb{Q}(a):\mathbb{Q}] \) is not a power of 2.
05

Conclude Non-Constructibility

Since constructible numbers must have a minimal polynomial leading to a field extension with a degree that is a power of 2, and \ \( a \) corresponds to a field extension whose degree is not a power of 2, \ \( a \) cannot be a constructible number.

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Key Concepts

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

Irreducible Polynomial
An irreducible polynomial is a polynomial that cannot be factored into the product of two non-constant polynomials in its given coefficient field. In the context of rational numbers, such a polynomial has coefficients from the field of rational numbers, denoted by \( \mathbb{Q} \). If \( p(x) \) is irreducible over \( \mathbb{Q} \), it has no factorization in \( \mathbb{Q}[x] \) other than 1 and itself.
  • Irreducibility ensures that the roots are not readily expressible as the values given by simpler polynomial expressions.
  • This characteristic is key to understanding why certain numbers like \( a \), which might be a root of such a polynomial, might not be constructible.
Irreducibility plays a pivotal role in determining the degree and the properties of the polynomials that create extensions of fields, especially over \( \mathbb{Q} \).
Field Extensions
A field extension is a larger field that contains a smaller field as a subfield. For example, if we start with the field \( \mathbb{Q} \), a field extension might include \( \mathbb{Q} \) along with some additional elements. Field extensions are essential for analyzing the complex roots and other solutions to polynomial equations.
  • The degree of a field extension indicates how much larger the new field is compared to the original.
  • Field extensions can be finite or infinite, but in the context of constructibility, we only consider finite extensions.
In proving non-constructibility, we look at whether the degree of the field extension, denoted \([\mathbb{Q}(a):\mathbb{Q}]\), is a power of 2. If it is not, this is a strong indicator that \( a \) is not a constructible number.
Degree of Polynomial
The degree of a polynomial is the highest power of \( x \) appearing in the polynomial when it is expressed in its standard form with coefficients in the field being considered. In mathematical terms, for a polynomial \( p(x) = a_n x^n + a_{n-1}x^{n-1} + ... + a_0 \), the degree is \( n \) if \( a_n eq 0 \).
  • The degree of the polynomial determines the degree of the corresponding field extension.
  • If \( a \) is a root of \( p(x) \), the degree of \( p(x) \) informs us about \([\mathbb{Q}(a):\mathbb{Q}]\).
If the degree of \( p(x) \) is not a power of 2, it gives us information crucial for ruling out the possibility of \( a \) being constructible, since constructible numbers arise from sequences of quadratic (degree 2) extensions.
Minimal Polynomial
The minimal polynomial is the polynomial of least degree among all polynomials having specific coefficients in which a given number or element is a root. For a number \( a \), its minimal polynomial over \( \mathbb{Q} \) is the lowest degree polynomial in \( \mathbb{Q}[x] \) for which \( a \) is a root.
  • The minimal polynomial is always irreducible over its coefficient field by definition.
  • It provides the simplest expression of the relationship between \( a \) and the other elements in the field.
The degree of the minimal polynomial determines the degree of the corresponding field extension. For \( a \) to be constructible, this polynomial must suggest a field extension degree that is a power of 2, otherwise, \( a \) cannot be constructed using a finite sequence of quadratic extensions.

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Most popular questions from this chapter

An angle \(\alpha\) is called constructible iff there exist constructible points \(A, B\), and \(C\) such that \(\angle A B C=\alpha\). Prove the following : If \(\alpha\) and \(\beta\) are constructible angles, so are \(\alpha+\beta, \alpha-\beta, \frac{1}{2} \alpha\), and \(n \alpha\) for any positive integer \(n\).

By de Moivre's theorem, $$ \omega=\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7} $$ is a complex seventh root of unity. Since $$ x^{7}-1=(x-1)\left(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\right) $$ \(\omega\) is a root of \(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\). Prove that $$ 8 \cos ^{3} \frac{2 \pi}{7}+4 \cos ^{2} \frac{2 \pi}{7}-4 \cos \frac{2 \pi}{7}-1=0 $$ (Use 1 and E1.) Conclude that \(\cos (2 \pi / 7)\) is a root of \(8 x^{3}+4 x^{2}-4 x-1\).

Prove that \(2 \pi / 5\) is a constructible angle.

By de Moivre's theorem, $$ \omega=\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7} $$ is a complex seventh root of unity. Since $$ x^{7}-1=(x-1)\left(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\right) $$ \(\omega\) is a root of \(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\). Prove that \(2 \pi / 7\) is not a constructible angle.

By de Moivre's theorem, $$ \omega=\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7} $$ is a complex seventh root of unity. Since $$ x^{7}-1=(x-1)\left(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\right) $$ \(\omega\) is a root of \(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\). Prove that \(\omega^{3}+\omega^{2}+\omega+1+\omega^{-1}+\omega^{-2}+\omega^{-3}=0\).
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