/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Let \(\zeta=\cos 2 \pi / 7+i \si... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(\zeta=\cos 2 \pi / 7+i \sin 2 \pi / 7,\) and let \(\alpha=\zeta+\zeta^{-1}\) (a) Find the minimal polynomial for \(\alpha\) over \(\mathbb{Q}\) (it is a cubic). (b) Show that \(\mathbb{Q}(\zeta)\) contains a unique subfield \(E\) of degree 2 over \(\mathbb{Q}\). Find an integer \(d\) for which \(E=\mathbb{Q}(\sqrt{d})\)

Short Answer

Expert verified
The minimal polynomial for \(\alpha\) over \(\mathbb{Q}\) is \(\alpha^3 - 3\alpha - 2\cos(6\pi / 7) = 0\). The unique subfield \(E\) of \(\mathbb{Q}(\zeta)\) of degree 2 over \(\mathbb{Q}\) is \(E = \mathbb{Q}(\sqrt{d})\), where \(d = 1 - \cos(6\pi / 7)\).

Step by step solution

01

Compute \(\zeta + \zeta^{-1}\)

We first compute the value of \(\zeta + \zeta^{-1} = \cos(2\pi / 7) + i\sin(2\pi / 7) + \cos(-2\pi / 7) + i\sin(-2\pi / 7) = 2\cos(2\pi / 7)\). So \(\alpha = 2\cos(2\pi / 7)\). In the next step we will use the formula for \(\cos(3\theta)\) to express \(\alpha\) in a nice form.
02

Find the minimal polynomial for \(\alpha\)

The minimal polynomial of \(\alpha\) over \(\mathbb{Q}\) is the lowest degree polynomial which has \(\alpha\) as a root. We use the expression for \(\cos(3\theta)\) in terms of \(\cos\theta\), substitute \(\theta = 2\pi / 7\), and rewrite to form a cubic polynomial. We get \(\cos(6\pi / 7) = 4\cos^3(2\pi / 7) - 3\cos(2\pi / 7) = 2\alpha^3 / 8 - 3\alpha / 2\) which simplifies to \(\alpha^3 / 2 - 3\alpha / 2 - \cos(6\pi / 7) = 0\). Hence, the minimal polynomial for \(\alpha\) over \(\mathbb{Q}\) is \(\alpha^3 - 3\alpha - 2\cos(6\pi / 7) = 0\).
03

Find the degree 2 subfield of \(\mathbb{Q}(\zeta)\)

To find the degree 2 subfield of \(\mathbb{Q}(\zeta)\), we need to find a quadratic polynomial that has \(\alpha / 2 = \cos(2\pi / 7)\) as a root. Note that \(\cos^2\theta = (1 + \cos(2\theta))/2\), hence we have \(\cos^2(2\pi / 7) = (1 + \cos(4\pi / 7)) / 2\). Using the triple-angle formula for cosine, the polynomial that has \(\cos(2\pi / 7)\) as a root is given by \(x^2 - x - (1 - \cos(6\pi / 7)) / 2 = 0\). Hence, the degree 2 subfield \(E\) of \(\mathbb{Q}(\zeta)\) is \(E = \mathbb{Q}(\sqrt{d})\), where \(d = 1 - \cos(6\pi / 7)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Cubic Polynomials
Cubic polynomials are equations of the form \(ax^3 + bx^2 + cx + d = 0\), where \(a\), \(b\), \(c\), and \(d\) are constants and \(a eq 0\). These types of polynomials are important because they can describe the roots of functions that involve third-degree expressions, as seen in algebraic equations. In the context of finding the minimal polynomial for \(\alpha = 2\cos(2\pi / 7)\), the polynomial that has \(\alpha\) as a root is cubic since it is expressed in terms of \(\cos(3\theta)\).
In our exercise, we use trigonometric identities to simplify the expression involving \(\alpha\). The specific identity for cosine triple angles, \(\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)\), allows us to rearrange and solve the equation, ultimately forming a cubic polynomial.
Identifying the minimal polynomial, especially when dealing with trigonometric functions, requires equating and manipulating expressions until the root is found. This provides insight into using trigonometric functions to solve algebraic expressions. Being able to simplify and factorize helps harness the full power of polynomials in mathematics and further applications in science and engineering.
Cyclotomic Fields
Cyclotomic fields are extensions of the field of rational numbers \(\mathbb{Q}\) obtained by adding a root of unity—specifically, a solution of \(z^n = 1\) for some integer \(n\). These fields are central in understanding the distribution of roots of polynomial equations surrounding the concept of roots of unity.
In this task, \(\zeta\) represents a primitive seventh root of unity described by \(\zeta = \cos(2\pi / 7) + i\sin(2\pi / 7)\). From this, \(\mathbb{Q}(\zeta)\) becomes a cyclotomic field. The significance lies in uncovering its subfields, such as degree two subfields, which simplify analysis of root structures and their algebraic relationships.
By finding \(\alpha/2\) as a root in a quadratic polynomial, we demonstrated that the subfield \(E\), containing \(\mathbb{Q}\), has degrees that can both unlock polynomial factorizations and explore relationships between imaginary units, complex numbers, and the roots of unity.
Algebraic Number Theory
Algebraic number theory explores the properties of algebraic numbers—numbers that are roots of polynomial equations with integer coefficients. This branch of mathematics provides a framework for understanding polynomials, field extensions, and offers tools for solving equations.
This exercise delves into the minimal polynomial, a concept rooted in algebraic number theory, which allows us to investigate the smallest degree polynomial with rational coefficients that has a given number, in this case \(\alpha\), as a root. By constructing polynomials in our filed extensions like \(\mathbb{Q}(\zeta)\), algebraic number theory offers a systematic approach to discerning the structure and properties of subfields.
One example is the degree 2 subfield in \(\mathbb{Q}(\zeta)\), \(E = \mathbb{Q}(\sqrt{d})\), without which it would be challenging to relate the complex numbers back to simple rational field extensions. This highlights algebraic number theory's relevance in understanding how complex numbers connect to their polynomial roots and how they integrate within field extensions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\mathbb{Q}(\zeta)\) be the cyclotomic field of 5 th roots of unity. (a) Show that \(\sqrt{5} \in \mathbb{Q}(\zeta)\) (b) Show that \(\mathbb{Q}(\zeta)=\mathbb{Q}(\delta),\) where \(\delta=\frac{i}{2} \sqrt{10+2 \sqrt{5}}\) (c) What is the minimal polynomial of \(\delta ?\)

Consider the polynomial \(f(x)=x^{4}+x-3\) over \(\mathbb{Q}\) (a) Find the cubic resolvent and show that it is irreducible. (b) Show that \(f(x)\) is irreducible. (c) By curve sketching, show that \(f(x)\) has two real roots and two complex roots. (d) Conclude that the Galois group of \(f(x)\) is \(S_{4}\) (e) If \(\alpha\) is a real root, then \(\mathbb{Q}(\alpha)\) is an extension of degree 4 of \(\mathbb{Q},\) but \(\alpha\) is not constructible by ruler and compass.

The discriminant. Let \(f(x)\) be an irreducible cubic polynomial with coefficients in the field \(F \subseteq \mathbb{R},\) and let its roots be \(\alpha_{1}, \alpha_{2}, \alpha_{3}\) in its splitting field. We define the \(d i s-\) criminant of \(f(x)\) to be $$\Delta=\left(\alpha_{1}-\alpha_{2}\right)^{2}\left(\alpha_{1} \alpha_{3}\right)^{2}\left(\alpha_{2}-\alpha_{3}\right)^{2}$$ (a) Show that \(\Delta \in F\) (b) Show that \(\Delta>0\) if and only if \(f(x)\) has 3 real roots, while \(\Delta<0\) if and only if \(f(x)\) has one real and two complex roots. (c) Show that the Galois group of \(f(x)\) is \(\mathbb{Z}_{3}\) if and only if \(\sqrt{\Delta} \in F\); otherwise, it is \(S_{3}\)

In the real Cartesian plane: (a) Show that the circle with center \((a, 1)\) passing through the origin intersects the parabola \(\Gamma\) given by \(y=\frac{1}{2} x^{2}\) at the point \((2 \sqrt[3]{a}, 2 \sqrt[3]{a^{2}})\) (b) Show that if we are given a single parabola drawn in the plane, then the problem of duplication of the cube becomes possible. (c) If a parabola is given by its focus and directrix, conclude that its intersection points with a circle are not always constructible.

Consider the polynomial \(x^{4}-2 x^{2}-7\) over \(\mathbb{Q}\). (a) Show that its Galois group is the dihedral group \(D_{4},\) defined by generators \(a, b\) and relations \(a^{4}=e, b^{2}=e, b a=a^{-1} b\) (b) Find the lattice of all subgroups of \(D_{4}\) (c) Find all the subfields of the splitting field, and explain their correspondence with the subgroups of \(D_{4}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.