Problem 3

Question

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 $8 x^{3}+4 x^{2}-4 x-1$ has no rational roots. Conclude that it is irreducible over $\mathbb{Q}$.

Step-by-Step Solution

Verified

Answer

The polynomial has no rational roots and is irreducible over $\mathbb{Q}$.

1Step 1: Understand Rational Roots Potential

To determine if the polynomial $8x^3 + 4x^2 - 4x - 1$ has rational roots, we apply the Rational Root Theorem. The theorem suggests that any rational solution $\frac{p}{q}$ must satisfy $p$ being a factor of the constant term and $q$ a factor of the leading coefficient.

2Step 2: Define Possible Rational Roots

For the polynomial $8x^3 + 4x^2 - 4x - 1$, the constant term is $-1$ with factors $\pm 1$ and the leading coefficient is $8$ with factors $\pm 1, \pm 2, \pm 4, \pm 8$. Hence, the possible rational roots are $\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}$.

3Step 3: Test Possible Rational Roots

Substitute each potential rational root into the polynomial $8x^3 + 4x^2 - 4x - 1$ and check if the polynomial evaluates to zero. None of these values $1, -1, \frac{1}{2}, -\frac{1}{2}, \frac{1}{4}, -\frac{1}{4}, \frac{1}{8}, -\frac{1}{8}$ actually satisfy the equation, as verified by substitution.

4Step 4: Conclude Polynomial Irreducibility over $\mathbb{Q}$

Since no rational roots were found, $8x^3 + 4x^2 - 4x - 1$ cannot be factored into polynomials with rational coefficients. Thus, the polynomial is irreducible over $\mathbb{Q}$, meeting the criterion of having no linear factors over the rationals.

Key Concepts

Rational Root TheoremRoots of UnityComplex NumbersIrreducibility over $\mathbb{Q}$

Rational Root Theorem

The Rational Root Theorem is a handy tool when dealing with polynomials. It offers a way to find possible rational solutions for a polynomial equation.
If a polynomial has a root that is a rational number, the theorem tells us that this root must be in the form $ \frac{p}{q} $, where $ p $ is a factor of the constant term and $ q $ is a factor of the leading coefficient.
Here's a quick rundown on how the theorem applies:

Consider a polynomial $ p(x) = a_n x^n + a_{n-1} x^{n-1} + \, ... \, + a_0 $.
For a rational root $ \frac{p}{q} $ to exist, $ p $ must divide the constant term $ a_0 $, while $ q $ divides the leading coefficient $ a_n $.
This can narrow down a list of potential rational candidates for roots.

The utility of the theorem lies in reducing a potentially infinite number of rational root trials into a manageable finite list. By testing only these candidates, we can determine if a polynomial can be simplified further.

Roots of Unity

In the realm of complex numbers, roots of unity play a significant role. They are complex numbers that satisfy the equation $ x^n = 1 $.
For example, the seventh roots of unity are the solutions to $ x^7 = 1 $. Such roots include powers of a complex number $ \omega = \cos \frac{2 \pi}{7} + i \sin \frac{2 \pi}{7} $, expressed using de Moivre's theorem.These roots represent equally spaced points on the unit circle in the complex plane, which has a radius of one:

Roots of unity are particularly useful for breaking down periodic functions and in polynomial factorization.
Graphically, these roots divide the circle into equal parts, akin to radians or trigonometric functions.
For every positive integer $ n $, the sum of all $ n $th roots of unity equals zero.

Understanding roots of unity helps in gaining deeper insights into the structure and symmetry of polynomials and their solutions.

Complex Numbers

Complex numbers extend the notion of one-dimensional real numbers to a two-dimensional number system by introducing the imaginary unit $ i $, where $ i^2 = -1 $.
A complex number is expressed as $ a + bi $, with $ a $ being the real part and $ b $ the imaginary part. This allows us to solve polynomial equations that have no real solutions.

Complex numbers reside on the complex plane, which consists of a real axis and an imaginary axis.
This plane provides a meaningful geometric representation where each point corresponds to a unique complex number.
They are indispensable in various fields like engineering and physics, particularly when addressing oscillatory systems.

Instead of being an obstacle, imaginary numbers offer a wider spectrum for solutions, broadening the scope of algebraic problems and solutions.

Irreducibility over $\mathbb{Q}$

A polynomial is irreducible over $ \mathbb{Q} $ if it cannot be factored into smaller polynomials with rational coefficients. The polynomial remains whole and cannot be broken into simpler polynomial products.
When a polynomial shows no rational roots, as determined through utilities like the Rational Root Theorem, it hints that the polynomial may be irreducible. In formal terms:

For a polynomial $ f(x) $, if no rational numbers satisfy $ f(x) = 0 $, it is a signal of irreducibility over $ \mathbb{Q} $.
Testing all possible rational roots strongly aids this discovery.
Exact mathematical demonstrations or related proofs (for example, Eisenstein's Criterion) can also establish irreducibility.

Irreducibility is vital in number theory and algebra, as it denotes a polynomial that stands indivisible by rational roots, presenting fundamental characteristics and properties on its own.

Problem 2

Problem 3