Question

Let \(R\) be a ring (commutative, with 1 ). (i) For \(p \in \mathbb{N}_{\geq 2}\), determine the quotient and remainder on division of \(f_{p}=x^{p-1}+x^{p-2}+\cdots+x+1\) by \(x-1\) in \(R[x]\). Conclude that \(x-1\) is invertible modulo \(f_{p}\) if \(p\) is a unit in \(R\) and that \(x-1\) is a zero divisor modulo \(f_{p}\) if \(p\) is a zero divisor in \(R\). (ii) Assume that 3 is a unit in \(R\), and let \(n=3^{k}\) for some \(k \in \mathbb{N}, D=R[x] /\left\langle x^{2 n}+x^{n}+1\right\rangle\), and \(\omega=\) \(x \bmod x^{2 n}+x^{n}+1 \in D\). Prove that \(\omega^{3 n}=1\) and \(\omega^{n}-1\) is a unit. Hint: Calculate \(\left(\omega^{n}+2\right)\left(\omega^{n}-1\right)\). Conclude that \(\omega\) is a primitive \(3 n\)th root of unity. (iii) Let \(p \in \mathbb{N}\) be prime and a unit in \(R, n=p^{k}\) for some \(k \in \mathbb{N}, \Phi_{p n}=f_{p}\left(x^{n}\right)=x^{(p-1) n}+x^{(p-2) n}+\) \(\cdots+x^{n}+1 \in R[x]\) the \(p n\)th cyclotomic polynomial, \(D=R[x] /\left\langle\Phi_{p n}\right\rangle\), and \(\omega=x \bmod \Phi_{p n} \in D\). Prove that \(\omega^{p n}=1\) and \(\omega^{n}-1\) is a unit. Hint: Calculate \(\left(\omega^{(p-2) n}+2 \omega^{(p-3) n}+\cdots+(p-2) \omega^{n}+(p-1)\right)\). \(\left(\omega^{n}-1\right)\). Conclude that \(\omega\) is a primitive \(p n\)th root of unity.

Short answer

(i) Quotient is geometric series, remainder is zero. (ii) \( \\\omega^{3 n} = 1 \), \( \\\omega^n - 1 \) is a unit. (iii) \( \\\omega^{pn} = 1 \), \( \\\omega^n - 1 \) is a unit.

Step-by-step solution

Division of Polynomial by x-1

We begin by dividing the polynomial \( f_{p} = x^{p-1} + x^{p-2} + \cdots + x + 1 \) by \( x-1 \). By using the formula for the sum of a geometric series, we have \( f_p = \frac{x^p - 1}{x - 1} \) when \( x eq 1 \). The quotient is \( x^{p-1} + x^{p-2} + \cdots + x + 1 \) and the remainder is zero since \( f_p = 0 \) when \( x = 1 \). Thus, \( f_p(x) = (x - 1)g(x) + 0 \).

Invertibility Conditions

Since the remainder is zero when dividing \( f_p \) by \( x-1 \), it follows that \( x-1 \mid f_p \). For \( x - 1 \) to be invertible modulo \( f_p \), the division should have no remainder. If \( p \) is a unit in \( R \), \( f_p(1) = p \equiv 0 \pmod{f_p} \), hence \( x-1 \) is invertible. If \( p \) is a zero divisor, \( f_p(1) = 0 \pmod{f_p} \), making \( x-1 \) a zero divisor modulo \( f_p \).

Set Up Initial Conditions for Part (ii)

Given that 3 is a unit in \( R \) and \( n = 3^k \), consider \( \omega = x \bmod (x^{2n} + x^n + 1) \) in \( D = R[x] / \langle x^{2n} + x^n + 1 \rangle \). We will prove that \( \omega^{3n} = 1 \) and \( \omega^n - 1 \) is a unit.

Compute \( \\omega^{3n} \)

Utilize the expression \( x^{2n} + x^n + 1 = 0 \) in \( D \). Powers of \( x \) such as \( \omega^{3n} = x^{3n} \equiv 1 \) mod (since \( x^{3n} - 1 = 0 \) modulo \( x^{2n} + x^n + 1 \)). So, \( \omega^{3n} = 1 \).

Check the Unit Condition for \( \\omega^{n} - 1 \)

Calculate \( (\omega^n + 2)(\omega^n - 1) = \omega^{2n} + 2\omega^n - \omega^n - 2 = \omega^{2n} - \omega^n - 2 \). From the relation \( x^{2n} + x^n + 1 = 0 \), replace \( x^{2n} \) with \( -x^n - 1 \) to find \( \omega^{2n} = -\omega^n - 1 \). Then, \( (\omega^n - 1) \equiv 1 \pmod{x^{2n}+x^n+1} \). Thus, \( \omega^n - 1 \) is a unit.

Conclusion for Primitive Root

Since \( \omega^{3n} = 1 \) and \( \omega^n - 1 \) is a unit, \( \omega \) is a primitive 3n-th root of unity. This satisfies the divisibility requirements for primitive roots.

Analyze Part (iii) Initial Setup

Given \( p \) as prime and unit, \( n = p^k \), and \( D = R[x]/\langle\Phi_{pn}\rangle \), with \( \Phi_{pn}(x) = f_p(x^n) \). Thus, \( \Phi_{pn} = x^{(p-1)n} + \cdots + x^n + 1 \). We will demonstrate \( \omega^{pn} = 1 \) and \( \omega^n - 1 \) is a unit.

Prove \( \\omega^{pn} \) and Unit in Part (iii)

Find \( (\omega^{(p-2)n} + 2\omega^{(p-3)n} + \cdots + (p-1)) \omega^n (\omega^n - 1) = f_p(\omega^n)(\omega^n - 1) = 1 \). With \( \omega^{pn} = 1 \), and applying similar arguments as before, \( \omega^n - 1 \) is unit.

Conclude Primitive Root in Part (iii)

Having established \( \omega^{pn} = 1 \) and \( \omega^n - 1 \) as a unit, deduce that \( \omega \) is a primitive \( pn \)-th root of unity.

Key concepts

Cyclotomic Polynomials

Cyclotomic polynomials are a fascinating class of polynomials intimately related to number theory and algebra. They play a crucial role in polynomial division within rings. These polynomials are defined as \[ \Phi_n(x) \] for a given integer \( n \), and are the minimal polynomials over the field of rational numbers that a primitive \( n \)-th root of unity will satisfy. A primitive \( n \)-th root of unity is an element that satisfies \[ \omega^n = 1 \]but no smaller power of it does (i.e., a non-trivial root of unity).Important properties of cyclotomic polynomials include:

• They are irreducible over the field of rational numbers.
• The degree of \( \Phi_n(x) \) is given by Euler's totient function \( \varphi(n) \), which counts how many integers up to \( n \) are coprime to \( n \).
• The roots of a cyclotomic polynomial are precisely the primitive \( n \)-th roots of unity.

Cyclotomic polynomials feature prominently in the exercise when considering the division of other polynomials and determining the units and zero divisors in the ring \( R[x] \). They connect directly to the concept of primitive roots, providing the groundwork for understanding polynomial roots over arbitrary rings.

Primitive Roots of Unity

Primitive roots of unity are central to understanding complex numbers and their powers in mathematical fields and rings. A primitive \( n \)-th root of unity, denoted by \( \omega \), is a special number satisfying:\[ \omega^n = 1 \]and for no smaller positive integer \( m \) is it true that \( \omega^m = 1 \).Every \( n \)-th root of unity can be expressed in the form:\[ e^{\frac{2\pi i k}{n}} \] where \( k \) is coprime to \( n \). The number \( n \) of these primitive roots is given by Euler's totient function \( \varphi(n) \).Features and properties include:

• They form a cyclic group under multiplication.
• Each \( n \)-th root of unity raised to its order results in 1, illustrating periodicity.
• In polynomials, primitive roots are used to evaluate division and root behavior in rings, as seen in showing that \( \omega \) is a primitive \( pn \)-th root of unity as in the exercise.

Understanding these roots is essential for polynomial division and revealing deeper algebraic structures in mathematics.

Geometric Series

The concept of geometric series is a fundamental topic in algebra and important for understanding polynomial operations. A geometric series is a sequence of the form:\[ a + ar + ar^2 + ar^3 + \, \cdots \, + ar^n \]where each term after the first is found by multiplying the previous term by a constant called the ratio, \( r \).The sum of a finite geometric series can be calculated using the formula:\[ S_n = a \frac{1 - r^{n+1}}{1 - r} \, \text{for} \, r eq 1. \]Characteristics of a geometric series include:

• The terms in the series increase or decrease geometrically depending on the value of \( r \).
• When \( r = 1 \), the series is simply \( (n+1)a \).
• This series is used in the exercise for dividing polynomials, such as when finding the quotient of \( f_p(x) \) by \( x-1 \), which results in a formula similar to that of an infinite geometric series but truncated.

In understanding polynomial division in rings, recognizing the role of geometric series helps simplify expressions and provides clarity to what can seem like complex algebraic manipulations.