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Chapter 12: Problem 2

Man zeige: In \(\mathbb{Z}_{n}\) ist jedes Element \(\neq 0\) entweder ein Nullteiler oder invertierbar.

Short Answer

Expert verified
Every nonzero element in \(\mathbb{Z}_n\) is either a zero divisor or invertible.

Step by step solution

01

Understand the Problem Statement

We are asked to show that in the set of integers modulo \( n \), denoted by \( \mathbb{Z}_n \), every nonzero element is either a zero divisor or invertible.
02

Define Zero Divisor and Invertible Element

An element \( a \in \mathbb{Z}_n \) is a zero divisor if there exists another element \( b eq 0 \) such that \( a \cdot b \equiv 0 \pmod{n} \). An element \( a \in \mathbb{Z}_n \) is invertible if there exists an element \( a^{-1} \) such that \( a \cdot a^{-1} \equiv 1 \pmod{n} \).
03

Consider the Euclidean Algorithm

For an element \( a \in \mathbb{Z}_n \), we use the Euclidean algorithm to check the greatest common divisor \( \text{gcd}(a,n) \). If \( \text{gcd}(a,n) = 1 \), then \( a \) is invertible.
04

Determine When an Element is a Zero Divisor

If \( \text{gcd}(a,n) eq 1 \), this means \( a \) and \( n \) are not coprime, implying there exists a common factor \( d > 1 \). This means \( a \) is a zero divisor because \( a \cdot (n/d) \equiv 0 \pmod{n} \).
05

Conclude the Argument

We have established that nonzero element \( a \in \mathbb{Z}_n \) is invertible if \( \text{gcd}(a,n) = 1 \), and it is a zero divisor if \( \text{gcd}(a,n) eq 1 \). Therefore, each nonzero element is either a zero divisor or invertible.

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Key Concepts

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

Integers Modulo n
The concept of integers modulo \( n \), often denoted as \( \mathbb{Z}_n \), is a fascinating aspect of abstract algebra. Imagine you have a set of whole numbers from 0 up to \( n-1 \). These numbers form the modular integers for any given positive integer \( n \). The idea is that when you perform arithmetic operations, you do them "modulo \( n \)". Think of it like a clock: once you go past the largest number, you circle back to zero.
  • When adding or multiplying numbers modulo \( n \), the result is the remainder when the true result is divided by \( n \).
  • This system creates a cyclical number line that wraps back around.
This system is very useful in computer science, cryptography, and other fields where cyclic operations are needed.
Zero Divisor
In the modular integer system, an interesting concept arises鈥攝ero divisors. Simply put, a zero divisor is an element that can multiply with another non-zero element to give a product of zero, but only within this modular system.
  • Let鈥檚 say \( a \) is an element in \( \mathbb{Z}_n \).
  • If there exists another non-zero \( b \) such that \( a \cdot b \equiv 0 \pmod{n} \), then \( a \) is a zero divisor.
Zero divisors are sometimes unexpected because, outside the modular system, multiplication of non-zero numbers typically never results in zero. But in modular arithmetic, common factors between the numbers and \( n \) can make this happen.
Invertibility
Invertibility is another intriguing aspect of modular arithmetic. An element is considered invertible or a unit if there exists another element that, when multiplied together, results in one, under modulo \( n \).
  • An element \( a \) in \( \mathbb{Z}_n \) is invertible if there exists an element \( a^{-1} \) such that \( a \cdot a^{-1} \equiv 1 \pmod{n} \).
  • This implies \( a \) and \( n \) must have no common divisors other than 1.
To find out if \( a \) is invertible, we usually calculate the greatest common divisor (GCD) of \( a \) and \( n \). If \( \text{gcd}(a, n) = 1 \), the element is invertible.
Greatest Common Divisor
The greatest common divisor (GCD) is a core concept in understanding modular arithmetic. It is the largest number that can exactly divide both numbers without leaving a remainder.
  • To determine if a number is invertible in \( \mathbb{Z}_n \), the GCD of the number and \( n \) play a pivotal role.
  • If \( \text{gcd}(a, n) = 1 \), then \( a \) is relatively prime to \( n \), and \( a \) is invertible.
Using the Euclidean algorithm helps in efficiently finding the GCD. This simple algorithm repeatedly applies the division algorithm until a remainder of zero is reached, thus revealing the largest common factor.

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Most popular questions from this chapter

Es sei \(R \neq\\{0\\}\) ein kommutativer Ring ohne Nullteiler, in dem jeder Teilring nur endlich viele Elemente enth盲lt. Zeigen Sie, dass \(R\) ein K枚rper ist.

Es sei \(d \in \mathbb{Z} \backslash\\{1\\}\) quadratfrei (d.h.: \(\left.x \in \mathbb{N}, x^{2} \mid d \Rightarrow x=1\right)\) und $$ \mathbb{Z}[\sqrt{d}]:=\\{a+b \sqrt{d} \mid a, b \in \mathbb{Z}\\}, \mathbb{Q}[\sqrt{d}]:=\\{a+b \sqrt{d} \mid a, b \in \mathbb{Q}\\} $$ Zeigen Sie: (a) \(\mathbb{Z}[\sqrt{d}]\) und \(\mathbb{Q}[\sqrt{d}]\) sind Teilringe von \(\mathbb{C}\). (b) Die Abbildung \(z=a+b \sqrt{d} \mapsto \bar{z}:=a-b \sqrt{d}(a, b \in \mathbb{Z}\) bzw. \(a, b \in \mathbb{Q})\) ist ein Automorphismus von \(\mathbb{Z}[\sqrt{d}]\) bzw. von \(\mathbb{Q}[\sqrt{d}]\). (c) Es ist \(N: z \mapsto z \bar{z}\) eine Abbildung von \(\mathbb{Q}[\sqrt{d}]\) in \(\mathbb{Q}\), die \(N\left(z z^{\prime}\right)=N(z) N\left(z^{\prime}\right)\) f眉r alle \(z, z^{\prime} \in \mathbb{Q}[\sqrt{d}]\) erf眉llt; und f眉r \(z \in \mathbb{Z}[\sqrt{d}]\) gilt: \(z \in \mathbb{Z}[\sqrt{d}]^{\times} \Leftrightarrow N(z) \in\\{1,-1\\}\). (d) \(\mathbb{Q}[\sqrt{d}]\) ist ein K枚rper, \(\mathbb{Z}[\sqrt{d}]\) jedoch nicht. (e) Ermitteln Sie die Einheiten von \(\mathbb{Z}[\sqrt{d}]\), falls \(d<0\). (f) Zeigen Sie, \(\mathbb{Z}[\sqrt{5}]^{\times}=\left\\{\pm(2+\sqrt{5})^{k} \mid k \in \mathbb{Z}\right\\}\).

Es sei \(K\) ein K枚rper, \(s \in K\) und \(K_{s}:=\left\\{\left(\begin{array}{cc}a & s b \\ b & a\end{array}\right) \mid a, b \in K\right\\} .\) Zeigen Sie: (a) \(K_{s}\) ist ein kommutativer Teilring von \(K^{2 \times 2} .\) Wann ist \(K_{s}\) ein K枚rper? (b) \(\mathbb{R}_{-1}\) ist zu \(\mathbb{C}\) isomorph. (c) F眉r jede Primzahl \(p \neq 2\) gibt es einen K枚rper mit \(p^{2}\) Elementen.

Zeigen Sie: \(\mid\) Aut \(\mathbb{Q} \mid=1\) und \(\mid\) Aut \(\mathbb{R} \mid=1\).

Es sei \(V\) ein endlichdimensionaler Vektorraum 眉ber einem K枚rper \(K\), und \(R:=\operatorname{End}(V)\) bezeichne den Endomorphismenring von \(V\). Man zeige f眉r \(\varphi \in R\) : (a) \(\varphi\) ist linksinvertierbar \(\Leftrightarrow \varphi\) ist rechtsinvertierbar \(\Leftrightarrow \varphi\) ist invertierbar. (Dabei hei脽t \(\varphi\) links- bzw. rechtsinvertierbar, wenn ein \(\psi \in R\) mit \(\psi \varphi=\) Id bzw. \(\varphi \psi=\) Id existiert.) (b) In \(R\) ist jeder Nichtnullteiler \(\varphi \neq 0\) invertierbar.
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