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Chapter 14: Problem 4

Prove that a unit in a ring \(R\) cannot be a proper divisor of zero.

Short Answer

Expert verified
A unit in a ring \(R\) cannot be a proper divisor of zero, because assuming it can be leads to a contradiction with it having a multiplicative inverse.

Step by step solution

01

Define and illustrate a unit

A unit in a ring \(R\) is an element that has a multiplicative inverse. If \(u\) is a unit, there exists \(v\) in \(R\) such that \(u \cdot v = v \cdot u = 1\). This means that \(u\) has an inverse, denoted by \(u^{-1}\), and this inverse is \(v\).
02

Define and illustrate a proper divisor of zero

An element of the ring \(R\) is a proper divisor of \(0\) if for it, say \(a\), there exists a non-zero element \(b\) in the ring \(R\) such that \(ab = 0\). But this \(a\) should not be a zero element.
03

Provide the proof that a unit cannot be a proper divisor of zero

To prove that a unit \(u\) cannot be a proper divisor of zero, we will assume the opposite and derive a contradiction. Suppose \(u\) is a unit and \(u\) is a proper divisor, i.e., there exists a non-zero \(b\) such that \(u \cdot b = 0\). Since \(u\) is a unit, we can multiply both sides of the equation by \(u^{-1}\). The equation becomes: \(u^{-1} \cdot u \cdot b = u^{-1} \cdot 0\). Following the ring's operations, the left side simplifies to \(b\) and the right to \(0\). So we have \(b = 0\), contradicting our assumption that \(b\) is non-zero. So our assumption is incorrect, and a unit in a ring \(R\) cannot be a proper divisor of zero.

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Most popular questions from this chapter

Let \(k, m\) be fixed integers. Find all values for \(k, m\) for which \((\mathbf{Z}, \oplus, \odot)\) is a ring under the binary operations \(x \oplus y=x+y-k, x \odot y=x+y-m x y\), where \(x, y \in \mathbf{Z}\).

If \(a, b\) are units in a ring \(R\), is \(a+b\) necessarily a unit in \(R ?\)

Write a computer program (or develop an algorithm) that reverses the order of the digits in a given positive integer. For example, the input 1374 should result in the output 4731 .

If \(a, b\), and \(c\) are any elements in a ring \(\left(R,+,^{\circ}\right)\), prove that \((a) a(b-c)=a b-(a c)=\) \(a b-a c\) and \((b)(b-c) a=b a-(c a)=b a-c a\).

For \(R=M_{2}(\mathbf{Z})\) let \(S\) be the subset of \(R\) where $$ s=\left\\{\left[\begin{array}{ll} x & x \\ y & y \end{array}\right] \mid x, y \in \mathbf{Z}\right\\} . $$ Prove that \(S\) is a subring of \(R\).
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