/*! 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 135 Let \(R\) be a commutative ring ... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 135

Let \(R\) be a commutative ring with at least one, but only finitely many, (nonzero) zero divisors. Prove that \(R\) is finite.

Short Answer

Expert verified
R is finite because it has finitely many zero divisors and every other element is a unit, generating R finitely.

Step by step solution

01

Understanding Zero Divisors

A zero divisor in a commutative ring is a nonzero element that, when multiplied by another nonzero element of the ring, results in zero. Formally, an element a in ring R is a zero divisor if there exists a nonzero element b in R such that ab = 0.
02

Given Information

The ring R is commutative and has at least one zero divisor. Moreover, the number of zero divisors is finite.
03

Properties of Zero Divisors

Since the ring R is commutative, the product ab = 0 implies that both a and b are zero divisors. Thus, if an element a annihilates more than one element, each of those elements would be a zero divisor, further limiting their number.
04

Consider Non-Zero Divisors

Elements in R are either zero divisors or not. Let A be the set of zero divisors, and B be the set of non-zero non-divisors. B is exactly the complement of A in R.
05

Finite Set of Elements in R

The set of zero divisors is finite by assumption. Any non-zero element of R that is not a zero divisor is a unit (invertible) because it cannot be a zero divisor or zero. A unit times any element of B remains in B.
06

Inverting Non-Zero Non-Divisors

Since there are finitely many zero divisors, and B is a set of elements that are not zero divisors and hence units, each element in R can be generated by multiplying zero divisors and units, making R finite.
07

Conclusion

Every element in R is either a zero divisor or a unit. Since there are finitely many zero divisors and each unit forms a finite group under multiplication, R is finite.

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Key Concepts

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

Commutative Ring Theory
A commutative ring is an algebraic structure where the multiplication operation between elements is commutative, meaning for any elements \(a\) and \(b\) in the ring \(R\), we have \(a \cdot b = b \cdot a\). This property simplifies many aspects of ring theory but still allows for a rich area of study. Important properties of commutative rings include having an additive identity (0), multiplicative identity (1), and the distributive property linking addition and multiplication.
Zero Divisors
Zero divisors are special elements in a commutative ring, which result in zero when multiplied by another nonzero element. Formally, an element \(a\) in \(R\) is a zero divisor if there exists a nonzero element \(b\) in \(R\) such that \(a \cdot b = 0\). Understanding zero divisors is crucial because their presence affects the ring's structure. For instance, they prevent the ring from being an integral domain, where no zero divisors exist. In the given exercise, the focus is on rings that have a finite number of zero divisors and showing that this implies the finiteness of the ring itself.
Finite Rings
Finite rings are rings with a finite number of elements. The problem under consideration deals with proving that a commutative ring \(R\) with finitely many zero divisors is itself finite. The key idea is that if there are finitely many zero divisors and the non-zero non-divisors are invertible (units), then combining these elements suitably can account for all elements in the ring. As such, the ring cannot have an infinite number of elements. Finite rings have applications in various areas, including coding theory and cryptography.
Invertible Elements
Invertible elements, or units, in a ring are those elements which have a multiplicative inverse. For an element \(a\) in a ring \(R\), \(a\) is a unit if there exists an element \(b\) in \(R\) such that \(a \cdot b = 1\). In the context of our exercise, non-zero elements that are not zero divisors must be invertible. This insight helps in proving the finiteness of the ring \(R\), as the presence of units alongside zero divisors accounts for all elements in \(R\). Each unit being able to generate other elements through multiplication further solidifies the structure of the ring.
Ring Elements Classification
Classifying elements in a ring means dividing them into distinct categories. In our exercise, elements of the ring \(R\) are classified as either zero divisors or units. Zero divisors lead to zero when multiplied by specific elements, while units can generate other elements and have inverses. This distinction is essential because it simplifies the understanding and structure of the ring. Since zero divisors and units cover all the elements in \(R\) and zero divisors are finite in number by assumption, proving that \(R\) is finite becomes feasible.

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Most popular questions from this chapter

Note that the integers \(a=1, b=5, c=7\) have the property that the square of \(b\) (namely, 25) is the average of the square of \(a\) and the square of \(c\) (1 and 49). Of course, from this one example we can get infinitely many examples by multiplying all three integers by the same factor. But if we don't allow this, will there still be infinitely many examples? That is, are there infinitely many triples \((a, b, c)\) such that the integers \(a, b, c\) have no common factors and the square of \(b\) is the average of the squares of \(a\) and \(c ?\)

It's not hard to see that in the plane, the largest number of nonzero vectors that can be chosen so that any two of the vectors make the same nonzero angle with each other is 3 (and the only possible nonzero angle for three such vectors to make is \(2 \pi / 3)\). Now suppose we have vectors in \(n\) dimensional space. What is the largest possible number of nonzero vectors in \(n\)-space so that the angle between any two of the vectors is the same (and not zero)? In that situation, what are the possible values for the angle?

Suppose all the integers have been colored with the three colors red, green and blue such that each integer has exactly one of those colors. Also suppose that the sum of any two (unequal or equal) green integers is blue, the sum of any two blue integers is green, the opposite of any green integer is blue, and the opposite of any blue integer is green. Finally, suppose that 1492 is red and that 2011 is green. Describe precisely which integers are red, which integers are green, and which integers are blue.

The Wohascum Center branch of Wohascum National Bank recently installed a digital time/temperature display which flashes back and forth between time, temperature in degrees Fahrenheit, and temperature in degrees Centigrade (Celsius). Recently one of the local college mathematics professors became concerned when she walked by the bank and saw readings of \(21^{\circ} \mathrm{C}\) and \(71^{\circ} \mathrm{F}\), especially since she had just taught her precocious five-year-old that same day to convert from degrees \(\mathrm{C}\) to degrees \(\mathrm{F}\) by multiplying by \(9 / 5\) and adding 32 (which yields \(21^{\circ} \mathrm{C}=69.8^{\circ} \mathrm{F}\), which should be rounded to \(70^{\circ} \mathrm{F}\) ). However, a bank officer explained that both readings were correct; the apparent error was due to the fact that the display device converts before rounding either Fahrenheit or Centigrade temperature to a whole number. (Thus, for example, \(21.4^{\circ} \mathrm{C}=70.52^{\circ} \mathrm{F}\).) Suppose that over the course of a week in summer, the temperatures measured are between \(15^{\circ} \mathrm{C}\) and \(25^{\circ} \mathrm{C}\) and that they are randomly and uniformly distributed over that interval. What is the probability that at any given time the display will appear to be in error for the reason above, that is, that the rounded value in degrees \(\mathrm{F}\) of the converted temperature is not the same as the value obtained by first rounding the temperature in degrees \(\mathrm{C}\), then converting to degrees \(\mathrm{F}\) and rounding once more?

Show that \(\sum_{k=0}^n \frac{(-1)^k}{2 n+2 k+1}\left(\begin{array}{l}n \\\ k\end{array}\right)=\frac{\left(2^n(2 n) !\right)^2}{(4 n+1) !}\).
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