Problem 6
Question
Prove that each of the following is true in a nontrivial ring with unity. \(A \times B\) always has divisors of zero.
Step-by-Step Solution
Verified Answer
The product ring \(A \times B\) always has zero divisors due to elements \((a, 0)\) and \((0, b)\) being zero divisors, where \(a\) and \(b\) are nonzero in \(A\) and \(B\).
1Step 1: Define a Nontrivial Ring with Unity
A nontrivial ring with unity is a ring that has at least one nonzero element and possesses an element called unity (1), such that for any element \( a \) in the ring, \( 1 \times a = a \times 1 = a \).
2Step 2: Consider the Ring Product \(A \times B\)
The ring product \(A \times B\) is the set of all ordered pairs \((a,b)\), where \(a\) is from ring \(A\), and \(b\) is from ring \(B\). Operations in \(A \times B\) are defined as \((a,b) + (c,d) = (a+c, b+d)\) for addition and \((a,b) \cdot (c,d) = (ac, bd)\) for multiplication.
3Step 3: Define Zero Divisors in Ring \(A \times B\)
Zero divisors in a ring \(A \times B\) are nonzero elements \((a, b)\) and \((c, d)\) such that \((a, b) \cdot (c, d) = (ac, bd) = (0, 0)\), where neither \((a, b)\) nor \((c, d)\) is the zero element \((0, 0)\).
4Step 4: Prove Existence of Zero Divisors
Assume \(a eq 0\) is an element of \(A\) for which there exists a zero divisor. Similarly, suppose \(b eq 0\) is an element of \(B\) such that it is also a zero divisor. In \(A \times B\), consider \((a, 0)\) and \((0, b)\). Then, \((a, 0) \cdot (0, b) = (a \cdot 0, 0 \cdot b) = (0, 0)\). Thus, \((a, 0)\) and \((0, b)\) are zero divisors.
5Step 5: Conclusion
Since \(A\) and \(B\) are nontrivial rings, they contain zero divisors. Therefore, in \(A \times B\), elements like \((a, 0)\) and \((0, b)\), where \(a\) and \(b\) are non-zero divisors in \(A\) and \(B\) respectively, guarantee that \(A \times B\) always has zero divisors.
Key Concepts
Ring with UnityZero DivisorsRing ProductOrdered Pairs in Rings
Ring with Unity
A ring with unity is a fundamental concept in ring theory. It is a ring that not only includes all the elements needed for a ring, but also has a special element called "unity." This unity element is denoted by "1" and it acts very much like the number 1 does in multiplication. For any element \( a \) in the ring, the unity element satisfies the equations \( 1 \times a = a \times 1 = a \). This means that when you multiply any element by the unity, you get the element itself back.
Having a unity element can be very important for certain mathematical constructions and proofs. It ensures that the ring behaves in a way that is familiar to us when we think of multiplication. Many rings you are familiar with, like the ring of integers \( \mathbb{Z} \), are rings with unity.
This characteristic makes understanding and proving properties about these rings easier. It is particularly useful when working with ring products and exploring their behaviors.
Having a unity element can be very important for certain mathematical constructions and proofs. It ensures that the ring behaves in a way that is familiar to us when we think of multiplication. Many rings you are familiar with, like the ring of integers \( \mathbb{Z} \), are rings with unity.
This characteristic makes understanding and proving properties about these rings easier. It is particularly useful when working with ring products and exploring their behaviors.
Zero Divisors
Zero divisors are special elements in a ring. They are non-zero elements that, when multiplied by another non-zero element, give you zero. Let's break this down:
In the context of \( A \times B \), you are guaranteed to find these zero divisors because of the nature of the element operations we've defined. The presence of zero divisors is a signal that the ring has much complexity, and this is what makes ring theory so intriguing.
- An element \( a \) is a zero divisor if there is another element \( b \) such that \( a \cdot b = 0 \).
- Neither \( a \) nor \( b \) is zero themselves.
In the context of \( A \times B \), you are guaranteed to find these zero divisors because of the nature of the element operations we've defined. The presence of zero divisors is a signal that the ring has much complexity, and this is what makes ring theory so intriguing.
Ring Product
The concept of the ring product merges two rings into a new structure. Imagine you have two rings, \( A \) and \( B \). The ring product \( A \times B \) encompasses all the possible ordered pairs \( (a, b) \), where \( a \) comes from \( A \), and \( b \) from \( B \). These ordered pairs form a new set, where we define specific rules for addition and multiplication.
For instance:
For instance:
- Addition: \( (a, b) + (c, d) = (a+c, b+d) \)
- Multiplication: \( (a, b) \cdot (c, d) = (ac, bd) \)
Ordered Pairs in Rings
Ordered pairs provide a structured way to explore relationships between elements from different sets. In ring theory, when we write \( (a, b) \), we are specifying exactly which element from ring \( A \) (\( a \)) and ring \( B \) (\( b \)) we're interested in.
These ordered pairs are pivotal in constructing new rings like \( A \times B \). They tell us not only about individual elements but also how they interact across different rings. Within this framework, each element of the ordered pair follows the operations of its original ring, but also adheres to the operational rules of the pair, as defined in the ring product:
These ordered pairs are pivotal in constructing new rings like \( A \times B \). They tell us not only about individual elements but also how they interact across different rings. Within this framework, each element of the ordered pair follows the operations of its original ring, but also adheres to the operational rules of the pair, as defined in the ring product:
- Addition and multiplication rules, as discussed with the ring product.
- Showcases cross-interactions, like zero divisor creation.
Other exercises in this chapter
Problem 5
The set \(S\) of all the invertible elements in a ring is a multiplicative group.
View solution Problem 5
Show that the commutative law for addition need not be assumed in defining a ring with unity: it may be proved from the other axioms. [HINT: Use the distributiv
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By part 5, the set of all the nonzero elements in a field is a multiplicative group. Now use Lagrange's theorem to prove that in a finite field with \(m\) eleme
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Let \(A\) be any ring. Prove that if the additive group of \(A\) is cyclic, then \(A\) is a commutative ring
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