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Chapter 6: Problem 42

Show that the center of a division ring is a field.

Short Answer

Expert verified
The center of a division ring is a field because it is closed under addition and multiplication, contains inverses, and is commutative.

Step by step solution

01

Define Key Terms

A division ring, also known as a skew field, is a ring in which division is possible, meaning for any non-zero elements \(a\) and \(b\), there exists an element \(c\) such that \(a \cdot c = b\). A field is a commutative division ring, meaning it additionally requires commutativity of multiplication. The center of a division ring consists of elements that commute with every other element in the ring.
02

Identify Elements of the Center

To determine if the center is a field, we must show it is commutative under multiplication. The center, \(Z(D)\), of a division ring \(D\) consists of elements \(z\) such that for all \(x \in D\), \(zx = xz\).
03

Verify Closure Properties

For \(Z(D)\) to be a field, it must be closed under addition and multiplication. Take any two elements \(z_1, z_2 \in Z(D)\). Since each commutes with every \(x \in D\), their sum \(z_1 + z_2\) and product \(z_1 \cdot z_2\) also commute with every \(x \in D\), hence \(z_1 + z_2, z_1 \cdot z_2 \in Z(D)\).
04

Verify Existence of Inverses

To be a field, \(Z(D)\) must have multiplicative inverses for non-zero elements. For any non-zero \(z \in Z(D)\), \(z^{-1}\) exists in \(D\) because \(D\) is a division ring. Since \(z\) commutes with all elements, so does \(z^{-1}\), which implies \(z^{-1} \in Z(D)\). This confirms the existence of inverses.
05

Demonstrate Commutativity of Multiplication

It's inherent that any two elements of \(Z(D)\) commute because of the definition of the center: they commute with all elements and, therefore, with each other. Thus, \(z_1 \cdot z_2 = z_2 \cdot z_1\) for all \(z_1, z_2 \in Z(D)\).
06

Conclusion

All conditions for \(Z(D)\) to be a field have been satisfied: closure, existence of multiplicative inverses, and commutativity. Therefore, the center of a division ring is a field.

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Key Concepts

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

Division Ring
A division ring is a special type of ring where division is possible, but it doesn't necessarily have to be commutative, like a field does. Let's break this down more simply:
  • Imagine you have numbers in a set where you can add, subtract, multiply, and divide them just like normal.
  • However, unlike fields, the multiplication in a division ring might not meet commutativity. This means if you multiply two elements, the order in which you multiply them can affect the result.
  • For any non-zero elements, there's always a partner that will result in the identity element when multiplied together.
An example of a division ring is the set of quaternions, which showcases that while division is feasible, the multiplication isn't commutative. The beauty of division rings lies in their allowance for versatility without strict adherence to the commutative property; a feature that distinctly separates them from fields.
Commutativity
Commutativity refers to a fundamental property in mathematics that indicates changing the order of operations does not affect the result. It's crucial in understanding fields and division rings.
  • In mathematics, a set is commutative under a specific operation if swapping the order of the numbers used does not change the outcome.
  • For instance, in a commutative set under multiplication, if you have two numbers, say 3 and 4, then multiplying them will give the same product regardless of their order: \(3 \times 4 = 4 \times 3\).
  • We learn early on that numbers we use in arithmetic are commutative for both addition and multiplication.
In division rings, however, multiplication is not always commutative. Thus, the center of a division ring, where any of its elements commute with every other element, becomes particularly interesting. This center forms something akin to a 'safe zone' where we reclaim the familiar comfort of commutativity. This forms the basis for transforming a division ring into a field.
Multiplicative Inverses
A multiplicative inverse is an essential feature of division rings and fields that makes division possible.
  • It is a concept where every element (except zero) has a corresponding element that, when multiplied together, equals the identity element (generally 1).
  • For example, the multiplicative inverse of 5 is \(\frac{1}{5}\) because \(5 \times \frac{1}{5} = 1\).
  • This principle allows you to 'undo' multiplication in a ring, leading to the availability of division.
In a division ring, having a multiplicative inverse for every non-zero element is obligatory. When we look at the center of a division ring, the presence of multiplicative inverses for its non-zero elements is one of the pillars supporting its structure as a field. Within this center, all elements behave "nicely" under multiplication, forming a cohesive network that satisfies the properties we come to expect from a field.
Closure Properties
Closure properties are a cornerstone concept in abstract algebra, crucial for identifying fields among other algebraic structures.
  • A set is said to be closed under an operation if performing that operation on elements of the set always produces an element also within the set.
  • For instance, consider the positive integers under addition: adding any two positive integers always results in a positive integer, showcasing closure under addition.
  • Similarly, for a division ring to have a center that is a field, it must be closed under both addition and multiplication.
In the center of a division ring (designated as a field), when you add or multiply any two elements, the result remains within the center. This ensures that you have a well-defined structure resembling familiar numeric systems, where operations do not lead you outside of the system's boundaries. Closure helps maintain consistency and predictability within the set, essential traits for the field.

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Most popular questions from this chapter

Let \(R_{1}\) and \(R_{2}\) be commutative rings with unity. Show that the group of units \(U\left(R_{1} \times R_{2}\right)=U\left(R_{1}\right) \times U\left(R_{2}\right)\)

Show that the indicated rings are integral domains. $$ Q(\sqrt{2}, \sqrt{3})=\\{a+b \sqrt{2}+c \sqrt{3}+d \sqrt{2} \sqrt{3} \mid a, b, c, d \in Q\\} $$

Show that if \(D\) is an integral domain, then 0 is the only nilpotent element in \(D\).

Let \(a\) be a nilpotent element in a commutative ring \(R\) with unity. Show that (a) \(a=0\) or \(a\) is a zero divisor. (b) \(a x\) is nilpotent for all \(x \in R\). (c) \(1+a\) is a unit in \(R\). (d) If \(u\) is a unit in \(R\), then \(u+a\) is also a unit in \(R\).

Determine whether the indicated sets form a ring under the indicated operations. $$ S=\left\\{\left[\begin{array}{cc} a & b \\ 0 & a \end{array}\right] \mid a, b \in \mathbb{R}\right\\} \text { , under matrix addition and multiplication } $$
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