Question

A ring (R, +, .), where all elements are idempotent is always:

• A. a commutative ring
• B. not an integral domain
• C. a field
• D. an integral domain with unity

Hint:

This is a classic result in ring theory often called Jacobson's Commutativity Theorem for Boolean rings. The proof involves two key steps: first showing \(xy+yx=0\) and second showing that the ring has characteristic 2 (i.e., \(x+x=0\)). Combining these proves \(xy=yx\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The question is about a specific type of ring known as a Boolean ring. A ring is called a Boolean ring if every element is idempotent, meaning \( x^2 = x \) for all \( x \in R \). We need to determine which property always holds for such rings.

Step 2: Key Formula or Approach:
We will use the property \( x^2 = x \) for all elements to prove commutativity. Consider the expression \( (x+y)^2 \).

Step 3: Detailed Explanation:
Let \( R \) be a ring where \( a^2 = a \) for all \( a \in R \). Let \( x, y \in R \). Since \( x+y \) is also an element of \( R \), it must be idempotent: \[ (x+y)^2 = x+y \] Using the distributive property in the ring: \[ (x+y)(x+y) = x^2 + xy + yx + y^2 \] Since every element is idempotent, \( x^2 = x \) and \( y^2 = y \). So, \[ x + xy + yx + y = x+y \] Using the cancellation law for addition in the ring, we can subtract \(x\) and \(y\) from both sides: \[ xy + yx = 0 \] This means \( xy = -yx \). Now, consider any element \( x \in R \). We have \( x^2 = x \). From the property \( xy = -yx \), let's set \( y=x \): \[ xx = -xx \implies x^2 = -x^2 \] Since \( x^2=x \), this means \( x = -x \). This implies \( x+x=0 \), or \( 2x=0 \), for all \( x \in R \). The ring has characteristic 2. Since \( x = -x \), we can substitute this back into \( xy = -yx \): \[ xy = (-y)x = y(-x) \] No, that is not helpful. Since \( yx = -yx \), adding \(yx\) to both sides gives \(yx+yx=0\), this is \(2yx=0\), which we already knew. Let's use \( -yx = yx \) because the characteristic is 2. From \( xy + yx = 0 \), we have \( xy = -yx \). Since the characteristic of the ring is 2, for any element \(a\), we have \(a = -a\). So, \( -yx = yx \). Therefore, \( xy = yx \). This proves that the ring is always commutative. Let's check other options: - (B, D) Is it an integral domain? A Boolean ring is an integral domain only if it is \( \mathbb{Z}_2 \). For example, the ring of subsets of a set with more than one element under symmetric difference and intersection is a Boolean ring, but not an integral domain. So it is not always an integral domain. - (C) Is it a field? Only the trivial Boolean ring and \( \mathbb{Z}_2 \) are fields. So not always a field.

Step 4: Final Answer:
A ring where all elements are idempotent is always a commutative ring.