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: Introduction to the Problem:
The problem concerns Boolean rings, which are rings where every element is idempotent (i.e., \( x^2 = x \) for all \( x \in R \)). We need to determine a property that *always* holds for these rings.

Step 2: Core Strategy:
We'll use the idempotent property (\( x^2 = x \)) to demonstrate commutativity. We'll start by examining \( (x+y)^2 \).

Step 3: Detailed Proof and Explanation:
Let \( R \) be a ring where \( a^2 = a \) for all \( a \in R \). Let \( x, y \in R \). Since \( x+y \) is in \( R \), it's also idempotent: \[ (x+y)^2 = x+y \] Expand using the distributive property: \[ (x+y)(x+y) = x^2 + xy + yx + y^2 \] Since \( x^2 = x \) and \( y^2 = y \): \[ x + xy + yx + y = x+y \] Cancel \(x\) and \(y\) from both sides (using the additive cancellation property): \[ xy + yx = 0 \] This means \( xy = -yx \). Consider any \( x \in R \). We know \( x^2 = x \). From \( xy = -yx \), set \( y=x \): \[ xx = -xx \implies x^2 = -x^2 \] Since \( x^2=x \), then \( x = -x \), which means \( x+x=0 \) or \( 2x=0 \), for all \( x \in R \). The ring has characteristic 2. Because \( x = -x \), substitute this into \( xy = -yx \): \[ xy = (-y)x = y(-x) \] Not helpful. Since \( yx = -yx \), adding \(yx\) to both sides gives \(yx+yx=0\), thus \(2yx=0\), which we already know. Use \( -yx = yx \) since the characteristic is 2. From \( xy + yx = 0 \), we have \( xy = -yx \). The ring has characteristic 2, so \(a = -a\) for any element \(a\). Hence, \( -yx = yx \). Therefore, \( xy = yx \). The ring is commutative. Checking other options: - (B, D) Is it an integral domain? A Boolean ring is an integral domain only if it is \( \mathbb{Z}_2 \). Examples such as the ring of subsets (with symmetric difference and intersection) demonstrate a Boolean ring that is *not* an integral domain. Thus, 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: Conclusion:
A Boolean ring (where all elements are idempotent) is always commutative.