Boolean Rings

# Boolean Rings

 Definition: A Boolean Ring is a ring $(R, +, \cdot)$ with the property that $a^2 = a$ for all $a \in R$.

The following proposition tells us that in Boolean rings, every $a \in R$ is its own additive inverse.

 Proposition 1: Let $(R, +, \cdot)$ be a Boolean ring. Then $a + a = 0$ for all $a \in R$.
• Proof: Let $a \in R$. Then $(a + a) \in R$ and since $R$ is a Boolean ring we have that:
(1)
\begin{align} \quad (a + a) &= (a + a)^2 \\ &= a^2 + a^2 + a^2 + a^2 \\ &= a + a + a + a \\ \end{align}
• So $a + a = 0$. $\blacksquare$

The next proposition characterizes every Boolean ring as a commutative ring.

 Proposition 2: Let $(R, +, \cdot)$ be a Boolean ring. Then $(R, +, \cdot)$ is a commutative ring.
• Proof: Let $a, b \in R$. Since $R$ is a Boolean ring we have that $(a + b)^2 = a + b$. Expanding the lefthand side of this equation yields:
(2)
\begin{align} \quad (a + b)^2 &= (a + b)(a + b) \\ &= a^2 + ab + ba + b^2 \\ &= a^2 + ab + ba + b^2 \\ &= a + ab + ba + b \end{align}
• Therefore:
(3)
\begin{align} \quad a + b = a + ab + ba + b \end{align}
• Or equivalently, $ab + ba = 0$. From Proposition 1 we also have that $ab + ab = 0$. So subtracting the equation $ab + ba = 0$ from $ab + ab = 0$ gives us:
(4)
\begin{align} 0 &= (ab + ab) - (ab + ba) &= ab - ba \end{align}
• Therefore $ab = ba$. Since this holds true for all $a, b \in R$ we conclude that $R$ is a commutative ring. Thus every Boolean ring is a commutative ring. $\blacksquare$

## Example 1

The simplest examples of a Boolean ring is the ring $\mathbb{Z}/2\mathbb{Z}$. Indeed, we see that:

(5)
\begin{align} \quad [0]^2 &= [0] \\ \quad [1]^2 &= [1][1] = [1] \\ \end{align}