Properties of Boolean rings

Detailed Solution

For any $x\in R$, apply idempotency to $x+x$: $(x+x)^2=x+x$. Expanding, $(x+x)(x+x)=4x^2=4x$ (using $x^2=x$), so $4x=2x$, giving $2x=0$ for every $x$ — hence $R$ has characteristic $2$. For commutativity, expand $(x+y)^2=x+y$ for any $x,y\in R$: $x^2+xy+yx+y^2=x+y$, and since $x^2=x$, $y^2=y$, this simplifies to $xy+yx=0$, i.e. $xy=-yx$. Since the characteristic is $2$, $-1=1$, so $xy=yx$ — proving $R$ is commutative. This confirms (C) and rules out (A) (characteristic is $2$, not $3$) and (B) (commutativity is forced, not merely "need not" hold). Option (D) is false: $R=\mathbb Z_2\times\mathbb Z_2$ is a Boolean ring (every element satisfies $x^2=x$) but is not a field, since $(1,0)(0,1)=(0,0)$ shows it has zero divisors. Thus the correct option is $\boxed{\mathrm{C}}$.