Characteristic of a Ring Satisfying $x^3=x$

Detailed Solution

We are given that $$ x^3=x $$ for every $x\in R$. We will prove that $$ 6\cdot1_R=0, $$ which implies that the characteristic of $R$ divides $6$. Step 1: Substitute $x+1$. Since every element satisfies the given identity, $$ (x+1)^3=x+1. $$ Expanding and using the fact that $1$ commutes with every element, $$ x^3+3x^2+3x+1=x+1. $$ Since $x^3=x$, we obtain $$ 3x^2+3x=0. $$ Thus, $$ 3x^2+3x=0. \tag{1} $$ Step 2: Substitute $x-1$. Similarly, $$ (x-1)^3=x-1. $$ Expanding gives $$ x^3-3x^2+3x-1=x-1. $$ Using $x^3=x$, we get $$ -3x^2+3x=0. \tag{2} $$ Step 3: Add equations (1) and (2). Adding yields $$ 6x=0 $$ for every $x\in R$. In particular, taking $x=1_R$, we obtain $$ 6\cdot1_R=0. $$ The characteristic of a ring with unity is the least positive integer $n$ such that $$ n\cdot1_R=0, $$ provided such an integer exists. Since $6\cdot1_R=0$, the characteristic must divide $6$. Therefore, $$ \boxed{\operatorname{char}(R)\mid6}. $$ Hence option B is correct. Why the other options are incorrect: Option A: A Boolean ring must satisfy $x^2=x$ for every element. The identity $x^3=x$ does not imply this. For example, in $\mathbb Z_3$, every element satisfies $x^3=x$, but $2^2=1\neq2$. Option C: The ring need not be a field. For example, $\mathbb Z_6$ satisfies $x^3=x$ for every element but is not a field. Option D: A definite conclusion follows from the identity: the characteristic divides $6$. Thus the correct option is $$ \boxed{\mathrm{B}}. $$