Exercise 9, § 9

Let $A$ be a ring such that $x^3 = x$ for all $x \in A$. It is proposed to prove that $A$ is commutative.

(a) Show that $6A = \{0\}$ and that $2A$ and $3A$ are two-sided ideals such that $2A + 3A = A$ and $2A \cap 3A = \{0\}$. Deduce that it can be assumed that either $2A = \{0\}$ or $3A = \{0\}$.

(b) If $2A = \{0\}$, calculate $(1 + x)^3$, deduce that $x^2 = x$ for all $x \in A$ and conclude by means of Exercise 8.

(c) If $3A = \{0\}$, calculate $(x + y)^3$ and $(x - y)^3$, show that $x^2y + xyx + yx^2 = 0$ and left multiply by $x$; deduce that $xy - yx = 0$.

(d) Let $A$ be a ring such that $3A = \{0\}$ and $x^3 = x$ for all $x \in A$; define a set $I$ and an injective homomorphism of $A$ into $(\mathbf{Z}/3\mathbf{Z})^I$ (same method as in Exercise 8).