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commutativity theorems on rings
Since Wedderburn proved his celebrated theorem that any finite division ring is commutative, the interest in studying properties on a ring that would render the ring commutative dramatically increased. Below is a list of some of the so-called ``commutativity theorems'' on a ring, showing how much one can generalize the result that Wedderburn first obtained. In the list below, $R$ is assumed to be unital ring.
Theorem 1 In each of the cases below, $R$ is commutative:
- (Wedderburn's theorem) $R$ is a finite division ring.
- (Jacobson) If for every element $a\in R$ , there is a positive integer $n>1$ (depending on $a$ ), such that $a^n=a$ .
- (Jacobson-Herstein) For every $a,b\in R$ , if there is a positive integer $n>1$ (depending on $a,b$ ) such that $$(ab-ba)^n=ab-ba.$$
- (Herstein) If there is an integer $n>1$ such that for every element $a\in R$ such that $a^n-a\in Z(R)$ , the center of $R$ .
- (Herstein) If for every $a\in R$ , there is a polynomial $p\in \mathbb{Z}[X]$ ($p$ depending on $a$ ) such that $a^2p(a)-a\in Z(R)$ .
- (Herstein) If for every $a,b\in R$ , such that there is an integer $n>1$ (depending on $a,b$ ) with $$(a^n-a)b=b(a^n-a).$$
Some of the commutativity problems can be derived fairly easily, such as the following examples:
Theorem 2 If $R$ is a ring with $1$ such that $(ab)^2=a^2b^2$ for all $a,b\in R$ , then $R$ is commutative.
Proof. Let $a, b \in R$ . From the assumption, we have $((a+1)b)^2=(a+1)^2b^2$ . Expanding the LHS, we get $(ab)^2+(ab)b+ b(ab)+b^2$ . Expanding the RHS, we get $a^2b^2+2ab^2+b^2$ . Equating both sides and eliminating common terms, we have \begin{equation} bab=ab^2 \end{equation}Similarly, from $(a(b+1))^2=a^2(b+1)^2$ , we expand the equations and get $$(ab)^2+(ab)a+a(ab)+a^2=a^2b^2+2a^2b+a^2.$$ Hence \begin{equation} aba=a^2b \end{equation}Finally, expanding out $((a+1)(b+1))^2=(a+1)^2(b+1)^2$ and eliminating common terms, keeping in mind Equations (1) and (2) from above, we get $ab=ba$ . 
Corollary 3 If each element of a ring $R$ is idempotent, then $R$ is commutative.
Proof. If $R$ contains $1$ , then we can apply Theorem 2: for $(st)^2=st=s^2t^2$ for any $s,t\in R$ . Otherwise, we do the following trick: first $2s = (2s)^2 = 4s^2 = 4s$ , so that $2s=0$ for all $s\in R$ . Next, $s+t = (s+t)^2 = s^2 + st + ts + t^2 = s+st+ts+t$ , so $0=st+ts$ , which implies $st=st+(st+ts)=2st + ts=ts$ , and the result follows.
The corollary also follows directly from part 2 of Theorem 1. 
Bibliography
- 1
- I. N. Herstein, Noncommutative Rings, The Mathematical Association of America (1968).
None.