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 socalled “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:

1.
(Wedderburn’s theorem) $R$ is a finite division ring.

2.
(Jacobson) If for every element^{} $a\in R$, there is a positive integer $n>1$ (depending on $a$), such that ${a}^{n}=a$.

3.
(JacobsonHerstein) For every $a,b\in R$, if there is a positive integer $n>1$ (depending on $a,b$) such that
$${(abba)}^{n}=abba.$$ 
4.
(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 $R$.

5.
(Herstein) If for every $a\in R$, there is a polynomial^{} $p\in \mathbb{Z}[X]$ ($p$ depending on $a$) such that ${a}^{2}p(a)a\in Z(R)$.

6.
(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 $\mathrm{1}$ such that ${\mathrm{(}a\mathit{}b\mathrm{)}}^{\mathrm{2}}\mathrm{=}{a}^{\mathrm{2}}\mathit{}{b}^{\mathrm{2}}$ for all $a\mathrm{,}b\mathrm{\in}R$, then $R$ is commutative.
Proof.
Let $a,b\in R$. From the assumption^{}, we have ${((a+1)b)}^{2}={(a+1)}^{2}{b}^{2}$. Expanding the LHS, we get ${(ab)}^{2}+(ab)b+b(ab)+{b}^{2}$. Expanding the RHS, we get ${a}^{2}{b}^{2}+2a{b}^{2}+{b}^{2}$. Equating both sides and eliminating common terms, we have
$$bab=a{b}^{2}$$  (1) 
Similarly, from ${(a(b+1))}^{2}={a}^{2}{(b+1)}^{2}$, we expand the equations and get
$${(ab)}^{2}+(ab)a+a(ab)+{a}^{2}={a}^{2}{b}^{2}+2{a}^{2}b+{a}^{2}.$$ 
Hence
$$aba={a}^{2}b$$  (2) 
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 ${(st)}^{2}=st={s}^{2}{t}^{2}$ for any $s,t\in R$. Otherwise, we do the following trick: first $2s={(2s)}^{2}=4{s}^{2}=4s$, 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. ∎
References
 1 I. N. Herstein, Noncommutative Rings, The Mathematical Association of America (1968).
Title  commutativity theorems on rings 

Canonical name  CommutativityTheoremsOnRings 
Date of creation  20130322 17:54:55 
Last modified on  20130322 17:54:55 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  12 
Author  CWoo (3771) 
Entry type  Theorem 
Classification  msc 16B99 