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:

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. (Jacobson-Herstein) For every $a,b\in R$, if there is a positive integer $n>1$ (depending on $a,b$) such that

   $${(ab-ba)}^n=ab-ba.$$

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

$${(ab)}^2+(ab)a+a(ab)+a^2=a^2{b}^2+2a^{2}b+a^2.$$

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 $2s={(2s)}^2=4s^2=4s$, 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. ∎

Title	commutativity theorems on rings
Canonical name	CommutativityTheoremsOnRings
Date of creation	2013-03-22 17:54:55
Last modified on	2013-03-22 17:54:55
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	12
Author	CWoo (3771)
Entry type	Theorem
Classification	msc 16B99