commutativity theorems on rings

Since Wedderburn proved his celebrated theorem that any finite division ring is commutativePlanetmathPlanetmathPlanetmath, 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. 1.

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

  2. 2.

    (Jacobson) If for every elementMathworldMathworld aR, there is a positive integer n>1 (depending on a), such that an=a.

  3. 3.

    (Jacobson-Herstein) For every a,bR, if there is a positive integer n>1 (depending on a,b) such that

  4. 4.

    (Herstein) If there is an integer n>1 such that for every element aR such that an-aZ(R), the center of R.

  5. 5.

    (Herstein) If for every aR, there is a polynomialMathworldPlanetmathPlanetmath p[X] (p depending on a) such that a2p(a)-aZ(R).

  6. 6.

    (Herstein) If for every a,bR, such that there is an integer n>1 (depending on a,b) with


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=a2b2 for all a,bR, then R is commutative.


Let a,bR. From the assumptionPlanetmathPlanetmath, we have ((a+1)b)2=(a+1)2b2. Expanding the LHS, we get (ab)2+(ab)b+b(ab)+b2. Expanding the RHS, we get a2b2+2ab2+b2. Equating both sides and eliminating common terms, we have

bab=ab2 (1)

Similarly, from (a(b+1))2=a2(b+1)2, we expand the equations and get



aba=a2b (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 idempotentMathworldPlanetmathPlanetmath, then R is commutative.


If R contains 1, then we can apply Theorem 2: for (st)2=st=s2t2 for any s,tR. Otherwise, we do the following trick: first 2s=(2s)2=4s2=4s, so that 2s=0 for all sR. Next, s+t=(s+t)2=s2+st+ts+t2=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