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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:

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 $1$ such that $(ab)^{2}=a^{2}b^{2}$ for all $a,b\in R$, then $R$ is commutative.

###### Proof.

Let $a,b\in R$. From the assumption, we have $((a+1)b)^{2}=(a+1)^{2}b^{2}$. Expanding the LHS, we get $(ab)^{2}+(ab)b+b(ab)+b^{2}$. Expanding the RHS, we get $a^{2}b^{2}+2ab^{2}+b^{2}$. Equating both sides and eliminating common terms, we have

$bab=ab^{2}$ | (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. ∎

# References

- 1
I. N. Herstein,
*Noncommutative Rings*, The Mathematical Association of America (1968).

## Mathematics Subject Classification

16B99*no label found*

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## Comments

## commutativity from nilpotence

Could someone please include commutativity theorems from nilpotence or infer from the existing theorems of this entry? It would be a nice complement to this entry!

Thank you!