# Monthly Archives: November 2016

## Simple cases of Jacobson’s theorem

A celebrated theorem of Jacobson states that

Theorem. Let $R$ be a ring, not necessarily containing $1$. If, for each $a\in R$ there exists a positive integer $n$ such that $a^n=a$, then $R$ is commutative.

This is a very strong and difficult result (although not very useful in practice). However, we can obtain some special cases via elementary means.

Proposition 1. Let $R$ be a ring such that for each $a\in R$ we have $a^2=a$. Then $R$ is commutative.

Proof. Let $a,b\in R$. Then $a+b=(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b$, i.e., $ab=-ba$. Again, $a-b=(a-b)^2=a^2-ab-ba+b^2=a-ab-ba+b$, i.e., $ab=-ba+2b$. Thus $2b=0$, i.e., $b=-b$ for each $b\in R$. Thus $ab=-ba=ba$, as desired. $\square$

The next case is already considerably harder.

Proposition 2. Let $R$ be a ring such that for each $a\in R$ we have $a^3=a$. Then $R$ is commutative.

Proof. Let $a,b\in R$. Then $a+b=(a+b)^3$ shows that

$(*)\qquad\qquad\qquad a^2b+aba+ba^2+ab^2+bab+b^2a=0$,

and $a-b=(a-b)^3$ shows that

$a^2b+aba+ba^2=ab^2+bab+b^2a$.

Hence

$(**)\qquad\qquad\qquad\qquad 2(a^2b+aba+ba^2)=0$

for all $a,b\in R$.

Plugging $a=b$ into $(**)$ gives $6a=0$, i.e., $3a=-3a$ for each $a\in R$.

Plugging $b=a^2$ into $(*)$ gives $3(a^2+a)=0$, i.e., $3a^2=3a$ for each $a\in R$. Replacing $a$ by $a+b$ gives $3(ab+ba)=0$, i.e., $3(ab-ba)=0$.

Also, multiplying $(**)$ by $a$ first on the left and then on the right and then subtracting the two gives $2(ab-ba)=0$.

From the last two paragraphs we conclude that $ab-ba=0$ for all $a,b\in R$. $\square$

Corollary. Let $R$ be a ring such that for each $a\in R$ we have $a^n=a$ for some $n\le 3$. Then $R$ is commutative.

Proof. Note that if $a^n=a$ for some $n\le 3$ then $a^3=a$. Hence the result follows by Proposition 2. $\square$