Monthly Archives: November 2016

25/11/2016 · 12:07 pm

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. $\square$

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Filed under Algebra

Tagged as commutativity, ring