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



(**)\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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