Tag Archives: commutativity

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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Commuting matrices

 

This post arose from an attempt to solve a question in this past Waterloo pure mathematics PhD comprehensive exam.

Let k be an algebraically closed field. Let \mathcal M_n(k) denote the set of all n\times n matrices with entries in k. Let

\displaystyle J_\lambda:=\left(\begin{matrix}\lambda & 1 & \cdots & 0\\ 0 & \lambda & \cdots & 0\\\vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda\end{matrix}\right)\in\mathcal M_n(k)

be a Jordan block. Let e_1,\dots,e_n be the standard basis vectors, i.e. the j-th component of e_i is \delta_{ij}. Note the action of J_0 on the basis vectors: J_0e_i=e_{i-1} for each i (where we take e_0=0).

Suppose P\in\mathcal M_n(k) commutes with J_\lambda. Then

Pe_{i-1}=PJ_0e_i=P(J_\lambda-\lambda I)e_i=(J_\lambda-\lambda I)Pe_i=J_0Pe_i

i.e. if P_i:=Pe_i is the i-th column of P, then P_{i-1}=J_0P_i for each i. Thus

P=(P_1\mid\cdots\mid P_n)=(J_0^{n-1}P_n\mid\cdots\mid P_n)=(J_0^{n-1}\mid\cdots\mid I)P_n

i.e. if P_n=(a_1,\dots,a_n)^T, then P=a_1J_0^{n-1}+\cdots+a_nI, a polynomial in J_0. Further, since J_0=J_\lambda-\lambda I, it follows that P is a polynomial in J_\lambda. So we deduce:

Fact 1. P commutes with J_\lambda iff P is a polynomial in J_0.

Fact 2. P commutes with J_\lambda iff P is a polynomial in J_\lambda.

If we denote

\mathcal C(A):=\{B\in\mathcal M_n(k): AB=BA\}

for A\in\mathcal M_n(k), then we’ve just shown

\begin{aligned}\mathcal C(J_\lambda)&=\{f(J_\lambda):f\in k[X]\}\\ &=\{f(J_0):f\in k[X]\}\\ &=\mathcal C(J_0)\end{aligned}

Now let A\in\mathcal M_n(k) have minimal and characteristic polynomial (X-\lambda)^n. This means the Jordan normal form of A is J_\lambda. So there exists an invertible matrix M such that A=MJ_\lambda M^{-1}. Thus

\begin{aligned}\mathcal C(A)&=\{P\in\mathcal M_n(k): PA=AP\}\\ &=\{P\in\mathcal M_n(k):PMJ_\lambda M^{-1}=MJ_\lambda M^{-1}P\}\\ &=\{P\in\mathcal M_n(k): M^{-1}PMJ_\lambda=J_\lambda M^{-1}PM\}\\ &=\{P\in\mathcal M_n(k): M^{-1}PM=f(J_\lambda)\text{ for some }f\in k[X]\}\\ &=\{P\in\mathcal M_n(k): P=Mf(J_\lambda)M^{-1}\text{ for some }f\in k[X]\}\\ &=\{P\in\mathcal M_n(k): P=f(MJ_\lambda M^{-1})\text{ for some }f\in k[X]\}\\ &=\{P\in\mathcal M_n(k): P=f(A)\text{ for some }f\in k[X]\}\\ &=\{f(A): f\in k[X]\}\end{aligned}

Thus

Fact 3. P commutes with A iff P is a polynomial in A.

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