# 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. $\square$

Filed under Algebra

## 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 $A$.