## Posts Tagged ‘Jacobson theorem’

Throughout this post, $R$ is a ring with $1.$

Theorem (Jacobson). If $x^n=x$ for some integer $n > 1$ and all $x \in R,$ then $R$ is commutative.

In fact $n,$ in Jacobson’s theorem, doesn’t have to be fixed and could depend on $x,$ i.e. Jacobson’s theorem states that if for every $x \in R$ there exists an integer $n > 1$ such that $x^n=x,$ then $R$ is commutative. But we are not going to discuss that here.
In this post, we’re going to prove Jacobson’s theorem. Note that we have already proved the theorem for $n=3, 4$ (see here and here) and we didn’t need $R$ to have $1,$ we didn’t need that much ring theory either. But to prove the theorem for any $n > 1,$ we need a little bit more ring theory.

Lemma. If Jacobson’s theorem holds for division rings, then it holds for all rings with $1.$

Proof. Let $R$ be a ring with $1$ such that $x^n=x$ for some integer $n > 1$ and all $x \in R.$ Then clearly $R$ is reduced, i.e. $R$ has no non-zero nilpotent element. Let $\{P_i: \ i \in I\}$ be the set of minimal prime ideals of $R.$
By the structure theorem for reduced rings, $R$ is a subring of the ring $\prod_{i\in I}D_i,$ where $D_i=R/P_i$ is a domain. Clearly $x^n=x$ for all $x \in D_i$ and all $i \in I.$ But then, since each $D_i$ is a domain, we get $x=0$ or $x^{n-1}=1,$ i.e. each $D_i$ is a division ring. Therefore, by our hypothesis, each $D_i$ is commutative and hence $R,$ which is a subring of $\prod_{i\in I}D_i,$ is commutative too. $\Box$

Example. Show that if $x^5=x$ for all $x \in R,$ then $R$ is commutative.

Solution. By the lemma, we may assume that $R$ is a division ring.
Then $0=x^5-x=x(x-1)(x+1)(x^2+1)$ gives $x=0,1,-1$ or $x^2=-1.$ Suppose that $R$ is not commutative and choose a non-central element $x \in R.$ Then $x+1,x-1$ are also non-central and so $x^2=(x+1)^2=(x-1)^2=-1$ which gives $1=0,$ contradiction! $\Box$

Remark 1. Let $D$ be a division ring with the center $F.$ If there exist an integer $n \ge 1$ and $a_i \in F$ such that $x^n+a_{n-1}x^{n-1}+ \cdots + a_1x+a_0=0$ for all $x \in D,$ then $F$ is a finite field. This is obvious because the polynomial $x^n+a_{n-1}x^{n-1}+ \cdots + a_1x+a_0 \in F[x]$ has only a finite number of roots in $F$ and we have assumed that every element of $F$ is a root of that polynomial.

Remark 2. Let $D$ be a domain and suppose that $D$ is algebraic over some central subfield $F.$ Then $D$ is a division ring and if $0 \ne d \in D,$ then $F[d]$ is a finite dimensional division $F$-algebra.

Proof. Let $0 \ne d \in D.$ So $d^m +a_{m-1}d^{m-1}+ \cdots + a_1d+a_0=0$ for some integer $m \ge 1$ and $a_i \in F.$ We may assume that $a_0 \ne 0.$ Then $d(d^{m-1} + a_{m-1}d^{m-2}+ \cdots + a_1)(-a_0^{-1})=1$ and so $d$ is invertible, i.e. $D$ is a division ring.
Since $F[d]$ is a subring of $D,$ it is a domain and algebraic over $F$ and so it is a division ring by what we just proved. Also, since $d^m \in \sum_{i=0}^{m-1} Fd^i$ for some integer $m \ge 1,$ we have $F[d]=\sum_{i=0}^{m-1} Fd^i$ and so $\dim_F F[d] \le m. \ \Box$

Proof of the Theorem. By the above lemma, we may assume that $R$ is a division ring.
Let $F$ be the center of $R.$ By Remark 1, $F$ is finite. Since $R$ is a division ring, it is left primitive. Since every element of $R$ is a root of the non-zero polynomial $x^n-x \in F[x], \ R$ is a polynomial identity ring.
Hence, by the Kaplansky-Amtsur theorem, $\dim_F R < \infty$ and so $R$ is finite because $F$ is finite. Thus, by the Wedderburn’s little theorem, $R$ is a field. $\Box$

Advertisements

## Rings satisfying x^3 = x are commutative

Posted: December 13, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , , , ,

Throughout $R$ is a ring.

Theorem (Jacobson). If for every $x \in R$ there exists some $n > 1$ such that $x^n=x,$ then $R$ is commutative.

The proof of Jacobson’s theorem can be found in any standard ring theory textbooks. Note that $n,$ in Jacobson’s theorem, doesn’t have to be fixed, i.e. it could depend on $x.$ See this post for the proof of the theorem when $n$ is fixed. Here we only discuss a very special case of the theorem, i.e. when $n=3.$

Definitions. An element $x \in R$ is called idempotent if $x^2=x.$ The center of $R$ is

$Z(R)=\{x \in R: \ xy=yx \ \text{for all} \ y \in R \}.$

It is easy to see that $Z(R)$ is a subring of $R.$ An element $x \in R$ is called central if $x \in Z(R).$ Obviously $R$ is commutative iff $Z(R)=R,$ i.e. every element of $R$ is central.

Problem. Prove that if $x^3=x$ for all $x \in R,$ then $R$ is commutative.

Solution.  First note that $R$ is reduced, i.e. $R$ has no nonzero nilpotent element. For every $x \in R$ we have $(x^2)^2=x^4 = x^2$ and so $x^2$ is idempotent for all $x \in R.$ Hence, by Remark 3 in this post, $x^2$ is central for all $x \in R.$ Now, since

$(x^2+x)^2=x^4+2x^3+x^2=2x^2+2x$

we have $2x=(x^2+x)^2-2x^2$ and thus $2x$ is central. Also, since

$x^2+x=(x^2+x)^3=x^6+3x^5+3x^4+x^3=4x^2+4x,$

we have $3x=-3x^2$ and so $3x$ is central. Thus $x = 3x-2x$ is central and so $R$ is commutative.  $\Box$

A similar argument shows that if $x^4=x$ for all $x \in R,$ then $R$ is commutative (see here!).