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

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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!).