Posts Tagged ‘division ring’

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


We saw in part (2) that von Neumann regular rings live somewhere between semisimple and semiprimitive rings. The goal in this post is to prove a theorem of Armendariz and others which gives a necessary and sufficient condition for a ring to be both regular and reduced. This result extends Kaplansky’s result for commutative rings (see the corollary at the end of this post). We remark that a commutative von Neumann regular ring R is necessarily reduced. That is because if x^2=0  for some x \in R, then choosing y \in R with x=xyx we will get x=yx^2=0.

Definition . A von Neumann regular ring R is called strongly regular if R is reduced.

Theorem 1. (Armendariz, 1974) A ring  R with 1 is strongly regular if and only if R_M is a division ring for all maximal ideals M of Z(R).

Proof. Suppose first that R is strongly regular and let M be a maximal ideal of Z(R). Let 0 \neq s^{-1}x \in R_M. So tx \neq 0 for all t \in Z(R) \setminus M. Since R is regular, there exists some y \in R such that xyx = x. Then xy=e is an idempotent and thus e \in Z(R) because in a reduced ring every idempotent is central.  Since (1-e)x=0 we have 1-e \in M and hence e \in Z(R) \setminus M. Thus e^{-1}sy is a right inverse of s^{-1}x. Similarly f=yx \in Z(R) \setminus M and f^{-1}sy is a left inverse of s^{-1}x. Therefore s^{-1}x is invertible and hence R_M is a division ring. Conversely, suppose that R_M is a division ring for all maximal ideals M of Z(R). If R is not reduced, then there exists 0 \neq x \in R such that x^2=0.
Let I=\{s \in Z(R): \ sx = 0 \}. Clearly I is a proper ideal of Z(R) and hence I \subseteq M for some maximal ideal M of Z(R). But then (1^{-1}x)^2=0 in R_M, which is a division ring. Thus 1^{-1}x=0, i.e. there exists some s \in Z(R) \setminus M such that sx = 0, which is absurd. To prove that R is von Neumann regular, we will assume, to the contrary, that R is not regular. So there exists x \in R such that xzx \neq x for all z \in R. Let J= \{s \in Z(R): \ xzx=sx \ \text{for some} \ z \in R \}. Clearly J is a proper ideal of Z(R) and so J \subseteq M for some maximal ideal M of Z(R). It is also clear that if sx = 0 for some s \in Z(R), then s \in J because we may choose z = 0. Thus 1^{-1}x \neq 0 in R_M and hence there exists some y \in R and t \in Z(R) \setminus M such that 1^{-1}x t^{-1}y = 1. Therefore u(xy-t)=0 for some u \in Z(R) \setminus M. But then x(uy)x=utx and so ut \in J, which is nonsense. This contradiction proves that R must be regular. \Box

Corollary. (Kaplansky) A commutative ring R is regular if and only if R_M is a field for all maximal ideals M of R. \ \Box

At the end let me mention a nice property of strongly regular rings.

Theorem 2. (Pere Ara, 1996) If R is strongly regular and Ra+Rb=R, for some a, b \in R, then a+rb is a unit for some r \in R.

Problem. Let R be an infinite ring with 1 and let U(R) be the set of units of R. Prove that if R \setminus U(R) is finite, then R is a division ring.

Solution. Suppose, to the contrary, that there exists some 0 \neq x \in R \setminus U(R). First note that if I \neq R is a left or right ideal of R, then I is finite because otherwise I \cap U(R) \neq \emptyset and so I=R. Therefore Rx and xR cannot both be infinite. Suppose that Rx is finite and let I=\{r \in R: \ rx = 0 \}. Then I is a left ideal of R and I \neq R because x \neq 0. Hence I is finite. Now the R-module isomorphism R/I \cong Rx implies that R is finite. Contradiction!  

Let R be a ring and let M be a left R-module. Recall the following definitions:

1) M is called faithful if rM=(0) implies r=0 for any r \in R. In other words, M is called faithful if \text{ann}_RM = \{r \in R : \ rM=(0) \}=(0).
2) M is called simple if (0) and M are the only left R submodules of M.

Faithful and simple right R-modules are defined analogously.

Definition. A ring R is called left (resp., right) primitive if there exists a left (resp., right) R-module M which is both faithful and simple.

Remark. We will show later that left and right primitivity are not equivalent.  From now on, I will only consider left primitive rings. If a statement is true for left but not for right, I will mention that.

Example 1. If R is a ring and M is a left simple R-module, then R_1=R/\text{ann}_RM is a left primitive ring.  This is clear because M would be a faithful simple left R_1-module.

Example 2. Every simple ring is left primitive. That’s because we can choose a maximal left ideal \mathbf{m} of R and then M=R/\mathbf{m} would be a faithful simple left R-module. The reason that M is faithful is that \text{ann}_R M is a two-sided ideal contained in \mathbf{m} and therefore \text{ann}_R M = (0), because R is simple. One special case of this example is M_n(D), the ring of n \times n matrices with entries from a division ring D. If V is an infinite dimensional vector space over a field F, then \text{End}_F V is an example of a left primitive ring which is not simple [see Example 4 and this post].

Example 3. If R is a left primitive ring and 0 \neq e \in R an idempotent, then R_1=eRe is left primitive: let M be a faithful simple R-module. The claim is that M_1=eM is a faithful simple left R_1-module. Note that M_1 \neq (0) because e \neq 0 and M is faithful. Clearly M_1 is a left R_1-module.  To see why it’s faithful, let r_1 = ere \in R_1 with r_1M_1=(0). Then (0)=ere^2M=ereM=r_1M. So r_1=0, because M is faithful. To prove that M_1 is a simple R_1-module let 0 \neq x_1 \in M_1. We need to show that R_1x_1=M_1. Well, since x_1 = ex, for some x \in M, we have ex_1=ex=x_1. Thus R_1x_1=eRex_1=eRx_1=eM=M_1.

Example 4. Let D be a division ring and let M be a right vector space over D. Then R=\text{End}_D M is a left primitive ring. Here is why: M is a left R-module if we define fx = f(x), for all f \in R and x \in M. It is clear that M is faithful as a left R-module. To see why it is simple, let x,y \in M with x \neq 0. Let B=\{x_i: \ i \in I \} be a basis for M over D such that x=x_k \in B, for some k \in I. Define f \in R by f(\sum_{i \in I}x_id_i) = yd_k. Then f(x)=f(x_k)=y. So, we’ve proved that Rx = M, which shows that M is a simple R-module. If \dim_D M = n < \infty, then R \cong M_n(D), which we already showed its primitivity in Example 2. Note that if M was a “left” vector space over D with \dim_D M = n < \infty, then R would be isomorphic to the ring M_n(D^{op}) rather than the ring M_n(D).