Posts Tagged ‘reduced 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

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Let R be a ring, which may or may not have 1. We proved in here that if x^3=x for all x \in R, then R is commutative.  A similar approach shows that if x^4=x for all x \in R, then R is commutative.

Problem. Prove that if x^4=x for all x \in R, then R is commutative.

Solution. Clearly R is reduced, i.e. R has no nonzero nilpotent element. Note that 2x=0 for all x \in R because x=x^4=(-x)^4=-x. Hence x^2+x is an idempotent for every x \in R because

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

Thus x^2+x is central for all x \in R, by Remark 3 in this post.  Therefore (x^2+y)^2+x^2+y is central for all x,y \in R. But

(x^2+y)^2+x^2+y=x^2+x+y^2+y+ x^2y+yx^2

and hence x^2y+yx^2 is central. Therefore (x^2y+yx^2)x^2=x^2(x^2y+yx^2) which gives us xy=yx. \ \Box

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!).

Problem. (McCoy) Let R be a commutative ring with identity and let f(x)= \sum_{i=0}^n a_ix^i \in R[x].

1) Prove that f(x) is a zero-divisor if and only if there exists some 0 \neq c \in R such that cf(x) = 0.

2) Prove that if R is reduced and f(x)g(x)=0 for some g(x)=\sum_{i=0}^m b_ix^i \in R[x], then a_ib_j=0 for all 0 \leq i \leq n and 0 \leq j \leq m.

Solution. 1) If there exists 0 \neq c \in R such that cf(x)=0, then clearly f(x) is a zero-divisor of R[x]. For the converse, let g(x)=\sum_{i=0}^m b_ix^i, \ b_m \neq 0, be a polynomial with minimum degree such that f(x)g(x)=0. I will show that m = 0. So, suppose to the contrary, that m \geq 1. If a_jg(x)=0, for all j, then a_jb_m=0, for all j, and so b_mf(x)=0 contradicting the minimality of m because \deg b_m = 0 < m. So we may assume that the set \{j: \ a_jg(x) \neq 0 \} is non-empty and so we can let

\ell=\max \{j : \ a_jg(x) \neq 0 \}.

Then

0=f(x)g(x)=(a_{\ell}x^{\ell} + \cdots + a_0)(b_mx^m + \cdots + b_0).

Thus

a_{\ell}b_m=0 and so a_{\ell}g(x)=a_{\ell}b_{m-1}x^{m-1} + \cdots + a_{\ell}b_0.

Hence \deg a_{\ell}g(x) < m=\deg g. But we have f(x)(a_{\ell}g(x))=a_{\ell}f(x)g(x)=0, which is impossible because g(x) was supposed to be a polynomial with minimum degree satisfying f(x)g(x)=0.

2) The proof of this part is by induction over i+j. It is obvious from f(x)g(x)=0 that a_0b_0=0. Now let 0 < \ell \leq m+n and suppose that a_rb_s=0 whenever 0 \leq r+s < \ell. We need to show that a_rb_s=0 whenever r+s=\ell. So suppose that r+s=\ell. The coefficient of x^{\ell} in f(x)g(x) is clearly

0=\sum_{i < r, \ i+j=\ell}a_ib_j+ a_rb_s + \sum_{i > r, \ i+j=\ell} a_ib_j,

which after multiplying both sides by a_rb_s gives us

\sum_{i < r, \ i+j=\ell} a_rb_sa_ib_j+ (a_rb_s)^2 + \sum_{i > r, \ i+j=\ell} a_rb_sa_ib_j=0.

Call this (1). Now in the first sum in (1), since i < r, we have i+s < r+s=\ell and hence by the induction hypothesis a_ib_s=0. Thus a_rb_sa_ib_j=0. So the first sum in (1) is 0. In the second sum in (1), since i > r and i+j=r+s=\ell, we have j < s. Therefore by the induction hypothesis a_rb_j=0 and hence a_rb_sa_ib_j=0. So the second sum in (1) is also 0. Thus (1) becomes (a_rb_s)^2=0 and so, since R is reduced, a_rb_s=0. \ \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.

Definition 1. A ring R is called Dedekind-finite if \forall a,b \in R: \ ab=1 \Longrightarrow ba=1.

Remark 1. Some trivial examples of Dedekind-finite rings: commutative rings, any direct product of Dedekind-finite rings, any subring of a Dedekind-finite ring.

Definition 2. A ring R is called reversible if \forall a,b \in R : \ ab = 0 \Longrightarrow ba = 0.

Example 1. Every reversible ring R is Dedekind-finite. In particular, reduced rings are Dedekind-finite.

Proof. Suppose that ab=1 for some a,b \in R. Then (ba-1)b=b(ab)-b=0 and thus b(ba-1)=0. So b^2a=b and hence ab^2a=ab=1. It follows that ba=(ab^2a)ba=(ab^2)(ab)a=ab^2a=1. So R is Dedekind-finite. Finally, note that every reduced ring is reversible because if ab=0, for some a,b \in R, then (ba)^2=b(ab)a=0 and thus ba=0. \Box

Example 2. Every (left or right) Noetherian ring R is Dedekind-finite.

Proof. We will assume that R is left Noetherian. Suppose that ab=1 for some a,b \in R. Define the map f: R \longrightarrow R by f(r)=rb. Clearly f is an R-module homomorphism and f is onto because f(ra)=(ra)b=r(ab)=r, for all r \in R. Now we have an ascending chain of left ideals of R

\ker f \subseteq \ker f^2 \subseteq \cdots.

Since R is left Noetherian, this chain stabilizes at some point, i.e. there exists some n such that \ker f^n = \ker f^{n+1}. Clearly f^n is onto because f is onto. Thus f^n(c)=ba-1 for some c \in R. Then

f^{n+1}(c)=f(ba-1)=(ba-1)b=b(ab)-b=0.

Hence c \in \ker f^{n+1}=\ker f^n and therefore ba-1=f^n (c) = 0. \Box

Example 3. Finite rings are obviously Noetherian and so Dedekind-finite by Example 2. More generally:

Example 4. If the number of nilpotent elements of a ring is finite, then the ring is Dedekind-finite. See here.

Note that Example 4 implies that every reduced ring is Dedekind-finite; a fact that we proved in Example 1.

Example 5. Let k be a field and let R be a finite dimensional k-algebra. Then R is Dedekind-finite.

Proof. Every left ideal of R is clearly a k-vector subspace of R and thus, since \dim_k R < \infty, any ascending chain of left ideals of R will stop at some point. So R is left Noetherian and thus, by Example 2, R is Dedekind-finite. \Box

Remark 2. Two important cases of Example 5 are M_n(R), the ring of n \times n matrices over a field, and, in general, semisimple rings. As a trivial result, M_n(R) is Dedekind-finite for any commutative domain R because M_n(R) is a subring of M_n(Q(R)), where Q(R) is the quotient field of R.
So the ring of n \times n matrices, where n \geq 2, over a field is an example of a Dedekind-finite ring which is not reversible, i.e. the converse of Example 1 is not true. Now let R_i = \mathbb{Z}, \ i \geq 1. Then R= \prod_{i=1}^{\infty} R_i is obviously Dedekind-finite but not Noetherian. So the converse of Example 2 is not true.

Example 6 and Example 7 are two generalizations of Example 5.

Example 6. Every algebraic algebra R over a field k is Dedekind-finite.

Proof. Suppose that ab=1 for some a,b \in R. Since R is algebraic over k, there exist integers n \geq m \geq 0 and some \alpha_i \in k with \alpha_n \alpha_m \neq 0 such that \sum_{i=m}^n \alpha_i b^i = 0. We will assume that n is as small as possible. Suppose that m \geq 1. Then, since ab=1, we have

\sum_{i=m}^n \alpha_i b^{i-1}=a \sum_{i=m}^n \alpha_i b^i = 0,

which contradicts the minimality of n. So m = 0. Let c = -\alpha_0^{-1}\sum_{i=1}^n \alpha_i b^{i-1} and see that bc=cb=1. But then a=a(bc)=(ab)c=c and therefore ba=bc=1. \ \Box

Remark 3. Regarding Examples 5 and 6, note that although any finite dimensional k-algebra R is algebraic over k, but R being algebraic over k does not necessarily imply that R is finite dimensional over k. For example, if \overline{\mathbb{Q}} is the algebraic closure of \mathbb{Q} in \mathbb{C}, then it is easily seen that \dim_{\mathbb{Q}} \overline{\mathbb{Q}}=\infty. Thus the matrix ring R = M_n(\overline{\mathbb{Q}}) is an algebraic \mathbb{Q}-algebra which is not finite dimensional over \mathbb{Q}. So, as a \mathbb{Q}-algebra, R is Dedekind-finite by  Example 6 not Example 5.

Example 7. Every PI-algebra R is Dedekind-finite.

Proof. Let J(R) be the Jacobson radical of R. If J(R)=\{0\}, then R is a subdirect product of primitive algebras R/P_i, where P_i are the primitive ideals of R. Since R is PI, each R/P_i is PI too and thus, by Kaplansky’s theorem, R/P_i is a matrix ring over some division algebra and thus Dedekind-finite by Example 2. Thus \prod R/P_i is Dedekind-finite and so R, which is a subalgebra of \prod R/P_i, is also Dedekind-finite. For the general case, let S=R/J(R). Now, S is PI, because R is PI, and J(S)=\{0\}. Therefore, by what we just proved, S is Dedekind-finite. Suppose that ab = 1 for some a,b \in R and let c,d be the image of a,b in S respectively. Clearly cd=1_S and so dc=1_S. Thus 1-ba \in J(R) and so ba=1-(1-ba) is invertible. Hence there exists e \in R such that e(ba)=1. But then eb=(eb)ab=e(ba)b=b and hence ba=(eb)a=e(ba)=1. \Box

Schur’s lemma states that if A is a simple R module, then \text{End}_R(A) is a division ring. A similar easy argument shows that:

Example 6. For simple R-modules A \ncong B we have \text{Hom}_R(A,B)=\{0\}.

Let’s generalize Schur’s lemma: let M be a finite direct product of simple R-submodules. So M \cong \bigoplus_{i=1}^k M_i^{n_i}, where each M_i is a simple R-module and M_i \ncong M_j for all i \neq j. Therefore, by Example 6 and Theorem 1, \text{End}_R(M) \cong \bigoplus_{i=1}^k \mathbb{M}_{n_i}(D_i), where D_i = \text{End}_R(M_i) is a division ring by Schur’s lemma. An important special case is when R is a semisimple ring. (Note that simple submodules of a ring are exactly minimal left ideals of that ring.)

Theorem 2. (Artin-Wedderburn) Let R be a semisimple ring. There exist a positive integer k and division rings D_i, \ 1 \leq i \leq , such that R \cong \bigoplus_{i=1}^k \mathbb{M}_{n_i}(D_i).

 Proof. Obvious, by Example 1 and the above discussion. \Box

Some applications of Theorem 2.

1. A commutative semisimple ring is a finite direct product of fields.

2. A reduced semisimple ring is a finite direct product of division rings.

3. A finite reduced ring is a finite direct product of finite fields.