Posts Tagged ‘reduced ring’

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


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 standard ring theory textbooks. Here we only discuss a very special case of the theorem, i.e. when x^3=x for all x \in R.

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.  Clearly 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


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


we have 3x=-3x^2 and hence 3x is central. Therefore x = 3x-2x is central. \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 \}.


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


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


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.

Definition. Let R, \ R_i, \ i \in I, be rings. For every j \in I we let \pi_j : \prod_{i \in I} R_i \longrightarrow R_j be the natural projection. Then R is called a subdirect product of R_i, \ i \in I, if the following conditions are satisfied:

1) There exists an injective ring homomorphism f: R \longrightarrow \prod_{i \in I} R_i,

2) For every j \in I the map \pi_j f: R \longrightarrow R_j is surjective.

Note. Suppose that A_i, \ i \in I, are some ideals of R and put R_i = R/A_i. Then we can define f: R \longrightarrow \prod_{i \in I} R/A_i by f(r)=(r+ A_i)_{i \in I}. Clearly the second condition in the above definition is satisfied. Thus R is a subdirect product of R/A_i, \ i \in I, if and only if f is injective, i.e. \bigcap_{i \in I} A_i = \{0\}.

Remark 6. If P is a minimal prime ideal of the ring R, then S=R \setminus P is multiplicatively closed iff s_1s_2 \cdots s_k \neq 0, for all s_i \in S, \ k \in \mathbb{N}.

Proof. Suppose that s_1s_2 \cdots s_k \neq 0, for any s_1,s_2, \cdots, s_k \in S and k \in \mathbb{N}. Let T be the set of all elements of R which are a finite product of some elements of S. Clearly T is multiplicatively closed, S \subseteq T and S is multiplicatively closed iff S=T. So we’ll be done if we show that S=T. Let \mathcal{C}=\{A \lhd R: \ A \cap T=\emptyset \}. We have \mathcal{C} \neq \emptyset because (0) \in \mathcal{C}. Therefore, by Zorn’s lemma, (\mathcal{C}, \subseteq) has a maximal element Q and Q is a prime ideal of R. Since Q \cap T = \emptyset, we have Q \cap S = \emptyset and thus Q \subseteq P. Thus Q=P because P is a minimal prime. So P \cap T= \emptyset, which means T \subseteq S. Hence T=S. \ \Box

 Remark 7. If R is reduced and P \lhd R is a minimal prime, then R/P is a domain.

Proof. Clearly R/P is a domain iff S = R \setminus P is multiplicatively closed. Let T be as defined in Remark 6. By that remark, we only need to show that 0 \notin T. So suppose that s_1s_2 \cdots s_k = 0, for some s_1, s_2, \cdots , s_k \in S, where the integer k \geq 2 is assumed to be minimal. Then by, Remark 1, we have s_k R s_1s_2 \cdots s_{k-1} = \{0\}. Now, since P is prime, s_k R s_1 cannot be a subset of P because otherwise we’d have either s_k \in P or s_1 \in P, which is clearly nonsense. Thus s_k Rs_1 \cap S \neq \emptyset. Let s \in s_kRs_1 \cap S. Then

ss_2 \cdots s_{k-1} \in s_kRs_1s_2 \cdots s_{k-1} = \{0\}.

Hence ss_2 \cdots s_{k-1}=0, which contradicts the minimality of k. \ \Box

The Structure Theorem For Reduced Rings. A ring R is reduced iff R is a subdirect product of domains.

Proof. If R is reduced, then, by Remarks 5 and 7, R is a subdirect product of the domains R/P_i, \ i \in I, where \{P_i \}_{i \in I} is the set of all minimal prime ideals of R. Conversely, suppose that R is a subdirect product of domains R_i, \ i \in I and f: R \longrightarrow \prod_{i \in I} R_i is an injective ring homomorphism. Suppose that x \in R and x^2=0. Let f(x)=(x_i)_{\in I}. Then (0_{R_i})_{i \in I} = f(x^2)=(f(x))^2=(x_i^2)_{i \in I}. Thus x_i^2=0, for all i \in I, and so x_i = 0, for all i \in I, because every R_i is a domain. Hence x=0 and so R is reduced. \Box