Triangular Matrix Representations and Triangular Matrix Extensions

Abstract

In this chapter, generalized triangular matrix representations are discussed by introducing the concept of a set of left triangulating idempotents. A criterion for a ring with a complete set of triangulating idempotents to be quasi-Baer is provided. A structure theorem for a quasi-Baer ring with a complete set of triangulating idempotents is shown using complete triangular matrix representations. A number of well known results follow as consequences of this useful structure theorem. The results which follow as a consequence include Levy’s decomposition theorem of semiprime right Goldie rings, Faith’s characterization of semiprime right FPF rings with no infinite set of central orthogonal idempotents, Gordon and Small’s characterization of piecewise domains, and Chatters’ decomposition theorem of hereditary noetherian rings. A result related to Michler’s splitting theorem for right hereditary right noetherian rings is also obtained as an application. The Baer, the quasi-Baer, the FI-extending, and the strongly FI-extending properties of (generalized) triangular matrix rings are discussed. A sheaf representation of quasi-Baer rings is obtained as an application of the results of this chapter.

A ring R is said to have a generalized triangular matrix representation if R is ring isomorphic to a generalized triangular matrix ring

$$\begin{bmatrix} R_1 & R_{12}&\cdots& R_{1n}\\ 0 & R_2 &\cdots&R_{2n}\\ \vdots& \vdots&\ddots&\vdots\\ 0 & 0 &\cdots& R_n \end{bmatrix} , $$

where each R _i is a ring and R _ij is an (R _i ,R _j )-bimodule for i<j, and the matrices obey the usual rules for matrix addition and multiplication. Generalized triangular matrix representations provide an effective tool in the investigation of the structures of a wide range of rings. In this chapter, these representations, in an abstract setting, are discussed by introducing the concept of a set of left triangulating idempotents.

The importance and applicability of the concept of a generalized triangular matrix representation can be seen from: (1) for any right R-module M, the generalized triangular matrix ring

$$\begin{bmatrix} S \, & \, M\\ 0 \,& \, R \end{bmatrix} , $$

where S=End(M), completely encodes the algebraic information of M into a single ring; (2) a ring R is ring isomorphic to

$$\begin{bmatrix} R_1 \, & \, R_{12}\\ 0 \,& \, R_2 \end{bmatrix} , $$

where R ₁≠0 and R ₂≠0 if and only if there exists e∈S _ℓ (R) with e≠0 and e≠1. From (2), we see that there is a natural connection between quasi-Baer rings and modules and generalized triangular matrix representation, since the “e” in Proposition 3.2.4(ii) is in S _ℓ (R) and the “f” in Proposition 4.6.3(ii) is in S _ℓ (End(M)).

In a manner somewhat analogous to determining a matrix ring by a set of matrix units (see 1.1.16), a generalized triangular matrix ring is determined by a set of left (or right) triangulating idempotents. The existence of a set of left triangulating idempotents does not depend on any specific conditions on a ring (e.g., {1} is a set of left triangulating idempotents); however, if the ring satisfies a mild finiteness condition, then such a set can be refined to a certain set of left triangulating idempotents in which each diagonal ring R _i has no nontrivial generalized triangular matrix representation. When this occurs, the generalized triangular matrix representation is said to be complete.

Complete triangular matrix representations and left triangulating idempotents are applied to get a structure theorem for a certain class of quasi-Baer rings (see Theorem 5.4.12). A number of well known results follow as consequences of this structure theorem. These include Levy’s decomposition theorem of semiprime right Goldie rings, Faith’s characterization of semiprime right FPF rings with no infinite set of central orthogonal idempotents, Gordon and Small’s characterization of piecewise domains, and Chatters’ decomposition theorem of hereditary Noetherian rings.

Further, a sheaf representation of quasi-Baer rings is studied as another application of our results of this chapter. Also the Baer, the quasi-Baer, the FI-extending, and the strongly FI-extending properties of (generalized) triangular matrix rings are discussed. Most results of Sects. 5.1, 5.2, and 5.3 are applicable to an algebra over a commutative ring.

5.1 Triangulating Idempotents

In this section, some basic properties of triangulating idempotents are discussed. Then a result showing the connection between triangulating idempotents and generalized triangular matrix rings is presented.

Definition 5.1.1

Let R be a ring. An ordered set {b ₁,…,b _n } of nonzero distinct idempotents in R is called a set of left triangulating idempotents of R if the following conditions hold:

• (i) 1=b ₁+⋯+b _n ;

• (ii) b ₁∈S _ℓ (R);

• (iii) b _k+1∈S _ℓ (c _k Rc _k ), where c _k =1−(b ₁+⋯+b _k ), for 1≤k≤n−1.

Similarly, we define a set of right triangulating idempotents of R by using part (i) in the preceding, b ₁∈S _r (R), and b _k+1∈S _r (c _k Rc _k ). By condition (iii) of Definition 5.1.1, a set of left (right) triangulating idempotents is a set of orthogonal idempotents.

Definition 5.1.2

A set {b ₁,…,b _n } of left (right) triangulating idempotents of R is said to be complete if each b _i is semicentral reduced.

Theorem 5.1.3

Let {b ₁,…,b _n } be an ordered set of nonzero idempotents of R. Then the following are equivalent.

• (i) {b ₁,…,b _n } is a set of left triangulating idempotents.

• (ii) b ₁+⋯+b _n =1 and b _j Rb _i =0, for all i<j≤n.

Proof

(i)⇒(ii) By definition, b ₁+⋯+b _n =1. As b ₂∈(1−b ₁)R(1−b ₁) and b ₁∈S _ℓ (R), b ₂ b ₁=0 and b ₂ Rb ₁=b ₂ b ₁ Rb ₁=0. Similarly we obtain b _j Rb ₁=0, for all j>1. By assumption b ₂∈S _ℓ ((1−b ₁)R(1−b ₁)) and {b ₁,…,b _n } is orthogonal, thus for j>2,

$$\begin{aligned} b_jRb_2 = & b_jR(1-b_1)b_2 =b_j(b_1R+(1-b_1)R)(1-b_1)b_2\\ =&b_j(1-b_1)R(1-b_1)b_2 = b_jb_2(1-b_1)R(1-b_1)b_2\\ =&0. \end{aligned}$$

Continue the process, using (1−b ₁−b ₂)R(1−b ₁−b ₂) in the next step, and so on, to get b _j Rb _i =0 for all i<j≤n.

(ii)⇒(i) Note that (1−b ₁)Rb ₁=(b ₂+⋯+b _n )Rb ₁=0. So b ₁∈S _ℓ (R) by Proposition 1.2.2. Now b ₂∈(1−b ₁)R(1−b ₁) as b ₂(1−b ₁)=b ₂−b ₂ b ₁=b ₂ and (1−b ₁)b ₂=b ₂. Also (1−b ₁−b ₂)(1−b ₁)=b ₃+b ₄+⋯+b _n . Therefore $(1-b_{1}-b_{2})[(1-b_{1})R(1-b_{1})]b_{2} = \sum_{i=3}^{n}b_{i}R(1-b_{1})b_{2} = \sum_{i=3}^{n}b_{i}Rb_{2} = 0$. So b ₂∈S _ℓ ((1−b ₁)R(1−b ₁)) by Proposition 1.2.2. Continuing this process yields the desired result. □

Theorem 5.1.4

R has a (resp., complete) set of left triangulating idempotents if and only if R has a (resp., complete) generalized triangular matrix representation.

Proof

Let {b ₁,…,b _n } be a set of left triangulating idempotents of R. Using Theorem 5.1.3 and a routine argument shows that the map

$$\theta: R\rightarrow\begin{bmatrix} b_1Rb_1 & b_1Rb_2&\cdots&b_1Rb_n\\ 0&b_2Rb_2&\cdots&b_2Rb_n\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&b_nRb_n\end{bmatrix} $$

defined by θ(r)=[b _i rb _j ] is a ring isomorphism, where [b _i rb _j ] is the matrix whose (i,j)-position is b _i rb _j . Conversely, assume that

$$\phi: R\rightarrow\begin{bmatrix} R_1 & R_{12}&\cdots&R_{1n}\\ 0&R_2&\cdots&R_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&R_n\end{bmatrix} $$

is a ring isomorphism. Then {ϕ ⁻¹(e ₁₁),…,ϕ ⁻¹(e _nn )} is a set of left triangulating idempotents of R by a routine calculation, where e _ii is the matrix with $1_{R_{i}}$ in the (i,i)-position and 0 elsewhere. □

Lemma 5.1.5

(i) S _ℓ (eRe)⊆S _ℓ (R) for e∈S _ℓ (R).

(ii) f S _ℓ (R)f⊆S _ℓ (fRf) for f ²=f∈R.

(iii) Let e∈S _ℓ (R). If f is a primitive idempotent of R such that efe≠0, then efe is a primitive idempotent in eRe and fef=f.

Proof

(i) For g∈S _ℓ (eRe), gRg=geReg=eReg=Rg. So g∈S _ℓ (R).

(ii) Let g∈S _ℓ (R) and r∈R. Then (fgf)(frf)(fgf)=(ff)(frf)(fgf). Thus (fgf)(frf)(fgf)=(frf)(fgf). So fgf∈S _ℓ (fRf).

(iii) Note that 0≠efe=fe=fefe, so fef≠0 and (fef)²=fef. As f is primitive, fef=f. To show that efe is a primitive idempotent of eRe, we note that (efe)(efe)=e(fef)e=efe. Let 0≠h ²=h∈(efe)(eRe)(efe). Since e∈S _ℓ (R), he=h,fh=h, so hf=fhf, and thus (hf)(hf)=hf. As hf=0 implies that h=hefe=hfe=0, hf is a nonzero idempotent in fRf. Thus, hf=f since f is a primitive idempotent. Note that (fe)²=fe and h∈(efe)(eRe)(efe), so h=hefe=hfe=fe=efe. Thus, efe is a primitive idempotent eRe. □

Lemma 5.1.6

(i) If h is a ring homomorphism from a ring R to a ring A, then h(S _ℓ (R))⊆S _ℓ (h(R)).

(ii) Assume that e∈S _ℓ (R)∪S _r (R) and f∈S _ℓ (eRe)∪S _r (eRe). Then the map h:R→fRf, defined by h(r)=frf for r∈R, is a ring epimorphism.

Proof

(i) The proof is routine.

(ii) Say x,y∈R. Since e∈S _ℓ (R)∪S _r (R) and f∈S _ℓ (eRe)∪S _r (eRe),

$$fxyf=fexyef=fexeyef = fexefeyef = fxfyf. $$

Therefore, h(xy)=h(x)h(y). □

Proposition 5.1.7

Let {b ₁,…,b _n } be a set of left triangulating idempotents of R. Then:

• (i) c _k ∈S _r (R),  k=1,…,n−1, where c _k =1−(b ₁+⋯+b _k ).

• (ii) b ₁+⋯+b _k ∈S _ℓ (R), k=1,…,n.

• (iii) The map h _j :R→b _j Rb _j , defined by h _j (r)=b _j rb _j for all r∈R, is a ring epimorphism.

Proof

(i) Recall that b ₁∈S _ℓ (R) implies c ₁=1−b ₁∈S _r (R) by Proposition 1.2.2. As b ₂∈S _ℓ (c ₁ Rc ₁), c ₂=1−b ₁−b ₂∈S _r (c ₁ Rc ₁) by Proposition 1.2.2. Therefore c ₂∈S _r (R) by the right-sided version of Lemma 5.1.5(i). Using this procedure, an induction proof completes the argument.

(ii) It is a direct consequence of part (i) and Proposition 1.2.2.

(iii) Put e=c _k and f=b _k+1. By part (i), e∈S _r (R), so f∈S _ℓ (eRe). From Lemma 5.1.6(ii), the map r→frf is a ring epimorphism. □

Corollary 5.1.8

The ordered set {b ₁,…,b _n } is a (complete) set of left triangulating idempotents of R if and only if the ordered set {b _n ,…,b ₁} is a (complete) set of right triangulating idempotents.

Proof

Let {b ₁,…,b _n } be a set of left triangulating idempotents of R. Then by Proposition 5.1.7(i), 1−(b ₁+⋯+b _n−1)=b _n ∈S _r (R). We next show that b _n−1∈S _r ((1−b _n )R(1−b _n )). For this, first it can be checked that {b ₁,…,b _n−1} is a set of left triangulating idempotents of (1−b _n )R(1−b _n ) and 1−b _n is the identity of (1−b _n )R(1−b _n ). By Proposition 5.1.7(ii), b ₁+⋯+b _n−2∈S _ℓ (R), and hence b ₁+⋯+b _n−2∈S _ℓ ((1−b _n )R(1−b _n )). Therefore by Proposition 1.2.2,

$$(1-b_n)-(b_1 + b_2+\cdots + b_{n-2}) = b_{n-1}\in \mathbf{S}_r((1-b_n)R(1-b_n)) $$

and so on. By this argument, the ordered set {b _n ,…,b ₁} is a set of right triangulating idempotents. Also, if {b ₁,…,b _n } is complete, then so is {b _n ,…,b ₁}.

The converse is proved similarly. Further, completeness is left-right symmetric since S _ℓ (b _i Rb _i )={0,b _i } if and only if S _r (b _i Rb _i )={0,b _i } (see Proposition 1.2.11). □

Exercise 5.1.9

• 1. Let R be a subdirectly irreducible ring (i.e., the intersection of all nonzero ideals of R is nonzero) and {b ₁,…,b _n } a set of left triangulating idempotents. Prove the following.

  • (i) For each i≠1 there exists j<i such that b _j Rb _i ≠0.

  • (ii) For each i≠n there exists j>i such that b _i Rb _j ≠0.

  • (iii) The heart of R (i.e., the intersection of all nonzero ideals of R) is contained in b ₁ Rb _n .

• 2. Let {b ₁,…,b _n } be a set of left triangulating idempotents of a ring R. Prove the following.

  • (i) b _i ∈S _ℓ (R) if and only if b _j Rb _i =0 for all j<i.

  • (ii) b _i ∈S _r (R) if and only if b _i Rb _j =0 for all j>i.

5.2 Generalized Triangular Matrix Representations

Rings with a complete generalized triangular matrix representation will be characterized. Then the uniqueness of a complete set of triangulating idempotents will be discussed. We shall see that if a ring R satisfies some mild finiteness conditions, then R has a generalized triangular matrix representation with semicentral reduced rings on the diagonal which satisfy the same finiteness condition as R. Thereby reducing the study of such rings to those which are semicentral reduced. Further, it will be shown that the condition of having a complete set of left triangulating idempotents is strictly between that of having a complete set of primitive idempotents and that of having a complete set of centrally primitive idempotents.

Lemma 5.2.1

Let 0≠f ²=f∈R. If fR=eR for every 0≠e∈S _ℓ (fRf), then f is semicentral reduced.

Proof

Let 0≠e∈S _ℓ (fRf). Then since fR=eR, f=ex for some x∈R, and so e=ef=eex=ex=f. Thus, f is semicentral reduced. □

Lemma 5.2.2

(i) A ring R has DCC on {bR∣b∈S _ℓ (R)} if and only if R has ACC on {Rc∣c∈S _r (R)}.

(ii) A ring R has ACC on {bR∣b∈S _ℓ (R)} if and only if R has DCC on {Rc∣c∈S _r (R)}.

(iii) If a ring R has DCC on {Rc∣c∈S _r (R)}, then R has DCC on {cR∣c∈S _r (R)}.

Proof

(i) Assume that R has DCC on {bR∣b∈S _ℓ (R)}. Consider a chain Rc ₁⊆Rc ₂⊆… , where c _i ∈S _r (R). Then (1−c ₁)R⊇(1−c ₂)R⊇… with 1−c _i ∈S _ℓ (R) (see Proposition 1.2.2). This descending chain becomes stationary, say with (1−c _n )R=(1−c _n+j )R for each j≥1. Then we have that ℓ _R ((1−c _n )R)=ℓ _R ((1−c _n+j )R) for each j>1. Thus, Rc _n =Rc _n+j for each j>1. The converse is proved similarly.

(ii) The proof is similar to that of part (i).

(iii) Assume that R has DCC on {Rc∣c∈S _r (R)}. Let c ₁ R⊇c ₂ R⊇… be a descending chain with c _i ∈S _r (R). Then c _i+1=c _i c _i+1. So it follows that c _i+1 c _i =c _i c _i+1 c _i =c _i c _i+1=c _i+1 because c _i ∈S _r (R). Therefore Rc _i ⊇Rc _i+1 for each i. Thus we have a descending chain Rc ₁⊇Rc ₂⊇… , so there is n with Rc _n =Rc _n+1=… . Therefore, (1−c _n )R=(1−c _n+1)R. Hence, we obtain that (1−c _n )Rc _n =(1−c _n+1)Rc _n =(1−c _n+1)Rc _n+1.

We observe that Rc _n =c _n Rc _n +(1−c _n )Rc _n =c _n R+(1−c _n )Rc _n and

$$Rc_{n+1} = c_{n+1}Rc_{n+1} + (1-c_{n+1})Rc_{n+1} = c_{n+1}R + (1-c_n)Rc_n $$

because c _n ,c _n+1∈S _r (R) and (1−c _n )Rc _n =(1−c _n+1)Rc _n+1. Therefore, we have that c _n R+(1−c _n )Rc _n =c _n+1 R+(1−c _n )Rc _n as Rc _n =Rc _n+1.

To show that c _n R=c _n+1 R, it suffices to check that c _n R⊆c _n+1 R because c _n+1 R⊆c _n R. Now c _n =c _n+1 y+α, where y∈R and α∈(1−c _n )Rc _n , as c _n R+(1−c _n )Rc _n =c _n+1 R+(1−c _n )Rc _n . Since c _n α=0 and c _n+1=c _n c _n+1 from c _n+1 R⊆c _n R, $c_{n} =c_{n}^{2}= c_{n}c_{n+1}y + c_{n}\alpha= c_{n+1}y\in c_{n+1}R$. Therefore c _n R⊆c _n+1 R, and hence c _n R=c _n+1 R=… . We conclude that R satisfies DCC on {cR∣c∈S _r (R)}. □

Lemma 5.2.3

Let e∈S _r (R). If R has DCC on {bR∣b∈S _ℓ (R)}, then eRe has DCC on {d(eRe)∣d∈S _ℓ (eRe)}.

Proof

First, we show that {(eRe)c∣c∈S _r (eRe)} has ACC. For this, assume that (eRe)c ₁⊆(eRe)c ₂⊆… is an ascending chain, where c _i ∈S _r (eRe) for i=1,2,… . By the right-sided version of Lemma 5.1.5(i), each c _i ∈S _r (R). Note that ec _i e∈(eRe)ec _i e⊆(eRe)ec _i+1 e.

So there exists x∈eRe such that ec _i e=xec _i+1 e. Thus,

$$\begin{aligned} (1-e)Rc_i = & (1-e)Rec_ie = (1-e)Rxec_{i+1}e\\ \subseteq&(1-e)Rec_{i+1}e = (1-e)Rc_{i+1}. \end{aligned}$$

Therefore, for each i,

$$\begin{aligned} Rc_i = & eRc_i + (1-e)Rc_i = (eRe)ec_ie + (1-e)Rc_i\\ \subseteq&(eRe)ec_{i+1}e + (1-e)Rc_{i+1} =eRc_{i+1}+(1-e)Rc_{i+1}\\ =& Rc_{i+1}. \end{aligned}$$

By assumption and Lemma 5.2.2(i), Rc _n =Rc _n+1=… for some n as each c _i is in S _r (R). Therefore, eRc _n =eRc _n+1=… , so (eRe)c _n =(eRe)c _n+1=… . From Lemma 5.2.2(i), eRe has DCC on {d(eRe)∣d∈S _ℓ (eRe)}. □

Lemma 5.2.4

Let {b ₁,…,b _n } be a complete set of left triangulating idempotents of R. If e∈S _ℓ (R), then eR=⨁ _i  b _i R, where the sum runs over a subset of {1,…,n}. Thus, |{eR∣e∈S _ℓ (R)}|≤2ⁿ.

Proof

Assume that 0≠e∈S _ℓ (R). Consider i such that b _i e≠0. We show that b _i eR=b _i R. For this, note that b _i eb _i e=b _i e≠0, so b _i eb _i ≠0. From Lemma 5.1.5(ii), b _i S _ℓ (R)b _i ⊆S _ℓ (b _i Rb _i ). Hence b _i eb _i ∈S _ℓ (b _i Rb _i ), but by hypothesis S _ℓ (b _i Rb _i )={0,b _i }. So b _i eb _i =b _i . Also b _i R=b _i eb _i R⊆b _i eR⊆b _i R, and thus b _i eR=b _i R. Recall that b _i are orthogonal. Hence, b _i eb _j e=b _i b _j e=0 yields that b ₁ e,…,b _n e are orthogonal idempotents. Let I={i ∣1≤i≤n and b _i e≠0}. Then eR=⊕ _i∈I b _i eR=⊕ _i∈I b _i R. □

The next result characterizes rings with a complete generalized triangular matrix representation.

Theorem 5.2.5

The following are equivalent for a ring R.

• (i) R has a complete set of left triangulating idempotents.

• (ii) {bR ∣b∈S _ℓ (R)} is a finite set.

• (iii) {bR ∣b∈S _ℓ (R)} satisfies ACC and DCC.

• (iv) {bR ∣b∈S _ℓ (R)} and {Rc ∣c∈S _r (R)} satisfy ACC.

• (v) {bR ∣b∈S _ℓ (R)} and {Rc ∣c∈S _r (R)} satisfy DCC.

• (vi) {bR ∣b∈S _ℓ (R)} and {cR ∣c∈S _r (R)} satisfy DCC.

• (vii) R has a complete set of right triangulating idempotents.

• (viii) R has a complete generalized triangular matrix representation.

Proof

Lemma 5.2.4 yields (i)⇒(ii), and (ii)⇒(iii) is trivial. From Lemma 5.2.2, (iii)⇒(iv)⇒(v)⇒(vi) follows immediately.

We show that (vi)⇒(i). If S _ℓ (R)={0,1}, then we are finished. Otherwise take e ₁ to be a nontrivial element of S _ℓ (R).

If e ₁ is not semicentral reduced, then there exists 0≠e ₂∈S _ℓ (e ₁ Re ₁) such that e ₁ R≠e ₂ R by Lemma 5.2.1, and so e ₁ R⊋e ₂ R. From Lemma 5.1.5(i), e ₂∈S _ℓ (R). If e ₂ is not semicentral reduced, then by Lemmas 5.2.1 and 5.1.5(i) again there exists 0≠e ₃∈S _ℓ (e ₂ Re ₂)⊆S _ℓ (R) such that e ₂ R≠e ₃ R. So we have that e ₂ R⊋e ₃ R. This process should be stopped within a finite steps. Thus, we obtain a semicentral reduced idempotent e _n ∈S _ℓ (R) for some positive integer n because {eR∣e∈S _ℓ (R)} has DCC.

Starting a new process, let b ₁=e _n . Then S _ℓ (b ₁ Rb ₁)={0,b ₁}. From Proposition 1.2.2, 1−b ₁∈S _r (R). If 1−b ₁ is semicentral reduced, then we see that {b ₁,1−b ₁} is a complete set of left triangulating idempotents.

Otherwise, we consider R ₁=(1−b ₁)R(1−b ₁). Note that by Lemma 5.2.3, R ₁ has DCC on {dR ₁∣d∈S _ℓ (R ₁)}. By a similar argument to that used to get b ₁, we obtain b ₂∈S _ℓ (R ₁) such that S _ℓ (b ₂ R ₁ b ₂)={0,b ₂}.

As 1−b ₁ is the identity of R ₁ and b ₂∈R ₁, it follows that b ₂ R ₁ b ₂=b ₂ Rb ₂, so S _ℓ (b ₂ Rb ₂)={0,b ₂}. Also, (1−b ₁)−b ₂∈S _r (R ₁). The right-sided version of Lemma 5.1.5(i) yields that S _r (R ₁)⊆S _r (R). Therefore, 1−b ₁−b ₂∈S _r (R). If 1−b ₁−b ₂ is semicentral reduced in R, then {b ₁,b ₂,1−b ₁−b ₂} is a complete set of left triangulating idempotents.

We continue the process to obtain a descending chain in {cR ∣ c∈S _r (R)}, which is (1−b ₁)R⊇(1−b ₁−b ₂)R⊇(1−b ₁−b ₂−b ₃)R⊇… . By the DCC hypothesis of {cR∣c∈S _r (R)}, this chain becomes stationary after a finite steps, yielding a complete set of left triangulating idempotents.

The equivalence (vii)⇔(i) follows from Corollary 5.1.8, while the equivalence (i)⇔(viii) follows from Theorem 5.1.4. □

Corollary 5.2.6

Let R be a ring with a complete set of left triangulating idempotents. Then for any 0≠e∈S _ℓ (R) (resp., 0≠e∈S _r (R)), the ring eRe also has a complete set of left (resp., right) triangulating idempotents.

Proof

Say 0≠e∈S _ℓ (R). Define

$$\lambda: \{bR\mid b\in\mathbf{S}_{\ell}(R)\}\rightarrow\{d(eRe)\mid d\in\mathbf{S}_{\ell}(eRe)\} $$

by λ(bR)=(ebe)(eRe). From Lemma 5.1.5(ii), ebe∈S _ℓ (eRe) for b∈S _ℓ (R). If bR=b ₁ R with b,b ₁∈S _ℓ (R), then bRe=b ₁ Re, and so ebeRe=eb ₁ eRe since e∈S _ℓ (R). Thus λ is well-defined. As S _ℓ (eRe)⊆S _ℓ (R) by Lemma 5.1.5(i), λ is onto. From Theorem 5.2.5, it follows that {bR∣b∈S _ℓ (R)} is finite. Furthermore, we get that {d(eRe)∣d∈S _ℓ (eRe)} is also finite. Again by Theorem 5.2.5, eRe has a complete set of left triangulating idempotents. Similarly, if 0≠e∈S _r (R), then eRe has also a complete set of right triangulating idempotents. □

In Theorem 5.2.8, the uniqueness of a complete generalized triangular matrix representation will be established. For the proof of this theorem, we need the following result due to Azumaya [32, Theorem 3].

Lemma 5.2.7

Let I be a quasi-regular ideal of a ring R. If {e ₁,…,e _n } and  {f ₁,…,f _n } are two sets of orthogonal idempotents of R such that $\overline{e}_{i} = \overline{f}_{i}$ for each i with images $\overline{e}_{i}$ and $\overline{f}_{i}$ in R/I, then there is an invertible element α∈R with f _i =α ⁻¹ e _i α for each i.

Proof

Let $e = \sum_{i=1}^{n}e_{i}$ and $f = \sum_{i=1}^{n}f_{i}$. Put $\beta= e + f - ef - \sum_{i=1}^{n}e_{i}f_{i}$. Then α=1−β is invertible and f _i =α ⁻¹ e _i α for each i. □

A nonzero central idempotent e of R is said to be centrally primitive if 0 and e are the only central idempotents in eRe. Let g be a nonzero central idempotent in R such that g=g ₁+⋯+g _t , where {g _i ∣1≤i≤t} is a set of centrally primitive orthogonal idempotents of R. Then t is uniquely determined (see Exercise 5.2.21.1). A ring R is said to have a complete set of centrally primitive idempotents if there exists a finite set of centrally primitive orthogonal idempotents whose sum is 1. It is routine to check that R has a complete set of centrally primitive idempotents if and only if R is a ring direct sum of indecomposable rings.

Theorem 5.2.8

(Uniqueness)

Let {b ₁,…,b _n } and {c ₁,…,c _k } each be a complete set of left triangulating idempotents of R. Then n=k and there exist an invertible element α∈R and a permutation σ on {1,…,n} such that b _σ(i)=α ⁻¹ c _i α for each i. Thus for each i, c _i R≅b _σ(i) R, as R-modules, and c _i Rc _i ≅b _σ(i) Rb _σ(i), as rings.

Proof

Let U=∑ _i<j b _i Rb _j . Then U⊴R and U ⁿ=0. Let $\overline{R} = R/U$ and denote by $\overline{x}$ the image of x∈R in R/U. Since b _i Rb _i  ∩ U=0, for i=1,…,n, $b_{i}Rb_{i}\cong\overline{b_{i}}\,\overline{R}\,\overline{b}_{i}$ as rings. So $\overline{R}$ is a direct sum of the $\overline{b}_{i}\,\overline{R}\,\overline{b}_{i}$, and consequently $\{\overline{b}_{1},\dots, \overline{b}_{n}\}$ is a complete set of centrally primitive idempotents of $\overline{R}$.

Clearly, $\overline{c}_{1}\in\mathbf{S}_{\ell}(\overline{R})$. Further, $\overline{c}_{1}\neq\overline{0}$. Indeed, if $\overline{c}_{1}=\overline{0}$, then c ₁∈U, and so $c_{1}=c_{1}^{n}\in U^{n}=0$, a contradiction. Because $\overline{b}_{i}$ is semicentral reduced, $\overline{c}_{1}\overline{b}_{i}\in\{\overline{0}, \,\overline{b}_{i}\}$. Therefore $\overline{c}_{1} = \sum_{i=1}^{n}\overline{c}_{1}\overline{b}_{i} = \sum\overline{b}_{k}$ for which $\overline{c}_{1}\overline{b}_{k}\neq\overline{0}$. So $\overline{c}_{1}\in\mathcal{B}(\overline{R})$. Now we note that $\overline{c}_{2}\in\mathbf{S}_{\ell}((\overline{1} - \overline{c}_{1})\,\overline{R}\,(\overline{1} - \overline{c}_{1}))$. As $\overline{1} - \overline{c}_{1}\in\mathcal{B}(\overline{R})$, $\overline{c}_{2}\in \mathbf{S}_{\ell}(\overline{R})$ by Lemma 5.1.5(i). Using the preceding argument, with $\overline{c}_{2}$ in place of $\overline{c}_{1}$, we obtain $\overline{c}_{2}\in\mathcal{B}(\overline{R})$.

Continuing this procedure, we obtain that $\{\overline{c}_{1},\dots, \overline{c}_{k}\}$ is a set of orthogonal nonzero central idempotents in $\overline{R}$. Hence $\overline{c}_{i}\,\overline{R}\,\overline{c}_{j} = \overline{0}$ for i<j. Thus c _i Rc _j ⊆U for all 1≤i<j≤k.

Let V=∑ _i<j c _i Rc _j . Then V ^k=0. By the preceding argument, b _i Rb _j ⊆V for all 1≤i<j≤n. Hence, U=V and so $\{\overline{b}_{1},\dots, \overline{b}_{n}\}$ and $\{\overline{c}_{1},\dots, \overline{c}_{k}\}$ are both complete sets of centrally primitive idempotents for $\overline{R}$. It is well known that for such sets of centrally primitive idempotents, n=k and there is a permutation σ on {1,…,n} such that $\overline{c}_{i} = \overline{b}_{\sigma(i)}$ (Exercises 5.2.21.1 and 5.2.21.2). As U ⁿ=0, U is a quasi-regular ideal of R.

From Lemma 5.2.7, there exists an invertible element α∈R such that b _σ(i)=α ⁻¹ c _i α for every i. Thus, c _i R≅b _σ(i) R as R-modules. We observe that End(c _i R _R )≅c _i Rc _i and End(b _j R _R )≅b _j Rb _j . So c _i Rc _i ≅b _σ(i) Rb _σ(i). □

The following example shows that the isomorphism c _i R≅b _σ(i) R, given in Theorem 5.2.8, cannot be sharpened to equality. This is in contrast to the result for a complete set of centrally primitive idempotents.

Example 5.2.9

Let $R = T_{2}(\mathbb{R})$. Consider

$$b_1 = \begin{bmatrix} 1 \,& \,0\\ 0 \,& \,0\end{bmatrix}, \,\,b_2 = \begin{bmatrix}0 \,& \,0\\ 0 \,& \,1 \end{bmatrix} $$

and let

$$c_1 = \begin{bmatrix} 1 \,& \,a\\ 0 \,& \,0 \end{bmatrix} , \,\,c_2 = \begin{bmatrix} 0 \,& \,-a\\ 0 \,& \,1 \end{bmatrix} , \,\,0\neq a\in\mathbb{R}. $$

Then {b ₁,b ₂} and {c ₁,c ₂} are complete sets of left triangulating idempotents for R. In this case, b ₁ R=c ₁ R and b ₂ R≅c ₂ R, but b ₂ R≠c ₂ R.

Kaplansky raised the following question: Let A and B be two rings. If Mat _n (A)≅Mat _n (B) as rings, does it follow that A≅B as rings? (See [261, p. 35].) It is known that there are nonisomorphic semicentral reduced rings (e.g., simple Noetherian domains) which have isomorphic matrix rings (see [260] and [378]). The next result shows that this cannot happen for n×n (n>1) upper triangular matrix rings over semicentral reduced rings.

Corollary 5.2.10

Let A and B be semicentral reduced rings. If T _m (A)≅T _n (B) as rings, then m=n and A≅B as rings.

Proof

Let e _ii be the matrix in T _m (A) with 1 _A in the (i,i)-position and 0 elsewhere. As A is semicentral reduced, {e ₁₁,…,e _mm } is a complete set of left triangulating idempotents for T _m (A). A similar fact holds for T _n (B). Because T _m (A)≅T _n (B), m=n by Theorem 5.2.8.

Next, say λ:T _n (A)→T _n (B) is an isomorphism. Then {λ(e ₁₁),…,λ(e _nn )} is a complete set of left triangulating idempotents of T _n (B). Let f _ii be the matrix in T _n (B) with 1 _B in the (i,i)-position and 0 elsewhere. Then because B is semicentral reduced, {f ₁₁,…,f _nn } is also a complete set of left triangulating idempotents of T _n (B).

By Theorem 5.2.8, f ₁₁ T _n (B)f ₁₁≅λ(e _jj )T _n (B)λ(e _jj ) for some j. Therefore, B≅f ₁₁ T _n (B)f ₁₁≅λ(e _jj )T _n (B)λ(e _jj )≅e _jj T _n (A)e _jj ≅A. □

From Theorem 5.2.8, the number of elements in a complete set of left triangulating idempotents is unique for a given ring R (which has such a set). This is also the number of elements in any complete set of right triangulating idempotents of R by Corollary 5.1.8. So we are motivated to give the following definition.

Definition 5.2.11

A ring R is said to have triangulating dimension n, written Tdim(R)=n, if R has a complete set of left triangulating idempotents with n elements. Note that R is semicentral reduced if and only if Tdim(R)=1. If R has no complete set of left triangulating idempotents, then we say that R has infinite triangulating dimension, denoted Tdim (R)=∞.

Lemma 5.2.12

Let {e ₁,…,e _n } be a complete set of primitive idempotents of R. If 0≠b∈S _ℓ (R)∪ S _r (R), then there exists a nonempty subset P of {e ₁,…,e _n } such that {be _j b ∣ e _j ∈P} forms a complete set of primitive idempotents of bRb.

Proof

Assume that b∈S _ℓ (R). From b=b(e ₁+⋯+e _n )b=be ₁ b+⋯+be _n b, some be _k b≠0. Let P be the set of all e _j such that the elements be _j b are nonzero. Without loss of generality, let P={e ₁,…,e _m }.

By Lemma 5.1.5(iii), the be _j b, j=1,…,m, are primitive idempotents in bRb. From b=be ₁ b+⋯+be _n b=be ₁ b+⋯+be _m b, {be _j b ∣ 1≤j≤m} is a complete set of primitive idempotents for bRb. The proof for b∈S _r (R) is a right-sided version of the preceding proof. □

The next two results may be useful for studying many well known classes of rings via complete generalized triangular matrix representations and semicentral reduced rings from the same respective class.

Proposition 5.2.13

Let a ring R satisfy any one of the following conditions.

• (i) R has a complete set of primitive idempotents.

• (ii) R is orthogonally finite.

• (iii) R has DCC on idempotent generated (resp., principal, or finitely generated) ideals.

• (iv) R has ACC on idempotent generated (resp., principal, or finitely generated) ideals.

• (v) R has DCC on idempotent generated (resp., principal, or finitely generated) right ideals.

• (vi) R has ACC on idempotent generated (resp., principal, or finitely generated) right ideals.

• (vii) R is a semilocal ring.

• (viii) R is a semiperfect ring.

• (ix) R is a right perfect ring.

• (x) R is a semiprimary ring.

Then Tdim(R)<∞ and

$$R \cong \begin{bmatrix} R_1 & R_{12} &\cdots& R_{1n}\\ 0 & R_2 & \cdots& R_{2n}\\ \vdots& \vdots& \ddots& \vdots\\ 0 & 0 & \cdots& R_n \end{bmatrix} , $$

where n=Tdim(R), each R _i is semicentral reduced, and satisfies the same condition as R. Further, each R _ij is an (R _i ,R _j )-bimodule, and the rings R ₁,…,R _n are uniquely determined by R up to isomorphism and permutation.

Proof

(i) Let {f ₁,…,f _k } be a complete set of primitive idempotents of R. Then for any 0≠b∈S _ℓ (R), b=f ₁ b+⋯+f _k b. Each f _i b is an idempotent, as b∈S _ℓ (R). Assume that j=1,…,m is the set of all indices for which f _j b≠0.

Now we have that bR⊆f ₁ bR+⋯+f _m bR=bf ₁ bR+⋯+bf _m bR⊆bR, hence bR=f ₁ bR+⋯+f _m bR. Primitivity of f _j implies that f _j bR=f _j R, whenever f _j b≠0. Hence, the total number of right ideals of the form bR, b∈S _ℓ (R) cannot exceed 2^k. Thus, by Theorem 5.2.5, R has a complete set of left triangulating idempotents. So Tdim(R)<∞.

Let {e ₁,…,e _n } be a complete set of left triangulating idempotents of R. Take R _i =e _i Re _i and R _ij =e _i Re _j for i<j. Then R _ij is an (R _i ,R _j )-bimodule for i<j. Since e ₁∈S _ℓ (R), R ₁=e ₁ Re ₁ has a complete set of primitive idempotents from Lemma 5.2.12. Also 1−e ₁∈S _r (R) by Proposition 1.2.2, (1−e ₁)R(1−e ₁) has a complete set of primitive idempotents by Lemma 5.2.12. Next we see that e ₂∈S _ℓ ((1−e ₁)R(1−e ₁)), again Lemma 5.2.12 yields that

$$R_2=e_2Re_2 = e_2((1-e_1)R(1-e_1))e_2 $$

has a complete set of primitive idempotents, and so on. The uniqueness of the R _i follows from Theorem 5.2.8.

(ii) By part (i) and Proposition 1.2.15, we have a unique generalized triangular matrix representation. Further, each R _i is orthogonally finite.

(iii) Assume that R has DCC on idempotent generated (resp., principal, or finitely generated) ideals. Then {eR∣e∈S _ℓ (R)} has DCC since eR=ReR for each e in S _ℓ (R). Consider {Rf∣f∈S _r (R)}. Then Rf=RfR for each f∈S _r (R). Thus {Rf∣f∈S _r (R)} also has DCC. By Theorem 5.2.5, R has a complete set of left triangulating idempotents. So Tdim(R)<∞.

Now say h ²=h∈R. Then hRh has DCC on idempotent generated (resp., principal, or finitely generated) ideals by using [259, Theorem 21.11].

(iv) By assumption, {eR∣e∈S _ℓ (R)} has ACC as eR=ReR. Also since Rf=RfR for any f∈S _r (R), {Rf∣f∈S _r (R)} has ACC. From Theorem 5.2.5, R has a complete set of left triangulating idempotents, so Tdim<∞. Say h ²=h∈R. By using [259, Theorem 21.11], hRh has ACC on idempotent generated (resp., principal, or finitely generated) ideals.

(v) By Proposition 1.2.13, R is orthogonally finite. By part (ii), R has a complete set of left triangulating idempotents, so Tdim(R)<∞. Next, let h ²=h∈R. Then hRh has DCC on idempotent generated (resp., principal or finitely generated) right ideals by using [259, Theorem 21.11].

(vi) The proof is similar to that of part (v) by Proposition 1.2.13 and using [259, Theorem 21.11].

(vii) and (viii) We note that, for each of these conditions, R is orthogonally finite. By part (ii), Tdim(R)<∞. Homomorphic images of a semilocal ring and a semiperfect ring are semilocal and semiperfect, respectively (see [259, Proposition 20.7] and [8, Corollary 27.9]). By Proposition 5.1.7(iii), if R is semilocal (resp., semiperfect), then each R _i is semilocal (resp., semiperfect).

(ix) If R is right perfect, then R is orthogonally finite. Thus part (ii) yields that Tdim(R)<∞. By 1.1.14, R has DCC on principal left ideals. Say h ²=h∈R. Then by the left-sided version of the proof for part (v), hRh also has DCC on principal left ideals. So hRh is right perfect, and hence each R _i is right perfect.

(x) If R is semiprimary, then also R is orthogonally finite. Hence by part (ii), Tdim(R)<∞. Say h ²=h∈R. It is well known that J(hRh)=hJ(R)h (see [259, Theorem 21.10]). Hence if R is semiprimary, then so is hRh. Thus each R _i is semiprimary. □

Proposition 5.2.14

Let P be a property of rings such that whenever a ring A satisfies P, then A/I (I⊴A) or eAe (e ²=e∈A) also satisfies P. Assume that R is a ring with Tdim(R)=n<∞ and satisfies P. Then

$$R \cong \begin{bmatrix} R_1 & R_{12} &\cdots& R_{1n}\\ 0 & R_2 & \cdots& R_{2n}\\ \vdots& \vdots& \ddots& \vdots\\ 0 & 0 & \cdots& R_n \end{bmatrix} , $$

where each R _i is semicentral reduced and satisfies the property P. Further, each R _ij is an (R _i ,R _j )-bimodule, and the rings R ₁,…,R _n are uniquely determined by R up to isomorphism and permutation.

Proof

Since Tdim(R)=n<∞, R has the indicated unique generalized triangular matrix representation by Theorems 5.1.4 and 5.2.8. Rings R _i have the form eRe, where e ²=e∈R, also R _i are ring homomorphic images of R by Proposition 5.1.7(iii). By assumption each R _i has the property P. □

We remark that the following classes of rings determined by property P indicated in Proposition 5.2.14: Baer rings, right Rickart rings, quasi-Baer rings, right p.q.-Baer rings, right hereditary rings, right semihereditary rings, π-regular rings, PI-rings, and rings with bounded index (of nilpotency), etc.

By the next result, if Tdim(R)<∞, central idempotents can be written as sums of elements in a complete set of left triangulating idempotents.

Proposition 5.2.15

Assume that {b ₁,…,b _n } is a complete set of left triangulating idempotents for a ring R. If $c\in\mathcal{B}(R)\setminus\{0, 1\}$, then there exists ∅≠Λ⊆̷{1,…,n} such that c=∑ _i∈Λ b _i .

Proof

Let $c\in\mathcal{B}(R)\setminus\{0, 1\}$. Then c=c(b ₁+⋯+b _n )=cb ₁+⋯+cb _n . We note that cb _i ∈S _ℓ (b _i Rb _i ) and S _ℓ (b _i Rb _i )={0,b _i } for each i. Therefore, there exists ∅≠Λ⊆̷{1,…,n} such that c=∑ _i∈Λ b _i . □

Theorem 5.2.16

Let R be a ring. Consider the following conditions.

• (i) R has a complete set of primitive idempotents.

• (ii) R has a complete set of left triangulating idempotents.

• (iii) R has a complete set of centrally primitive idempotents.

Then (i)⇒(ii)⇒(iii).

Proof

Proposition 5.2.13(i) yields the implication (i)⇒(ii). For (ii)⇒(iii), assume that R has a complete set of left triangulating idempotents for R. By Proposition 5.2.15, $\mathcal{B}(R)$ is a finite set. Now a standard argument yields that R has a complete set of centrally primitive idempotents. □

We remark that when R is commutative, conditions (i), (ii), and (iii) of Theorem 5.2.16 are equivalent. The next example shows that the converse of each of the implications in Theorem 5.2.16 does not hold.

Example 5.2.17

(i) There is a ring R with a complete set of left triangulating idempotents (i.e., Tdim (R)<∞), but R does not have a complete set of primitive idempotents. Indeed, let V be an infinite dimensional right vector space over a field F and let R=End _F (V). Then R is a prime ring, so Tdim(R)=1. Since R is a regular ring which is not semisimple Artinian, R cannot have a complete set of primitive idempotents.

(ii) There is a ring R with a complete set of centrally primitive idempotents, but R does not have a complete set of left triangulating idempotents. For this, let R be the ℵ₀×ℵ₀ upper triangular row finite matrix ring over a field. Then {1} is a complete set of centrally primitive idempotents of R, where 1 is the identity of R. Let e _ii be the matrix in R with 1 in the (i,i)-position and 0 elsewhere. Then for any positive integer n, e ₁₁+⋯+e _nn ∈S _ℓ (R). As

$$(e_{11} +\cdots + e_{nn})R\subsetneq (e_{11} + \cdots + e_{nn} + e_{n+1 \,n+1})R $$

for each n, Theorem 5.2.5 yields that R cannot have a complete set of left triangulating idempotents.

We need the next lemma for investigating Tdim(R) of a ring R.

Lemma 5.2.18

Let {b ₁,…,b _n } be a set of left triangulating idempotents of a ring R and $\{b_{(i, 1)},\dots, b_{(i, k_{i})}\}$ a set of left triangulating idempotents of b _i Rb _i . Then $\{b_{(1,1)},\dots, b_{(1, k_{1})}, b_{(2,1)},\dots, b_{(2, k_{2})},\dots, b_{(n, 1)},\dots, b_{(n, k_{n})}\}$ is a set of left triangulating idempotents of R.

Proof

Clearly $1 = \sum_{i=1}^{k_{1}}b_{(1, i)} +\cdots+ \sum_{i=1}^{k_{n}}b_{(n, i)}$. Also b _(1,1)∈S _ℓ (R) by Lemma 5.1.5(i). Let $c_{(i, j)} = 1 - \sum_{\alpha= 1}^{i-1}b_{\alpha} - \sum_{\gamma = 1}^{j}b_{(i, \gamma)}$, where 1≤j<k _i . Then $b_{(i, j+1)}(\sum _{\alpha=1}^{i-1}b_{\alpha}+\sum_{\gamma=1}^{j}b_{(i, \gamma)})=0$, and so b _(i,j+1) c _(i,j)=b _(i,j+1). Similarly, c _(i,j) b _(i,j+1)=b _(i,j+1). So b _(i,j+1)∈c _(i,j) Rc _(i,j). Note that $c_{(i,\, j)}^{2}=c_{(i,\, j)}$.

We claim that b _(i,j+1)∈S _ℓ (c _(i,j) Rc _(i,j)). Put $c_{j} = b_{i} - \sum_{\gamma= 1}^{j}b_{(i, \gamma)}$. Then b _(i,j+1)∈S _ℓ (c _j (b _i Rb _i )c _j )=S _ℓ (c _j Rc _j ) and $c_{(i, j)} = 1 - \sum_{\alpha= 1}^{i}b_{\alpha} + c_{j}$. Note that b _(i,j+1)∈b _i Rb _i , $(\sum_{\alpha=1}^{i-1}b_{\alpha})b_{(i, j+1)}=0$, and {b ₁,…,b _n } is a set of orthogonal idempotents. Hence,

$$\begin{aligned} b_{(i,j+1)} =&c_{(i, j)}b_{(i, j+1)} = (1 - \sum_{\alpha= 1}^ib_{\alpha} + c_j)b_{(i, j+1)}\\ = & b_{(i, j+1)}-b_ib_{(i,j+1)}+c_jb_{(i, j+1)}=c_jb_{(i, j+1)} \end{aligned}$$

as b _i b _(i,j+1)=b _(i,j+1). Similarly, b _(i,j+1)=b _(i,j+1) c _(i,j)=b _(i,j+1) c _j . For r∈R,

$$\begin{aligned} (c_{(i, j)}rc_{(i, j)})b_{(i, j+1)} = & (1-\sum_{\alpha= 1}^ib_{\alpha} + c_j)rc_jb_{(i, j+1)}\\ = & (1 - \sum_{\alpha=1}^i b_{\alpha})rc_jb_{(i, j+1)} + c_jrc_jb_{(i, j+1)}. \end{aligned}$$

From Proposition 5.1.7(i), $1 - \sum_{\alpha=1}^{i}b_{\alpha}\in\mathbf{S}_{r}(R)$. Therefore, we now obtain that $(1 - \sum_{\alpha=1}^{i}b_{\alpha})rc_{j}b_{(i, j+1)} = (1 - \sum_{\alpha=1}^{i}b_{\alpha})r(1 - \sum_{\alpha=1}^{i}b_{\alpha})c_{j}b_{(i, j+1)} = 0$ since

$$\begin{aligned} (1 - \sum_{\alpha=1}^ib_{\alpha})c_jb_{(i, j+1)} =&(1 - \sum_{\alpha=1}^ib_{\alpha})b_{(i, j+1)}=(1 - \sum_{\alpha=1}^{i-1}b_{\alpha}-b_i)b_{(i, j+1)}\\ =&b_{(i, j+1)}-b_ib_{(i, j+1)}=0. \end{aligned}$$

Thus,

$$\begin{aligned} (c_{(i,j)}rc_{(i,j)})b_{(i,j+1)} = &(c_jrc_j)b_{(i,j+1)} = b_{(i, j+1)}(c_jrc_j)b_{(i,j+1)}\\ =& b_{(i,j+1)}(c_{(i, j)}rc_{(i, j)})b_{(i,j+1)}. \end{aligned}$$

So b _(i,j+1)∈S _ℓ (c _(i,j) Rc _(i,j)). Now routinely we obtain the desired result. □

Theorem 5.2.19

Let {b ₁,…,b _n } be a set of left triangulating idempotents of a ring R. Then $\mathrm{Tdim}(R) = \sum_{i=1}^{n}\mathrm{Tdim}(b_{i}Rb_{i})$.

Proof

If Tdim(R)=∞, then Tdim(b _j Rb _j )=∞ for some 1≤j≤n, otherwise Lemma 5.2.18 yields a contradiction.

Let Tdim(R)<∞. By Corollary 5.2.6, Tdim(b ₁ Rb ₁)<∞. From Proposition 1.2.2, 1−b ₁∈S _r (R). By Corollary 5.2.6, Tdim((1−b ₁)R(1−b ₁))<∞. We see that b ₂∈S _ℓ ((1−b ₁)R(1−b ₁)). Hence, Corollary 5.2.6 yields that Tdim(b ₂ Rb ₂)<∞. This procedure, by using Corollary 5.2.6, can be continued to show that Tdim(b _i Rb _i )<∞ for all 1≤i≤n. Lemma 5.2.18 yields that $\text{Tdim}(R) = \sum_{i=1}^{n}\text{Tdim}(b_{i}Rb_{i})$. □

Corollary 5.2.20

Let R be a ring with a generalized triangular matrix representation

$$\begin{bmatrix} R_1 & R_{12} &\cdots& R_{1n}\\ 0 & R_2 & \cdots& R_{2n}\\ \vdots& \vdots& \ddots& \vdots\\ 0 & 0 & \cdots& R_n \end{bmatrix} . $$

Then $\mathrm{Tdim}(R) = \sum_{i=1}^{n}\mathrm{Tdim}(R_{i})$. So, Tdim(T _n (A))=nTdim(A), where A is a ring and n is a positive integer.

Exercise 5.2.21

• 1. Assume that R is a ring and $0\neq g\in\mathcal{B}(R)$ such that g=g ₁+⋯+g _t , where {g _i  ∣1≤i≤t} is a set of orthogonal centrally primitive idempotents in R. Show that t is uniquely determined.

• 2. Let R be a ring, and let {e ₁,…,e _m } and {f ₁,…,f _n } be two complete sets of centrally primitive idempotents of R. Show that m=n and there exists a permutation σ on {1,…,n} such that e _i =f _σ(i).

• 3. Assume that M _R is a right R-module and S=End(M _R ). Show that the following are equivalent.

  • (i) S has a complete set of left triangulating idempotents.

  • (ii) There exists a positive integer n such that:

    • (1) M=M ₁⊕⋯⊕M _n .

    • (2) Hom(M _i ,M _j )=0 for i<j.

    • (3) Each M _i has no nontrivial fully invariant direct summands.

• 4. ([93, Birkenmeier, Park and Rizvi]) Assume that S is an overring of a ring R such that R _R ≤^ess S _R . (The ring S is called a right essential overing of R. See Chap. 7 for right essential overrings for more details.) Show that if R is right FI-extending, then Tdim(S)≤Tdim(R).

• 5. ([93, Birkenmeier, Park and Rizvi]) Let S be an overring of a ring R such that R _R ≤^ess S _R . Prove that if R is right extending and {e ₁,…,e _n } is a complete set of primitive idempotents for R, then {e ₁,…,e _n } is a complete set of primitive idempotents for S.

• 6. ([79, Birkenmeier, Kim, and Park]) Show that a ring R is left perfect if and only if R has a complete generalized triangular matrix representation, where each diagonal ring R _i is simple Artinian or left perfect with $(\text{Soc}({R_{i}}_{R_{i}}))^{2}=0$.

5.3 Canonical Representations

We show that if a ring R has a set of left triangulating idempotents, then it has a canonical generalized triangular matrix representation, where the diagonal subrings are organized into blocks of square diagonal matrix rings. This canonical representation is then used to obtain a result on the right global dimension of rings with a set of left triangulating idempotents.

Let {b ₁,…,b _n } be a set of left triangulating idempotents of R. If J is a subset of {1,…,n}, we denote σ _J =∑ _i∈J b _i . Our first result shows that under certain conditions the ordering in a set of left triangulating idempotents can be changed to obtain a new set of left triangulating idempotents.

Proposition 5.3.1

Let j and m be in {1,…,n} with j<m≤n. If {b ₁,…,b _n } is a set of left triangulating idempotents of a ring R such that b _i Rb _m =0 for each i with j≤i<m, then

$$\{b_1,\dots, b_{j-1}, b_m, b_j, b_{j+1},\dots, b_{m-1}, b_{m+1},\dots, b_n\} $$

is a set of left triangulating idempotents of R.

Proof

The proof follows routinely from Theorem 5.1.3. □

Proposition 5.3.1 is applied to obtain a canonical form for a generalized triangular matrix representation of R. Let {b ₁,…,b _n } be a set of left triangulating idempotents. Recursively define the sets I _k and J(k) as follows:

$$I_1 = \{i\,\mid\, b_i\in\mathbf{S}_{\ell}(R)\}\text{ and }J(1) = I_1; $$

and let

$$I_{k+1} = \{i\,\mid\ b_i\in\mathbf{S}_{\ell}((1-\sigma_{J(k)})R(1-\sigma_{J(k)}))\}\text{ and }J(k+1) = J(k)\,\cup\,I_{k+1}, $$

whenever I _k and J(k) are defined. This process terminates within n steps.

Let S _j ={b _i  ∣ i∈I _j }. Then S ₁,…,S _q is a partition for {b ₁,…,b _n } (we will show in the proof of Theorem 5.3.2 that this always occurs). Then reorder {1,…,n} so that each I _j has any (fixed) ordering and so that elements of I _j always precede elements in I _j+1. This can be thought of in terms of a permutation ψ on {1,…,n}. Then the ordered set {b _ψ(1),…,b _ψ(n)} is called a canonical form  for {b ₁,…,b _n }.

Theorem 5.3.2

Let {b ₁,…,b _n } be a set of left triangulating idempotents. Then a canonical form for {b ₁,…,b _n } exists, and any such canonical form is a set of left triangulating idempotents of R.

Proof

The proof involves repeated use of Propositions 5.3.1, as in the following discussion. We note that b ₁∈S ₁=S _ℓ (R). If b _m ∈S ₁ and m≠1, then b _i Rb _m =b _i b _m Rb _m =0 for all i≠m. We use Proposition 5.3.1 to get that {b _m ,b ₁,…,b _m−1,b _m+1,…,b _n } is a set of left triangulating idempotents of R. Continue this process using elements of S ₁ until they are exhausted.

Following the procedure given in Proposition 5.3.1, there exists a permutation α on {1,…,n} such that $S_{1}=\{b_{\alpha(1)},\dots, b_{\alpha(n_{1})}\}$. Also, the ordered set {b _α(1),b _α(2),…,b _α(n)} is a set of left triangulating idempotents of R.

If n ₁=n, then we are finished. So consider n ₁<n and let q=α(n ₁+1), where α(n ₁+1) is the smallest positive integer i such that $b_{i}\not\in S_{1}$. Observe that b _q is the first element in this new ordering which is not in S ₁.

We show that b _q ∈S ₂. For this, let y be the sum of all elements in S ₁. Thus, $y=b_{\alpha(1)}+\cdots+b_{\alpha(n_{1})}$. Let g be the sum of all elements in {b _α(1),…,b _α(n)} which are not in $\{b_{q}, b_{\alpha(1)},\dots, b_{\alpha(n_{1})}\}$. Then 1=y+b _q +g. Thus 1−y=b _q +g, and therefore b _q ∈(1−y)R(1−y). Now for every a∈R, we can see that

$$\begin{aligned} (1-y)a(1-y)b_q =& (1-y)ab_q=(b_q+g)ab_q\\ =&b_qab_q=b_q(1-y)a(1-y)b_q \end{aligned}$$

as b _q (1−y)=b _q , (1−y)b _q =b _q , and gab _q =0. So b _q ∈S _ℓ ((1−y)R(1−y)).

Consequently, q∈I ₂ and hence b _q ∈S ₂. Either this exhausts the elements in S ₂ or (in the ordering given by α) there is an element b _p ∈S ₂ beyond b _q . Use Proposition 5.3.1 as before to obtain a set of left triangulating idempotents of R of the form $\{b_{\alpha(1)},\dots, b_{\alpha(n_{1})}, b_{p}, b_{q}, b_{\alpha(n_{1}+2)},\dots, b_{\alpha(n)}\}$.

Repeat this process using elements of S ₂ until they are exhausted. Then there exists a permutation γ on {1,…,n} such that

$$\{b_{\gamma(1)},\dots, b_{\gamma(n_1)}, b_{\gamma(n_1+1)},\dots, b_{\gamma(n_2)},\dots, b_{\gamma(n)}\} $$

forms a set of left triangulating idempotents, where γ(i)=α(i) for $1\leq i\leq n_{1}, b_{\gamma(n_{2})} = b_{q}$, and $\{b_{\gamma(n_{1}+1)},\dots, b_{\gamma(n_{2})}\} = S_{2}$.

Now either S ₁ ∪ S ₂={b ₁,…,b _n } or we can continue the process on S ₃, and so on. After k steps, k≤n, the process terminates in a set of left triangulating idempotents of R in a canonical form. So we obtain a permutation ψ so that S ₁,…,S _k is our desired partition of {b ₁…,b _n }. □

Theorems 5.1.4 and 5.3.2 provide a tool for a generalized triangular matrix representation of R in a special canonical form, which we give next.

Corollary 5.3.3

(Canonical Representation)

Let {b ₁,…,b _n }, S ₁,…,S _k , and ψ be as before. Then using 0=n ₀<n ₁<⋯<n _k , we have that $S_{j+1} = \{b_{\psi(n_{j}+1)},\dots, b_{\psi(n_{j+1})}\}, \,j = 0, 1,\dots, k-1$, and R is isomorphic to the n×n matrix [A(i,j)], where the A(i,j) are n _i ×n _j block matrices

$$A(i+1, i+1) = \begin{bmatrix} b_{\psi(n_i+1)}Rb_{\psi(n_i+1)} & 0&\cdots&0\\ 0 & \ddots& { } & 0\\ \vdots& { } &\ddots& \vdots\\ 0 & 0 &\cdots& b_{\psi(n_{i+1})}Rb_{\psi(n_{i+1})} \end{bmatrix} ; $$

$$A(i+1, j+1) = \begin{bmatrix} b_{\psi(n_i+1)}Rb_{\psi(n_j+1)} &\cdots & b_{\psi(n_i+1)}Rb_{\psi(n_{j+1})}\\ \vdots&\ddots&\vdots\\ b_{\psi(n_{i+1})}Rb_{\psi(n_j+1)}& \cdots& b_{\psi(n_{i+1})}Rb_{\psi(n_{j+1})} \end{bmatrix} , $$

for i<j; and A(i,j)=0 for i>j, where i,j=0,1,…,k−1.

For the proof of Theorem 5.3.5, we need the following lemma.

Lemma 5.3.4

Let A and B be rings, and let M be an (A,B)-bimodule. Set $R = \begin{bmatrix} A & M\\ 0 & B \end{bmatrix}$, a generalized triangular matrix ring. Then

$$\begin{aligned} &\max\{\mathrm{r.gl.dim}(A), \mathrm{r.gl.dim}(B)\}\leq\mathrm{r.gl.dim} (R)\\ \leq& \max\{\mathrm{r.gl.dim}(A) + \mathrm{pd}(M_B) + 1, \,\mathrm{r.gl.dim}(B)\}, \end{aligned}$$

where pd(M _B ) is the projective dimension of M _B .

Proof

See [295, Proposition 7.5.1] for the proof. □

In Lemma 5.3.4, if M=0, then R=A⊕B (ring direct sum). Also

$$\text{r.gl.dim}(R)\leq \text{max}\{\text{r.gl.dim}(A)+\text{pd}(A_R), \text{r.gl.dim}(B)+\text{pd}(B_R)\} $$

from the proof of [295, Proposition 7.5.1]. As A _R and B _R are projective, it follows that pd(A _R )=0 and pd(B _R )=0, so

$$\text{r.gl.dim}(R)\leq \text{max}\{\text{r.gl.dim}(A), \text{r.gl.dim}(B)\}. $$

Thus, r.gl.dim(A⊕B)=max{r.gl.dim(A),r.gl.dim(B)} by Lemma 5.3.4.

As an application of canonical representation, we discuss the following result which exhibits a connection between the right global dimension of R and that of the sum of diagonal subrings.

Theorem 5.3.5

Let {b ₁,…,b _n } be a set of left triangulating idempotents of R, and S ₁,…,S _k be as in Corollary 5.3.3. Then

$$\mathrm{r.gl.dim}(D)\leq\mathrm{r.gl.dim}(R)\leq k\,(\mathrm{r.gl.dim}(D)) + k -1, $$

where D=b ₁ Rb ₁+⋯+b _n Rb _n . Thereby, $\mathrm{r.gl.dim}(R) < \infty$ if and only if $\mathrm{r.gl.dim}(D) < \infty$.

Proof

The proof is given by induction on k. If k=1, then R=D by Theorem 5.3.2 and we are finished. Assume that k≥2. We take $A = \sum_{b_{i}\in S_{1}}b_{i}Rb_{i}$, $M = \sum_{b_{i}\in S_{1}, b_{j}\in S_{2}\cup\cdots\cup S_{k}}b_{i}Rb_{j}$, and $B = (1 - \sum_{b_{i}\in S_{1}}b_{i}) \, R \, (1-\sum_{b_{i}\in S_{1}}b_{i})$. Then obviously $B = (\sum_{b_{j}\in S_{2}\cup\cdots\cup S_{k}}b_{j}) \, R \, (\sum_{b_{j}\in S_{2}\cup\cdots\cup S_{k}}b_{j})$.

We note that S ₂∪⋯∪S _k is a set of left triangulating idempotents of B and {S ₂,…,S _k } is a partition which establishes a canonical generalized triangular matrix representation for B. Let $D_{1} = \sum_{b_{j}\in S_{2}\cup\cdots\cup S_{k}}b_{j}Rb_{j}$. Then by induction r.gl.dim(D ₁)≤r.gl.dim(B)≤(k−1)(r.gl.dim(D ₁)) + k−2.

Because D=A⊕D ₁ from Theorem 5.3.2 or Corollary 5.3.3, it follows that r.gl.dim(D)=max {r.gl.dim(A),r.gl.dim(D ₁)}. Observe that $R = \begin{bmatrix} A & M\\ 0 & B \end{bmatrix}$ and M is an (A,B)-bimodule. Hence,

$$\begin{aligned} &\text{max}\,\{\text{r.gl.dim}(A), \text{r.gl.dim}(B)\}\leq\text{r.gl.dim} (R)\\ \leq& \text{max}\{\text{r.gl.dim}(A) + \text{pd}(M_B) + 1, \,\text {r.gl.dim}(B)\} \end{aligned}$$

from Lemma 5.3.4. Because r.gl.dim(D ₁)≤r.gl.dim(B),

$$\begin{aligned} \text{r.gl.dim}(D) =& \text{max}\,\{\text{r.gl.dim}(A), \,\text{r.gl.dim}(D_1)\}\\ \leq& \text{max}\,\{\text{r.gl.dim}(A), \,\,\,\text{r.gl.dim}(B)\}\\ \leq& \text{r.gl.dim}(R). \end{aligned}$$

We observe that pd(M _B )≤r.gl.dim(B). Therefore,

$$\begin{aligned} \text{r.gl.dim}(R) \leq&\text{max}\,\{\text {r.gl.dim}(A) + \text{pd}(M_B) + 1, \, \text{r.gl.dim}(B)\}\\ \leq& \text{max}\{\text{r.gl.dim}(A) + \text{r.gl.dim}(B) + 1, \, \text{r.gl.dim}(B)\}\\ = & \text{r.gl.dim}(A) + \text{r.gl.dim}(B) + 1\\ \leq& \text{r.gl.dim}(D) + [(k-1)\,(\text{r.gl.dim}(D_1)) + (k-2)] + 1\\ \leq& \text{r.gl.dim}(D) + (k-1)\,(\text{r.gl.dim}(D)) + k-1\\ = & k\,(\text{r.gl.dim}(D)) + k-1. \end{aligned}$$

Therefore, r.gl.dim(D)≤r.gl.dim(R)≤k (r.gl.dim(D))+k−1. Thereby, r.gl.dim(R)<∞ if and only if r.gl.dim(D)<∞. □

5.4 Piecewise Prime Rings and Piecewise Domains

In this section, a criterion for a ring with a complete set of triangulating idempotents to be quasi-Baer is provided. Also a structure theorem for a quasi-Baer ring with a complete set of triangulating idempotents is shown. Among the applications of this structure theorem, several well-known results are obtained as its consequences. These include Levy’s decomposition theorem of semiprime right Goldie rings, Faith’s characterization of semiprime right FPF rings with no infinite set of central orthogonal idempotents, Gordon and Small’s characterization of piecewise domains, and Chatters’ decomposition theorem of hereditary Noetherian rings. A result related to Michler’s splitting theorem for right hereditary right Noetherian rings is also obtained as an application.

The next result provides a criterion for a ring with a complete set of left triangulating idempotents to be quasi-Baer.

Theorem 5.4.1

Assume that a ring R has a complete set of left triangulating idempotents with Tdim(R)=n. Then the following are equivalent.

• (i) R is quasi-Baer.

• (ii) For any complete set of left triangulating idempotents {b ₁,…,b _n } of R, if b _i xb _j Rb _j yb _k =0 for some x,y∈R and some 1≤i,j,k≤n, then either b _i xb _j =0 or b _j yb _k =0.

• (iii) There is a complete set of left triangulating idempotents {c ₁,…,c _n } of R such that if c _i xc _j Rc _j yc _k =0 for some x,y∈R and some 1≤i,j,k≤n, then either c _i xc _j =0 or c _j yc _k =0.

• (iv) For any complete set of left triangulating idempotents {b ₁,…,b _n }, assume that Kb _j V=0 for some ideals K and V of R and some b _j , 1≤j≤n. Then either Kb _j =0 or b _j V=0.

Proof

(i)⇒(ii) Assume that b _i xb _j Rb _j yb _k =0 for some x,y∈R and some 1≤i,j,k≤n. Since R is quasi-Baer, r _R (b _i xb _j R)=fR for some f∈S _ℓ (R). By Lemma 5.1.5(ii), b _j fb _j ∈S _ℓ (b _j Rb _j ). As {b ₁,…,b _n } is a complete set of left triangulating idempotents, S _ℓ (b _j Rb _j )={0, b _j }. So either b _j fb _j =0 or b _j fb _j =b _j . If b _j fb _j =0, then since b _j yb _k ∈r _R (b _i xb _j R)=fR, we have that b _j yb _k =fb _j yb _k . So b _j yb _k =b _j fb _j yb _k =0. On the other hand, if b _j fb _j =b _j , then b _i xb _j =b _i xb _j fb _j =0 as b _i xb _j f=0.

(ii)⇒(iii) It follows immediately because R has a complete set of left triangulating idempotents.

(iii)⇒(i) Say L is a left ideal of R. First, assume that Rc _i ∩ℓ _R (L)≠0 for some i. Then we may assume that

$$Rc_1\cap \ell_R(L)\neq 0,\dots,Rc_m\cap \ell_R(L)\neq 0, $$

and

$$Rc_{m+1}\cap \ell_R(L) = 0,\dots, Rc_n\cap \ell_R(L) = 0. $$

Thus ℓ _R (L)Rc _m+1=0,… , and ℓ _R (L)Rc _n =0. Put T=Rc ₁+⋯+Rc _m .

Say v∈ℓ _R (L). Then v=v(c ₁+⋯+c _n )=vc ₁+⋯+vc _m ∈T. Therefore, ℓ _R (L)⊆T. To show that c ₁∈ℓ _R (L), take y∈L. Since Rc ₁∩ℓ _R (L)≠0, there exists x∈R such that 0≠xc ₁∈Rc ₁∩ℓ _R (L). So xc ₁ Rc ₁ y=0. Now there is c _k xc ₁≠0 for some c _k because 1=c ₁+⋯+c _n . Thus, c _k xc ₁ Rc ₁ yc _j =0 for all j. Therefore c ₁ yc _j =0 for all j, and so c ₁ y=0. Hence, c ₁∈ℓ _R (L). Thus, Rc ₁⊆ℓ _R (L). Similarly, Rc ₂,…,Rc _m ⊆ℓ _R (L). So T⊆ℓ _R (L). Therefore, ℓ _R (L)=T=Rc ₁+⋯+Rc _m =R(c ₁+⋯+c _m ). Put e=c ₁+⋯+c _m . Then e ²=e∈R and so ℓ _R (L)=Re.

Next, assume that Rc _i ∩ℓ _R (L)=0 for all i. Then ℓ _R (L)Rc _i =0 for all i. So ℓ _R (L)=ℓ _R (L)(Rc ₁+⋯+Rc _n )=0. Therefore, R is quasi-Baer.

(ii)⇒(iv) Let Kb _j V=0 and b _j V≠0 for some b _j . Say y∈V with b _j y≠0. So $0\neq b_{j}y=\sum_{t=1}^{n}b_{j}yb_{t}$, hence b _j yb _k ≠0 for some b _k .

Let x∈K. Then xb _j Rb _j y=0. Hence b _i xb _j Rb _j yb _k =0 for each b _i . As b _j yb _k ≠0, b _i xb _j =0 for all b _i . Thus $xb_{j}=\sum_{i=1}^{n}b_{i}xb_{j}=0$, so Kb _j =0. If Kb _j ≠0, similarly b _j V=0.

(iv)⇒(ii) If b _i xb _j Rb _j yb _k =0, then (Rb _i xb _j R)b _j (Rb _j yb _k R)=0. By assumption Rb _i xb _j R=0 or Rb _j yb _k R=0, so b _i xb _j =0 or b _j yb _k =0. □

Corollary 5.4.2

If R has a complete set of primitive idempotents, then the following are equivalent.

• (i) R is quasi-Baer.

• (ii) For any given complete set of primitive idempotents {e ₁,…,e _n }, if e _i xe _j Re _j ye _k =0 for some x,y∈R and some 1≤i,j,k≤n, then either e _i xe _j =0 or e _j ye _k =0.

• (iii) There is a complete set of primitive idempotents {f ₁…,f _m } of R such that if f _i xf _j Rf _j yf _k =0 for some x,y∈R and some 1≤i,j,k≤m, then either f _i xf _j =0 or f _j yf _k =0.

• (iv) For any complete set of primitive idempotents {g ₁,…,g _ℓ }, assume that Kg _j V=0 for some ideals K and V of R and for some g _j , 1≤j≤ℓ. Then either Kg _j =0 or g _j V=0.

Proof

Let f∈S _ℓ (R) and 0≠e ²=e∈R. Then efe∈S _ℓ (eRe) by Lemma 5.1.5(ii). In particular, if e is primitive, then S _ℓ (eRe)={0,e}. So either efe=0 or efe=e. The proof can then be completed by using a similar argument as in the proof of Theorem 5.4.1. □

Definition 5.4.3

A ring R is called a piecewise domain (or simply, PWD) if there is a complete set of primitive idempotents {e ₁,…,e _n } such that xy=0 implies x=0 or y=0 whenever x∈e _i Re _j and y∈e _j Re _k , for 1≤i, j, k≤n.

To avoid ambiguity, we sometimes say that R is a PWD with respect to a complete set $\{e_{i}\}_{i=1}^{n}$ of primitive idempotents. In light of Theorem 5.4.1 and Corollary 5.4.2, it is interesting to compare quasi-Baer rings having a complete set of left triangulating (or primitive) idempotents with PWDs. In fact, Definition 5.4.3 and the equivalence of (i) and (iii) in Theorem 5.4.1 and Corollary 5.4.2 suggest the following definition.

Definition 5.4.4

A quasi-Baer ring with a complete set of triangulating idempotents is called a piecewise prime ring (or simply, PWP ring).

The following result is somewhat of a right p.q.-Baer analogue of Theorem 3.1.25.

Proposition 5.4.5

Let R be a right p.q.-Baer ring with Tdim(R)<∞. Then R is a PWP ring.

Proof

Let I be a right ideal of R, and say I=∑ _i∈Λ x _i R with x _i ∈R. Then r _R (I)=∩ _i∈Λ r _R (x _i R)=∩ _i∈Λ e _i R with e _i ∈S _ℓ (R) for each i∈Λ because R is right p.q.-Baer. By Theorem 5.2.5 and Proposition 1.2.4(i), there exists e∈S _ℓ (R) such that ∑ _i∈Λ e _i R=eR. Therefore R is a PWP ring. □

The next question was posed by Gordon and Small (see [187, p. 554]): Can a PWD R possess a complete set $\{f_{i}\}_{i=1}^{m}$ of primitive idempotents for which it is not true that xy=0 implies x=0 or y=0 for some x∈f _i Rf _k and y∈f _k Rf _j ? Theorem 5.4.1 and Corollary 5.4.2 show that if R is a PWP ring, then it is a PWP ring with respect to any complete set of left triangulating idempotents. Thereby for the case of PWP rings it provides an answer to the above question.

Proposition 5.4.6

Any PWD is a PWP ring.

Proof

The result follows from Proposition 5.2.13 and Corollary 5.4.2. □

The following example illustrates that the converse of Proposition 5.4.6 does not hold true.

Example 5.4.7

(i) Let R be the ring in Example 3.2.7(ii). Then R is a PWP ring, but it is not a PWD.

(ii) Let R be the ring of Example 5.2.17(i). Then R is a prime ring, so it is a PWP ring. But R does not have a complete set of primitive idempotents. Thus, R is not a PWD.

Example 5.4.8

There is a PWD which is not Baer. Let R be a commutative domain which is not semihereditary (e.g., $\mathbb{Z}[x]$). Then Mat _n (R) is a PWD for any positive integer n>1, but it is not a Baer ring (see Theorem 6.1.4).

Proposition 5.4.9

Let R be a ring and {e ₁,…,e _n } be a complete set of primitive idempotents of R. Then the following are equivalent.

• (i) R is a PWD with respect to {e ₁,…,e _n }.

• (ii) Every nonzero element of Hom(e _i R _R ,e _j R _R ) is a monomorphism for all i,j, 1≤i, j≤n.

• (iii) Every nonzero element of Hom(e _i R _R ,R _R ) is a monomorphism for all i, 1≤i≤n.

Proof

Exercise. □

Example 5.4.10

(i) It is routine to check that the ring of n×n matrices over a PWD is a PWD.

(ii) The polynomial ring over a PWD is a PWD. Indeed, say R is a PWD with respect to a complete set of primitive idempotent {e ₁,…,e _n }. Then {e ₁,…,e _n } is a complete set of primitive idempotents of R[x], and R[x] is a PWD with respect to {e ₁,…,e _n }.

(iii) A right Rickart ring with a complete set of primitive idempotents is a PWD. In fact, say R is a right Rickart ring with a complete set {e ₁,…,e _n } of primitive idempotents.

Suppose that e _i xe _j e _j ye _k =0, where x,y∈R and 1≤i,j,k≤n. Since R is right Rickart, r _R (e _i xe _j )=fR for some f ²=f∈R. So 1−e _j =f(1−e _j ) since 1−e _j ∈r _R (e _i xe _j ). Note that $1-e_{j}=\sum_{k\neq j}^{n}e_{k}$, thus

$$\sum_{k\neq j}e_k=1-e_j=f(1-e_j)=\sum_{k\neq j}fe_k. $$

Hence e _k =fe _k for k≠j and 1≤k≤n. Therefore,

$$f=\sum_{k=1}^nfe_k=\sum_{k\neq j}^nfe_k+fe_j= \sum_{k\neq j}^ne_k+fe_j. $$

Thus $1-f=1-\sum_{k\neq j}^{n}e_{k}-fe_{j}=e_{j}-fe_{j}=(1-f)e_{j}$, so R(1−f)⊆Re _j . Hence, it follows that R(1−f)=Re _j or R(1−f)=0 as e _j is a primitive idempotent.

If R(1−f)=Re _j , then e _j f=0. Because e _i xe _j e _j ye _k =e _i xe _j ye _k =0, we get that ye _k ∈r _R (e _i xe _j )=fR, and ye _k =fye _k . Hence, e _j ye _k =e _j fye _k =0. Finally, assume that R(1−f)=0. Then f=1, and thus e _i xe _j =0. So R is a PWD.

If R(1−f)=Re _j , then e _j f=0. Because e _i xe _j e _j ye _k =e _i xe _j ye _k =0, we get ye _k ∈r _R (e _i xe _j )=fR, and therefore ye _k =fye _k . Hence e _j ye _k =e _j fye _k =0.

Further, if R(1−f)=0, then f=1, and thus e _i xe _j =0. So R is a PWD.

(iv) There exists a PWD which is not right Rickart. Let $R=\text{Mat}_{2}(\mathbb{Z}[x])$. Then R is a PWD by part (i), but R is not (right) Rickart (see Example 3.1.28).

(v) A right nonsingular ring which is a direct sum of uniform right ideals is a PWD. Indeed, let R be a right nonsingular ring such that $R=\oplus_{i=1}^{n}I_{i}$, where each I _i is a uniform right ideal of R. Then there is a complete set of primitive idempotents {e ₁,…,e _n } with I _i =e _i R for each i. As Z(R _R )=0, by Corollary 1.3.15 E(R _R )=Q(R). Now Q(R) is a regular ring from Theorem 2.1.31 and Q(R)=e ₁ Q(R)⊕⋯⊕e _n Q(R). Also each e _i Q(R) _Q(R) is uniform, so {e ₁,…,e _n } is a complete set of primitive idempotents in Q(R). Thus, Q(R) is semisimple Artinian. Say e _i xe _j e _j ye _k =0, where x,y∈R and 1≤i,j,k≤n. Then since Q(R) is a PWD with respect to {e ₁,…,e _n } by part (iii), either e _i xe _j =0 or e _j ye _k =0. So R is a PWD.

Proposition 5.4.11

Let {b ₁,…,b _n } be a set of left triangulating idempotents of a ring R. Then the following are equivalent.

• (i) P is a (minimal) prime ideal of R.

• (ii) There exist m, 1≤m≤n, and a (minimal) prime ideal P _m of the ring b _m Rb _m such that P=P _m +∑ _k≠m b _k Rb _k +∑ _i≠j b _i Rb _j .

Proof

The proof is routine. □

Theorem 5.4.12

Let R be a PWP ring with Tdim(R)=n. Then R=A⨁B (ring direct sum) such that:

• (i) $A = \bigoplus_{i=1}^{k}A_{i}$ is a direct sum of prime rings A _i .

• (ii) There exists a ring isomorphism

  $$B\cong \begin{bmatrix} B_1 & B_{12} &\cdots& B_{1m}\\ 0 & B_2 &\cdots& B_{2m}\\ \vdots& \vdots& \ddots& \vdots\\ 0 & 0& \cdots& B_m \end{bmatrix} , $$

  where each B _i is a prime ring, and B _ij is a (B _i ,B _j )-bimodule.

• (iii) n=k+m.

• (iv) For each i∈{1,…,m} there is j∈{1,…,m} such that B _ij ≠0 or B _ji ≠0.

• (v) The rings B ₁,…,B _m are uniquely determined by B up to isomorphism and permutation.

• (vi) B has exactly m minimal prime ideals P ₁,…,P _m , R has exactly n minimal prime ideals of the form A⊕P _i or C _i ⊕B where C _i =⨁ _j≠i A _j . Further, P ₁,…,P _m are comaximal, P(R)=P(B), and P(R)^m=0.

Proof

Say E={b ₁,b ₂,…,b _n } is a complete set of left triangulating idempotents of R.

(i) Let $\{e_{1},\dots, e_{k}\} = E\cap\, \mathcal{B}(R)$. Take A _i =e _i R. By Proposition 3.2.5 and Theorem 3.2.10, each A _i is a prime ring.

(ii) Let {f ₁,…,f _m }=E∖{e ₁,…,e _k }, where the f _i are maintained in the same relative order as they were in E. Let B _i =f _i Bf _i and B _ij =f _i Bf _j . Then each B _i is a prime ring by Proposition 3.2.5 and Theorem 3.2.10. Define ϕ by ϕ(b)=[f _i bf _j ] for b∈B, as in the proof of Theorem 5.1.4. Then ϕ is a ring isomorphism.

(iii) The proof follows immediately from the proof of part (ii).

(iv) It is evident since {f ₁,…,f _m }=E∖{e ₁,,…,e _k }.

(v) This is a consequence of Theorem 5.2.8.

(vi) The proof follows from a routine argument using Lemma 5.4.11. □

Corollary 5.4.13

(i) Any semiprime PWP ring is a finite direct sum of prime rings.

(ii) Any biregular ring R with Tdim(R)<∞ is a finite direct sum of simple rings.

Proof

The proof follows from Theorems 5.4.12 and 3.2.22(ii). □

The next corollary is related to Michler’s splitting theorem [299, Theorem 2.2] for right hereditary right Noetherian rings.

Corollary 5.4.14

Let R be a right hereditary right Noetherian ring. Then

$$R\cong \begin{bmatrix} R_1 & R_{12} & \cdots& R_{1n}\\ 0 & R_2 &\cdots& R_{2n}\\ \vdots& \vdots& \ddots& \vdots\\ 0 & 0 &\cdots& R_n \end{bmatrix} , $$

where each R _i is a prime right hereditary, right Noetherian ring, and each R _ij is an (R _i ,R _j )-bimodule.

Proof

As R is right hereditary right Noetherian, R is Baer by Theorem 3.1.25. Thus the proof follows from Theorem 5.4.12 and Proposition 5.2.14. □

We will now see that Levy’s decomposition theorem [279] for semiprime right Goldie right hereditary rings, follows as a consequence of Theorem 5.4.12.

Corollary 5.4.15

Any semiprime right Goldie, right hereditary ring is a finite direct sum of prime right Goldie, right hereditary rings.

Proof

Let R be a semiprime right Goldie, right hereditary ring. Then R is orthogonally finite, so R is Baer by Theorem 3.1.25 and Tdim(R)<∞ from Proposition 5.2.13(ii). Corollary 5.4.13(i) and a routine verification yield that R is a finite direct sum prime right Goldie, right hereditary rings. □

A ring R is called right FPF if every faithful finitely generated right R-module generates the category Mod-R of right R-modules (see [156]). We may note that a semiprime right FPF ring is quasi-Baer (see [78, Corollary 1.19]). By Theorem 5.4.12, Faith’s characterization of semiprime right FPF rings with no infinite set of central orthogonal idempotents (see [156, Theorem I.4]) is provided as follows.

Corollary 5.4.16

Let R be a ring with no infinite set of central orthogonal idempotents. Then R is semiprime right FPF if and only if R is a finite direct sum of prime right FPF rings.

Proof

Let R be a semiprime right FPF ring with no infinite set of central orthogonal idempotents. Because R is semiprime, $\mathcal {B}(R)=\mathbf{ S}_{\ell}(R)$ by Proposition 1.2.6(ii). Since R has no infinite set of central orthogonal idempotents, we see that

$$\{eR\mid e\in\mathbf{S}_{\ell}(R)\}=\{eR\mid e\in\mathcal{B}(R)\} $$

has ACC and DCC. By Theorem 5.2.5, Tdim(R)<∞, so R is a PWP ring. By Corollary 5.4.13(i), R is a finite direct sum of prime rings. Since ring direct summands of right FPF rings are right FPF, these prime rings are right FPF. The converse is immediate. □

A ring R for which the diagonal rings R _i in a complete generalized triangular matrix representation are simple Artinian, is called a TSA ring. Recall from 1.1.14 that if R is a right (or left) perfect ring, then J(R)=P(R). Thus any prime right (or left) perfect ring is simple Artinian.

By Theorem 5.4.12, every quasi-Baer right (or left) perfect ring is a TSA ring. So Teply’s result [391] given next follows from Theorem 5.4.12 since an orthogonally finite right Rickart ring is Baer by Theorem 3.1.25.

Corollary 5.4.17

A right (or left) perfect right Rickart ring is a semiprimary TSA ring.

For a π-regular Baer ring with only countably many idempotents, we obtain the following.

Corollary 5.4.18

A π-regular Baer ring with only countably many idempotents is a semiprimary TSA ring.

Proof

Theorems 3.1.11, 3.1.26, and 5.4.12 yield the result. □

Corollary 5.4.19

Assume that R is a PWP ring with Tdim(R)=n. Then the following are equivalent.

• (i) $\mathrm{r.gl.dim}(R)<\infty$.

• (ii) $\mathrm{r.gl.dim}(R/P(R))<\infty$.

• (iii) $\mathrm{r.gl.dim}(R_{1}+\cdots+ R_{n})<\infty$, where the R _i are the diagonal rings in the complete generalized triangular matrix representation of R.

Proof

(i)⇔(iii) is a direct consequence of Theorem 5.3.5. From Theorem 5.4.12, R/P(R)≅R ₁⊕⋯⊕R _n . Hence, (ii)⇔(iii) follows immediately. □

Theorem 5.4.20

Let R be a right p.q.-Baer ring. Then Tdim(R)=n if and only if R has exactly n minimal prime ideals.

Proof

Assume that Tdim (R)=n. By Proposition 5.4.5, R is a PWP ring. Thus from Theorem 5.4.12, R has exactly n minimal prime ideals.

Conversely, let R have exactly n minimal prime ideals. We proceed by induction on n. First, say n=1. If Tdim(R)≠1, then R is not semicentral reduced. So there is 0≠b∈S _ℓ (R) with b≠1. Then bRb and (1−b)R(1−b) each have at least one minimal prime ideal. Note that {b,1−b} is a set of left triangulating idempotents of R. Thus, by Proposition 5.4.11, R has at least two minimal prime ideals, a contradiction. Hence, Tdim(R)=1.

Suppose that n>1. If R is semicentral reduced, then R is prime by Proposition 3.2.25. So n=1, a contradiction. Thus R is not semicentral reduced, hence there is 0≠d∈S _ℓ (R) and d≠1. By Theorem 3.2.34(i), both dRd and (1−d)R(1−d) are right p.q.-Baer rings. We note that {d,1−d} is a set of left triangulating idempotents. From Proposition 5.4.11, there are some positive integers k ₁ and k ₂ such that dRd and (1−d)R(1−d) have exactly k ₁ and k ₂ number of minimal prime ideals, respectively, where k ₁+k ₂=n.

By induction, Tdim(dRd)+Tdim((1−d)R(1−d))=k ₁+k ₂=n. From Theorem 5.2.19, Tdim(R)=n. □

Corollary 5.4.21

The PWP property is Morita invariant.

Proof

Assume that R and S are Morita equivalent rings. Suppose that R is a PWP ring and let Tdim(R)=n. By Theorem 5.4.20, R has exactly n minimal prime ideals. Since R is quasi-Baer, S is also quasi-Baer from Theorem 3.2.11. Now S has also exactly n minimal prime ideals because R and S are Morita equivalent (see [262, Proposition 18.44 and Corollary 18.45]). Thus Tdim(S)=n by Theorem 5.4.20, so S is also a PWP ring. □

The next example illustrates that the right p.q.-Baer condition is not superfluous in Theorem 5.4.20.

Example 5.4.22

There exists a ring R such that:

• (i) R has only two minimal prime ideals.

• (ii) Tdim(R)=1.

Indeed, we let F{X,Y} be the free algebra over a field F, and we put R=F{X,Y}/I, where I is the ideal of F{X,Y} generated by YX. Say x=X+I and y=Y+I in R. Then R/RxR≅F[y] and R/RyR≅F[x], so RxR and RyR are prime ideals of R. As yx=0, we see that (RyR)(RxR)=0. So, if P is a prime ideal, then either RyR⊆P or RxR⊆P. Thus RxR and RyR are the only two minimal prime ideals of R. We can verify that all idempotents of R are only 0 and 1. In particular, R is semicentral reduced, so Tdim(R)=1.

Let R be a quasi-Baer (resp., Baer) ring with Tdim (R)<∞. Then P(R) is nilpotent and R/P(R) is a finite direct sum of prime (resp., Baer) rings from Theorem 5.4.12, so R/P(R) is a quasi-Baer (resp., Baer) ring (cf. Example 3.2.42). There is a quasi-Baer ring R with P(R) nilpotent, but Tdim(R) is infinite. Let $R=T_{2}(\prod_{n=1}^{\infty}F_{n})$, where F is a field, and F _n =F,n=1,2,… . In this case, P(R)²=0, but Tdim(R)=∞.

An R-module M is said to satisfy the restricted minimum condition if, for every essential submodule N of M, the module M/N is Artinian.

Lemma 5.4.23

Let R be a hereditary Noetherian ring. Then both R _R and _R R satisfy the restricted minimum condition.

Proof

Assume that J _R ≤^ess R _R . Then J _R is finitely generated projective because R is right hereditary and right Noetherian. From Dual Basis lemma (see [262, Lemma 2.9]), there are a ₁,…,a _n ∈J and f ₁,…,f _n ∈Hom(J _R ,R _R ) such that x=a ₁ f ₁(x)+⋯+a _n f _n (x) for each x∈J. Because Z(R _R )=0 from Proposition 3.1.18, J _R ≤^den R _R by Proposition 1.3.14. Thus, it follows that f _i ∈Q(R) for i=1,…,n, so a ₁ f ₁+⋯+a _n f _n ∈Q(R). We note that a ₁ f ₁+⋯+a _n f _n =1 in Q(R) as a ₁ f ₁+⋯+a _n f _n is the identity map of J.

Put D(J)=Hom (J _R ,R _R ). Then Rf ₁+⋯+Rf _n ⊆D(J) because D(J) is a left R-module. Let q∈D(J). Then qJ⊆R and so

$$q = q(a_1f_1+\cdots+ a_nf_n) = qa_1f_1 +\cdots+qa_nf_n\in Rf_1+\cdots+ Rf_n $$

since each a _i ∈J. So D(J)=Rf ₁+⋯+Rf _n .

Furthermore, J={r∈R∣D(J)r⊆R}. Indeed, first obviously we have that J⊆{r∈R∣D(J)r⊆R}. Next, we take r∈R such that D(J)r⊆R. Then

$$r = a_1f_1r+\cdots + a_nf_nr\in a_1D(J)r +\cdots+ a_nD(J)r\subseteq JR\subseteq J $$

since 1=a ₁ f ₁+⋯+a _n f _n in Q(R). So J={r∈R∣D(J)r⊆R}.

We show that R _R satisfies the restricted minimum condition. For this, we now let I ₁⊇I ₂⊇… be a descending chain of right ideals of R all containing a fixed essential right ideal I of R. Then D(I ₁)⊆D(I ₂)⊆… and all D(I _i ) are contained in the left R-module D(I). By the preceding argument, D(I) is finitely generated as a left R-module.

Since R is left Noetherian, D(I) is Noetherian as a left R-module. So there exists a positive integer n such that D(I _n )=D(I _n+1)=… . Therefore, we have that {r∈R∣D(I _n )r⊆R}={r∈R∣D(I _n+1)r⊆R}=… . Hence I _n =I _n+1=… , so R _R satisfies the restricted minimum condition. Similarly, _R R has the restricted minimum condition. □

As another application of Theorem 5.4.12, Chatters’ decomposition theorem [117] for hereditary Noetherian rings is shown as follows.

Theorem 5.4.24

If R is a hereditary Noetherian ring, then R=A⊕B (ring direct sum), where A is a finite direct sum of prime rings and B is an Artinian TSA ring.

Proof

Note that a hereditary Noetherian ring is Baer by Theorem 3.1.25. Thus R is a PWP ring. Therefore, R=A⊕B as in Theorem 5.4.12.

We claim that B is an Artinian TSA ring. For this, say {f ₁,…,f _m } is a complete set of left triangulating idempotents of B as in the proof of Theorem 5.4.12. We need to show that each B _i is simple Artinian. By Theorem 5.4.12, for given i, 1≤i≤m there exists j, 1≤j≤m such that either B _ij ≠0 or B _ji ≠0. We may assume that B _ij ≠0 and i<j. Now B _i =f _i Bf _i , B _ij =f _i Bf _j , and B _j =f _j Bf _j . Consider

$$S = (f_i + f_j)B(f_i + f_j)\cong\begin{bmatrix} B_i & B_{ij}\\ 0 & B_j\end{bmatrix}. $$

Then S is a hereditary Noetherian ring. Also {f _i ,f _j } is a complete set of left triangulating idempotents of S. Since B is Baer, so is S by Theorem 3.1.8. Therefore, S is a PWP ring.

We show that B _ij is a faithful left B _i -module. For this, let f _i bf _i ∈B _i with b∈B such that f _i bf _i B _ij =0. Since f _i Bf _j =B _ij ≠0, there exists y∈B such that f _i yf _j ≠0. Now (f _i bf _i )(f _i Bf _i yf _j )⊆(f _i bf _i )(f _i Bf _j )=0, and so we have that f _i bf _i Bf _i yf _j =(f _i bf _i )(f _i Bf _i yf _j )=0. Since f _i yf _j ≠0, f _i bf _i =0 from Theorem 5.4.1. Therefore, B _ij is a faithful left B _i -module. Similarly, B _ij is a faithful right B _j -module. Let

$$V_1 = \begin{bmatrix} 0 & B_{ij}\\ 0 & B_j \end{bmatrix} \text{ and } V_2 = \begin{bmatrix} B_i & B_{ij}\\ 0 & 0 \end{bmatrix} . $$

The ideal V ₁ of S is right essential in S since B _ij is a faithful left B _i -module. Also the ideal V ₂ of S is left essential in S. Since both S _S and _S S satisfy the restricted minimum condition by Lemma 5.4.23, S/V ₁ is a right Artinian S-module, while S/V ₂ is a left Artinian S-module. Now to show that B _i is a right Artinian ring, we let I ₁⊇I ₂⊇… be a descending chain of right ideals of B _i . Put

$$K_{\ell}=\left\{ \begin{bmatrix} \alpha \,& \,0\\ 0 \,& \,0 \end{bmatrix} + V_1\in S/V_1\mid\alpha\in I_{\ell}\right\} $$

for ℓ=1,2,… . Then we see that each K _ℓ is a right S-submodule of (S/V ₁) _S and K ₁⊇K ₂⊇… . Since (S/V ₁) _S is Artinian, K _t =K _t+1=… for some positive integer t. So I _t =I _t+1=… . Therefore, B _i is a right Artinian ring. Similarly, B _j is a left Artinian ring. Since B _i and B _j are prime rings by Theorem 5.4.12, B _i and B _j are simple Artinian rings.

The preceding argument is applied to show that all B _i are simple Artinian rings. Now J(B)=∑ _i≠j B _ij is nilpotent and B/J(B)=B ₁⊕⋯⊕B _m . Hence, B is semiprimary Noetherian. So B is an Artinian TSA ring. □

To obtain a structure theorem for PWDs, we need the next lemma.

Lemma 5.4.25

If R is a PWD and 0≠e∈S _ℓ (R)∪S _r (R), then the ring eRe is also a PWD.

Proof

Say e∈S _ℓ (R). Let R be a PWD with respect to a complete set of primitive idempotents {e ₁,…,e _n }. Since e∈S _ℓ (R), e _i e=ee _i e is an idempotent for each i. As e _i is primitive and e _i eR⊆e _i R, either e _i e=0 or e _i eR=e _i R. If necessary, rearrange {e ₁,…,e _n } so that J={1,…,r} is the set of all indices such that e _i e≠0 for all i∈J. Then e=(e ₁+⋯+e _n )e=e ₁ e+⋯+e _r e and

$$eR = e_1eR +\cdots + e_reR = e_1R+\cdots + e_rR. $$

Further, by Lemma 5.2.12, {ee ₁ e,…,ee _r e} is a complete set of primitive idempotents in eRe.

Assume that x∈(ee _i e)(eRe)(ee _j e) and y∈(ee _j e)(eRe)(ee _k e) with xy=0 for 1≤i, j, k≤r. Put x=(ee _i e)(eae)(ee _j e) and y=(ee _j e)(ebe)(ee _k e) with a,b∈R. Then x=e _i ae _j e since e∈S _ℓ (R). Similarly, y=e _j be _k e. Thus xy=e _i ae _j ee _j be _k e=e _i ae _j be _k e=0. So e _i ae _j e _j be _k eR=e _i ae _j e _j be _k R=0 since e _k eR=e _k R. Hence (e _i ae _j )(e _j be _k )=0, so e _i ae _j =0 or e _j be _k =0 as R is a PWD. Thus x=0 or y=0. Therefore, eRe is a PWD with respect to the complete set of primitive idempotents {ee ₁ e,…,ee _r e}. Similarly, when e∈S _r (R), we see that eRe is a PWD. □

As yet another application of Theorem 5.4.12, we obtain the next theorem, due to Gordon and Small [187], which describes the structure of a PWD.

Theorem 5.4.26

Assume that R is a PWD. Then

$$R\cong \begin{bmatrix} R_1 & R_{12} & \cdots& R_{1n}\\ 0 & R_2 &\cdots& R_{2n}\\ \vdots& \vdots& \ddots& \vdots\\ 0 & 0 &\cdots& R_n \end{bmatrix} , $$

where each R _i is a prime PWD and each R _ij is an (R _i ,R _j )-bimodule. The integer n is unique and the ring R _i is unique up to isomorphism. Furthermore,

$$R_i\cong\begin{bmatrix} D_1 &\cdots& D_{1n_{i}}\\ \vdots& \ddots& \vdots\\ D_{n_{i}1} & \cdots& D_{n_i}\end{bmatrix}, $$

where each D _i is a domain and each D _jk is isomorphic as a right D _k -module to a nonzero right ideal in D _k , and as a left D _j -module to a nonzero left ideal in D _j .

Proof

Let R be a PWD. By Proposition 5.4.6, R is a PWP ring. The uniqueness of n and that of the ring R _i up to isomorphism follow from Theorem 5.2.8 or Theorem 5.4.12.

Say {b ₁,…,b _n } is a complete set of left triangulating idempotents of R. By Theorem 5.4.12, each R _i =b _i Rb _i is a prime ring. From Lemma 5.4.25, R ₁=b ₁ Rb ₁ and (1−b ₁)R(1−b ₁) are PWDs.

We observe that 0≠b ₂∈S _ℓ ((1−b ₁)R(1−b ₁)). Thus, Lemma 5.4.25 yields that R ₂=b ₂ Rb ₂=b ₂(1−b ₁)R(1−b ₁)b ₂ is a PWD. By the same method, we see that each R _i =b _i Rb _i is a PWD. Hence, there exists a complete set of primitive idempotents $\{c_{1},\dots, c_{n_{i}}\}$ for R _i such that c _j xc _k yc _q =0 implies that c _j xc _k =0 or c _k yc _q =0, for x,y∈R _i . Put D _jk =c _j R _i c _k and D _i =D _ii . Then each D _i is a domain.

As R _i is a prime ring and 0≠c _k ,0≠c _j ∈R _i , it follows that c _k R _i c _j ≠0. We let 0≠x∈c _k R _i c _j . Then c _j R _i c _k is isomorphic to a nonzero right ideal xc _j R _i c _k of c _k R _i c _k as a right c _k R _i c _k -module since R _i is a PWD with respect to the complete set of primitive idempotents $\{c_{1},\dots, c_{n_{i}}\}$. Similarly c _j R _i c _k is isomorphic to a nonzero left ideal of c _j R _i c _j as a left c _j R _i c _j -module. □

Exercise 5.4.27

• 1. Prove Propositions 5.4.9 and 5.4.11.

• 2. Show that if R is a PWD, then Mat _n (R) is a PWD for every positive integer n (see Example 5.4.10(i)).

• 3. ([66, Birkenmeier and Park]) Assume that R is a ring and X is a nonempty set of not necessarily commuting indeterminates. Show that R is quasi-Baer with Tdim(R)=n if and only if Γ is quasi-Baer with Tdim(Γ)=n, where Γ is any of the following ring extensions of R.

  • (i) R[X]. (ii) R[x,x ⁻¹]. (iii) R[[x,x ⁻¹]]. (iv) Mat _k (R) for every positive integer k.

• 4. ([82, Birkenmeier, Kim, and Park]) Prove that the following conditions are equivalent for a ring R.

  • (i) R is a TSA ring.

  • (ii) R is a left perfect ring such that there exists a numbering of all the distinct prime ideals P ₁,P ₂,…,P _n of R such that P ₁ P ₂⋯P _n =0.

  • (iii) R is a left perfect ring such that some product of distinct prime ideals, without repetition, is zero.

• 5. Let R be a quasi-Baer ring such that S _ℓ (R) is a countable set. Show that R is a PWP ring. Additionally, if R is also biregular, then R is a direct sum of simple rings (cf. Corollary 5.4.13(ii)).

5.5 A Sheaf Representation of Piecewise Prime Rings

After a brief discussion on certain ideals in a quasi-Baer ring, PWP rings with a sheaf representation will be studied in this section. Quasi-Baer rings with a nontrivial subdirect product representation will also be discussed.

The set of all prime ideals and the set of all minimal prime ideals of a ring R is denoted by Spec (R) and MinSpec(R), respectively. For a subset X of R, let $\text{supp}(X) = \{P\in\text{Spec}(R)\mid X\not\subseteq P\}$, which is called the support of X. In case, X={s}, we write supp(s).

For any P∈Spec(R), there is s∈R∖P and so P∈supp(s). Thus the family {supp(s)∣s∈R} covers Spec(R). Also for P∈supp(x)∩supp(y), $d = xcy\not\in P$ for some c∈R. So P∈supp(d)⊆supp(x)∩supp(y). Therefore, {supp(s)∣s∈R} forms a base (for open sets) on Spec(R). This induced topology on Spec(R) is called the hull-kernel topology on Spec(R).

For P∈Spec(R), let O(P)={a∈R∣aRs=0 for some s∈R∖P}. Then O(P) is an ideal of R, O(P)=∑ _s∈R∖P ℓ _R (Rs), and O(P)⊆P. We let

$$\mathfrak{K}(R) = \bigcup_{P\in\text{Spec}(R)}R/O(P) $$

be the disjoint union of the rings R/O(P), where P ranges through Spec(R).

For a∈R, define $\widehat{a}:\text{Spec}(R)\rightarrow \mathfrak{K}(R)$ by $\widehat{a}(P)=a + O(P)$. Then it can be verified that $\mathfrak{K}(R)$ is a sheaf of rings over Spec(R) with the topology on $\mathfrak{K}(R)$ generated by $\{\widehat{a}(\text{supp}(s))\mid a, s\in R\}$. By a sheaf representation of a ring R, we mean a sheaf representation whose base space is Spec(R) and whose stalks are the R/O(P), where P∈Spec(R). Let $\varGamma (\text{Spec}(R), \,\mathfrak{K}(R))$ be the set of all global sections. We remark that $\varGamma (\text{Spec}(R), \,\mathfrak{K}(R))$ becomes a ring (see [345, 3.1], [209], and [369] for more details).

It is well-known that $\widehat{a}$ is a global section for a∈R. Next, for a,b∈R and P∈Spec(R), $(\widehat{a}+ \widehat{b})(P) = a+b+O(P)$ and $(\widehat{a} \,\widehat{b})(P)=ab+O(P)$. Therefore we see that the map

$$\theta: R\rightarrow \varGamma (\text{Spec}(R), \,\mathfrak{K}(R)) $$

defined by $\theta(a) = \widehat{a}$ is a ring homomorphism, which is called the Gelfand homomorphism. Furthermore, Ker(θ)=⋂ _P∈Spec(R) O(P), which is 0 (see Proposition 5.5.7). Thus θ is a monomorphism.

We discuss some relevant properties of O(P) and R/O(P) for the previously mentioned sheaf representation of PWP rings.

Proposition 5.5.1

Let R be a quasi-Baer ring and P a prime ideal of R. Then O(P)=∑Rf, where the sum is taken for all f∈S _r (R)∩P.

Proof

Note that O(P)=∑ _s∈R∖P ℓ _R (Rs). As R is quasi-Baer, ℓ _R (Rs)=Rf with f∈S _r (R). Then f∈P because fRs=0 and $s\not\in P$. Next let f∈S _r (R)∩P. Then f∈O(P) since fR(1−f)=0 (Proposition 1.2.2) and 1−f∈R∖P. Thus, we get the desired result. □

Corollary 5.5.2

Let R be a quasi-Baer ring. If P and Q are prime ideals such that P⊆Q, then O(P)=O(Q).

Proof

From the definition, we see that O(Q)⊆O(P). Proposition 5.5.1 yields that O(P)⊆O(Q), so O(P)=O(Q). □

We remark that Proposition 5.5.1 and Corollary 5.5.2 hold true when R is a left p.q.-Baer ring.

Proposition 5.5.3

Assume that R is a PWP ring and P is a prime ideal. Then O(P)=Re for some e∈S _r (R).

Proof

As R has a complete set of triangulating idempotents, {Rb∣b∈S _r (R)} is a finite set by the left-sided version of Theorem 5.2.5. From Proposition 5.5.1, O(P)=∑Rf, where the sum is taken for all f∈S _r (R)∩P. Therefore, O(P)=Rf ₁+⋯+Rf _k with f _i ∈S _r (R). By Proposition 1.2.4(ii), O(P)=Re for some e∈S _r (R). □

Let R be a ring and S be a multiplicatively closed subset of R (i.e., 1∈S and s,t∈S implies st∈S). A ring RS ⁻¹ is called a right ring of fractions of R with respect to S together with a ring homomorphism ϕ:R→RS ⁻¹ if the following are satisfied:

• (i) ϕ(s) is invertible for every s∈S.

• (ii) Each element in RS ⁻¹ has the form ϕ(a)ϕ(s)⁻¹ with a∈R and s∈S.

• (iii) ϕ(a)=0 with a∈R if and only if as=0 for some s∈S.

Proposition 5.5.4

Let R be a ring and S a multiplicatively closed subset of R. Then RS ⁻¹ exists if and only if S satisfies:

• S1. If s∈S and a∈R, then there exist t∈S and b∈R with sb=at.

• S2. If sa=0 with a∈R and s∈S, then at=0 for some t∈S.

Proof

See [382, Proposition 1.4, p. 51] for the proof. □

When RS ⁻¹ exists, it has the form RS ⁻¹=(R×S)/∼, where ∼ is the equivalence relation defined as (a,s)∼(b,t) if there exist c,d∈R such that sc=td∈S and ac=bd. A multiplicatively closed subset with S1 and S2 is called a right denominator set. In particular, if R is a right Ore ring and S is the set of all nonzero-divisors in R, then S is a right denominator set. Thus RS ⁻¹ exists by Proposition 5.5.4 and $Q_{c\ell}^{r}(R) = RS^{-1}$ (see 1.1.17).

Proposition 5.5.5

Assume that P is a prime ideal of a ring R and let $S_{P} = \{e\in\mathbf{S}_{\ell}(R)\mid e\not\in P\}$. Then $RS_{P}^{-1}$ exists.

Proof

Obviously 1∈S _P . To see that S _P is a multiplicatively closed subset, let e,f∈S _P . Then ef∈S _ℓ (R) by Proposition 1.2.4(i). If ef∈P, then efRf⊆P. Therefore eRf=efRf⊆P, a contradiction. Thus, $ef\not\in P$. So ef∈S _P and hence S _P is a multiplicatively closed subset of R.

For e∈S _P and a∈R, we have that e(ae)=ae. So the condition S1 is satisfied. Next for S2, take e∈S _P and a∈R such that ea=0. Then

$$ae = (1-e)ae = (1-e)eae = 0, $$

so the condition S2 is satisfied. Hence S _P is a denominator set. Thus, $RS_{P}^{-1}$ exists from Proposition 5.5.4. □

When R is a quasi-Baer ring, we obtain the next result for stalks R/O(P).

Theorem 5.5.6

Assume that R is a quasi-Baer ring and P is a prime ideal of R. Then $RS_{P}^{-1}\cong R/O(P)$.

Proof

First we show that O(P)={a∈R∣ae=0 for some e∈S _P }. Indeed, if a∈R such that ae=0 with e∈S _P , then aRe=aeRe=0 and so a∈O(P). Thus I:={a∈R∣ae=0 for some e∈S _P }⊆O(P). To see that O(P)⊆I, first we prove that I⊴R. For this, say a ₁,a ₂∈I with a ₁ e ₁=0 and a ₂ e ₂=0 for some e ₁,e ₂∈S _P . Then (a ₁+a ₂)e ₁ e ₂=a ₂ e ₁ e ₂=a ₂ e ₂ e ₁ e ₂=0. By Proposition 5.5.5, S _P is a multiplicatively closed set, hence e ₁ e ₂∈S _P . So a ₁+a ₂∈I. Let a∈I and r∈R. Clearly ra∈I. Say e∈S _P such that ae=0. Then are=aere=0, so ar∈I. Therefore I⊴R.

Now say f∈S _r (R)∩P. Then $1-f\not\in P$ and 1−f∈S _ℓ (R). Hence 1−f∈S _P , so f∈I. By Proposition 5.5.1, O(P)⊆I. Thus O(P)=I.

From Proposition 5.5.5, $RS_{P}^{-1}$ exists and there is a ring homomorphism ϕ from R to $RS_{P}^{-1}$, where $RS_{P}^{-1}= \{\phi(a)\phi(e)^{-1}\mid a\in R\text{ and } e\in S_{P}\}$. Now we observe that O(P)=I, so Ker(ϕ)=O(P).

Further, for each e∈S _P , note that $\phi(e)^{2} = \phi(e)\in RS_{P}^{-1}$, which is invertible. Thus ϕ(e)=1 for every e∈S _P . So $RS_{P}^{-1}=\phi(R)$ and Ker(ϕ)=O(P). Hence we get that $RS_{P}^{-1}\cong R/O(P)$. □

Recall that a ring R is a subdirect product of rings S _i ,i∈Λ, if S _i ≅R/K _i , where K _i ⊴R and ∩ _i∈Λ K _i =0. A subdirect product is nontrivial if K _i ≠0 for all i∈Λ. Otherwise, it is trivial.

Proposition 5.5.7

Let R be a ring. Then ⋂ _P∈Spec(R) O(P)=0. Thus R has a subdirect product representation of {R/O(P)∣P∈Spec(R)}.

Proof

Assume that ∩ _P∈Spec(R) O(P)≠0. Let 0≠a∈∩ _P∈Spec(R) O(P). Then r _R (aR) is a proper ideal of R. Let P ₀ be a prime ideal such that r _R (aR)⊆P ₀. Because a∈∩ _P∈Spec(R) O(P)⊆O(P ₀), aRs=0 with s∈R∖P ₀. Therefore s∈r _R (aR)⊆P ₀, a contradiction. So ∩ _P∈Spec(R) O(P)=0. □

The following example shows that the subdirect product representation in Proposition 5.5.7 may be trivial.

Example 5.5.8

For a field F, let R=T ₂(F). Then R is quasi-Baer. Let e _ij ∈T ₂(F) be the matrix with 1 in the (i,j)-position and 0 elsewhere. Put P=Fe ₁₁+Fe ₁₂ and Q=Fe ₁₂+Fe ₂₂. Then we see that R has only two prime ideals which are P and Q (see Proposition 5.4.11). Hence, O(P)=0 and O(Q)=Q by using Proposition 5.5.1.

Next, we consider the subdirect product representation of Proposition 5.5.7 for quasi-Baer rings. Corollary 5.5.2 suggests that we may be able to improve the subdirect product representation by reducing the number of components through using only the minimal prime ideals. So it is natural to consider suitable conditions under which ∩ _{P∈MinSpec(R)} O(P)=0. The next example illustrates that there is a ring R such that ∩ _{P∈MinSpec(R)} O(P)≠0.

Example 5.5.9

Assume that R is the Dorroh extension of $S = \begin{bmatrix} \mathbb{Z}_{2} \,\,& \,\,\mathbb{Z}_{2}\\ 0 \,\,& \,\,0 \end{bmatrix}$ by $\mathbb{Z}$ (i.e., the ring formed from $S\times\mathbb{Z}$ with componentwise addition and multiplication given by (x,k)(y,m)=(xy+mx+ky,km)). Let e _ij be the matrix in S with 1 in the (i,j)-position and 0 elsewhere.

Put e=(e ₁₁,0)∈R. Then e∈S _ℓ (R), so (1−e)Re=0 by Proposition 1.2.2. Also $eRe = (\mathbb{Z}_{2}e_{11}, 0)$, $(1_{R}-e)R(1_{R}-e) = \{(me_{11}, m)\mid m\in\mathbb{Z}\}$, and $P(R) = eR(1-e) = (\mathbb{Z}_{2}e_{12}, 0)$ (note that 1:=1 _R =(0,1)∈R). Since

$$R\cong \begin{bmatrix} eRe \,\,& \,\,eR(1-e)\\ 0 \,\,& \,\,(1-e)R(1-e) \end{bmatrix} , $$

all the minimal prime ideals of R are P ₁:=Q ₁+eR(1−e)+(1−e)R(1−e) and P ₂:=eRe+eR(1−e)+Q ₂, where Q ₁ and Q ₂ are minimal prime ideals of eRe and (1−e)R(1−e), respectively by Proposition 5.4.11.

As $eRe\cong\mathbb{Z}_{2}$ and $(1-e)R(1-e)\cong\mathbb{Z}$, Q ₁=0 and Q ₂=0. So

$$P_1 = \{(me_{11} + ne_{12}, m)\mid m, n\in\mathbb{Z}\}\text{ and }P_2 = (\mathbb{Z}_2e_{11} + \mathbb{Z}_2e_{12}, 0). $$

Take α=(e ₁₂,0)∈R. Then $\alpha R = (\mathbb{Z}_{2}e_{12}, 0)$. Now say s ₁=e=(e ₁₁,0) and s ₂=(0,2). Then αRs ₁=0 with s ₁∈R∖P ₁, and αRs ₂=0 with s ₂∈R∖P ₂. Hence, 0≠α∈O(P ₁)∩O(P ₂)=∩ _{P∈MinSpec(R)} O(P).

In spite of Example 5.5.9, we have the following.

Lemma 5.5.10

If R is a quasi-Baer ring, then ⋂ _{P∈MinSpec(R)} O(P)=0.

Proof

For a minimal prime ideal P of R, O(P)=O(Q) for every prime ideal Q of R containing P by Corollary 5.5.2. Thus, ∩ _{P∈MinSpec(R)} O(P)=0 by Proposition 5.5.7. □

Theorem 5.5.11

Let R be a semiprime ring, which is not prime. If R is quasi-Baer, then R has a nontrivial representation as a subdirect product of R/O(P), where P ranges through all minimal prime ideals.

Proof

As R is a nonprime quasi-Baer ring, R is not semicentral reduced by Proposition 3.2.5. So there is e∈S _ℓ (R) with e≠0 and e≠1. By Proposition 1.2.6(ii), $e\in\mathcal{B}(R)$ since R is semiprime. Suppose that there exists a minimal prime ideal P with O(P)=0. Since R is not prime, P≠0. As (1−e)Re=0, e∈P or 1−e∈P. If e∈P, then $1-e\not\in P$ and eR(1−e)=0, so e∈O(P), a contradiction. Similarly, if 1−e∈P, then we get a contradiction. Thus O(P)≠0 for every minimal prime ideal P of R. Lemma 5.5.10 yields the desired result. □

Corollary 5.5.12

Let R be a semiprime ring, which is not prime. If R is quasi-Baer, then R has a nontrivial representation as a subdirect product of $RS_{P}^{-1}$, where P ranges through all minimal prime ideals.

Proof

It is a direct consequence of Theorems 5.5.6 and 5.5.11. □

Definition 5.5.13

For a ring R, a left (resp., right) semicentral idempotent e (≠1) is called maximal if eR⊆fR (resp., Re⊆Rf) with f∈S _ℓ (R) (resp., f∈S _r (R)), then fR=eR or fR=R (resp., Rf=Re or Rf=R).

Hofmann showed in [209, Theorem 1.17] that $\theta: R\cong \varGamma (\text{Spec}(R), \mathfrak{K}(R))$ when R is a semiprime ring. This result motivates the following question: If a quasi-Baer ring R has such the sheaf representation, then is R semiprime? Theorem 5.5.14 provides an affirmative partial answer to the question by giving a characterization of a certain class of quasi-Baer rings having such the sheaf representation.

Theorem 5.5.14

The following are equivalent for a ring R.

• (i) R is a PWP ring and $\theta: R\cong \varGamma (\mathrm{Spec}(R), \mathfrak{K}(R))$.

• (ii) R is a finite direct sum of prime rings.

• (iii) R is a semiprime PWP ring.

Proof

(i)⇒(ii) Let Tdim(R)=n. If n=1, then R is semicentral reduced, so R is prime by Proposition 3.2.5, and hence we are done. So suppose that n≥2. By Theorem 5.4.20, there are exactly n minimal prime ideals of R, say P ₁,P ₂,…,P _n and from Theorem 5.4.12 these are comaximal (i.e., P _i +P _j =R for i≠j).

For each i=1,2,…,n, we let $\mathfrak{A}_{i} = \{P\in\text{Spec}(R) \,\mid\,P_{i}\subseteq P\}$. Then it follows that $\text{Spec}(R) = \mathfrak{A}_{1}\cup\mathfrak{A}_{2}\cup\cdots\cup \mathfrak{A}_{n}$ since {P ₁,P ₂,…,P _n } is the set of all minimal prime ideals. Also because P _i +P _j =R for i≠j, $\mathfrak{A}_{i}\cap\mathfrak{A}_{j} = \emptyset$ for i≠j. By the hull-kernel topology on Spec(R), each $\mathfrak{A}_{i}$ is a closed subset of Spec(R). Hence for i=1,2,…,n, $\mathfrak{A}_{1}\cup\cdots\cup\mathfrak{A}_{i-1}\cup\mathfrak {A}_{i+1}\cup\cdots\cup \mathfrak{A}_{n}$ is closed, and so each $\mathfrak{A}_{i}$ is open.

Define $f:\text{Spec}(R)\rightarrow\mathfrak{K}(R)$ such that f(P)=1+O(P) for $P\in\mathfrak{A}_{1}$, and f(P)=0+O(P) for $P\in\mathfrak{A}_{k}$ with k≠1. We claim that f is a continuous function. For this, first take $P\in\mathfrak{A}_{1}$. Then $f(P) = 1 + O(P)\in\mathfrak{K}(R)$. Consider a basic neighborhood $\widehat{r}(\text{supp}(s))$ (with r,s∈R) containing f(P)=1+O(P) in $\mathfrak{K}(R)$. Then $\text{supp}(s) \,\cap\, \mathfrak{A}_{1}$ is an open subset of Spec(R) with $P\in\text{supp}(s) \,\cap \,\mathfrak{A}_{1}$.

For $M\in\text{supp}(s)\cap \mathfrak{A}_{1}$, f(M)=1+O(M)∈R/O(M). Hence we obtain that 1+O(P)=r+O(P) and so r−1∈O(P) as $1 + O(P)\in \widehat{r}(\text{supp}(s))$. Now we note that O(P ₁)=O(P)=O(M) from Corollary 5.5.2, hence r−1∈O(M). Thus,

$$f(M)=1 +O(M)=r + O(M)\in \widehat{r}(\text{supp}(s)). $$

So $f(\text{supp}(s)\cap \mathfrak{A}_{1})\subseteq \widehat{r}(\text{supp}(s))$.

For $P\in\mathfrak{A}_{k}$ with k≠1, assume that $f(P) = 0 + O(P)\in \widehat{r}(\text{supp}(s))$ for some r,s∈R. Then we also see that $f(\text{supp}(s)\cap \mathfrak{A}_{k})\subseteq\widehat{r}(\text{supp}(s))$. Therefore, f is a continuous function.

Next, consider $\pi: \mathfrak{K}(R)\rightarrow\text{Spec}(R)$ defined by π(r+O(P))=P for r∈R and P∈Spec(R). Then we see that π(f(P))=P for all P∈Spec(R). Thus, it follows that $f\in \varGamma (\text{Spec}(R), \,\mathfrak{K}(R))$ as f is a continuous function.

Since $R\cong \varGamma (\text{Spec}(R), \,\mathfrak{K}(R))$, there exists a∈R with $f = \widehat{a}$. Therefore

$$a + O(P_1) = 1 + O(P_1)\,\,\text{ and }\,\,a + O(P_k) = 0 + O(P_k)\,\,\text{ for each }\,\,k\neq 1. $$

So 1−a∈O(P ₁) and a∈O(P _k ) for each k≠1. Thus O(P ₁)+O(P _k )=R for each k≠1. Similarly, O(P _i )+O(P _j )=R for i≠j, 1≤i,j≤n. By Lemma 5.5.10, we obtain that O(P ₁)∩⋯∩O(P _n )=0, hence

$$R\cong R/O(P_1)\oplus\cdots\oplus R/O(P_n) $$

by Chinese Remainder Theorem. From Proposition 5.5.3, O(P ₁)=Re with e∈S _r (R), so eR(1−e)=0. Hence R/O(P ₁)≅(1−e)R(1−e).

Our claim is that (1−e)R(1−e) is semicentral reduced. For this, assume on the contrary that (1−e)R(1−e) is not semicentral reduced. By Theorem 3.2.10, (1−e)R(1−e) is a quasi-Baer ring. Hence, (1−e)R(1−e) is a PWP ring by Theorem 5.2.19.

From Theorem 5.2.5, there is a maximal right semicentral idempotent in the ring (1−e)R(1−e), say (1−e)b(1−e). Because (1−e)R(1−e) is not semicentral reduced,

$$[(1-e)R(1-e)](1-e)b(1-e) $$

is a nonzero proper ideal of (1−e)R(1−e). Since e∈S _r (R),

$$e+(1-e)b(1-e)\in\mathbf{S}_r(R). $$

Put g=e+(1−e)b(1−e). We show that g is a maximal right semicentral idempotent of R. Take α∈S _r (R) such that Rg⊆Rα and α≠1. Because

$$R = \begin{bmatrix} eRe \,\,& \,\,0\\ (1-e)Re \,\,& \,\,(1-e)R(1-e)\end{bmatrix} $$

and g=e+(1−e)b(1−e), we have that α=e+k+h with k∈(1−e)Re and h∈S _r ((1−e)R(1−e)).

Since Rg⊆Rα, (1−e)R(1−e)(1−e)b(1−e)⊆(1−e)R(1−e)h. From the maximality of (1−e)b(1−e) and h≠1−e (because α≠1), we have that (1−e)R(1−e)(1−e)b(1−e)=(1−e)R(1−e)h, and thus h(1−e)b(1−e)=h. Further, ke=k since k∈(1−e)Re. Hence,

$$\alpha g = \begin{bmatrix} e \,\,& \,\,0\\ k \,\,& \,\,h \end{bmatrix} \begin{bmatrix} e \,\,& \,\,0\\ 0 \,\,& \,\,(1-e)b(1-e) \end{bmatrix} = \begin{bmatrix} e \,\,& \,\,0\\ k \,\,& \,\,h \end{bmatrix} = \alpha. $$

Thus, Rα⊆Rg. Therefore, g is a maximal right semicentral idempotent of R.

Next, note that {1,1−g} forms a multiplicatively closed subset of R. By Zorn’s lemma, there is an ideal Q of R maximal with respect to being disjoint with {1,1−g}. Then Q is a prime ideal of R. Since gR(1−g)=0 and $1-g\not\in Q$, it follows g∈O(Q). Also, since g is a maximal right semicentral idempotent of R and g∈O(Q), O(Q)=Rg from Proposition 5.5.3. We observe that O(P ₁)=Re≠Rg as (1−e)b(1−e)≠0. Hence, $Q\not\in \mathfrak{A}_{1}$ by Corollary 5.5.2. So $Q\in\mathfrak{A}_{k}$ for some k≠1. So O(Q)=O(P _k ) from Corollary 5.5.2. Now R=O(P ₁)+O(P _k )=Re+Rg, a contradiction since (1−e)b(1−e)≠1−e. Thus, the ring (1−e)R(1−e) is a semicentral reduced quasi-Baer ring. So (1−e)R(1−e) is a prime ring by Proposition 3.2.5, thus R/O(P ₁) is a prime ring because R/O(P ₁)≅(1−e)R(1−e). Similarly, R/O(P _i ) is a prime ring for each i=2,…,n. Therefore R≅R/O(P ₁)⊕⋯⊕R/O(P _n ), which is a finite direct sum of prime rings. Further, note that O(P _i )=P _i for each i=1,…,n, so R≅R/P ₁⊕⋯⊕R/P _n .

(ii)⇒(iii) It is evident.

(iii)⇒(i) The proof follows from [209, Theorem 1.17]. □

We obtain the next corollary from Proposition 5.4.6, Lemma 5.4.25, and Theorem 5.5.14.

Corollary 5.5.15

The following are equivalent.

• (i) R is a PWD with $\theta: R\cong \varGamma (\mathrm{Spec}(R), \,\mathfrak{K}(R))$.

• (ii) R is a finite direct sum of prime PWDs.

• (iii) R is a semiprime PWD.

Exercise 5.5.16

• 1. ([74, Birkenmeier, Kim, and Park]) Assume that R is a (quasi-)Baer ring with Tdim(R)<∞ and P is a prime ideal of R. Prove that R/O(P) is a (quasi-)Baer ring.

• 2. ([74, Birkenmeier, Kim, and Park]) Let R be a Baer ring and P be a prime ideal of R. Show that R/O(P) is a right Rickart ring.

• 3. ([74, Birkenmeier, Kim, and Park]) Assume that R is a quasi-Baer ring and P is a prime ideal of R. Prove that r.gl.dim(R/O(P))≤r.gl.dim(R).

5.6 Triangular Matrix Ring Extensions

Our focus in this section is the study of the Baer, the quasi-Baer, and the (strongly) FI-extending properties of upper triangular and generalized triangular matrix ring extensions. The study of full matrix ring extensions will be considered in Chap. 6.

Theorem 5.6.1

Let R be a ring. Then the following are equivalent.

• (i) R is regular and right self-injective.

• (ii) T _n (R) is right nonsingular right extending for every positive integer n.

• (iii) T _k (R) is right nonsingular right extending for some integer k>1.

• (iv) T ₂(R) is right nonsingular right extending.

Proof

(i)⇒(ii) The proof follows from [3, Corollary 2.8(3)] and [1, Proposition 1.8(ii)].

(ii)⇒(iii) It is evident.

(iii)⇒(i) [3, Corollary 2.8(2) and Proposition 1.6(2)] yield this implication.

(i)⇔(iv) This equivalence follows from [393, Theorem 3.4] (see also Theorem 5.6.9). □

Theorem 5.6.2

Let R be an orthogonally finite Abelian ring. Then the following are equivalent.

• (i) R is a direct sum of division rings.

• (ii) T _n (R) is a Baer (resp., right Rickart) ring for every positive integer n.

• (iii) T _k (R) is a Baer (resp., right Rickart) ring for some integer k>1.

• (iv) T ₂(R) is a Baer (resp., right Rickart) ring.

Proof

(i)⇒(ii) The proof follows from Theorems 5.6.1, 3.3.1, and 3.1.25.

(ii)⇒(iii) It is evident.

(iii)⇒(iv) The proof follows from Theorems 3.1.8 and 3.1.22(i).

(iv)⇒(i) Let T ₂(R) be Baer (resp., right Rickart). By Proposition 1.2.15, R has a complete set of primitive idempotents. As R is Abelian, $R=\oplus_{i=1}^{m}R_{i}$ (ring direct sum), for some positive integer m, where each R _i is indecomposable as a ring. Then each T ₂(R _i ) is a Baer (resp., right Rickart) ring by Proposition 3.1.5(i) (resp., Proposition 3.1.21). From Theorem 3.1.8 (resp., Theorem 3.1.22(i)), each R _i is a Baer (resp., right Rickart) ring. If R _i is Baer or right Rickart, R _i is a domain (see Example 3.1.4(ii)). From [246, Exercise 2, p. 16] or [262, Exercise 25, p. 271], each R _i is a division ring. □

Notation 5.6.3

Let S and R be rings, and let _S M _R be an (S,R)-bimodule. For the remainder of this section, we let

$$T = \begin{bmatrix} S & M\\ 0 & R \end{bmatrix} $$

denote a generalized triangular matrix ring.

Lemma 5.6.4

Let T be the ring as in Notation 5.6.3. Say

$$e = \begin{bmatrix} e_1 & k\\ 0 & e_2 \end{bmatrix} \in\mathbf{S}_{\ell}(T)\textit{ and } f = \begin{bmatrix} e_1 & 0\\ 0 & e_2 \end{bmatrix} . $$

Then we have the following.

• (i) e ₁∈S _ℓ (S), e ₂∈S _ℓ (R), and f∈S _ℓ (T)

• (ii) eT=fT.

Proof

(i) It can be easily checked that e ₁∈S _ℓ (S) and e ₂∈S _ℓ (R). Also we see that e ₁ me ₂=me ₂ for all m∈M. Thus, f∈S _ℓ (T).

(ii) Since e ₁ me ₂=me ₂ for all m∈M, in particular e ₁ ke ₂=ke ₂ and so $f=e \begin{bmatrix} e_{1} & -ke_{2}\\ 0 & e_{2} \end{bmatrix}$. Hence fT⊆eT. As e∈S _ℓ (T), $\begin{bmatrix} 1 \,& \,0\\ 0 \,& \,0 \end{bmatrix} e = e \begin{bmatrix} 1 \,& \,0\\ 0 \,& \,0 \end{bmatrix} e$, so k=e ₁ k. Thus, $e = f \begin{bmatrix} 1 \,& \,k \\ 0 \,& \,1 \end{bmatrix} \in fT$. Therefore eT⊆fT, and so eT=fT. □

Next, we characterize the quasi-Baer property for the ring T.

Theorem 5.6.5

Let T be the ring as in Notation 5.6.3. Then the following are equivalent.

• (i) T is a quasi-Baer ring.

• (ii)

  • (1) R and S are quasi-Baer rings.

  • (2) r _M (I)=r _S (I)M for all I⊴S.

  • (3) For any _S N _R ≤  _S M _R , r _R (N)=gR for some g ²=g∈R.

Proof

(i)⇒(ii) By Theorem 3.2.10, R and S are quasi-Baer. Let I⊴S. Then $A:= \begin{bmatrix} I & M\\ 0 & 0 \end{bmatrix} \trianglelefteq T$. Hence, r _T (A)=eT for some e ²=e∈T. Because A⊴T, e∈S _ℓ (T) by Proposition 1.2.2. Put $e= \begin{bmatrix} e_{1} & k\\ 0 & e_{2} \end{bmatrix} \text{ and }f= \begin{bmatrix} e_{1} & 0\\ 0 & e_{2} \end{bmatrix}$. From Lemma 5.6.4, e ₁∈S _ℓ (S), e ₂∈S _ℓ (R), f∈S _ℓ (T), and eT=fT. Thus it is routine to check that r _S (I)=e ₁ S and r _M (I)=e ₁ M=e ₁ SM=r _S (I)M.

Next, let _S N _R ≤  _S M _R . Then $K:= \begin{bmatrix} 0 & N\\ 0 & 0 \end{bmatrix} \trianglelefteq T$. So r _T (K)=hT for some h∈S _ℓ (T). Say $h= \begin{bmatrix} g_{1} & m\\ 0 & g_{2} \end{bmatrix}$. Then r _R (N)=g ₂ R, where g ₂∈S _ℓ (R). Take g=g ₂. Then r _R (N)=gR and g ²=g∈R.

(ii)⇒(i) Let K⊴T. Then we see that $K= \begin{bmatrix} I \,& \,N\\ 0 \,& \,J \end{bmatrix}$, where I⊴S, J⊴R, _S N _R ≤  _S M _R , and IM+MJ⊆N. Because S and R are quasi-Baer, there are e ₁∈S _ℓ (S), f∈S _ℓ (R) satisfying r _S (I)=e ₁ S and r _R (J)=fR. By assumption, r _M (I)=r _S (I)M=e ₁ M and r _R (N)=gR for some g ²=g∈R. As r _R (N)=gR⊴R, g∈S _ℓ (R) by Proposition 1.2.2. From Proposition 1.2.4(i), gf∈S _ℓ (R). Put $e= \begin{bmatrix} e_{1} & 0\\ 0 & gf \end{bmatrix} \in T$. Then e ²=e and r _T (K)=eT. Thus, T is quasi-Baer. □

Corollary 5.6.6

Let S=End(M _R ) and let T be the ring as in Notation 5.6.3. Then the following are equivalent.

• (i) T is a quasi-Baer ring.

• (ii)

  • (1) R is a quasi-Baer ring.

  • (2) M _R is a quasi-Baer module.

  • (3) If N _R ⊴M _R , then r _R (N)=gR for some g ²=g∈R.

Proof

(i)⇒(ii) Assume that T is a quasi-Baer ring. Then M _R is a quasi-Baer module by Proposition 4.6.3 and Theorem 5.6.5. So we get (ii).

(ii)⇒(i) As M _R is a quasi-Baer module, S is a quasi-Baer ring by Theorem 4.6.16. Let I⊴S. Then r _S (I)=fS for some f ²=f∈S. Also r _M (I)=hM for some h ²=h∈S by Proposition 4.6.3. Since If=0, IfM=0, and so fM⊆r _M (I)=hM. As IhM=0, Ih=0, and hence h∈r _S (I)=fS. Thus, hM⊆fSM=fM. Therefore hM=fM=fSM=r _S (I)M. So T is a quasi-Baer ring by Theorem 5.6.5. □

We observe that in contrast to Theorem 5.6.2, the next two results hold true without any additional assumption on R.

Theorem 5.6.7

The following are equivalent for a ring R.

• (i) R is a quasi-Baer ring.

• (ii) T _n (R) is a quasi-Baer ring for every positive integer n.

• (iii) T _k (R) is a quasi-Baer ring for some integer k>1.

• (iv) T ₂(R) is a quasi-Baer ring.

Proof

(i)⇒(ii) We use induction on n. As R is quasi-Baer, T ₂(R) is quasi-Baer by applying Corollary 5.6.6.

Let T _n (R) be quasi-Baer. We show that T _n+1(R) is quasi-Baer. Write

$$T_{n+1}(R)= \begin{bmatrix} R \, & \,M\\ 0 \, & \,T_n(R) \end{bmatrix} , $$

where M=[R,…,R] (n-tuple). To apply Theorem 5.6.5, let I⊴R. Then r _R (I)=eR for some e ²=e∈R. Also r _M (I)=eM=r _R (I)M.

Next, say $_{R}N_{T_{n}(R)}\leq\,_{R}M_{T_{n}(R)}$. Note that $\begin{bmatrix} 0 \,& \,N\\ 0 \,& \,0 \end{bmatrix} \trianglelefteq T_{n+1}(R)$. Therefore, we have that N=[N ₁,…,N _n ], where N _i ⊴R for each i and N ₁⊆⋯⊆N _n . As R is quasi-Baer, r _R (N _i )=f _i R with $f_{i}^{2}=f_{i}\in R$ for each i.

Let e _ij ∈T _n (R) be the matrix with 1 in the (i,j)-position and 0 elsewhere. Put g=f ₁ e ₁₁+⋯+f _n e _nn ∈T _n (R). Then g ²=g and $r_{T_{n}(R)}(N)=gT_{n}(R)$. By Theorem 5.6.5, T _n+1(R) is a quasi-Baer.

(ii)⇒(iii) is obvious. For (iii)⇒(iv), let e _ij ∈T _k (R) be the matrix with 1 in the (i,j)-position and 0 elsewhere. Set f=e ₁₁+e ₂₂. Then f ²=f∈T _k (R) and T ₂(R)≅fT _k (R)f. By Theorem 3.2.10, T ₂(R) is quasi-Baer. Similarly, (iv)⇒(i) follows from Theorem 3.2.10. □

Proposition 5.6.8

The following are equivalent for a ring R.

• (i) R is a right p.q.-Baer ring.

• (ii) T _n (R) is a right p.q.-Baer ring for every positive integer n.

• (iii) T _k (R) is a right p.q.-Baer ring for some integer k>1.

• (iv) T ₂(R) is a right p.q.-Baer ring.

Proof

(i)⇒(ii) Put T=T _n (R). Let e _ij be the matrix in T with 1 in the (i,j)-position and 0 elsewhere. Say [a _ij ]∈T and consider the right ideal [a _ij ]T. Take α=[α _ij ]∈r _T ([a _ij ]T). Since R is right p.q.-Baer, for i≤j, r _R (a _ij R)=f _ij R with $f_{ij}^{2}=f_{ij}\in R$. Then f _ij ∈S _ℓ (R) from Proposition 1.2.2 because f _ij R⊴R.

Now observe that α _1ℓ ∈r _R (a ₁₁ R)=f ₁₁ R for ℓ=1,…,n. Also we see that α _2ℓ ∈r _R (a ₁₁ R)∩r _R (a ₁₂ R)∩r _R (a ₂₂ R)=f ₁₁ R∩f ₁₂ R∩f ₂₂ R=f ₁₁ f ₁₂ f ₂₂ R for ℓ=2,…,n, and f ₁₁ f ₁₂ f ₂₂∈S _ℓ (R) (see Proposition 1.2.4(i)). In general, α _kℓ ∈(f ₁₁⋯f _1k )(f ₂₂⋯f _2k )⋯(f _k−1k−1 f _k−1k )f _kk R for ℓ=k,…,n.

Put g _k =(f ₁₁⋯f _1k )(f ₂₂⋯f _2k )⋯(f _{k−1 k−1} f _k−1 k )f _kk for k=1,…,n. Then g _k ∈S _ℓ (R) by Proposition 1.2.4(i). Note that g _k α _kℓ =α _kℓ for ℓ=k,…,n.

Let e=g ₁ e ₁₁+⋯+g _n e _nn ∈T. Then e ²=e and r _R ([a _ij ]T)=eT. Therefore, T=T _n (R) is right p.q.-Baer.

(ii)⇒(iii) It is evident.

(iii)⇒(iv) Let f=e ₁₁+e ₂₂∈T _k (R). Then we see that f ²=f∈T _k (R) and T ₂(R)≅fT _k (R)f, so T ₂(R) is right p.q.-Baer by Theorem 3.2.34(i).

(iv)⇒(i) It follows also from Theorem 3.2.34(i). □

The following result, due to Tercan in [393], characterizes the generalized triangular matrix ring T (see Notation 5.6.3) to be a right nonsingular right extending ring (hence T is Baer and right cononsingular by Theorem 3.3.1) when _S M is faithful.

Theorem 5.6.9

Let T be the ring as in Notation 5.6.3 and _S M be faithful. Then the following are equivalent.

• (i) T is right nonsingular and right extending.

• (ii)

  • (1) For each complement K _R in M _R there is e ²=e∈S with K=eM.

  • (2) R is right nonsingular and right extending.

  • (3) M _R is nonsingular and injective.

In the next result, a characterization for T to be right FI-extending is presented. This will be used to consider the FI-extending triangular matrix ring extensions.

Theorem 5.6.10

Let T be the ring as in Notation 5.6.3. Then the following are equivalent.

• (i) T _T is FI-extending.

• (ii)

  • (1) For _S N _R ≤  _S M _R and I⊴S with IM⊆N, there is f ²=f∈S such that I⊆fS, N _R ≤^ess fM _R , and (I∩ℓ _S (M)) _S ≤^ess(fS∩ℓ _S (M)) _S .

  • (2) R _R is FI-extending.

Proof

Throughout the proof, we let $e_{11} = \begin{bmatrix} 1 \,& \,0\\ 0 \,& \,0 \end{bmatrix} \in T$.

(i)⇒(ii) First, we claim that ℓ _S (M)=eS for some e ²=e∈S. Observe that T _T =e ₁₁ T _T ⊕(1−e ₁₁)T _T and e ₁₁∈S _ℓ (T). From Proposition 2.3.11(i), $e_{11}T_{T} = \begin{bmatrix} S \,& \,M\\ 0 \, & \,0 \end{bmatrix} _{T}$ is FI-extending. First, to see that ℓ _S (M)=eS for some e ²=e∈S, put $U= \begin{bmatrix} \ell_{S}(M) & 0\\ 0 & 0 \end{bmatrix}$. Then U _T ⊴e ₁₁ T _T because ℓ _S (M)⊴S and $\text{End}(e_{11}T_{T})\cong e_{11}Te_{11} = \begin{bmatrix} S \,& \,0\\ 0 \,& \,0 \end{bmatrix}$. Because e ₁₁ T _T is FI-extending, we have that $U_{T}\leq^{\text{ess}} \begin{bmatrix} e \,& \,0\\ 0 \,& \,0 \end{bmatrix} e_{11}T_{T}$ for some e ²=e∈S. So $\begin{bmatrix} \ell_{S}(M) \,& \,0\\ 0 \,& \,0 \end{bmatrix} _{T}\leq^{\text{ess}} \begin{bmatrix} eS \,& \,eM\\ 0 \,& \,0 \end{bmatrix} _{T}$. Thus, ℓ _S (M)⊆eS. For any m∈M, em=0 because $U\cap \begin{bmatrix} 0 & em\\ 0 & 0 \end{bmatrix} T = 0$. Hence eM=0, so e∈ℓ _S (M). Thus eS⊆ℓ _S (M), and hence ℓ _S (M)=eS.

For condition (1), let _S N _R ≤  _S M _R and I⊴S such that IM⊆N. Then $V: = \begin{bmatrix} I & N\\ 0 & 0 \end{bmatrix} _{T}\trianglelefteq e_{11}T_{T} = \begin{bmatrix} S \,& \,M\\ 0 \,& \,0 \end{bmatrix} _{T}$. Since e ₁₁ T _T is FI-extending, we have that $\begin{bmatrix} I & N\\ 0 & 0 \end{bmatrix} _{T}\leq^{\text{ess}} \begin{bmatrix} fS \,& \,fM\\ 0 \,& \,0 \end{bmatrix} _{T}$ for some f ²=f∈S, therefore I⊆fS and N _R ≤^ess fM _R . Next, for 0≠fs∈fS∩eS=fS∩ℓ _S (M) with s∈S, we see that $V\cap \begin{bmatrix} fs & 0\\ 0 & 0 \end{bmatrix} T = V\cap \begin{bmatrix} fsS & 0\\ 0 & 0 \end{bmatrix} \neq0$. Hence, fsS∩(I∩eS)=fsS∩I≠0 because fsS⊆eS. Therefore, we have that (I∩eS) _S ≤^ess(fS∩eS) _S .

Since e ₁₁∈S _ℓ (T), Proposition 2.3.11(ii) yields condition (2) immediately.

(ii)⇒(i) By condition (2), (1−e ₁₁)T _T is FI-extending. To show that e ₁₁ T _T is FI-extending, let V _T ⊴e ₁₁ T _T . Since e ₁₁∈S _ℓ (T), e ₁₁ T _T ⊴T _T from Proposition 1.2.2, and so V _T ⊴T _T by Proposition 2.3.3(ii). Thus $V = \begin{bmatrix} I \, & \,N\\ 0 \, & \,0 \end{bmatrix}$ with I⊴S,    _S N _R ≤  _S M _R , and IM⊆N. By condition (1), there is f ²=f∈S such that I⊆fS,N _R ≤^ess fM _R , and (I∩ℓ _S (M)) _S ≤^ess(fS∩ℓ _S (M)) _S . Thus, it follows that $V\subseteq \begin{bmatrix} f & 0\\ 0 & 0 \end{bmatrix} \begin{bmatrix} S & M\\ 0 & 0 \end{bmatrix} = \begin{bmatrix} fS & fM\\ 0 & 0 \end{bmatrix}$. Let $W = \begin{bmatrix} fS & fM\\ 0 & 0 \end{bmatrix}$. Then W _T is a direct summand of e ₁₁ T _T because f ²=f∈S≅End(e ₁₁ T _T ).

We prove that V _T ≤^ess W _T . For this, take $0\neq w=\begin{bmatrix} fs & fm\\ 0 & 0 \end{bmatrix} \in W$, where s∈S and m∈M. If fm≠0, then V∩wT≠0 since N _R ≤^ess fM _R . Next, assume that fm=0. Then fs≠0. Hence $wT = \begin{bmatrix} fsS & fsM\\ 0 & 0 \end{bmatrix}$.

If fsM≠0, clearly V∩wT≠0 since N _R ≤^ess fM _R . If fsM=0, then

$$fs\in \ell_S(M),\,\,\text{ so }\,\,0\neq fs\in fS\cap \ell_S(M). $$

Since (I∩ℓ _S (M)) _S ≤^ess(fS∩ℓ _S (M)) _S , fsS∩(I∩ℓ _S (M))≠0, so V∩wT≠0. Therefore V _T ≤^ess W _T , thus e ₁₁ T _T is FI-extending. Hence T _T is FI-extending by Theorem 2.3.5. □

Corollary 5.6.11

Let T be the ring as in Notation 5.6.3. Assume that _S M is faithful. Then the following are equivalent.

• (i) T _T is FI-extending.

• (ii)

  • (1) For _S N _R ≤  _S M _R , there is f ²=f∈S with N _R ≤^ess fM _R .

  • (2) R _R is FI-extending.

Proof

(i)⇒(ii) Assume that T _T is FI-extending. As _S M is faithful, ℓ _S (M)=0. By taking I=0 in Theorem 5.6.10, we obtain part (ii).

(ii)⇒(i) Let _S N _R ≤  _S M _R and I⊴S such that IM⊆N. By (1), there exists f ²=f∈S such that N _R ≤^ess fM _R . Since IM⊆N⊆fM, fn=n for all n∈N, in particular fsm=sm for any s∈I and m∈M. Therefore, (s−fs)M=0, so s−fs=0 for any s∈I because _S M is faithful. Hence, I=fI⊆fS. Thus, T _T is FI-extending by Theorem 5.6.10. □

Corollary 5.6.12

Let M _R be a right R-module. Then the ring

$$T = \begin{bmatrix} \mathrm{End}_R(M) & M\\ 0 & R \end{bmatrix} $$

is right FI-extending if and only if M _R and R _R are FI-extending.

Proof

It follows immediately from Corollary 5.6.11. □

We remark that if R is a right FI-extending ring, then T ₂(R) is right FI-extending by taking M=R _R in Corollary 5.6.12. When n≥2, we obtain the FI-extending property of T _n (R) in Theorem 5.6.19 precisely when R is right FI-extending. By our previous results, we establish a class of rings which are right FI-extending, but not left FI-extending as the next example illustrates.

Example 5.6.13

Let R be a right self-injective ring with J(R)≠0. Put

$$T = \begin{bmatrix} R/J(R) & R/J(R)\\ 0 & R \end{bmatrix} . $$

Then the ring R/J(R) is right self-injective by Corollary 2.1.30. Further, End _R (R/J(R))≅R/J(R). Also R/J(R) is an FI-extending right R-module. Thus the ring T is right FI-extending by Corollary 5.6.12. If T is left FI-extending, then r _R ((R/J(R)) _R )=J(R)=Rf for some f∈S _r (R) from the left-sided version of the proof for (i)⇒(ii) of Theorem 5.6.10. Thus f=0 and hence J(R)=0, a contradiction. Thus, T cannot be left FI-extending.

Definition 5.6.14

Let N _R ≤M _R . We say that N _R has a direct summand cover $\mathcal{D}(N_{R})$ if there is e ²=e∈End _R (M) with $N_{R}\leq^{\text{ess}}eM_{R} = \mathcal{D}(N_{R})$.

If M _R is a strongly FI-extending module, then every fully invariant submodule has a unique direct summand cover from Lemma 2.3.22. For N _R ≤M _R , let (N _R :M _R )={a∈R∣Ma⊆N}. Then (N _R :M _R )⊴R.

We use $\mathcal{D}[(N_{R} : M_{R})_{R}]$ to denote a direct summand cover of the right ideal (N _R :M _R ) in R _R . Let M be an (S,R)-bimodule and _S N _R ≤  _S M _R . If there exists e ²=e∈S _ℓ (S) such that N _R ≤^ess eM _R , then we write $\mathcal{D}_{S}(N_{R}) = eM$.

In the next result, we obtain a necessary and sufficient condition for a 2×2 generalized triangular matrix ring to be right strongly FI-extending. Some applications of this characterization will also be presented.

Theorem 5.6.15

Let T be as in Notation 5.6.3. Then the following are equivalent.

• (i) T _T is strongly FI-extending.

• (ii)

  • (1) For _S N _R ≤  _S M _R and I⊴S with IM⊆N, there is e∈S _ℓ (S) such that I⊆eS,  N _R ≤^ess eM _R and  (I∩ℓ _S (M)) _S ≤^ess(eS∩ℓ _S (M)) _S .

  • (2) R _R is strongly FI-extending.

  • (3) $\mathcal{D}_{S}(N_{R})\mathcal{D}[(N_{R}: M_{R})_{R}] = M\mathcal{D}[(N_{R}: M_{R})_{R}]$ for _S N _R ≤  _S M _R .

Proof

(i)⇒(ii) We let $e_{11} = \begin{bmatrix} 1 \,& \,0\\ 0 \,& \,0 \end{bmatrix} \in T$. Assume that T _T is strongly FI-extending. By Theorem 2.3.19, (1−e ₁₁)T _T is strongly FI-extending, so R _R is strongly FI-extending, which is condition (2).

For condition (1), let _S N _R ≤  _S M _R and I⊴S with IM⊆N. Then $V: = \begin{bmatrix} I & N\\ 0 & 0 \end{bmatrix} _{T}\trianglelefteq e_{11}T_{T} = \begin{bmatrix} S \,& \,M\\ 0 \,& \,0 \end{bmatrix} _{T}$. Since e ₁₁ T _T is strongly FI-extending, there exists e ²=e∈S _ℓ (S) such that $V_{T}\leq^{\text{ess}}\begin{bmatrix} eS \,& \,eM\\ 0 \,& \,0\end{bmatrix}_{T}$. So I⊆eS and N _R ≤^ess eM _R .

Next, say 0≠es∈eS∩ℓ _S (M) with s∈S. There is $\begin{bmatrix} s_{1} \,& \,m_{1}\\ 0 \,& \,r_{1}\end{bmatrix}\in T$ such that

$$0\neq \begin{bmatrix} es \,& \,0\\ 0 \,& \,0\end{bmatrix}\begin{bmatrix}s_1 \,& \,m_1\\ 0 \,& \,r_1\end{bmatrix}=\begin{bmatrix}ess_1 \,& \,0\\ 0 \,& \,0\end{bmatrix}\in V. $$

Thus 0≠ess ₁∈I∩ℓ _S (M). Therefore (I∩ℓ _S (M)) _S ≤^ess(eS∩ℓ _S (M)) _S .

For condition (3), let _S N _R ≤  _S M _R and put A=(N _R :M _R ). Take I=0 in condition (1). There exists e∈S _ℓ (S) with $\mathcal{D}_{S}(N_{R}) = eM$. By condition (2), $\mathcal{D}(A_{R}) = fR$ for some f∈S _ℓ (R). Since MA⊆N,  $W:= \begin{bmatrix} 0 \,\,& \,\,N\\ 0 \,\,& \,\,A \end{bmatrix} \trianglelefteq T$, and W _T ≤^ess wT _T for some w∈S _ℓ (T). By Lemma 5.6.4, there exist e ₀∈S _ℓ (S) and f ₀∈S _ℓ (R) such that $wT = \begin{bmatrix} e_{0} & 0\\ 0 & f_{0} \end{bmatrix} T$. We put $w_{0}= \begin{bmatrix} e_{0} & 0\\ 0 & f_{0} \end{bmatrix} \in\textbf{S}_{\ell}(T)$. Hence N _R ≤^ess e ₀ M _R and A _R ≤^ess f ₀ R _R . So $\mathcal{D}_{S}(N_{R}) = eM = e_{0}M$ by Lemma 2.3.22 as e ₀∈S _ℓ (S). Also $\mathcal{D}(A_{R}) = fR = f_{0}R$.

Note that Mf ₀=e ₀ Mf ₀ since w ₀∈S _ℓ (T). Thus, e ₀ Mf ₀ R=Mf ₀ R. Therefore, $\mathcal{D}_{S}(N_{R})\mathcal{D}[(N_{R} : M_{R})_{R}] = M\mathcal{D}[(N_{R}: M_{R})_{R}]$.

(ii)⇒(i) Assume that K⊴T. Then

$$K=\begin{bmatrix} I & N\\ 0 & B\end{bmatrix}\trianglelefteq T, $$

where _S N _R ≤  _S M _R , I⊴S,  IM+MB⊆N, and B⊴R.

From condition (1), there exists e∈S _ℓ (S) with

$$I\subseteq eS,\,\, \mathcal{D}_S(N_R) = eM,\,\,\text{ and }\,\,(I\cap \ell_S(M))_S\leq^{\text{ess}}(eS\cap \ell_S(M))_S. $$

Since B⊴R, by condition (2), there exists f∈S _ℓ (R) with $\mathcal{D}(B_{R}) = fR$. Also, from condition (2), $\mathcal{D}[(N_{R} : M_{R})_{R}] =f_{0}R$ for some f ₀∈S _ℓ (R).

As MB⊆N, B⊆(N _R :M _R ). Thus,

$$B_R\leq^{\text{ess}}(fR\cap f_0R)_R = f_0fR $$

with f ₀ f∈S _ℓ (R) (see Proposition 1.2.4(i)). So $\mathcal{D}(B_{R}) = f_{0}fR$. By Lemma 2.3.22, we get that fR=f ₀ fR.

By condition (3), eMf ₀ R=Mf ₀ R. Because f∈S _ℓ (R) and f ₀ fR=fR, eMf ₀ Rf=eMf ₀ fRf=eMfRf=eMRf=eMf. Similarly, we have that Mf ₀ Rf=Mf. As eMf ₀ R=Mf ₀ R, eMf ₀ Rf=Mf ₀ Rf and so eMf=Mf.

Since (I∩ℓ _S (M)) _S ≤^ess(eS∩ℓ _S (M)) _S and N _R ≤^ess eM _R , we see that $\begin{bmatrix} I \,\,& \,\,N\\ 0 \,\,& \,\,0 \end{bmatrix} _{T}\leq^{\text{ess}} \begin{bmatrix} e \,\,& \,\,0\\ 0 \,\,& \,\,0 \end{bmatrix} T_{T}$. So $\begin{bmatrix} 0 \,\,& \,\,0\\ 0 \,\,& \,\,B \end{bmatrix} _{T}\leq^{\text{ess}} \begin{bmatrix} 0 \,\,& \,\,0\\ 0 \,\,& \,\,f \end{bmatrix} T_{T}$ because B _R ≤^ess fR _R .

Thus $K_{T}\leq^{\text{ess}} \begin{bmatrix} e \,\,& \,\,0\\ 0 \,\,& \,\,f \end{bmatrix} T_{T}$. As Mf=eMf, mf=emf for each m∈M. Hence $\begin{bmatrix} e \,\,& \,\,0\\ 0 \,\,& \,\,f \end{bmatrix} \in\mathbf{S}_{\ell}(T)$. Therefore, T _T is strongly FI-extending. □

Corollary 5.6.16

Let T be the ring as in Notation 5.6.3 with _S M faithful. Then the following are equivalent.

• (i) T _T is strongly FI-extending.

• (ii)

  • (1) For _S N _R ≤  _S M _R , there is e∈S _ℓ (S) with N _R ≤^ess eM _R .

  • (2) R _R is strongly FI-extending.

  • (3) $\mathcal{D}_{S}(N_{R})\mathcal{D}[(N_{R} : M_{R})_{R}] = M\mathcal{D}[(N_{R} : M_{R})_{R}]$ for _S N _R ≤  _S M _R .

Proof

(i)⇒(ii) The proof follows from Theorem 5.6.15 by taking I=0. For (ii)⇒(i), let _S N _R ≤  _S M _R and I⊴S such that IM⊆N. By condition (1), there is e∈S _ℓ (S) with N _R ≤^ess eM _R . As IM⊆N⊆eM,  n=en for all n∈N, in particular sm=esm for any s∈I and m∈M. Thus (s−es)M=0, so s−es=0 for any s∈I, as _S M is faithful. So I=eI⊆eS. Thus T _T is strongly FI-extending by Theorem 5.6.15. □

Corollary 5.6.17

Let M _R be a right R-module and $T = \begin{bmatrix} \mathrm{End}_{R}(M) & M\\ 0 & R \end{bmatrix}$. Then the following are equivalent.

• (i) T _T is strongly FI-extending.

• (ii)

  • (1) M _R is strongly FI-extending.

  • (2) R _R is strongly FI-extending.

  • (3) For any N _R ⊴M _R , $\mathcal{D}(N_{R})\mathcal{D}[(N_{R} : M_{R})_{R}] = M\mathcal{D}[(N_{R} : M_{R})_{R}]$.

Proof

It follows immediately from Corollary 5.6.16. □

Theorem 5.6.18

Let R be a ring. Then the following are equivalent.

• (i) R is right strongly FI-extending.

• (ii) T _n (R) is right strongly FI-extending for every positive integer n.

• (iii) T _k (R) is right strongly FI-extending for some integer k>1.

• (iv) T ₂(R) is right strongly FI-extending.

Proof

(i)⇒(ii) Assume that R is right strongly FI-extending. We proceed by induction on n. Let n=2. Take M=R in Corollary 5.6.17. Let N _R ⊴M _R . Since R _R is strongly FI-extending, there exists e ²=e∈S _ℓ (R) such that N _R ≤^ess eM _R . We observe that (N _R :M _R )=N _R ≤^ess eR _R . Therefore we have that

$$\mathcal{D}_R(N_R)\mathcal{D}[(N_R : M_R)_R]=eReR=ReR=M\mathcal{D}[(N_R: M_R)_R]. $$

Hence, T ₂(R) is a right strongly FI-extending ring by Corollary 5.6.17.

Assume that T _n (R) is right strongly FI-extending. Then we show that T _n+1(R) is right strongly FI-extending. Now

$$T_{n+1}(R) = \begin{bmatrix} R & M\\ 0 & T_n(R) \end{bmatrix} , $$

where M=[R,…,R] (n-tuple). Let $_{R}N_{T_{n}(R)}\leq \,_{R}M_{T_{n}(R)}$. As in the proof of Theorem 5.6.7, N=[N ₁,…,N _n ], where N _i ⊴R for each i and N ₁⊆…⊆N _n . As R _R is strongly FI-extending, there is e∈S _ℓ (R) with N _{n R} ≤^ess eR _R , so $N = [N_{1},\dots, N_{n}]_{T_{n}(R)}\leq^{\text{ess}}e[R,\dots, R]_{T_{n}(R)} = eM$. Thus,

$$(N_{T_n(R)} : M_{T_n(R)}) = \begin{bmatrix} N_1 & N_2 &\cdots& N_n\\ 0 & N_2 &\cdots& N_n\\ \vdots& \vdots&\ddots& \vdots\\ 0 & 0 & \cdots& N_n \end{bmatrix} _{T_n(R)}\leq^{\text{ess}}(e\mathbf{1})T_n(R)_{T_n(R)}, $$

where 1 is the identity matrix in T _n (R). Hence, we have that

$$\mathcal{D}_R(N_{T_n(R)})\mathcal{D}[(N_{T_n(R)} : M_{T_n(R)})_{T_n(R)}] = eM(e\mathbf{1})T_n(R)=M(e\mathbf{1})T_n(R), $$

since e∈S _ℓ (R). Note that $M\mathcal{D}[(N_{T_{n}(R)}:M_{T_{n}(R)})_{T_{n}(R)}] = M(e\mathbf{1})T_{n}(R)$. So $M\mathcal {D}[(N_{T_{n}(R)}:M_{T_{n}(R)})_{T_{n}(R)}]= \mathcal{D}_{R}(N_{T_{n}(R)})\mathcal {D}[(N_{T_{n}(R)}:M_{T_{n}(R)})_{T_{n}(R)}]$. Thus by Corollary 5.6.16, T _n+1(R) is a right strongly FI-extending ring.

(ii)⇒(iii) is obvious, and (iii)⇒(i) is a consequence of Theorem 5.6.15.

(i)⇒(iv) follows from the proof of (i)⇒(ii) for the case when n=2, and (iv)⇒(i) follows from Theorem 5.6.15. □

Theorem 5.6.19

Let R be a ring. Then the following are equivalent.

• (i) R is right FI-extending.

• (ii) T _n (R) is right FI-extending for every positive integer n.

• (iii) T _k (R) is right FI-extending for some integer k>1.

• (iv) T ₂(R) is right FI-extending.

Proof

The proof follows by using Corollary 5.6.11 and an argument similar to that used in the proof of Theorem 5.6.18. □

Theorem 5.6.19 provides a full characterization of T _n (R) to be right FI-extending for any positive integer n. Let R be a commutative domain which is not a field. Say n is an integer such that n>1. Then T _n (R) is right strongly FI-extending (hence right FI-extending) by Theorem 5.6.18. Observe that T _n (R) is not Baer from Theorem 5.6.2. Thus by Corollary 3.3.3, T _n (R) is neither right nor left extending. Corollary 5.6.16 and Theorem 5.6.18 are now applied to show that the strongly FI-extending property for rings is not left-right symmetric.

Example 5.6.20

Let R be a commutative domain and let $M = \begin{bmatrix} 0 \,& \,R\\ 0 \,& \,0 \end{bmatrix}$. Then naturally M can be considered as an (R,T ₂(R))-bimodule. We show that the generalized triangular matrix ring $T = \begin{bmatrix} R & M\\ 0 & T_{2}(R) \end{bmatrix}$ is right strongly FI-extending, but it is not left strongly FI-extending. For this, note that _R M is faithful. Because R is right strongly FI-extending, T ₂(R) is right strongly FI-extending from Theorem 5.6.18. Say $_{R}N_{T_{2}(R)}\leq\,_{R}M_{T_{2}(R)}$. If N=0, then $\mathcal{D}_{R}(N_{T_{2}(R)})\mathcal {D}[(N_{T_{2}(R)}:M_{T_{2}(R)})_{T_{2}(R)}]=0= M\mathcal {D}[(N_{T_{2}(R)}:M_{T_{2}(R)})_{T_{2}(R)}]$. So assume that N≠0. Then there is 0≠I⊴R with $N = \begin{bmatrix} 0 \,& \,I\\ 0 \,& \,0 \end{bmatrix}$. Then I _R ≤^ess R _R , hence $\mathcal{D}_{R}(N_{T_{2}(R)})= \begin{bmatrix} 0 \,& \,R\\ 0 \,& \,0 \end{bmatrix} =M$. Therefore,

$$\mathcal{D}_R(N_{T_2(R)}) \mathcal{D}[(N_{T_2(R)} : M_{T_2(R)})_{T_2(R)}] = M\mathcal{D}[(N_{T_2(R)} : M_{T_2(R)})_{T_2(R)}]. $$

Thus, T _T is strongly FI-extending by Corollary 5.6.16.

We may note that $r_{T_{2}(R)}(M)$ is not generated, as a left ideal, by an idempotent in T ₂(R). Thus, _T T is not FI-extending by the left-sided version of the proof for (i)⇒(ii) of Theorem 5.6.10. So _T T is not strongly FI-extending.

Exercise 5.6.21

• 1. Assume that R is a PWP ring. Show that T _n (R) is a PWP ring for each positive integer n.

• 2. ([85, Birkenmeier, Park, and Rizvi]) Let R be a prime ring with P a nonzero prime ideal. Prove that the ring $\begin{bmatrix} R/P \,& \,R/P\\ 0 \,& \,R \end{bmatrix}$ is right FI-extending, but not left FI-extending.

• 3. ([85, Birkenmeier, Park, and Rizvi]) Let R be a commutative PID and let I be a nonzero proper ideal of R. Show that the ring $\begin{bmatrix} R/I \,& \,R/I\\ 0 \,& \,R \end{bmatrix}$ is right FI-extending, but not left FI-extending.

• 4. ([64, Birkenmeier and Lennon]) Let T be the ring as in Notation 5.6.3. Prove that T _T is FI-extending if and only if the following conditions hold.

  • (1) ℓ _S (M)=eS, where e∈S _ℓ (S), and eS _S is FI-extending.

  • (2) For _S N _R ≤  _S M _R , there is f ²=f∈S with N _R ≤^ess fM _R .

  • (3) R _R is FI-extending.

• 5. Let T be the ring as in Notation 5.6.3. Characterize T being right p.q.-Baer in terms of conditions on S, M, and R. (Hint: see [78, Birkenmeier, Kim, and Park].)

Historical Notes

Some of the diverse applications associated with generalized triangular matrix representations appear in the study of operator theory [212], quasitriangular Hopf algebras [113], and various Lie algebras [303]. Also many authors have studied a variety of conditions on generalized triangular matrix rings (e.g., [37, 189–191, 196, 228, 280], and [416]). Most results from Sects. 5.1, 5.2, and 5.3 are due to Birkenmeier, Heatherly, Kim, and Park [70]. Results 5.2.18–5.2.20 appear in [66]. Some of the motivating ideas for defining triangulating idempotents originated with [55]. Lemma 5.3.4 is due to Fields [164].

Theorem 5.4.1, Corollary 5.4.2, and Definition 5.4.4 appear in [70]. Piecewise domains (PWDs) were defined and investigated by Gordon and Small [187]. Proposition 5.4.6 is in [70]. Proposition 5.4.9 and Example 5.4.10(i)–(iii) and (v) are taken from [187]. Theorem 5.4.12 from [70] is a structure theorem for a PWP ring. Results 5.4.13–5.4.16 and Corollary 5.4.19 appear in [70]. Theorem 5.4.20 and Corollary 5.4.21 are taken from [66]. Examples 5.4.22 appears in [103] and [68]. In [118], Theorem 5.4.24 has been improved to the case when R is a Noetherian Rickart ring. Lemma 5.4.25 is in [70].

Results 5.5.1–5.5.3, Proposition 5.5.5, and Theorem 5.5.6 appear in [74]. Proposition 5.5.7 is in [369]. Examples 5.5.8, 5.5.9, Results 5.5.10–5.5.12 are taken from [74]. Theorem 5.5.14 is due to Birkenmeier, Kim, and Park [74]. Koh ([255] and [256]), Lambek [265], Shin [369], and Sun [388] showed that the Gelfand homomorphism θ is an isomorphism for various classes of rings.

Theorem 5.6.1 is due to Akalan, Birkenmeier, and Tercan (see [1, 3], and [393]). Theorem 5.6.2 appears to be a new result which is due to the authors. Results 5.6.4–5.6.6 appear in [85]. Theorem 5.6.7 was obtained by Pollingher and Zaks in [347], but we give the proof in a different way by applying Theorem 5.6.5. Proposition 5.6.8 is from [78]. Theorem 5.6.9 is completely generalized in [3]. Results 5.6.10–5.6.13 and Definition 5.6.14 appear in [85]. A characterization of generalized triangular right FI-extending rings are also considered in [64] (see Exercise 5.6.21.4). Results 5.6.15–5.6.18 appear in [85]. Theorem 5.6.19 was shown in [83], while Example 5.6.20 was given in [85]. Further related references include [51, 81, 91, 116, 122, 125, 135, 160], and [387].