Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 38 (1997), No. 2, 233-239.

On Rings with Zero Total

K. I. Beidar

Department of Mathematics, National Cheng-Kung University, Tainan, Taiwan

Abstract. Rings with zero total and bounded indices are characterized. It is proved that essential left (right) ideals of such rings contain essential two sided ideals generated by central idempotents. In particular, a prime ring with bounded indices has zero total if and only if it is a matrix ring over a skew field. It is shown that the maximal right ring of quotients of a such ring is equal to the maximal left ring of quotients and is isomorphic to finite direct sum of matrix rings over abelian regular selfinjective rings.

MSC 1991: 16E50 (primary), 16S90 (secondary) Keywords: idempotent, total, index of nilpotency, ring of quotients

All rings considered will be associative, not necessarily with unity element. Given a ring A, I ◁ A will denote the fact that I is an ideal of A. Given two right ideals I ⊆ J of a ring A, I is said to be essential in J, if I ∩ K ≠ 0 for every nonzero right ideal K of A contained in J. A nonzero element a ∈ A is said to be partially invertible, if there exists an element b ∈ A such that ab is a nonzero idempotent. Following Kasch [12] the total Tot(A) of a ring A may be defined as the set

Tot(A) = {a ∈ A | aA does not contain a nonzero idempotent}
= {a ∈ A | a is not partially invertible}.

One can check that (a) Tot(A) is left and right symmetric,

(b) Tot(A) is a semi-ideal: aTot(A)b ⊆ Tot(A) for every a, b ∈ A,

(c) J(A) + Tot(A) ⊆ Tot(A) where J(A) is the Jacobson radical of A.

Recently [6] it was shown that the class M of all rings with zero total forms a weakly special class (in particular, M is closed under taking of ideals and essential extensions) and

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 1997 Heldermann Verlag

an upper radical K = UM is a supernilpotent normal radical. Note that the class M contains the class of all strongly π-regular rings having no nonzero right nil ideals [7]. In particular, M contains all von Neumann regular rings and all semisimple rings with dcc for principal right ideals (see also [13, Chapter XIV]). Moreover, M contains Rickart C^{∗}-algebras [2, Chapter 1, §8, Proposition 4]. The condition Tot(A) = 0 is proved to be useful in study of Baer ∗-rings [9], [10] and [14].

The structure of semiprime rings with bounded indices of nilpotent elements was studied in [3], [1], [8] and [4]. In the present paper we study the structure of rings with zero total and bounded indices of nilpotent elements.

Let A be a ring. A finite subset {e_{ij} | 1 ≤ i, j ≤ n} ⊆ A of nonzero elements such that e_{ij}e_{kl} = δ_{jk}e_{il} for all i, j, k, = 1, 2, …, n is called a system of n^{2} matrix units. Given a nilpotent element a ∈ A, we set In(a) = min{n | a^{n} = 0}. If a ∈ A is not nilpotent, we set In(a) = 1. Let S ⊆ A be a subset. Set

In(S) = sup{In(a) | a ∈ S}.

Recall that a ring A is called I-ring if every non-nil right ideal of A contains a nonzero idempotent. Noting, that every ring in M is a Jacobson semisimple I-ring, we can state the following two theorems due to Levitzki as follows.

Theorem 1. [11, Theorem 1, Chapter X, Section 11] Let A be a ring with zero total and let a ∈ A be a nilpotent element of index n. Then the ideal (a) of A generated by a contains a system of n^{2} matrix units.

Theorem 2. [11, Theorem 2, Chapter X, Section 11] Let A be a ring with zero total. Suppose that A contains a system {e_{ij}} of n^{2} matrix units such that either e_{11}Ae_{11} contains nonzero nilpotent elements or u = ∑_{i=1}^{n} e_{ii} is not in the center of A. Then there exists a system of ≥ (n + 1)^{2} matrix units {f_{kl}} such that f_{11} ∈ e_{11}Ae_{11}.

The following result is a direct consequence of Theorem 1.

Corollary 3. If A is a ring with zero total, then either, for any natural number n, A contains systems of n^{2} matrix units or In(A) = n for some natural number n.

Lemma 4. Let A be a ring with In(A) = n for some natural number n and {e_{ij} | 1 ≤ i, j ≤ n} ⊆ A a system of n^{2} matrix units. Then e_{11}Ae_{11} is a reduced ring and e_{11} + e_{22} + … + e_{nn} a central idempotent of A.

Proof. Suppose that e_{11}Ae_{11} is not a reduced ring. Choose 0 ≠ b ∈ e_{11}Ae_{11} such that b^{2} = 0. Letting c = b + e_{12} + e_{23} + … + e_{n−1,n}, we note that c^{n} = be_{1n} and c^{n+1} = 0. Since In(A) = n, we conclude that be_{1n} = c^{n} = 0. Then b = be_{1n}e_{n1} = 0 contradicting b ≠ 0. Thus e_{11}Ae_{11} is a reduced ring.

Next, let u = e_{11}+e_{22}+…+e_{nn}. Assume that (1−u)Au ≠ 0 and choose 0 ≠ b ∈ (1−u)Au. Clearly be_{ii} ≠ 0 for some 1 ≤ i ≤ n. After a suitable renumbering of e_{ij}’s we may assume that be_{11} ≠ 0. Set c = b+e_{12} +e_{23} +…+e_{n−1,n}. Then c^{n} = be_{1n} and c^{n+1} = 0. It follows that c^{n} = 0 which implies that b = 0 contradicting b ≠ 0. Therefore (1 − u)Au = 0 and au = uau for all a ∈ A. Analogously one can show that ua = uau for all a ∈ A. Thus au = ua for all a ∈ A and so u is a central idempotent of A.

Given a subset S of a ring A, we set

l_{A}(S) = {a ∈ A | aS = 0} and
r_{A}(S) = {a ∈ A | Sr = 0}.

Let A be a ring and E ⊆ A. The set E is called an orthogonal subset of A if it consists of central idempotents and uv = 0 for all u, v ∈ E, u ≠ v. A subset E of A is said to be a dense orthogonal subset of A if E is an orthogonal subset of A and l_{A}(E) = 0. Given a right ideal L of A and a ∈ A, we set

(L : a) = {x ∈ A | ax ∈ L}.

Clearly (L : a) is a right ideal of A. Next, if L is an essential right ideal of A, then one can easily show that (L : a) is an essential right ideal of A as well.

Theorem 5. Let A be a ring with In(A) = n for some natural number n. Then the following two conditions are equivalent:

1. Tot(A) = 0;

2. A contains an essential ideal I which is a direct sum of ideals I_{k} = M_{n_{k}}(D_{k}), k = 1, 2, …, t, where

(a) n_{1} < n_{2} < … < n_{t} = n.

(b) Each M_{n_{k}}(D_{k}) is the n_{k} × n_{k} matrix ring over a reduced ring D_{k}.

(c) If L is a nonzero right ideal of D_{k}, then the set of all central idempotents of D_{k} belonging to L generates an ideal K essential in L.

If Tot(A) = 0, then every essential right (left) ideal of A contains a dense orthogonal subset of A.

Proof. (1) ⟹ (2). We proceed by induction on n. If n = 1, the A is a reduced ring and A = M_{1}(A). It is well-known that each idempotent of a reduced ring is central. Let L be any nonzero right ideal of A and I the ideal generated by all the central idempotents belonging to L. Next, let 0 ≠ a ∈ L. Since Tot(A) = 0, aA contains a nonzero idempotent and so I ∩ aA ≠ 0.

Now suppose that n > 1. Choose a ∈ A with In(a) = n. Then by Theorem 1 the ideal (a) contains a system {e_{ij}} of n^{2} matrix units. By Lemma 4, e_{11}Ae_{11} is a reduced ring and e_{11} + e_{22} + … + e_{nn} is a central idempotent of A. According to [11, Proposition 6, Chapter III, Section 8], Au = M_{n}(K) where

K = {a ∈ Au | ae_{ij} = e_{ij}a for all i, j = 1, 2, …, n}

and K ≅ e_{11}Aue_{11} = e_{11}Ae_{11}. Therefore K is a reduced ring. Let 0 ≠ a ∈ e_{11}Ae_{11}. Then (ab)^{2} = ab ≠ 0 for some b ∈ A. Set c = e_{11}be_{11}. Since e_{11}a = a = ae_{11},

(ac)^{2} = (ab)^{2}e_{11} = abe_{11} = ac.

Further, (ac)(ab) = (ab)^{2} = ab ≠ 0 and so ac ≠ 0. It follows that e_{11}Ae_{11} ∈ M and hence K ∈ M. By the case n = 1 we have that K satisfies (c).

From the Kuratowski-Zorn Lemma it follows that there exists a maximal family of pairwise orthogonal central idempotents {u_{j} | j ∈ J} such that Au_{j} = M_{n}(K_{j}) where K_{j} is a reduced subring of A belonging to M. We set

M = ∑_{j∈J} Au_{j} = ⊕_{j∈J}M_{n}(K_{j}) = M_{n}(⊕_{j∈J}K_{j})

and note that D = ⊕_{j∈J}K_{j} ∈ M. If M is an essential ideal of A, then there is nothing to prove. Suppose that N = r_{A}(M) ≠ 0. Clearly N is an ideal of A belonging to M and M + N = M ⊕ N is an essential ideal of A. We claim that In(N) < n. If not, then by the above result N contains a central idempotent u′ such that u′N = M_{n}(K′) where K′ is a reduced ring belonging to M. Obviously u′ is a central idempotent of A and u′N = u′A. As NM = 0, we have that u′u_{j} = 0 for all j ∈ J, contradicting to the maximality of the family {u_{j} | j ∈ J}. Therefore In(N) < n. Applying the induction assumption we complete the proof.

(2) ⟹ (1). It follows from (c) that each D_{k} ∈ M. Let B = M_{n_{k}}(D_{k}) and

R = \{( \begin{matrix}d_{11} & d_{12} & … & d_{1n_{k}} \\ 0 & 0 & … & 0 \\ … \\ 0 & 0 & … & 0 \end{matrix} ) ∈ M_{n_{k}}(D_{k}) | d_{1i} ∈ D_{k}, 1 ≤ i ≤ n_{k}\} ,

S = \{( \begin{matrix}0 & d_{12} & … & d_{1n_{k}} \\ 0 & 0 & … & 0 \\ … \\ 0 & 0 & … & 0 \end{matrix} ) ∈ M_{n_{k}}(D_{k}) | d_{1i} ∈ D_{k}, 2 ≤ i ≤ n_{k}\} .

Clearly R is a right ideal of B such that r_{B}(R) = 0, S = l_{R}(R) and R/S ≅ D_{k} ∈ M, and hence the conditions of [6, Lemma 2.3] are fulfilled which implies that B ∈ M. Therefore I = ⊕_{k=1}^{t}M_{n_{k}}(D_{k}) ∈ M. Since I is an essential ideal of A and M is closed under taking essential extensions, we conclude that A ∈ M and so Tot(A) = 0.

Finally, let L be an essential right ideal of A. Clearly L ∩ I is an essential right ideal of the ring I and every dense orthogonal subset of the ring I is a dense orthogonal subset of A. Therefore it is enough to show that L ∩ I contains a dense orthogonal subset of I. Since I = ⊕_{k=1}^{t}M_{n_{k}}(D_{k}), we can reduce the problem to the case when A = M_{n}(D) and D is a reduced ring belonging to M. By Kuratowski-Zorn lemma L contains a maximal orthogonal subset E. Suppose that P = l_{A}(E) ≠ 0. Clearly there exists a nonzero ideal J of D such that M_{n}(J) ⊆ P. As D ∈ M, J contains a nonzero central idempotent u. Therefore P contains a nonzero central idempotent e. Obviously eA = M_{n}(eD) and eL = L ∩ eA is an essential right ideal of the ring eA. Clearly eE = 0. It follows from the choice of E that eL contains no nonzero central idempotents of A. We conclude that it is enough to show that L contains a nonzero central idempotent of A provided that A = M_{n}(D) where D ∈ M is a reduced ring with unity 1 and L is an essential right ideal of A. Let {e_{ij} | 1 ≤ i, j ≤ n} be a system of n^{2} matrix units of A. Clearly M = ∩_{i=1}^{n}(L : e_{i1}) is an essential right ideal of A and e_{i1}M ⊆ L for all i = 1, 2, …, n. As M is an essential right ideal, e_{11}M ≠ 0. Taking into account that A is semiprime, we obtain that (e_{11}M)^{2} ≠ 0 and so e_{11}Me_{11} ≠ 0. Choose any 0 ≠ x ∈ e_{11}Me_{11}. Then e_{i1}xe_{1i} ∈ e_{i1}M ⊆ L and so a = ∑_{i=1}^{n} e_{i1}xe_{1i} ∈ L. Clearly 0 ≠ a ∈ D. Therefore there exists b ∈ D such that ab is a nonzero central idempotent. Since ab ∈ L, we conclude that L contains a nonzero central idempotent. The proof is completed.

Corollary 6. Let A be a ring with In(A) = n < ∞. Then the following conditions are equivalent:

1. A is a prime ring with zero total;

2. A = M_{n}(D) where D is a skew field.

Proof. The implication (2) ⟹ (1) is obvious. Suppose that A ∈ M. Since A is a prime ring, it follows from Theorem 5 that A contains an nonzero ideal I of the form I = M_{n}(D) where D is a reduced ring belonging to M. Clearly I is a prime ring and so D is a prime ring as well. It follows that D is a domain and hence every central idempotent of D is either zero or the unity of D. According to Theorem 5, every nonzero right ideal of D contains a nonzero central idempotent. Therefore D has no proper nonzero right ideals and thus D is a skew field which implies A = I = M_{n}(D).

Corollary 7. Let A be a ring with In(A) = n for some natural number n. Suppose that Tot(A) = 0. Then:

1. A left ideal L of A is essential if and only if r_{A}(L) = 0.

2. A right ideal R of A is essential if and only if l_{A}(R) = 0.

Proof. By symmetry it is enough to prove (1). If L is an essential left ideal, then L contains a dense orthogonal subset of A and so r_{A}(L) = 0. Next, let L be a left ideal such that r_{A}(L) = 0. Suppose that L is not essential. Choose K to be a left ideal maximal with respect to the property K ∩ L = 0. It is well-known that K ⊕ L is an essential left ideal of A. By Theorem 5, K ⊕ L contains a dense orthogonal subset E of A. If E ⊆ L, then L contains the ideal ⟨E⟩ of A generated by E. Since r_{A}(E) = 0, ⟨E⟩ is an essential left ideal of A. Hence L is an essential left ideal of A contradicting to our assumption. Therefore E ⊈ L. Choose any e ∈ E \ L. Then

Ae = eAe ⊆ e(K ⊕ L) = eK ⊕ eL.

Clearly Ae is a ring with unity e and eK, eL are left ideals of Ae. Therefore there exist two orthogonal idempotents u and v of Ae such that e = u + v, eK = Aeu, and eL = Aev. As e ∉ L, u ≠ 0. Since eu = u, we have that

Lu = Leu = (eL)u = (Aev)u = Ae(vu) = 0,

contradicting to r_{A}(L) = 0. Therefore L is an essential ideal of A.

The reader is referred to [5, Chapter 2] and [15] for basic results of the theory of rings of quotients. Let Tot(A) = 0 and 0 ≠ a ∈ A. Then there exists b ∈ A such that ba is a nonzero idempotent. Therefore r_{A}(a) ⊆ r_{A}(ba) = (1 − ba)A and so r_{A}(a) is not an essential right ideal of A. Therefore the condition Tot(A) = 0 implies that A is a right nonsingular ring. Analogously one can show that A is a left nonsingular ring. Recall that a ring A is called abelian regular (or strongly regular) if it is a von Neumann regular reduced ring.

Corollary 8. Let A be a ring with In(A) = n for some natural number n. Suppose that Tot(A) = 0. Then:

1. The left maximal ring of quotients Q_{ml}(A) of A is equal to the right maximal ring of quotients Q_{mr}(A) of A.

2. For each q ∈ Q_{mr}(A) there exists a dense orthogonal subset E_{q} of A such that eq ∈ A for all e ∈ E_{q}.

3. Q_{mr}(A) = ⊕_{i=1}^{t}M_{n_{i}}(R_{i}) where each D_{i} is an abelian regular left and right selfinjective ring.

Proof. According to [16, Theorem 3.3], given a right and left nonsingular ring D, Q_{mr}(D) = Q_{ml}(D) if and only if a right (left) ideal R of D is essential if and only if l_{D}(R) = 0 (respectively, r_{D}(R) = 0). Therefore Corollary 7 implies that Q_{mr}(A) = Q_{ml}(A). The second statement follows from the facts that each dense right ideal of A is an essential right ideal and so contains a dense orthogonal subset of A by Theorem 5. Finally, by Theorem 5 the ring A contains an essential ideal I = ⊕_{i=1}^{t}M_{n_{i}}(D_{i}) where each D_{i} is a reduced ring belonging to M. Clearly Q_{mr}(A) = ⊕_{i=1}^{t}M_{n_{i}}(Q_{mr}(D_{i})). Therefore it is enough to show that Q_{mr}(D_{i}) is an abelian regular ring. Since D_{i} ∈ M, we conclude that D_{i} is a right and left nonsingular ring and Q_{mr}(D_{i}) = Q_{ml}(D_{i}). It is well-known that the maximal right ring of quotients of a right nonsingular ring is a right selfinjective von Neumann regular ring (see [5, Theorem 2.1.15]). Therefore Q_{mr}(D_{i}) is a left and right selfinjective von Neumann regular ring because Q_{mr}(D_{i}) = Q_{ml}(D_{i}). Suppose that q ∈ Q_{mr}(D_{i}) and q^{2} = 0. Then (eq)^{2} = eq^{2} = 0 for all e ∈ E_{q}. Since D_{i} is reduced, we obtain that eq = 0 for all e ∈ E_{q}. Therefore q⟨E_{q}⟩ = 0 where ⟨E_{q}⟩ is the ideal of D_{i} generated by E_{q}. As E_{q} is a dense orthogonal subset of D_{i}, we conclude that ⟨E_{q}⟩ is a dense ideal of D_{i} and so the equality q⟨E_{q}⟩ = 0 implies that q = 0. Thus Q_{mr}(D_{i}) is a reduced ring and the proof is complete.

I would like to express my gratitude to R. Wiegandt for bringing my attention to rings with zero total and for his valuable comments.

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Received July 11, 1996