STABLE RANK OF CORNER RINGS

P. Ara and K. R. Goodearl

Abstract. B. Blackadar recently proved that any full corner pAp in a unital C*-algebra A has K-theoretic stable rank greater than or equal to the stable rank of A. (Here p is a projection in A, and fullness means that ApA = A.) This result is extended to arbitrary (unital) rings A in the present paper: If p is a full idempotent in A, then sr(pAp) ≥ sr(A). The proofs rely partly on algebraic analogs of Blackadar’s methods, and partly on a new technique for reducing problems of higher stable rank to a concept of stable rank one for skew (rectangular) corners pAq. The main result yields estimates relating stable ranks of Morita equivalent rings. In particular, if B ≅ End_{A}(P) where P_{A} is a finitely generated projective generator, and P can be generated by n elements, then sr(A) ≤ n· sr(B) − n + 1.

Introduction

The theory of stable range of rings was developed by H. Bass [2] and L. N. Vaserstein [10]. As is now common, we define the stable rank of a ring A, denoted sr(A), to be the least positive integer n such that A satisfies Bass’s n-th stable range condition, or ∞ if no such n exists. It is well known that stable rank is not Morita invariant. In fact, Vaserstein [10] computed the stable rank of a matrix ring M_{n}(A), obtaining the following amazing formula

sr(M_{n}(A)) = ⌈\frac{sr(A) − 1}{n}⌉ + 1,

where ⌈r⌉ denotes the least integer greater than or equal to a real number r. If B is a ring Morita equivalent to A, then B ≅ pM_{n}(A)p for some full idempotent p ∈ M_{n}(A). Thus, to understand the behavior of stable rank under Morita equivalence, it remains to see what happens to stable rank under the passage from a ring to a full corner. Vaserstein’s formula already contains some information in this direction, namely that sr(A) ≥ sr(M_{n}(A)) for all n ∈ ℕ. Since A is isomorphic to a corner ring in M_{n}(A), corresponding to the full idempotent e_{11}, this suggests the inequality sr(pAp) ≥ sr(A) for any full corner pAp of

2000 Mathematics Subject Classification. Primary 19B10; Secondary 16S50. Key words and phrases. Stable range, stable rank, corner ring, matrix ring. The first-named author was partially supported by the DGI and European Regional Development

Fund, jointly, through Project BFM2002-01390, and by the Comissionat per Universitats i Recerca de la Generalitat de Catalunya. The second-named author was partially supported by an NSF grant. Part of this work was done during his research stay at the Centre de Recerca Matemàtica (Barcelona) in Spring 2003; he thanks the CRM for its hospitality and support.

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A. Such a formula was conjectured by Blackadar [3, Remark A7] to hold for the topological stable rank introduced by Rieffel in [9]. It was subsequently proved by Herman and Vaserstein [5] that the Rieffel topological stable rank and the Bass stable rank agree for any C*-algebra. Blackadar has recently verified the corner conjecture in [4, Theorem 4.5]. His methods are focussed on the topological stable rank, and rely on norm estimates for differences of row vectors.

Previous work on stable rank of corners gave weaker inequalities of the following form. If p is a full projection in a unital C*-algebra A, then, for some n, there exist n pairs (a_{i}, b_{i}) ∈ A^{2} such that ∑_{i=1}^{n} a_{i}pb_{i} = 1. Blackadar showed in [3, Lemma A6] that in this situation, sr(A) ≤ sr(pAp)+n−1. That this result extends to full idempotents in arbitrary rings was noted by the present authors in [1, Remark 1.4]. In particular, this inequality suffices to show that finiteness of the stable rank is Morita invariant.

Here we prove that the inequality sr(A) ≤ sr(pAp) holds for any full corner pAp in any unital ring A (Theorem 7). The structure of the proof has been modelled after Blackadar’s paper [4], but we have had to replace his topological methods with purely algebraic ones. Of crucial importance is the notion of stable rank one for skew, or rectangular, corners pAq, where p and q are distinct idempotents of A. This allows us to work only with stable rank one conditions, thus avoiding higher rank conditions. By combining our main result with Vaserstein’s formula, we obtain estimates comparing the stable ranks of Morita equivalent rings (Theorem 9).

We note that Lam and Dugas [8] have recently shown that the reverse inequality sr(A) ≥ sr(eAe) holds for any quasi-duo ring A and any idempotent e in A. By definition, a quasiduo ring is a ring in which every maximal one-sided ideal is an ideal. It is clear that the only full idempotent in a quasi-duo ring is 1, so our result does not give any further insight into Lam and Dugas’s, nor vice versa.

Stable rank and skew corners

Throughout, let A be a unital ring. We start by recalling the definition of the (Bass) stable rank:

Definition. An n-row (a_{1}, …, a_{n}) ∈ A^{n} is said to be right unimodular if ∑_{i=1}^{n} a_{i}A = A. An (n+1)-row (a_{1}, …, a_{n}, b) ∈ A^{n+1} is reducible in case there is an n-row (c_{1}, …, c_{n}) ∈ A^{n} such that the n-row (a_{1} + bc_{1}, …, a_{n} + bc_{n}) is right unimodular. The stable rank of A, denoted sr(A), is the least positive integer n such that every right unimodular (n + 1)-row in A^{n+1} is reducible, or ∞ if no such n exists.

We next recall some useful terminology. Two idempotents p and q in A are said to be orthogonal, written p ⊥ q, in case pq = qp = 0. The set of all idempotents of A is partially ordered by declaring p ≤ q if and only if p = pq = qp. The idempotents p and q are equivalent, written p ∼ q, in case there are elements a ∈ pAq and b ∈ qAp such that p = ab and q = ba. (Note that p and q are equivalent if and only if the right (respectively, left) ideals generated by p and q are isomorphic as a right (respectively, left) R-modules [6, Proposition 21.20].) We write p ≲ q in case there is an idempotent p′ such that p′ ≤ q and p ∼ p′; this occurs if and only if there exist elements a ∈ pAq and b ∈ qAp such that ab = p. For any idempotents p, q ∈ A, we write p ⊕ q for the idempotent diag(p, q) in M_{2}(A). Accordingly, the notation n·p is used for the idempotent diag(p, p, …, p) in M_{n}(A).

For all m, n ∈ ℕ, identify M_{n}(A) with the n × n upper left corner subring of M_{n+m}(A). In particular, A = M_{1}(A) is then identified with a subring of each M_{n}(A). With this identification, 1_{A} equals the matrix unit e_{11}, and n·1_{A} equals the identity matrix in M_{n}(A). These identifications allow us to work in as large a matrix ring as is convenient. When the size of the matrices is not relevant, we write M_{•}(A) to stand for M_{n}(A) with n unspecified.

We say that an idempotent p in A is full in case p generates A as a two-sided ideal, that is, ApA = A. It is standard (and an easy exercise) that p is a full idempotent if and only if 1_{A} ≲ t·p for some t ∈ ℕ. A corner of A is any subring of the form pAp, where p is an idempotent, and we say that pAp is a full corner in case p is a full idempotent. A skew (rectangular) corner in A is any subset of the form pAq, for idempotents p, q ∈ A. Note that pAq is a (pAp, qAq)-bimodule, and that the ring multiplication in A induces bimodule homomorphisms pAq ⊗_{qAq} qAp → pAp and qAp ⊗_{pAp} pAq → qAq.

Definition. Let p, q ∈ A be idempotents. We say that the skew corner pAq has (right) stable rank 1, abbreviated sr(pAq) = 1, provided the following condition holds: Whenever a ∈ pAq, x ∈ qAp, and b ∈ pAp such that ax+b = p, there exist y ∈ pAq and z ∈ qAp such that (a + by)z = p. Note that in case p = q, there is no conflict between this definition and the statement that the stable rank of the ring pAp is 1.

The key to our methods is the following lemma, which reduces stable rank calculations to questions of stable rank 1 for skew corners. Note that 1_{A}M_{n}(A) is a skew corner, since it equals 1_{A}M_{n}(A)(n·1_{A}).

Lemma 1. Let n ∈ ℕ. Then sr(A) ≤ n if and only if sr(1_{A}M_{n}(A)) = 1.

Proof. (⟹): Let α ∈ 1_{A}M_{n}(A), χ ∈ M_{n}(A)1_{A}, and β ∈ 1_{A}M_{n}(A)1_{A} such that αχ + β = 1_{A}. Then

(†) α = [ \begin{matrix}a_{1} & a_{2} & ⋯ & a_{n} \\ 0 & 0 & ⋯ & 0 \\ ⋮ & ⋮ & & ⋮ \\ 0 & 0 & ⋯ & 0 \end{matrix} ] , χ = [ \begin{matrix}x_{1} & 0 & ⋯ & 0 \\ x_{2} & 0 & ⋯ & 0 \\ ⋮ & ⋮ & & ⋮ \\ x_{n} & 0 & ⋯ & 0 \end{matrix} ] , β = [ \begin{matrix}b & 0 & ⋯ & 0 \\ 0 & 0 & ⋯ & 0 \\ ⋮ & ⋮ & & ⋮ \\ 0 & 0 & ⋯ & 0 \end{matrix} ]

for some a_{i}, x_{i}, b ∈ A such that a_{1}x_{1} + ⋯ + a_{n}x_{n} + b = 1. Since sr(A) ≤ n, there exist y_{1}, …, y_{n} ∈ A such that the row (a_{1} + by_{1}, …, a_{n} + by_{n}) is right unimodular, that is, (a_{1} + by_{1})z_{1} + ⋯ + (a_{n} + by_{n})z_{n} = 1 for some z_{i} ∈ A. Setting

(‡) ζ = [ \begin{matrix}y_{1} & y_{2} & ⋯ & y_{n} \\ 0 & 0 & ⋯ & 0 \\ ⋮ & ⋮ & & ⋮ \\ 0 & 0 & ⋯ & 0 \end{matrix} ] ∈ 1_{A}M_{n}(A),  ξ = [ \begin{matrix}z_{1} & 0 & ⋯ & 0 \\ z_{2} & 0 & ⋯ & 0 \\ ⋮ & ⋮ & & ⋮ \\ z_{n} & 0 & ⋯ & 0 \end{matrix} ] ∈ M_{n}(A)1_{A},

we have (α + βζ)ξ = 1_{A}.

(⟸): If (a_{1}, …, a_{n}, b) ∈ A^{n+1} is a right unimodular row, there exist x_{1}, …, x_{n}, x ∈ A such that a_{1}x_{1} + ⋯ + a_{n}x_{n} + bx = 1. After replacing b by bx, we may assume that x = 1. Define matrices α ∈ 1_{A}M_{n}(A), χ ∈ M_{n}(A)1_{A}, and β ∈ 1_{A}M_{n}(A)1_{A} as in (†) above, and observe that αχ + β = 1_{A}. Since sr(1_{A}M_{n}(A)) = 1, there exist matrices ζ and ξ as in (‡) such that (α + βζ)ξ = 1_{A}. It follows that the row (a_{1} + by_{1}, …, a_{n} + by_{n}) is right unimodular. □

We next show how stable rank 1 can be transferred from certain skew corners to others.

Lemma 2. Let p, q ∈ A be idempotents.

(a) If sr(pAq) = 1, then p ≲ q.

(b) If p′, q′ ∈ A are idempotents such that p′ ∼ p and q′ ∼ q, then sr(pAq) = 1 if and only if sr(p′Aq′) = 1.

Proof. (a) Consider the equation 0·0 + p = p, where we view the first 0 ∈ pAq, the second 0 ∈ qAp, and p ∈ pAp. The hypothesis sr(pAq) = 1 then gives us y ∈ pAq and z ∈ qAp such that (0 + py)z = p. Hence, pyqzp = p, and it follows that p ≲ q.

(b) There are elements u ∈ pAp′ and u′ ∈ p′Ap such that uu′ = p and u′u = p′, and elements v ∈ qAq′ and v′ ∈ q′Aq such that vv′ = q and v′v = q′.

Assume that sr(pAq) = 1, and consider elements a ∈ p′Aq′, x ∈ q′Ap′, and b ∈ p′Ap′ such that ax + b = p′. Then we have uav′ ∈ pAq, vxu′ ∈ qAp, and ubu′ ∈ pAp satisfying (uav′)(vxu′) + (ubu′) = p. Since sr(pAq) = 1, there exist y ∈ pAq and z ∈ qAp such that (uav′ + ubu′y)z = p. Then u′yv ∈ p′Aq′ and v′zu ∈ q′Ap′ are elements satisfying the equation (a + bu′yv)v′zu = p′, which proves that sr(p′Aq′) = 1. The converse follows by symmetry. □

Lemma 3. Let p, q, s ∈ A be idempotents such that s ⊥ q. If sr(pAq) = 1, then also sr(pA(q + s)) = 1.

Proof. Let a ∈ pA(q + s), x ∈ (q + s)Ap, and b ∈ pAp such that ax + b = p. Rewrite this equation as (aq)(qx) + (b + asx) = p, where aq ∈ pAq, qx ∈ qAp, and b + asx ∈ pAp. Since sr(pAq) = 1, there exist y ∈ pAq and z ∈ qAp such that

[(aq) + (b + asx)y]z = p.

Note that sz = 0 because s ⊥ q, which allows us to rewrite the equation above as

[a(q + s + sxy) + by]z = p.

Since ys = 0, the element sxy is nilpotent, and so the element u := q + s + sxy is a unit in the ring (q + s)A(q + s). Now

(a + byu^{−1})(uz) = (au + by)z = p

with yu^{−1} ∈ pA(q + s) and uz ∈ (q + s)Ap, which completes the proof. □

Lemma 4. Let p, q, r ∈ A be idempotents such that p, q ⊥ r. If sr((p + r)A(q + r)) = 1, then sr(pAq) = 1.

Proof. Let a ∈ pAq, x ∈ qAp, and b ∈ pAp such that ax + b = p. Then the elements a + r ∈ (p + r)A(q + r) and x + r ∈ (q + r)A(p + r) satisfy (a + r)(x + r) + b = p + r. Since sr((p + r)A(q + r)) = 1, there exist y ∈ (p + r)A(q + r) and z ∈ (q + r)A(p + r) such that (a + r + by)z = p + r. Observe that

rz = r(a + r + by)z = r  (a + by)z = p(a + r + by)z = p.

Then z = (q + r)z = qz + r, and so zp = qzp ∈ qAp. Since

(a + bpyq)(zp) = (a + by)zp = p,

we have shown that sr(pAq) = 1. □

The main results

The final ingredient needed to prove our main theorem is a (partial) converse to Lemma 4, which holds when r ∈ ApA. The following observations will be helpful.

Observation 5. If we are trying to establish sr(pAq) = 1 for some idempotents p, q ∈ A, then we are given ax+b = p for some a ∈ pAq, x ∈ qAp, and b ∈ pAp, and we seek y ∈ pAq and z ∈ qAp such that (a + by)z = p. Several reduction steps are possible, in which we may do any of the following:

(1) Replace a by a + bc for any c ∈ pAq;

(2) Replace a by au for any unit u of qAq;

(3) Replace a by va for any unit v of pAp;

(4) Replace A by M_{n}(A) for any n ∈ ℕ. In cases (1)–(3), the replacement of a by another element of pAq must be accompanied by corresponding replacements for x and b.

To see why (1) is allowed, for instance, observe that

(a + bc)x + b(p − cx) = p

with a + bc ∈ pAq and b(p − cx) ∈ pAp; if there exist y′ ∈ pAq and z′ ∈ qAp such that

[(a + bc) + b(p − cx)y′]z′ = p,

then [a+b(c+y′ −cxy′)]z′ = p, with c+y′ −cxy′ ∈ pAq. For (2), we have (au)(u^{−1}x)+b = p, and if (au + by′)z′ = p, then (a + by′u^{−1})(uz′) = p. In the case of (3), we have (va)(xv^{−1}) + (vbv^{−1}) = p, and if (va + vbv^{−1}y′)z′ = p, then (a + bv^{−1}y′)(z′v) = p.

Finally, we address (4). Because of our identification of A with the corner e_{11}M_{n}(A)e_{11}, we have pAq = pM_{n}(A)q, and similarly for qAp and pAp. Thus, if there exist y ∈ pM_{n}(A)q and z ∈ qM_{n}(A)p satisfying (a + by)z = p, then y ∈ pAq and z ∈ qAp, and the equation holds in A.

Proposition 6. Let p, q, r ∈ A be idempotents such that p, q ⊥ r. If sr(pAq) = 1 and r ≲ n·p for some n ∈ ℕ, then sr((p + r)A(q + r)) = 1.

Proof. By Lemma 2, p ∼ p′ for some idempotent p′ ≤ q, and we may replace p by p′. Thus, we may assume that p ≤ q.

We claim that it suffices to establish the proposition under the additional hypothesis r ∼ p. Given that case, it follows by induction that sr((2^{m}·p)M_{•}(A)(q ⊕ (2^{m} − 1)·p)) = 1 for all m ∈ ℕ. In the general case, we choose m large enough that 2^{m} − 1 ≥ n, so that (2^{m} − 1)·p = r′ ⊕ r′′ for some orthogonal idempotents r′ and r′′ with r′ ∼ r. In a lower right corner of a suitably large matrix ring, we can find an idempotent s such that s ⊥ 1_{A} and s ∼ r′′. Thus, p, q, r ⊥ s and (2^{m} − 1)·p ∼ r + s. Consequently, sr((p+r+s)M_{•}(A)(q+r+s)) = 1, and the desired result follows from Lemma 4. Therefore the claim holds, and we may assume that p ∼ r. In particular, sr(rAq) = 1.

To prove that sr((p + r)A(q + r)) = 1, we may work within the ring (q + r)A(q + r). Hence, we may assume, for convenience, that q +r = 1. Now elements a ∈ A can be viewed as formal matrices of the form

[ \begin{matrix}qaq & qar \\ raq & rar \end{matrix} ] .

We mimic the proof that stable rank 1 passes from a ring to its 2 × 2 matrix ring, which would be exactly our present situation in case p = q. In order to allow for the possibility that p < q, we must be careful to modify our matrices starting at the lower right corner (where the entries come from rAr), rather than starting at the upper left corner, where we can control pAq but not qAq.

Let a ∈ (p + r)A, x ∈ A(p + r), and b ∈ (p + r)A(p + r) such that ax + b = p + r. Note that qaq = paq and qar = par. Now

(raq)(qxr) + (rarxr + rbr) = r(ax + b)r = r.

Since sr(rAq) = 1, there exist y_{1} ∈ rAq and z_{1} ∈ qAr such that

[(raq) + (rarxr + rbr)y_{1}]z_{1} = r.

The factor q in raq can be dropped from the last equation because qz_{1} = z_{1}. Hence,

r[a(1 + rxy_{1}) + by_{1}]z_{1} = r.

Since y_{1}r = 0, the element 1 + rxy_{1} is a unit in A. In view of Observation 5, we may replace a by a(1 + rxy_{1}) + by_{1}. Thus, we may now assume that there exists z_{1} ∈ qAr such that raz_{1} = r.

Next, we replace a by the element

[ \begin{matrix}paq & par \\ raq & rar \end{matrix} ] [ \begin{matrix}q & z_{1}(r − rar) \\ 0 & r \end{matrix} ] = [ \begin{matrix}paq & ∗ \\ raq & r \end{matrix} ] ,

which is allowed since [ \begin{matrix}q & z_{1}(r − rar) \\ 0 & r \end{matrix} ] is a unit in A. At this stage, we have rar = r.

After replacing a by

[ \begin{matrix}p & −par \\ 0 & r \end{matrix} ] [ \begin{matrix}paq & par \\ raq & r \end{matrix} ] = [ \begin{matrix}∗ & 0 \\ raq & r \end{matrix} ]

(note that [ \begin{matrix}p & −par \\ 0 & r \end{matrix} ] is a unit in (p + r)A(p + r)), we may assume in addition that

par = 0.

Now return to the equation ax + b = p + r, and observe that

(paq)(qxp) + (pbp) = p(ax + b)p = p,

because pa = paq. Since sr(pAq) = 1, there exist y_{2} ∈ pAq and z_{2} ∈ qAp such that (paq + pbpy_{2})z_{2} = p. Consequently, p(a + by_{2})z_{2} = p. Note that (a + by_{2})r = ar, and so neither of the conditions rar = r and par = 0 is lost on replacing a by a + by_{2}. Thus, we may assume that there exists z_{2} ∈ qAp with paz_{2} = p, whence

[ \begin{matrix}paq & 0 \\ raq & r \end{matrix} ] [ \begin{matrix}z_{2} & 0 \\ −raz_{2} & r \end{matrix} ] = [ \begin{matrix}p & 0 \\ 0 & r \end{matrix} ] .

In other words, we have found an element z = [ \begin{matrix}z_{2} & 0 \\ −raz_{2} & r \end{matrix} ] in A(p+r) such that az = p+r,

and therefore we have proved that (p + r)A has stable rank 1. □

Theorem 7. If p is a full idempotent in A, then sr(A) ≤ sr(pAp).

Proof. Assume that sr(pAp) = n < ∞. By Lemma 1, pM_{n}(pAp) = pM_{n}(A)(n·p) has stable rank 1. Since p is full, 1_{A} ≲ t·p for some t ∈ ℕ. Working in a suitably large matrix ring R = M_{•}(A), we have sr(pR(n·p)) = 1 and we have room for an idempotent r which is equivalent to (t − 1)·p and orthogonal to both p and n·p. Proposition 6 now implies that sr((p + r)R(n·p + r)) = 1, that is, sr((t·p)R((n + t − 1)·p)) = 1. We also have t·p = e + f and (n + t − 1)·p = e + f + g for some orthogonal idempotents e, f, g with e ∼ 1_{A} and g ∼ (n − 1)·p. Since sr((e + f)R(e + f + g)) = 1, we can use Lemma 4 to see that sr(eR(e + g)) = 1. Finally, using Lemma 3 to increase e + g by an orthogonal idempotent equivalent to (n − 1)·(1_{A} − p), we conclude that sr(eR(e ⊕ (n − 1)·1_{A})) = 1. Since e ∼ 1_{A}, we thus have sr(1_{A}R(n·1_{A})) = 1, and so sr(1_{A}M_{n}(A)) = 1. Therefore sr(A) ≤ n, by Lemma 1. □

An upper bound for sr(pAp) in terms of sr(A) can be obtained from Theorem 7 and Vaserstein’s formula, as follows.

Theorem 8. If p is a full idempotent in A and ∑_{i=1}^{n} a_{i}pb_{i} = 1 for some a_{i}, b_{i} ∈ A, then sr(pAp) ≤ n· sr(A) − n + 1.

Proof. We may clearly assume that each a_{i} ∈ Ap and each b_{i} ∈ pA. Set

α = [ \begin{matrix}a_{1} & a_{2} & ⋯ & a_{n} \\ 0 & 0 & ⋯ & 0 \\ ⋮ & ⋮ & & ⋮ \\ 0 & 0 & ⋯ & 0 \end{matrix} ] ∈ 1_{A}M_{n}(A)(n·p)

β = [ \begin{matrix}b_{1} & 0 & ⋯ & 0 \\ b_{2} & 0 & ⋯ & 0 \\ ⋮ & ⋮ & & ⋮ \\ b_{n} & 0 & ⋯ & 0 \end{matrix} ] ∈ (n·p)M_{n}(A)1_{A},

and observe that αβ = 1_{A}. Then the matrix q := βα is an idempotent in the ring (n·p)M_{n}(A)(n·p), which we identify with M_{n}(pAp). Since (n·p)αqβ(n·p) = p, we see that q is full in M_{n}(pAp). Moreover, 1_{A} ∼ q in M_{n}(A), and so

A ≅ 1_{A}M_{n}(A)1_{A} ≅ qM_{n}(A)q = qM_{n}(pAp)q.

Hence, sr(M_{n}(pAp)) ≤ sr(A) by Theorem 7. According to Vaserstein’s formula [10, Theorem 3],

sr(M_{n}(pAp)) ≥ \frac{sr(pAp) − 1}{n} + 1,

and the theorem follows. □

To conclude, we derive the following estimates for the stable ranks of Morita equivalent rings.

Theorem 9. Let A and B be Morita equivalent rings; then B ≅ End_{A}(P) for some finitely generated projective generator P_{A}. If P can be generated by n elements as a right A-module, then

sr(A) ≤ n· sr(B) − n + 1.

If there are t homomorphisms f_{i} ∈ Hom_{A}(P, A) such that ∑_{i=1}^{t} f_{i}(P) = A, then

sr(B) ≤ t· sr(A) − t + 1.

Proof. There exists a split epimorphism A^{n} ↠ P, so that P ≅ pA^{n} for some idempotent p ∈ M_{n}(A), and B ≅ pM_{n}(A)p. Since P is a generator, p is full. Thus, by Vaserstein’s formula and Theorem 7,

\frac{sr(A) − 1}{n} + 1 ≤ sr(M_{n}(A)) ≤ sr(B).

We now identify B with End_{A}(P), and view P as a left B-module. Then _{B}P is a finitely generated projective generator, and End_{B}(P) ≅ A (e.g., [7, Propositions 18.17, 18.22]). There exist elements x_{i} ∈ P such that ∑_{i=1}^{t} f_{i}(x_{i}) = 1 in A. Given any element x ∈ P, there are endomorphisms xf_{i} ∈ B (sending any y ↦ xf_{i}(y)) such that ∑_{i=1}^{t}(xf_{i})x_{i} = x ∑_{i=1}^{t} f_{i}(x_{i}) = x. This shows that P is generated as a left B-module by x_{1}, …, x_{t}. Therefore the inequality sr(B) ≤ t· sr(A) − t + 1 follows from the first part of the theorem, on replacing A and B by B^{op} and A^{op}, respectively. □

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P. Ara: Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellaterra (Barcelona), Spain

 E-mail address: para@mat.uab.es

K. R. Goodearl: Department of Mathematics, University of California, Santa Barbara, CA 93106, USA

 E-mail address: goodearl@math.ucsb.edu