In this paper, several equivalent conditions on the Drazin invertibility of product and difference of idempotents are obtained in a ring. Some results in Banach algebra are extended to the ring case. Many authors extended the ordinary invertibility to the (generalized) Drazin invertibility. For example, Deng and Wei [8] proved that if p, q are idempotents in A = B(X), the ring of all bounded linear operators in a complex Banach space X. Then, (1) p − q ∈ A D if and only if p + q ∈ A D if and only if 1 − pq ∈ A D ; (2) pq − qp ∈ A D if and only if pq + qp ∈ A D if and only if pq ∈ A D and p − q ∈ A D . These results are generalizedto Banach algebra in [10]. More results on the Drazin invertibility of sum, difference and product of idempotents can be found in [3-8, 10, 13].In this paper, we consider the Drazin invertibility of p − q, pq, pq − qp (commutator) and pq + qp (anti-commutator), where p and q are idempotents in a ring. Some results in [3,4] for Banach algebra are extended to the ring case.
Key lemmasIn this section, we begin with some elementary and known results which will be useful in section 3.Lemma 2.5. Let a, b ∈ R D and p 2 = p ∈ R. If ap = pa and bp = pb, then ap + b(1 − p) ∈ R D andProof. Since p 2 = p, we have p D = p. Thus, a, b, p ∈ R D . Note that ap = pa and bp = pb. We obtain (ap) D = a D p and (b(1 − p)) D = b D (1 − p) by Lemma 2.2. As apb(1 − p) = b(1 − p)ap = 0, according to Lemma 2.1(2), it follows that (ap2 Lemma 2.6. Let a ∈ R, p 2 = p ∈ R, b = pa(1 − p) and c = (1 − p)ap. The following statements are equivalent:As p(bc + cb) = (bc + cb)p, we obtain that px = xp by Lemma 2.1(1).Next, we prove that (bc) D = pxp.Since x = (bc + cb)x 2 , we get pxp = p(bc + cb)x 2 p = bcx 2 p = bc(px 2 p) = bc(pxp) 2 .By (bc + cb)x = x(bc + cb), we obtain p(bc + cb)xp = px(bc + cb)p. It follows that bc(pxp) = (pxp)bc.Because (bc + cb) n+1 x = (bc + cb) n for some n, we have ((bc) n+1 + (cb) n+1 )x = (bc) n + (cb) n .Multiplying the equation above by p on two sides yields p(bc) n+1 xp = p(bc) n p, i.e., (bc) n+1 pxp = (bc) n . So, bc is Drazin invertible and (bc) D = pxp.(2) ⇒ (1) According to Lemma 2.3, bc ∈ R D is equivalent to cb ∈ R D . Note that bc · cb = cb · bc = 0. We have (b + c) 2 = (bc + cb) ∈ R D by Lemma 2.1(2). It follows that b + c ∈ R D .(2) ⇔ (3) Its proof is similar to (1) ⇔ (2).Lemma 2.7. Let a ∈ R with a − a 2 ∈ R D or a + a 2 ∈ R D . Then a ∈ R D .Proof. We only need to prove the situation when a − a 2 ∈ R D with x = (a − a 2 ) D .By Lemma 2.1(1), it is clear ax = xa since a(a − a 2 ) = (a − a 2 )a.Since a − a 2 ∈ R D , we get (a − a 2 ) n = (a − a 2 ) n+1 x for some integer n ≥ 1, that is, a n (1 − a) n = a n+1 (1 − a) n+1 x.