Abstract
Abstract. If a ring element α = 1 − ab is Drazin invertible, so is β = 1 − ba. We show that the Drazin inverse of β is expressible by a simple formula (in generalization of "Jacobson's Lemma") in terms of the Drazin inverse and the spectral idempotent of α. The commutation rules α a = a β and b α = β b are extended to the Drazin inverses, and the spectral idempotents of α and β are shown to be isomorphic. In a related direction, we also prove that Jacobson's Lemma holds for π-regular elements and unit π-regular elements in arbitrary rings.
Introduction
For any ring R (with identity), "Jacobson's Lemma" states that, if α = 1 − ab is a unit, so is β = 1 − ba; indeed, (1.1)
Introduction
Halmos observed in [11] that this can be "formally" checked by writing and pointed out that this nice trick is "usually ascribed to Jacobson." Kaplansky, widely known for his gift with words, once remarked to one of us that, using this trick, you can always reproduce formula (1.1) without fail "even if you are thrown up on a desert island, with all your books and papers lost" [16, p. 4] .
Introduction
Resonating with Kaplansky's witty comment, some of our students and close associates have sometimes fondly referred to (1.1) as the "Desert Island Formula". Over the years, it has been realized that Jacobson's Lemma has suitable analogues for various kinds of "generalized inverses". An element r ∈ R is said to be (von Neumann) regular if r = rxr for some x ∈ R (called an inner inverse of r), and r is said to be unit-regular if it has a unit inner inverse. It is known that if α = 1 − ab is regular (resp. unit-regular), then so is β = 1 − ba, and the formula (1.1) continues to hold if "inverse" is replaced by "an inner inverse". See, for example, [5] .
Introduction
In [9] , Drazin introduced a new notion of invertibility, which is now commonly known as Drazin invertibility. An element r ∈ R is called Drazin invertible if there exists s ∈ R such that rs = sr, s = srs, and r n = r n+1 s for some n ≥ 0. If s exists, it is unique. It is called the Drazin inverse of r, and is denoted by r (following Drazin). The smallest integer n for which the equation r n = r n+1 r holds is called the Drazin index of r, and is denoted by i (r). In case i (r) ≤ 1, r is said to be group invertible (or strongly regular), and its Drazin inverse is called the group inverse of r. According to [9, Theorem 4] , r ∈ R is Drazin invertible iff r is strongly π-regular in the sense that r n ∈ r n+1 R∩R r n+1 for some integer n ≥ 1. In this case, we say that r has associated idempotent r r and spectral idempotent 1 − r r.
Introduction
In 2009, Patrício and Veloso da Costa [23] asked if the Drazin invertibility of α = 1 − ab implies that of β = 1 − ba. In [22] and [7] , Patrício-Hartwig and Cvetković-Ilić-Harte answered this question affirmatively, showing in fact that α and β always have the same Drazin index. In their papers, however, no formula was given that expresses β in terms of α . Subsequently, such a formula was found by Castro-González, Mendes-Araújo, and Patrício; see Theorem 3.6 in their recent paper [4] . In [4] , the Drazin inverse formula for β was derived from a preliminary group inverse formula, so the proof was long and indirect. (See also [24] which deals with a more general notion of Drazin inverses.) In Theorem 2.1 of the present note, we offer a formally simpler formula (2.2) that computes β directlywithout going through (but of course subsuming) the group inverse case. While (2.2) does completely recapture the formula in Theorem 3.6 in [4] , its proof is short, self-contained, and does not assume knowledge of any parts of [22] , [7] , [4] , or [24] . It also contains more information on the associated idempotent of β as well as its "Azumaya realization" as a Drazin invertible element.
Introduction
In §3, we use our techniques to derive commutation rules for the Drazin inverses α , β , and the spectral idempotents e, f of α and β, proving in particular the hitherto unknown fact that e and f are isomorphic idempotents. We then place our computations in the general context of "Jacobson pairs" (pairs of the form (1 − xy, 1 − yx) over rings), and show that several pairs naturally associated with (α, β) are also Jacobson pairs. In Theorem 3.8, we further demonstrate the universal nature of Jacobson's Lemma by proving that it holds for π-regular elements and unit π-regular elements in arbitrary rings. While we show in §4 that Jacobson's Lemma does not hold for "clean elements" and "suitable elements" in the sense of Nicholson [18] , this work paves the way to a further study of Jacobson's Lemma for other significant classes of ring elements, which we hope to present in the forthcoming work [17] .
Introduction
Throughout this paper, R denotes an arbitrary ring with 1 ∈ R, and U(R) denotes the group of units in R. Other standard notations in ring theory follow those in [15] .
Formula for Drazin Inverse
We begin this section by proving the theorem below, where new compact expressions are given for the Drazin inverse and associated idempotent of 1 − ba in terms of the Drazin inverse and spectral idempotent of 1 − ab. Theorem 2.1. Suppose α := 1 − ab ∈ R is Drazin invertible with index k, Drazin inverse α , and spectral idempotent e = 1 − α α. Then β := 1 − ba is also Drazin invertible with index k, and has Drazin inverse
Formula for Drazin Inverse
The associated idempotent for β is (1 − b ea) n for any n ≥ k.
Formula for Drazin Inverse
This means that, to get the new formula from (1.1), all we need to do is to interpret "inverse" as the Drazin inverse, and simply replace the "1" on the right-hand-side of (1.1) by the associated idempotent of β. In this way, (2.2) retains the simplicity and memorability of a "Desert Island Formula".
Formula for Drazin Inverse
To prove Theorem 2.1, let γ := 1 + b α a (an expression inspired by the formula (1.1)) and ε := b ea, recalling that e = 1 − α α = 1 − α α . Since β b = b α, iteration shows that β i b = b α i for all i ≥ 0. Similarly, we have a β i = α i a for all i ≥ 0. These two important commutation rules will be used freely (without further mention) below.
Formula for Drazin Inverse
Proof. Keeping in mind that α k = α k+1 α , and e α = (1 − α α) α = 0, we have
Formula for Drazin Inverse
(We'll show later that actually γ ∈ U(R) ; see the proof of Theorem 3.10(D).)
Formula for Drazin Inverse
Proof of Theorem 2.1. The point of (C) and (D) above is that they give an explicit "Azumaya realization" of β as a strongly π-regular element, in the sense of [2, Theorem 3] . With this, the rest of the proof tharper@example.com. First, (C) and (D) imply that β is Drazin invertible with i (β) ≤ k = i (α), and hence equality must hold on account of symmetry. Next, the Drazin inverse of β can be computed via Drazin's Theorem 4 in [9] ; namely, for any n ≥ k, β = β n γ n+1 . Therefore, by (C) in Lemma 2.3,
Formula for Drazin Inverse
where, toward the end, we have used the fact that ε (γ − 1) = 0 (by (B) in Lemma 2.3). Since the computation above is valid for any n ≥ k, it follows that the powers (1 − ε) n are all equal for n ≥ k. Thus, by (C) again, the associated idempotent for β is
Formula for Drazin Inverse
The next theorem gives two explicit expressions for the spectral idempotent f of β (in case α is Drazin invertible of index k). The first expression for f shows that our formula (2.2) above completely recaptures [4, Theorem 3.6] (which also appeared in [24, Corollary 2.4] . The second expression for f will provide the groundwork for some further generalizations of Theorem 2.1 in [17] .
Formula for Drazin Inverse
Theorem 2.4. Keep the notations and assumptions in Theorem 2.1, and let u := α − e. Then u ∈ U(R), and the spectral idempotent f for β is given by b era = −b eu −1 a, where
Formula for Drazin Inverse
Proof. For δ := 1 − ε ∈ R, Theorem 2.1 shows that
Formula for Drazin Inverse
Since e commutes with ab, we have a δ = a − ab ea = (1 − eab) a. Therefore,
Formula for Drazin Inverse
From (2.5) and (2.6), it follows that f = b era where
Formula for Drazin Inverse
, r is to be interpreted as 0.] In particular, R f R ⊆ R eR, so equality holds by symmetry.
Formula for Drazin Inverse
For the rest of the theorem, recall that u := α − e ∈ U(R) is a standard consequence of Fitting's Decomposition Theorem. To prove the second formula f = −b eu −1 a, it suffices to check that er = −eu −1 . Since ur = ru, this amounts to eur = −e. Now
Formula for Drazin Inverse
As (e α) k = 0 (again from Fitting's Decomposition Theorem), we get eur = −e.
Formula for Drazin Inverse
For k = i (α) ≤ 1, we have the following special case of Theorems 2.1 and 2.4.
Formula for Drazin Inverse
Corollary 2.7. If α = 1 − ab ∈ R is group invertible with spectral idempotent e, then β = 1 − b a is also group invertible, with group inverse β = 1 + b (α − e) a and spectral idempotent f = b ea. (By symmetry, we also have e = af b.)
Formula for Drazin Inverse
In the case where α is nilpotent, Theorem 2.1 can be refined a bit as follows.
Formula for Drazin Inverse
is Drazin invertible with index k, and its associated idempotent is
Formula for Drazin Inverse
If R is a Dedekind-finite ring, then β is similar to α. In this case, β is also nilpotent of index k.
Formula for Drazin Inverse
Proof. (A) follows from Theorem 2.1 since in the notations there, α = 0 and e = 1. Indeed, the equation β k = β k+1 can be directly seen as follows. From
Formula for Drazin Inverse
. Finally, for (C), assume that R is Dedekind-finite. Since α is nilpotent, we have ab = 1 − α ∈ U(R), and hence b ∈ U(R) too. Then β = b α b −1 is similar to α. In particular, β is also nilpotent of index k.
Formula for Drazin Inverse
Remark 2.9. In the case k = 1, we have α = 0, so ab = 1. Then (b a) 2 = b a, and (1 − b a) 2 = 1 − b a too (consistently with all conclusions of (A) in Corollary 2.8). In general, however, b a may not be 1, so β = 1 − b a may not be 0. This shows that the assumption on b cannot be dropped in (B) of Corollary 2.8.
Commutation Rules, and Jacobson Pairs
In this section, we present another main result in this paper relating the Drazin inverses and the spectral idempotents of α and β. These relations are "formal extensions" of the basic commutation rules β b = b α and α a = a β. However, they have not been noticed in the literature before, and their proofs do require some work. 
Commutation Rules, and Jacobson Pairs
is a standard fact on any Drazin invertible element α with index k. It can be proved formally by expanding (α (e + α )) k = (α e + (1 − e)) k into α k e + (1 − e). More conceptually, it can be seen easily by thinking of α as a Fitting endomorphism on a right module M (over some ring). If M = I ⊕ K is the Fitting decomposition of α on M (so that α | I is an automorphism and α | K is nilpotent of index k), then α k (e + α ) k+1 is zero on K and is the inverse of α on I, so it agrees precisely with α on M .
Commutation Rules, and Jacobson Pairs
Using (A), (B), and the earlier equation β = β k γ k+1 , we have now
Commutation Rules, and Jacobson Pairs
Finally, to deal with the "double" Drazin inverses, recall from [9] that β = β 2 β . From this, we have
Commutation Rules, and Jacobson Pairs
Note that there is no need to discuss further "higher-order" Drazin inverses, since α = α and a similar equation holds for β, as long as α is Drazin invertible.
Commutation Rules, and Jacobson Pairs
The referee pointed out that the theorem above can be seen as a consequence of a very general commutation rule proved by Drazin in his recent paper [10] . Indeed, for any x, y ∈ R, let us define the "intertwining set" of the ordered pair (x, y) to be (3.4) I (x, y) := { r ∈ R : x r = r y}.
Commutation Rules, and Jacobson Pairs
For any pair of Drazin invertible elements x, y ∈ R, Drazin's commutation rule in [10, Theorem 2.2] states that I (x, y) ⊆ I (x , y ). In the notations of Theorem 3.1, since we know that a ∈ I (α, β) and b ∈ I (β, α), Drazin's result gives immediately that a ∈ I (α , β ), b ∈ I (β , α ), and therefore the same holds for the double Drazin inverses (and spectral idempotents). While this gives a quicker and more conceptual proof of Theorem 3.1 (assuming Drazin's new result), we have chosen to preserve our original proof of the theorem above as it gives a good illustration of the Fitting decomposition technique, and it will also be applicable to the setting of the remark below.
Commutation Rules, and Jacobson Pairs
Remark 3.5. In the recent paper [24] , Jacobson's Lemma was studied for a generalized notion of Drazin invertibility introduced by Koliha and Patrício. In the setting of [14] (see also [13] ), the role of nilpotent elements in one of the standard characterizations of Drazin inverses is replaced by that of quasi-nilpotent elements. (An element q ∈ R is quasi-nilpotent if 1 − qx ∈ U(R) for every x ∈ R such that qx = xq.) If we interpret α and β as the generalized Drazin inverses of α = 1 − ab and β = 1 − ba, Theorem 2.3 and Corollary 2.4 in [24] give explicit formulas expressing β and the spectral idempotent f of β in terms of α and the spectral idempotent e of α. Assuming these two results, it is relatively routine to check that the formulas β b = b α and f b = b e proved in Theorem 3.1 do remain valid in the setting of generalized Drazin inverses introduced in [14] . The details of these verifications are left to the reader.
Commutation Rules, and Jacobson Pairs
To facilitate the further study and generalization of Jacobson's Lemma, it is now convenient to introduce the following terminology. Definition 3.6. We say that (α, β) ∈ R 2 is a Jacobson pair if there exist a, b ∈ R such that α = 1−ab and β = 1−ba. In this case, of course (β, α) is also a Jacobson pair. The diagonal elements (r, r) ∈ R 2 are always Jacobson pairs. In the case of a commutative ring R, these are the only Jacobson pairs.
Commutation Rules, and Jacobson Pairs
Example 3.7. Two idempotents e, f ∈ R are said to be isomorphic (written e ∼ = f ) if eR ∼ = f R as right R-modules (or equivalently, R e ∼ = R f as left R-modules). A criterion for e ∼ = f is that there exist x, y ∈ R such that e = xy and f = yx : see [15, (21.20) ]. From this, it follows that e ∼ = f iff (1 − e, 1 − f ) is a Jacobson pair.
Commutation Rules, and Jacobson Pairs
Recall that an element x ∈ R is said to be unit-regular if x = xux for some u ∈ U(R). Also, y ∈ R is said to be π-regular (resp. unit π-regular) if y n is regular (resp. unit-regular) for some integer n ≥ 1.
Commutation Rules, and Jacobson Pairs
Theorem 3.8. If (α, β) ∈ R 2 is a Jacobson pair, so is (α n , β n ) for any integer n ≥ 1. If α is π-regular (resp. unit π-regular), then so is β.
Commutation Rules, and Jacobson Pairs
Proof. Say α = 1−ab and β = 1−ba. Then by [4, Lemma 2.3], we have (for any n ≥ 1) α n = 1−(ra) b and β n = 1 − b (ra), where r := 1 + α + · · · + α n−1 ∈ R. Thus, (α n , β n ) is a Jacobson pair. If α is π-regular, then α n is regular for some n ≥ 1. By what we have said at the beginning of the paper, this implies that β n is also regular, so β is π-regular. If α is unit π-regular, the same argument works, by using Chen's result [5, Lemma 2.1] that the unit-regularity of α n implies that of β n (as long as (α n , β n ) is a Jacobson pair).
Commutation Rules, and Jacobson Pairs
Corollary 3.9. Let α ∈ U(R) and β ∈ R. If (α, β) is a Jacobson pair, so is (α n , β n ) for any n ∈ Z.
Commutation Rules, and Jacobson Pairs
Proof. By the "original" Jacobson's Lemma, β ∈ U(R) too, so the notation (α n , β n ) makes sense for any n ∈ Z. In view of Theorem 3.8, it suffices to prove that (α −1 , β −1 ) is a Jacobson pair. Again, write α = 1 − ab, and β = 1 − ba. By Jacobson's Lemma,
Commutation Rules, and Jacobson Pairs
is indeed a Jacobson pair.
Commutation Rules, and Jacobson Pairs
Theorem 3.10. Let (α, β) ∈ R 2 be a Jacobson pair such that α (and hence also β ) is Drazin invertible. Let e and f be the spectral idempotents of α and β. Then (A) e and f are also the spectral idempotents of α and β , and we have e ∼ = f . 
Commutation Rules, and Jacobson Pairs
Similarly, β has spectral idempotent f . To see that e ∼ = f , write α = 1 − ab and β = 1 − ba. By Theorem 2.4, we have f = b era and (by symmetry) e = af sb, where r = 1 + α + · · · + α k−1 and
Commutation Rules, and Jacobson Pairs
. From α a = a β, we have ra = as, so the relation ea = af in Theorem 3.1 gives
Commutation Rules, and Jacobson Pairs
(B) By Example 3.7, e ∼ = f implies that (1 − e, 1 − f ) is a Jacobson pair. To show the synergy of ideas, here is another proof. From Theorem 2.1, we know that 1 − e = (1 − b ea) k , and similarly,
Commutation Rules, and Jacobson Pairs
(This, in turn, implies that e ∼ = f .) (C) By Theorems 2.1 and 2.4, we have (3.11) β = 1 − b (er − α ) a, and by symmetry,
Commutation Rules, and Jacobson Pairs
Using again the relations ra = as, α a = a β , and ea = a f , we may define (3.12)
Commutation Rules, and Jacobson Pairs
Then, by (3.11), (α , β ) = (1 − b a 1 , 1 − a 1 b) is a Jacobson pair (and it is well known that α , β are group invertible, as long as they exist).
Commutation Rules, and Jacobson Pairs
(D) To prove (D), we go back to the element γ = 1 + b α a used in the proofs of both Lemma 2.3 and Theorem 3.1. Using the rule α a = a β in Theorem 3.1, we can rewrite
Commutation Rules, and Jacobson Pairs
(As the reader might have noticed, this new expression for γ was partly inspired by the earlier relation γ b = b (e + α ) in the proof of Theorem 3.1.) Now γ = 1 + b (α a) forms a Jacobson pair with
Commutation Rules, and Jacobson Pairs
Therefore, (e + α , f + β ) is a Jacobson pair. A straightforward calculation shows that e + α and f + β are units in R, with inverses given respectively by (3.14) e + (1 − e) α = e + α and f + (1 − f ) β = f + β .
Commutation Rules, and Jacobson Pairs
Applying Corollary 3.9, we can then get other Jacobson pairs by forming the powers (e + α ) n , (f + β ) n (n ∈ Z ). For n > 0, we get e + (α ) n , f + (β ) n (since eα = α e = 0), and for n = −m < 0, we get e + (α ) m , f + (β ) m , so all of these powers are Jacobson pairs, with unit entries. (Note that the latter case can also be deduced from the former case by using part (A) above.) Remark 3.15. Of course, direct applications of Theorem 3.1 also lead to other Jacobson pairs arising from (α, β). Some examples are, say, (1 − a α, 1 − a β), (1 − a α , 1 − a β ) , and (1 − ae, 1 − af ).
Commutation Rules, and Jacobson Pairs
However, these Jacobson pairs are somewhat less interesting, since their entries may no longer be Drazin invertible (when α itself is).
Commutation Rules, and Jacobson Pairs
Remark 3.16. In general, in part (B) of Theorem 3.10, the associated idempotents 1 − e and 1 − f of α and β need not be isomorphic. For instance, if a, b ∈ R are such that ab = 1 = ba, then α = 1 − ab = 0 has associated idempotent 0 (and spectral idempotent e = 1). If β = 1 − ba has an isomorphic associated idempotent, then the latter is 0 too, and hence f = 1. But Theorem 2.4 implies that f = b ea = b a = 1, a contradiction. If R happens to be an IC ring (a ring with internal cancellation in the sense of [12] ), then of course Theorem 3.10(A) would imply that the associated idempotents of α and β are isomorphic (and hence similar). This is the case, for instance, when R is a unit-regular ring. More deeply, it is also the case when R is a strongly π-regular ring (that is, when all elements in R are Drazin invertible), since Ara has proved in [1] that such a ring R has stable range one, which (in view of the work in [12] ) is equivalent to R being an IC ring.
Clean Elements and Suitable Elements
The idea behind Jacobson's Lemma suggests that, if C R is a class of elements in a ring R (defined, say for all rings, by some specific ring-theoretic property), it is of interest to investigate whether α ∈ C R ⇒ β ∈ C R for every Jacobson pair (α, β) ∈ R 2 . If this is the case, let us say that C R satisfies Jacobson's Lemma (or Jacobson's Lemma holds for C R ). We have seen in the previous sections that this is indeed the case if C R is, say, the set of units, regular or unit-regular elements, π-regular or strongly π-regular elements, or Drazin invertible elements of a fixed Drazin index n in R. Another case in which Jacobson's Lemma is satisfied (by all rings) is where we define (4.1)
Clean Elements and Suitable Elements
Here, if 1 − ab ∈ C R , then (ab) n = (ab) n+1 for some n ≥ 1, so left-multiplying by b and rightmultiplying by a gives (ba) n+1 = (ba) n+2 . This shows that 1 − ba ∈ C R . On the other hand, there are families {C R } such that Jacobson's Lemma is satisfied only for some specific classes of rings R. In such cases, the Lemma would "define" these classes of rings. The two simplest examples are as follows. If C R := {0} for all R, then C R satisfies Jacobson's Lemma iff R is Dedekind-finite, and if C R := {1} for all R, then C R satisfies Jacobson's Lemma iff R is reversible (ab = 0 ⇒ ba = 0).
Clean Elements and Suitable Elements
Note that there do exist some families {C R } which are in spirit close to those named in the first paragraph of this section, but for which Jacobson's Lemma fails to hold. Each of the four families C R below has the "complement property" that α ∈ C R ⇒ 1 − α ∈ C R , so to say that Jacobson's Lemma fails for C R (for a given ring R ) is tantamount to saying that there exist a, b ∈ R such that ab ∈ C R but ba / ∈ C R .
Clean Elements and Suitable Elements
The first example is where C R is the set of all idempotents in R. [21] . Let suit (R) be the set of all suitable elements in R ; that is, elements α ∈ R for which there exists an idempotent e ∈ R α such that 1 − e ∈ R (1 − α). (By [19] , "left suitable" is equivalent to "right suitable", so we can simply refer to such α as being suitable.) Next, let cn (R) (resp. ucn (R)) be the set of clean (resp. uniquely clean) elements of R ; that is, elements α ∈ R that are expressible (resp. expressible uniquely) in the form e + u where e is an idempotent and u is a unit in R.
Clean Elements and Suitable Elements
1 An equation α = e + u of this form is called a clean decomposition of α. It is easy to check that all three sets introduced above have the complement property. Somewhat surprisingly, we have the following "negative" result. in R, we have AB = 0, which is easily seen to be in ucn (R). However, the following distinct clean decompositions show that BA / ∈ ucn (R):
Clean Elements and Suitable Elements
For the other two sets, recall Nicholson's important observation (in [18] ) that cn (R) ⊆ suit (R) (for any ring R ). To show that cn (R) and suit (R) do not satisfy Jacobson's Lemma for our test ring R = M 2 (Z), it suffices to name two matrices C, D ∈ R such that CD ∈ cn (R), but DC / ∈ suit (R).
Clean Elements and Suitable Elements
Let C = in R, where n ∈ Z \ {−1, 0, 1, 2}. Then CD ∈ cn (R) since
Clean Elements and Suitable Elements
where the RHS is a clean decomposition (that is, the first matrix is idempotent, and the second is invertible). We finish by showing that P := DC = n 0 0 0 / ∈ suit (R). Indeed, if otherwise, there would exist an idempotent matrix E ∈ R · P such that I 2 − E ∈ R · (I 2 − P ). Then E has the form nx 0 ny 0
Clean Elements and Suitable Elements
, and E = E 2 implies that n 2 x 2 = nx and n 2 xy = ny, so x = y = 0; that is, E = 0. But then (4.5)
Clean Elements and Suitable Elements
(Note that, if n ∈ {−1, 0, 1, 2} instead, the matrix P = DC would have been in cn (R).)
Clean Elements and Suitable Elements
What about the implication " ab ∈ C R ⇒ ba ∈ C R " for π-regular and strongly π-regular elements? If we take C R to be the subset R π of π-regular elements in R, we see easily that the complement property does not always hold. By Theorem 3.8, Jacobson's Lemma does hold for R π . However, this no longer guarantees that AB ∈ R π ⇒ BA ∈ R π . For instance, taking the idempotents A = is not π-regular, since C n = C n (x ij ) C n ∈ R for some n ≥ 1 would have led to an impossible equation 2 n = 2 2n x 11 in Z. As for the set of Drazin invertible elements in R (which is often denoted by R D in the literature), the classical "Cline's Formula" in [6] (see also [3] ) does guarantee that ab ∈ R D ⇒ ba ∈ R D for all a, b in any ring R. However, this in itself does not imply Jacobson's Lemma for R D (or vice versa), since the set R D does not have the complement property in general. [For instance, in the ring Z, r = −1 is (invertible and hence) Drazin invertible, but 1 − r = 2 is not.] Indeed, the proof of Jacobson's Lemma for R D for a general ring (as given in Theorem 2.1) is considerably harder than the classical proof for Cline's Formula in [6] or in [3] .
Clean Elements and Suitable Elements
While Proposition 4.2 might have been a bit of a disappointment, it turns out that there is a good "reason" behind the failure of Jacobson's Lemma for cn (R) and ucn (R); namely, the lack of a "commuting property" in the clean decompositions α = e + u used in the definitions of these sets. In a forthcoming work [17] , we will show that, as soon as a commuting property is imposed on the intervening decompositions, a strong form of Jacobson's Lemma will hold for several classes of ring elements of interest to current researchers in ring theory, including the strongly clean elements of Nicholson [20] , the strongly nil-clean elements of Diesl [8] , and the quasipolar elements of Zhuang, Chen, and Cui [24] , etc.

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