Clean Matrices and Unit-Regular Matrices

Dinesh Khurana and T. Y. Lam

Abstract

An element in a ring R is said to be clean (resp. unit-regular) if it is the sum (resp. product) of an idempotent element and an invertible element. If all elements in R are unit-regular, it is known that all elements in R are clean. In this note, we show that a single unit-regular element in a ring need not be clean. More generally, a criterion is given for a matrix (\begin{matrix}a & b \\ 0 & 0 \end{matrix}) to be clean in a matrix ring 𝕄_{2}(K) over any commutative ring K. For K = ℤ, this criterion shows, for instance, that the unit-regular matrix (\begin{matrix}12 & 5 \\ 0 & 0 \end{matrix}) is not clean. Also, this turns out to be the “smallest” such example.

§1. Introduction

The notion of a unit-regular element was introduced by G. Ehrlich. According to [Eh], an element x in a ring R is unit-regular if x = xux for some u ∈ U(R) (the group of units of R). It is easy to see that x is unit-regular iff x is an idempotent times a unit, iff x is a unit times an idempotent [La_{2}: Ex. (4.14B)]. As their name suggests, unit-regular elements are regular (in the sense of J. von Neumann). Ehrlich called a ring unit-regular if all of its elements are unit-regular. Rings of this kind have been extensively studied in the literature on von Neumann regular rings; see, e.g. [Go: Sec. 4].

In parallel, clean elements in rings were introduced by W. K. Nicholson. In [Ni_{1}], a ring element x ∈ R is said to be clean if x is the sum of an idempotent and a unit in R. If all elements in R are clean, Nicholson called R a clean ring. Such rings are of interest since they constitute a subclass of the so-called exchange rings in the theory of noncommutative rings.

The relationship between cleanness and unit-regularity seems to be rather subtle. In [CY], or more correctly, in [CK], Camillo, Yu, and Khurana showed that any unit-regular ring is clean. This answered a question of Nicholson, but it does not say whether a single unit-regular element in a ring R need to be clean. In general, if an element x ∈ R has the form eu where e is an idempotent and u is a unit commuting with e, then, writing f = 1 − e, we see that x = f + (eu − f) is clean since f is an idempotent, and eu − f is a unit (with inverse eu^{−1} − f). This shows that, in any ring in which idempotents are central (e.g. a reduced ring, a local ring, or a commutative ring), any unit-regular element is indeed clean.^{1} More generally, in [Ni_{2}: Theorem 1], Nicholson has shown that, if x ∈ R is such that some power x^{n} (n ≥ 1) has a factorization eu = ue with e = e^{2} and u ∈ U(R), then x is clean. This theorem implies that any strongly π-regular ring^{2} is clean; in particular, any right artinian ring (e.g. finite ring) is clean.

The initial goal of this work was to show that, in a noncommutative ring, unit-regular elements need not be clean. A natural place to look for examples is the family of various kinds of matrix rings over a commutative ring K. Our first attempt working with the ring 𝕋_{n}(K) of upper triangular matrices over K did not bring about the desired examples, as it turned out, curiously enough, that unit-regular elements are always clean in 𝕋_{n}(K). Then, moving on to the case of full matrix rings 𝕄_{n}(K), we found our first

examples from matrices of the form A = (\begin{matrix}a & b \\ 0 & 0 \end{matrix}) (over suitable commutative rings K).

Here, we solve the problem at hand by proving a general criterion (in Theorem 3.2) for the above matrix A to be clean in the ring 𝕄_{2}(K). As a consequence of this criterion,

we show in (3.12) that (\begin{matrix}1 + xy & x^{2} \\ 0 & 0 \end{matrix}) (a derivative of the Cohn matrix in [Co]) is unit-

regular but not clean over K = k[x, y] for any commutative domain k. Specializing the cleanness criterion to the case K = ℤ, we also obtain in §4 an algorithmically very

simple method (Theorem 4.7) for deciding the cleanness of matrices of the form (\begin{matrix}a & b \\ 0 & 0 \end{matrix})

over ℤ. In particular, we see that the choices (a, b) = (12, 5), (13, 5), (12, 7), etc. lead to 2 × 2 unit-regular matrices over ℤ that are not clean. As it turns out, these are the “smallest” such examples.

As a by-product of this work, we obtain the most general form of a 2 × 2 idempotent matrix over a commutative ring with a prescribed (idempotent) determinant that has a unimodular second row. This result constitutes Theorem 5.1 below.

§2. Preliminary Results

In this section, we assemble a few results that will be needed in the subsequent sections. Throughout this paper, K denotes a commutative ring, and R denotes the 2 × 2 matrix ring 𝕄_{2}(K) over K.

To begin with, we determine all unit-regular matrices in R with a zero second row. This turned out to be fairly easy.

Proposition 2.1. A matrix A = (\begin{matrix}a & b \\ 0 & 0 \end{matrix}) is unit-regular in R = 𝕄_{2}(K) iff there exists

an idempotent e ∈ K and a unimodular row (a′, b′) ∈ K^{2} such that (a, b) = e (a′, b′). In particular, if K is connected (that is, K has only trivial idempotents), then the only nonzero unit-regular matrices in R with a zero second row are (\begin{matrix}a & b \\ 0 & 0 \end{matrix}) with (a, b)

1On the other hand, the example 2 ∈ ℤ shows that a clean element in a commutative domain need not be unit-regular.

2A ring R is strongly π-regular if, for any x ∈ R, x^{n} ∈ x^{n+1}R for some n.

unimodular.

Proof. If (a, b) has the form e (a′, b′) as above, then a′y − b′x = 1 for some x, y ∈ K, and the equation

(\begin{matrix}a & b \\ 0 & 0 \end{matrix}) = (\begin{matrix}e & 0 \\ 0 & 0 \end{matrix}) (\begin{matrix}a′ & b′ \\ x & y \end{matrix})

shows that A is unit-regular. Conversely, assume that A is unit-regular; say A = EU,

where E = E^{2} and U ∈ GL_{2}(K). Let E = (\begin{matrix}e & r \\ s & t \end{matrix}), and U = (\begin{matrix}w & x \\ y & z \end{matrix}). Then

(s, t) = (0, 0) · U^{−1} = (0, 0), and so E = E^{2} = (\begin{matrix}e^{2} & er \\ 0 & 0 \end{matrix}) shows that e = e^{2}, and r = er.

Therefore, (a, b) = e (a′, b′), where (a′, b′) = (w, x) + r (y, z) is a unimodular row since (w, x) and (y, z) are the rows of a matrix in GL_{2}(K).

The problem of deciding the cleanness of A = (\begin{matrix}a & b \\ 0 & 0 \end{matrix}) is considerably harder. Nat-

urally, the clean elements of K should play a key role in this problem. Here, we first dispose of the easy case where a itself is a clean element in K.

Proposition 2.2. If a ∈ K is clean, then for any b ∈ K, A = (\begin{matrix}a & b \\ 0 & 0 \end{matrix}) is clean in R.

Proof. If a = e + u where e = e^{2} and u ∈ U(R), then A = (\begin{matrix}e & 0 \\ 0 & 1 \end{matrix}) + (\begin{matrix}u & b \\ 0 & −1 \end{matrix}) shows

that A is clean.^{3}

Corollary 2.3. let K be connected. If a ∈ K \ {0} is clean but not a unit in K, then

(\begin{matrix}a & a \\ 0 & 0 \end{matrix}) is clean but not unit-regular in R = 𝕄_{2}(K).

As was pointed out to us by Professor P. Ara, this Corollary is not true if K is not assumed to be connected. A counterexample is given by a (commutative) von Neumann regular ring K that is not a field. In this case, it is known that K and 𝕄_{2}(K) are both unit-regular rings, so the last conclusion of (2.3) cannot possibly hold.

We shall now introduce a matrix that will be crucial for the work in the rest of the paper. For any three elements e, x, k ∈ K such that e = e^{2} and ex = 0, we define the matrix

(2.4)  E(e, x, k) := (\begin{matrix}e − kx & ke − k(kx + 1) \\ x & kx + 1 \end{matrix}) ∈ R.

The basic properties of E(e, x, k) are summarized as follows.

3A somewhat more sophisticated version of this result will be given in (3.4) below.

Proposition 2.5. E := E(e, x, k) is an idempotent matrix over K with tr(E) = e + 1 and det(E) = e.

Proof. The trace equation is clear, and by direct computation:

det(E) = e(kx + 1) − kx(kx + 1) − xke + xk(kx + 1) = e.

Furthermore, using the equation ex = 0, we get e · E = e · I_{2}. Thus, by the Cayley-Hamilton Theorem,

E^{2} = tr(E) · E − det(E) · I_{2} = (e + 1)E − eE = E,

as desired.

For more information about E(e, x, k), we might point out that, if we let

(2.6)  E_{1} := (\begin{matrix}e & ke \\ 0 & 0 \end{matrix}) , and E_{2} := (\begin{matrix}−kx & −k(kx + 1) \\ x & kx + 1 \end{matrix}) ,

then E = E_{1} + E_{2}, and (as we can easily check) E_{1}, E_{2} are orthogonal idempotents in R. This shows that, if we think of E(e, x, k) as an idempotent linear operator acting on the right of K^{2}, then its cokernel is isomorphic to the projective module K/eK. A strong uniqueness property for E(e, x, k) will be proved in Theorem 5.1 below.

The following proposition will only be needed for 2 × 2 matrices, but we shall prove it for the n × n case since this does not call for any additional work.

Proposition 2.7. Let E = (a_{ij}) be any n × n idempotent matrix over K with determinant e. Then e^{2} = e, and ea_{ij} = δ_{ij}e (where {δ_{ij}} are the Kronecker deltas). If the last row of E happens to be unimodular, then a_{nn} ≡ 1 (mod a_{n1}K + ⋯ + a_{n,n−1}K).

Proof. First, we have e = det(E) = det(E^{2}) = (det(E))^{2} = e^{2}. Let f = 1 − e be the complementary idempotent of e. Over the factor ring K/fK, \overline{E} has determinant \overline{1}, and is thus invertible. But then \overline{E} = \overline{E}^{2} implies that \overline{E} is the identity matrix. This means that a_{ii} ∈ 1 + fK for all i, and a_{ij} ∈ fK for i ≠ j. Multiplying these by e, we see that ea_{ii} = e for all i, and ea_{ij} = 0 for i ≠ j. If, in addition, the last row of E is unimodular, then, over the factor ring K/(a_{n1}K + ⋯ + a_{n,n−1}K), a_{nn} becomes a unit, and E becomes an (idempotent) block-upper triangular matrix. The latter implies that the image of a_{nn} in K/(a_{n1}K + ⋯ + a_{n,n−1}K) is an idempotent, and thus we must have a_{nn} ≡ 1 (mod a_{n1}K + ⋯ + a_{n,n−1}K).

§3. Criterion for Some Clean Matrices

Throughout this section, A denotes the matrix (\begin{matrix}a & b \\ 0 & 0 \end{matrix}). In order to solve the prob-

lem of deciding the cleanness of A, it will be convenient to introduce a small technical variation of the notion of cleanness.

Definition 3.1. Let e be a given idempotent in K. If a matrix M ∈ 𝕄_{n}(K) can be written in the form E + U, for some E = E^{2} of determinant e and some U ∈ GL_{n}(K), we shall say that M is e-clean. In particular, for n = 1, the e-clean elements of K = 𝕄_{1}(K) are just those of the form e + u where u ∈ U(K).

Since (by (2.7)) an idempotent matrix has an idempotent determinant, a matrix M is clean iff it is e-clean for some idempotent e ∈ K. Although this idempotent is not uniquely determined by M in general, the definition above helps us to “catalogue” the clean matrices in a fashion. In view of these remarks, we could then transform the problem of deciding the cleanness of matrices into the problem of deciding their e-cleanness for any given idempotent e ∈ K. As it turns out, it is indeed in this form that the cleanness problem admits a reasonable solution, at least for matrices of the

form (\begin{matrix}a & b \\ 0 & 0 \end{matrix}). The precise statement is as follows.

Theorem 3.2. Let e be a given idempotent in K. Then A = (\begin{matrix}a & b \\ 0 & 0 \end{matrix}) is e-clean iff

there exist x, y ∈ K with ex = 0 and y ≡ 1 (mod xK) such that ay − bx is e-clean.

Proof. For the “if” part, write y in the form kx+1 with ex = 0, and let ay−bx = e+u, where u ∈ U(K). We can then form the idempotent matrix E := E(e, x, k) in (2.4), with det(E) = e by (2.5). Letting U := A − E, we have

det(U) = −ay + bx + det(E) = −(e + u) + e = −u ∈ U(K).

Thus, U ∈ GL_{2}(K), and A = E + U shows that A is e-clean.

For the “only if” part, assume that A = E + U, with E = E^{2} = (\begin{matrix}p & q \\ x & y \end{matrix}) of

determinant e, and U ∈ GL_{2}(K). Since U = (\begin{matrix}a − p & b − q \\ −x & −y \end{matrix}), we have xK + yK = K.

Thus, by (2.7), we have ex = 0, and y ≡ 1 (mod xK). Now let u := −det(U) ∈ U(K). Then

−u = det (\begin{matrix}a − p & b − q \\ −x & −y \end{matrix}) = −ay + bx + e,

so ay − bx = e + u ∈ K is e-clean, as desired.

Corollary 3.3. Let k ∈ K and v ∈ U(K). For any idempotent e ∈ K, A = (\begin{matrix}a & b \\ 0 & 0 \end{matrix})

is e-clean iff A′ = (\begin{matrix}a & vb + ka \\ 0 & 0 \end{matrix}) is e-clean.

Proof. It suffices to prove the “only if” part. Assuming A is e-clean, there exist (by Theorem 3.2) x, y ∈ K such that

ex = 0, y ≡ 1 (mod xK), and ay − bx is e-clean.

Let b′ := vb + ka. Then ay − bx = ay − v^{−1}(b′ − ka)x = ay_{1} − b′x_{1} is e-clean for x_{1} = v^{−1}x and y_{1} := y + v^{−1}kx. Since ex_{1} = ev^{−1}x = 0 and y_{1} ≡ y ≡ 1 modulo the ideal xK = x_{1}K, A′ is e-clean again by (3.2).

Remarks. (1) We could have also proved (3.3) by observing that A′ = V^{−1}AV, where

V = (\begin{matrix}1 & k \\ 0 & v \end{matrix}) ∈ GL_{2}(K) (since the conjugate of an e-clean matrix remains e-clean).

However, the proof given above is a good illustration of the use of (3.2).

(2) The roles of a and b in Theorem 3.2 are by no means symmetrical. Thus, the

conclusion of (3.3) applies to A′ only, and not to (for instance) A′′ = (\begin{matrix}a + kb & b \\ 0 & 0 \end{matrix}). This

can be seen easily, for instance, from (3.5)(1) below, or from the many examples in §4.

Corollary 3.4. Let r ∈ K be e-clean, where e is an idempotent of K. Then for any

b, k ∈ K, the matrix B = (\begin{matrix}(1 − e)kb + r & b \\ 0 & 0 \end{matrix}) is e-clean.

Proof. For a := (1 − e)kb + r, y = 1, and x = (1 − e)k, we have

ay − bx = (1 − e)kb + r − b(1 − e)k = r.

Since ex = e(1 − e)k = 0, y ≡ 1 (mod xK), and r is e-clean, (3.2) implies that the matrix B is e-clean. (Note that (2.2) is essentially the special case of this result for k = 0.)

For later use let us record the following more explicit version of (3.2) corresponding to the case of trivial idempotents e = 0, 1.

Corollary 3.5. (1) A is 1-clean iff a ∈ 1 + U(K). (2) A is 0-clean iff there exist x_{0}, y_{0} ∈ K such that ay_{0} − bx_{0} = 1 and y_{0} + x_{0}K contains a unit of K. (In this case, A is also unit-regular, according to (2.1).)

Proof. (1) follows by letting e = 1 in (3.2) and noting that this implies that x = 0 and y = 1. With these uniquely determined values of x and y, the 1-clean criterion for A boils down to a ∈ 1 + U(K). (This criterion is easy to see in any case, since the only idempotent matrix of determinant 1 is the identity matrix.)

For (2), let e = 0, in which case the condition ex = 0 is automatic. The 0-clean criterion for A then requires the existence of x, y ∈ K with y ≡ 1 (mod xK) such that u := ay −bx ∈ U(K). Upon dividing everything by u, this transforms into the existence of x_{0}, y_{0} ∈ K such that ay_{0} − bx_{0} = 1 and (y_{0} + x_{0}K) ∩ U(K) ≠ ∅.

In the case where K is connected, a clean matrix is either 0-clean or 1-clean. Here, the cleanness criterion for A can be more conveniently stated as follows (via (3.5)).

Corollary 3.6. Let K be a connected (commutative) ring. (1) If a ∈ 1 + U(K), then A is always clean. (2) If otherwise, then A is clean iff there exist x_{0}, y_{0} ∈ K such that ay_{0} −bx_{0} = 1 and y_{0} +x_{0}K contains a unit of K. (In this case, A is also unit-regular.)

Another consequence of (3.3), (3.4), and (3.5)(2) useful for the next section is the following (for any commutative ring K).

Corollary 3.7. Let a, b, k ∈ K, and u, v ∈ U(K).

(1) If A = (\begin{matrix}a & b \\ 0 & 0 \end{matrix}) is 0-clean, so is (\begin{matrix}ua & vb + ka \\ 0 & 0 \end{matrix}).

(2) The matrix (\begin{matrix}kb + u & b \\ 0 & 0 \end{matrix}) is always 0-clean.

Proof. (1) If u = 1, this is covered by (3.3) (even in the e-clean case). Thus, it

only remains to make the passage from A to (\begin{matrix}ua & b \\ 0 & 0 \end{matrix}). This can be done (albeit only

in the 0-clean case) by rewriting the equation ay_{0} − bx_{0} = 1 in (3.5)(2) in the form au(u^{−1}y_{0}) − bx_{0} = 1, and noting that

u^{−1}y_{0} + x_{0}K = u^{−1}(y_{0} + x_{0}uK) = u^{−1}(y_{0} + x_{0}K).

Finally, (2) follows from (3.4) by letting e = 0.

There are also a few interesting connections between (3.5) and notions in algebraic K-theory that are worth mentioning. Recall that a ring K is said to have stable range 1 if, for any x, y ∈ K, yK + xK = K implies that y + xK contains a unit of K (see [La_{1}: §20]). In this case, the 0-clean criterion in (3.5)(2) above can be further simplified as follows.

Corollary 3.8. If K has stable range 1, A is 0-clean iff its first row (a, b) is unimodular over K.

A second connection between 0-cleanness and the K_{1}-functor in algebraic K-theory is provided by E_{n}(K), the subgroup of SL_{n}(K) generated by the elementary matrices. According to a theorem of Suslin [GM: p. 14], E_{n}(K) is a normal subgroup of SL_{n}(K) whenever n ≥ 3 (and K is a commutative ring). For any unimodular row (a, b), the Mennicke symbol [ a, b] is defined to be the class in the factor group SL_{3}(K)/E_{3}(K)

given by any matrix (\begin{matrix}a & b & 0 \\ x & y & 0 \\ 0 & 0 & 1 \end{matrix}) where ay − bx = 1. (See [GM: p. 43].) The following

consequence of (3.5)(2) gives some interesting necessary conditions for 0-cleanness.

Proposition 3.9. If A = (\begin{matrix}a & b \\ 0 & 0 \end{matrix}) is 0-clean, then

(1) (a, b) can be brought to (1, 0) by a finite sequence of elementary transformations;

(2) (a, b) is completable to a matrix in E_{2}(K); and

(3) any completion of (a, b) to a matrix in SL_{2}(K) belongs to E_{2}(K). In particular, the Mennicke symbol [ a, b] ∈ SL_{3}(K)/E_{3}(K) is trivial.

Proof. For any unimodular row (a, b), it is easy to show that the three properties (1), (2), (3) above are equivalent. If A is 0-clean, (a, b) is unimodular by (3.5)(2), so we are done if we can prove (2). By (3.5)(2), there exist x_{0}, y_{0} ∈ K such that ay_{0} − bx_{0} = 1

and u := y_{0} + x_{0}k is a unit for some k ∈ K. Then B := (\begin{matrix}a & b \\ x_{0} & y_{0} \end{matrix}) ∈ SL_{2}(K).

For C_{1} = (\begin{matrix}1 & k \\ 0 & 1 \end{matrix}), we have BC_{1} = (\begin{matrix}a & b + ka \\ x_{0} & u \end{matrix}). Further right-multiplying this by

C_{2} = (\begin{matrix}1 & 0 \\ −x_{0}u^{−1} & 1 \end{matrix}), we get a matrix of the form C_{3} = (\begin{matrix}c & d \\ 0 & u \end{matrix}). Since this matrix

remains in SL_{2}(K), we have c = u^{−1}. Now Whitehead’s Lemma (as in [LS: p. 344]) implies that C_{3} ∈ E_{2}(K). Thus, B = C_{3}C_{2}^{−1}C_{1}^{−1} ∈ E_{2}(K), proving (2). In particular,

we have (\begin{matrix}B & 0 \\ 0 & 1 \end{matrix}) ∈ E_{3}(K), so the Mennicke symbol [ a, b] is trivial.

Corollary 3.10. Suppose there exists a matrix (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ SL_{2}(K) \ E_{2}(K), where K is

a connected ring. If a ∉ 1 + U(K), then the matrix A = (\begin{matrix}a & b \\ 0 & 0 \end{matrix}) is unit-regular but not

clean. In particular, this conclusion on A always holds for any unimodular row (a, b) with a ∉ 1 + U(K) that gives a nontrivial Mennicke symbol [ a, b].

Proof. By (3.9), A is not 0-clean, and by (3.5)(1), A is not 1-clean. Since K is connected, A is not clean. But A is unit-regular by (2.1).

With (3.10) at our disposal (and assuming some results in the literature), we can now give some quick examples of 2 × 2 unit-regular matrices that are not clean.

Example 3.11. Let K be the real coordinate ring of the circle S^{1}; that is, K = ℝ [a, b],

with the relation a^{2} + b^{2} = 1. It is known that, for the matrix B = (\begin{matrix}a & b \\ −b & a \end{matrix}) ∈ SL_{2}(K),

the suspension (\begin{matrix}B & 0 \\ 0 & I_{n} \end{matrix}) is not in E_{n+2}(K) for any n; see, e.g. [LS: p. 345]. In particular,

the Mennicke symbol [ a, b] is nontrivial. Since K is connected and a ∉ 1 + U(K), the

last part of (3.10) implies that A = (\begin{matrix}a & b \\ 0 & 0 \end{matrix}) is unit-regular but not clean. Note that A

here is pretty close to being a “generic” counterexample. The “generic” one would be

(\begin{matrix}a & b \\ 0 & 0 \end{matrix}) over the commutative ℝ-algebra ℝ [a, b, c, d] with the relation ad − bc = 1. In

retrospect, we now know that this is indeed a counterexample!

Example 3.12. We can also give an example where the Mennicke symbol is trivial. Let K be the polynomial ring k [x, y], where k is any commutative domain. According to Cohn [Co] and Gupta-Murthy [GM: p. 16], the Mennicke symbol [ 1 + xy, x^{2} ] is trivial, although the “Cohn matrix”

(3.13)  B = (\begin{matrix}1 + xy & x^{2} \\ −y^{2} & 1 − xy \end{matrix}) ∈ SL_{2}(K)

does not lie in E_{2}(K). Using (3.10) again, we see as in the last example that the matrix

A = (\begin{matrix}1 + xy & x^{2} \\ 0 & 0 \end{matrix}) is unit-regular but not clean in 𝕄_{2}(K).

Finally, we note that the converse of (3.9) is false in general; that is, even if (a, b) is

the first row of a matrix in E_{2}(K), the matrix A = (\begin{matrix}a & b \\ 0 & 0 \end{matrix}) may still be non-clean. For

instance, over the Euclidean domain ℤ, we have SL_{2}(ℤ) = E_{2}(ℤ), so any unimodular

row (a, b) is completable to a matrix in E_{2}(ℤ). But a matrix such as (\begin{matrix}12 & 5 \\ 0 & 0 \end{matrix}) turns

out to be non-clean over ℤ, as we shall see in (4.5) below.

§4. Clean and Non-Clean Matrices Over ℤ

In this section, we shall specialize the results of §3 to the ring of integers; that is, we let K = ℤ. In this case, it turns out that the criterion for the cleanness of a unit-

regular matrix of the form A = (\begin{matrix}a & b \\ 0 & 0 \end{matrix}) can be further simplified. In fact, the simplified

criterion is in such a form that the cleanness of A can be quickly tested by carrying out the Euclidean algorithm. The results in this section were worked out in collaboration with Tom Dorsey and Alex Dugas, whose contributions are gratefully acknowledged here.

Throughout this section, A continues to denote the matrix (\begin{matrix}a & b \\ 0 & 0 \end{matrix}). In studying the

cleanness of A over ℤ, we are free to change the signs of a and b (by (3.7)), so we may assume that a, b ≥ 0 whenever it is convenient to do so. We begin with the following useful observation on cleanness versus 0-cleanness in the case K = ℤ.

Proposition 4.1. The matrix A is 0-clean iff it is clean and (a, b) ∈ ℤ^{2} is unimodular.

Proof. The “only if” part is clear from (3.5)(2). Conversely, assume that A is clean with (a, b) unimodular. If A is not 0-clean, then it must be 1-clean (since ℤ has only trivial idempotents). Therefore, by (3.5)(1), a ∈ {0, 2}. If a = 0, then b = ±1; in this case, the 0-clean criterion for A in (3.5)(2) is fulfilled by choosing (say) y_{0} = 1 and x_{0} = ∓1. If a = 2, then b = 2n + 1 for some n; here, the 0-clean criterion in (3.5)(2) is fulfilled by choosing y_{0} = −n and x_{0} = −1.

In view of (2.1), there is no real loss if we restrict ourselves to the case where (a, b) is unimodular. Then “clean” becomes synonymous with “0-clean”, according to (4.1). We start by giving a couple of numerical examples to illustrate the process of testing cleanness and computing “clean decompositions” by the constructive proof of (3.2).

Example 4.2. To test by (3.2) if (\begin{matrix}9 & 7 \\ 0 & 0 \end{matrix}) is clean (and hence 0-clean), we deal with

the diophantine equations 9y − 7x = u ∈ U(ℤ) = {±1}. For u = 1, this has a solution y = −3, x = −4 satisfying y ≡ 1 (mod x). Therefore, by Theorem 3.2 (and its proof), A is 0-clean, with an idempotent summand E = E(0, −4, 1). Thus, we arrive at a clean representation:

(4.3)  (\begin{matrix}9 & 7 \\ 0 & 0 \end{matrix}) = (\begin{matrix}4 & 3 \\ −4 & −3 \end{matrix}) + (\begin{matrix}5 & 4 \\ 4 & 3 \end{matrix}) .

On the other hand, we could have also solved the equation 9y−7x = u for u = −1 with y = −4 and x = −5 again satisfying y ≡ 1 (mod x). This gives a different idempotent summand E(0, −5, 1) for A, leading to a second explicit clean representation:

(4.4)  (\begin{matrix}9 & 7 \\ 0 & 0 \end{matrix}) = (\begin{matrix}5 & 4 \\ −5 & −4 \end{matrix}) + (\begin{matrix}4 & 3 \\ 5 & 4 \end{matrix}) .

Example 4.5. Let us show that A = (\begin{matrix}12 & 5 \\ 0 & 0 \end{matrix}) is not clean (although A is unit-

regular by (2.1)). Applying (3.5)(2), we want to show that, for any (x_{0}, y_{0}) ∈ ℤ^{2} solving 12y_{0} − 5x_{0} = 1, y_{0} + x_{0}ℤ contains neither 1 nor −1. Now the general solution for the diophantine equation 12y_{0} − 5x_{0} = 1 is y_{0} = −2 + 5t, x_{0} = −5 + 12t (with t ∈ ℤ). Clearly, neither −1 + 5t nor −3 + 5t can be a multiple of −5 + 12t, for any t ∈ ℤ. Therefore, the matrix A is not clean in 𝕄_{2}(ℤ). The same work applies to the

matrix (\begin{matrix}13 & 5 \\ 0 & 0 \end{matrix}), and an application of (3.3) generates two more examples: (\begin{matrix}12 & 7 \\ 0 & 0 \end{matrix}) and

(\begin{matrix}13 & 8 \\ 0 & 0 \end{matrix}).

Of course, for a given matrix A, the cleanness criterion in (3.5) is explicit enough to be checked by a computer. A program written by Tom Dorsey showed that the nonnegative unimodular rows (a, b) ∈ ℤ_{+}^{2} leading to “dirty” unit-regular matrices are the following, listed in increasing values^{4} of a + b:

(∗)  \begin{matrix}(12, 5), (13, 5), (12, 7), (13, 8), (17, 5), (16, 7), (18, 5), (17, 7), \\ (16, 9), (18, 7), (19, 7), (17, 10), (19, 8), (22, 5), (23, 5), (12, 17), \\ (17, 12), (18, 11), (20, 9), (21, 8), (19, 11), (23, 7), (12, 19), etc. \end{matrix}

Thus, at least in the initial range 1 ≤ a + b ≤ 30, examples of non-clean unit-regular matrices of the form A occur rather scarcely, and the examples mentioned in (4.5) correspond, in fact, to the four smallest choices for s := a + b, each occurring uniquely for the given row sum s.

The fact that all rows (a, b) listed in (∗) have a ≥ 12 and b ∈ {5}∪[ 7, ∞) led us to the following general observations, which will be proved independently of the tabulation (∗) above.

Proposition 4.6. Assume that the row (a, b) ∈ ℤ^{2} is unimodular. Then, (1) the matrix A is clean if b is congruent (mod a) to ±1, ±2, ±3, ±4 or ±6; and (2) A is always clean if |a| < 12, or if |a| + |b| ∈ [1, 16] ∪ {20}.

4For rows (a, b) with the same row sum a + b, we list them in increasing values of a.

Proof. (1) Applying (3.3), and changing b to −b if necessary, we may assume that b is one of 1, 2, 3, 4 or 6. We shall only treat the case b = 6, the other cases being completely similar. Since U(ℤ/6ℤ) = {±1}, we can write a in the form 6n ± 1. Then we can solve the equation ay_{0} − bx_{0} = 1 with x_{0} = ±n, y_{0} = ±1 (so in particular y_{0} ≡ ±1 (mod x_{0})). Therefore, by (3.5)(2), the matrix A is 0-clean.

(2) If |a| < 12, the hypothesis on b above is clearly satisfied, so A is always clean in this case. Next, assume that |a| + |b| ∈ [1, 16] ∪ {20}. Changing signs if necessary, we may also assume that b > 0 and a ≥ 12. If a + b ≤ 16, then b ≤ 4, so A is clean by (1). If a + b = 20 and b > 4, the only possibility is (a, b) = (13, 7). But this is not in (∗) by (3.3) since (13, 13 − 7) = (13, 6) is not in (∗) by (1) above. Thus, A is clean in all the cases claimed.

In the case a = 12 (resp. a = 13), the two cases not covered by the Proposition above are b = 5, 7 (resp. b = 5, 8). And indeed, these choices are responsible for the first four rows in the tabulation (∗).

The idea of using the transformation (a, b) ↦ (a, b − ka) in the proof of (4.6) also suggests the best way to analyze the general data in (∗). Note, for instance, that (12, 17) is the first occurrence of a row in (∗) with a < b. But (12, 17) could have been obtained from the first row (12, 5) in (∗) by the transformation (a, b) ↦ (a, b + a). Thus, the presence of (12, 5) “induces” that of (12, 17). In general, if (a, b) is present in (∗) with b = ka + r where k ∈ ℤ and 0 < r < a, then (3.3) implies that (a, r) must be an “earlier” entry in (∗) (whose presence induces that of (a, b)). Once we are down to a row (a, r) with 0 < r < a, we may also assume that a ≥ 2r, for if otherwise, we may perform the transformations (a, r) ↦ (a, r − a) ↦ (a, a − r) to achieve our goal.

Let us say that a unimodular row (a, b) ∈ ℤ^{2} is reduced if |a| ≥ 2 |b|. By the considerations in the paragraph above, any row in the tabulation (∗) can be brought in a canonical fashion to an earlier reduced row in (∗). Therefore, for the purposes of deciding which unimodular rows (a, b) ∈ ℤ_{+}^{2} are in (∗), there is no loss of generality in working with the reduced rows (a, b). Thus, the problem of deciding the cleanness of A will be solved in an algorithmically very simple fashion as soon as we prove the following result.

Theorem 4.7. Let (a, b) ∈ ℤ_{+}^{2} be a reduced unimodular row. Then A is clean iff a ≡ ±1 (mod b).

Proof. If b = 0, then a = 1, in which case both statements are true. Therefore, in the following, we may assume that b > 0.

The “if” part follows from (3.7)(2) (even without the reducedness assumption on (a, b)). For the “only if” part, assume that a ≢ ±1 (mod b). By the Euclidean algorithm, we can write a = kb + r, where 0 < r < b and k ∈ ℤ. Since a ≥ 2b, k ≥ 2. Applying (3.5)(2), it suffices to check that, for any integers x, y solving

(4.8)  1 = ay − bx = (kb + r)y − bx,

we have y ≢ ±1 (mod x). From (4.8), we have ay ≡ ry ≡ 1 (mod b), so a ≢ ±1 (mod b) implies that y ≢ ±1 (mod b). In particular, b ≠ 1 and |y| ≥ 2. Using the equation x = ky + (ry − 1)/b, we argue in two cases.

Case 1. y < 0. Since (ry − 1)/b < 0, x < ky ≤ 2y < y < −1. This clearly implies that y ≢ ±1 (mod x). Case 2. y > 0. Here, (ry − 1)/b > 0, so x ≥ 1 + ky ≥ 1 + 2y. Since y ≥ 2, this again implies that y ≢ ±1 (mod x), as desired.

While the reducedness assumption on (a, b) is not needed for the “if” part above, it is essential for the “only if” part. For instance, in (4.1), the matrix A with a (nonreduced) top row (9, 7) is clean, but a = 9 is not congruent to ±1 (mod b) for b = 7. In fact, the “only if” part in (4.7) is false even if we try to relax the reducedness assumption b ≤ a/2 to b < 3a/4, as the example (28, 19) shows.

The new method given in Theorem 4.7 for testing the cleanness of A (over the integers) is algorithmically very easy to implement since the decision procedure involves nothing more than the Euclidean algorithm. Let us illustrate this with a couple of examples.

Examples 4.9. To determine all (a, b) in (∗) with first entry a = 36, we may assume (36, b) is unimodular and reduced. Thus, b ∈ {1, 5, 7, 11, 13, 17}. Among these choices of b, only the first three satisfy 36 ≡ ±1 (mod b). Therefore, by (4.7), the reduced rows in (∗) with a = 36 are (36, 11), (36, 13), and (36, 17). Similarly, we can check, for instance, that a reduced row of the form (69, b) occurs in (∗) iff

b ∈ {8, 11, 13, 16, 19, 20, 22, 25, 26, 28, 29, 31, 32}.

Theorem 4.7 can also be used in various ways to give more precise quantitative information about the rows in (∗) corresponding to unit-regular matrices A that are non-clean. Its first natural consequence is the following.

Corollary 4.10. Let b > 0. Then there exists a row of the form (a, b) in the tabulation (∗) iff b ∉ {1, 2, 3, 4, 6}.

Proof. The “only if” part is a special case of (4.6)(1). For the converse, assume that b ∉ {1, 2, 3, 4, 6}. It is well-known that in this case U(ℤ/b ℤ) ⊋ {±1}. Thus, there is a positive integer r < b relatively prime to b such that r ≢ ±1 (mod b). Then for any integer k ≥ 2, (kb+r, b) is a reduced unimodular row, and the matrix A with first row (kb + r, b) is non-clean by (4.7). Therefore, the row (kb + r, b) occurs in the tabulation (∗) for every integer k ≥ 2.

For a row (a, b) ∈ ℤ^{2}, let s = |a| + |b| denote its (absolute) row sum (also known as its “L^{1}-norm”). We saw in the tabulation (∗) (and proved by hand in (4.6)(2)) that the row sums 1, 2, …, 16 and 20 are “missing”. A missing sum s means, of course, that the matrix A is clean whenever (a, b) is unimodular with |a| + |b| = s. An obvious question to ask is whether there exist in (∗) other missing row sums beyond 20. With the help of Alex Dugas and Michael Filaseta, we shall give a negative answer to this question below.

It all comes down to using (4.7) efficiently. If we use the proof of its corollary (4.10), we may start with any b = 2n + 1 ≥ 5, and take r = k = 2 in that proof to generate a row (2b + 2, b) in (∗). This gives a row sum 3b + 2 = 6n + 5 for any n ≥ 2. Similar constructions would generate other infinite sequences of row sums. If we want to get all row sums (beyond 20) in one stroke, we may proceed as follows. By standard results in prime number theory,^{5} there exists a bound n_{0} such that, for any integer n ≥ n_{0}, there is a prime p satisfying

(4.11)  \frac{n + 1}{4} < p < \frac{n − 1}{3}.

For such a prime p, we have 3p < n − 1 < n < n + 1 < 4p. Therefore, n = 3p + r for some r ∈ [ 2, p − 2]. Taking b = p and a = 2p + r, we have a ≡ r ≢ ±1 (mod b), and so by (4.7), the (reduced) unimodular row (a, b) must show up in the tabulation (∗), with row sum a + b = 3p + r = n. This construction produces all row sums n ≥ n_{0}, thus proving that there can only be finitely many “missing” row sums in (∗).

Now we are reduced to showing that there are no missing sums > 20 in the range [1, n_{0}). Of course this would be an easier problem only if we can name an explicit bound n_{0} ! Fortunately, help is available from the classical number theory literature. In [Sch: Hilfssatz 1], Schur proved that, for any real number x ≥ 29, there exists a prime p such that x < p ≤ 5x/4. This can be applied to our situation since 5/4 is (if only a little) less than 4/3: letting x = (n + 1)/4, we see that for n ≥ 115, there is always a prime p > (n + 1)/4 such that

p ≤ \frac{5}{4} · \frac{n + 1}{4} = \frac{5(n + 1)}{16} < \frac{n − 1}{3}.

Thus, we may take n_{0} = 115, although we certainly would not expect this to be the best bound. In fact, a quick check on Maple shows that the required prime p in (4.11) exists also for every n in the range [ 53, 115). On the other hand, if n = 52, there is no prime in the open interval ((n + 1)/4, (n − 1)/3) = (13.25, 17). Therefore, the best bound for our analytic problem is n_{0} = 53: this was pointed out to us by M. Filaseta. Finally, a quick hand computation (or direct inspection of the computer data for (∗)) shows that all sums from 21 to 52 are also “realized”! Therefore, the missing sums in (∗) are precisely: 1, 2, …, 16, and 20. Moreover, any sum s that is not one of these exceptional values can be realized by some (a, b) in (∗) with b a prime number.

5More details on this point will be given in the next paragraph.

The method above actually gives all missing sums for the reduced rows in (∗) as well. We may drop all non-reduced rows from (∗) to form a new list (∗)_{red} , and ask for the missing sums herein:

(∗)_{red}  \begin{matrix}(12, 5), (13, 5), (17, 5), (16, 7), (18, 5), (17, 7), (18, 7), \\ (19, 7), (19, 8), (22, 5), (23, 5), (20, 9), (21, 8), (23, 7), etc. \end{matrix}

Since we have dropped (12, 7) and (13, 8), the sums 19 and 21 are now also missing (along with 20). However, the construction in the last paragraph used only reduced rows, and it can be checked easily that the sums in the interval [ 22, 52] can each be realized by reduced rows. We may, therefore, conclude that, in the new list (∗)_{red} of reduced rows giving rise to non-clean matrices A, the missing sums are precisely 1, 2, …, 16, and 19, 20, 21.

We close this section by pointing out that the examples of non-clean integral matrices obtained in this section naturally induce many examples of the same type over other

rings. For instance, we can infer that (\begin{matrix}1 + xy & x^{2} \\ 0 & 0 \end{matrix}) is (unit-regular and) non-clean over

ℤ [x, y] by specializing x to 3 and y to 5 (and noting that (16, 9) is in (∗)).^{6} Similarly,

since (1 + x)^{2} = 1 + x(2 + x), we see that (\begin{matrix}(1 + x)^{2} & x^{2} \\ 0 & 0 \end{matrix}) is unit-regular and non-clean

over the polynomial ring ℤ [x].

§5. Idempotent Matrices with a Unimodular Row

In this final section, we return to the case of an arbitrary commutative base ring K. In §3, we were able to prove our main result Theorem (3.2) by making use of the matrices E(e, x, k) defined in (2.4). However, we have not given any explanation as to how we arrived at these idempotent matrices. In this section, we shall prove a strong characterization theorem on the matrices E(e, x, k), which, in particular, shows how they are constructed from three specific defining properties. We should, however, stress the fact that, while the matrices E(e, x, k) seemed crucial for the proof of Theorem 3.2, the following characterization result on such matrices was not needed in that proof.

Theorem 5.1. Let e be a given idempotent in K. Then the most general idempotent matrix E ∈ R = 𝕄_{2}(K) with a unimodular second row and with det(E) = e is E(e, x, k), where x, k ∈ K, with ex = 0.

Proof. If ex = 0, we have shown in (2.5) that E(e, x, k) is idempotent, has determinant

e, and its second row (x, kx + 1) is obviously unimodular. Conversely, let E = (\begin{matrix}p & q \\ x & y \end{matrix})

be an idempotent matrix with det(E) = e, and with (x, y) unimodular. By (2.7), we have

(5.2)  eq = ex = 0, and ey = e,

6This, however, would not work over k [x, y] where k is an arbitrary commutative domain. For this case, we have to go back to (3.12)!

and we can write y in the form k_{0}x + 1 for some k_{0} ∈ K. Consider the two linear equations

−qx + py = e, and − k_{0}x + y = 1,

and let k := det (\begin{matrix}−q & p \\ −k_{0} & 1 \end{matrix}). By Cramer’s Rule, we have kx = det (\begin{matrix}e & p \\ 1 & 1 \end{matrix}) = e − p,

and ky = det (\begin{matrix}−q & e \\ −k_{0} & 1 \end{matrix}) = −q + k_{0}e. Therefore, p = e − kx, and q = k_{0}e − ky. Let

r := tr(E) − 1 ∈ K. By the Cayley-Hamilton Theorem (and the fact that E = E^{2}), we have r · E = e · I_{2}, so rx = 0, and ry = e. This yields e = r(k_{0}x + 1) = r. Therefore,

e = tr(E) − 1 = p + y − 1 = e − kx + k_{0}x,

which implies that kx = k_{0}x. It follows that

y = kx + 1, and q = k_{0}e − k(kx + 1).

The final step is to prove that k_{0}e = ke, for, if this is the case, then indeed E equals E(e, x, k). Now from (5.2), we have

0 = eq = e(k_{0}e − ky) = k_{0}e − k(ey) = k_{0}e − ke,

so k_{0}e = ke, as desired.

References

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[Co] P. M. Cohn: On the structure of GL_{2} of a ring, Publ. I.H.E.S. 30(1966), 5–53.

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[Go] K. R. Goodearl: Von Neumann Regular Rings, Second Edition, Krieger Publ. Co., Malabar, Florida, 1991.

[GM] S. K. Gupta and M. P. Murthy: Suslin’s Work on Linear Groups over Polynomial Rings and Serre Problem, ISI Lecture Notes, Vol. 8, Macmillan Co. of India, Ltd., New Delhi, 1980.

[La_{1}] T. Y. Lam: A First Course in Noncommutative Rings, 2nd edition, Graduate Texts in Math., Vol. 131, Springer-Verlag, Berlin-Heidelberg-New York, 2001.

[La_{2}] T. Y. Lam: Exercises in Classical Ring Theory, 2nd ed., Problem Books in Mathematics, Springer-Verlag, Berlin-Heidelberg-New York, 2003.

[LS] T. Y. Lam and M. K. Siu: K_{0} and K_{1} — an introduction to algebraic K-theory, Amer. Math. Monthly 82(1975), 329–364.

[Ni_{1}] W. K. Nicholson: Lifting idempotents and exchange rings, Trans. A.M.S. 229(1977), 269–278.

[Ni_{2}] W. K. Nicholson: Strongly clean rings and Fitting’s Lemma, Comm. Algebra 27(1999), 3583–3592.

[Sch] I. Schur: Einige Sätze über Primzahlen mit Anwendungen auf Irreduzibilitätsfragen, Sitz.-ber. Preuss. Akad. Wiss., Physik.-Math. Klasse, 1929, 125– 136. (Ges. Abh., Vol. 3, pp. 140–152, Springer-Verlag, 1973.)

Department of Mathematics Panjab University Chandigarh-160014, India dkhurana@pu.ac.in

Department of Mathematics University of California Berkeley, Ca 94720 lam@math.berkeley.edu