The Kaplansky condition and rings of almost stable range 1

We present some variants of the Kaplansky condition for a K-Hermite ring $R$ to be an elementary divisor ring; for example, a commutative K-Hermite ring $R$ is an EDR iff for any elements $x,y,z\in R$ such that $(x,y)=(1)$, there exists an element $\lambda\in R$ such that $x+\lambda y=uv$, where $(u,z)=(v,1-z)=(1)$. We present an example of a a B\'ezout domain that is an elementary divisor ring, but it does not have almost stable range 1, thus answering a question of Warren Wm. McGovern.


Introduction
First we recall some basic definitions and known results. All rings here are commutative with unity. A ring R is Bézout if each finitely generated ideal of R is principal.
Two rectangular matrices A and B in M m,n (R) are equivalent if there exist invertible matrices P ∈ M m,m (R) and Q ∈ M n,n (R) such that B = P AQ.
The ring R is K-Hermite if every rectangular matrix A over R is equivalent to an upper or a lower triangular matrix (following [9, Appendix to §4] we use the term 'K-Hermite' rather then 'Hermite' as in [8]). From [8] it follows that this definition is equivalent to the definition given there. See also [6,Theorem 3]: by this theorem, a ring is K-Hermite iff for every two comaximal elements a, b ∈ R, there are two comaximal elements a 1 , b 1 ∈ R such that (a, b) = (a 1 a + b 1 b) in R. Parentheses () are used to denote the ideal generated by the specified elements.
A ring R is an elementary divisor ring (EDR) iff every rectangular matrix A over R is equivalent to a diagonal matrix. It follows from [8] that this definition is equivalent to the definition given there.
An EDR is K-Hermite and a K-Hermite ring is Bézout. An integral domain is Bézout iff it is K-Hermite.
In section 2, we elaborate on the Kaplansky condition. A row [r 1 , . . . , r n ] over a ring R is unimodular if the elements r 1 , . . . , r n generate the ideal R. The stable range sr(R) of a ring R is the least integer n ≥ 1 (if it exists) such that for any unimodular row [r 1 , ..., r n+1 ] over R, there exist t 1 , ..., t n ∈ R such that the row [r 1 + t 1 r n+1 , ..., r n + t n r n+1 ] is unimodular (see comments on [9, Theorem 5.2, Ch. VIII]). For background on stable range see [1, §3, Ch. V].
The ring R has almost stable range 1 if every proper homomorphic image of R has stable range 1 (see [10]). By [10, Theorem 3.7] a Bézout ring with almost stable range 1 is an EDR. We elaborate on the almost stable range 1 condition in §3. In particular, we present an elementary divisor domain (so Bézout) that does not have almost stable range 1, thus answering the question of Warren Wm. McGovern in [10] (Example 13). By Remark 12 below, a ring of stable range 1 is of almost stable range 1. On the other hand, Z is of almost stable range 1, but is not of stable range 1 (the stable range of Z is 2. More generally, the stable range of any K-Hermite ring is ≤ 2 [11,Proposition 8]. Also by [12, Theorem 1], a Bézout ring is K-Hermite iff is of stable range ≤ 2).
For general background see [8], [ For any three elements a, b, c in R that generate the ideal R, (K) there exist elements p, q ∈ R so that (pa, pb + qc) = (1) in R.

Remark 1.
If R is a ring of stable range 1, then R satisfies Kaplansky's condition with p = 1. Thus, if R is a ring satisfying Kaplansky's condition, then R is not necessarily K-Hermite, even if R is a Noetherian local domain.
Indeed, a local ring R is of stable range 1. If R is also a Noetherian domain, then R is K-Hermite iff R is a principal ideal ring. Hence a Noetherian local domain that is not a principal ideal ring is of stable range 1, but it does not satisfy Kaplansky's condition.
In the proof of Lemma 3 below, we will use the following well-known fact: Remark 2. Let R be a ring, let A be a matrix in M m,n (R), let r be a row in in M 1,n (R), and let 1 ≤ k ≤ n. Then r belongs to the submodule of R n generated by the rows of the matrix A iff there exists a matrix C ∈ M k,m (R) such that r is the first row of the matrix CA.
Lemma 3. Let A be a 2 × 2-matrix over a ring R, and let u be a unimodular row of length 2 over R.
Then u belongs to the submodule of R 2 generated by the rows of A ⇐⇒ there exists an invertible matrix P so that u is the first row of P A.

Proof.
=⇒ : By Remark 2, there exists a 2-row r over R so that u = rA. Since the row u is unimodular, the row r is also unimodular. Since r is unimodular of length 2, there exists an invertible matrix P with first row equal to r. Thus u is the first row of the matrix P A.
Lemma 4. Let A be a 2 × 2 matrix over a ring R so that its entries generate the ideal R. Then A is equivalent to a diagonal matrix ⇐⇒ the submodule of R 2 generated by the rows of A contains a unimodular row.
In this case A is equivalent to a matrix of the ⇐= By Lemma 3, the matrix A is equivalent over R to a matrix B with first row unimodular. Hence the submodule generated by the columns of B contains a column of the form 1 * . By Lemma 3 again (for columns), we obtain that A is equivalent to a matrix 1 r * * . By subtracting the first column of the matrix 1 r * * multiplied by r from its second column and by a similar elementary row transformation, we obtain a diagonal matrix of the form 1 0 0 * . Let R be a ring. Let A = a b 0 c a triangular 2 × 2-matrix over R so that (a, b, c) = (1). Then A is equivalent to a diagonal matrix over R iff there exist elements p, q in R so (pa, pb + qc) = (1).
Proof. Since p[a, b] + q[0, c] = [pa, pb + qc] for any elements p, q ∈ R, the theorem follows from Lemma 4. Proposition 7. Let R be a K-Hermite ring, and let a, b and c be elements of R that generate the ideal R. Then the following four conditions are equivalent: (1) The matrix a b 0 c is equivalent to a diagonal matrix; (2) There exist elements p, q in R so (pa, pb + qc) = (1); (3) There exist elements p and q in R so that (pb+qc, a) = (p, c) = (1); (4) For some elements λ, u, v ∈ R we have b + λc = uv, and (u, a) = (v, c) = (1). Moreover, in (4) we may choose the elements u and v such that (u, v) = (1).
In the proof of Proposition 7, we have used the assumption that R is K-Hermite just for the implication (3) =⇒ (2).

Remark 8.
If R is a Bézout domain, then the following condition is equivalent to the conditions of Proposition 7: ( * ) For some elements λ, a 1 , Indeed, assume condition ( * ). Let u ∈ R so that with (a, b, c) = (1) has a diagonal reduction [8], Proposition 7 provides necessary and sufficient conditions for a K-Hermite ring to be an EDR.
We present an additional condition in the next proposition.
Moreover, the elements u and v can be chosen such that (u, v) = (1).

On rings of almost stable range 1
Proposition 11. Let R be any ring. The following conditions are equivalent: (1) R is of almost stable range 1; (2) For each nonzero element z ∈ R, the ring R/zR is of stable range 1; (3) For each three elements x, y, z ∈ R so that (x, y) = (1) and z = 0, there exists an element λ ∈ R so that (x + λy, z) = (1).
Proof. Cf. Let I be a nonzero ideal of R and let z be a nonzero element in I. Let x + I, y + I be two comaximal elements in R/I. Hence there exist elements r, s ∈ R such that 1 − rx − sy ∈ I. By assumption, there exists an element λ ∈ R such that (x + λ(1 − rx), z)R = R. Thus x + λsy is invertible modulo the ideal I. It follows that R/I is of stable range 1, so R is almost of stable range 1. As we have seen in §2, the stable range 1 property implies Kaplansky's condition for an arbitrary ring. As for the converse, even if R is an elementary divisor domain, thus R satisfies Kaplansky's condition, then R does not necessarily has even almost stable range 1: Example 13. An elementary divisor domain (and so Bézout) that does not have almost stable range 1 (this example answers the question in [10,Remark 4.7]).
We use a well-known example of a Bézout domain, namely, R = Z + XQ[X] (for a general theorem on pullbacks of Bézout domains see [7,Theorem 1.9]). R is an elementary divisor ring by [2,Theorem 4.61]. However, R/XQ[X] is isomorphic to Z and sr Z = 2 (clearly, there is no integer m such that 2 + 5m = ±1. Thus sr Z > 1. As mentioned in the introduction, the stable range of any Bézout domain is ≤ 2, hence sr Z = 2).
We conjecture that a Bézout domain that is a pullback of type (as defined in [7]) of elementary divisor domains is again an EDR. In this case the conditions of [7, Theorem 1.9] must be satisfied. If this conjecture proves to be false, thiwill yield a negative answer to the question in [6] whether a Bézout domain is an EDR.