2405/2405.01234.md

Title: Matrix invertible extensions over commutative rings. Part III: Hermite rings

URL Source: https://arxiv.org/html/2405.01234

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Why HTML? Report Issue Back to Abstract Download PDF Abstract 1Introduction 2Test matrices 3Proof of Theorem 1.10 4Proof of Theorem 1.12 and applications 5Proof of Theorem 1.15 6On ( 𝑊 ​ 𝑆 ​ 𝑈 ′ ) 𝑛 and ( 𝑊 ​ 𝑆 ​ 𝑈 ) 𝑛 rings 7Proof of Theorem 1.17 and applications 8Proof of Theorem 1.18 9Proof of Criteria 1.19 and 1.20 10Proof of Criterion 1.22 References License: CC BY-NC-ND 4.0 arXiv:2405.01234v2 [math.AC] 27 Jul 2025 Matrix invertible extensions over commutative rings. Part III: Hermite rings Grigore Călugăreanu, Horia F. Pop, Adrian Vasiu Abstract.

We reobtain and often refine prior criteria due to Kaplansky, McGovern, Roitman, Shchedryk, Wiegand, and Zabavsky–Bilavska and obtain new criteria for a Hermite ring to be an EDR. We mention three criteria: (1) a Hermite ring 𝑅 is an EDR iff for all pairs ( 𝑎 , 𝑐 ) ∈ 𝑅 2 , the product homomorphism 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑐 ) × 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ​ ( 1 − 𝑎 ) ) → 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) between groups of units is surjective; (2) a reduced Hermite ring 𝑅 is an EDR iff it is a pre-Schreier ring and for each 𝑎 ∈ 𝑅 , every zero determinant unimodular 2 × 2 matrix with entries in 𝑅 / 𝑅 ​ 𝑎 lifts to a zero determinant matrix with entries in 𝑅 ; (3) a Bézout domain 𝑅 is an EDD iff for all triples ( 𝑎 , 𝑏 , 𝑐 ) ∈ 𝑅 3 there exists a unimodular pair ( 𝑒 , 𝑓 ) ∈ 𝑅 2 such that ( 𝑎 , 𝑒 ) and ( 𝑏 ​ 𝑒 + 𝑎 ​ 𝑓 , 1 − 𝑎 − 𝑏 ​ 𝑐 ) are unimodular pairs. We use these criteria to show that each Bézout ring 𝑅 that is an ( 𝑆 ​ 𝑈 ) 2 ring (as introduced by Lorenzini) such that for each nonzero 𝑎 ∈ 𝑅 there exists no nontrivial self-dual projective 𝑅 / 𝑅 ​ 𝑎 -module of rank 1 generated by 2 elements (e.g., all its elements are squares), is an EDR.

Key words and phrases: ring, matrix, projective module, unimodular. 2020 Mathematics Subject Classification: Primary: 15A83, 13F07, 19B10. Secondary: 13A05, 13G05, 15B33. 1.Introduction

Let 𝑅 be a commutative ring with identity element 1 . Let 𝑈 ​ ( 𝑅 ) , 𝑁 ​ ( 𝑅 ) , 𝐽 ​ ( 𝑅 ) , 𝑍 ​ ( 𝑅 ) and 𝑃 ​ 𝑖 ​ 𝑐 ​ ( 𝑅 ) be its group of units, its nilpotent radical, its Jacobson radical, its set of zero divisors, and its Picard group (respectively). For 𝑎 ∈ 𝑅 , let Ann 𝑅 ​ ( 𝑎 ) be its annihilator in 𝑅 . For 𝑛 ∈ ℕ

{ 1 , 2 , … } , let 𝕄 𝑛 ​ ( 𝑅 ) be the 𝑅 -algebra of 𝑛 × 𝑛 matrices with entries in 𝑅 . Let 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) be the general linear group of units of 𝕄 𝑛 ​ ( 𝑅 ) . Let 𝑆 ​ 𝐿 𝑛 ​ ( 𝑅 ) := { 𝑀 ∈ 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) | det ( 𝑀 )

1 } be the special linear subgroup of 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) . Let 𝑈 ​ 𝑚 ​ ( 𝐹 ) be the set of unimodular elements of a free 𝑅 -module 𝐹 ; so we have an identity 𝑈 ​ 𝑚 ​ ( 𝑅 )

𝑈 ​ ( 𝑅 ) of sets. We say that 𝐴 , 𝐵 ∈ 𝕄 2 ​ ( 𝑅 ) are congruent modulo an ideal 𝐼 of 𝑅 if 𝐴 − 𝐵 ∈ 𝕄 2 ​ ( 𝐼 ) .

In this paper we study Bézout rings, i.e., rings whose finitely generated ideals are principal. Each Bézout ring is an arithmetical ring, i.e., its lattice of ideals is distributive and all its localizations at prime ideals are valuation rings (see [10], Thms. 1 and 2). The ring 𝑅 is a Bézout ring iff each diagonal matrix with entries in 𝑅 admits diagonal reduction (see [11], Thm. (3.1)).

Recall that: (i) 𝑅 is a Hermite ring in the sense of Kaplansky if 𝑅 2

𝑅 ​ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , equivalently if each 1 × 2 matrix with entries in 𝑅 admits diagonal reduction; (ii) 𝑅 is an elementary divisor ring, to be abbreviated as EDR, if for each ( 𝑚 , 𝑛 ) ∈ ℕ 2 with 𝑚 ≤ 𝑛 , every matrix of size either 𝑚 × 𝑛 or 𝑛 × 𝑚 with entries in 𝑅 admits diagonal reduction, i.e., is equivalent to a matrix whose off diagonal entries are 0 and whose diagonal entries 𝑎 1 , 1 , … , 𝑎 𝑚 , 𝑚 are such that 𝑎 𝑖 , 𝑖 divides 𝑎 𝑖 + 1 , 𝑖 + 1 for all 𝑖 ∈ { 1 , … , 𝑚 − 1 } ; (iii) 𝑅 is a pre-Schreier ring if each 𝑥 ∈ 𝑅 is primal, i.e., if 𝑥 divides 𝑦 ​ 𝑧 , with ( 𝑦 , 𝑧 ) ∈ 𝑅 2 , then there exists ( 𝑢 , 𝑣 ) ∈ 𝑅 2 such that 𝑥

𝑢 ​ 𝑣 , 𝑢 divides 𝑦 and 𝑣 divides 𝑧 .

Each Hermite ring 𝑅 is a Bézout ring but the converse does not hold (see [6], Ex. 3.4, [22], Ex. 3.3, or [2], Prop. 8). However, a Hermite domain is the same as a Bézout domain. Bézout domains are GCD (greatest common divisors exist) domains and hence pre-Schreier domains with trivial Picard groups. Clearly, each EDR is a Hermite ring.

We use the tools of Parts I and II in order to obtain necessary and sufficient criteria for a Hermite ring (or domain) to be an EDR (or an EDD, i.e., an elementary divisor domain).

Let 𝐴 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) . Recall that 𝐴 is called extendable if it is obtained from a matrix 𝐴 + ∈ 𝑆 ​ 𝐿 3 ​ ( 𝑅 ) by removing its third row and its third column (see [3], Def. 1.1). If we can choose 𝐴 + such that its ( 3 , 3 ) entry is 0 , then 𝐴 is called simply extendable. Recall that 𝐴 is called determinant liftable (resp. weakly determinant liftable) if there exists 𝐵 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) (resp.  𝐵 ∈ 𝕄 2 ​ ( 𝑅 ) ) congruent to 𝐴 modulo 𝑅 ​ det ( 𝐴 ) and with det ( 𝐵 )

0 (see [4], Def. 1.1). The only general implications between these 4 notions on 𝐴 are the ones in the diagram

(I) simply extendable extendable determinant liftable weakly determinant liftable

(see [4], Thm. 1.3 for the vertical ones; for the fact that their converses do not hold and that there exist no implications between extendable and determinant liftable see [3], Exs. 5.3 and 6.1 and [4], Exs. 5.1 and 1.9). Recall that a matrix 𝐵 ∈ 𝕄 2 ​ ( 𝑅 ) is called non-full if it is the product of 2 × 1 with 1 × 2 matrices with entries in 𝑅 . An integral domain 𝑅 is a pre-Schreier domain iff each matrix in 𝕄 2 ​ ( 𝑅 ) of zero determinant is non-full by [13], Lem. 1.

Recall from [3], Def. 1.2 that 𝑅 is called a Π 2 ring if each matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) of zero determinant is extendable, equivalently it is simply extendable by [3], Lem. 4.1(1), and that 𝑅 is called an 𝐸 2 (resp. an 𝑆 ​ 𝐸 2 ) ring if each matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is extendable (resp. simply extendable). If 𝑃 ​ 𝑖 ​ 𝑐 ​ ( 𝑅 ) is trivial (say, 𝑅 is a Bézout domain), then 𝑅 is a Π 2 ring (see [3], paragraph after Thm. 1.4).

See [3], Def. 1.5 for stable ranges notation. If 𝑅 is a Hermite ring, then 𝑠 ​ 𝑟 ​ ( 𝑅 ) ≤ 2 (see [16], Prop. 8(i) or [23], Cor. 2.1.1). In fact, a Bézout ring 𝑅 is a Hermite ring iff 𝑠 ​ 𝑟 ​ ( 𝑅 ) ≤ 2 (see [23], Thm. 2.1.2). Bézout domains that have stable range 1 or 1.5 were also studied in [18] and respectively [20], [21] and [1].

For Hermite rings, Parts I and II get summarized in an example and a theorem as follows.

Example 1.1.

(1) For Hermite rings 𝑅 , simply extendable and extendable properties on a matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) are equivalent (see [3], Thm. 1.6).

(2) For Hermite rings 𝑅 that are Π 2 rings, simply extendable, extendable and determinant liftable properties on a matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) are equivalent (see [4], Cor. 1.5).

(3) For rings 𝑅 such that each zero determinant matrix in 𝕄 2 ​ ( 𝑅 ) is non-full (e.g., if 𝑅 is a product of pre-Schreier domains), extendable and weakly determinant liftable properties on a matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) are equivalent (see [4], Thm. 1.7).

(4) From parts (1) to (3) it follows that for Hermite rings 𝑅 with the property that zero determinant matrices in 𝕄 2 ​ ( 𝑅 ) are non-full, the simply extendable, extendable, determinant liftable, and weakly determinant liftable properties on a matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) are equivalent (cf. [4], Cor. 1.8).

Theorem 1.2.

Let 𝑅 be a Hermite ring. Then the following statements are equivalent.

(1) The ring R is an EDR.

(2) The ring R is an S ​ E 2 ring.

(3) The ring R is an E 2 ring.

(4) For each a ∈ R , R / R ​ a is a Π 2 ring (equivalently, every projective R / R ​ a -module of rank 1 generated by 2 elements is free).

(5) Each matrix in U ​ m ​ ( 𝕄 2 ​ ( R ) ) is determinant liftable and R is a Π 2 ring.

If all zero determinant matrices in 𝕄 2 ​ ( 𝑅 ) are non-full, then these five statements are also equivalent to:

(6) Each matrix in U ​ m ​ ( 𝕄 2 ​ ( R ) ) is weakly determinant liftable.

Referring to Theorem 1.2, see [3], Cor. 1.8 for ( 1 ) ⇔ ( 2 ) ⇔ ( 3 ) , see [3], Cor. 4.2 for ( 3 ) ⇔ ( 4 ) , see [3], Thm. 1.4 for the equivalence of statement ( 4 ) , and see [4], Thm. 1.4 for ( 2 ) ⇔ ( 5 ) . See [4], Thm. 1.7 for ( 3 ) ⇔ ( 6 ) .

The equivalence ( 1 ) ⇔ ( 4 ) of Theorem 1.2 can be viewed as a more practical way to check that a Hermite ring is an EDR than [22], Thm. 2.1 (which does not restrict to only 2 generators) or [25], Thms. 3 and 5 (which work with finitely generated projective modules over all quotients of 𝑅 ).

Recall that 𝑅 is a Hermite ring iff for all ( 𝑚 , 𝑛 ) ∈ ℕ 2 and each 𝑚 × 𝑛 matrix 𝐵 with entries in 𝑅 , there exists ( 𝑀 , 𝑁 ) ∈ 𝐺 ​ 𝐿 𝑚 ​ ( 𝑅 ) × 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) such that 𝑀 ​ 𝐵 and 𝐵 ​ 𝑁 are both lower (equivalently, upper) triangular (see [8], Thm. 3). Thus, in Theorem 1.2(4) or (5) it suffices to consider only upper triangular matrices, and it is part of the goals of this paper to show that in fact one can restrict to specific upper triangular matrices (e.g., see Propositions 2.2 and 2.3).

Example 1.3.

For ( 𝑎 , 𝑏 , 𝑐 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 3 ) , let 𝐴 := [ 𝑎

𝑏

0

𝑐 ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) .

(1) The matrix 𝐴 is determinant liftable iff there exists ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ 𝑅 4 such that 𝑎 ​ 𝑥 + 𝑏 ​ 𝑦 + 𝑐 ​ 𝑤

1 and 𝑥 ​ 𝑤

𝑦 ​ 𝑧 (see [4], Thm. 1.2).

(2) If 𝑅 is arbitrary (resp. if 𝑁 ​ ( 𝑅 )

0 ), then the matrix 𝐴 is weakly determinant if (resp. iff) there exists ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ 𝑅 4 such that 1 − 𝑎 ​ 𝑥 − 𝑏 ​ 𝑦 − 𝑐 ​ 𝑤 + 𝑎 ​ 𝑐 ​ ( 𝑥 ​ 𝑤 − 𝑦 ​ 𝑧 )

0 by Thm. 6.1(2) (resp. by Thm. 6.1(2) and (3)). As 1 − 𝑎 ​ 𝑥 − 𝑏 ​ 𝑦 − 𝑐 ​ 𝑤 + 𝑎 ​ 𝑐 ​ ( 𝑥 ​ 𝑤 − 𝑦 ​ 𝑧 ) can be rewritten as ( 1 − 𝑎 ​ 𝑥 ) ​ ( 1 − 𝑐 ​ 𝑤 ) − 𝑦 ​ ( 𝑏 + 𝑎 ​ 𝑐 ​ 𝑧 ) (cf. [4], Eq. (I)), it follows that 𝐴 is weakly determinant liftable if (resp. iff) there exists ( 𝑥 , 𝑧 , 𝑤 ) ∈ 𝑅 3 such that 𝑏 + 𝑎 ​ 𝑐 ​ 𝑧 divides ( 1 − 𝑎 ​ 𝑥 ) ​ ( 1 − 𝑐 ​ 𝑤 ) .

Example 1.3(2) and the equivalence ( 1 ) ⇔ ( 5 ) of Theorem 1.2 (with (5) stated for upper triangular matrices in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) ) imply directly the following result.

Corollary 1.4.

Assume 𝑅 is a Hermite ring and a Π 2 ring. Then 𝑅 is an EDR iff for each ( 𝑎 , 𝑏 , 𝑐 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 3 ) there exists ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ 𝑅 4 such that 𝑎 ​ 𝑥 + 𝑏 ​ 𝑦 + 𝑐 ​ 𝑤

1 and 𝑥 ​ 𝑤

𝑦 ​ 𝑧 (equivalently, there exists ( 𝑥 , 𝑦 , 𝑤 ) ∈ 𝑅 3 such that 𝑎 ​ 𝑥 + 𝑏 ​ 𝑦 + 𝑐 ​ 𝑤

1 and 𝑥 ​ 𝑤 ∈ 𝑅 ​ 𝑦 ).

For Hermite rings, to get necessary and/or sufficient conditions to be EDRs that involve units, one would like to get unit interpretations that would characterize the inequality 𝑠 ​ 𝑟 ​ ( 𝑅 ) ≤ 2 in some way similar to [3], Prop. 2.4. As a progress in this direction we introduce using units the following class of rings.

Definition 1.5.

We say that 𝑅 is a 𝑈 2 ring if for each ( 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and 𝑐 ∈ 𝑅 , the natural product homomorphism

𝜐 𝑎 , 𝑏 , 𝑐 : 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑐 ) × 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑏 ​ 𝑐 ) → 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 )

is surjective, i.e., the functorial commutative diagram

(II) 𝑈 ​ ( 𝑅 / ( 𝑅 ​ 𝑎 ​ 𝑐 ∩ 𝑅 ​ 𝑏 ​ 𝑐 ) )     𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑐 ) 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑏 ​ 𝑐 )     𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) ,

whose arrows are natural reductions, is not only a pullback but also a pushout.

For ( 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , if ( 𝑎 ′ , 𝑏 ′ ) ∈ 𝑅 2 is such that 𝑎 ​ 𝑎 ′ + 𝑏 ​ 𝑏 ′

1 , then 𝜐 ( 𝑎 ​ 𝑎 ′ , 1 − 𝑎 ​ 𝑎 ′ , 𝑐 ) factors through 𝜐 𝑎 , 𝑏 , 𝑐 . Hence 𝑅 is a 𝑈 2 ring iff 𝜐 𝑎 , 1 − 𝑎 , 𝑐 is surjective for all ( 𝑎 , 𝑐 ) ∈ 𝑅 2 . The fact that a unit in 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) does or does not belong to Im ​ ( 𝜐 𝑎 , 1 − 𝑎 , 𝑐 ) relates to a certain zero determinant matrix being or not being non-full (see Example 3.1).

Example 1.6.

Assume 𝑎 ​ 𝑠 ​ 𝑟 ​ ( 𝑅 )

1 , i.e., 𝑅 has almost stable range 1 . We show that 𝑅 is a 𝑈 2 ring. Let ( 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and 𝑐 ∈ 𝑅 . If 𝑐 ∈ 𝐽 ​ ( 𝑅 ) , then 𝑈 ​ ( 𝑅 ) surjects onto 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) . If 𝑐 ∉ 𝐽 ​ ( 𝑅 ) , then, as 𝑐 is a linear combination of 𝑎 ​ 𝑐 and 𝑏 ​ 𝑐 , either 𝑎 ​ 𝑐 ∉ 𝐽 ​ ( 𝑅 ) or 𝑏 ​ 𝑐 ∉ 𝐽 ​ ( 𝑅 ) . Hence either 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑐 ) → 𝑈 ​ ( 𝑅 / ( 𝑅 ​ 𝑎 ​ 𝑐 + 𝑅 ​ 𝑐 ) )

𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) or 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑏 ​ 𝑐 ) → 𝑈 ​ ( 𝑅 / ( 𝑅 ​ 𝑏 ​ 𝑐 + 𝑅 ​ 𝑐 ) )

𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) is a surjective homomorphisms (see [3], Prop. 2.4(3)). We incur that 𝜐 𝑎 , 𝑏 , 𝑐 is surjective. So 𝑅 is a 𝑈 2 ring.

The next two examples underline the relevance of pre-Schreier rings.

Example 1.7.

Let 𝑅 be such that each zero determinant matrix 𝐵 ∈ 𝕄 2 ​ ( 𝑅 ) is non-full. We show that 𝑅 is a pre-Schreier ring. Let ( 𝑥 , 𝑦 , 𝑧 ) ∈ 𝑅 3 be such that 𝑥 divides 𝑦 ​ 𝑧 . Let 𝑤 ∈ 𝑅 be such that 𝑥 ​ 𝑤

𝑦 ​ 𝑧 . Let 𝐶 := [ 𝑥

𝑦

𝑧
𝑤 ] ∈ 𝕄 2 ​ ( 𝑅 ) . As det ( 𝐶 )

0 , there exists ( 𝑙 , 𝑚 , 𝑜 , 𝑞 ) ∈ 𝑅 4 such that 𝐶

[ 𝑙

𝑚 ] ​ [ 𝑜
𝑞 ] . Hence 𝑥

𝑙 ​ 𝑜 with 𝑙 dividing 𝑦

𝑙 ​ 𝑞 and 𝑜 dividing 𝑧

𝑚 ​ 𝑜 . Thus 𝑅 is a pre-Schreier ring.

Example 1.8.

Assume 𝑁 ​ ( 𝑅 )

0 . Let 𝐵 ∈ 𝕄 2 ​ ( 𝑅 ) with det ( 𝐵 )

0 . We check that if 𝐵 admits diagonal reduction, then 𝐵 is non-full. Let 𝑀 , 𝑁 ∈ 𝐺 ​ 𝐿 2 ​ ( 𝑅 ) be such that 𝑀 ​ 𝐵 ​ 𝑁

[ 𝑎

0

0

𝑎 ​ 𝑏 ] with ( 𝑎 , 𝑏 ) ∈ 𝑅 2 . As

𝑎 2 ​ 𝑏

det ( 𝑀 ​ 𝐵 ​ 𝑁 )

det ( 𝑀 ) ​ det ( 𝐵 ) ​ det ( 𝑁 )

0 ,

it follows that ( 𝑎 ​ 𝑏 ) 2

0 and thus, as 𝑁 ​ ( 𝑅 )

0 , we have 𝑎 ​ 𝑏

0 . Therefore, 𝐵

𝑀 − 1 ​ [ 𝑎

0

0
0 ] ​ 𝑁 − 1

𝑀 − 1 ​ [ 𝑎

0 ] ​ [ 1

0 ] ​ 𝑁 − 1 is non-full.

Examples 1.7 and 1.8 imply directly the following result.

Corollary 1.9.

If 𝑅 is an EDR with 𝑁 ​ ( 𝑅 )

0 , then 𝑅 is a pre-Schreier ring.

Section 3 proves the following theorem.

Theorem 1.10.

We consider the following conditions.

(1) The ring R is a pre-Schreier ring and for each ( a , b , c ) ∈ U ​ m ​ ( R 3 ) there exists ( x , y , z , w ) ∈ R 4 such that 1 − a ​ x − b ​ y − c ​ w + a ​ c ​ ( x ​ w − y ​ z )

0 .

(2) The ring R is a Π 2 ring and each upper triangular matrix in U ​ m ​ ( 𝕄 2 ​ ( R ) ) is simply extendable (e.g., R is an S ​ E 2 ring).

(3) Each upper triangular matrix in U ​ m ​ ( 𝕄 2 ​ ( R ) ) is extendable and R is an integral domain.

If any one of these conditions holds, then 𝑅 is a 𝑈 2 ring.

Example 1.11.

Assume each upper triangular matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is weakly determinant liftable (e.g., this holds if 𝑅 is a 𝑊 ​ 𝐽 2 , 1 ring in the sense of [4], Def. 1.10(1), see [4], Thm. 1.11(1)). If 𝑅 is also a pre-Schreier ring with 𝑁 ​ ( 𝑅 )

0 , then condition (1) of Theorem 1.10 holds by Example 1.3(2), hence 𝑅 is a 𝑈 2 ring.

Each Dedekind domain is a 𝑈 2 ring by Example 1.6. Hence Dedekind domains which are not principal ideal domains are 𝑈 2 rings which (by [3], Thm. 1.7(4)) are not Π 2 rings.

In Section 4 we prove the following ‘units supplement’ to Theorem 1.2.

Theorem 1.12.

For a Hermite ring 𝑅 the following statements are equivalent.

(1) The ring R is an EDR.

(2) The ring R / N ​ ( R ) is a pre-Schreier ring and each matrix in U ​ m ​ ( 𝕄 2 ​ ( R ) ) is weakly determinant liftable.

(3) The ring R is a U 2 ring.

(4) Given ( ( a , b ) , ( c , d ) ) ∈ U ​ m ​ ( R 2 ) 2 , there exists t ∈ R such that we can factor d + c ​ t

d 1 ​ d 2 with ( ( a , d 1 ) , ( b , d 2 ) ) ∈ U ​ m ​ ( R 2 ) 2 .

(5) Given ( a , d ) ∈ R 2 and c ∈ 1 + R ​ d , there exists t ∈ R such that we can factor d + c ​ t

d 1 ​ d 2 with ( ( a , d 1 ) , ( 1 − a , d 2 ) ) ∈ U ​ m ​ ( R 2 ) 2 .

Example 1.13.

Let 𝑅 be a Hermite ring which is not an EDR (see [6], Sect. 4 and Ex. 4.11 and [2], Prop. 8). From Theorems 1.2 and 1.12 it follows that 𝑅 is neither an 𝐸 2 ring nor a 𝑈 2 ring and there exists 𝑎 ∈ 𝑅 such that the Hermite ring 𝑅 / 𝑅 ​ 𝑎 is not a Π 2 ring.

Theorem 1.12 implies directly the following result.1

Corollary 1.14.

Assume 𝑅 is a Hermite ring and a pre-Schreier ring such that 𝑁 ​ ( 𝑅 )

0 . Then 𝑅 is an EDR iff each matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is weakly determinant liftable and iff for each ( 𝑎 , 𝑏 , 𝑐 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 3 ) there exists ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ 𝑅 4 such that

(III) ( 1 − 𝑎 ​ 𝑥 ) ​ ( 1 − 𝑐 ​ 𝑤 )

𝑦 ​ ( 𝑏 + 𝑎 ​ 𝑐 ​ 𝑧 ) .

If 𝑎 ​ 𝑠 ​ 𝑟 ​ ( 𝑅 )

1 , then 𝑅 is an 𝑈 2 ring (see Example 1.6); so McGovern’s theorem that each Hermite ring 𝑅 with 𝑎 ​ 𝑠 ​ 𝑟 ​ ( 𝑅 )

1 is an EDR (see [14], Thm. 3.7 or [3], Cor. 1.9(2)) also follows from the equivalence ( 1 ) ⇔ ( 3 ) of Theorem 1.12.

In Section 5 we prove the following ‘ Π 2 supplement’ to Theorem 1.2.

Theorem 1.15.

Assume 𝑅 is a Hermite ring with 𝑁 ​ ( 𝑅 )

0 . Then 𝑅 is an EDR iff for each 𝑎 ∈ 𝑅 the quotient ring 𝑅 / Ann 𝑅 ​ ( 𝑎 ) is a Π 2 ring and each matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is weakly determinant liftable (equivalently, and for every ( 𝑎 , 𝑏 , 𝑐 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 3 ) there exists ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ 𝑅 4 such that Equation (III) holds).

Lorenzini introduced 3 classes of rings that are ‘between’ Hermite rings and EDRs (see [12], Prop. 4.11). For the first class, called 𝐽 2 , 1 (see [12], Def. 4.6), it was proved in [4], Thm. 1.11(2) that each 𝐽 2 , 1 ring which is a Π 2 ring is an EDR. We recall the last 2 classes of rings (see [12], Def. 2.1) in a slightly different way.

Definition 1.16.

For 𝑛 ≥ 2 we say that 𝑅 is:

(1) a ( W ​ S ​ U ′ ) n (resp.  ( W ​ S ​ U ) n ) ring if for each A ∈ U ​ m ​ ( 𝕄 n ​ ( R ) ) there exists N ∈ S ​ L n ​ ( R ) (resp.  N ∈ G ​ L n ​ ( R ) ) such that A ​ N is symmetric;

(2) an ( S ​ U ′ ) n (resp.  ( S ​ U ) n ) ring if it is a Hermite ring and a ( W ​ S ​ U ′ ) n (resp.  ( W ​ S ​ U ) n ) ring.

In Definition 1.16(1), the existence of 𝑁 ∈ 𝑆 ​ 𝐿 𝑛 ​ ( 𝑅 ) (resp.  𝑁 ∈ 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) ) is equivalent to the existence of 𝑀 ∈ 𝑆 ​ 𝐿 𝑛 ​ ( 𝑅 ) (resp.  𝑀 ∈ 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) ) such that 𝑀 ​ 𝐴 is symmetric and is equivalent to the existence of a pair ( 𝑀 , 𝑁 ) ∈ 𝑆 ​ 𝐿 𝑛 ​ ( 𝑅 ) 2 (resp.  ( 𝑀 , 𝑁 ) ∈ 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) 2 ) such that 𝑀 ​ 𝐴 ​ 𝑁 is symmetric. This follows via conjugation from the fact that for each symmetric matrix 𝑂 ∈ 𝕄 𝑛 ​ ( 𝑅 ) and every 𝑀 ∈ 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) , 𝑀 ​ 𝑂 ​ 𝑀 𝑇 is symmetric.

Each ( 𝑊 ​ 𝑆 ​ 𝑈 ′ ) 𝑛 ring is a ( 𝑊 ​ 𝑆 ​ 𝑈 ) 𝑛 ring. As the extendable (resp. simply extendable) property of a unimodular 2 × 2 matrix depends only on the equivalence class of the matrix (see [3], Lem. 4.1(3)), it follows that a ( 𝑊 ​ 𝑆 ​ 𝑈 ) 2 ring is an 𝐸 2 (resp.  𝑆 ​ 𝐸 2 ring) iff each symmetric matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is extendable (resp. simply extendable). Section 6 proves basic properties of ( 𝑊 ​ 𝑆 ​ 𝑈 ′ ) 𝑛 and ( 𝑊 ​ 𝑆 ​ 𝑈 ) 𝑛 rings; e.g., our definition of ( 𝑆 ​ 𝑈 ′ ) 𝑛 or ( 𝑆 ​ 𝑈 ) 𝑛 rings is equivalent to the one in [12] by Proposition 6.1.

In Section 7 we prove the following theorem.

Theorem 1.17.

Let 𝑅 be a ( 𝑊 ​ 𝑆 ​ 𝑈 ) 2 ring. Then the following properties hold.

(1) For each ( a , b ) ∈ U ​ m ​ ( R 2 ) and c ∈ R , Coker ​ ( υ a , b , c ) is a Boolean group, i.e., we have an inclusion { x 2 | x ∈ U ​ ( R / R ​ c ) } ⊂ Im ​ ( υ a , b , c ) .

(2) If s ​ r ​ ( R ) ≤ 4 , then for each d ∈ R , every projective R / R ​ d -module of rank 1 generated by 2 elements is self-dual.

(3) Assume that each element of R is the square of an element of R (e.g., this holds if R is an integrally closed domain with an algebraically closed field of fractions or is a perfect ring of characteristic 2 ). Then R is a U 2 ring. If moreover R is a Hermite ring, then R is an EDR.

In Section 8 we prove the following theorem.

Theorem 1.18.

Let 𝑅 be a Hermite ring such that for each ( 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and 𝑐 ∈ 𝑅 , Coker ​ ( 𝜐 𝑎 , 𝑏 , 𝑐 ) is a Boolean group. Then for each 𝑑 ∈ 𝑅 , every projective 𝑅 / 𝑅 ​ 𝑑 -module of rank 1 generated by 2 elements is self-dual.

To connect with Pell-type equations and provide more examples of ( 𝑊 ​ 𝑆 ​ 𝑈 ) 2 rings that are EDRs, we first prove in Section 9 the following non-full Pell-type criterion.

Criterion 1.19.

Let 𝐴

[ 𝑎

𝑏

𝑏

𝑐 ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) be symmetric and with zero determinant. Then the following properties hold.

(1) The matrix A is simply extendable if there exists ( e , f ) ∈ R 2 such that a ​ e 2 − c ​ f 2 ∈ U ​ ( R ) , and the converse holds if R has characteristic 2 .

(2) Assume that R is a Hermite ring of characteristic 2 with N ​ ( R )

0 . If b is not a zero divisor, then A is simply extendable.

By combining Theorem 1.17(1) with Criterion 1.19, we obtain the following Pell-type criterion proved in Section 9.

Criterion 1.20.

Let 𝑅 be an ( 𝑆 ​ 𝑈 ) 2 ring. Then 𝑅 is an EDR if for all triples ( 𝑎 , 𝑏 , 𝑐 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 3 ) there exists ( 𝑒 , 𝑓 ) ∈ 𝑅 2 such that ( 𝑎 ​ 𝑒 2 − 𝑐 ​ 𝑓 2 , 𝑎 ​ 𝑐 − 𝑏 2 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , and the converse holds if 𝑅 has characteristic 2 .

We recall that, if the characteristic of 𝑅 if a prime, then its perfection 𝑅 𝑝 ​ 𝑒 ​ 𝑟 ​ 𝑓 is the inductive limit of the inductive system indexed by 𝑛 ∈ ℕ whose all transition homomorphisms are the Frobenius endomorphism of 𝑅 . If 𝑅 is a Hermite (or  Bézout) ring, then so is 𝑅 𝑝 ​ 𝑒 ​ 𝑟 ​ 𝑓 ≅ ( 𝑅 / 𝑁 ​ ( 𝑅 ) ) 𝑝 ​ 𝑒 ​ 𝑟 ​ 𝑓 .

Example 1.21.

Assume 𝑅 has characteristic 2 . If for all ( 𝑎 , 𝑐 ) ∈ 𝑅 2 , Coker ​ ( 𝜐 𝑎 , 1 − 𝑎 , 𝑐 ) is a Boolean group, then 𝑅 𝑝 ​ 𝑒 ​ 𝑟 ​ 𝑓 is a 𝑈 2 ring. From this and Theorem 1.17(3) it follows that perfections of ( 𝑊 ​ 𝑆 ​ 𝑈 ) 2 rings of characteristic 2 are 𝑈 2 rings. Thus perfections of ( 𝑆 ​ 𝑈 ) 2 rings of characteristic 2 are EDRs by (the implication ( 3 ) ⇒ ( 1 ) of) Theorem 1.12.

As an application of Corollary 1.14 in Section 10 we prove the following criterion.

Criterion 1.22.

For a Bézout domain 𝑅 the following statements are equivalent.

(1) The ring R is an EDD.

(2) For each ( a , b , s ) ∈ R 3 , there exists a pair ( q , r ) ∈ R 2 such that by defining y := r + s − a ​ s ​ q − b ​ q ​ r and t := 1 + q − a ​ q − b ​ r we have t ∈ R ​ y + R ​ a ​ t (equivalently, there exists ( y 1 , y 2 ) ∈ R 2 such that y

y 1 ​ y 2 , y 1 divides t , and ( a , y 2 ) ∈ U ​ m ​ ( R 2 ) ).

(3) For each ( a , b , s ) ∈ R 3 there exists a pair ( e , f ) ∈ U ​ m ​ ( R 2 ) such that ( ( a , e ) , ( b ​ e + a ​ f , 1 − b ​ s − a ) ) ∈ U ​ m ​ ( R 2 ) 2 .

E.g., statement (3) of Criterion 1.22 holds if ( 𝑎 , 𝑠 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) as we can take ( 𝑒 , 𝑓 ) := ( 𝑠 , 1 ) , if ( 1 − 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) as we can take ( 𝑒 , 𝑓 ) := ( 1 , 0 ) , or if there exists 𝑞 ∈ 𝑅 such that ( 𝑏 + 𝑎 ​ 𝑞 , 1 − 𝑏 ​ 𝑠 − 𝑎 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) as we can take ( 𝑒 , 𝑓 ) := ( 1 − 𝑎 , 𝑞 + 𝑏 ) . Hence directly from Criterion 1.22 we get the following consequence.

Corollary 1.23.

Assume 𝑅 is a Bézout domain with the property that for all ( 𝑎 , 𝑏 , 𝑠 ) ∈ 𝑅 3 with ( ( 𝑎 , 𝑠 ) , ( 𝑏 , 1 − 𝑎 ) ) ∉ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) 2 there exists 𝑞 ∈ 𝑅 such that ( 𝑏 + 𝑎 ​ 𝑞 , 1 − 𝑏 ​ 𝑠 − 𝑎 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) . Then 𝑅 is an EDD.

The implicit and explicit questions raised in the literature, such as, “Is a Bézout domain of finite Krull dimension [at least 2 ] an EDD?” (see [5], Ch. III, Probl. 5, p. 122), and, ‘What classes of Bézout domains which are not EDDs exist?’, remain unanswered. However, the above results reobtain or can be easily used to reobtain multiple other criteria existing in the literature of when a Hermite ring is an EDR. Moreover, the equivalence ( 1 ) ⇔ ( 4 ) of Theorem 1.12 was proved for Bézout domains in [19], Thm. 5 and a variant of it was proved for Hermite rings in [17], Prop. 2.9. Equation (III) generalizes and refines the equation one would get based on [17], Rm. 2.8. The last two references were reinterpreted in terms of neat stable range 1 (see [24], Def. 21) in [24], Thms. 31 and 33; see [15] for clean and neat rings.

2.Test matrices

In this section we introduce companion and universal test matrices for unimodular 2 × 2 matrices and prove a few properties of them that will be often used in the subsequent sections to streamline proofs. We begin with companion test matrices.

Definition 2.1.

Let 𝐴 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) . An upper triangular matrix 𝐵 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is called a companion test matrix for 𝐴 if there exists ( 𝑎 , 𝑏 , 𝑐 , 𝑎 ′ , 𝑐 ′ ) ∈ 𝑅 5 such that 𝐴 is equivalent to [ 𝑎

𝑏

0
𝑐 ] and 𝐵

[ 𝑎 ​ 𝑎 ′

𝑏

0

𝑐 ​ 𝑐 ′ ] .

Let 𝒫 be one of the 4 notions extendable, simply extendable, determinant liftable and weakly determinant liftable. Definition 2.1 is justified by the next proposition.

Proposition 2.2.

If 𝒫 is weakly determinant liftable we assume that 𝑁 ​ ( 𝑅 )

0 . If 𝐴 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) has a companion test matrix 𝐵 which is 𝒫 , then 𝐴 is 𝒫 .

Proof.

As 𝒫 property depends only on equivalence classes (see [3], Lem. 4.1(3) for the case of extendable and simply extendable notions), it suffices to show that if ( 𝑎 , 𝑏 , 𝑐 , 𝑎 ′ , 𝑐 ′ ) ∈ 𝑅 5 is such that 𝐵 := [ 𝑎 ​ 𝑎 ′

𝑏

0

𝑐 ​ 𝑐 ′ ] is 𝒫 , then 𝐶 := [ 𝑎

𝑏

0

𝑐 ] is 𝒫 .

First we assume that 𝒫 is extendable (resp. simply extendable). Let ( 𝑒 , 𝑔 ) ∈ 𝑅 2 (resp.  ( 𝑒 , 𝑔 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) ) be such that ( 𝑎 ​ 𝑎 ′ ​ 𝑒 , 𝑏 ​ 𝑒 + 𝑐 ​ 𝑐 ′ ​ 𝑔 , 𝑎 ​ 𝑎 ′ ​ 𝑐 ​ 𝑐 ′ ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 3 ) (resp.  ( 𝑎 ​ 𝑎 ′ ​ 𝑒 , 𝑏 ​ 𝑒 + 𝑐 ​ 𝑐 ′ ​ 𝑔 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) ) by [3], Cor. 4.7(1) (resp. [3], Thm. 4.3) applied to 𝐵 . Let 𝑓 := 𝑐 ′ ​ 𝑔 ∈ 𝑅 . As ( 𝑎 ​ 𝑎 ′ ​ 𝑒 , 𝑏 ​ 𝑒 + 𝑐 ​ 𝑓 , 𝑎 ​ 𝑎 ′ ​ 𝑐 ​ 𝑐 ′ ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 3 ) (resp. ( 𝑎 ​ 𝑎 ′ ​ 𝑒 , 𝑏 ​ 𝑒 + 𝑐 ​ 𝑓 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) ), it follows that ( 𝑎 ​ 𝑒 , 𝑏 ​ 𝑒 + 𝑐 ​ 𝑓 , 𝑎 ​ 𝑐 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 3 ) (resp.  ( 𝑎 ​ 𝑒 , 𝑏 ​ 𝑒 + 𝑐 ​ 𝑓 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) ). Thus 𝐶 is 𝒫 by [3], Cor. 4.7(1) (resp. [3], Thm. 4.3).

Next we assume that 𝒫 is determinant liftable (resp. weakly determinant liftable). From Example 1.3(1) (resp. 1.3(2)) applied to 𝐵 it follows that there exists a quadruple ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ 𝑅 4 such that 1 − 𝑎 ​ 𝑎 ′ ​ 𝑥 − 𝑏 ​ 𝑦 − 𝑐 ​ 𝑐 ′ ​ 𝑤

0 and 𝑥 ​ 𝑤

𝑦 ​ 𝑧 (resp. such that 1 − 𝑎 ​ 𝑎 ′ ​ 𝑥 − 𝑏 ​ 𝑦 − 𝑐 ​ 𝑐 ′ ​ 𝑤 + 𝑎 ​ 𝑎 ′ ​ 𝑐 ​ 𝑐 ′ ​ ( 𝑥 ​ 𝑤 − 𝑦 ​ 𝑧 )

0 ). For the quadruple ( 𝑥 ′ , 𝑦 ′ , 𝑧 ′ , 𝑤 ′ ) := ( 𝑎 ′ ​ 𝑥 , 𝑦 , 𝑎 ′ ​ 𝑐 ′ ​ 𝑧 , 𝑐 ′ ​ 𝑤 ) ∈ 𝑅 4 , we compute that 1 − 𝑎 ​ 𝑥 ′ − 𝑏 ​ 𝑦 ′ − 𝑐 ​ 𝑤 ′

0 and 𝑥 ′ ​ 𝑤 ′

𝑦 ′ ​ 𝑧 ′ (resp.  1 − 𝑎 ​ 𝑥 ′ − 𝑏 ​ 𝑦 ′ − 𝑐 ​ 𝑤 ′ + 𝑎 ​ 𝑐 ​ ( 𝑥 ′ ​ 𝑤 ′ − 𝑦 ′ ​ 𝑧 ′ )

0 ), so 𝐶 is 𝒫 by Example 1.3(1) (resp. 1.3(2)). ∎

For universal test matrices we consider two cases as follows.

Case 1: Hermite rings. In this and the next paragraph we assume that 𝑅 is a Hermite ring, i.e., for every pair ( 𝑝 , 𝑞 ) ∈ 𝑅 2 there exists ( 𝑟 , 𝑠 , 𝑡 ) ∈ 𝑅 3 such that 𝑝

𝑟 ​ 𝑠 , 𝑞

𝑟 ​ 𝑡 and ( 𝑠 , 𝑡 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) . If moreover 𝑅 is an integral domain (i.e., if 𝑅 is a Bézout domain), then 𝑟 is unique up to a multiplication with a unit of 𝑅 and is called the greatest common divisor of 𝑝 and 𝑞 and one writes 𝑟

gcd ⁡ ( 𝑥 , 𝑦 ) . Our convention is gcd ⁡ ( 0 , 0 )

0 (so that we can still write 0

0 ⋅ 1 with gcd ⁡ ( 1 , 1 )

1 ).

Let 𝐴 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) and 𝑀 ∈ 𝑆 ​ 𝐿 2 ​ ( 𝑅 ) be such that 𝐵 := 𝑀 ​ 𝐴

[ 𝑔

𝑢

0
ℎ ] is upper triangular. We write 𝑔

𝑎 ​ 𝑐 and ℎ

𝑏 ​ 𝑐 with ( 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and 𝑐 ∈ 𝑅 . As 𝐵 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) and 𝑅 ​ 𝑔 + 𝑅 ​ ℎ

𝑅 ​ 𝑐 , we have ( 𝑐 , 𝑢 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) . Let ( 𝑎 ′ , 𝑏 ′ , 𝑐 ′ , 𝑢 ′ ) ∈ 𝑅 4 be such that 𝑎 ​ 𝑎 ′ + 𝑏 ​ 𝑏 ′

𝑐 ​ 𝑐 ′ + 𝑢 ​ 𝑢 ′

1 . For 𝑑 ∈ 𝑅 , 𝐴 and 𝐵 are equivalent to

𝐶 := [ 1

𝑑 ​ 𝑏 ′

0

1 ] ​ 𝐵 ​ [ 1

𝑑 ​ 𝑎 ′

0
1 ]

[ 𝑎 ​ 𝑐

𝑢 + 𝑐 ​ 𝑑 ​ ( 𝑎 ​ 𝑎 ′ + 𝑏 ​ 𝑏 ′ )

0
𝑏 ​ 𝑐 ]

[ 𝑎 ​ 𝑐

𝑢 + 𝑐 ​ 𝑑

0

𝑏 ​ 𝑐 ] .

Here the role of 𝑢 + 𝑐 ​ 𝑑 is that of an arbitrary element of 𝑅 whose reduction modulo 𝑅 ​ 𝑐 is the ‘fixed’ unit 𝑢 + 𝑅 ​ 𝑐 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) . Let 𝐷 𝑎 ′ , 𝑏 ′ , 𝑑 := [ 𝑎 ​ 𝑎 ′ ​ 𝑐 ​ 𝑐 ′

𝑢 + 𝑐 ​ 𝑑

0

𝑏 ​ 𝑏 ′ ​ 𝑐 ​ 𝑐 ′ ] . If 𝐷 𝑎 ′ , 𝑏 ′ , 𝑑 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) , then 𝐷 𝑎 ′ , 𝑏 ′ , 𝑑 is a companion test matrix for 𝐴 (or 𝐵 ). Note that 𝐷 𝑎 ′ , 𝑏 ′ , 0 is in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) and is the image of the ‘first universal test matrix for Hermite rings’

𝒟 := [ 𝑥 ​ ( 1 − 𝑦 ​ 𝑧 )

𝑦

0

( 1 − 𝑥 ) ​ ( 1 − 𝑦 ​ 𝑧 ) ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( ℤ ​ [ 𝑥 , 𝑦 , 𝑧 ] ) )

via the ring homomorphism ℤ ​ [ 𝑥 , 𝑦 , 𝑧 ] → 𝑅 that maps 𝑥 , 𝑦 , and 𝑧 to 𝑎 ​ 𝑎 ′ , 𝑢 , and 𝑢 ′ (respectively); so 1 − 𝑥 maps to 𝑏 ​ 𝑏 ′

1 − 𝑎 ​ 𝑎 ′ and 1 − 𝑦 ​ 𝑧 maps to 1 − 𝑢 ​ 𝑢 ′

𝑐 ​ 𝑐 ′ . We also note that 𝒟 is equivalent to the matrix

ℰ := [ 1

0

𝑧 ​ ( 𝑥 − 1 ) ​ ( 1 − 𝑦 ​ 𝑧 )

1 ] ​ 𝒟 ​ [ 1

0

𝑥 ​ 𝑧
1 ]

[ 𝑥

𝑦

0

( 1 − 𝑥 ) ​ ( 1 − 𝑦 ​ 𝑧 ) 2 ]

which is the image of the ‘second universal test matrix for Hermite rings’

ℱ := [ 𝑥

𝑦

0

( 1 − 𝑥 ) ​ ( 1 − 𝑦 ​ 𝑧 ) ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( ℤ ​ [ 𝑥 , 𝑦 , 𝑧 ] ) )

via the endomorphism of ℤ ​ [ 𝑥 , 𝑦 , 𝑧 ] that fixes 𝑥 and 𝑦 and maps 𝑧 to 2 ​ 𝑧 − 𝑦 ​ 𝑧 2 . See Corollary 4.2 for the usage of universal test matrix in this paragraph.

Case 2: all rings. The ‘universal test upper triangular matrix for all rings’ is

𝒢 := [ 𝑥

𝑦

0

1 − 𝑥 − 𝑦 ​ 𝑧 ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( ℤ ​ [ 𝑥 , 𝑦 , 𝑧 ] ) ) .

Proposition 2.3.

If 𝒫 is weakly determinant liftable we assume that 𝑁 ​ ( 𝑅 )

0 . Then the following properties hold.

(1) Each upper triangular matrix in U ​ m ​ ( 𝕄 2 ​ ( R ) ) is 𝒫 iff for each homomorphism ϕ : ℤ ​ [ x , y , z ] → R , the image of 𝒢 ∈ U ​ m ​ ( 𝕄 2 ​ ( ℤ ​ [ x , y , z ] ) ) in U ​ m ​ ( 𝕄 2 ​ ( R ) ) via ϕ is 𝒫 .

(2) Each zero determinant upper triangular matrix in U ​ m ​ ( 𝕄 2 ​ ( R ) ) is 𝒫 iff for each homomorphism ϕ : ℤ ​ [ x , y , z ] → R with x ​ ( 1 − x − y ​ z ) ∈ Ker ​ ( ϕ ) , the image of 𝒢 ∈ U ​ m ​ ( 𝕄 2 ​ ( ℤ ​ [ x , y , z ] ) ) in U ​ m ​ ( 𝕄 2 ​ ( R ) ) via ϕ is 𝒫 .

Proof.

The ‘only if’ parts are clear. For the ‘if’ part of (1), we consider an upper triangular matrix 𝐴

[ 𝑎

𝑏

0
𝑐 ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) . Let ( 𝑎 ′ , 𝑏 ′ , 𝑐 ′ ) ∈ 𝑅 3 be such that 𝑎 ​ 𝑎 ′ + 𝑏 ​ 𝑏 ′ + 𝑐 ​ 𝑐 ′

1 . Let 𝜙 : ℤ ​ [ 𝑥 , 𝑦 , 𝑧 ] → 𝑅 be the homomorphism that maps 𝑥 , 𝑦 and 𝑧 to 𝑎 ​ 𝑎 ′ , 𝑏 and 𝑏 ′ (respectively). The image [ 𝑎 ​ 𝑎 ′

𝑏

0
𝑐 ​ 𝑐 ′ ] of 𝒢 via 𝜙 is 𝒫 , so 𝐴 is 𝒫 by Proposition 2.2. The ‘if’ part of (2) is proved similarly as 𝑎 ​ 𝑐

0 implies 𝑎 ​ 𝑎 ′ ​ 𝑐 ​ 𝑐 ′

0 and 𝑥 ​ ( 1 − 𝑥 − 𝑦 ​ 𝑧 ) ∈ Ker ​ ( 𝜙 ) . ∎

3.Proof of Theorem 1.10

Let ( 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and 𝑐 ∈ 𝑅 . It suffices to prove that 𝜐 𝑎 , 𝑏 , 𝑐 is surjective if any one of conditions (1) to (3) of Theorem 1.10 holds.

For a fixed unit 𝑢 ¯ ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) , let 𝑢 ∈ 𝑅 be such that 𝑢 ¯

𝑢 + 𝑅 ​ 𝑐 and consider the matrix 𝐴

[ 𝑎 ​ 𝑐

𝑢

0

𝑏 ​ 𝑐 ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) .

Assume condition (1) of Theorem 1.10 holds. Thus 𝑅 is a pre-Schreier ring and there exists ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ 𝑅 4 such that ( 1 − 𝑎 ​ 𝑐 ​ 𝑥 ) ​ ( 1 − 𝑏 ​ 𝑐 ​ 𝑤 )

𝑦 ​ ( 𝑢 + 𝑎 ​ 𝑏 ​ 𝑐 2 ​ 𝑧 ) by Example 1.3(2). As 𝑢 + 𝑎 ​ 𝑏 ​ 𝑐 2 ​ 𝑧 divides ( 1 − 𝑎 ​ 𝑐 ​ 𝑥 ) ​ ( 1 − 𝑏 ​ 𝑐 ​ 𝑤 ) and 𝑅 is a pre-Schreier ring, we only use the existence of a nonuple ( 𝑥 , 𝑦 , 𝑧 1 , 𝑧 2 , 𝑤 , 𝑞 , 𝑟 , 𝑠 , 𝑡 ) ∈ 𝑅 9 , e.g., 𝜈 := ( 𝑥 , 𝑦 , 𝑏 ​ 𝑐 ​ 𝑧 , 0 , 𝑤 , 1 , 1 , 1 , 1 ) ∈ 𝑅 9 , such that ( 𝑎 ​ 𝑐 , 𝑠 ​ 𝑞 ) , ( 𝑏 ​ 𝑐 , 𝑟 ​ 𝑡 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and we can factor 𝑞 ​ 𝑟 ​ 𝑢 + 𝑎 ​ 𝑐 ​ 𝑧 1 + 𝑏 ​ 𝑐 ​ 𝑧 2

𝑢 𝑎 ​ 𝑢 𝑏 , with 𝑢 𝑎 ∈ 𝑅 dividing 𝑠 − 𝑎 ​ 𝑐 ​ 𝑥 and 𝑢 𝑏 ∈ 𝑅 dividing 𝑡 − 𝑏 ​ 𝑐 ​ 𝑤 . Hence the two units ( 𝑞 + 𝑅 ​ 𝑎 ​ 𝑐 ) − 1 ⋅ ( 𝑢 𝑎 + 𝑅 ​ 𝑎 ​ 𝑐 ) ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑐 ) and ( 𝑟 + 𝑅 ​ 𝑏 ​ 𝑐 ) − 1 ⋅ ( 𝑢 𝑏 + 𝑅 ​ 𝑏 ​ 𝑐 ) ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑏 ​ 𝑐 ) are such that the product of their images in 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) is 𝑢 ¯ . Thus 𝜐 𝑎 , 𝑏 , 𝑐 is surjective.

Assume condition (2) of Theorem 1.10 holds. Thus 𝑅 is a Π 2 ring and 𝐴 is simply extendable. Hence 𝐴 is determinant liftable by Diagram (I). Let the quadruple ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ 𝑅 4 be such that ( 1 − 𝑎 ​ 𝑐 ​ 𝑥 ) ​ ( 1 − 𝑏 ​ 𝑐 ​ 𝑤 )

𝑦 ​ ( 𝑢 + 𝑎 ​ 𝑏 ​ 𝑐 2 ​ 𝑧 ) and 𝑥 ​ 𝑤

𝑦 ​ 𝑧 (see Example 1.3(1)). If there exists 𝔪 ∈ Max  ​ 𝑅 which contains 1 − 𝑎 ​ 𝑐 ​ 𝑥 , 1 − 𝑏 ​ 𝑐 ​ 𝑤 and 𝑦 , then it follows that 𝑥 ​ 𝑤 + 𝔪

𝑦 ​ 𝑧 + 𝔪 ∈ 𝑅 / 𝔪 is both zero and non-zero, a contradiction. Thus ( 1 − 𝑎 ​ 𝑐 ​ 𝑥 , 1 − 𝑏 ​ 𝑐 ​ 𝑤 , 𝑦 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 3 ) . We only use the existence of a nonuple ( 𝑥 , 𝑦 , 𝑧 1 , 𝑧 2 , 𝑤 , 𝑞 , 𝑟 , 𝑠 , 𝑡 ) ∈ 𝑅 9 , e.g., the 𝜈 above, such that

𝐷 := [ 𝑠 − 𝑎 ​ 𝑐 ​ 𝑥

𝑞 ​ 𝑟 ​ 𝑢 + 𝑎 ​ 𝑐 ​ 𝑧 1 + 𝑏 ​ 𝑐 ​ 𝑧 2

𝑦

𝑡 − 𝑏 ​ 𝑐 ​ 𝑤 ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) )

has zero determinant, ( 𝑎 ​ 𝑐 , 𝑠 ​ 𝑞 ) , ( 𝑏 ​ 𝑐 , 𝑟 ​ 𝑡 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and ( 𝑠 − 𝑎 ​ 𝑐 ​ 𝑥 , 𝑡 − 𝑏 ​ 𝑐 ​ 𝑤 , 𝑦 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 3 ) . As 𝑅 is a Π 2 ring, 𝐷 is non-full (see [3], Thm. 1.4) and therefore we can factor 𝑞 ​ 𝑟 ​ 𝑢 + 𝑎 ​ 𝑐 ​ 𝑧 1 + 𝑏 ​ 𝑐 ​ 𝑧 2

𝑢 𝑎 ​ 𝑢 𝑏 , with 𝑢 𝑎 ∈ 𝑅 dividing 𝑠 − 𝑎 ​ 𝑐 ​ 𝑥 and 𝑢 𝑏 ∈ 𝑅 dividing 𝑡 − 𝑏 ​ 𝑐 ​ 𝑤 . As in the previous paragraph we argue that 𝜐 𝑎 , 𝑏 , 𝑐 is surjective.

Assume condition (3) of Theorem 1.10 holds. Thus 𝑅 is an integral domain and 𝐴 is extendable. To prove that 𝜐 𝑎 , 𝑏 , 𝑐 is surjective we can assume that 𝑎 ​ 𝑏 ​ 𝑐 ≠ 0 . As 𝐴 is extendable, its reduction modulo 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 2 is non-full by [3], Prop. 5.1(2). So the system of congruences

𝑥 ​ 𝑦 ≡ 𝑎 ​ 𝑐 ​ ( mod ​ 𝑎 ​ 𝑏 ​ 𝑐 2 ) , 𝑥 ​ 𝑤 ≡ 𝑢 ​ ( mod ​ 𝑎 ​ 𝑏 ​ 𝑐 2 ) , 𝑦 ​ 𝑧 ≡ 0 ​ ( mod ​ 𝑎 ​ 𝑏 ​ 𝑐 2 ) , 𝑧 ​ 𝑤 ≡ 𝑏 ​ 𝑐 ​ ( mod ​ 𝑎 ​ 𝑏 ​ 𝑐 2 )

has a solution ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ 𝑅 4 . As ( 𝑐 , 𝑢 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , from the second congruence it follows that ( 𝑐 , 𝑥 ​ 𝑤 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) . From this and the first and fourth congruences, it follows that there exists ( 𝑦 ′ , 𝑧 ′ ) ∈ 𝑅 2 such that 𝑦

𝑐 ​ 𝑦 ′ and 𝑧

𝑐 ​ 𝑧 ′ . So, as 𝑅 is an integral domain, the above first system of congruences is equivalent to a second one

𝑥 ​ 𝑦 ′ ≡ 𝑎 ​ ( mod ​ 𝑎 ​ 𝑏 ​ 𝑐 ) , 𝑥 ​ 𝑤 ≡ 𝑢 ​ ( mod ​ 𝑎 ​ 𝑏 ​ 𝑐 2 ) , 𝑦 ′ ​ 𝑧 ′ ≡ 0 ​ ( mod ​ 𝑎 ​ 𝑏 ) , 𝑧 ′ ​ 𝑤 ≡ 𝑏 ​ ( mod ​ 𝑎 ​ 𝑏 ​ 𝑐 ) .

As ( 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , the last two congruences imply firstly that ( 𝑎 , 𝑧 ′ ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and secondly that there exists 𝑦 ′′ ∈ 𝑅 such that 𝑦 ′

𝑎 ​ 𝑦 ′′ . Thus the second system of congruences is equivalent to a third one

𝑥 ​ 𝑦 ′′ ≡ 1 ​ ( mod ​ 𝑏 ​ 𝑐 ) , 𝑥 ​ 𝑤 ≡ 𝑢 ​ ( mod ​ 𝑎 ​ 𝑏 ​ 𝑐 2 ) , 𝑦 ′′ ​ 𝑧 ′ ≡ 0 ​ ( mod ​ 𝑏 ) , 𝑧 ′ ​ 𝑤 ≡ 𝑏 ​ ( mod ​ 𝑎 ​ 𝑏 ​ 𝑐 ) .

From the first and third congruences it follows firstly that ( 𝑏 , 𝑦 ′′ ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and secondly that there exists 𝑧 ′′ ∈ 𝑅 such that 𝑧 ′

𝑏 ​ 𝑧 ′′ . Thus the third system of congruences is equivalent to a fourth one that has only three congruences

𝑥 ​ 𝑦 ′′ ≡ 1 ​ ( mod ​ 𝑏 ​ 𝑐 ) , 𝑥 ​ 𝑤 ≡ 𝑢 ​ ( mod ​ 𝑎 ​ 𝑏 ​ 𝑐 2 ) , 𝑧 ′′ ​ 𝑤 ≡ 1 ​ ( mod ​ 𝑎 ​ 𝑐 ) .

Thus 𝑥 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑏 ​ 𝑐 ) and 𝑤 + 𝑅 ​ 𝑎 ​ 𝑐 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑐 ) are such that the product of their images in 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) is 𝑢 ¯ . Thus 𝜐 𝑎 , 𝑏 , 𝑐 is surjective.

Hence Theorem 1.10 holds.

Example 3.1.

For ( 𝑎 , 𝑐 , 𝑢 ) ∈ 𝑅 3 , let 𝐴 := [ 𝑎 ​ 𝑐

𝑢

0
( 1 − 𝑎 ) ​ 𝑐 ] ∈ 𝕄 2 ​ ( 𝑅 ) . We assume that 𝑅 ​ 𝑎 ​ 𝑐 ∩ 𝑅 ​ ( 1 − 𝑎 ) ​ 𝑐

0 and ( 𝑐 , 𝑢 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) ; so 𝑎 ​ ( 1 − 𝑎 ) ​ 𝑐

0 and 𝐴 is unimodular with zero determinant. We consider the ideals 𝐽 1 := 𝑅 ​ 𝑎 ​ 𝑐 and 𝐽 2 := 𝑅 ​ ( 1 − 𝑎 ) ​ 𝑐 of 𝑅 . For 𝑖 ∈ { 1 , 2 } , let 𝑅 𝑖 := 𝑅 / 𝐽 𝑖 , 𝑐 𝑖 := 𝑐 + 𝐽 𝑖 ∈ 𝑅 𝑖 , and 𝑢 𝑖 := 𝑢 + 𝐽 𝑖 ∈ 𝑅 𝑖 . The reduction of 𝐴 modulo 𝐽 1 is 𝐴 1 := [ 0

𝑢 1

0

𝑐 1 ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 1 ) ) and modulo 𝐽 2 is 𝐴 2 := [ 𝑐 2

𝑢 2

0
0 ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 2 ) ) . Both 𝐴 1 and 𝐴 2 are non-full. For 𝑖 ∈ { 1 , 2 } , to solve the matrix equation 𝐴 𝑖

[ 𝑙 𝑖

𝑚 𝑖 ] ​ [ 𝑜 𝑖
𝑞 𝑖 ] in 𝕄 2 ​ ( 𝑅 𝑖 ) is the same as solving the system of equations 𝑙 𝑖 ​ 𝑜 𝑖

𝑐 𝑖 ​ 𝛿 𝑖 ​ 2 , 𝑙 𝑖 ​ 𝑞 𝑖

𝑢 𝑖 , 𝑚 𝑖 ​ 𝑜 𝑖

0 , 𝑚 𝑖 ​ 𝑞 𝑖

𝑐 𝑖 ​ 𝛿 𝑖 ​ 1 with ( 𝑙 𝑖 , 𝑚 𝑖 , 𝑜 𝑖 , 𝑞 𝑖 ) ∈ 𝑅 𝑖 4 , where 𝛿 𝑖 ​ 1 and 𝛿 𝑖 ​ 2 are Kronecker deltas. As ( 𝑐 , 𝑢 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , we have 𝑞 1 ∈ 𝑈 ​ ( 𝑅 1 ) and 𝑙 2 ∈ 𝑈 ​ ( 𝑅 2 ) . So the solution sets are

{ ( 𝑢 1 ​ 𝑞 1 − 1 , 𝑐 1 ​ 𝑞 1 − 1 , 0 , 𝑞 1 ) | 𝑞 1 ∈ 𝑈 ​ ( 𝑅 1 ) } ⊂ 𝑅 1 4

for 𝑖

1 and

{ ( 𝑙 2 , 0 , 𝑐 2 ​ 𝑙 2 − 1 , 𝑢 2 ​ 𝑙 2 − 1 ) | 𝑙 2 ∈ 𝑈 ​ ( 𝑅 2 ) } ⊂ 𝑅 2 4

for 𝑖

2 . As we have an identity 𝑅 ​ 𝑐

𝐽 1 + 𝐽 2 , the matrix 𝐴 is non-full iff there exists a pair ( 𝑞 1 , 𝑙 2 ) ∈ 𝑈 ​ ( 𝑅 1 ) × 𝑈 ​ ( 𝑅 2 ) such that the images of 𝑢 1 ​ 𝑞 1 − 1 and 𝑙 2 in 𝑅 / 𝑅 ​ 𝑐 are equal, i.e., 𝑢 + 𝑅 ​ 𝑐

𝜐 𝑎 , 1 − 𝑎 , 𝑐 ​ ( 𝑞 1 , 𝑙 2 ) , and hence iff 𝑢 + 𝑅 ​ 𝑐 ∈ Im ​ ( 𝜐 𝑎 , 1 − 𝑎 , 𝑐 ) .

4.Proof of Theorem 1.12 and applications

To prove Theorem 1.12, let 𝑆 := 𝑅 / 𝑁 ​ ( 𝑅 ) .

We show that ( 1 ) ⇒ ( 2 ) . As 𝑅 is an EDR, each matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is determinant liftable by (the implication ( 1 ) ⇒ ( 5 ) of) Theorem 1.2 and so it is weakly determinant liftable. Also, from [9], Thm. 3 it follows that 𝑆 is an EDR and hence it is a pre-Schreier ring by Example 1.8. So ( 1 ) ⇒ ( 2 ) holds.

We show that ( 2 ) ⇒ ( 3 ) if 𝑁 ​ ( 𝑅 )

0 . As 𝑁 ​ ( 𝑅 )

0 , from the ‘iff’ part of Example 1.3(2) it follows that condition (1) of Theorem 1.10 holds, hence 𝑅 is a 𝑈 2 ring.

We have ( 3 ) ⇔ ( 4 ) , as for ( ( 𝑎 , 𝑏 ) , ( 𝑐 , 𝑑 ) ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) 2 , ( 4 ) only translates what means that the unit 𝑑 + 𝑅 ​ 𝑐 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) is in the image of 𝜐 𝑎 , 𝑏 , 𝑐 .

Clearly, ( 4 ) ⇒ ( 5 ) by taking 𝑏 := 1 − 𝑎 .

To check that ( 5 ) ⇒ ( 4 ) , let ( ( 𝑎 , 𝑏 ) , ( 𝑐 , 𝑑 ) ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) 2 . If the quadruple ( 𝑒 , 𝑓 , 𝑠 , 𝑡 ) ∈ 𝑅 4 is such that 𝑎 ​ 𝑒 + 𝑏 ​ 𝑓

𝑐 ​ 𝑠 + 𝑑 ​ 𝑡

1 , then by applying (5) to ( 𝑎 ​ 𝑒 , 𝑑 ) and 𝑠 ​ 𝑐 ∈ 1 + 𝑅 ​ 𝑑 it follows that there exists 𝑡 1 ∈ 𝑅 such that we can factor 𝑑 + 𝑡 1 ​ 𝑠 ​ 𝑐

𝑑 1 ​ 𝑑 2 with ( 𝑎 ​ 𝑒 , 𝑑 1 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and ( 1 − 𝑎 ​ 𝑒 , 𝑑 2 )

( 𝑏 ​ 𝑓 , 𝑑 2 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and thus ( ( 𝑎 , 𝑑 1 ) , ( 𝑏 , 𝑑 2 ) ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and statement (4) holds by taking 𝑡 := 𝑡 1 ​ 𝑠 .

We show that ( 3 ) ⇒ ( 1 ) . As 𝑅 is an EDR iff it is an 𝐸 2 ring by (the implication ( 1 ) ⇒ ( 3 ) of) Theorem 1.2, it suffices to show that if 𝑅 is a 𝑈 2 ring, then each 𝐴 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is extendable. Up to matrix equivalence (see [3], Lem. 4.1(3)), we can assume that 𝐴 is upper triangular, and hence we can write 𝐴

[ 𝑎 ​ 𝑐

𝑢

0
𝑏 ​ 𝑐 ] where ( 𝑎 , 𝑏 , 𝑐 , 𝑢 ) ∈ 𝑅 4 is such that det ( 𝐴 )

𝑎 ​ 𝑏 ​ 𝑐 2 , ( 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and 𝑢 in 𝑅 is (up to matrix equivalence, see Case 1 of Section 2) an arbitrary representative of a fixed unit 𝑢 + 𝑅 ​ 𝑐 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) . As 𝑅 is a 𝑈 2 ring, based on the arbitrariness part we can assume there exists ( 𝑑 , 𝑒 ) ∈ 𝑅 2 such that 𝑢

𝑑 ​ 𝑒 , 𝑑 + 𝑅 ​ 𝑎 ​ 𝑐 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑐 ) and 𝑒 + 𝑅 ​ 𝑏 ​ 𝑐 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑏 ​ 𝑐 ) . Let 𝑑 ′ , 𝑒 ′ ∈ 𝑅 be such that 𝑑 ′ + 𝑅 ​ 𝑎 ​ 𝑐 is the inverse of 𝑑 + 𝑅 ​ 𝑎 ​ 𝑐 and 𝑒 ′ + 𝑅 ​ 𝑎 ​ 𝑐 is the inverse of 𝑒 + 𝑅 ​ 𝑏 ​ 𝑐 . Let ( 𝑓 , 𝑔 ) ∈ 𝑅 2 be such that 𝑑 ​ 𝑑 ′

1 + 𝑎 ​ 𝑐 ​ 𝑓 and 𝑒 ​ 𝑒 ′

1 + 𝑏 ​ 𝑐 ​ 𝑔 . The non-full matrix

𝐵 := [ 𝑎 ​ 𝑐 ​ ( 1 + 𝑏 ​ 𝑐 ​ 𝑔 )

𝑢

𝑎 ​ 𝑏 ​ 𝑐 2 ​ 𝑑 ′ ​ 𝑒 ′
𝑏 ​ 𝑐 ​ ( 1 + 𝑎 ​ 𝑐 ​ 𝑓 ) ]

[ 𝑎 ​ 𝑐 ​ 𝑒 ​ 𝑒 ′

𝑑 ​ 𝑒

𝑎 ​ 𝑏 ​ 𝑐 2 ​ 𝑑 ′ ​ 𝑒 ′

𝑏 ​ 𝑐 ​ 𝑑 ​ 𝑑 ′ ] ∈ 𝕄 2 ​ ( 𝑅 )

is congruent to 𝐴 modulo 𝑅 ​ det ( 𝐴 ) . So 𝐴 modulo 𝑅 ​ det ( 𝐴 ) is non-full, thus 𝐴 is extendable by [3], Prop. 5.1(2).

We are left to show that ( 2 ) ⇒ ( 3 ) in general (i.e., without assuming that 𝑁 ​ ( 𝑅 )

0 ). For 𝐴 ¯ ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑆 ) ) , let 𝐴 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) be such that its reduction modulo 𝑁 ​ ( 𝑅 ) is 𝐴 ¯ . As 𝐴 is weakly determinant liftable, there exists 𝐶 ∈ 𝕄 2 ​ ( 𝑅 ) congruent to 𝐴 modulo 𝑅 ​ det ( 𝐴 ) and det ( 𝐶 )

0 . As the reduction 𝐶 ¯ ∈ 𝕄 2 ​ ( 𝑆 ) of 𝐶 modulo 𝑁 ​ ( 𝑅 ) is congruent to 𝐴 ¯ modulo 𝑆 ​ det ( 𝐴 ¯ ) and has zero determinant, 𝐴 ¯ is weakly determinant liftable. So 𝑆 has the same properties as 𝑅 (note that 𝑁 ​ ( 𝑆 )

0 and 𝑆

𝑆 / 𝑁 ​ ( 𝑆 )

𝑅 / 𝑁 ​ ( 𝑅 ) is a pre-Schreier ring). As we proved that ( 2 ) ⇒ ( 3 ) if 𝑁 ​ ( 𝑅 )

0 , it follows that 𝑆 is a 𝑈 2 ring. As ( 3 ) ⇒ ( 1 ) , we incur that 𝑆 is an EDR. It is well-known that this implies that 𝑅 is an EDR: e.g., see [9], Thm. 3; this also follows from Theorem 1.2 via the fact that 𝐴 is (simply) extendable iff 𝐴 ¯ is so, as one can easily check based on [3], Thm. 4.3 or Cor. 4.7. As 𝑅 is an EDR, it is an 𝑆 ​ 𝐸 2 ring by (the implication ( 1 ) ⇒ ( 2 ) of) Theorem 1.2 and hence a 𝑈 2 ring by Theorem 1.10(2). Thus ( 2 ) ⇒ ( 3 ) .

Hence Theorem 1.12 holds.

Corollary 4.1.

Assume 𝑅 is a Bézout domain such that for each ( 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and 𝑐 ∈ 𝑅 , the image of the functorial homomorphism

𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 ) → 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑏 ) ≅ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ) × 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑏 )

is the product of the images of the following two functorial homomorphisms 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 ) → 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ) and 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 ) → 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑏 ) . Then 𝑅 is an EDD.

Proof.

Based on (the implication ( 3 ) ⇒ ( 1 ) of) Theorem 1.12, it suffices to show that 𝑅 is a 𝑈 2 ring. Given a unit 𝑢 + 𝑅 ​ 𝑐 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) and ( 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , let 𝑎 ′ := gcd ⁡ ( 𝑎 , 𝑐 ) and 𝑏 ′ := gcd ⁡ ( 𝑏 , 𝑐 ) . As gcd ⁡ ( 𝑎 , 𝑏 )

1 , 𝑎 ′ ​ 𝑏 ′ divides 𝑐 and hence there exists 𝑐 ′ ∈ 𝑅 such that 𝑐

𝑎 ′ ​ 𝑏 ′ ​ 𝑐 ′ . From our hypothesis applied to ( 𝑎 ′ , 𝑏 ′ ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and 𝑐 ′ ∈ 𝑅 it follows that there exist units 𝑒 + 𝑅 ​ 𝑐 , 𝑓 + 𝑅 ​ 𝑐 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) such that their images in 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ′ ) × 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑏 ′ ) are ( 𝑢 + 𝑅 ​ 𝑎 ′ , 1 + 𝑅 ​ 𝑏 ′ ) and ( 1 + 𝑅 ​ 𝑎 ′ , 𝑢 + 𝑅 ​ 𝑏 ′ ) (respectively). Then ( 𝑢 + 𝑅 ​ 𝑎 , 1 + 𝑅 ​ 𝑏 ) ∈ 𝑅 / 𝑅 ​ 𝑎 ​ 𝑏 ≅ 𝑅 / 𝑅 ​ 𝑎 × 𝑅 / 𝑅 ​ 𝑏 and 𝑒 + 𝑅 ​ 𝑐 ∈ 𝑅 / 𝑅 ​ 𝑐 (resp.  ( 1 + 𝑅 ​ 𝑎 , 𝑢 + 𝑅 ​ 𝑏 ) ∈ 𝑅 / 𝑅 ​ 𝑎 ​ 𝑏 ≅ 𝑅 / 𝑅 ​ 𝑎 × 𝑅 / 𝑅 ​ 𝑏 and 𝑓 + 𝑅 ​ 𝑐 ∈ 𝑅 / 𝑅 ​ 𝑐 ) map to the same element in 𝑅 / 𝑅 ​ 𝑎 ′ ​ 𝑏 ′ ≅ 𝑅 / 𝑅 ​ 𝑎 ′ × 𝑅 / 𝑅 ​ 𝑏 ′ and hence, as 𝑎 ​ 𝑏 ​ 𝑐 ′ is the least common multiple of 𝑐 and 𝑎 ​ 𝑏 , there exists a pair ( 𝑔 , ℎ ) ∈ 𝑅 2 such that 𝑔 + 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 ′ (resp.  ℎ + 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 ′ ) reduces to both of them. As we have 𝑔 + 𝑅 ​ 𝑏 ​ 𝑐 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑏 ​ 𝑐 ) and ℎ + 𝑅 ​ 𝑎 ​ 𝑐 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑐 ) , it follows that ( 𝑒 + 𝑅 ​ 𝑐 ) ​ ( 𝑓 + 𝑅 ​ 𝑐 ) ∈ Im ​ ( 𝜐 𝑎 , 𝑏 , 𝑐 ) . Hence to show that 𝜐 𝑎 , 𝑏 , 𝑐 is surjective, by replacing 𝑢 + 𝑅 ​ 𝑐 with ( 𝑢 + 𝑅 ​ 𝑐 ) ​ [ ( 𝑒 + 𝑅 ​ 𝑐 ) ​ ( 𝑓 + 𝑅 ​ 𝑐 ) ] − 1 we can assume that 𝑢 − 1 ∈ 𝑅 ​ 𝑎 ′ ​ 𝑏 ′ , i.e., the images of 𝑢 + 𝑅 ​ 𝑐 and 1 + 𝑅 ​ 𝑎 ​ 𝑏 in 𝑅 ​ 𝑎 ′ ​ 𝑏 ′ are equal, which implies that 𝑢 + 𝑅 ​ 𝑐 ∈ Im ​ ( 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 ′ ) → 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) ) . As 𝑎 ​ 𝑏 ​ 𝑐 ′ + 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 ∈ 𝑅 / 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 has square 0 , the homomorphism 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 ) → 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 ′ ) is surjective and thus 𝑢 + 𝑅 ​ 𝑐 ∈ Im ​ ( 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 ) → 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) ) ⊂ Im ​ ( 𝜐 𝑎 , 𝑏 , 𝑐 ) . So Im ​ ( 𝜐 𝑎 , 𝑏 , 𝑐 ) is surjective. Thus 𝑅 is a 𝑈 2 ring.∎

Corollary 4.2.

Let 𝑅 be a Hermite ring. Then 𝑅 is an EDR iff for all homomorphisms 𝜙 : ℤ ​ [ 𝑥 , 𝑦 , 𝑧 ] → 𝑅 , the image of 𝒟 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( ℤ ​ [ 𝑥 , 𝑦 , 𝑧 ] ) ) (equivalently, ℱ ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( ℤ ​ [ 𝑥 , 𝑦 , 𝑧 ] ) ) , see Case 1 of Section 2) in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) via 𝜙 is extendable.

Proof.

The ‘only if’ part is obvious. To prove the ‘if’ part, based on (the implication ( 3 ) ⇒ ( 1 ) of) Theorem 1.12, it suffices to prove that 𝑅 is a 𝑈 2 ring, i.e., for each ( 𝑎 , 𝑐 ) ∈ 𝑅 2 and every ( 𝑢 , 𝑐 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , we have 𝑢 + 𝑅 ​ 𝑐 ∈ Im ​ ( 𝜐 𝑎 , 1 − 𝑎 , 𝑐 ) . Let ( 𝑠 , 𝑡 ) ∈ 𝑅 2 be such that 𝑐 ​ 𝑠 + 𝑢 ​ 𝑡

1 . As 𝑢 + 𝑅 ​ 𝑐 ​ 𝑠 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ​ 𝑠 ) , it suffices to prove that 𝑢 + 𝑅 ​ 𝑐 ​ 𝑠 ∈ Im ​ ( 𝜐 𝑎 , 1 − 𝑎 , 𝑐 ​ 𝑠 ) . Thus by replacing 𝑐 with 𝑐 ​ 𝑠 we can assume that 𝑐

1 − 𝑢 ​ 𝑡 . Hence 𝐴

[ 𝑎 ​ ( 1 − 𝑢 ​ 𝑡 )

𝑢

0

( 1 − 𝑎 ) ​ ( 1 − 𝑢 ​ 𝑡 ) ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is the image of 𝒟 via the homomorphism 𝜙 : ℤ ​ [ 𝑥 , 𝑦 , 𝑧 ] → 𝑅 that maps 𝑥 , 𝑦 , and 𝑧 to 𝑎 , 𝑢 , and 𝑡 (respectively) and so 𝐴 is extendable. As 𝒟 is equivalent to ℰ (see Case 1 of Section 2), it follows that 𝐴 is equivalent to the matrix 𝐵 := [ 𝑎

𝑢

0
( 1 − 𝑎 ) ​ ( 1 − 𝑢 ​ 𝑤 ) ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) , where 𝑤 := 2 ​ 𝑡 − 𝑢 ​ 𝑡 2 . From [3], Lem. 4.1(3) it follows that 𝐵 is extendable and hence its reduction modulo det ( 𝐵 )

𝑅 ​ 𝑎 ​ ( 1 − 𝑎 ) ​ ( 1 − 𝑢 ​ 𝑤 )

𝑅 ​ 𝑎 ​ ( 1 − 𝑎 ) ​ 𝑐 2 is non-full (see [3], Prop. 5.1(2)). We denote by ∗ ¯ the reduction modulo 𝑅 ​ 𝑎 ​ ( 1 − 𝑎 ) ​ 𝑐 2 of ∗ , where ∗ is either 𝑅 or an element of 𝑅 . As 𝐵 is non-full, we have a product decomposition 𝑢 ¯

𝑢 1 ¯ ​ 𝑢 ¯ 2 with 𝑢 ¯ 1

𝑢 1 + 𝑅 ​ 𝑎 ​ ( 1 − 𝑎 ) ​ 𝑐 2 ∈ 𝑅 ¯ dividing 𝑎 ¯ and 𝑢 ¯ 2

𝑢 2 + 𝑅 ​ 𝑎 ​ ( 1 − 𝑎 ) ​ 𝑐 2 ∈ 𝑅 ¯ dividing ( 1 − 𝑎 ¯ ) ​ ( 1 − 𝑤 ¯ ​ 𝑢 ¯ ) and hence also 1 − 𝑎 ¯ ; here ( 𝑢 1 , 𝑢 2 ) ∈ 𝑅 2 . Thus there exists a triple ( 𝑑 , 𝑒 , 𝑓 ) ∈ 𝑅 3 such that 𝑢 1 ​ 𝑢 2

𝑢 + 𝑎 ​ ( 1 − 𝑎 ) ​ 𝑐 2 ​ 𝑑 , 𝑢 1 divides 𝑎 + 𝑎 ​ ( 1 − 𝑎 ) ​ 𝑐 2 ​ 𝑒 and 𝑢 2 divides 1 − 𝑎 + 𝑎 ​ ( 1 − 𝑎 ) ​ 𝑐 2 ​ 𝑓 . It follows that ( ( 𝑢 1 , ( 1 − 𝑎 ) ​ 𝑐 ) , ( 𝑢 2 , 𝑎 ​ 𝑐 ) ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) 2 , and thus we have an identity 𝑢 + 𝑅 ​ 𝑐

𝑢 1 ​ 𝑢 2 + 𝑅 ​ 𝑐

𝜐 𝑎 , 1 − 𝑎 , 𝑐 ​ ( 𝑢 2 + 𝑅 ​ 𝑎 ​ 𝑐 , 𝑢 1 + 𝑅 ​ ( 1 − 𝑎 ) ​ 𝑐 ) . ∎

Corollary 4.3.

Let 𝑅 be a Bézout domain. Then 𝑅 is an EDD iff for all triples ( 𝑎 , 𝑢 , 𝑡 ) ∈ 𝑅 3 with 𝑢 ≠ 0 there exists ( 𝑠 , 𝑙 , 𝑧 ) ∈ 𝑅 3 such that

(IV) ( 1 − 𝑢 ​ 𝑠 − 𝑎 ​ 𝑙 ) 2 + 𝑙 − 𝑢 ​ 𝑠 ​ 𝑙 − 𝑎 ​ 𝑙 2 − ( 𝑠 + 𝑡 − 𝑢 ​ 𝑠 ​ 𝑡 ) ​ 𝑧

0 .

Proof.

The Hermite ring 𝑅 is an EDR iff for all triples ( 𝑎 , 𝑢 , 𝑡 ) ∈ 𝑅 3 , the matrix 𝐴

[ 𝑎 ​ ( 1 − 𝑢 ​ 𝑡 )

𝑢

0

( 1 − 𝑎 ) ​ ( 1 − 𝑢 ​ 𝑡 ) ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is extendable (see Corollary 4.2), equivalently it is determinant liftable (see Example 1.1(1)), and hence equivalently (see [4], Thm. 1.2) there exists ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ 𝑅 4 such that

1 − 𝑎 ​ ( 1 − 𝑢 ​ 𝑡 ) ​ 𝑥 − 𝑢 ​ 𝑦 − ( 1 − 𝑎 ) ​ ( 1 − 𝑢 ​ 𝑡 ) ​ 𝑤

0

𝑥 ​ 𝑤 − 𝑦 ​ 𝑧 .

If 𝑢

0 , then 𝐴 is diagonal and hence simply extendable (see [3], Ex. 4.9(3)). Thus we can assume 𝑢 ≠ 0 . The equation 1 − 𝑎 ​ ( 1 − 𝑢 ​ 𝑡 ) ​ 𝑥 − 𝑢 ​ 𝑦 − ( 1 − 𝑎 ) ​ ( 1 − 𝑢 ​ 𝑡 ) ​ 𝑤

0 can be rewritten as 1 − 𝑢 ​ 𝑦

( 1 − 𝑢 ​ 𝑡 ) ​ [ 𝑎 ​ 𝑥 − ( 1 − 𝑎 ) ​ 𝑤 ] and working modulo 𝑅 ​ 𝑢 with 𝑢 ≠ 0 it follows that its general solution is 𝑎 ​ 𝑥 + ( 1 − 𝑎 ) ​ 𝑤

1 − 𝑢 ​ 𝑠 and 𝑦

𝑠 + 𝑡 − 𝑢 ​ 𝑠 ​ 𝑡 with 𝑠 ∈ 𝑅 . As 𝑅 is an integral domain, the general solution of 𝑎 ​ 𝑥 + ( 1 − 𝑎 ) ​ 𝑤

1 − 𝑢 ​ 𝑠 is 𝑥

1 − 𝑢 ​ 𝑠 + ( 1 − 𝑎 ) ​ 𝑙 and 𝑤

1 − 𝑢 ​ 𝑠 − 𝑎 ​ 𝑙 with 𝑙 ∈ 𝑅 and the equation 𝑥 ​ 𝑤 − 𝑦 ​ 𝑧

0 becomes [ 1 − 𝑢 ​ 𝑠 + ( 1 − 𝑎 ) ​ 𝑙 ] ​ ( 1 − 𝑢 ​ 𝑠 − 𝑎 ​ 𝑙 ) − ( 𝑠 + 𝑡 − 𝑢 ​ 𝑠 ​ 𝑡 ) ​ 𝑧

0 and the corollary follows. ∎

5.Proof of Theorem 1.15

We first prove two general lemmas.

Lemma 5.1.

Let ( 𝑑 , 𝑒 ) ∈ 𝑅 2 be such that 𝑅 ​ 𝑑

𝑅 ​ 𝑒 . Then there exists 𝑁 ∈ 𝑆 ​ 𝐿 2 ​ ( 𝑅 ) such that 𝑁 ​ [ 𝑑

0

0
0 ]

[ 𝑒

0

0

0 ] (so [ 𝑑

0

0

0 ] and [ 𝑒

0

0

0 ] are equivalent).

Proof.

Let ( 𝑢 , 𝑣 ) ∈ 𝑅 2 be such that ( 𝑑 , 𝑒 )

( 𝑒 ​ 𝑢 , 𝑑 ​ 𝑣 ) ; thus 1 − 𝑢 ​ 𝑣 ∈ Ann 𝑅 ​ ( 𝑑 ) . For 𝑁 := [ 𝑣

− 1

1 − 𝑢 ​ 𝑣

𝑢 ] ∈ 𝑆 ​ 𝐿 2 ​ ( 𝑅 ) , one computes 𝑁 ​ [ 𝑑

0

0
0 ]

[ 𝑒

0

0

0 ] . ∎

Lemma 5.2.

Assume 𝑁 ​ ( 𝑅 )

0 . Let 𝑒 ∈ 𝑅 be such that for 𝑚 ∈ { 2 , 4 } the reduction function 𝑈 𝑚 ( 𝑅 𝑚 ) → 𝑈 𝑚 ( ( 𝑅 / Ann 𝑅 ( 𝑒 ) ) 𝑚 ) is surjective2. Then all zero determinant matrices in 𝑒 ​ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) admit diagonal reduction iff 𝑅 / Ann 𝑅 ​ ( 𝑒 ) is a Π 2 ring.

Proof.

For ∗ ∈ 𝑅 ∪ 𝕄 2 ( 𝑅 ) , let ∗ ¯ be the reduction of ∗ modulo Ann 𝑅 ​ ( 𝑒 ) . For two matrices 𝑀

[ 𝑚 11

𝑚 12

𝑚 21
𝑚 22 ] and 𝐿

[ 𝑙 11

𝑙 12

𝑙 21

𝑙 22 ] in 𝕄 2 ​ ( 𝑅 ) , the product

[ 𝑚 11

𝑚 12

𝑚 21

𝑚 22 ] ​ [ 𝑒

0

0

0 ] ​ [ 𝑙 11

𝑙 12

𝑙 21
𝑙 22 ]

[ 𝑒 ​ 𝑚 11 ​ 𝑙 11

𝑒 ​ 𝑚 11 ​ 𝑙 12

𝑒 ​ 𝑚 21 ​ 𝑙 11

𝑒 ​ 𝑚 21 ​ 𝑙 12 ]

depends only on the first column of 𝑀 and the first row of 𝐿 , i.e., it depends only on the quadruple ( 𝑚 11 , 𝑚 21 , 𝑙 11 , 𝑙 12 ) ∈ 𝑅 4 . Given such a quadruple in 𝑅 4 , we have ( ( 𝑚 11 , 𝑚 21 ) , ( 𝑙 11 , 𝑙 12 ) ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) 2 iff there exists ( 𝑚 12 , 𝑚 22 , 𝑙 21 , 𝑙 22 ) ∈ 𝑅 4 such that the resulting matrices 𝑀 and 𝐿 are in 𝑆 ​ 𝐿 2 ​ ( 𝑅 ) .

Let 𝐴

[ 𝑎

𝑏

𝑐
𝑑 ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) be such that det ( 𝑒 ​ 𝐴 )

0 ; so 𝑒 2 ​ det ( 𝐴 )

0 and as 𝑁 ​ ( 𝑅 )

0 it follows that 𝑒 ​ det ( 𝐴 )

0 . Thus det ( 𝐴 ) ∈ Ann 𝑅 ​ ( 𝑒 ) and hence det ( 𝐴 ¯ )

0 . The matrix 𝑒 ​ 𝐴 admits diagonal reduction iff it is equivalent to ( 𝑓 , 0 ) (see Example 1.8) with 𝑅 ​ 𝑒

𝑅 ​ 𝑓 and hence iff it is equivalent to ( 𝑒 , 0 ) by Lemma 5.1. From this and the previous paragraph it follows that 𝑒 ​ 𝐴 admits diagonal reduction iff there exists ( ( 𝑚 11 , 𝑚 21 ) , ( 𝑙 11 , 𝑙 12 ) ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) 2 such that ( 𝑒 ​ 𝑎 , 𝑒 ​ 𝑏 , 𝑒 ​ 𝑐 , 𝑒 ​ 𝑑 )

( 𝑒 ​ 𝑚 11 ​ 𝑙 11 , 𝑒 ​ 𝑚 11 ​ 𝑙 12 , 𝑒 ​ 𝑚 21 ​ 𝑙 11 , 𝑒 ​ 𝑚 21 ​ 𝑙 12 ) and, as the map 𝑈 𝑚 ( 𝑅 2 ) → 𝑈 𝑚 ( ( 𝑅 / Ann 𝑅 ( 𝑒 ) ) 2 ) is surjective, iff 𝐴 ¯

[ 𝑚 11 ¯

𝑚 21 ¯ ] ​ [ 𝑙 11 ¯
𝑙 12 ¯ ] is non-full. Each 𝐶 ∈ 𝑈 𝑚 ( 𝕄 2 ( 𝑅 / Ann 𝑅 ( 𝑒 ) ) ) is 𝐵 ¯ for some 𝐵 ∈ 𝑈 𝑚 ( 𝕄 2 ( 𝑅 ) ) by hypotheses; if det ( 𝐶 )

0 , then det ( 𝐵 ) ∈ Ann 𝑅 ​ ( 𝑒 ) and hence det ( 𝑒 ​ 𝐵 )

0 . Thus we conclude that all zero determinant matrices in 𝑒 ​ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) admit diagonal reduction iff all zero determinant matrices in 𝑈 𝑚 ( 𝕄 2 ( 𝑅 / Ann 𝑅 ( 𝑒 ) ) ) are non-full (equivalently, are simply extendable by [3], Prop. 5.1(1)) and iff 𝑅 / Ann 𝑅 ​ ( 𝑒 ) is a Π 2 ring.∎

We prove Theorem 1.15. If 𝑅 is an EDR then all zero determinant matrices in 𝕄 2 ​ ( 𝑅 ) admit diagonal reduction. If for each 𝑎 ∈ 𝑅 , 𝑅 / 𝐴 ​ 𝑛 ​ 𝑛 𝑅 ​ ( 𝑎 ) is a Π 2 ring, then all zero determinant matrices in 𝑅 ​ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) )

𝕄 2 ​ ( 𝑅 ) admit diagonal reduction by Lemma 5.2. As 𝑁 ​ ( 𝑅 )

0 , in the last two sentences we can replace ‘admit diagonal reduction’ with ‘are non-full’ by Example 1.8. Thus to prove Theorem 1.15 we can assume that the Hermite 𝑅 is such that all zero determinant matrices in 𝕄 2 ​ ( 𝑅 ) are non-full, in which case Theorem 1.15 follows from (the equivalence ( 1 ) ⇔ ( 6 ) of) Theorem 1.2.

6.On ( 𝑊 ​ 𝑆 ​ 𝑈 ′ ) 𝑛 and ( 𝑊 ​ 𝑆 ​ 𝑈 ) 𝑛 rings

For 𝐹 ∈ 𝕄 𝑛 ​ ( 𝑅 ) , let 𝐹 𝑇 ∈ 𝕄 𝑛 ​ ( 𝑅 ) be its transpose and let Ker 𝐹 and Im 𝐹 be the kernel and the image (respectively) of the 𝑅 -linear map 𝐿 𝐹 : 𝑅 𝑛 → 𝑅 𝑛 defined by it. Note that, as usual, the elements of the domain or codomain of 𝐿 𝐹 are written as column vectors.

Proposition 6.1.

A ring 𝑅 is an ( 𝑆 ​ 𝑈 ′ ) 𝑛 (resp.  ( 𝑆 ​ 𝑈 ) 𝑛 ) ring in the sense of Definition 1.16(2) iff it is so in the sense of [12], Def. 2.1.

Proof.

For the ‘only if’ part, let 𝐵 ∈ 𝕄 𝑛 ​ ( 𝑅 ) . As 𝑅 is a Hermite ring, there exist 𝑒 ∈ 𝑅 and 𝐶 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 𝑛 ​ ( 𝑅 ) ) such that 𝐵

𝑒 ​ 𝐶 . As 𝑅 is an ( 𝑆 ​ 𝑈 ′ ) 𝑛 (resp.  ( 𝑆 ​ 𝑈 ) 𝑛 ) ring, there exists 𝑁 ∈ 𝑆 ​ 𝐿 𝑛 ​ ( 𝑅 ) (resp.  𝑁 ∈ 𝐺 𝐿 𝑛 ( 𝑅 ) ) such that 𝐶 ​ 𝑁 is symmetric. Hence 𝐵 ​ 𝑁

𝑒 ​ ( 𝐶 ​ 𝑁 ) is symmetric, thus 𝑅 is an ( 𝑆 ​ 𝑈 ′ ) 𝑛 (resp.  ( 𝑆 ​ 𝑈 ) 𝑛 ) ring in the sense of [12], Def. 2.1.

For the ‘if’ part, clearly 𝑅 is a ( 𝑊 ​ 𝑆 ​ 𝑈 ′ ) 𝑛 (resp.  ( 𝑊 ​ 𝑆 ​ 𝑈 ) 𝑛 ) ring and it is a Hermite ring by [12], Prop. 3.1. ∎

Directly from Proposition 6.1 and [12], Prop. 3.3 we get the following result.

Corollary 6.2.

Let 𝑛 ∈ ℕ ∖ { 1 , 2 } . Assume 𝑅 is an integral domain and an ( 𝑆 ​ 𝑈 ) 𝑛 ring. Then 𝑅 is an ( 𝑆 ​ 𝑈 ) 𝑚 ring for each 𝑚 ∈ { 2 , … , 𝑛 − 1 } .

As in [12], Sect. 2, if 𝑈 ​ ( 𝑅 ) is a 2 -divisible abelian group, 𝑅 is a ( 𝑊 ​ 𝑆 ​ 𝑈 ′ ) 𝑛 ring iff it is a ( 𝑊 ​ 𝑆 ​ 𝑈 ) 𝑛 ring. For ( 𝑊 ​ 𝑆 ​ 𝑈 ) 𝑛 rings we have the following general result.

Theorem 6.3.

Assume 𝑅 is a ( 𝑊 ​ 𝑆 ​ 𝑈 ) 𝑛 ring. Then the following properties hold.

(1) Each unimodular matrix A ∈ U ​ m ​ ( 𝕄 n ​ ( R ) ) is equivalent to its transpose A T .

(2) If n

2 , then every projective R -module of rank 1 and generated by 2 elements is self-dual.

(3) If s ​ r ​ ( R ) ≤ n 2 , then for each d ∈ R , R / R ​ d is a ( W ​ S ​ U ) n ring.

Proof.

If 𝑁 ∈ 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) is such that 𝐴 ​ 𝑁 is symmetric, i.e., 𝐴 ​ 𝑁

𝑁 𝑇 ​ 𝐴 𝑇 then ( 𝑁 𝑇 ) − 1 ​ 𝐴 ​ 𝑁

𝐴 𝑇 and hence part (1) holds.

To check part (2), let 𝑃 be a projective 𝑅 -module of rank 1 generated by 2 elements. We consider an isomorphism 𝑃 ⊕ 𝑄 ≅ 𝑅 2 to be viewed as an identification. Taking determinants it follows that 𝑄 is the dual of 𝑃 . Let 𝐵 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) be such that Ker 𝐵

𝑄 and Im 𝐵

𝑃 . As 𝐵 𝑇 and 𝐵 are equivalent by part (1), the 𝑅 -modules 𝑃 and 𝑄 are isomorphic, so 𝑃 is self-dual. Thus part (2) holds.

To check part (3) for 𝑑 ∈ 𝑅 , let 𝐴 ¯ ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 𝑛 ​ ( 𝑅 / 𝑅 ​ 𝑑 ) ) . As 𝑠 ​ 𝑟 ​ ( 𝑅 ) ≤ 𝑛 2 , there exists 𝐴 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 𝑛 ​ ( 𝑅 ) ) such that its reduction modulo 𝑅 ​ 𝑑 is 𝐴 ¯ (see [3], Prop. 2.4(1)). Let 𝑁 ∈ 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) be such that 𝐴 ​ 𝑁 is symmetric. If 𝑁 ¯ ∈ 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 / 𝑅 ​ 𝑑 ) is the reduction of 𝑁 modulo 𝑅 ​ 𝑑 , then 𝐴 ¯ ​ 𝑁 ¯ , being the reduction of 𝐴 ​ 𝑁 modulo 𝑅 ​ 𝑑 , is symmetric. Thus 𝑅 / 𝑅 ​ 𝑑 is a ( 𝑊 ​ 𝑆 ​ 𝑈 ) 𝑛 ring; so part (3) holds. ∎

Remark 6.4.

Assume 𝑅 is an integral domain such that for each 𝑑 ∈ 𝑅 , every projective 𝑅 / 𝑅 ​ 𝑑 -module of rank 1 and generated by 2 elements is self-dual and the functorial homomorphism 𝐺 ​ 𝐿 2 ​ ( 𝑅 ) → 𝐺 ​ 𝐿 2 ​ ( 𝑅 / 𝑅 ​ 𝑑 ) is surjective. Then each matrix 𝐴 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is equivalent to its transpose 𝐴 𝑇 .

Definition 6.5.

Let ( 𝑚 , 𝑛 ) ∈ ℕ 2 be such that 𝑚 < 𝑛 . We say that 𝑅 is:

(1) a 𝑊 ​ 𝐻 𝑛 , 𝑚 ′ (resp.  𝑊 ​ 𝐻 𝑛 , 𝑚 ) ring if for each ( 𝐴 1 , … , 𝐴 𝑚 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 𝑛 ​ ( 𝑅 ) ) 𝑚 there exists 𝑁 ∈ 𝑆 ​ 𝐿 𝑛 ​ ( 𝑅 ) (resp.  𝑁 ∈ 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) ) such that the trace of 𝐴 𝑖 ​ 𝑁 is 0 for every 𝑖 ∈ { 1 , … , 𝑚 } .

(2) an 𝑆 ​ 𝐻 𝑛 , 𝑚 ′ (resp.  𝑆 ​ 𝐻 𝑛 , 𝑚 ) ring if it is a Hermite ring and a 𝑊 ​ 𝐻 𝑛 , 𝑚 ′ (resp.  𝑊 ​ 𝐻 𝑛 , 𝑚 ) ring.

(3) an 𝐻 𝑛 , 𝑚 ′ (resp.  𝐻 𝑛 , 𝑚 ) ring if for each ( 𝐴 1 , … , 𝐴 𝑚 ) ∈ 𝕄 𝑛 ​ ( 𝑅 ) 𝑚 there exists 𝑁 ∈ 𝑆 ​ 𝐿 𝑛 ​ ( 𝑅 ) (resp.  𝑁 ∈ 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) ) such that the trace of 𝐴 𝑖 ​ 𝑁 is 0 for every 𝑖 ∈ { 1 , … , 𝑚 } .

Definition 6.5(3) was introduced by Lorenzini im [12], Def. 2.4. In Definition 6.5(1) or (3), the existence of 𝑁 is equivalent to the existence of 𝑀 ∈ 𝑆 ​ 𝐿 𝑛 ​ ( 𝑅 ) (resp.  𝑀 ∈ 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) ) such that the trace of 𝑀 ​ 𝐴 𝑖 is 0 for every 𝑖 ∈ { 1 , … , 𝑚 } and equivalent to the existence of ( 𝑀 , 𝑁 ) ∈ 𝑆 ​ 𝐿 𝑛 ​ ( 𝑅 ) 2 (resp.  ( 𝑀 , 𝑁 ) ∈ 𝐺 ​ 𝐿 𝑛 ​ ( 𝑅 ) 2 ) such that the trace of 𝑀 ​ 𝐴 𝑖 ​ 𝑁 is 0 for every 𝑖 ∈ { 1 , … , 𝑚 } . This follows via conjugation from the fact that traces are preserved under transposition.

Clearly, if 𝑛 − 1 ≥ 𝑚 ≥ 2 and 𝑙 ∈ { 1 , … , 𝑚 − 1 } , then each 𝑊 ​ 𝐻 𝑛 , 𝑚 ′ (or 𝑊 ​ 𝐻 𝑛 , 𝑚 or 𝑆 ​ 𝐻 𝑛 , 𝑚 ′ or 𝑆 ​ 𝐻 𝑛 , 𝑚 ) ring is a 𝑊 ​ 𝐻 𝑛 , 𝑙 ′ (or 𝑊 ​ 𝐻 𝑛 , 𝑙 or 𝑆 ​ 𝐻 𝑛 , 𝑙 ′ or 𝑆 ​ 𝐻 𝑛 , 𝑙 ) ring.

The trace of the product matrix [ 𝑎

𝑏

𝑐

𝑑 ] ​ [ 0

− 1

1

0 ] is 𝑏 − 𝑐 . Based on this it follows easily that 𝑅 is a 𝑊 ​ 𝐻 2 , 1 ′ (resp.  𝑊 ​ 𝐻 2 , 1 ) ring iff it is a ( 𝑊 ​ 𝑆 ​ 𝑈 ′ ) 2 (resp.  ( 𝑊 ​ 𝑆 ​ 𝑈 ) 2 ) ring (cf. [12], Prop. 4.11).

Proposition 6.6.

Let 𝑛 ∈ ℕ ∖ { 1 } . Then the following properties hold.

(1) Let 𝑚 ∈ { 1 , … , 𝑛 − 1 } . If 𝑅 is an 𝑆 ​ 𝐻 𝑛 , 𝑚 ′ (resp.  𝑆 ​ 𝐻 𝑛 , 𝑚 ) ring, then it is an 𝐻 𝑛 , 𝑚 ′ (resp.  𝐻 𝑛 , 𝑚 ) ring.

(2) A ring 𝑅 is an 𝑆 ​ 𝐻 𝑛 , 𝑛 − 1 ′ (resp.  𝑆 ​ 𝐻 𝑛 , 𝑛 − 1 ) ring iff it is an 𝐻 𝑛 , 𝑛 − 1 ′ (resp.  𝐻 𝑛 , 𝑛 − 1 ) ring.

(3) A ring 𝑅 is an 𝑆 ​ 𝐻 2 , 1 ′ ring iff it is an 𝐻 2 , 1 ′ ring and iff it is an ( 𝑆 ​ 𝑈 ′ ) 2 ring.

(4) A ring 𝑅 is an 𝑆 ​ 𝐻 2 , 1 ring iff it is an 𝐻 2 , 1 ring and iff it is an ( 𝑆 ​ 𝑈 ) 2 ring.

Proof.

For part (1), let ( 𝐵 1 , … , 𝐵 𝑚 ) ∈ 𝕄 𝑛 ​ ( 𝑅 ) 𝑚 . As 𝑅 is a Hermite ring, for each 𝑖 ∈ { 1 , … , 𝑚 } there exists ( 𝑒 𝑖 , 𝐶 𝑖 ) ∈ 𝑅 × 𝑈 ​ 𝑚 ​ ( 𝕄 𝑛 ​ ( 𝑅 ) ) such that 𝐵 𝑖

𝑒 𝑖 ​ 𝐶 𝑖 . As 𝑅 is an 𝑊 ​ 𝐻 𝑛 , 𝑚 ′ (resp.  𝑊 ​ 𝐻 𝑛 , 𝑚 ) ring, there exists 𝑁 ∈ 𝑆 ​ 𝐿 𝑛 ​ ( 𝑅 ) (resp.  𝑁 ∈ 𝐺 𝐿 𝑛 ( 𝑅 ) ) such that the trace of 𝐶 𝑖 ​ 𝑁 is 0 for every 𝑖 ∈ { 1 , … , 𝑚 } . Hence the trace of 𝐵 𝑖 ​ 𝑁

𝑒 𝑖 ​ ( 𝐶 𝑖 ​ 𝑁 ) is 0 for every 𝑖 ∈ { 1 , … , 𝑚 } . Thus 𝑅 is an 𝐻 𝑛 , 𝑚 (resp.  𝐻 𝑛 , 𝑚 ′ ) ring. So part (1) holds.

Based on part (1), to prove part (2) it suffices to prove its ‘if’ part. As 𝑅 is an 𝐻 𝑛 , 𝑛 − 1 ′ (resp.  𝐻 𝑛 , 𝑛 − 1 ) ring, it is also a 𝑊 ​ 𝐻 𝑛 , 𝑛 − 1 ′ (resp.  𝑊 ​ 𝐻 𝑛 , 𝑛 − 1 ) and it is a Hermite ring by [12], Lem. 4.3. So 𝑅 is an 𝑆 ​ 𝐻 𝑛 , 𝑛 − 1 ′ (resp.  𝑆 ​ 𝐻 𝑛 , 𝑛 − 1 ) ring. Thus part (2) holds.

Parts (3) and (4) follow from part (2) applied to 𝑛

2 and the paragraph before this proposition. ∎

Remark 6.7.

(1) For ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 4 ) , let

ℒ ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) := { ( 𝑎 ​ 𝑥 + 𝑏 ​ 𝑦 + 𝑐 ​ 𝑧 + 𝑑 ​ 𝑤 , 𝑥 ​ 𝑤 − 𝑦 ​ 𝑧 ) | ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ 𝑅 4 } ⊂ 𝑅 2 .

We have ( 1 , 0 ) ∈ ℒ ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) iff the matrix [ 𝑎

𝑏

𝑐
𝑑 ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is determinant liftable. Moreover, 𝑅 is a 𝑊 ​ 𝐽 2 , 1 ring in the sense of [4], Def. 1.10(1) iff for each ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 4 ) we have ℒ ( 𝑎 , 𝑏 , 𝑐 , 𝑑 )

𝑅 2 . Similarly, 𝑅 is a 𝑊 ​ 𝐻 2 , 1 ′ (resp.  𝑊 ​ 𝐻 2 , 1 ) ring iff for each ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 4 ) we have ( 0 , 1 ) ∈ ℒ ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) (resp.  [ { 0 } × 𝑈 ​ ( 𝑅 ) ] ∩ ℒ ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) ≠ ∅ ), cf. [12], Lem. 4.1.

(2) The sets ℒ ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) of part (1) have many properties. For instance, if 𝑢 ∈ 𝑈 ​ ( 𝑅 ) and ( Ψ , Δ ) ∈ ℒ ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) , then ( 𝑢 ​ Ψ + 2 ​ ( 𝑎 ​ 𝑑 − 𝑏 ​ 𝑐 ) , 𝑢 2 ​ Δ + 𝑢 ​ Ψ + 𝑎 ​ 𝑑 − 𝑏 ​ 𝑐 ) ∈ ℒ ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) ; this is so as for ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ 𝑅 4 with ( 𝑎 ​ 𝑥 + 𝑏 ​ 𝑦 + 𝑐 ​ 𝑧 + 𝑑 ​ 𝑤 , 𝑥 ​ 𝑤 − 𝑦 ​ 𝑧 )

( Ψ , Δ ) for ( 𝑥 ′ , 𝑦 ′ , 𝑧 ′ , 𝑤 ′ ) := ( 𝑢 ​ 𝑥 + 𝑑 , 𝑢 ​ 𝑦 − 𝑐 , 𝑢 ​ 𝑧 − 𝑏 , 𝑢 ​ 𝑤 + 𝑎 ) we have

( 𝑎 ​ 𝑥 ′ + 𝑏 ​ 𝑦 ′ + 𝑐 ​ 𝑧 ′ + 𝑑 ​ 𝑤 ′ , 𝑥 ′ ​ 𝑤 ′ − 𝑦 ′ ​ 𝑧 ′ )

( 𝑢 ​ Ψ + 2 ​ ( 𝑎 ​ 𝑑 − 𝑏 ​ 𝑐 ) , 𝑢 2 ​ Δ + 𝑢 ​ Ψ + 𝑎 ​ 𝑑 − 𝑏 ​ 𝑐 ) .

A similar argument gives that if ( ( 1 , 𝑡 ) , ( 0 , 𝑢 ) ) ∈ ℒ ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) 2 , then there exists 𝑠 ∈ 𝑅 such that for each 𝑥 ∈ 𝑅 we have ( 1 , 𝑢 ​ 𝑥 2 + 𝑠 ​ 𝑥 + 𝑡 ) ∈ ℒ ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) 2 .

7.Proof of Theorem 1.17 and applications

Let Spec  ​ 𝑅 be the spectrum of 𝑅 and let Max  ​ 𝑅 be its subset of maximal ideals.

To prove part (1), it suffices to show that for all ( 𝑢 , 𝑐 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and 𝑎 ∈ 𝑅 , we have 𝑢 2 + 𝑅 ​ 𝑐 ∈ Im ​ ( 𝜐 𝑎 , 1 − 𝑎 , 𝑐 ) . To check this we can assume that 𝑐 ≠ 0 . As in the proof of Corollary 4.2, we can assume that there exists 𝑡 ∈ 𝑅 such that 𝑐

1 − 𝑢 ​ 𝑡 . As we are assuming that 𝑅 is a ( 𝑊 ​ 𝑆 ​ 𝑈 ) 2 ring, there exists 𝑀

[ 𝑥

𝑦

𝑧

𝑤 ] ∈ 𝐺 ​ 𝐿 2 ​ ( 𝑅 ) such that 𝑀 ​ [ 𝑎 ​ 𝑐

𝑢

0

( 1 − 𝑎 ) ​ 𝑐 ] is a symmetric matrix, i.e., we have an identity

(V) 𝑎 ​ 𝑐 ​ 𝑧

𝑢 ​ 𝑥 + ( 1 − 𝑎 ) ​ 𝑐 ​ 𝑦 .

As 𝑐 divides 𝑢 ​ 𝑥 and ( 𝑢 , 𝑐 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , 𝑐 divides 𝑥 . Let 𝑠 ∈ 𝑅 be such that 𝑥

𝑐 ​ 𝑠 .

Let 𝜇 := det ( 𝑀 )

𝑐 ​ 𝑠 ​ 𝑤 − 𝑦 ​ 𝑧 ∈ 𝑈 ​ ( 𝑅 ) . Thus ( 𝑐 ​ 𝑠 , 𝑦 ​ 𝑧 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) .

We first assume that 𝑐 ∉ 𝑍 ​ ( 𝑅 ) . Dividing Equation (V) by 𝑐 it follows that

(VI) 𝑎 ​ 𝑧 + ( 𝑎 − 1 ) ​ 𝑦

𝑢 ​ 𝑠 .

We check that the assumption ( 𝑎 , 𝑠 ) ∉ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) leads to a contradiction. This assumption implies that there exists 𝔪 ∈ Max  ​ 𝑅 such that 𝑅 ​ 𝑎 + 𝑅 ​ 𝑠 ⊆ 𝔪 . From this and Equation (VI) it follows that ( 𝑎 − 1 ) ​ 𝑦 ∈ 𝔪 . As 𝑎 − 1 ∉ 𝔪 , it follows that 𝑦 ∈ 𝔪 , hence ( 𝑠 , 𝑦 ​ 𝑧 ) ∉ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , a contradiction. As ( 𝑎 , 𝑠 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , 𝑠 + 𝑅 ​ 𝑎 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ) .

From Equation (VI) it follows that the images of ( − 𝑦 + 𝑅 ​ 𝑐 ) ​ ( 𝑢 + 𝑅 ​ 𝑐 ) − 1 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) and 𝑠 + 𝑅 ​ 𝑎 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ) in 𝑈 ​ ( 𝑅 / ( 𝑅 ​ 𝑎 + 𝑅 ​ 𝑐 ) ) are equal. Thus there exists a unit of the quotient ring 𝑅 / ( 𝑅 ​ 𝑎 ∩ 𝑅 ​ 𝑐 ) that maps to ( − 𝑦 + 𝑅 ​ 𝑐 ) ​ ( 𝑢 + 𝑅 ​ 𝑐 ) − 1 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) . As the ideal ( 𝑅 ​ 𝑎 ∩ 𝑅 ​ 𝑐 ) / 𝑅 ​ 𝑎 ​ 𝑐 of 𝑅 / 𝑅 ​ 𝑎 ​ 𝑐 has square 0 , the functorial homomorphism 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑐 ) → 𝑈 ​ ( 𝑅 / ( 𝑅 ​ 𝑎 ∩ 𝑅 ​ 𝑐 ) ) is surjective. From the last two sentences it follows that ( − 𝑦 + 𝑅 ​ 𝑐 ) ​ ( 𝑢 + 𝑅 ​ 𝑐 ) − 1 ∈ Im ​ ( 𝜐 𝑎 , 1 − 𝑎 , 𝑐 ) .

Due to the symmetry of Equation (VI) in 𝑎 ​ 𝑧 and ( 𝑎 − 1 ) ​ 𝑦 , similar arguments give 𝑠 + 𝑅 ​ ( 1 − 𝑎 ) ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ ( 1 − 𝑎 ) ) and ( 𝑧 + 𝑅 ​ 𝑐 ) ​ ( 𝑢 + 𝑅 ​ 𝑐 ) − 1 ∈ Im ​ ( 𝜐 𝑎 , 1 − 𝑎 , 𝑐 ) . Clearly 𝜇 + 𝑅 ​ 𝑐 ∈ Im ​ ( 𝜐 𝑎 , 1 − 𝑎 , 𝑐 ) . As 𝜇 + 𝑅 ​ 𝑐

( − 𝑦 + 𝑅 ​ 𝑐 ) ​ ( 𝑧 + 𝑅 ​ 𝑐 ) , it follows that

( 𝑢 + 𝑅 ​ 𝑐 ) 2

[ ( − 𝑦 + 𝑅 ​ 𝑐 ) ​ ( 𝑢 + 𝑅 ​ 𝑐 ) − 1 ] − 1 ​ [ ( 𝑧 + 𝑅 ​ 𝑐 ) ​ ( 𝑢 + 𝑅 ​ 𝑐 ) − 1 ] − 1 ​ ( 𝜇 + 𝑅 ​ 𝑐 ) ∈ Im ​ ( 𝜐 𝑎 , 1 − 𝑎 , 𝑐 ) .

We use a trick with annihilators to show that an analogue of Equation (VI) holds even if 𝑐 ∈ 𝑍 ​ ( 𝑅 ) . As ( ( 𝑐 , 𝑢 ) , ( 𝑐 , 𝑦 ​ 𝑧 ) ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) 2 , it follows that ( 𝑐 , 𝑢 ​ 𝑦 ​ 𝑧 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) . This implies that Ann 𝑅 ​ ( 𝑐 ) ⊂ 𝑅 ​ 𝑢 ​ 𝑦 ​ 𝑧 . As 𝑎 ​ 𝑧 + ( 𝑎 − 1 ) ​ 𝑦 − 𝑢 ​ 𝑠 ∈ Ann 𝑅 ​ ( 𝑐 ) by Equation (V), it follows that there exists 𝑞 ∈ 𝑅 such that 𝑎 ​ 𝑧 + ( 𝑎 − 1 ) ​ 𝑦 − 𝑢 ​ 𝑠

𝑢 ​ 𝑦 ​ 𝑧 ​ 𝑞 . If 𝑠 1 := 𝑠 + 𝑦 ​ 𝑧 ​ 𝑞 , then ( 𝑠 1 , 𝑦 ​ 𝑧 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and the identity 𝑎 ​ 𝑧 + ( 𝑎 − 1 ) ​ 𝑦

𝑢 ​ 𝑠 1 holds. Thus, using 𝑠 1 and the last identity instead of 𝑠 and Equation (VI), as above we argue that ( 𝑢 + 𝑅 ​ 𝑐 ) 2 ∈ Im ​ ( 𝜐 𝑎 , 1 − 𝑎 , 𝑐 ) . So part (1) holds.

Part (2) follows from Theorem 6.3(2) and (3).

The first part of part (3) follows from part (1) and definitions. Based on it, the second part of part (3) follows from (the implication ( 3 ) ⇒ ( 1 ) of) Theorem 1.12.

Thus Theorem 1.17 holds.

Remark 7.1.

If we have ( 𝑎 , 𝑦 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) (resp.  ( 1 − 𝑎 , 𝑧 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) ), then − 𝑦 + 𝑅 ​ 𝑎 ​ 𝑐 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑐 ) (resp.  𝑧 + 𝑅 / 𝑅 ​ ( 1 − 𝑎 ) ​ 𝑐 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ ( 1 − 𝑎 ) ​ 𝑐 ) ), and therefore 𝑢 + 𝑅 ​ 𝑐 ∈ Im ​ ( 𝜐 𝑎 , 1 − 𝑎 , 𝑐 ) .

Corollary 7.2.

For a Bézout domain the following statements are equivalent.

(1) The ring R is a ( W ​ S ​ U ) 2 (resp.  ( W ​ S ​ U ′ ) 2 ) ring.

(2) For all ( ( a , b ) , ( c , u ) ) ∈ U ​ m ​ ( R 2 ) 2 with a ​ b ​ c ≠ 0 , if a pair ( a ′ , b ′ ) ∈ R 2 such that a ​ a ′ + b ​ b ′

1 is given, then there exists a triple ( s , l , w ) ∈ R 3 with the property that c ​ s ​ w + ( a ′ ​ u ​ s − b ​ l ) ​ ( b ′ ​ u ​ s + a ​ l ) ∈ U ​ ( R ) (resp.  c ​ s ​ w + ( a ′ ​ u ​ s − b ​ l ) ​ ( b ′ ​ u ​ s + a ​ l )

1 ).

Proof.

If 𝐴 1 , 𝐴 2 ∈ 𝕄 2 ​ ( 𝑅 ) are equivalent, then there exists 𝑀 1 ∈ 𝐺 ​ 𝐿 2 ​ ( 𝑅 ) such that 𝑀 1 ​ 𝐴 1 is symmetric iff there exists 𝑀 2 ∈ 𝐺 ​ 𝐿 2 ​ ( 𝑅 ) such that 𝑀 2 ​ 𝐴 2 is symmetric. This is so, as for ( 𝑀 , 𝑁 , 𝑀 1 ) ∈ 𝐺 ​ 𝐿 2 ​ ( 𝑅 ) 3 such that 𝐴 2

𝑀 ​ 𝐴 1 ​ 𝑁 and 𝑀 1 ​ 𝐴 1 is symmetric, by denoting 𝑀 2 := 𝑁 𝑇 ​ 𝑀 1 ​ 𝑀 − 1 ∈ 𝐺 ​ 𝐿 2 ​ ( 𝑅 ) , 𝑀 2 ​ 𝐴 2

𝑁 𝑇 ​ ( 𝑀 1 ​ 𝐴 1 ) ​ 𝑁 is symmetric. Also, if ( 𝑀 , 𝑁 , 𝑀 1 ) ∈ 𝑆 ​ 𝐿 2 ​ ( 𝑅 ) 3 , then 𝑀 2 ∈ 𝑆 ​ 𝐿 2 ​ ( 𝑅 ) .

Based on the previous paragraph and Case 1 of Section 2, the ring 𝑅 is a ( 𝑊 ​ 𝑆 ​ 𝑈 ) 2 (resp.  ( 𝑊 ​ 𝑆 ​ 𝑈 ′ ) 2 ) ring iff for all ( ( 𝑎 , 𝑏 ) , ( 𝑐 , 𝑢 ) ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) 2 there exists 𝑀

[ 𝑥

𝑦

𝑧

𝑤 ] in 𝐺 ​ 𝐿 2 ​ ( 𝑅 ) (resp. in 𝑆 ​ 𝐿 2 ​ ( 𝑅 ) ) such that the matrix 𝑀 ​ [ 𝑎 ​ 𝑐

𝑢

0
𝑏 ​ 𝑐 ] is symmetric, i.e., the equation 𝑎 ​ 𝑐 ​ 𝑧

𝑢 ​ 𝑥 + 𝑏 ​ 𝑐 ​ 𝑦 holds. If 𝑎 ​ 𝑏 ​ 𝑐

0 , then [ 𝑎 ​ 𝑐

𝑢

0
𝑏 ​ 𝑐 ] is simply extendable (see [3], Ex. 4.9(3)) and hence equivalent to a diagonal matrix (see [3], Thm. 4.3). From this and the previous paragraph we get that 𝑀 exists. Thus we can assume that 𝑎 ​ 𝑏 ​ 𝑐 ≠ 0 . So 𝑐 ∉ 𝑍 ​ ( 𝑅 )

{ 0 } and hence as in the proof of Theorem 1.17, we write 𝑥

𝑐 ​ 𝑠 with 𝑠 ∈ 𝑅 and the identity 𝑎 ​ 𝑧

𝑢 ​ 𝑠 + 𝑏 ​ 𝑦 holds. The general solution of the equation 𝑎 ​ 𝑧 − 𝑏 ​ 𝑦

𝑢 ​ 𝑠 in the indeterminates 𝑦 , 𝑧 is 𝑧

𝑎 ′ ​ 𝑢 ​ 𝑠 − 𝑏 ​ 𝑙 and 𝑦

− 𝑏 ′ ​ 𝑢 ​ 𝑠 − 𝑎 ​ 𝑙 with 𝑙 ∈ 𝑅 . As det ( 𝑀 )

𝑥 ​ 𝑤 − 𝑦 ​ 𝑧

𝑐 ​ 𝑠 ​ 𝑤 + ( 𝑎 ′ ​ 𝑢 ​ 𝑠 − 𝑏 ​ 𝑙 ) ​ ( 𝑏 ′ ​ 𝑢 ​ 𝑠 + 𝑎 ​ 𝑙 ) , the corollary holds. ∎

Remark 7.3.

Referring to Corollary 7.2(2) with 𝑏

1 − 𝑎 , we can take 𝑎 ′

1

𝑏 ′ , and one is searching for ( 𝑠 , 𝑙 , 𝑤 ) ∈ 𝑅 3 such that 𝑐 ​ 𝑠 ​ 𝑤 + ( 𝑢 ​ 𝑠 + 𝑎 ​ 𝑙 − 𝑙 ) ​ ( 𝑢 ​ 𝑠 + 𝑎 ​ 𝑙 ) ∈ 𝑈 ​ ( 𝑅 ) (resp. and 𝑐 ​ 𝑠 ​ 𝑤 + ( 𝑢 ​ 𝑠 + 𝑎 ​ 𝑙 − 𝑙 ) ​ ( 𝑢 ​ 𝑠 + 𝑎 ​ 𝑙 )

1 , which for 𝑐

1 − 𝑢 ​ 𝑡 with 𝑡 ∈ 𝑅 is closely related to Equation (IV) via a substitution of the form 𝑧 := − 𝑤 ).

Remark 7.4.

A symmetric matrix 𝐴

[ 𝑎

𝑏

𝑏
𝑑 ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is determinant liftable iff there exists ( 𝑥 , 𝑤 ) ∈ 𝑅 2 such that 𝑏 divides 1 − 𝑎 ​ 𝑥 − 𝑑 ​ 𝑤 and for a suitable 𝑠 ∈ 𝑅 such that 𝑏 ​ 𝑠

1 − 𝑎 ​ 𝑥 − 𝑐 ​ 𝑤 , the equation 𝑋 2 + 𝑠 ​ 𝑋 + 𝑥 ​ 𝑤

0 has two solutions 𝑦 , 𝑧 ∈ 𝑅 (so 𝑥 ​ 𝑤 − 𝑦 ​ 𝑧

0 and 1 − 𝑎 ​ 𝑥 − 𝑏 ​ 𝑦 − 𝑏 ​ 𝑧 − 𝑑 ​ 𝑤

0 ).

Recall from [26], Sect. 1 that 𝑅 is said to have square stable range 1 , and one writes 𝑠 ​ 𝑠 ​ 𝑟 ​ ( 𝑅 )

1 , if for each ( 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , there exists 𝑟 ∈ 𝑅 such that 𝑎 2 + 𝑏 ​ 𝑟 ∈ 𝑈 ​ ( 𝑅 ) . Clearly, we have 𝑠 ​ 𝑠 ​ 𝑟 ​ ( 𝑅 )

1 iff for each 𝑏 ∈ 𝑅 , the cokernel of the reduction homomorphism 𝑈 ​ ( 𝑅 ) → 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑏 ) is a Boolean group. From this and [3], Prop. 2.4(1) it follows that if 𝑠 ​ 𝑟 ​ ( 𝑅 )

1 , then 𝑠 ​ 𝑠 ​ 𝑟 ​ ( 𝑅 )

1 .

Proposition 7.5.

Assume 𝑠 ​ 𝑠 ​ 𝑟 ​ ( 𝑅 )

1 . Then the following properties hold.

(1) For ( ( 𝑎 , 𝑏 ) , ( 𝑐 , 𝑢 ) ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) 2 , let 𝐴 := [ 𝑎 ​ 𝑐

𝑢

0

( 1 − 𝑎 ) ​ 𝑐 ] . Then there exists 𝑀 ∈ 𝐺 ​ 𝐿 2 ​ ( 𝑀 ) such that 𝑀 ​ 𝐴 is symmetric.

(2) Assume also that 𝑅 is a Hermite ring. Then each matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) has a companion test matrix equivalent to a symmetric matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) . Moreover, 𝑅 is an EDR iff each symmetric matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is extendable.

Proof.

For part (1), as 𝑠 ​ 𝑠 ​ 𝑟 ​ ( 𝑅 )

1 , there exists 𝑤 ∈ 𝑅 be such that 𝑢 2 + 𝑐 ​ 𝑤 ∈ 𝑈 ​ ( 𝑅 ) . For 𝑀 := [ 𝑐

− 𝑢

𝑢

𝑤 ] ∈ 𝐺 ​ 𝐿 2 ​ ( 𝑅 ) , the product 𝑀 ​ 𝐴 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is symmetric. So part (1) holds.

For part (2), each matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) has a companion test matrix of the form 𝐴 of part (1) by Case 1 of Section 2 and hence which is equivalent to a symmetric matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) by part (1). Based on this, (the equivalence ( 1 ) ⇔ ( 3 ) of) Theorem 1.2 and Proposition 2.2 applied to 𝒫 being extendable, it follows that the last sentence of part (2) holds as well. ∎

8.Proof of Theorem 1.18

Let 𝑃 be a projective 𝑅 / 𝑅 ​ 𝑑 -module of rank 1 generated by 2 elements. We consider an isomorphism 𝑃 ⊕ 𝑄 ≅ ( 𝑅 / 𝑅 ​ 𝑑 ) 2 . Let 𝐷 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 / 𝑅 ​ 𝑑 ) ) be such that Ker 𝐷

𝑄 and Im 𝐷

𝑃 . As 𝑠 ​ 𝑟 ​ ( 𝑅 ) ≤ 2 , there exists 𝐴 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) such that its reduction modulo 𝑅 ​ 𝑑 is 𝐷 (see [3], Prop. 2.4(1)). Let 𝑅 ¯ := 𝑅 / 𝑅 ​ det ( 𝐴 ) and let 𝐴 ¯ ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ¯ ) ) be the reduction of 𝐴 modulo 𝑅 ​ det ( 𝐴 ) . We have det ( 𝐴 ) ∈ 𝑅 ​ 𝑑 . We know that Ker 𝐴 ¯ and Im 𝐴 ¯ are dual projective 𝑅 ¯ -modules of rank 1 generated by 2 elements (see [3], Lem. 3.1) and their reductions modulo the ideal 𝑅 ​ 𝑑 / 𝑅 ​ det ( 𝐴 ) of 𝑅 ¯ are 𝑄 and 𝑃 . Thus, to complete the proof of Theorem 1.18, it suffices to show that the classes of Ker 𝐴 ¯ and Im 𝐴 ¯ in Pic  ​ ( 𝑅 ¯ ) are equal (equivalently, have orders 1 or 2 or equivalently, Ker 𝐴 ¯ 2 and Im 𝐴 ¯ 2 are free 𝑅 -modules of rank 2 ). So, 𝑑 and 𝐷 will not be mentioned again.

Based on Case 1 of Section 2, we can assume that 𝐴

[ 𝑎 ​ 𝑐

𝑢

0
𝑏 ​ 𝑐 ] ∈ 𝕄 2 ​ ( 𝑅 ) with ( 𝑎 , 𝑏 , 𝑐 ) ∈ 𝑅 3 such that ( ( 𝑎 , 𝑏 ) , ( 𝑐 , 𝑢 ) ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) 2 and we only use that Coker ​ ( 𝜐 𝑎 , 𝑏 , 𝑐 ) is a Boolean group. As Coker ​ ( 𝜐 𝑎 , 𝑏 , 𝑐 ) makes sense for 𝑅 / 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 and as two finitely generated projective 𝑅 / 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 2 -modules are isomorphic if and only if their reductions modulo the square 0 ideal 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 / 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 2 are isomorphic, by reducing modulo 𝑅 ​ 𝑎 ​ 𝑏 ​ 𝑐 we can assume that 𝑎 ​ 𝑏 ​ 𝑐

0 . Hence ( 𝑅 , 𝐴 )

( 𝑅 ¯ , 𝐴 ¯ ) and we only use ( 𝑅 , 𝐴 ) .

Let ( 𝑠 , 𝑡 ) ∈ 𝑅 2 be such that 𝑐 ​ 𝑠 + 𝑢 ​ 𝑡

1 . Recall that Ker 𝐴

𝑅 ​ 𝑣 1 + 𝑅 ​ 𝑣 2 , where 𝑣 1 := [ − 𝑢

𝑎 ​ 𝑐 ] and 𝑣 2 := [ − 𝑏 ​ 𝑐

0 ] , by [3], Ex. 5.2. As ( 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , we have Spec  ​ 𝑅

Spec  ​ 𝑅 𝑎 ∪ Spec  ​ 𝑅 𝑏 . Note ( Ker 𝐴 ) 𝑎

𝑅 𝑎 ​ 𝑣 1 and ( Ker 𝐴 ) 𝑏

𝑅 𝑏 ​ ( 𝑡 ​ 𝑣 1 + 𝑠 𝑏 ​ 𝑣 2 ) . Over 𝑅 𝑎 ​ 𝑏 we have 𝑐

0 and hence 𝑣 1

𝑢 ​ ( 𝑡 ​ 𝑣 1 + 𝑠 𝑏 ​ 𝑣 2 ) . Thus we have identities ( Ker 𝐴 ⊗ 2 ) 𝑎

𝑅 𝑎 ​ ( 𝑣 1 ⊗ 𝑣 1 ) , ( Ker 𝐴 ⊗ 2 ) 𝑏

𝑅 𝑏 ​ [ ( 𝑡 ​ 𝑣 1 + 𝑠 𝑏 ​ 𝑣 2 ) ⊗ ( 𝑡 ​ 𝑣 1 + 𝑠 𝑏 ​ 𝑣 2 ) ] , and over 𝑅 𝑎 ​ 𝑏 we have an identity 𝑣 1 ⊗ 𝑣 1

𝑢 2 ​ [ ( 𝑡 ​ 𝑣 1 + 𝑠 𝑏 ​ 𝑣 2 ) ⊗ ( 𝑡 ​ 𝑣 1 + 𝑠 𝑏 ​ 𝑣 2 ) ] . As 𝑢 2 + 𝑅 ​ 𝑐 ∈ Im ​ ( 𝜐 𝑎 , 𝑏 , 𝑐 ) , there exists 𝑢 𝑎 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑏 ​ 𝑐 ) and 𝑢 𝑏 ∈ 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑎 ​ 𝑐 ) such that the product of their images in 𝑈 ​ ( 𝑅 / 𝑅 ​ 𝑐 ) is 𝑢 2 + 𝑅 ​ 𝑐 . As 𝑅 𝑎 is a localization of 𝑅 / 𝑅 ​ 𝑏 ​ 𝑐 and 𝑅 𝑏 is a localization of 𝑅 / 𝑅 ​ 𝑎 ​ 𝑐 , we denote also by 𝑢 𝑎 and 𝑢 𝑏 their images in 𝑅 𝑎 and 𝑅 𝑏 (respectively). Thus 𝑢 𝑎 − 1 ​ ( 𝑣 1 ⊗ 𝑣 1 ) and 𝑢 𝑏 − 1 ​ [ ( 𝑡 ​ 𝑣 1 + 𝑠 𝑏 ​ 𝑣 2 ) ⊗ ( 𝑡 ​ 𝑣 1 + 𝑠 𝑏 ​ 𝑣 2 ) ] coincide over 𝑅 𝑎 ​ 𝑏 and hence there exists an element 𝑣 ∈ Ker 𝐴 ⊗ 2 whose images in ( Ker 𝐴 ⊗ 2 ) 𝑎 and ( Ker 𝐴 ⊗ 2 ) 𝑏 are equal to 𝑢 𝑎 − 1 ​ ( 𝑣 1 ⊗ 𝑣 1 ) and 𝑢 𝑏 − 1 ​ [ ( 𝑡 ​ 𝑣 1 + 𝑠 𝑏 ​ 𝑣 2 ) ⊗ ( 𝑡 ​ 𝑣 1 + 𝑠 𝑏 ​ 𝑣 2 ) ] (respectively). Thus 𝑅 ​ 𝑣

Ker 𝐴 ⊗ 2 , hence Ker 𝐴 ⊗ 2 ≅ 𝑅 .

So Theorem 1.18 holds.

Corollary 8.1.

Let 𝑅 be a Hermite ring such that for each 𝑎 ∈ 𝑅 , every self-dual projective 𝑅 / 𝑅 ​ 𝑎 -module of rank 1 generated by 2 elements is free. Then the following properties hold.

(1) The ring R is an EDR iff for all ( a , b ) ∈ U ​ m ​ ( R 2 ) and c ∈ R , Coker ​ ( υ a , b , c ) is a Boolean group.

(2) If R is a ( W ​ S ​ U ) 2 ring, then R is an EDR.

Proof.

The ‘only if’ of part (1) follows from (the implication ( 1 ) ⇒ ( 3 ) of) Theorem 1.12. To check the ‘if’ of part (1), from Theorem 1.18 and hypotheses it follows that for each 𝐴 ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) , the reduction 𝐴 ¯ of 𝐴 modulo 𝑅 ​ det ( 𝐴 ) is such that Im 𝐴 ¯ ≅ 𝑅 / 𝑅 ​ det ( 𝐴 ) , hence 𝐴 ¯ is non-full and thus extendable (see [3], Prop. 5.1(2)). Hence 𝑅 is an 𝐸 2 ring and thus an EDR by (the implication ( 3 ) ⇒ ( 1 ) of) Theorem 1.2. Thus part (1) holds.

Part (2) follows from part (1) and Theorem 1.17(1) (it also follows from Theorems 6.3(2) and 1.2).∎

9.Proof of Criteria 1.19 and 1.20

We first prove Criterion 1.19(1). If there exists a pair ( 𝑒 , 𝑓 ) ∈ 𝑅 2 such that 𝑎 ​ 𝑒 2 − 𝑐 ​ 𝑓 2 ∈ 𝑈 ​ ( 𝑅 ) , then from the identity 𝑎 ​ 𝑒 2 − 𝑐 ​ 𝑓 2

𝑒 ​ ( 𝑎 ​ 𝑒 + 𝑏 ​ 𝑓 ) − 𝑓 ​ ( 𝑏 ​ 𝑒 + 𝑐 ​ 𝑓 ) it follows that ( 𝑎 ​ 𝑒 + 𝑏 ​ 𝑓 , 𝑏 ​ 𝑒 + 𝑐 ​ 𝑓 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , hence 𝐴 is simply extendable by [3], Thm. 4.3. For the converse, assume 𝑅 has characteristic 2 and 𝐴 is simply extendable. Let ( 𝑒 , 𝑓 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) be such that ( 𝑎 ​ 𝑒 + 𝑏 ​ 𝑓 , 𝑏 ​ 𝑒 + 𝑐 ​ 𝑓 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) by loc. cit. To prove that 𝑎 ​ 𝑒 2 − 𝑐 ​ 𝑓 2 ∈ 𝑈 ​ ( 𝑅 ) it suffices to show that the assumption that there exist 𝔪 ∈ Max  ​ 𝑅 such that 𝑎 ​ 𝑒 2 − 𝑐 ​ 𝑓 2 ∈ 𝔪 , leads to a contradiction. By replacing 𝑅 with 𝑅 / 𝔪 , we can assume that 𝑅 is a field and we know that either 𝑎 ≠ 0 or 𝑐 ≠ 0 , either 𝑎 ​ 𝑒 + 𝑏 ​ 𝑓 ≠ 0 or 𝑏 ​ 𝑒 + 𝑐 ​ 𝑓 ≠ 0 , 𝑎 ​ 𝑐

𝑏 2 and 𝑎 ​ 𝑒 2

𝑐 ​ 𝑓 2 ; hence either 𝑒 ≠ 0 or 𝑓 ≠ 0 . If 𝑎 ​ 𝑐 ​ 𝑒 ​ 𝑓

0 , then by the symmetry in ( 𝑎 , 𝑐 ) and ( 𝑒 , 𝑓 ) we can assume that 𝑎 ​ 𝑒

0 ; if 𝑎

0 , then 𝑏

0 , 𝑐 ≠ 0 , and hence 𝑓

0 , thus 𝑎 ​ 𝑒 + 𝑏 ​ 𝑓

𝑏 ​ 𝑒 + 𝑐 ​ 𝑓

0 , a contradiction, and if 𝑒

0 , then 𝑓 ≠ 0 , hence 𝑐

0 , and by the symmetry in the pair ( 𝑎 , 𝑐 ) , we similarly reach a contradiction. Thus we can assume that 𝑎 ​ 𝑐 ​ 𝑒 ​ 𝑓 ≠ 0 , hence 𝑏 2

𝑎 ​ 𝑐

𝑒 − 2 ​ 𝑐 2 ​ 𝑓 2

𝑓 − 2 ​ 𝑎 2 ​ 𝑒 2 ; as 𝑅 is a field of characteristic 2 , it follows that 𝑏

𝑒 − 1 ​ 𝑐 ​ 𝑓

𝑓 − 1 ​ 𝑎 ​ 𝑒 , thus 𝑎 ​ 𝑒 + 𝑏 ​ 𝑓

𝑏 ​ 𝑒 + 𝑐 ​ 𝑓

0 , a contradiction. So part (1) holds.

To prove Criterion 1.19(2), as 𝑏 ∉ 𝑍 ​ ( 𝑅 ) , for each 𝔪 ∈ Max  ​ 𝑅 , 𝑏 is a nonzero element of 𝑅 𝔪 . As 𝑅 is a Hermite ring and 𝑁 ​ ( 𝑅 )

0 , each 𝑅 𝔪 is a valuation domain (e.g., see footnote of [4], Thm. 11.3(3) for a proof). Let 𝑑 ∈ 𝑅 and ( 𝑎 ′ , 𝑏 ′ ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) be such that 𝑎

𝑑 ​ 𝑎 ′ and 𝑏

𝑑 ​ 𝑏 ′ . As 𝑑 divides 𝑎 and ( 𝑎 , 𝑐 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , we have ( 𝑑 , 𝑐 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) . Thus also ( 𝑑 2 , 𝑐 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) . From this, as 𝑑 2 divides 𝑎 ​ 𝑐 , it follows that 𝑑 2 divides 𝑎 ; let 𝑢 ∈ 𝑅 be such that 𝑎

𝑑 2 ​ 𝑢 . As 𝑏 ∈ 𝑅 𝔪 ∖ { 0 } , it is easy to see that the image of 𝑢 in 𝑅 𝔪 is a unit for all 𝔪 ∈ Max  ​ 𝑅 . Hence 𝑢 ∈ 𝑈 ​ ( 𝑅 ) . By symmetry, there exist 𝑔 ∈ 𝑅 and 𝑣 ∈ 𝑈 ​ ( 𝑅 ) such that 𝑐

𝑔 2 ​ 𝑣 . As 𝑏 ∉ 𝑍 ​ ( 𝑅 ) and 𝑏 2

𝑎 ​ 𝑐

𝑑 2 ​ 𝑔 2 ​ 𝑢 ​ 𝑣 , we have 𝑑 ∉ 𝑍 ​ ( 𝑅 ) and 𝑔 ∉ 𝑍 ​ ( 𝑅 ) .

To complete the proof that 𝐴 is extendable, based on Criterion 1.19(1) it suffices to show that there exists ( 𝑒 , 𝑓 ) ∈ 𝑅 2 such that 𝑎 ​ 𝑒 2 − 𝑐 ​ 𝑓 2

𝑎 ​ 𝑒 2 + 𝑐 ​ 𝑓 2 ∈ 𝑈 ​ ( 𝑅 ) . Replacing 𝐴 by 𝑢 − 1 ​ 𝐴 (see [3], Lem. 4.1(3)), we can assume that 𝑢

1 . Thus 𝑎

𝑑 2 . From this and the identities 𝑎 ​ 𝑐

𝑏 2 and 𝑐

𝑔 2 ​ 𝑣 , the element 𝑤 := 𝑏 𝑔 ​ 𝑑 of the total ring of fractions of 𝑅 has the property that 𝑤 2

𝑣 in each valuation domain 𝑅 𝔭 with 𝔭 ∈ Spec  ​ 𝑅 , hence 𝑤 ∈ 𝑅 𝔭 , and we conclude that 𝑤 ∈ 𝑅 and 𝑣

𝑤 2 . By replacing ( 𝑣 , 𝑔 ) with ( 1 , 𝑔 ​ 𝑤 ) , we can assume that 𝑣

1 and 𝑐

𝑔 2 . As ( 𝑎 , 𝑐 )

( 𝑑 2 , 𝑔 2 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , it follows that ( 𝑑 , 𝑔 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , hence there exists ( 𝑒 , 𝑓 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) such that 𝑑 ​ 𝑒 + 𝑔 ​ 𝑓

1 , and so 𝑎 ​ 𝑒 2 + 𝑐 ​ 𝑓 2

1 ∈ 𝑈 ​ ( 𝑅 ) . Thus Criterion 1.19(2) holds. This completes the proof of Criterion 1.19.

We prove Criterion 1.20. As 𝑅 is an ( 𝑆 ​ 𝑈 ) 2 ring, it is a Hermite ring, and, as the extendability is preserved under matrix equivalence (see [3], Lem. 4.1(3)), it is an 𝐸 2 ring iff each symmetric matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is extendable. Thus 𝑅 is an EDR iff each symmetric matrix in 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) is extendable by Theorem 1.2. Let 𝐴

[ 𝑎

𝑏

𝑏

𝑐 ] ∈ 𝑈 ​ 𝑚 ​ ( 𝕄 2 ​ ( 𝑅 ) ) . Based on [3], Lem. 4.1(1) and Criterion 1.19(1) applied to 𝐴 modulo 𝑅 ​ ( 𝑎 ​ 𝑐 − 𝑏 2 ) , 𝐴 is extendable if there exists ( 𝑒 , 𝑓 ) ∈ 𝑅 2 such that 𝑎 ​ 𝑒 2 − 𝑐 ​ 𝑓 2 + 𝑅 ​ ( 𝑎 ​ 𝑐 − 𝑏 2 ) ∈ 𝑈 ​ ( 𝑅 / [ 𝑅 ​ ( 𝑎 ​ 𝑐 − 𝑏 2 ) ] ) , i.e., ( 𝑎 ​ 𝑒 2 − 𝑐 ​ 𝑓 2 , 𝑎 ​ 𝑐 − 𝑏 2 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , and the converse holds if 𝑅 has characteristic 2 ; hence Criterion 1.20 holds.

10.Proof of Criterion 1.22

As 𝑅 is a Bézout domain, 𝑅 is an EDD iff for all ( 𝑎 , 𝑏 , 𝑠 ) ∈ 𝑅 3 , the equation

(VII) [ 1 − ( 1 − 𝑎 − 𝑏 ​ 𝑠 ) ​ 𝑤 − 𝑏 ​ 𝑦 ] ​ ( 1 − 𝑎 ​ 𝑥 )

𝑎 ​ 𝑦 ​ [ ( 1 − 𝑎 − 𝑏 ​ 𝑠 ) ​ 𝑧 + 𝑏 ​ 𝑥 ]

in the indeterminates 𝑥 , 𝑦 , 𝑧 and 𝑤 has a solution in 𝑅 4 (see Corollary 1.14 and Proposition 2.3(1)). To solve this equation, we note that, as ( 𝑎 , 1 − 𝑎 ​ 𝑥 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , 𝑎 divides 1 − ( 1 − 𝑎 − 𝑏 ​ 𝑠 ) ​ 𝑤 − 𝑏 ​ 𝑦 . Thus, as 𝑅 is an integral domain, Equation (VIII) has a solution in 𝑅 4 iff the system of two equations

(VIII) 1 − ( 1 − 𝑎 − 𝑏 ​ 𝑠 ) ​ 𝑤 − 𝑏 ​ 𝑦

𝑡 ​ 𝑎 ​ 𝑎 ​ 𝑛 ​ 𝑑 ​ 𝑡 ​ ( 1 − 𝑎 ​ 𝑥 )

𝑦 ​ [ ( 1 − 𝑎 − 𝑏 ​ 𝑠 ) ​ 𝑧 + 𝑏 ​ 𝑥 ]

in the indeterminates 𝑡 , 𝑥 , 𝑦 , 𝑧 and 𝑤 has a solution in 𝑅 5 (this holds even if 𝑎

0 !).

The first equation of System (VIII), rewritten as 𝑤

1 + 𝑎 ​ ( 𝑤 − 𝑡 ) + 𝑏 ​ ( 𝑠 ​ 𝑤 − 𝑦 ) , has the general solution given by 𝑤

1 − 𝑎 ​ 𝑞 − 𝑏 ​ 𝑟 , 𝑡

𝑞 + 𝑤

1 + 𝑞 − 𝑎 ​ 𝑞 − 𝑏 ​ 𝑟 and 𝑦

𝑟 + 𝑠 ​ 𝑤

𝑟 + 𝑠 ​ ( 1 − 𝑎 ​ 𝑞 − 𝑏 ​ 𝑟 ) , where ( 𝑞 , 𝑟 ) ∈ 𝑅 2 can be arbitrary.

The second equation of System (VIII), rewritten as 𝑡

( 𝑎 ​ 𝑡 + 𝑏 ​ 𝑦 ) ​ 𝑥 + ( 1 − 𝑎 − 𝑏 ​ 𝑠 ) ​ 𝑦 ​ 𝑧 , has a solution iff 𝑡 ∈ 𝑅 ​ ( 𝑎 ​ 𝑡 + 𝑏 ​ 𝑦 ) + 𝑅 ​ ( 1 − 𝑎 − 𝑏 ​ 𝑠 ) ​ 𝑦 . As 1

( 1 − 𝑎 − 𝑏 ​ 𝑠 ) ​ 𝑤 + ( 𝑎 ​ 𝑡 + 𝑏 ​ 𝑦 ) , we have 𝑡 ∈ 𝑅 ​ ( 𝑎 ​ 𝑡 + 𝑏 ​ 𝑦 ) + 𝑅 ​ ( 1 − 𝑎 − 𝑏 ​ 𝑠 ) ​ 𝑦 iff 𝑡 ∈ 𝑅 ​ ( 𝑎 ​ 𝑡 + 𝑏 ​ 𝑦 ) + 𝑅 ​ 𝑦

𝑅 ​ 𝑎 ​ 𝑡 + 𝑅 ​ 𝑦 .

From the last two paragraphs it follows that System (VIII) has a solution in 𝑅 5 iff there exists ( 𝑞 , 𝑟 ) ∈ 𝑅 2 as mentioned in statement (2). Thus ( 1 ) ⇔ ( 2 ) .

If we have ( 1 − 𝑎 , 𝑏 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , then we can choose ( 𝑞 , 𝑟 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) such that 𝑡

1 + 𝑞 ​ ( 1 − 𝑎 ) + 𝑟 ​ 𝑏

0 , hence 𝑡

0 ∈ 𝑅 ​ 𝑎 ​ 𝑡 + 𝑅 ​ 𝑦

𝑅 ​ 𝑦 , and for 𝑒 := 1 and 𝑓 := 0 , we have ( 𝑎 , 𝑒 ) , ( 𝑏 ​ 𝑒 + 𝑎 ​ 𝑓 , 1 − 𝑏 ​ 𝑠 − 𝑎 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) . Thus to prove that ( 2 ) ⇔ ( 3 ) we can assume that ( 1 − 𝑎 , 𝑏 ) ∉ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , and hence always 𝑡 ≠ 0 .

For 𝑡 1 := gcd ⁡ ( 𝑟 − 𝑞 ​ 𝑠 , 𝑡 ) we write 𝑡

𝑡 1 ​ 𝑡 2 and 𝑟 − 𝑞 ​ 𝑠

𝛼 ​ 𝑡 1 with ( 𝛼 , 𝑡 2 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) ; thus 𝑦

𝑟 + 𝑠 ​ ( 𝑡 − 𝑞 )

𝑠 ​ 𝑡 + 𝛼 ​ 𝑡 1 . We would like to find ( 𝑞 , 𝑟 ) ∈ 𝑅 2 such that there exists ( 𝑔 , ℎ ) ∈ 𝑅 2 with the property that 𝑔 ​ 𝑦 + 𝑎 ​ ℎ ​ 𝑡

𝑡 , i.e., 𝑔 ​ 𝑠 ​ 𝑡 + 𝑔 ​ 𝛼 ​ 𝑡 1 + 𝑎 ​ ℎ ​ 𝑡

𝑡 , and thus we must have 𝑔 ∈ 𝑅 ​ 𝑡 2 . Writing 𝑔

𝑡 2 ​ 𝛽 with 𝛽 ∈ 𝑅 , we want to find ( 𝑞 , 𝑟 ) ∈ 𝑅 2 such that there exists ( 𝛽 , ℎ ) ∈ 𝑅 2 satisfying 1

𝛽 ​ ( 𝑠 ​ 𝑡 2 + 𝛼 ) + 𝑎 ​ ℎ , equivalently, such that ( 𝑎 , 𝑠 ​ 𝑡 2 + 𝛼 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) . Replacing 𝑟

𝑞 ​ 𝑠 + 𝛼 ​ 𝑡 1 in 𝑡 , it follows that 𝑡

1 + 𝑞 − 𝑏 ​ 𝑞 ​ 𝑠 − 𝑏 ​ 𝛼 ​ 𝑡 1 − 𝑎 ​ 𝑞 , hence there exists 𝛾 ∈ 𝑅 such that 1 + 𝑞 ​ ( 1 − 𝑏 ​ 𝑠 − 𝑎 )

𝑡 1 ​ 𝛾 ; we have 𝑡 2

𝛾 − 𝑏 ​ 𝛼 and thus 𝑒 := 𝑠 ​ 𝑡 2 + 𝛼

𝑠 ​ 𝛾 + ( 1 − 𝑏 ​ 𝑠 ) ​ 𝛼 . As ( 𝛼 , 𝑡 2 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) , it follows that ( 𝛼 , 𝛾 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) . There exists a unique 𝑓 ∈ 𝑅 such that 𝛼

𝑒 − 𝑓 ​ 𝑠 and 𝛾

𝑏 ​ 𝑒 + 𝑓 ​ ( 1 − 𝑏 ​ 𝑠 ) ; in fact we have 𝑓

𝑡 2

𝛾 − 𝑏 ​ 𝛼 and thus 𝑅

𝑅 ​ 𝛼 + 𝑅 ​ 𝛾

𝑅 ​ 𝑒 + 𝑅 ​ 𝑓 , i.e., ( 𝑒 , 𝑓 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) . There exists 𝑞 ∈ 𝑅 such that 𝑏 ​ 𝑒 + 𝑓 ​ ( 1 − 𝑏 ​ 𝑠 ) divides 1 + 𝑞 ​ ( 1 − 𝑏 ​ 𝑠 − 𝑎 ) iff ( 𝑏 ​ 𝑒 + 𝑓 ​ ( 1 − 𝑏 ​ 𝑠 ) , 1 − 𝑏 ​ 𝑠 − 𝑎 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) and hence iff ( 𝑏 ​ 𝑒 + 𝑎 ​ 𝑓 , 1 − 𝑏 ​ 𝑠 − 𝑎 ) ∈ 𝑈 ​ 𝑚 ​ ( 𝑅 2 ) . Thus ( 2 ) ⇔ ( 3 ) . So Criterion 1.22 holds.

Acknowledgement. The second author acknowledges the support of the research project “Romanian Hub for Artificial Intelligence - HRIA”, Smart Growth, Digitization and Financial Instruments Program, 2021–2027, MySMIS no. 334906. The third author would like to thank SUNY Binghamton for good working conditions. Competing interests: the authors declare none.

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Grigore Călugăreanu   E-mail: calu@math.ubbcluj.ro Address: Department of Mathematics, Babeş-Bolyai University, 1 Mihail Kogălniceanu Street, Cluj-Napoca 400084, Romania.

Horia F. Pop   E-mail: horia.pop@ubbcluj.ro Address: Department of Computer Science, Babeş-Bolyai University, 1 Mihail Kogălniceanu Street, Cluj-Napoca 400084, Romania.

Adrian Vasiu,   E-mail: avasiu@binghamton.edu Address: Department of Mathematics and Statistics, Binghamton University, P. O. Box 6000, Binghamton, New York 13902-6000, U.S.A.

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