Does every commutative Bézout domain R have the property that every matrix A∈ R^m× n admits a Smith normal form; that is, do there exist invertible matrices P∈ operatornameGL_m(R) and Q∈ operatornameGL_n(R) such that PAQ is diagonal, with nonzero diagonal entries d_1,…,d_r satisfying d_1| d_2|…| d_r? Equivalently, is every commutative Bézout domain an elementary divisor domain?

Bézout-domain elementary-divisor problem

The problem asks whether every commutative Bézout domain is an elementary divisor domain, meaning that matrices over it admit Smith normal form. Earlier literature treated this as open; a recent paper claims a counterexample arising from the Möbius line bundle.

Known results

Bézout domains of Gelfand range $1$ are elementary divisor domains, including $PM^{*}$ -domains and local Gelfand domains (2015).
For semihereditary Bézout rings, the Shores criterion reduces the question to the factor rings $R/aR$ (2012).
Bézout domains with all maximal ideals principal are elementary divisor domains; further criteria use $R/\operatorname{rad}(aR)$ (2015).
Feckly zero-adequate Bézout rings are elementary divisor rings (2015).

September 28, 2026 counterexample claim

Christian Hägg and Anders Mörtberg report that a topological obstruction from the Möbius line bundle produces a Bézout domain that is not an elementary divisor domain. If correct, this settles the unrestricted question negatively and explains a concrete failure of Smith normal form; the claim is presently unverified.

Current status (as of September 2026): A counterexample has been claimed, while the earlier positive results for restricted classes remain established and independent verification of the claimed counterexample is pending.

Bézout-domain elementary-divisor problem

Primary source

Additional references

Progress summary

Solutions 0