# Reduction matrices over Bezout rings

- Written by Andrij Sagan
- Be the first to comment!

*Department of Algebra and Logic** Faculty of Mechanics and Mathematics** Ivan Franko National University of L'vivemail: andrij.sagan at gmail.com*

* *

**Abstract:**

*A ring R is understood as a associative ring with nonzero unit element.*

*A ring R is said to have stable range 1, if for any a, b є R satisfying aR + bR = R, there exists t є R such that a + bt is an invertible element in R.*

*A ring R is called a exchange ring if for any element a є R there exists an idempotent e є R such that e є aR and 1-e є (1-a)R.*

*We will denote the Jacobson radical of a ring R by J(R). A ring R is said to be a semiexchange ring if the factor ring R/J(R) is an exchange ring.*

*Let R be a ring and a, b є R. The pair (a,b) is said to be an e-atomic pair if there exist Q є GE _{2}(R) and an atom element q є R such that (a,b)Q=(q,m) for a m є R. Then R is said to be e-atomic if for any a,b є R such that aR+bR=R and 0≠ c є R, there exists y є R such that (a+by,c) is an e-atomic pair.*

*A matrix ring of order n over a ring R is denoted by M _{n}(R). We say that a matrix A є M_{n}(R) is full if M_{n}(R) A M_{n}(R) = M_{n}(R). *

*We denote by F(M*

_{n}(R)) the class of all full matrices of the ring M_{n}(R).**Theorem 1.*** A Bezout duo ring with stable range 1 is a ring with elementary reduction of matrices. *

**Theorem 2.*** Let R be a semiexchange quasi-duo Bezout ring. Then R is a ring with elementary reduction of matrices if and only if it is a duo.*

**Theorem 3.*** Let R be a commutative e-atomic ring. The following statements are equivalent:** (1) R - ring with elementary reduction of matrices;** (2) R - Bezout ring.*

**Theorem 4.*** Let R be a commutative elementary divisor ring and n є Z _{>1}. Then for any full matrices A,B є F(M_{n}(R)), B≠O, there exists a right (left) 2(n-1)-stage terminating division chain in M_{n}(R).*

**Theorem 5.*** Let R is a PID and n є N _{>1}. Then M_{n}(R) is right (left) 2-stage euclidean set.*

## Коментарі (0)