Let $R$ be an exchange ring in which all regular elements are one-sided unit-regular. Then every regular element in $R$ is the sum of an idempotent and a one-sided unit. Furthermore, we extend this result to exchange rings satisfying related comparability.

In this paper we investigate the related comparability over exchange rings. It is shown that an exchange ring R satisfies the related comparability if and only if for any regular x C R, there exists a related unit w C R and a group G in R such that wx C G.

We characterize exchange rings having stable range one. An exchange ring $R$ has stable range one if and only if for any regular $a\in R$, there exist an $e\in E\left(R\right)$ and a $u\in U\left(R\right)$ such that $a=e+u$ and $aR\cap eR=0$ if and only if for any regular $a\in R$, there exist $e\in r.ann\left(a^{+}\right)$ and $u\in U\left(R\right)$ such that $a=e+u$ if and only if for any $a,b\in R$, $R/aR\cong R/bR⟹aR\cong bR$.

By a rotation in a Euclidean space V of even dimension we mean an orthogonal linear operator on V which is an orthogonal direct sum of rotations in 2-dimensional linear subspaces of V by a common angle α ∈ [0,π]. We present a criterion for the existence of a 2-dimensional subspace of V which is invariant under a given pair of rotations, in terms of the vanishing of a determinant associated with that pair. This criterion is constructive, whenever it is satisfied. It is also used to prove that every...

An $R$-module $M$ is said to be an extending module if every closed submodule of $M$ is a direct summand. In this paper we introduce and investigate the concept of a type 2 $\tau$-extending module, where $\tau$ is a hereditary torsion theory on $\text{Mod}$-$R$. An $R$-module $M$ is called type 2 $\tau$-extending if every type 2 $\tau$-closed submodule of $M$ is a direct summand of $M$. If ${\tau }_I$ is the torsion theory on $\text{Mod}$-$R$ corresponding to an idempotent ideal $I$ of $R$ and $M$ is a type 2 ${\tau }_I$-extending $R$-module, then the question of whether or not $M/MI$ is...