Papers
Topics
Authors
Recent
Search
2000 character limit reached

Einfach-teilbare und einfach-torsionsfreie R-Moduln

Published 14 Nov 2019 in math.AC and math.RA | (1911.06141v1)

Abstract: Let $(R, \mathfrak{m})$ be a commutative Noetherian local ring with total quotient ring $K$. An $R$-module $M$ is called simple divisible, if $M$ is divisible $\neq 0$, but every proper submodule $0 \neq U \subsetneqq M$ is not divisible. Dually, $M$ is called simple torsion free, if $M$ ist torsion free $\neq 0$, but, for every proper submodule $0 \neq U \subsetneqq M$, the factor module $M/U$ is not torsion free. Our first result is that $M \neq 0$ is simple torsion free iff $M$ is a submodule of $\kappa(\mathfrak{p}) = R_{\mathfrak{p}}/\mathfrak{p} R_{\mathfrak{p}}$ for a maximal element $\mathfrak{p}$ in $\operatorname{Ass}(R)$. The structure of simple divisible modules is more complicated and was examined primarily by E. Matlis (1973) over 1-dimensional local $CM$-rings and by A. Facchini (1989) over any integral domain. Our main results are: If the injective hull $E(R/\mathfrak{q})$ is simple divisible ($\mathfrak{q} \in \operatorname{Spec}(R)$), then the ring $R_{\mathfrak{q}}$ is analytically irreducible and essentially complete. Especially for $\mathfrak{q} = \mathfrak{m}$, the simple divisible submodules of $E(R/\mathfrak{m})$ correspond exactly to the maximal ideals of the ring $\hat{R} \otimes_R K$, and $E(R/\mathfrak{m})$ itself is simple divisible iff $\hat{R} \otimes_R K$ is a field.

Summary

Paper to Video (Beta)

Whiteboard

No one has generated a whiteboard explanation for this paper yet.

Open Problems

We haven't generated a list of open problems mentioned in this paper yet.

Continue Learning

We haven't generated follow-up questions for this paper yet.

Authors (1)

Collections

Sign up for free to add this paper to one or more collections.