Conch maximal subrings

Abstract

It is shown that if R is a ring, p a prime element of an integral domain $D\le R$ with ${\cap }_{n=1}^{\infty }{p}^{n}D=0$ and $p\in U(R),$ then R has a conch maximal subring (see [14]). We prove that either a ring R has a conch maximal subring or $U(S)=S\cap U(R)$ for each subring S of R (i.e., each subring of R is closed with respect to taking inverse, see [25]). In particular, either R has a conch maximal subring or U(R) is integral over the prime subring of R. We observe that if R is an integral domain with $\left|R\right|=2^{2^{{\aleph }_0}},$ then either R has a maximal subring or $|\mathit{Max}(R)|=2^{{\aleph }_0},$ and in particular if in addition dim(R) = 1, then R has a maximal subring. If $R\subseteq T$ is an integral ring extension, $Q\in \mathit{Spec}(T),P:=Q\cap R,$ then we prove that whenever R has a conch maximal subring S with $(S:R)=P,$ then T has a conch maximal subring V such that $(V:T)=Q$ and $V\cap R=S.$ It is shown that if K is an algebraically closed field which is not algebraic over its prime subring and R is affine ring over K, then for each prime ideal P of R with $ht(P)\ge \mathit{dim}(R)-1,$ there exists a maximal subring S of R with $(S:R)=P.$ If R is a normal affine integral domain over a field K, then we prove that R is an integrally closed maximal subring of a ring T if and only if dim(R) = 1 and in particular in this case $(R:T)=0.$

Keywords:

• Conductor ideal
• Conch subring
• integral element
• Maximal subring

2020 Mathematics Subject Classification:

• 13B02
• 13B21
• 13B30
• 13A05
• 13A15
• 13A18

Acknowledgement

The author would like to thank the referee for helpful comments which improve the paper.

Additional information

Funding

We are grateful to the Research Council of Shahid Chamran University of Ahvaz (Ahvaz-Iran) for financial support (Grant Number: SCU.MM99.721).