Bull. Korean Math. Soc. 2021; 58(3): 711-720
Online first article January 5, 2021 Printed May 31, 2021
https://doi.org/10.4134/BKMS.b200491
Copyright © The Korean Mathematical Society.
Reza Naghipour, Somayeh Vosughian
Institute for Research in FundamentalSciences (IPM), Institute for Advanced Studies in Basic Sciences
Let $R$ denote a commutative noetherian ring, and let ${\bf x} :=x_1,\ldots,x_d$ be an $R$-regular sequence. Suppose that $\mathfrak a$ denotes a monomial ideal with respect to ${\bf x}$. The first purpose of this article is to show that $\mathfrak a$ is irreducible if and only if $\mathfrak{a}$ is a generalized-parametric ideal. Next, it is shown that, for any integer $n\geq 1$, $(x_1,\ldots,x_d)^{n}=\bigcap {\bf P}(f),$ where the intersection (irredundant) is taken over all monomials $f=x_1^{e_1}\cdots x_d^{e_d}$ such that ${\rm deg}(f)=n-1$ and ${\bf P}(f):=(x_{1}^{e_{1}+1},\dots,x_{d}^{e_{d}+1})$. The second main result of this paper shows that if $\mathfrak q:=(\bf x)$ is a prime ideal of $R$ which is contained in the Jacobson radical of $R$ and $R$ is $\mathfrak q$-adically complete, then $\mathfrak{a}$ is a parameter ideal if and only if $\mathfrak{a}$ is a monomial irreducible ideal and ${\rm Rad}(\mathfrak{a})=\mathfrak q$. In addition, if $\mathfrak{a}$ is generated by monomials $m_{1},\dots, m_{r},$ then ${\rm Rad}(\mathfrak{a})$, the radical of $\mathfrak a$, is also monomial and ${\rm Rad}(\mathfrak{a})=(\omega_{1},\dots, \omega_{r})$, where $\omega_i={\rm rad}(m_i)$ for all $i=1, \dots, r$.
Keywords: Monomial ideal, parameter ideal, generalized-parametric ideal, monomial irreducible ideal, regular sequence
MSC numbers: 13A15, 13E05
Supported by: This work was financially supported by the Institute for Advanced Studies in Basic Sciences
2024; 61(1): 147-160
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