Powers of Ideals : Primary Decompositions , Artin-Rees Lemma and Regularity

Powers of Ideals : Primary Decompositions , Artin-Rees Lemma and Regularity
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理想的力量:一次分解,Artin-Rees引理和正则性

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发表时间:
2002
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通讯作者:
I. Swanson
I. Swanson
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作者:
I. Swanson

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正则性部分是由初级分解部分衍生而来的,因此本文的核心是对初级分解的分析。在[S]中,证明了高度的主要分量不超过理想值的1。本文证明了这样一个k的存在性,但没有给出它的公式。在论文[SS]中,Karen E. Smith和我找到了域上多项式环上单项式理想的普通幂和Frobenius幂的显式k,以及Katzman首先研究的特殊理想的Frobenius幂。Chandler [C]和Geramita, Gimigliano和Pitteloud [GGP]给出了特殊理想下Castelnuovo-Mumford正则性的显式k。Heinzer和Swanson [HS]给出了另一种证明局部形式等维解析非分枝Noetherian环上理想幂次初等分解k存在的方法。初等分解结果并非对所有初等分解都有效。举个例子:设I为域k上两个变量X和Y中的多项式环k[X,Y]中的理想(X, XY)。对于每一个正整数m, I = (X)∩(X, XY, Y)是I的一个无冗余初等分解。然而,对于每一个整数k,存在一个整数m, m = k + 1,使得(X,Y) 6≥(X, XY, Y)。于是结果只能
The regularity part follows from the primary decompositions part, so the heart of this paper is the analysis of the primary decompositions. In [S], this was proved for the primary components of height at most one over the ideal. This paper proves the existence of such a k but does not provide a formula for it. In the paper [SS], Karen E. Smith and myself find explicit k for ordinary and Frobenius powers of monomial ideals in polynomial rings over fields modulo a monomial ideal and also for Frobenius powers of a special ideal first studied by Katzman. Explicit k for the Castelnuovo-Mumford regularity for special ideals is given in the papers by Chandler [C] and Geramita, Gimigliano and Pitteloud [GGP]. Another method for proving the existence of k for primary decompositions of powers of an ideal in Noetherian rings which are locally formally equidimensional and analytically unramified is given in the paper by Heinzer and Swanson [HS]. The primary decomposition result is not valid for all primary decompositions. Here is an example: let I be the ideal (X, XY ) in the polynomial ring k[X,Y ] in two variables X and Y over a field k. For each positive integer m, I = (X) ∩ (X, XY, Y ) is an irredundant primary decomposition of I. However, for each integer k there exists an integer m, say m = k + 1, such that (X,Y ) 6⊆ (X, XY, Y ). Hence the result can only