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Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.

0 votes

Faithfully flat ring homomorphism and annihilator

The following counterexample might be interesting too. There exists an unmixed d-dimensional commutative Noetherian local domain $(R,\mathfrak{m})$ such that its completion is not unmixed. Then $H^d_\ …
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1 vote
0 answers
217 views

Computing the bourbaki ideals

By virtue the Griffith's paper and subsequently e.g. Goto's paper several examples of several desired class of Noetherian normal domains with specific finite length local cohomologies are constructed …
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  • 591
0 votes
1 answer
227 views

Canonical module of rees algebra

[Example 4.27, Integral Closure, Rees Algebras, Multiplicities, Algorithms] by Vasconcelos, says that if $I=(f_1,\ldots,f_g)$ is an ideal generated by a regular sequence with $g\ge 2$ then the canoni …
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  • 591
0 votes
0 answers
110 views

$F$-pure threshold of an $F$-pure ideal

According to this reference an elliptic curve is $F$-pure if and only if the $F$-pure threshold of its defining ideal is $1$. Does there exist an $F$-pure local ring $R=A/\mathfrak{a}$ such that $\tex …
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  • 591
3 votes
0 answers
300 views

Hilbert-Samuel multiplicity

Let $G$ be a group of order $p$ acting linearly on $A=\mathbb{Z}/p\mathbb{Z}[x_1,\ldots,x_r]$. Does there exist a formula for the Hilbert-Samuel multiplicity of the unique homogeneous maximal ideal of …
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  • 591
0 votes
0 answers
371 views

Completion of commutative rings.

Assume that $(R,\mathfrak{m})$ is a commutative local ring of equal characteristic zero. So $R$ contains the field of rationals. The well known $\mathfrak{m}$-adic completion of $R$ provides a comple …
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  • 591
8 votes
2 answers
784 views

Faithfully flat ring homomorphism and annihilator

Suppose that the ring homomorphism $R\rightarrow S$ is faithfully flat ($R$ and $S$ are Noetherian commutative rings). Let $A$ be an Artinian $R$-module. Do we have $0:_S(A\bigotimes_R S)=(0:_RA)S$. …
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  • 591
5 votes
1 answer
1k views

discrete valuation ring and ring of witt vectors

Given a perfect field $F$ of prime characteristic the ring of Witt Vectors $W(F)$ is a discrete valuation ring. For example, $W(\mathbb{F}_p)$ is the ring of $p$-adic integers. Is it possible to embed …
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2 votes
0 answers
225 views

why do we care about the irreducibility of parameter ideals?

It is well known that a local commutative unital ring $R$ is Gorenstein if and only if every parameter ideal is irreducible. Why the irreducibility of parameter ideals in a Gorenstein local ring is …
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1 vote
Accepted

M is an R-module which is not finitely generated. is it true that $\inf \{ i| H^i_I(M)\neq 0...

For the concept of the grade of an arbitrary module with respect to a finitely generated ideal see the section 9.1 (entitled as "Grade and acyclicity") of the book entitled as "Cohen-Macaulay Rings" w …
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  • 591
3 votes

Definitions of the module $R/(x_0^\infty,x_1^\infty,\ldots,x_{n-1}^\infty)$

I strongly guess that, you mean, the maps of your direct system, $$R/I_t\rightarrow R/I_{t+1},$$ whose colimit is, $$R/I^\infty,$$ are the multiplication by $x_0\ldots x_{n-1}$. So I think that your …
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  • 591
3 votes
1 answer
224 views

Canonical module of a Buchsbaum ring

Is the canonical module of a Buchsbaum ring a Buchsbaum module?
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  • 591
2 votes
1 answer
268 views

A question about Complete Intersections

Suppose that $(A,\mathfrak{m})$ is a complete intersection and $\mathbf{x}$ is a minimal basis for $\mathfrak{m}$. Consider the Koszul homologies $H_\bullet(\mathbf{x},A)$. It is well-known that $\tex …
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2 votes

Almost complete intersection ideal and $d$-sequence

Not, necessarily. For example, in $K[x,y]$, $x^3,xy^2$ is not a $d$-sequence, but it generates an almost complete intersection. However, in some cases, the answer is positive. For example, see (5) of …
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1 vote

Example of a locally complete intersection ideal

Let ,$R=\mathbb{Z}/3\mathbb{Z}[x,y,z,w]/(x^4+y^3+z^4)$. Then, $y\in (x,z)^F\subseteq (x,z)^*\subseteq \overline{(x,z)}$, and thence $(x,z)$ is not integrally closed, however $x,z$ is a regular sequenc …
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