# Questions tagged [krull-dimension]

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### What is the dimension of this subvariety of the Grassmannian?

Well, actually, what are the dimensions of the following two subvarieties of the Grassmannian. Let $N$ be a positive integer. Let $V \subseteq \mathbb{C}^N$ be a linear subspace of dimension $N-k$ ...
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### Dimension of the associated graded module at an ideal

Let $I$ be an ideal of a Noetherian local ring $(R, \mathfrak m)$. Define the associated graded ring $G_I(R):=\bigoplus_{n=0}^\infty I^n/I^{n+1}$. Then $G_I(R)$ is a Noetherian ring of the same ...
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A text I am following uses of the following (probably basic) commutative algebraic lemma, omitting its proof. Lemma: Let $n\in\mathbb{N}_{>0}$, and let $P_0\subsetneq P_1\subsetneq\cdots\... 7 votes 0 answers 112 views ### Catenarity and epimorphisms of rings Let$R$be a commutative ring. The following are well-known: If$R$is catenary and$\mathfrak{a}\subseteq R$is an ideal, then$R/\mathfrak{a}$is catenary. If$R$is catenary and$S\subseteq R$is ... 3 votes 0 answers 212 views ### Krull dimension of schemes locally of finite type over PID Let$R$be a commutative unital ring that is a PID. Assume that$R$is not a DVR. Let$X$be an integral scheme locally of finite type over$\mathrm{Spec}\,R$. Can the Krull dimension of$\mathcal{O}... 168 views

### Is every universally catenary ring a going-between ring?

This question asks for the necessity of a noetherian hypothesis in a certain relation between properties of rings concerning chains of prime ideals. We use the following definitions. A ring $R$ is ...
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### Computing the codimension of the variety defined by a system of quadratic forms

Suppose I have an $m \times n$ matrix $L$, where $m \leq n$ and each entry is $L_{i,j}(x_1, ..., x_s)$ which is a linear form over $\mathbb{C}$. Let $\mathbf{y} = (y_1, \ldots, y_n)$. Let us consider ...
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### Valuation ring satisfying either a.c.c. or d.c.c. on prime ideals

If a commutative ring with unity has finite Krull dimension, then it satisfies a.c.c. and d.c.c. on prime ideals. The converse is not true in general, as can be seen from here An infinite dimensional ...
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I am trying to understand if a certain condition on quotient rings is sufficient for a sequence to be regular. Here is the setting: Let $\mathbb{C}[u_1,...,u_n]$ be the ring of regular functions on $\... 7 votes 1 answer 343 views ### An infinite dimensional local domain whose chains of primes are finite Does there exist a local domain of infinite dimension in which every chain of prime ideals is finite? Of course, such a ring must be neither noetherian nor catenary. (This question arose while ... 1 vote 1 answer 101 views ### Decomposing Noetherian hereditary rings of Krull dimension$1$into product of hereditary domains (i.e. Dedekind domains) Let$R$be a commutative Noetherian hereditary ring (https://en.wikipedia.org/wiki/Hereditary_ring) of Krull dimension$1$. Then is it true that$R$is a finite direct product of Dedekind domains ? 3 votes 1 answer 383 views ### Local ring of infinite dimension Short version: Let$R$be a commutative ring such that all chains of primes of$R$with the same extremities have the same finite cardinality. Is$R$locally finite-dimensional? Longer version: Let$R$... 2 votes 0 answers 343 views ### Intersection of all positive powers of prime ideal in an integral domain with all ideals of finite height Let$R$be an integral domain with every prime ideal having finite height. Then is$\bigcap_{n>1} P^n$a prime ideal of$R$for every prime ideal$P$of$R$? If that is not true in general, then ... 1 vote 1 answer 365 views ### Valuation ring whose maximal ideal and every ideal of finite height are principal Let$(R, \mathfrak m)$be a valuation ring such that$\mathfrak m$and every ideal of finite height is principal. Then is$R$Noetherian , i.e. a discrete valuation ring ? 3 votes 0 answers 281 views ### On rings for which given an ideal , over it every minimal prime ideal is finitely generated Let$R$be a commutative ring with unity. If for every ideal of$R$, the minimal prime ideals over it are all finitely generated, then there are finitely many minimal prime ideals over every ideal of$... 1 vote
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### Integral domain satisfying a.c.c. on radical ideals and with algebraically closed fraction field

If $R$ is an integral domain satisfying acc on radical ideals (i.e. Noetherian spectrum) and if the fraction field of $R$ is algebraically closed, then is $R$ a field ? If $R$ is normal (integrally ... 317 views

Let $R$ be an integral domain with Krull dimension $1$. If $0\ne a \in R$ is such that for every $b \in R$ , the ideal $Ra \cap Rb$ is principal , then is it true that for every $b\in R$, the ideal $... 4 votes 1 answer 285 views ### torsion free modules$M$over Noetherian domain of dimension$1$for which$l(M/aM) \le (\dim_K K \otimes_R M) \cdot l(R/aR), \forall 0 \ne a \in R$Let$R$be a Noetherian domain of Krull-dimension$1$(i.e. every non-zero prime ideal maximal). Let$M$be a torsion free$R$-module . Let$K$be the fraction-field of$R$and let$r=\dim_K S^{-1}M=\... 556 views

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### Generalization of Krull dimension for commutative rings

In the paper How to construct huge chains of prime ideals in power series rings by B. Kang and P. Toan the Krull dimension of a commutative ring with $1$ is defined as follows: Let $R$ be a ...
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### Dimension of Ext modules [closed]

Let $(R,m)$ be a noetherian local ring, and $M$ and $N$ be two finitely generated $R$-module. Then is it true that $\dim \text{Ext}^k(M,N)\leq \dim M-k$? If not does the reversed inequality hold?
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Let $(R,m)$ be a commutative unital noetherian local ring (with $m$ as its maximal ideal), $I$ an ideal of $R$, and $M$ a finite $R$-module with $\dim M\gt 0$. $f_I(M) = \inf\ \{i : H_I^i(M)\ ... 6 votes 3 answers 682 views ### Algebraic characterization of commutative rings of Krull dimension 1,2, or 3 A commutative ring$R$(with$1$) is$0$-dimensional if and only if$R/\sqrt 0$is von Neumann regular. Besides this result, there is a wealth of information about the algebraic structure of zero-... 5 votes 2 answers 367 views ### Integral domains with totally ordered spectra In my research I ended up trying to prove some properties of integral domains such that their spectrum is a totally ordered poset. Are there some nice (ubiqitous/natural) examples of such domains, ... 101 votes 1 answer 14k views ### A proof for$\dim(R[T])=\dim(R)+1$without prime ideals? If$R$is a commutative ring, it is easy to prove$\dim(R[T]) \geq \dim(R)+1$, where$\dim$denotes the Krull dimension. If$R$is Noetherian, we have equality. Every proof I'm aware of uses quite a ... 0 votes 0 answers 614 views ### valuation ring of dimension 2 I was looking the valuation ring of dimension$2$. Then I found, it has two number of non-zero prime ideals and localization at prime is a valuation domain again. Moreover, there is a one-to-one ... -3 votes 1 answer 1k views ### An elementary question about the Krull dimension of modules [closed] Let$R$be a commutative ring. If$0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$is an exact sequence of modules, we have that$\operatorname{Supp}M=\operatorname{Supp}M'\cup \operatorname{...
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For all $n$, I need to find examples of rings $A\subset B$ such that: i) $\dim A-\dim B\gt n$ ii) $\dim B-\dim A\gt n$ (where $\dim$ is the Krull dimension)
A totally ordered finite set $\quad \mathcal P_0 \varsubsetneq \mathcal P_1\varsubsetneq \dots \mathcal \varsubsetneq \mathcal P_n \quad$ of prime ideals of a ring $A$ is said to be a chain of ...