Questions tagged [krull-dimension]

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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 ...
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Does a surjection between polynomial rings lose one Krull dimension per generator quotiented?

Let $k$ be a Noetherian commutative ring with unity (I am happy adding hypotheses, as I will ultimately want $k = \mathbb Z$) and $$\varphi\colon A = k[x_1,\ldots,x_n] \to k[y_1,\ldots,y_p] = B$$ a ...
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Generating the annihilator ideal up to finite index

Let $\Lambda_2 = \mathbb{Z}_p[[T_1,T_2]]$ be the power series ring over $\mathbb{Z}_p$ in two variables, i.e., $\Lambda_2$ is a regular local ring of dimension 3. Let $M$ be the quotient of an ...
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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 ...
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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 ?
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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 ...
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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 $\dim M\gt 0$. $f_I(M) = \inf\ \{i : H_I^i(M)\ ... 3answers 521 views Algebraic characterization of commutative rings of Krull dimension 1,2, or 3 A commutative ring$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-... 2answers 318 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, ... 1answer 11k views A short proof for$\dim(R[T])=\dim(R)+1$For a commutative ring$R$we clearly have$\dim(R[T]) \geq \dim(R)+1$. If$R$is noetherian, we have equality. Every proof I'm aware of uses quite a bit of commutative algebra and non-trivial ... 0answers 539 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 ... 1answer 752 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{...
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 ...