# 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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### 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 $R$...
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### 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 ...
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### 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 ?
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### Krull-dimension of local domain

Let $(R,{\frak m}_R)$ be a local domain (not necessarily Noetherian). That is, $R$ is integral and ${\frak m}_R$ is the unique maximal ideal of $R$. Suppose that ${\frak m}_R$ is finitely generated. ...
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### Elementary characterization of Krull dimension

I was reading the following paper: "A Short Proof for the Krull Dimension of a Polynomial Ring. Thierry Coquand and Henri Lombardi" and came across this corollary. (This is present with a better ...
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### 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-...
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### 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, ...
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### A proof of $\dim(R[T])=\dim(R)+1$ without prime ideals?

Please read this first before answering. This question is only concerned with a proof of the dimension formula using the Coquand-Lombardi characterization below. If you post something that doesn't ...
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### Does every regular Noetherian domain have finite Krull dimension?

Does every regular Noetherian domain have finite Krull dimension? Background: A Noetherian ring is said to be regular if its localizations at all prime (or maximal) ideals are regular local rings. ...
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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{...