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Questions tagged [krull-dimension]

For questions about or related to the Krull dimension, which counts the length of the longest chain of prime ideals of a ring under inclusion.

The Krull dimension of a commutative ring $R$ is defined to be the supremum of the lengths of chains of prime ideals in $R$. Given a chain of prime ideals

$$p_0 \subsetneq p_1 \subsetneq \dots \subsetneq p_n$$

we define the length of this chain to be $n$ (that is, $n$ is the number of strict inclusions). The Krull dimension is the supremum of the quantity $n$ over all such chains.

A field has Krull dimension $0$, and any principal ideal domain that is not a field has Krull dimension $1$. It is not necessary that a ring has finite Krull dimension, even if the ring is Noetherian.

If $M$ is an $R$-module, we define the Krull dimension of $M$ to be

$$\dim_R M = \dim(R/\operatorname{Ann}_R(M))$$

where $\operatorname {Ann}_R(M)$ is the annihilator in $R$ of $M$.

Reference: Krull dimension.

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A proof for $\dim(R[T])=\dim(R)+1$ without prime ideals?

Please read this first before answering. This is not the right place for you to advertise your favorite proof of the dimension formula. This question is only concerned with a proof using the Coquand-Lombardi characterization below. If you post…
Martin Brandenburg
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Is $\operatorname{height} \mathfrak{p} + \dim A / \mathfrak{p} = \dim A$ true?

Let $A$ be an integral domain of finite Krull dimension. Let $\mathfrak{p}$ be a prime ideal. Is it true that $$\operatorname{height} \mathfrak{p} + \dim A / \mathfrak{p} = \dim A$$ where $\dim$ refers to the Krull dimension of a ring?…
Zhen Lin
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Noetherian ring with infinite Krull dimension (Nagata's example).

I just started to read about the Krull dimension (definition and basic theory), at first when I thought about the Krull dimension of a noetherian ring my idea was that it must be finite, however this turned out to be wrong. I am looking for an…
user117449
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Examples of rings whose polynomial rings have large dimension

If $A$ is a commutative ring with unity, then a fact proved in most commutative algebra textbooks is: $$\dim A + 1\leq\dim A[X] \leq 2\dim A + 1$$ Idea of proof: each prime of $A$ in a chain can arise from at most two prime ideals of $A[X]$. The…
Andrew Dudzik
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What is the "dimension" of a locally ringed space?

Let $(X,\mathscr{O}_X)$ be a locally ringed space. If it is a scheme, the natural notion of dimension is the dimension of the subjacent topological space (the size of the biggest chain of irreducible closed subsets). But if $X$ is a manifold, I…
Gabriel
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Why are Artinian rings of Krull dimension 0?

Why are Artinian rings of Krull dimension 0? As in the example of $\mathbb{Z}/(6)$, the ideal $\mathbb{Z}/(2)$ is prime, I think. So, Artinian rings may contain prime ideals. But why does not the primes ideal contain other prime ideals…
ShinyaSakai
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Krull dimension of $\mathbb{C}[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>$

Krull dimension of a ring $R$ is the supremum of the number of strict inclusions in a chain of prime ideals. Question 1. Considering $R = \mathbb{C}[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>$, how does one…
math-visitor
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Krull Dimension of a scheme

Can someone give a hint or a solution for showing that a scheme has Krull dimension $d$ if and only if there is an affine open cover of the scheme such that the Krull dimension of each affine scheme is at most $d$, with equality for at least one…
only
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Krull dimension of polynomial rings over noetherian rings

I want to prove the following theorem concerning Krull dimension: Theorem If $A$ is a noetherian ring then $$\dim(A[x_1,x_2, \dots , x_n]) = \dim(A) + n$$ where $\dim$ stands for the Krull dimension of the rings. Thus, $\dim(K[x_1,x_2, \dots ,…
Pipicito
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Krull dimension and localization

If we have a commutative ring with 1, what can we say about $\operatorname{dim} S^{-1}A$ for some multiplicative subset $S$, or more specifically, what happens if $S = A \setminus \mathfrak{p}$ for a prime ideal $\mathfrak{p}$? Do the dimensions of…
argon
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Krull dimension of quotient by principal ideal

Let $R$ be a commutative unitary ring of finite Krull dimension and let $x \in R$. Is it true that $\dim R/(x) \ge \dim R -1$ ?
user120513
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Finding a space $X$ such that $\dim C(X)=n$.

Let $n\in \mathbb{N}$ . Is there some topological space $X$ such that $C(X)$ is a finite dimensional ring with $\dim C(X) = n$? Here, $C(X):=\{ f:X \to \mathbb{R} \mid f$ is continuous$\}$ and $\dim C(X)$ means Krull dimension.
number
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The Krull dimension of a module

Let $R$ be a ring, $M$ is a $R$-module. Then the Krull dimension of $M$ is defined by $\dim (R/\operatorname{Ann}M)$. I can understand the definition of an algebra in a intuitive way, since the definition by chain of prime ideals agrees with the…
Arsenaler
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Noetherian ring with finitely many height $n$ primes

If $R$ is a Noetherian commutative ring with unity having finitely many height one prime ideals, one could derive from the "Principal Ideal Theorem", due to Krull, that $R$ has finitely many prime ideals (all of height less than or equal to $1$). It…
karparvar
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Why is every Noetherian zero-dimensional scheme finite discrete?

In the book The geometry of schemes by Eisenbud and Harris, at page 27 we find the exercise asserting that Exercise I.XXXVI. The underlying space of a zero-dimensional scheme is discrete; if the scheme is Noetherian, it is finite. Thoughts I…
awllower
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