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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?

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 bit of commutative algebra and non-trivial…
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…
Slade
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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 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.
numberwat
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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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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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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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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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