Talk:Krull dimension

Some edits to improve the readability of the article

• No symbol has been specified for the Krull dimension of a ring, the definition should be changed to something like:

We define the Krull dimension of R to be the supremum of the lengths of all chains of prime ideals in R and we denote it by $\operatorname {dim} _{\rm {K}}(R)$ (or simply $\operatorname {dim} (R)$ when there is no risk of confusion).

• Incorrect use of punctuation: (geometers call it the ring of the normal cone of I.) should be changed to (geometers call it the ring of the normal cone of I).
• Add some links:

• the space of prime ideals of R equipped with the Zariski topology --> the space of prime ideals of R equipped with the Zariski topology
• The equality holds if R is finitely generated as algebra (for instance by the noether normalization lemma). ---> The equality holds if R is finitely generated as an algebra (for instance by the Noether normalization lemma).
• In the language of schemes, finitely generated modules are interpreted as coherent sheaves, or generalized finite rank vector bundles. ---> In the language of schemes, finitely generated modules are interpreted as coherent sheaves, or generalized finite rank vector bundles.

• Rename the Notes section as References;
• Introduce a Notes section for remarks and clarifications on the many facts listed in the article.
• Add the following note concerning the fact the height of ${\mathfrak {p}}$ is the Krull dimension of the localization of $R$ at ${\mathfrak {p}}$ to the Note section

This follows from the following observation: for any prime ideal ${\mathfrak {p}}\varsubsetneq R$ consider the localization of $R$ to the multiplicative system $S=R\setminus {\mathfrak {p}}$ which we denote by $S^{-1}R=:R_{\mathfrak {p}}$; the natural map $j\colon R\to R_{\mathfrak {p}}$ induces a bijection

$\lbrace {\text{prime ideals }}I\subseteq {\mathfrak {p}}\varsubsetneq R\rbrace \rightleftharpoons \lbrace {\text{prime ideals }}I\varsubsetneq R_{\mathfrak {p}}\rbrace ,$

defined by $I\mapsto I_{\mathfrak {p}}=\lbrace x/t\mid x\in I,\,t\in S\rbrace$, with inverse $J\mapsto j^{-1}(J)=\lbrace x\in R\mid x/1\in J\rbrace$.

• I am concerned with the use of both I and I to denote an ideal: not only is this confusing for the reader, but the symbol I (or $\mathbb {I}$) is also commonly used to denote either the set of imaginary numbers or the compact $[0,1]\subset \mathbb {R}$ (especially in algebraic topology).

Please, let me know what you think.--Ale.rossi91 (talk) 23:08, 1 February 2020 (UTC)

Example

It seems to me that there is an error in computation of the Krull dimension of (Z/8Z)[x,y,z] : we get a chain of prime ideals of length four by adding the (0) ideal to the chain that is given : $(0)\subsetneq (2)\subsetneq (2,x)\subsetneq (2,x,y)\subsetneq (2,x,y,z)$. Thus I think that the dimension is 4.

129.199.2.17 (talk) 11:38, 13 February 2009 (UTC)

It's correct, because we don't count (0)? (Otherwise, the field would have the dimension 1.) -- Taku (talk) 21:25, 13 February 2009 (UTC)

In fact, we are both wrong, and the article was correct. The ideal (0) is prime if and only if the ring is a domain. The example is not a domain, so (0) is not prime. In the case of a field, the only prime ideal is (0), because the whole ring (field) is never a prime ideal. Thus the dimension of a field is still 0.82.67.178.125 (talk) 22:33, 14 February 2009 (UTC)

Eh?

Eh?

Mr. Billion 08:47, 12 Jan 2005 (UTC)

More precise in the case of infinite Krull dimension

The Krull dimension of a commutative ring $R$ should be defined as

$\dim R:=\sup\{{\text{cardinality of totally-ordered subsets of }}\operatorname {Spec} R\},$

where $\operatorname {Spec} R$ is the set of prime ideals of $R$, partially ordered by inclusion. We see clearly that the Krull dimension is in general a cardinal. By the way, the zero ring should have Krull dimension $0-1=-1$ since it has no prime ideals. 129.104.241.225 (talk) 18:51, 15 November 2024 (UTC)