Talk:Krull dimension

message: * 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)
thread_title:  Some edits to improve the readability of the article 

message: 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)
replies: text: 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)
replies: text: 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)
thread_title:  Example 

message: Eh? 
Mr. Billion 08:47, 12 Jan 2005 (UTC)
thread_title:  Eh?