Talk:Localization (commutative algebra)

message: It seems that integers are being embedded into dyadic fractions, contrary to what is stated in the article.
of course you're right on the dyadic fractions.
Concerning etymology, does the word come from turning rings into local rings? - 80.143.125.195 14:57, 18 January 2007 (UTC)
thread_title:  Dyadic and etymology 

message: I think the author has the wrong definition of "total ring of fractions," which I believe is a very specific localization, namely the localization of a ring with respect to the multiplicatively closed set of all non-zero-divisors in that ring.
--66.92.4.19 (talk) 15:23, 23 October 2008 (UTC)
thread_title:  Total ring of fractions 

message: A recent edit suggested inverting a multiplicative system containing an ideal.  However, every ideal contains 0, so the localization at any multiplicative system containing an ideal is the zero ring.  I think they just meant the semigroup containing a specific element, which might as well be written {f^n:n=0,1,...}. JackSchmidt (talk) 07:07, 31 December 2008 (UTC)
thread_title:  Inverting an ideal 

message: I have a strange feeling the citation at the bottom should be Lang's Algebraic Number Theory. I know of no book by him entitled "Analytic Number Theory," and furthermore, the information in this article falls under algebraic number theory, not analytic.  —Preceding unsigned comment added by 141.154.116.219 (talk) 23:23, 7 September 2008 (UTC)
thread_title:  Citation correction 

message: First of the properties listed ($S^{-1}R = 0$ iff $0 \in S$) appears to be wrong. S may also contain nilpotent element.  — Preceding unsigned comment added by 188.123.231.34 (talk) 23:20, 29 September 2011 (UTC)
replies: text: No, the property is correct. Note that S contains a nilpotent element if and only if 0\in S For example, if s\in S is nilpotent, then there exists a positive integer n such that sn = 0 and since S is multiplicatively closed, it follows that 0\in S.--PST 00:27, 2 October 2011 (UTC)
thread_title:  Wrong property 

message: The example section says that when R is a commutative ring, and p is a prime ideal, then localizing against R-p yields a local ring Rp with maximal ideal p. But isn't the ideal really pRp? Even when localization is 1-1, p is not in general an ideal of Rp under the inclusion.
Thomaso (talk) 19:26, 14 January 2013 (UTC)
replies: text: You are right. I have corrected the sentence. D.Lazard (talk) 20:24, 14 January 2013 (UTC)
thread_title:  Maximal ideal when localizing versus (the complement of ) a prime ideal 

message: The use of the term "annihilator" for the ideal {a ∈ R; ∃s ∈ S : as = 0} in the section "For general commutative rings" is wrong. Any element of Ann(S) annihilates all of S (compare with Annihilator (ring theory)), but that's not what is needed in this situation.
2A02:810D:980:1704:4446:96B8:E70F:85A8 (talk) 08:43, 17 October 2014 (UTC)
replies: text: There isn't technically anything wrong with the use of the word annihilator here, but the explanation is a little muddled. I'll take a look at it tonight. Rschwieb (talk) 20:14, 17 October 2014 (UTC)
replies: text:  Also btw,  {a ∈ R; ∃s ∈ S : as = 0}  is rarely an ideal, and I don't believe the text implies this at all. The problem seems to be a disconnect between this statement and the use of Ann(S) in the rest of the paragraph. Rschwieb (talk) 00:11, 19 October 2014 (UTC)
replies: text:  Oh wait, upon rereading it I see of course the author did make that mistaken claim about the annihilator. I went ahead and took it out. Rschwieb (talk) 00:18, 19 October 2014 (UTC)
replies: text:  You confused ∃ with ∀. The annihilator is {a ∈ R; ∀s ∈ S : as = 0}. GeoffreyT2000 (talk) 18:01, 9 June 2015 (UTC)
thread_title:  Use of the term "annihilator" 

message: I propose the article be renamed 'Localisation (commutative algebra)' or similar Joel Brennan (talk) 16:14, 24 May 2019 (UTC)
replies: text: I agree. I'll be bold and doing it. D.Lazard (talk) 17:19, 24 May 2019 (UTC)
thread_title:  Title of article 

message: It is stated in the article that :
"If  and  are two multiplicative sets, then S^{-1}R and T^{-1}R are isomorphic if and only if they have the same saturation, or, equivalently, if  belongs to one of the multiplicative set, then there exists t\in R such that  belongs to the other."
This appears to be false : take S = \{0, 1\} and T = R, then the localizations are isomorphic (to the zero ring) but the saturation of  is merely the set of zero divisors. Another example is R = K \times K, and letting  and  be the complements of the maximal ideals K \times 0 and 0 \times K. Then the localizations are isomorphic (to ) but  and  are already saturated and different from each other.
Did I miss something ? I think the usual theorem states that this is supposed to be an isomorphism of -algebras. EHecky (talk) 17:20, 20 March 2023 (UTC)
replies: text: By definition of the saturation, if 0\in S, then the saturation of S is R, since x\cdot 0=0 for every x\in R. In other words, x divides 0, even when it is not a zero divisor. For zero divisors, the second factor must be nonzero. D.Lazard (talk) 17:49, 20 March 2023 (UTC)
thread_title:  Isomorphic localization iff same saturation ? 

message: I'm not an algebraic geometer, so I offer somebody who is one to take a look. Svennik (talk) 14:24, 17 April 2024 (UTC)
thread_title:  Filled in Zariski open sets section