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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)
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)
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)
I believe that the end of the article contains an error. Micro local analysis has nothing to do with (micro) localization, as far as I understand.
This has now been corrected.
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)
Thank you for pointing this out: this has now been corrected.
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)
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)
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)
I propose the article be renamed 'Localisation (commutative algebra)' or similar Joel Brennan ( talk) 16:14, 24 May 2019 (UTC)
In the 'Properties' section (1.3), before the bullet point about the bijection between prime ideals in the ring and in the localisation, there should be a bullet point about the bijection between ordinary ideals of the ring and the localisation; the bijection of the prime ideals is then a restriction of this bijection. There should also be a bullet point afterwards saying that the bijection between ideals does not restrict to a bijection of maximal ideals. The reason I have not made this edit myself is because I do not know a counterexample for the maximal ideals. Joel Brennan ( talk) 21:15, 29 May 2019 (UTC)
Two points.
It is stated in the article that : "If S and T are two multiplicative sets, then and are isomorphic if and only if they have the same saturation, or, equivalently, if s belongs to one of the multiplicative set, then there exists such that st belongs to the other."
This appears to be false : take and , then the localizations are isomorphic (to the zero ring) but the saturation of S is merely the set of zero divisors. Another example is , and letting S and T be the complements of the maximal ideals and . Then the localizations are isomorphic (to K) but S and T 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 R-algebras. EHecky ( talk) 17:20, 20 March 2023 (UTC)
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)
![]() | This article is rated C-class on Wikipedia's
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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)
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)
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)
I believe that the end of the article contains an error. Micro local analysis has nothing to do with (micro) localization, as far as I understand.
This has now been corrected.
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)
Thank you for pointing this out: this has now been corrected.
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)
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)
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)
I propose the article be renamed 'Localisation (commutative algebra)' or similar Joel Brennan ( talk) 16:14, 24 May 2019 (UTC)
In the 'Properties' section (1.3), before the bullet point about the bijection between prime ideals in the ring and in the localisation, there should be a bullet point about the bijection between ordinary ideals of the ring and the localisation; the bijection of the prime ideals is then a restriction of this bijection. There should also be a bullet point afterwards saying that the bijection between ideals does not restrict to a bijection of maximal ideals. The reason I have not made this edit myself is because I do not know a counterexample for the maximal ideals. Joel Brennan ( talk) 21:15, 29 May 2019 (UTC)
Two points.
It is stated in the article that : "If S and T are two multiplicative sets, then and are isomorphic if and only if they have the same saturation, or, equivalently, if s belongs to one of the multiplicative set, then there exists such that st belongs to the other."
This appears to be false : take and , then the localizations are isomorphic (to the zero ring) but the saturation of S is merely the set of zero divisors. Another example is , and letting S and T be the complements of the maximal ideals and . Then the localizations are isomorphic (to K) but S and T 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 R-algebras. EHecky ( talk) 17:20, 20 March 2023 (UTC)
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)