Talk:Localization (commutative algebra)

Dyadic and etymology

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)[reply]

Citation correction

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)[reply]

Total ring of fractions

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)[reply]

Inverting an ideal

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)[reply]

Microlocalization

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.

Wrong property

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)[reply]

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 s^{n&=&0 and since $S$ is multiplicatively closed, it follows that $0\in S$ .--PST 00:27, 2 October 2011 (UTC)}[reply]