Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a set $X$, we can construct the Boolean ring whose elements are the power set of $X$. The multiplication therein is intersection, and the addition is symmetric difference. I am interested in finding information on the study of such rings, for instance, information about the spectrum of such a ring. In that case, I am specifically interested in trying to find a systematic description of what the localization of such a ring at an object (i.e. a subset of $X$) would look like. Doing some basic calculations, strange things seem to happen. For instance, if $X$ is the natural numbers, and we localize at (i.e. invert all powers of) the set $\{1,2\}$, we get constructions like $\{1,2\}+1/\{1,2\}$, which is an object that when multiplied by (intersected with) the set $\{1,2\}$, yields $X-\{1,2\}$, which is pretty wild. This is a pretty algebraically simple subject, but I don't recall ever seeing anything written down about it.

Thanks for any help!

share|improve this question

1 Answer 1

up vote 3 down vote accepted

The spectrum is the space $\beta X$ of ultrafilters on $X$. More generally, the spectrum of any Boolean ring is a Stone space, and this actually gives a contravariant equivalence of categories. For more details, see this blog post. For way more details, see Johnstone's Stone Spaces.

Algebraically, the Boolean ring you are interested in is $\mathbb{F}_2^X$. Inverting the indicator function $\chi_S$ of a subset $S \subseteq X$ just gets you $\mathbb{F}_2^S$ (it kills of all the copies of $\mathbb{F}_2$ at which $\chi_S$ is equal to $0$). If $X = \mathbb{N}$ and $S = \{ 1, 2 \}$ then the localization is just $\mathbb{F}_2^{ \{ 1, 2 \} }$ and $\chi_S + \frac{1}{\chi_S} = 2 \chi_S = 0$ in the localization.

share|improve this answer
    
Hmmmmm that's really helpful. I must have made some mistake in trying to figure out what these localizations look like. –  Jon Beardsley Aug 22 '12 at 0:42
1  
@JBeardz: you haven't made a mistake. In your notation, $X \setminus \{ 1, 2 \}$ is equal to $0$ in the localization. (Note that the only way you can force an idempotent to be invertible is to force it to be the identity.) –  Qiaochu Yuan Aug 22 '12 at 0:43
    
so i guess basically what happens if, in terms of just straight up Boolean algebras instead of indicator functions, to localize at $S$ means that you get a new Boolean ring whose elements are just the subsets of $S$....? –  Jon Beardsley Aug 22 '12 at 0:53
1  
@JBeardz: yes, that's right. –  Qiaochu Yuan Aug 22 '12 at 0:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.