# Localizations of the Boolean Ring P(X)

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!

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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.

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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
@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
@JBeardz: yes, that's right. –  Qiaochu Yuan Aug 22 '12 at 0:54