Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $A$ be a fixed boolean ring. Is there a sort of classification of boolean rings $B$ with $A \subseteq B$? For example, if $A=\mathbb{F}_2$, the answer would be Stone duality: $B=C(Spec(B),\mathbb{F}_2)$. Also in the general case, Stone duality tells us that the inclusion $A \subseteq B$ corresponds to a surjective map of Stone spaces $Spec(B) \to Spec(A)$. Remark that $B=C(T,A)$ for some Stone space $T$ if and only if $Spec(A)$ is a direct factor of $Spec(B)$, which does not hold in general. Anyway, is there another description of $B$ which involves $A$ and perhaps a kind of relative spectrum "$Spec(B/A)$"?

For example if $A$ is finite, then $X=Spec(A)$ is finite discrete and it's not hard to see that $B \to (B/\mathfrak{p}B)_{\mathfrak{p} \in X}$ and $(B^{\mathfrak{p}})_{\mathfrak{p} \in X} \mapsto \prod_{\mathfrak{p} \in X} B^{\mathfrak{p}}$ is an equivalence of categories between those boolean rings $B$ with $A \subseteq B$ and $X$-tuples $(B^{\mathfrak{p}})$ of boolean rings such that $A/\mathfrak{p} \subseteq B^{\mathfrak{p}}$. And this category is dual to the category of $X$-tuples of Stone Spaces.

What happens when $A$ is infinite? I hope it's clear what I'm looking for. Otherwise feel free to comment. Please don't answer "There is no such description".

share|improve this question
    
I am confused about your initial example, because every Boolean algebra includes the two-element Boolean algebra as a subalgebra. Are you asking the question you intend to ask? What am I misunderstanding? –  Joel David Hamkins Jun 23 '10 at 14:48
    
Yes of course $\mathbb{F}_2 \subseteq B$, but this description then does not involve $A$. I want a description which depends on $A$ (and, of course, another data). –  Martin Brandenburg Jun 24 '10 at 8:56

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.