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If B is a Boolean ring is of uncountable cardinality c, does B have 2^c distinct maximal ideals ? Can you please give me a reference where this question is answered (hopefully) positively ? Thanks

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3 Answers 3

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This particular question is easy to answer. Many related questions about the relationships between various cardinals associated with Boolean algebra (or Boolean rings), such as: cardinality, number of ultrafilters, density, cellularity, distributivity etc, can be found in

  • Monk, J. Donald: Cardinal invariants on Boolean algebras. Progress in Mathematics, 142. Birkhäuser Verlag, Basel, 1996. 3-7643-5402-X. MR1393943
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  • $\begingroup$ Since this is the accepted answer, it may be useful to say that the answer is: no. $\endgroup$
    – YCor
    Commented Nov 9, 2019 at 9:32
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This is false. We might as well count the number of ultrafilters, since an ultrafilter is the complement of a maximal ideal.

Let $X$ be a set of cardinality $c$. If $B$ is the set of subsets $S \subseteq X$ such that $S$ or its complement is finite, then $B$ is a Boolean algebra under the usual set-theoretic operations. If an ultrafilter $U \subset B$ contains a finite set $S$, then it is easy to see it must be principal. If it contains no finite set $S$, then it must be the ultrafilter of cofinite sets. So here the cardinality of the set of ultrafilters or of maximal ideals is $c$ again.

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  • $\begingroup$ Thanks for your answer and the example. I understand that the statement is true for atomless countable Boolean rings. $\endgroup$
    – Rangaswamy
    Commented Dec 24, 2012 at 20:55
  • $\begingroup$ Yes, there is only one such Boolean algebra up to isomorphism, and it is the free Boolean algebra on countably many generators. It is immediate that this has $c$ many ultrafilters in it. $\endgroup$
    – Todd Trimble
    Commented Jan 1, 2013 at 21:19
  • $\begingroup$ Equivalent topological description: $Y$ is the 1-point compactification of a discrete set of cardinal $c$ and $B$ is the Boolean algebra of continuous functions $Y\to\mathbf{Z}/2\mathbf{Z}$. $\endgroup$
    – YCor
    Commented Nov 25, 2016 at 8:05
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I think this is false—is it supposed to be the other way around?

The Stone representation theorem embeds $B$ in $\prod_{\mathfrak{p} \in \operatorname{Spec} B} \mathbb{F}_2$, where $\operatorname{Spec} B$ is the set of prime (or maximal) ideals of $B$. In particular, $c = \lvert B \rvert \le 2^{\lvert \operatorname{Spec} B\rvert}$.

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  • $\begingroup$ For infinite Boolean algebras you have $|B|\leq|Spec(B)|$, so the other inequality is not sharp except for finite Boolean algebras. $\endgroup$
    – Joseph Van Name
    Commented Dec 22, 2012 at 21:06

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