Cardinality of the set of maximal ideals in a Boolean ring/algebra 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
 A: 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. 
A: 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 


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*Monk, J. Donald: Cardinal invariants on Boolean algebras.
Progress in Mathematics, 142. Birkhäuser Verlag, Basel, 1996. 3-7643-5402-X.  MR1393943

A: 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, $c = \lvert B \rvert \le 2^{\lvert \operatorname{Spec} B\rvert}$.
