Is every quotient algebra of a Boolean algebra isomorphic to a subalgebra?

Question

Is every (non-trivial) quotient of a Boolean algebra isomorphic to a subalgebra of that Boolean algebra? And conversely is every subalgebra isomorphic to a quotient algebra?

Answer 1

For the first question, the answer is no, there are quotient algebras that cannot be embedded as subalgebras. Let's take $\newcommand{\powerset}{\mathcal{P}}\newcommand{\N}{\mathbb{N}}A = \powerset(\N)$, the ideal $I \subseteq A$ to be the set of finite subsets, and $B = A/I$. We need to show that $B$ is not a subalgebra of $A$.

We start by observing that $A$ has what is known as the countable chain condition, which is that if $(S_i)_{i \in I}$ is a pairwise disjoint family of non-zero elements of $A$, then $I$ is countable. This follows from the countability of $\N$.

We can therefore show that $B$ is not a subalgebra of $A$ by finding an uncountable pairwise disjoint family in $B$. We do this as follows. For each $n \in \N$, we write $\mathbf{digits}(n)$ for the finite sequence expressing $n$ in binary. For each infinite binary sequence $a \in 2^\N$, define $S_a \subseteq \N$ as: $$ S_a = \{ n \in \N \mid \mathbf{digits}(n) \text{ is a prefix of } a\} $$ As is well known, $2^\N$ is uncountable. Each $S_a$ is infinite, because each finite prefix of $a$ comes from some $n \in \N$, so the equivalence class $[S_a]$ is non-zero in $B$. Finally, if $a \neq b$ are elements of $2^\N$, they have only finitely many prefixes that are the same, so $S_a \cap S_b$ is finite, so the equivalence classes $[S_a]$ and $[S_b]$ are disjoint in $B$. So we have shown that $([S_a])_{a \in 2^\N}$ is an uncountable disjoint family in $B$, which would be a contradiction if $B$ could be embedded as a subalgebra of $A$.

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For your second question, the answer is also no, there are subalgebras that are not isomorphic to any quotient. However, I was not able to make this example as elementary as the other one, so I will be using Stone duality and a bit of general topology.

Let's start again with $A = \powerset(\N)$, and define the subalgebra $B$ to consist of finite sets and cofinite sets, i.e. complements of finite sets. Assume for a contradiction that $B$ is representable as a quotient of $A$, so there is a surjective Boolean homomorphism $f : A \rightarrow B$. The Stone space of $A$ is $\beta \N$, the Stone-Čech compactification of $\N$, and the Stone space of $B$ is the "free convergent sequence", i.e. it is homeomorphic to the subspace $S = \{ 0, 1, \frac{1}{2}, \cdots, \frac{1}{n}, \cdots \} \subseteq \mathbb{R}$. The Stone dual of $f$ is a continuous injective map $g : S \rightarrow \beta(X)$. But as $\beta(X)$ is extremally disconnected, every convergent sequence is eventually constant, which is a contradiction.

Answer 2

This question gets right to the inner workings of Boolean algebras so I'm surprised nobody has answered it. I don't know the answer in general and would love to see a proof of a positive answer, or a counterexample!

For what it's worth I think I can show the answer is "yes" in both directions for finite Boolean algebras, and that something stronger than isomorphism holds. If $A$ is a Boolean algebra, and $Q$ is a quotient algebra of $A$ by some ideal, and $S$ is a subalgebra of $A$, let us say, for lack of a better word, that $Q$ "subsumes" $S$ when each $Q$-element (which is of course an equivalence class of $A$-elements) has exactly one element from $S$. If $Q$ subsumes $S$, the map taking each $Q$-member to its respective $S$-member is a (boolean) isomorphism onto $S$.

Claim: when $A$ is finite, every quotient algebra $Q$ subsumes some subalgebra $S$, and conversely every $S$ is subsumed by some $Q$.

First note that any finite boolean algebra $A$ is isomorphic to the power set of its atoms: the function mapping each $p \in A$ to the set of atoms $\leq p$ is a bijection, and the boolean operations on $A$ agree respectively with union, intersection, and complement on the power set of $A$'s atoms. Thus the finite boolean algebras can be characterized up to isomorphism as $\mathcal{P}(\{1, ..., n\})$ where $n$ can be any natural number $> 0$. (I am ignoring the one-element "degenerate" algebra.)

Second, note that a subalgebra of $\mathcal{P}(\{1,...,n\})$ is uniquely determined by its atoms, which together necessarily form a partition of $\{1,...,n\}$; and any such partition can serve as atoms for a subalgebra. A quotient of $\mathcal{P}(\{1,...,n\})$ by a nontrivial ideal is uniquely determined by that ideal's greatest element, which can be any nonempty subset of $\{1,...,n\}$.

Now let $Q$ be the quotient of $\mathcal{P}(\{1,...,n\})$ by some given ideal, which by the preceding paragraph must have a greatest element which can by any nonempty subset $X \subseteq \{1,...,n\}$. Let $Z$ be any partition of $\{1,...,n\}$ such that each $e \in Z$ has exactly one member that is not in $X$. Then I think it should be easy to show that $Z$ is the set of atoms of a subalgebra $S$ that is subsumed by $Q$. Note that since $Z$ need not be unique in general for a given $Q$, $S$ need not be unique in general.

Conversely, when $S$ is any subalgebra of $\mathcal{P}(\{1,...,n\})$, and $Z$ is its set of atoms (which again can be any partition of $\{1,...,n\}$), let $X$ be any subset of $\{1,...,n\}$ such that for all $e \in Z$, $|X \cap e| = |e| - 1$. Then when $Q$ is the quotient by the ideal whose greatest element is $X$, $Q$ subsumes $S$.