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I'm trying to study universal algebra and lattice theory by myself. Just got stuck with an exercise from Gratzer's "General Lattice Theory" and it seems to me that I don't fully understand the notion of the WAP (Weak Amalgamation Property) and may be of the variety of algebras.

Here is the definition from the book:

A class $\mathbb{K}$ of algebras is said to have the Weak Amalgamation Property iff for any $B_0, B_1 \in \mathbb{K}$ there exists a $C \in \mathbb{K}$ into which both $B_0$ and $B_1$ can be embedded.

And then it's said that any variety of lattices has this property since we can form the direct product $C = B_0 \times B_1$ with the embeddings $\psi_0 \colon x \mapsto (x, b_1)$ and $\psi_1 \colon x \mapsto (b_0, x)$, where $b_0 \in B_0$ and $b_1 \in B_1$ are the fixed elements.

Problem: Show that the class of all Boolean algebras doesn't have the WAP.

Questions:

  1. Why can't we follow the same argument as for lattices: form the direct product with the embeddings? It seems to me that this can be done for any class of algebras closed under the direct products. Where am I mistaken?
  2. Do the Boolean algebras form a variety (and in particular is closed under the direct products by the Birkhoff's HSP theorem)? I thought so, because this class can be axiomatized, but now I doubt it. Can someone give me an example of two Boolean algebras such that their direct product is not a Boolean algebra?

If the answer on the second question is negative then I should look for the counterexample to WAP among infinite Boolean algebras, since I'm sure that for finite ones their direct product is a Boolean algebra.

Updated.

Attempted answers:

Now it seems to me that the main reason for which we can't follow the same argument as for lattices is that in the case of the Boolean algebras the maps $\psi_0 \colon x \mapsto (x, b_1)$ and $\psi_1 \colon x \mapsto (b_0, x)$, where $b_0 \in B_0$ and $b_1 \in B_1$ are fixed elements, aren't the homomorphims of Boolean algebras because of the constants $0$ and $1$ and their images ($0$ should be mapped to $(0, 0)$ and $1$ should be mapped to $(1, 1)$, hence we can't choose such fixed $b_1 \in B_1$). So now I'm sure that the second question has positive answer: the class of Boolean algebras is a variety and it's closed under the direct products.

But I still have no ideas about the counterexample to WAP. I think that my argument about finite algebras is valid, since it seems to me that any Boolean algebra $B^n$ on $n$ generators contains a subalgebra $B^m$ for any $m = \overline{1, n}$ (I haven't yet tried to prove it formally, but it seems that if we take an apropriate set of atoms as a generating set we'll get the desired subalgebra) so there is no need in the direct product, and hence the class of finite Boolean algebras has the WAP.

So at least one of the algebras $B_0$ or $B_1$ must be infinite. But I don't know how to proceed unless looking over various pairs (infinite\infinite or infinite\finite) of the Boolean algebras. Thanks!

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  • $\begingroup$ Please avoid making trivial edits just to keep the question on the front page. Either wait until you have a significant improvement to make or offer a bounty to get more attantion to your question. $\endgroup$
    – robjohn ♦
    Commented Jul 12, 2014 at 13:05

1 Answer 1

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The class of Boolean algebras fails to have the Weak Amalgamation Property for an extremely silly reason: The trivial Boolean algebra in which $0 = 1$ does not embed into any nontrivial Boolean algebra.

The class of all nontrivial Boolean algebras (this is no longer a variety of algebras) has the WAP: Given $B_1$ and $B_2$, choose an ultrafilter $U$ on $B_1$. Consider the map $f_U: B_1\rightarrow B_2$ given by $b\mapsto 1$ if $b\in U$, and $b\mapsto 0$ if $b\notin U$. The properties of an ultrafilter guarentee that $f_U$ is a homormorphism. Then the map $B_1\mapsto B_1\times B_2$ by $b\rightarrow (b,f_U(b))$ is an embedding.

By the way, the Weak Amalgamation Property is often (more commonly? well, at least among logicians...) called the Joint Embedding Property.

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  • $\begingroup$ The reason is really silly. It shames me that I forgot about trivial algebra, although every variety contains one. By the way, the proof style for non-trivial BA seems very model-theoretic to me. That is really interesting. Thank you a lot! $\endgroup$
    – Random Jack
    Commented Jul 28, 2014 at 21:37
  • $\begingroup$ Don't be ashamed - your question got me confused for a little while there, before the answer occurred to me. I also like the simple application of ultrafilters here. $\endgroup$
    – Alex Kruckman
    Commented Jul 28, 2014 at 21:43
  • $\begingroup$ I've taken the propositional calculus course a year ago and as I recall exactly the same function was used in the proof of completness theorem for Hilbert system. In other words, every ultrafilter defines the interpretation on Lindenbaum-Tarski algebra. Is this true? $\endgroup$
    – Random Jack
    Commented Jul 28, 2014 at 21:51
  • $\begingroup$ Yes, one can view a Boolean algebra $B$ as a collection of propositions modulo provable equivalence for some theory $T$ (this is equivalent to picking a presentation of $B$ as an algebraic structure - for example, name every element of $B$ with a proposition symbol and take $T$ to be the collection of all equalities between terms in $B$). Then an ultrafilter on $B$ is the same thing as a completion of $T$ is the same thing as a homomorphism from $B$ to the $2$-element Boolean algebra. $\endgroup$
    – Alex Kruckman
    Commented Jul 28, 2014 at 23:34
  • $\begingroup$ So in my answer I'm using the ultrafilter to get a map $B\rightarrow \{0,1\}$, and then using the fact that the $2$-element Boolean algebra embeds in every nontrivial Boolean algebra. $\endgroup$
    – Alex Kruckman
    Commented Jul 28, 2014 at 23:36

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