In the general sense of an algebra (a set with some operations, as in Universal Algebra courses), is it always possible to construct any full algebra (up to isomorphism) just from its finitely generated subalgebras, by the taking of suitable direct products?

  • $\begingroup$ What do you mean "construct"? It's hard to answer this question without it being more precise. $\endgroup$
    – Jim Belk
    Jan 19, 2014 at 23:18
  • $\begingroup$ @JimBelk I thought that would come up, which is why I added the last7 words. I meant take "smartly chosen" direct products and get something isomorphic. This comes from a question, in which I need to prove whether a given algebra is in a variety (of its same type) iff all of its finitely generated subalgebras are in the variety to begin with. I am exploring the idea of accomplishing this only with direct producs, I feel it'll be enough. $\endgroup$
    – FPP
    Jan 19, 2014 at 23:46

1 Answer 1


It is not possible to do it with direct product. There are many counterexamples, but I will take one that is right now on the top of my head: any atomless Boolean algebra is not a direct product of finite (that is the same as finitely generated) Boolean algebras, because every direct product of finite Boolean algebras is an atomic Boolean algebra.

However, every algebra is a direct limit of its finitely generated subalgebras.


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