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On the grounds that I'm currently teaching a linear algebra class and I enjoy making my students furious, let a linear algebra be an algebra $\mathcal{A}$ in the sense of universal algebra equipped with a linear ordering. More interestingly, given a linear algebra $\mathfrak{A}=(\mathcal{A},\le)$, say that a linear subpower of $\mathfrak{A}$ is a subalgebra of some power structure $\mathcal{A}^\kappa$ which is linearly ordered by the corresponding product relation $\le^\kappa=\{(a,b): \forall \eta<\kappa, a(\eta)\le b(\eta)\}$ (which is a priori only a partial order).

For example, let ${\bf 2}$ be the two-element Boolean algebra ordered as usual. While (being nontrivial) it has arbitrarily large powers, it has no linear subpowers with more than two elements.

I'm curious how subtle the bounds on sizes of linear subpowers can be:

Suppose $\mathfrak{A}$ is a finite linear algebra which has an infinite linear subpower. Must $\mathfrak{A}$ have arbitrarily large finite linear subpowers?

This is related to an old question of mine, which in retrospect seems unlikely to have a good answer.

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Yes, and you can find them included in your infinite linear subpower. The variety $V$ generated by a finite algebra $\mathcal A$ (which includes all subpowers) is locally finite, i.e., finitely generated algebras in $V$ are finite, whence any algebra in $V$ is a directed union of finite subalgebras, whence any infinite algebra in $V$ has arbitrarily large finite subalgebras.

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  • $\begingroup$ Possibly a stupid question: is it obvious that $V(\mathcal{A})$ is locally finite if $\mathcal{A}$ is finite? I'm not immediately seeing that ... $\endgroup$
    – Noah Schweber
    Commented Nov 22, 2023 at 20:08
  • $\begingroup$ $V(\mathcal A)$ is locally finite if $\mathcal A$ is finite. I don’t think it holds in general if $\mathcal A$ is only locally finite. $\endgroup$
    – Emil Jeřábek
    Commented Nov 22, 2023 at 20:11
  • $\begingroup$ Sorry, I meant to ask the relevant version (comment now edited). $\endgroup$
    – Noah Schweber
    Commented Nov 22, 2023 at 20:11
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    $\begingroup$ This follows easily from the fact that for any $n<\omega$, there are only finitely many term functions $\mathcal A^n\to\mathcal A$, thus the $n$-generated free algebra of $V$ is finite. See e.g. Theorem 10.16 in Burris and Sankappanavar. $\endgroup$
    – Emil Jeřábek
    Commented Nov 22, 2023 at 20:15
  • $\begingroup$ Oh, yes. Derp - not my finest moment! $\endgroup$
    – Noah Schweber
    Commented Nov 22, 2023 at 20:17

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