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Describe the free algebra on one generator in the variety of all algebras with two unary operations $f, g$. Do the same for the subvariety axiomatized by $f(g(x)) = g(f(x))$.

I'm not sure how to get started with this. Any insight would be greatly appreciated.

wasatar
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1 Answers1

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I believe this hint will give you insight enough; otherwise, I can expand this to a complete answer.

Let $\mathcal B$ be the variety of bi-unary algebras (say, with operations $f$ and $g$)
and $\mathcal V$ be its sub-variety given by the identity $fg(x)=gf(x)$.

The free algebra on one generator, say $1$, over $\mathcal B$ is represented by an infinite binary tree (say $f$ corresponds to the left hand branch, from each node, and $g$ to the right one);
The free algebra on one generator, say $(0,0)$, over $\mathcal V$ is isomorphic to $(\mathbb N, \sigma)^2$ (where $\sigma$ is the successor operation), where $f$ corresponds to increments on the first coordinate and $g$ on the second.

Prove each of these by checking these algebras have the universal mapping property for the correspondent varieties over a singleton.

amrsa
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    Probably, $\mathbf F_{\mathcal V}({(0,0)})$ is not $(\mathbb N, \sigma)^2$, but rather $(\mathbb N^2, \sigma_1,\sigma_2)$, where $\sigma_i$ is the increment in the $i$-th coordinate. – amrsa Oct 22 '19 at 11:44
  • So there can't be any finite free algebra in $\mathcal{V}$ since one element already generates an infinite algebra or am I wrong? – GEO Oct 23 '19 at 18:53
  • Thanks for responding, though I do not quite get your response: I was talking about finite FREE algebras in my comment. The "What we can conclude, along those lines" seems to exactly confirm what I stated, so am I missing something? – GEO Oct 23 '19 at 22:41