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Call a Lawvere theory $T$ dimensive iff, letting $F_T : \mathbf{Set} \rightarrow \mathbf{Mod}(T)$ denote the free functor, we have the following.

  • Every finitely generated $T$-algebra is free.
  • From $F_T(J) \cong F_T(I)$ we may deduce $J \cong I$, for all finite sets $I,J$.

Motivation. If $T$ is dimensive, then every finitely-generated $T$-algebra has a well-defined dimension.

Examples.

  • The initial Lawvere theory (whose models are sets).
  • For each field $K$, the Lawvere theory of $K$-modules.

Question. Is this an exhaustive list of dimensive Lawvere theories?

goblin GONE
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1 Answers1

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This is getting too long for a comment, but here are some more examples:

  • Modules over a division ring $K$ work.
  • Also, pointed sets work.
  • Affine spaces over a field, or more generally a division ring, work.

These examples come from here where Lawvere theories with all algebras free are considered.

Some observations from over there still apply. Your Lawvere theory must have at most one constant. If you want all algebras to be free (not just finitely-generated ones), then affine spaces over a division ring and sets themselves are the only examples with no constants, and it seems likely that pointed sets and vector spaces over a division ring are the only examples with constants.

Update If you follow the link above, you'll see that Keith Kearnes and collaborators now have a paper out showing that these are the only Lawvere theories for which every finitely generated algebra is free, regardless of the second condition! They all satisfy the second condition, too.

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tcamps
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  • Glad to help! I'd be interested to see an example of a Lawvere theory where every finitely-generated algebra is free, but not every algebra is free. I suspect there are examples well-known to the experts... – tcamps Jul 16 '15 at 14:42
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    Isn't every finitely generated boolean algebra free? – Zhen Lin Jul 16 '15 at 16:26
  • @ZhenLin, nice example; see here for proofs. However, I think that the Lawvere theory of $\mathbb{F}_2$-modules is isomorphic to the Lawvere theory of Boolean algebras; if so, then this isn't really a new example. – goblin GONE Jul 17 '15 at 04:01
  • No, it is not. Boolean algebras are $\mathbb{F}_2$-algebras. – Zhen Lin Jul 17 '15 at 05:38
  • @ZhenLin, hmm yeah that makes sense. $\mathbb{F}_2$-algebras are just Boolean rings. Just out of curiosity, how do we prove $xy = 0$ in the $\mathbb{F}_2$-algebra freely generated by ${x,y}$? – goblin GONE Jul 17 '15 at 06:55
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    It isn't true. Boolean algebras are $\mathbb{F}_2$-algebras but not every $\mathbb{F}_2$-algebra is a boolean algebra. (There's an extra equation.) – Zhen Lin Jul 17 '15 at 10:35
  • @ZhenLin, I see. So it seems that "boolean algebra" equals "boolean ring", and that "$\mathbb{F}_2$-algebra" equals "ring whose characteristic is $2$", but the latter strictly generalizes the former. Nonetheless it seems that the $\mathbb{F}_2$-module freely generated by a set $X$ becomes a Boolean ring in a canonical way. Have I got anything wrong this time? – goblin GONE Jul 17 '15 at 11:57
  • I don't believe that. What's the canonical algebra structure on the $\mathbb{F}_2$-vector space of countable dimension? – Zhen Lin Jul 17 '15 at 12:13
  • @ZhenLin, excepting that we don't have a multiplicative identity, I think we can simply define that $xy = 0$ for all distinct generators $x$ and $y$, and that $x^2=x$ for all generators $x$, and then extend using bilinearity. – goblin GONE Jul 17 '15 at 12:15
  • That's not a boolean algebra, however. – Zhen Lin Jul 17 '15 at 12:30
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    @ZhenLin Don't free Boolean algebras on $n$ elements have cardinality $2^{2^n}$, whereas finitely generated Booleans can have more general cardinalities $2^k$? – user43208 Aug 20 '15 at 05:08
  • Ah, yes – you're right. – Zhen Lin Aug 20 '15 at 06:20