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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?

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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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  • $\begingroup$ 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... $\endgroup$
    – tcamps
    Commented Jul 16, 2015 at 14:42
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    $\begingroup$ Isn't every finitely generated boolean algebra free? $\endgroup$
    – Zhen Lin
    Commented Jul 16, 2015 at 16:26
  • $\begingroup$ @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. $\endgroup$ Commented Jul 17, 2015 at 4:01
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    $\begingroup$ 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.) $\endgroup$
    – Zhen Lin
    Commented Jul 17, 2015 at 10:35
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    $\begingroup$ @ZhenLin Don't free Boolean algebras on $n$ elements have cardinality $2^{2^n}$, whereas finitely generated Booleans can have more general cardinalities $2^k$? $\endgroup$
    – user43208
    Commented Aug 20, 2015 at 5:08

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