# IBN for algebraic theories

Let us say that a finitary algebraic theory $\tau$ has IBN (invariant basis number) if the free functor $F : \mathsf{Set} \to \mathsf{Mod}(\tau)$ reflects the isomorphism relation: If $S,T$ are sets with $F(S) \cong F(T)$, then $S \cong T$. When $\tau$ is the theory of $R$-modules for some ring $R$, then this is the usual IBN property of $R$ (at least when we restrict to finite sets).

If $\tau \to \sigma$ is a homomorphism and $\sigma$ satisfies IBN, then also $\tau$ satisfies IBN. Besides the classical example of vector spaces, this gives lots of examples for IBN theories (abelian groups, modules over commutative rings $\neq 0$, groups and Lie algebras (using abelianization), monoids, semigroups, quasigroups, loops, magmas, commutative variants of them, etc.). One can show IBN for (commutative) $R$-algebras, where $R \neq 0$ is a commutative ring. Benjamin Steinberg has remarked in the comments that $\tau$ has IBN when there is a $\tau$-module with finite cardinality $>1$. This gives lots of further examples.

A directed colimit of IBN theories $\tau = \mathrm{colim}_i \tau_i$ is IBN for finite sets and therefore IBN by E: If $F_\tau(S) \to F_\tau(T)$ is a homomorphism, it is given by a map $S \to F(T)$, which factors through some $S \to F_{\tau_i}(T)$. Similarily the other way round. That $F_\tau(S) \to F_\tau(T) \to F_\tau(S)$ is the identity, already holds for the factorizations if $i$ is big enough. Therefore it suffices to consider finitely presented theories.

Now I have several questions:

A. Has the IBN property for algebraic theories in general been studied in the literature?

B. What are further interesting examples of IBN or $\neg$ IBN (beyond module categories)?

C. What about the infinitary theory of compact Hausdorff spaces? If $X,Y$ are sets such that their Stone-Čech compactifications $\beta(X),\beta(Y)$ are homeomorphic, does it follow $X \cong Y$? (answered by Benjamin Steinberg: Yes)

D. Do nontrivial commutative algebraic theories satisfy IBN? In other words, is the rank of a free module over a nontrivial generalized ring à la Durov well-defined? This should be crucial for the theory of generalized schemes, right?

E. Is there some algebraic theory which satisfies IBN for finite sets, but not IBN for arbitrary sets? (answered by Joseph Van Name: No).

F. When $|S| < \kappa$, then $F(S)$ is $\kappa$-presentable. Is there some $\tau$ which satisfies IBN, but $F(S)$ is $\kappa$-presentable for some $\kappa < |S|$?

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Shall I ask D for example separately? –  Martin Brandenburg Apr 9 '13 at 19:11

I believe the answer to C is yes. By Stone duality $\beta X$ and $\beta Y$ are homomorphic if and ony if $2^X\cong 2^Y$ as Boolean algebras. But $2^X$ has the singletons as the atoms so you can recover $X$ up to bijection.

Alternatively, the points of X are the only clopen points of $\beta X$.

For B, the free Jonsson-Tarski algebra on any non-empty finite set is the same.

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Agreed, I should have seen this. –  Martin Brandenburg Apr 7 '13 at 1:59
Alternatively, one can use E and reduce to the finite case, but then it's trivial. –  Martin Brandenburg Apr 8 '13 at 12:40
His argument is for varieties of algebras with finitary operations. I don't believe that compact spaces form a variety in this sense. I think that it cannot be defined with just finitary operations. For instance, one has operations of taking limits along ultrafilters built in I believe. Of course E does cover boolean algebras. –  Benjamin Steinberg Apr 8 '13 at 16:41

E. The answer to question E is no for varieties by a simple cardinality argument. Indeed, we can say much more. We claim that if $\mathcal{A}$ is an algebra with arbitrarily finitary operations and $X,Y\subseteq\mathcal{A}$ are minimal generating sets(i.e. no subset of $X$ generates $\mathcal{A}$), then either $X$ and $Y$ are both finite or $X$ and $Y$ are both infinite and of the same cardinality. The proof is by a simple cardinality argument. If $X,Y$ are minimal generating sets, then for each $x\in X$ there is a finite subset $Y_{x}\subseteq Y$ with $x\in\langle Y_{x}\rangle$. Therefore $X$ is generated by $\bigcup_{x\in X}Y_{x}$. However, since $X$ and $Y$ are minimal generating sets, we have $Y=\bigcup_{x\in X}Y_{x}$. Therefore, if $X$ is finite, then $Y=\bigcup_{x\in X}Y_{x}$ is finite as well. Similarly, if $Y$ is finite, then $X$ is finite, so $X$ is finite if and only if $Y$ is finite. If $X$ is infinite, then $|Y|=|\bigcup_{x\in X}Y_{x}|\leq\aleph_{0}\cdot|X|=|X|$ and similarly $|X|\leq|Y|$, so $|X|=|Y|$.

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Also note that if a variety has a non-trivial finite algebra then it has IBN for finite sets by counting homomorphisms to this object. –  Benjamin Steinberg Apr 7 '13 at 2:43
@Joseph: Great! I had already seen this proof for modules, but of course it works for every algebraic theory. I wonder if this is already written down somewhere in the literature? @Benjamin: Oh, that is interesting, and gives lots of examples of algebraic theories with IBN. –  Martin Brandenburg Apr 7 '13 at 13:10

A. Has the IBN property for algebraic theories in general been studied in the literature?

Yes, mostly by many polish authors. See Gr\"atzer, universal algebra, Chap. 5.

For $v$ and $v^*$-algebras, one can prove that any two bases have the same number of elements by using the same technique as for vector spaces, namely, by use of thef EIS property. It is rather surprising that, although the EIS property is not valid for $v^{**}$-algebras, the invariance of the number of elements in a basis can still be proved.