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Following the case of groups, I asked in this MSE question for a quick proof that given a free-forgetful adjunction $F\dashv U$ for some algebraic theory, we have $X\not\cong Y\implies FX\not\cong FY$. I was really surprised to find out this isn't true at all.

On the other hand, it seems that for the usual theories, i.e of (abelian) groups, (commutative) rings, modules, monoids, this is true, usually by finding a canonical way to vector spaces and using the dimension theorem. What is the underlying property of such theories that makes this theorem true? Can it be stated without reference to vector spaces, using only the axiom of choice? Is there a nice sufficient condition?

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    $\begingroup$ The same question was asked here: mathoverflow.net/questions/126747 $\endgroup$ – HeinrichD Sep 23 '16 at 10:10
  • $\begingroup$ @HeinrichD ah, whoops. I didn't know the term 'IBN'. $\endgroup$ – Arrow Sep 23 '16 at 10:12
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It is not true for "usual theories", at least if you accept that left $R$-modules for noncommutative $R$ is a usual theory. Take $R=\mathrm{End(}V)$ for some infinite-dimensional vector space $V$, then $R \cong R^2$ as left $R$-modules. Rings for which $\forall n,m : \mathbb{N}. R^n \cong R^m \Rightarrow n=m$ holds (the case of infinite bases follows by cardinality arguments) are called IBN-rings; they are studied for example in the beginning of Lam's Lectures on modules and rings.

As for the general question, the following criterion is very useful: Assume that there is some algebra $A$ with a finite underlying set $U(A)$ which has at least $2$ elements. Then, if $F(X) \cong F(Y)$ holds for two finite sets $X,Y$, we get $$\hom(X,U(A)) \cong \hom(F(X),A) \cong \hom(F(X),A) \cong \hom(Y,U(A)).$$ Counting yields $X \cong Y$.

This was also remarked here, an almost identical question.

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Jónsson and Tarski (Math. Scand. 9 (1961), 95-101, link) proved the following:

Consider a variety $\mathcal{V}$ of algebras of a certain signature. If $\mathcal{V}$ contains a finite algebra of cardinal $\ge 2$, then the free algebra $F_n(\mathcal{V})$ in $\mathcal{V}$ on $n$ elements cannot be generated by less than $n$ elements. In particular, $F_n(\mathcal{V})$ and $F_m(\mathcal{V})$ are not isomorphic for any $0\le n<m$ for $n$ finite.

This applies in many cases, for which the result itself is trivial or not: set with no law, free lattices, free magmas, free semigroups, free groups, free associative commutative unital ring, free associative ring, etc.

They also provide a counterexample without the assumption that $\mathcal{V}$ contains a finite algebra of cardinal $\ge 2$. Namely, consider the variety $\mathcal{V}_2$ with signature $(2,1,1)$, encoding a set $M$ with a bijection $M^2\to M$. Then the free algebras $F_n(\mathcal{V}_2)$ for finite $n\ge 1$ are all isomorphic. (This is called the variety of Jónsson-Tarski algebras, or pairing functions.)

(It was later checked by Higman that if we consider the variety $\mathcal{V}_k$ encoding a bijection $M^k\to M$, then $F_n(\mathcal{V}_k)$ and $F_m(\mathcal{V}_k)$, for finite $n,m\ge 1$, are isomorphic if and only if $k-1$ divides $m-n$.)

They also mention the variety of $M$-sets, where $M$ is the monoid $\langle u,v\mid uv=1\rangle$: although it has the given property, the free algebra on 1 generator in this algebra can be non-freely generated by a single element. (They also give a sufficient condition in variety ensuring that every generating $n$-tuple in the free $n$-generated algebra is free, namely residual finiteness of free algebras.)

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  • $\begingroup$ [Of course another counterexample for the first fact is for degenerate varieties, namely consisting only of sets of cardinal $\le 1$, for which all free algebras on nonempty sets are singletons. A nondegenerate variety has models of arbitrary large infinite cardinals (consider a fixed model of cardinal $\ge 2$ and its powers). So $\mathcal{V}_2$ is interesting because it's also nondegenerate.] $\endgroup$ – YCor Feb 5 at 17:30

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