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Suppose $F\dashv U$ is a free-forgetful adjunction of some algebraic theory. I thought there should be a one line proof that if $X\not\cong Y$ then $FX\not\cong FY$, i.e free objects on sets of different cardinality are not isomorphic. I am nowhere near finding one.

Having poked around on MSE, I don't see any hint of such a proof. The case of groups uses a lot of special structure which doesn't seem applicable to general algebraic theories. Now I also kind of doubt whether my intuitive claim is even true.

Is the claim true? Is there a (slick) proof?

Community
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Arrow
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  • Special case: If the signature of the algebraic theory is finite, I am pretty sure a cardinality argument works. – Mees de Vries Sep 20 '16 at 11:24
  • @MeesdeVries It'd be great if you could post a full solution as an answer. Could this fail for infinite signatures even if we allow choice (as choice is used in the linked case of groups)? – Arrow Sep 20 '16 at 11:31
  • Apparently I was wrong either way, seeing the answer! Actually, the idea I had in mind works only for finite signatures and infinite base sets. – Mees de Vries Sep 20 '16 at 11:38

2 Answers2

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Counterexample: The Jónsson–Tarski algebra.

Let $\mathbf K$ be the class of algebras $\mathfrak A=(A,f,g,h)$ of signature $(2,1,1)$ satisfying the identities $g(f(x,y))=x,\ h(f(x,y))=y,\ f(g(z),h(z))=z.$ The free $\mathbf K$-algebra on one generator is isomorphic to the free $\mathbf K$-algebra on $n$ generators for every positive integer $n.$

On the other hand, for a variety containing a finite algebra with more than one element, free algebras with different numbers of generators are nonisomorphic.

Reference: Bjarni Jónsson and Alfred Tarski, On two properties of free algebras, Math. Scand. 9 (1961), 95–101.

bof
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The standard counterexample is to take a take $R=\mathrm{End}(V)$ for some infinite-dimensional vector space $V$. Then $R \cong R^2$ as left $R$-modules.

HeinrichD
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