6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Special case: If the signature of the algebraic theory is finite, I am pretty sure a cardinality argument works. $\endgroup$ – Mees de Vries Sep 20 '16 at 11:24
  • $\begingroup$ @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)? $\endgroup$ – Arrow Sep 20 '16 at 11:31
  • $\begingroup$ 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. $\endgroup$ – Mees de Vries Sep 20 '16 at 11:38
11
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.