# Can nonisomorphic algebras generate isomorphic relatively free algebras?

Suppose that we have two varieties of algebras $A$ and $B$, whose operators all have arities less than some regular cardinal, and such that every $B$-algebra (please correct me if this is not the usual notation) is also an $A$-algebra. It is known that the forgetful functor from the category of $B$-algebras to that of $A$-algebras has a left adjoint $F$, which takes every $A$-algebra $a$ to the relatively free $B$-algebra $Fa$ generated by $a$. My question is, if $a$ and $b$ are $A$-algebras such that $Fa$ is isomorphic (as a $B$-algebra) to $Fb$, are $a$ and $b$ necessarily isomorphic? If not in general, how about the special case where $A$-algebras are Boolean algebras and $B$-algebras are countably complete Boolean algebras?

-
@Rotwang: I don't think your left adjoint is correct. For instance, if $A$ is the variety of all groups, and $B$ is the variety of abelian groups. The forgetful functor from $\mathcal{A}b$ to $\mathcal{G}roup$ indeed has a left adjoint, but the left adjoint $\mathcal{F}$ takes a group $G$ to $\mathcal{F}(G)=G^{\rm ab}$, the abelianization of $G$; it does not take an arbitrary group to a free abelian group. – Arturo Magidin Dec 9 '10 at 19:23
@Rotwang: The example of abelian groups inside groups will also show you that you can have $\mathcal{F}(a)\cong\mathcal{F}(b)$ even with $a\not\cong b$ in general. – Arturo Magidin Dec 9 '10 at 19:29
@Arturo: I wrote relatively free, not just free; I picked up this terminology from answers I received in this MO thread and the links therein, but perhaps I'm using it incorrectly: mathoverflow.net/questions/47278/… – Rotwang Dec 9 '10 at 19:37
Also, thanks for your answer. – Rotwang Dec 9 '10 at 19:38
@Rotwang: "Relatively free" is used to denote "free inside the given variety", especially when considering a subvariety. Thus, free abelian groups are 'relatively free' (as opposed to the "free group" that are 'absolutely free'). You have "relatively free nilpotent groups of rank $c$", "relatively free Burnside groups of exponent $n$", etc. What you can do to construct $Fa$ is to first take the free $A$-algebra on the underlying set of $a$, and then mod out by the congruence generated by the relations $a$ satisfied as a $B$-algebra. – Arturo Magidin Dec 9 '10 at 19:44

For example, take the case where $A$ is the variety of all groups, and $B$ is the variety of abelian groups. The forgetful functor from $\mathcal{A}b$ to $\mathcal{G}roup$ indeed has a left adjoint, but the left adjoint $\mathcal{F}$ takes a group $G$ to $G^{\rm ab}$, the abelianization of $G$. This easily yields examples of objects $a$ and $b$ such that $\mathcal{F}(a)\cong \mathcal{F}(b)$, even though $a\not\cong b$.

For the specific case of Boolean algebras I think the answer is still negative as stated: take a boolean algebra $a$ that is not [countably] complete and such that its [countable] completion has strictly larger cardinality than $a$ (I assume such things exist, but if this is not the case, then of course I'm falling flat on my case here). Let $b$ be the image of the [countable] completion under the forgetful functor. Since $a$ and $b$ have distinct cardinality, $a\not\cong b$, but $\mathcal{F}(b) = \mathcal{F}(\mathcal{U}(\mathcal{F}(a)))\cong \mathcal{F}(a)$, with the last equality using the unit and counits of the adjunction.

One way to construct $\mathcal{F}(a)$ if your varieties have a nice underlying set functor with an adjoint is to first consider $U(a)$, the underlying set of $a$, then take $\mathfrak{F}_B(U(a))$, the free $B$-algebra on the underlying set of $a$, and then let $\mathcal{F}(a) = \mathfrak{F}_B(U(a))/\Phi$, where $\Phi$ is the $B$-congruence generated by the relations corresponding to the $A$-operations on $a$. For instance, if $B$ is $\mathcal{G}roups$ and $A$ is $\mathcal{M}onoids$, given a monoid $M$, you first take $U(M)=\{[m]\mid m\in M\}$, the underlying set of $M$ (I'm using $[m]$ to denote the element $m$ in the set $U(M)$). Then you take the free group on $U(M)$, and you mod out by the the normal subgroup generated by the relations $[m][n]=[mn]$ (and since $[mn]=[nm]$, you end up with an abelian group).

Added: I need to think a bit more about the Boolean algebra case; you are correct that my assertion about $\mathcal{F}(\mathcal{U}(\mathcal{F}(a)))$ does not hold in general, and likely not in this case either. My gut feeling is that you can have $\mathcal{F}(a)\cong\mathcal{F}(b)$ even with $a\not\cong b$, though.

Added 2: Heh; reading the MO question you link to in the comments suggests that I am facing exactly what the long discussion between Joel David Hamkins and Todd Trimble was about, and I was making the same kind of assumptions Joel made to reach my conclusion that my construction would necessarily yield an example.

-
Sorry if I'm missing something, but I don't see how $\mathcal{F}(\mathcal{U}(\mathcal{F}(a)))$ can be isomorphic to $\mathcal{F}(a)$ in general - surely there can be homomorphisms $f: \mathcal{U}(\mathcal{F}(a)) \to \mathcal{U}(b)$ which fail to commute with countable sups? If the unit $\eta_{\mathcal{U}(\mathcal{F}(a))}$ were iso then it would necessarily commute with countable sups (since it would preserve the order relation), so how can $f = g \circ \eta_{\mathcal{U}(\mathcal{F}(a))}$ for some homomorphism $g$ of countably complete algebras? – Rotwang Dec 9 '10 at 20:40
But surely that only shows that $\eta_{\mathcal{F}(b)}$ is a split monic, not that it's iso? – Rotwang Dec 9 '10 at 20:50
@Rotwang. Hmmm... You may be right there and I misrememberd the unit and counit conditions. My modus operandi here is to always fall back to the free-group-underlying-set adjunction; and now that I think about it, $F(U(F(a)))$ is not isomorphic to $F(a)$ in that case, there is only a natural epimorphism. Sorry about that. Edited the answer. – Arturo Magidin Dec 9 '10 at 20:50
@Arturo: I think that the completion whose existence is proved in Neil's link is not the same as the one I'm talking about, since it is not left adjoint to the forgetful functor: any homomorphism from $a$ to a complete BA can be extended to a homomorphism from the completion of $a$, but the homomorphism in question will not commute with infinite joins in general. – Rotwang Dec 9 '10 at 21:20
@Rotwang: Fair enough. I'll try to think more about it, but I don't often think about Boolena algebras (complete or otherwise), so I don't have much of a feel for them. As I said, my gut feeling is that the answer is very likely to be "you can have $a\not\cong b$ with $\mathcal{F}(a)\cong\mathcal{F}(b)$", just on the grounds that this is what seems to happen in almost all other cases I know. – Arturo Magidin Dec 9 '10 at 21:38