Suppose $A$ is an algebra of signature $\mathcal{L}$ and $V=Var(A)$ is the variety generated by $A$. I want to know is it possible to classify relatively free elements of $V$? As a special case, for a group $G$, under what conditions $G$ is free in $Var(G)$?
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$\begingroup$ In what sense do you wish to classify these? $\endgroup$– Benjamin SteinbergCommented Jan 6, 2014 at 19:47
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$\begingroup$ @Benjamin Steinberg: For example I want to know is $A$ free relative to $V$? From one side it seems that the answer is YES because $A$ satisfies exactly the same identities as the defining identities of $V$. On the other side, I have a sense that says this may be not true. $\endgroup$– Sh.M1972Commented Jan 6, 2014 at 20:13
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$\begingroup$ A will not in general be free in V. For example, the group S_3 is not free in the variety. If it were it would be free on 2-generators. Both generators would have to have the same order and hence must be of order 2. But then the map sending these generators to a 3-cycle and a 2-cycle does not extend. $\endgroup$– Benjamin SteinbergCommented Jan 6, 2014 at 20:43
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$\begingroup$ Another counterexample: The variety of Boolean algebras is generated by any non-trivial boolean algebra $A$ (because the 2-element Boolean algebra is a subalgebra of $A$), but many non-trivial Boolean algebras fail to be free. $\endgroup$– Andreas BlassCommented Jan 6, 2014 at 20:47
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$\begingroup$ You might think about nonfree algebras to understand the situation. Take a variety V generated by A. In imteresting cases V will have many other algebras B which generate V, such as those with a subalgebra isomorphic to A. Many of these B will not be free in V. There will be relations revealed by looking at proper subvarieties of V. This may help you find those B which generate V and are not free in V $\endgroup$– The Masked AvengerCommented Jan 6, 2014 at 20:57
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3 Answers
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Suppose that $A$ is a finite universal algebra with minimal cardinality of a generating set $d$. Then $A$ is relatively free in some variety iff it is relatively free on $d$ generators in the variety it generates, in which case it is free on any generating set of $d$ elements. Moreover, this occurs iff each map from a fixed $d$-element set generating $A$ to $A$ extends to an endomorphism.
It is easy to check that if $A$ is relatively free in some variety then it is relatively free in $Var(A)$. Next note that if $A$ is relatively free on $k$ generators, then the cardinality of $Hom(A,A)$ is $|A|^k$. On the other hand, $|Hom(A,A)\leq |A|^d$ and so $k\leq d$. But $d$ was minimal, so $A$ is free on $d$ generators. Also note that if $X$ is a free set of $d$-generators, then we can map $X$ to any other set of $d$-generators and this extends to a surjective homomorphism which is injective by finiteness. So all $d$-element generating sets are free generating sets.
Clearly if $A$ is relatively free on $d$ generators, then any map from a $d$-element generating set (necessarily a free generating set) to $A$ extends to an endomorphism. Suppose the converse holds. Let $B$ be the free algebra on $d$ generators in the variety generated by $A$. It is well known that $B$ embeds in a finite product $A^m$ where $m=|A|^d$. Let $g_i\colon B\to A$ be the projection to the $i^{th}$ factor. If $f$ is the map taking the $d$ generators of $A$ to the free generators of $B$, then we can extend $g_if$ to an endomorphism $h_i$ of $A$ by hypothesis on $A$. The product $h$ of these $h_i$ gives a homomorphism of $A$ to $A^m$ sending the generators of $A$ to the generators of $B$. Hence $h$ splits the canonical surjection $B\to A$.
Added. This last argument doesn't require finiteness. A universal algebra $A$ is relatively free if and ony if it has a generating set $X$ such that each map from $X$ to $A$ extends to an endomorphism.
answered Jan 6, 2014 at 21:05
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$\begingroup$ Let me to summarize the result above as I understand: A finite algebra $A$ with $d$ generator ($d=\min$) is relatively free if and only if any map from the set of $d$ generators to $A$ extends to an endomorphism. Am I right? $\endgroup$– Sh.M1972Commented Jan 8, 2014 at 5:14
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More general than my comment above, but still only a partial answer: Notice that $\text{Var}(A)$ is generated by any $B\in\text{Var}(A)$ that has $A$ as a subalgebra (or quotient, or subquotient). It would seem that only in very special varieties would all such $B$'s be free.
answered Jan 6, 2014 at 20:49
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Free algebras of $V(\textbf{A})$ lie in $SP(\textbf{A})$ (subalgebra of product). Unless your algebra is trivial or very special, I do not see a better characterization of free algebras forthcoming.
answered Jan 6, 2014 at 21:20