Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $Var(G)$?

share|improve this question
    
In what sense do you wish to classify these? –  Benjamin Steinberg Jan 6 at 19:47
    
@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. –  M. Shahryari Jan 6 at 20:13
    
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. –  Benjamin Steinberg Jan 6 at 20:43
    
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. –  Andreas Blass Jan 6 at 20:47
    
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 –  The Masked Avenger Jan 6 at 20:57

3 Answers 3

up vote 3 down vote accepted

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.

share|improve this answer
    
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? –  M. Shahryari Jan 8 at 5:14
    
Yes you are right. –  Benjamin Steinberg Jan 8 at 14:32

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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