Re: sub-structures of free structures
Date: Sun, 23 Oct 1994 15:35:43 +0000 (GMT) From: "John G. Stell" <john@cs.keele.ac.uk>
What results of the form `A sub-X of a free X is free' are known? I am aware of groups, magmas (= set with binary operation), and modules over a principal ideal domain. I'd be interested to hear of structures for which this is either known to be true, or examples where it fails.
John Stell
It is true for abelian groups, but is false for both monoids and commutative monoids. The free (commutative) monoid on one element x has a submonoid consisting of 1,x^3,x^5,x^6 and all x^n for n > 7 is not free. Throwing away 1, you get counter-examples for semigroups and commutative semigroups. By taking the free abelian group on this example, you get counter-examples for rings and commutative rings. I am sure that it is a fairly rare occurrence. Michael Barr
John Stell asked when to expect subobjects of free objects to be free and observed that such is the case for modules over prinicpal ideal domains. For commutative rings that's the only case: if every ideal is free as a module then the ring is a PID. He points out that subalgebras of free algebras are free if the theory consists of just one binary operation and no equations. Isn't that the case for any equationless theory?
There is certainly a good bit more known than in Michael's response. The varieties in which subalgebras of free algebras are free are called Schreier varieties. A paper of mine in the early 70's -- Epimorphisms and dominions, V, probably in Alg. Univ. -- answered someone's question (Sabbagh?): Is every Schreier variety balanced? I remember the answer was No. John Isbell
The discussion and responses on sub-structures of free structures got me thinking about a problem that bothered to me some time a go. Suppose an equational class V has the following property: (1) if A and B are FINITELY GENERATED V-agebras whose coproduct is free, then A and B are both free. Does this imply that V has the prorerty that (2) if A and B are ANY V-agebras whose coproduct is free, then A and B are both free? I proved in 1983 that the variety of vector-lattices (= abelian lattice-ordered groups with order-preserving action by positive real numbers) has (1), using PL topology. (In the variety of vector-lattices, projective does not imply free, so this result has some content.) In fact, if V = vector-lattices, the following is true: (1a) if A and B are ANY V-agebras whose coproduct is FINITELY GENERATED and free, then A and B are both free. BUT, I was never able to show that the variety of vector-lattices has (2). Does (2) follow from (1) or from (1a) by any general theorems of universal algebra, model-theory, etc.? James Madden
participants (4)
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James Madden -
Michael Barr -
MTHISBEL＠ubvms.cc.buffalo.edu -
Peter Freyd