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

I wonder whether this is true in the categories of groups, monoids, commutative algebras, associative algebras, Lie algebras?

share|improve this question
2  
It is not true in the variety of groups generated by $S_3$. –  Mark Sapir Mar 9 '12 at 17:32
2  
There's a literature on rectracts of polynomial rings, which is referenced in this answer: mathoverflow.net/questions/55931/… –  Charles Rezk Mar 9 '12 at 17:33
1  
In any category $\mathcal{C}$ on $Set$, If "any projactive object is free" then "any retract of a a free object is free" (where an object $X$ is free on $S$ if represent the co-presheav $X\mapsto \mathcal{C}(S, |X|)$ where $X\mapsto |X|$ is the canonical functor on $Set$. we have the inverse implication if the funtor $X\mapsto |X|$ has a left adjoint $L$ and the counit $L(|X|)\to X$ is (puntually) a epimorphism. –  Buschi Sergio Mar 9 '12 at 20:34
2  
So the question basically asks when "free = projective". –  Martin Brandenburg Mar 9 '12 at 20:59
2  
@Charles. One can see from Costa, Douglas L. Retracts of polynomial rings. J. Algebra 44 (1977), no. 2, 492–502. that in 1977 it was unknown whether every retract of $K[X_1,\ldots,X_n]$ is a polynomial ring, where $K$ is a field: The author shows that an affirmative answer to this question would solve the well-known cancellation problem for polynomial rings over fields. Is it still unknown?!! –  Victor Mar 9 '12 at 21:53
show 6 more comments

2 Answers

up vote 9 down vote accepted

A retract of a finitely generated free monoid is free even though submonoids need not be free. I don't know about the infinitely generated case.

Edit: infinitely generated seems ok. The fg case I saw in an automata theory book but I see a general proof.

Added: here is the proof. Let P be a projective monoid (retract of free). Since it is a submonoid of a free monoid it has a unique minimal generating set Y consisting of the elements which are irreducible. Consider the map from the free monoid on Y to P sending generator to generator. Since P is projective it must split. But since elements of Y are irreducible their only preimages are the corresponding generators in the free monoid. Thus the splitting is an inverse to the projection.

Added: It seems to me the above proof works verbatim for free commutative monoids and more generally relatively free monoids in varieties containing all commutative monoids.

Added: Theorem 7 of http://arxiv.org/pdf/math/9711202.pdf seems to imply retracts of free non-associative algebras are free.

share|improve this answer
1  
About your last paragraph (free associative case): as the beginning of Section 2.2 says, it is applicable in various non-associative case only, unfortunately. –  Vladimir Dotsenko Mar 10 '12 at 16:51
    
@Vladimir, thanks! I should have read it more carefully. I will fix the entry. –  Benjamin Steinberg Mar 10 '12 at 18:46
add comment

The answer is yes in the category of groups. Suppose that $f: G \to H$ is a retraction with $G$ a free group. Then there is a homomorphism $g: H \to G$ such that $fg = \mathrm{id}_H$. Thus $g$ is injective and hence embeds $H$ isomorphically as a subgroup of $G$. But any subgroup of a free group is free, so $H$ must be free. The same proof works in the category of abelian groups also.

Edit: the same proof will work anytime you have the theorem that a subobject of a free object is free, I think. I don't know if that is true in the other categories that you mention.

Further edit: this property can fail even in very nice categories. For example, let $k$ be a field and consider the matrix algebra $M_n(k)$. In the category of finitely generated modules over $M_n(k)$, $M_n(k)$ itself is a direct sum of $n$ copies of $k^n$, but $k^n$ is not free over $M_n(k)$.

share|improve this answer
3  
The same argument works for Lie algebras. –  Mariano Suárez-Alvarez Mar 9 '12 at 17:37
5  
The last example can be generalized as follows: in the category of $R$-modules ($R$ a unital ring), the retracts of free objects are precisely the projective modules. –  Qiaochu Yuan Mar 9 '12 at 20:48
add comment

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.