I wonder whether this is true in the categories of groups, monoids, commutative algebras, associative algebras, Lie algebras?
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3$\begingroup$ It is not true in the variety of groups generated by $S_3$. $\endgroup$– user6976Mar 9, 2012 at 17:32
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2$\begingroup$ There's a literature on rectracts of polynomial rings, which is referenced in this answer: mathoverflow.net/questions/55931/… $\endgroup$– Charles RezkMar 9, 2012 at 17:33
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5$\begingroup$ So the question basically asks when "free = projective". $\endgroup$– Martin BrandenburgMar 9, 2012 at 20:59
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2$\begingroup$ @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?!! $\endgroup$– VictorMar 9, 2012 at 21:53
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2$\begingroup$ I think that for commutative algebras this problem is very hard (description of all retracts is related to both the cancellation conjecture and the Jacobian conjecture). For associative algebras, the only source I know uses the results in the commutative case (arxiv.org/pdf/math/9701210v1.pdf), maybe it's possible to do better. $\endgroup$– Vladimir DotsenkoMar 10, 2012 at 10:21
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3 Answers
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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.
answered Mar 9, 2012 at 18:42
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1$\begingroup$ 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. $\endgroup$ Mar 10, 2012 at 16:51
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$\begingroup$ @Vladimir, thanks! I should have read it more carefully. I will fix the entry. $\endgroup$ Mar 10, 2012 at 18:46
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$\begingroup$ That's great! Do you remember the precise reference for the automata theory book? $\endgroup$ Jul 18 at 4:42
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1$\begingroup$ @JonasFrey, this was over 11 years ago. I have no idea which book. If I had to make a guess I would day the theory of codes by Berstel, perrin, reutenauer $\endgroup$ Jul 18 at 10:06
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1$\begingroup$ I don’t remember what book but apparently the result was first proved in Head, T. 1982. Expanded subalphabets in the theories of languages and semigroups. Intern. J. Comput. Math, 12: 113–123. $\endgroup$ Jul 19 at 20:52
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A few months after the last activity on this question, Neena Gupta gave a proof that over a field $k$ of positive characteristic, a retract of a polynomial algebra need not be a polynomial algebra: http://arxiv.org/abs/1208.0483.
In fact, she gives a counterexample to the cancellation problem: there is an algebra $A$ such that $A[t]$ is isomorphic to $k[x_1,x_2,x_3,x_4]$ but $A$ is not isomorphic to $k[y_1,y_2,y_3]$. Composing the isomorphism $k[x_1,x_2,x_3,x_4]\to A[t]$ with the evaluation map $A[t]\to A$ at $t=0$ expresses $A$ as a retract of $k[x_1,x_2,x_3,x_4]$.
answered Jun 5, 2015 at 8:58
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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 $M_n(k)$.
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3$\begingroup$ The same argument works for Lie algebras. $\endgroup$ Mar 9, 2012 at 17:37
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6$\begingroup$ 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. $\endgroup$ Mar 9, 2012 at 20:48