# Are projective rings over $\mathbb{Z}$ free?

Let $R$ be a ring. Say that an $R$-algebra $A$ is $R$-projective if it has the left lifting property with respect to surjections of $R$-algebras: that is, whenever $B \to C$ is a surjection of $R$-algebras, then $\hom_R(A, B) \to \hom_R(A, C)$ is a surjection. A polynomial algebra is an example of a projective $R$-algebra, and conversely any projective $R$-algebra has to be a retract of a polynomial algebra (by the same argument as for modules). It is known that any projective module over a PID is free. Conversely, is it true that any projective algebra over a PID is free?

A bit of background: The Lazard ring $L$ is a ring with the universal property that $\hom(L, R)$ is in natural bijection with formal group laws over $R$. It is a non-obvious fact that $L$ is a polynomial ring on a countable set of generators, and is thus a "projective ring"; thus if $A \to B$ is a surjection of rings then any formal group law on $B$ can be lifted to one on $A$. I don't know how to prove this directly, but I'm curious whether, if I did, it would then be clear that $L$ has to be free.

(Incidentally, what happens in the noncommutative case, where "polynomial" is replaced by "free associative"?)

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Even over a field this is a difficult problem! It is related, among other things, to the Jacobian conjecture and the constellation of problems around it.

In [D. Costa, Retracts of polynomial rings, J. Algebra 44 (1977), 492-502.] Costa deals with the two variable case (that is, with projective algebras generated by two elements) (he works with coefficient rings which are zero-dimensional and noetherian, for extra fun) He also deals with the non-commutative polynomial algebra.

Then [Picavet, Gabriel. Algebraically flat or projective algebras. J. Pure Appl. Algebra 174 (2002), no. 2, 163--185. MR1921819 (2003i:13009)] shows that over a field, a projective algebra of finite type is either a polynomial algebra or the coordinate algebra of a complete intersection.

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Thanks. I hadn't realized this problem was so difficult... –  Akhil Mathew Jan 12 '12 at 0:11