# 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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Do you mean $\hom_R(A,B)\rightarrow \hom_R(A,C)$? –  Thomas Andrews Jan 11 '12 at 17:12
@Thomas: Thanks for the correction. –  Akhil Mathew Jan 11 '12 at 17:13