# An analogue of the Bass-Quillen conjecture with power or Laurent series

The famous Quillen-Suslin theorem (formerly known as Serre's problem/conjecture) states that every projective module over $k[x_1,\dots, x_n]$ is free for $k$ a field. Replacing $k$ by a more general ring, we get the Bass-Quillen conjecture:

Let $R$ be a regular ring and $P$ a projective module over $R[x_1,\dots, x_n]$, then $P \cong Q \otimes_R R[x_1,\dots, x_n]$ for a projective $R$-module $Q$.

This has been proven in many cases, for example for $R$ of Krull dimension $\leq 2$ or if $R$ is a localization of an affine $k$-algebra for $k$ a field.

Now one could put forward similar conjectures replacing polynomial variables by power or Laurent series variables:

(Power) Let $R$ be a regular ring and $P$ a projective module over $R[[x_1,\dots, x_n]]$, then $P \cong Q \otimes_R R[[x_1,\dots, x_n]]$ for a projective $R$-module $Q$.

(Laurent) Let $R$ be a regular ring and $P$ a projective module over $R((x_1,\dots, x_n))$, then $P \cong Q \otimes_R R((x_1,\dots, x_n))$ for a projective $R$-module $Q$.

If I understand the (affine) Horrocks Theorem correctly, then the Laurent series version of the conjecture actually implies the original Bass-Serre conjecture for a ring $R$ if $n=1$. Actually, Horrocks proves the Laurent series version for $R$ regular local either of dimension $\leq 1$ or of dimension $\leq 2$ and containing a field already in 1964. My question is now:

What is known about the power and Laurent series versions of the Bass-Quillen conjecture?

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You may want to have a look at Lam's book titled "Serre's problem on projective modules", Section V.4 and V.5. – Oblomov Jul 23 '13 at 8:30
I have the impression that these sections study the ordinary Bass-Quillen conjecture for $R$ a power series or Laurent polynomial ring. – Lennart Meier Jul 23 '13 at 9:44
Indeed, sorry for my previous comment. – Oblomov Jul 23 '13 at 9:56

Here is an attempt at the power series question. One can easily reduce to the one variable case. So, I will attempt to prove that if $R$ is any Noetherian ring and $A=R[[x]]$ and $P$ a projective module, then $P\cong P/xP\otimes_R A =P'$, First, note that if $K$ is any finitely generated $A$ module, then it is complete with respect to the $x$-adic topology and so if $K=xK$, then $K=0$. Since $P'$ is $A$-projective, we can lift the surjective map $P'\to P/xP$ to a map $P'\to P$ which is an isomorphism mod $x$. Then the cokernel $K$ of $P'\to P$ has the property $K/xK=0$ and thus $K=0$. So, $P'\to P$ is surjective and so it splits. If $C$ is its kernel, then $C/xC=0$ and so $C=0$. Thus the map $P'\to P$ is an isomorphism.

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That's great! I didn't expect this to have a short and general proof. – Lennart Meier Jul 23 '13 at 7:28