# 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?

• 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.

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

I believe it should be possible to remove the "Noetherian" hypothesis in Mohan's answer, using the following slight generalization of the Nakayama-style fact:

Lemma: Let $$f : R \to A$$ be a ring map admitting a section $$g : A \to R$$ and such that the inclusion $$A^{\times} \subseteq g^{-1}(R^{\times})$$ is an equality. Let $$K$$ be a finitely presented $$A$$-module such that $$K \otimes_{A,g} R = 0$$. Then $$K = 0$$.

Proof: Let $$A^{\oplus s} \stackrel{\varphi}{\to} A^{\oplus r} \to K \to 0$$ be a presentation of $$K$$ as a $$A$$-module, and let $$M \in \mathrm{Mat}_{r \times s}(A)$$ be the matrix corresponding to $$\varphi$$. Since $$\varphi \otimes_{A,g} R$$ is surjective, there exists $$N \in \mathrm{Mat}_{s \times r}(R)$$ such that $$g(M) \cdot N = \operatorname{id}_{r}$$, hence $$g(M \cdot f(N)) = \operatorname{id}_{r}$$, which means $$M \cdot f(N)$$ is invertible (since $$A^{\times} = g^{-1}(R^{\times})$$).

Remarks: I guess in this question we only consider finitely generated projective modules. To check that $$K,C$$ are finitely presented, we may use e.g. part (4) of [SP, 0519].

• Nice lemma! In fact, the power series case of the question follows from the following general result that is a special case of Thm. 5.8.14 (or of the earlier Cor. 5.4.41) from Gabber--Ramero "Almost ring theory" (take $t = 1$ there): for a Henselian pair $(A, I)$ (such as, for instance, $(R[[x_1, \dotsc, x_n]], (x_1, \dotsc, x_n)R[[x_1, \dotsc, x_n]])$), base change induces a bijection between the set of isomorphism classes of vector bundles on $\mathrm{Spec}(A)$ to that of vector bundles on $\mathrm{Spec}(A/I)$. – Kestutis Cesnavicius Oct 19 '18 at 7:25
• @KestutisCesnavicius Great, thanks for the reference. – Minseon Shin Oct 19 '18 at 20:23