Finitely generated projective modules over polynomial rings with integral coefficients

Question

There is famous Quillen-Suslin theorem which states that every finitely generated projective module over a ring of polynomials $k[x_1,...,x_n]$, where $k$ is a field, is free.

I have never carefully read a proof of this theorem, which is for example in the Lang's Algebra. Probably it is based on Quillen's original ideas.- this is not true as it was pointed out in the answer below.

Questions: Is every finitely generated projective modules over $\mathbb{Z}[x_1,...,x_n]$ free?

If yes, then is the proof modification of the one given in Lang's Algebra?

And if yes, then how about polynomial rings over other Dedekind domains or number rings?

Answer 1

Ok Slup, here goes.

Let $R$ be any commutative ring and let $A$ be a polynomial ring over $R$. Let $P$ be any projective module over $R$. Then Quillen (and Suslin a bit later in this generality) proved that if for every maximal ideal $\mathfrak{m}$ of $R$, $P_{\mathfrak{m}}$ is of the form $Q\otimes_{R_{\mathfrak{m}}} A_{\mathfrak{m}}$ for some projective $R_{\mathfrak{m}}$ module $Q$, then there exists a projective module $Q$ over $R$ such that $P=Q\otimes_R A$. Using this, they deduced that this always happens when $R$ is a Dedekind domain. The last part is done by the following observation. If for such a $P$, $P_f$ is free for a polynomial which is monic in one of the variables, then $P$ is free. As you can see, since any non-zero polynomial after a change of variables can be made into a monic polynomial in one of the variables if $R$ is a field, one immediately deduces Serre conjecture from this.

Many years later, Lindel generalized this for $R$ any regular ring containing a field.

Answer 2

It should be true for any PID. See the book by Lam, Serre's Conjecture

Answer 3

Let us say that a ring $A$ satisfies condition $(S)$ if every finite type projective $A$-module is free. Clearly a necessary condition for $(S)$ to hold is $K_0(A)=\mathbf{Z}$.

Example. (i) If $A$ is local Noetherian then $(S)$ holds. (ii) If $A$ is a Dedekind domain then $K_0(A)=\mathbf{Z}\oplus\mathrm{Pic}(A)$, and thus a necessary condition for $(S)$ is $\mathrm{Pic}(A)=0$, which is the case iff $A$ is a pid.

Theorem.(Grothendieck) If $A$ is a regular ring then we have a canonical isomorphism $K_0(A)\rightarrow K_0(A[(T_i)_{i\in I}])$.

Thus a necessary condition for $(S)$ to hold for $A[T_1,\ldots,T_n]$ is $K_0(A)=\mathbf{Z}$.

Corollary. Let $A$ be a Dedekind domain. Then $(S)$ holds for $A[T_1,\ldots,T_n]$ if and only if $\mathrm{Pic}(A)=0$, i.e. iff $A$ is a pid.

(This follows from what we said previously and Quillen's theorem stating that if $A$ is a pid, then $A[T_1,\ldots,T_n]$ satisfies $(S)$.)

(Remark. Quillen's original proof is different from the one given in Lang!)