# Finitely generated projective modules over polynomial rings with integral coefficients

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?

## 3 Answers

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.

• What do you mean by 'this' in your sentence 'Lindel generalized this for $R$ any regular ring containing a field'? – Slup Jun 25 '16 at 10:10
• Lindel proved that any projective module over $A$ (notation as above) comes from $R$ by tensoring as above. – Mohan Jun 25 '16 at 14:11
• Serre's conjecture is false for Dedekind domains – syzygy Jun 25 '16 at 23:25
• I do not know what you mean. The theorem that is proved is that projective modules come from the base in these cases (for example Dedekind domains). Of course they are not free, since there are non-free projective modules over Dedekind domains, unless it is a pid. – Mohan Jun 25 '16 at 23:48
• I meant that if $A$ is a dedekind domain, then in general not every finite type projective $A[T_1,\ldots,T_n]$-module is free. (This was the OP's initial question.) I think the OP (and me too) would like to know some more rings $A$ (not pids, is being a pid necessary?) such that every finite type projective $A[T_1,\ldots,T_n]$-module is free. Do you know of a survey paper that discusses such $A$?. I think it is true for $A$ Bezout. Aren't you actually a reasearch mathematician in CA? I think I heard your name somewhere – syzygy Jun 26 '16 at 0:21

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!)

• Thank you for answer and pointing out to that the original proof is different. Is there any research in the direction of number rings? – Slup Jun 24 '16 at 10:44
• It is true in greater generality. To specilaize to your case, if $P$ is a projective module over a polynomial ring $A$ over a ring of integers $R$ (the set of all algebraic integers in a number field), and more generally for any Dedekind domain, then there exists a projective module $Q$ over $R$ such that $P=Q\otimes_R A$. In particular, $P=F\oplus L$ where $F$ is free and $L$ is a projective module of rank one. Further, $L$ comes from $R$ and thus if $R$ is a pid, then $L$ too is free. – Mohan Jun 24 '16 at 13:19
• It has something to do with $K$-theory? Please turn your comment into answer, other two answers are very valuable, but your is the most complete one. – Slup Jun 24 '16 at 18:33
• Thanks for adding this reformulation in terms of $K$-theory. I have already up-voted your answer, so there is now way I can award you for that except for expressing my gratitude. – Slup Jun 25 '16 at 16:40

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

• Thank you for your answer and for the reference. – Slup Jun 24 '16 at 10:45