[Apologies in the delay in posts on the Segal paper — there are a couple of things I’m confused on that are preventing me from proceeding.]

A classical problem (posed by Serre) was to determine whether there were any nontrivial algebraic vector bundles over affine space , for an algebraically closed field. In other words, it was to determine whether a finitely generated projective module over the ring is necessarily free. The topological analog, whether (topological) vector bundles on are trivial is easy because is contractible. The algebraic case is harder.

The problem was solved affirmatively by Quillen and Suslin. In this post, I would like to describe an elementary proof, due to Vaserstein, of the Quillen-Suslin theorem.

**1. Stable freeness**

An initial step, already taken by Serre, was to show that any finitely generated projective module over a polynomial ring (for a field) is stably free. Recall that a finitely generated module is said to be **stably free** if it becomes free after adding a finitely generated free module.

**Remark:** Given a projective module , there is always a free module such that is free. To see this, first choose a*projective* module such that is free, and then take . It is easy to see that and that is free (if one appropriately groups the terms); this is the Eilenberg swindle. So, the finiteness conditions are really necessary here.

By the Serre-Swan theorem, one should think of projective modules as vector bundles, and, in particular, if is a compact Hausdorff space, we can actually identify (via an equivalence of categories) vector bundles on with finitely generated projective modules over the ring of continuous functions . Then, it follows that:

Proposition 1A stably free module over is the same thing as a stably trivial vector bundle on : that is, a vector bundle that becomes trivial after adding a trivial vector bundle.

This observation allows one to get a simple example of a stably free module which is not free. The tangent bundle to is stably trivial (in fact, its one-dimensional normal bundle is trivial), but it is not trivial unless by a famous theorem (which is in fact a consequence of the Hopf invariant one theorem).

The first part of the proof of the Quillen-Suslin theorem is accomplished by:

Theorem 2Let be a noetherian ring such that every finitely generated projective module over is stably free. Then the same property holds true for .

By induction, we see:

Corollary 3Every finitely generated projective module over , for any field , is necessarily stably free.

This result is actually a special case of a theorem of Grothendieck. Given a ring , we can form the group , which is defined to be the Grothendieck group of the category of finitely generated projective -modules. Two projective modules map to the same element of if and only if there is a finite free module such that . Consequently, if and only if every projective -module is stably free.

The next result of Grothendieck is thus a generalization of the previous theorem:

Theorem 4For a ring , extension of scalars induces an isomorphism .

The same is actually true of the higher K-groups, by a theorem of Quillen. I won’t describe the proof here.

**2. Unimodular vectors**

The main step is to go from “stably free” to free. Equivalently, we have to show that if we let , then any split injection

has a free cokernel. Let us start with the case ; this will turn out to be sufficient. We are interested in a condition such that any split injection will have a free cokernel, which is to say that is isomorphic to the canonical imbedding sending an element to .

We can reformulate the problem in a possibly more intuitive way. To give a split injection is the same as giving a vector whose components generate the unit ideal in . To say that the injection induced is isomorphic to the standard inclusion is to say that there is an isomorphism of taking to the vector . Alternatively, it is to say that the element can be completed to a basis for .

Definition 5Let be any ring. A vector isunimodularif its components generate the unit ideal in . For two unimodular vectors , we write

if there is a matrix such that . This is clearly an equivalence relation.

So, the problem we are faced with now is to show that, for the rings of the form , any two unimodular vectors are equivalent. Alternatively, we have to check when one is equivalent to the standard one . Stated another way, we have to check whether there is an automorphism of carrying onto . If we can show this, then it will follow that any split injection has a free cokernel.

Here is an easy first step:

Proposition 6Over a principal ideal domain , any two unimodular vectors are equivalent.

*Proof:* In fact, unimodular vectors correspond to imbeddings which are split injections. But if we have a split injection in this way, the cokernel is free (as we are over a PID), and consequently there is a basis for one of whose elements is . This implies that is conjugate to .

In a similar manner, if we use the fact that a finitely generated projective module over a *local* ring is free, then we obtain:

Corollary 7Over a local ring , any two unimodular vectors are equivalent.

**3. Polynomial rings over a local ring**

The proof of the Quillen-Suslin theorem is essentially to induct on the number of variables. To do this, we’ll need an auxiliary result which states that, under mild hypotheses, a unimodular vector in a polynomial ring is equivalent to a unimodular vector in the *base* ring. This will be proved locally—one prime at a time. So, we start with:

Theorem 8 (Horrocks)Let for a local ring. Then any unimodular vector in one of whose elements has leading coefficient one is equivalent to .

*Proof:* Let be a unimodular vector. Suppose without loss of generality that the leading coefficient of is one, so that . If , then is a unit and there is nothing to prove. We will induct on .

Then, by making elementary row operations (which don’t change the equivalence class of ), we can assume that all have degree . Consider the coefficients of these elements. At least one of them must be a unit. In fact, if we reduce mod , then not all the can go to zero or the would not generate the unit ideal mod . So let us assume that contains a unit among its coefficients.

The claim is now that we can make elementary row operations so as to find another unimodular vector, in the same equivalence class, one of whose elements is monic of degree . If we can show this, then induction on will easily complete the proof.

Now, here is a lemma: If we have two polynomials , with and monic, and of degree containing at least one coefficient which is a unit, there is a polynomial of degree whose leading coefficient is one. This is easy to see with a bit of explicit manipulation.

This means that there are , such that has degree and leading coefficient a unit. If we keep this fact in mind, we can, using row and column operations, modify the vector such that it contains a monic element of degree . We just add appropriate multiples of to to make the leading coefficient a unit. This works if . If or , the lemma can be checked directly.

Consider the ring , and let be a unimodular vector. We want a condition to conclude that , where is the vector obtained by pointwise substitution. This will be the inductive argument we need for the Quillen-Suslin theorem. We already have a good criterion for when this is true in the case local.

Corollary 9If is local and is a unimodular vector one of whose elements is monic, then .

In fact, is a unimodular vector in , hence equivalent to . We have also seen that is equivalent to .

The goal of the next step is to generalize this to the case where is not assumed local.

**4. Localization**

\newtheorem{lemma}{Lemma} Let be a domain. We start by observing that if in , then over . In fact, by hypothesis there is a matrix such that

which means that

We just have to then observe that

so we can take as the relevant matrix taking into .

The next lemma will be the required step to reduce to the case of local.

Lemma 10Suppose over the localization . Then there exists a such that over .

*Proof:* As before, we can choose a matrix such that , and then the matrix has the property that

It follows that if we substitute for , then we have

The claim is that we can choose such that actually has -coefficients. In fact, this is because , which implies that for some matrix with values in . If we replace with for an element of , then we can clear the denominators in and arrange it so that .

Here, now, is the promised result which will be the crucial inductive step:

Corollary 11Suppose is any ring, and is a unimodular vector one of whose leading coefficients is one. Then .

*Proof:* Let us consider the set of such that in . If we can show that , then we will be done, because after applying the homomorphism , we will get that in .

We start by observing that is an ideal. In fact, suppose and . Then, substituting in the first leads to

and since , we get easily by transitivity that . Similarly, we have to observe that if and , then . But this is true because one can substitute .

Since is an ideal, to show that we just need to show that is contained in no maximal ideal. Let be a maximal ideal. We then note that, by what we have already done for local rings, we have that

By the lemma, this means that there is a such that ; this means that . So cannot be contained in . Since this applies to any maximal ideal , it follows that must be the unit ideal.

**5. The Quillen-Suslin theorem**

With all these preliminaries, it will be relatively straightforward to establish the main result; the first step is to show that unimodular vectors over a polynomial ring are all equivalent.

Theorem 12Let be a polynomial ring over a principal ideal domain , and let be a unimodular vector. Then .

*Proof:* We can now prove this by induction on . When , it is immediate.

Suppose . Then we can treat as , where we replace by to make it stand out. We can think of as a vector of polynomials in with coefficients in the smaller ring .

If has a term with leading coefficient one, then the previous results enable us to conclude that , and as lies in we can use induction to work downwards. The claim is that, possibly after a change of variables , we can always arrange it so that the leading coefficient in is one. The relevant change of variables leaves constant and

If is chosen very large, one makes by this substitution the leading term of each of the elements of a unit. So, without loss of generality we can assume that this is already the case. Thus, we can apply the inductive hypothesis on to complete the proof.

Theorem 13 (Quillen-Suslin)A finitely generated projective module over for a principal ideal domain is free.

In fact, we have to show that a *stably* free module over is free. That is, if is such a finitely generated module such that , then is free. By induction on , one reduces to the case . In this case we have an exact sequence

and we have to conclude that the cokernel is free.

But the injection corresponds to a unimodular vector, and we have seen that this is isomorphic to the standard embedding , whose cokernel is obviously free. Thus is free.

January 24, 2012 at 10:55 am

[…] days I read Akhil Mathew's post on Vaserstein's proof of the Quillen-Suslin theorem, once known as Serre's Problem. This inspired the […]

February 10, 2012 at 5:26 pm

I’ve always found this theorem fascinating. Why do you say “say noetherian” in your definition of K_0 though? Also I think there’s a missing latex at the start of section 4. Nice post!

February 10, 2012 at 5:39 pm

Thanks! You’re right that noetherianness is not necessary. In any event, we can always write any ring as a filtered colimit of noetherian rings and commutes with filtered colimits — so if we know something related to about noetherian rings, it probably works for general commutative rings.

February 14, 2012 at 3:13 pm

Well, provided that the “property” also commutes with colimits 🙂 In some sense this is why Q is flat, but not projective (as a Z module)!

August 21, 2012 at 1:09 pm

[…] Mathew: Delooping and the bar construction, Categories and cohomology theories, The Quillen-Suslin theorem, Homotopy is not […]

January 12, 2020 at 12:29 pm

There’s a typo in the lemma header right at the start of Section 4.