QUILLEN-SUSLIN THEOREM
Abstract. These notes on the Quillen-Suslin Theorem are based
on lectures given by T. E. Venkata Balaji at the NBHM AIS on
Commutative Algebra held at IIT Madras in May 2012. The proof
follows the material in Serge Lang’s book Algebra. The notes were
taken and LATEXed by Shuddhodan Kadattur Vasudevan, Research
Scholar, School of Mathematics, TIFR, Mumbai.
1. The Statement
Serre’s problem or the Quillen-Suslin (QS) theorem has the following
equivalent formulations,
• (Algebraic) Any finitely-generated projective module over
k[X1 , X2 , · · · Xn ] is free.1
• (Geometric) Any vector bundle over the affine space Ank is
trivial.
Before building towards the proof we formulate the question in an
equivalent way using objects which are defined subsequently.
2. Some Definitions
Definition 1. A finitely generated R-module M is said to be stably free
if there exist R-modules F1 and F2 both finite free such that M ⊕ F1 ∼
=
F2 .
Equivalently there exists finite free modules F1 and F2 such that the
following short exact sequence (ses),
0
F1
F2
M
0
splits. Observe that stably free modules are projective, since they are
a direct summand of a free module. Clearly all finite free modules are
stably free. This leads to the obvious question of under what conditions are stably free modules necessarily free. This is what we concern
ourselves with during the proof of QS.
Definition 2. An element r = (r1 , r2 , · · · rn ) ∈ Rn is said to be unimodular if the elements ri , 1 ≤ i ≤ n generate the whole ring.
1Through
these notes by k we mean an arbitrary field and by R a commutative
ring with 1.
1
Further an unimodular element r = (r1 , r2 , · · · rn ) ∈ Rn is said to
have unimodular extension property (uep) iff there exists an invertible
n × n matrix with entries in R such that the first column of the matrix
is r. Equivalently r can be extended to a basis of Rn with precisely n
elements.
Also a ring R is said to have uep if for any n every unimodular element of Rn has uep.
3. Stably Free To Finite Free
In this section under suitable conditions on the ring R we prove that
every stably free module is finite free.
Theorem 1. Let R be Noetherian with uep then any stably free module
M over R is actually finite free.
Stably Free
uep
Free
Proof. First we prove the above claim for modules M such that M ⊕R is
finite free. Consider the natural projection p from M ⊕R onto R. Extend
this to a map from Rm onto R by the isomorphism M ⊕ R ≡ Rm . By
abuse of notation we also call that map as p.
M ⊕ R ≡ Rm
p
R
Let e1 = (r1 , r2 · · · , rm ) ∈ Rm be such that
p(r1 , r2 · · · rm ) = r1 p(1, 0, · · · 0) + r2 p(0, 1, · · · 0) + · · · rm p(0, 0 · · · 1) =
1. This implies the element e1 is unimodular. Thus it can be extended
to e1 , e2 , · · · em an R-basis for Rm . Let φ ∈ GLm (R) be the matrix
with columns as ei . Let ei be the standard basis for Rm . Hence, φei =
ei . The isomorphism φ induces a map p̃ from Rm onto R. Observe that
if p̃(ej ) = cp̃(e1 ) for some j ≥ 2 then replace ej by ej − ce1 . This
amounts to a row operation on the invertible matrix φ thus preserves
invertibility. By repeated use of the above procedure one has p̃(ej ) =
2
0, j ≥ 2. Thus, ej , m ≥ j ≥ 2 lie in M which is the kernel of the
map p̃. Thus, one has maps p̃ and p from isomorphic modules Rm
and M ⊕ R respectively onto a fixed module R. Thus, the respective
kernels are isomorphic. Namely, the module generated by ej , j ≥ 2 is
isomorphic to M .
In general when M ⊕ Rn ≡ Rm one has by the above proof M ⊕ Rn−1
is free. Proceeding iductively one gets the desired result.
4. Free Resolutions
Let M be any R-module. Then there is an obvious surjective map
α from ⊕m∈M R onto M. Similarly there is a obvious surjective map
β from ⊕m∈Ker(α) R onto Ker(α). Thus we have the following exact
sequence which can be continued on the left,
L
···
m∈Ker(α)
β
R
L
m∈M
R
α
M
0
We say a module a R-module M has a finite free resolution (FFR) iff
it has a finite length resolution by finite free modules. Similarly we say
a module a R-module M has a stably free resolution (SFR) iff it has
a finite length resolution by stably free modules. A stably free module
M trivially has a FFR,
0
F1
F2
M
where M ⊕ F1 ∼
= F2 .
Theorem 2. M has a FFR ⇔ M has a SFR.
Proof. ⇒
is obvious for every finite free module is trivially stably free.
⇐
Consider a SFR of M of length n. That is there exists stably free
modules Ei , 1 ≤ i ≤ n such that the following sequence is exact,
0
En
En−1 · · ·
E1
M
0
Let Ei ⊕ Fi ∼
= Fi where Fi and Fi are finite free. The result immediatley follows once we observe the following fact,
0
0
···
Ei
Ei−1
is exact implies,
3
···
···
0
0
0
Ei ⊕ (Fi−1 ⊕ Fi )
0
Ei−1 ⊕ (Fi−1 ⊕ Fi )
···
0
i
is exact for free modules Fi and Fi−1
. Thus for every consecutive pair
in the SFR of M append the sum of free mnodules which make each of
them free. This process terminates in n steps and we have a FFR for
M.
Now the basic idea behind the proof of QS can be stated diagrammatically as,
FFR
projective
Stably Free
uep
Free
We carry out the necessary steps one by one.
5. Finite Free Resolutions
Lemma 3. (Schanuel’s Lemma) If P and P 0 are projective modules
with P/K ∼
= P 0 /K 0 then P ⊕ K 0 ∼
= P 0 ⊕ K.
Proof. Let γ be the isomorphism inducing map between P/K and
P 0 /K 0 . Consider the following commutative diagram,
0
K
i
β
0
K0
P
P/K
α̂
j
P0
α
0
γ
P 0 /K 0
0
Here α is the lift of γ and α̂ the lift of α. These lifts are an immediate
consequence of P and P 0 being projective and the map γ being surjective. By restricting α̂ to the image of K we get a map β from K to K 0 .
Now consider the following sequence,
0
K
P ⊕ K0
4
P0
0
obtained by sending x → (β(x), i(x)) for all x ∈ K and (y, z) →
α̂(y) − j(z) for all (y, z) ∈ P ⊕ K 0 . It is trivial that the above sequence
is actually exact. Since p0 is projective it immediately follows that the
exact sequence splits giving the required isomorphism.
Now similar to the notion of stably free modules the notion of stable
isomorphism is defined.
Definition 3. R-modules M and M 0 are said to be stably isomorphic
if there exists finite free modules F and F 0 such that M ⊕ F ∼
= M 0 ⊕ F 0.
Lemma 4. If P/K and P 0 /K 0 are stably isomorphic with P and P 0
stably free then K and K 0 are stably isomorphic.
Proof. Let F and F 0 be such that P/K ⊕ F ∼
= P 0 /K 0 ⊕ F 0 . One has
the the following exact sequences then,
K
P ⊕F
P/K ⊕ F
0
K0
P0 ⊕ F0
P 0 /K 0 ⊕ F 0
0
∼
0
0
Since P ⊕ F/K ∼
= P 0 ⊕ F 0 /K 0 sing Schanuel’s lemma on projective
modules P ⊕ F and P 0 ⊕ F 0 one obtains K 0 ⊕ (P ⊕ F ) ∼
= K ⊕ (P 0 ⊕
0
0
F ). Thus K and K are stably isomorphic since P and P 0 are stably
free.
Definition 4. A module M is said to be of finite stably free dimension
if it has a SFR. Its stably free dimension is the length of the least such
SFR.
In what follows in this section we assume that R is Noetherian and
that all R-modules are finitely generated to simplify the proofs though
strictly speaking this assumption is not necessary.
Theorem 5. If M has a stably free dimension n then any partial SFR
of length less than n can be continued to a SFR of length n.
Proof. Let,
Fm−1 · · ·
Fm
F1
M
0
be a partial SFR for M of length m < n. Further let,
0
En
En−1 · · ·
E1
5
M
0
be a resolution of length n. Insert the kernels and cokernels in each of
the sequences. Let, Km = Ker(Fm → Fm−1 ) if m 6= 1 else let K1 =
0
. By the above two lemmas Km is
Ker(F1 → M ). Similarly define Km
0
0
stably isomorphic to Km . Let, Km ⊕ F ∼
⊕ F 0 with F and F 0 finite
= Km
free.
Then we consider the following two cases,
a)m = n − 1
0
0
In this case Kn = En . Thus Kn ⊕ F 0 is stably free thus Kn ⊕ F is stably
free which trivially implies Kn is stably free. Thus one can set Fn = Kn
and get a SFR of length n.
b)m < n − 1
Here we use the Noetherian condition to trivially extend the partial
SFR. Observe that stably free modules being quotients of finite free
modules are finitely generated. Thus the kernel Km is finitely generated
as an R-module. Say it is generated by p elements then set Fm+1 =
Rp .
Corollary 6. If
0
M1
E
M
is exact with E being stably free then, M has stably free dimension less
≤ n implies M1 has stably free dimension ≤ n − 1
Lemma 7. Consider the following short exact sequence of R-modules,
0
M0
M 00
M
0
00
0
then there exists R-modules M1 , M1 , M1 and stably free modules E, E 0 , E 00
such that,
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0
0
0
M1
0
E
0
M
0
0
0
0
0
M1
M1
E
E
M
M
0
0
00
00
00
0
0
0
is exact along the rows and columns and is commutative.
Proof. The proof is quite simple because of the Noetherian condition
inherent to the problem at hand. We first observe the obvious fact that
suppose,
0
M0
M
M 00
0
is exact then the image of the generators of M as a R-module and
the pre-images of the generators of M 00 as an R-module in M generate M as a R-module. Further, by the Noetherian and finiteness
00
conditions we have, M 0 = Rn /I 0 , M = Rn+m /I, M 00 = Rm /I for some
integers n and m. I, I 0 , I 00 being sub-modules of Noetherian R-modules
Rn+m , Rn , Rm respectively are finitely generated. Thus it is easy to
see that Rn+m , Rn , Rm , I, I 0 , I 00 satisfy the necessary conditions for
0
00
E, E 0 , E 00 , M1 , M1 , M1 .
Theorem 8. Let,
0
M0
M
M 00
0
be exact. If any two these modules have a FFR then so does the third.
Proof. Though the theorem is about FFR, without loss of generality
we prove the theorem for SFR. Suppose M and M 0 have SFR. Then M
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is finitely generated which implies M 00 is finitely generated. We induct
on the stably free dimension of M . Using the notation of the lemma
0
7 both M1 and M1 have SFR with the stably free dimension of M1
being ≤ n − 1. Thus, inductively we reduce to the case where stably
free dimension of M is 0. That is it is stably free. In this scenario one
can trivially continue the SFR of M 0 to that of a SFR for M 00 using M .
Next assume M 0 and M 00 have stably free resolutions. Then M being
the quotient of a finitely generated module is itself finite. If both M 0
and M 00 have stably free dimensions 0 then they are projective and one
has M ∼
= M 0 ⊕ M 00 . Thus,M is itself stably free and has the trivial
SFR. Now one inducts on the maximum of stably free dimensions of
M 0 and M 00 . Say it is n. Then in the notation of lemma 7 the maximum
0
00
of the stably free dimensions of M1 and M1 is ≤ n − 1. Thus one is
reduced to the case when the stably free dimensions of both M 0 and
M 00 is equal to 0.
Finally let M 00 and M have SFR. This implies M 00 and M are finitely
generated. Thus are Noetherian modules which implies M 0 is Noetherian and hence finitely generated. Now, the result follows by inducing
on the stably free dimension of M 00 and lemma 7.
Now we come to the central result of this section,
Theorem 9. Let R be a Noetherian ring. If any finitely generated Rmodule has FFR then any finitely generated R[X]-module has FFR.
Proof. Let M be finitely generated as a R[X]-module. R Noetherian
implies R[x] is Noetherian. Thus there is a filtration for M ,
0 = Mn ⊂ Mn−1 ⊂ · · · M1 ⊂ M0 = M
such that Mi /Mi+1 ∼
= R[X]/℘i for some prime ℘i ⊂ R[X]. Since one
has exact sequences of the form,
0
Mi+1
Mi
R[X]/℘i
0
and Mn−1 ∼
= R[X]/℘n−1 by theorem 8 it sufices to prove the above
theorem for modules of the form R[X]/℘ with ℘ prime in R[X].
Also observe that,
0
℘
R[X]
R[X]/℘
0
Thus R[X]/℘ has a FFR as R[X]-module iff ℘ has a FFR as R[X]module.
Suppose R[X]/℘ does not have a FFR as R[X]-module for some ℘
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prime. Consider the collection S of all ℘ and SR the collection of contracted prime ideals ℘c of primes ℘ in R[X] such that R[X]/℘ does not
have a FFR as R[X]-module. Here the contraction is carried out with
respect to the obvious injective map R ,→ R[X]. By our assumption S
and SR are both non-empty and by the Noetherian property of R, SR
has a maximal element. Let ℘ ∈ S be such that ℘c is maximal in SR .
Let ℘c R[X] ⊂ ℘ be the ideal generated by ℘c in R[X]. Further, let
R0 = R/℘c . Consider the following short exact sequences,
℘c R[X]
0
0
℘c R[X]
R[X]/℘c R[X] = R0 [X]
R[X]
℘/℘c R[X] = ℘0
℘
0
0
Let f1 (X), f2 (X) · · · fm (X) generate ℘0 over R0 [X]. Let f (X) be a
polynomial of least degree in ℘0 . Also observe that R0 [X] ∼
= R/℘c [X]. Identify R0 [X] with R/℘c [X] and let k0 be the fraction field of the integral
domain R/℘c . Thus one has,
℘0 ⊂ R0 [X] ⊂ k0 [X]
Using division algorithm in k0 [X] one obtains fi = qi f + ri , 1 ≤ i ≤ m.
If ℘ = ℘c R[X] then ℘c being an ideal of R has FFR as a R-module. Tensor every element of the exact sequence by the flat module R[X] and
extend by scalars so that ℘ has a FFR as R[X]-module which contradicts ℘c ∈ SR . Thus ℘ strictly contains ℘c R[X].
Let d0 ∈ R0 be the common denominator of ri and fi . multiplying by
d0 6= 0 throughout one obtains,
0
0
d0 fi = qi f + ri ∈ R0 [X]
(1)
0
By our choice of f the degree of f is the least in ℘0 [X]. This forces ri
to be zero. Further, choose d ∈ R such that d mod ℘c = d0 . Then, d0 6=
0 =⇒ d ∈
/ ℘c . Hence,
d0 ℘0 ⊂ (f )
Let N0 = ℘0 /(f ). Consider a filtration of N0 as an R0 [X]-module,
0 = (N0 )n ⊂ (N0 )n−1 ⊂ · · · (N0 )1 ⊂ (N0 )0 = N0
0
such that there exist primes Qi ⊂ R0 [X] containing the set of associated primes of N0 as a R0 [x]-module such that (N0 )i /(N0 )i−1 =
R0 [X]/Qi . Another crucial observation is that the minimal elements
among Qi are precisely the minimal elements among the associated
primes. Thus every Qi contains atleast one associated prime. Observe
that d0 kills N0 and hence lies in all the associated primes of N0 . By
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extension of scalars look at N0 as a R[X] module and let Qi be the
0
inverse image of Qi in R[X]. Thus all the Qi ’s in particular the associated primes of N0 (which is a non-empty-set by the virtue of N0 being
Noetherian) and ℘ contain ℘c R[X]. Thus, Qci contains ℘c . In particular, whenever Qi is an associated prime of N0 (and hence for all i)
the containment is proper because d ∈
/ ℘c belongs to every associated
0
prime of N0 . Hence, R[X]/Qi ∼
= R0 [X]/Qi has a FFR as R[X]-module.
In what follows we repeatedly use theorem 8. Proceeding along the
chain obtained by the filtration of N0 one obtains that N0 has a FFR
as R[X] module. By the hypothesis of the theorem ℘c has a FFR as
R-module. Tensoring the FFR with the flat module R[X] one obtains
a FFR for ℘c R[X] as a R[X] module. Trivially R[X] has a FFR as
R[X]-module. Thus, implying R0 [X] has a FFR as R[X]-module. (f )
being isomorphic R0 [X] also has a FFR as R[X]-module. This along
with the fact that N0 has a FFR as R[X]-module implies that ℘0 has
FFR as R[X]-module. Finally, since ℘0 = ℘/℘c R[X], one concludes
that ℘ has a FFR as R[X]-module.
6. k[X1 , X2 , · · · Xn ] has uep
This is the last section in the proof of QS theorem where the uep
property is proved for k[X1 , X2 , · · · Xn ]. The proof is achieved in steps
starting from an apparently specific case for local rings. Before starting
with the proofs we make certain simple observations.
(f1 , f2 · · · fn )t has uep iff the vector obtained by any permutation of the
fi ’s does. Further one can add or subtract any multiple of fj from fi
as long as i 6= j without affecting the uep of the vector. Now consider
the following theorem by Horrocks,
Theorem 10 (Horrocks). Let (E, m) be a local ring with m being
the unique maximal ideal in E. Let A = E[X] and f be an unimodular
vector in E[X]n such that atleast one of the components of f has leading
coefficient 1. Then f has uep.
Proof. The theorem is trivially true if n is equal to 1 or 2. Hence assume
that n ≥ 3. The proof is obtained by inducting on the smallest degree
d of the polynomial having leading coefficient 1. By repeated use of
Euclidean algorithm and row operations one can assume that f1 is the
polynomial with degree d and leading coefficient 1 and that fj , j > 1
have degrees strictly
less than d.
P
Further since
fi gi = 1 for some polynomials gi it is easy to see
that not all coefficients of the polynomials fj , j > 2 lie in m. Because
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otherwise reading the equality mod m in the field E/m we have f1 g1 ∼
=
1 mod E/m which is not possible. Thus there exists a polynomial such
that atleast one of its coefficients does not belong to m and hence by
the virtue of E being local is invertible. Without loss of generality let
that polynomial be f2 .
Now consider the ideal generated by the leading coefficients of the
polynomials of the form f1 p1 + f2 p2 for arbitrary polynomials p1 and p2
with degree strictly less than d. It is easy to see that all the coefficients
of f2 belong to this ideal hence, the ideal is the full ring E. In particular
this means that there exists a polynomial of the form f1 p1 + f2 p2 with
leading coefficient 1 and degree less than d. Now by row operations one
can have entries in f with leading coefficient 1 and degree less than
d. Thus, by induction we are reduced to the case where d = 0. Then
by repeated used of row operations one reduces to the case where the
vector f just has constants as its entries with atleast one of them being
1. In this case uep is trivial.
Now we define the a notion of equivalence of vectors in Rn for an
arbitrary ring R and any n. Two vectors f, g ∈ Rn are said to be
equivalent if there exists an element M ∈ GLn (R) such that f = M g
and the equivalence is denoted by f ∼ g over R. Using the above
language we obtain the following corollary whose proof is implicit in
the proof of Horrocks theorem.
Corollary 11. Let E be a local ring. Let f be a unimodular vector
in E[X]n with some component having leading coefficient 1. Then f ∼
f (0) over E[X].
Now consider the following lemma which will help us prove a statement similar to the above corollary for not necessarily local rings.
Lemma 12. Let R be an integral domain and S a multiplicative subset of R. Let X, Y be two independent variables. If f ∼ f (0) over
S −1 R[X], then there exists c ∈ S such that f (X + cY ) ∼ f (X) over
R[X, Y ].
Proof. Let M ∈ GLn (S −1 R[X]) be such that f (X) = M (X)f (0). Then,
M (X)−1 f (X) = f (0) is constant. Define,
G(X, Y ) = M (X)M (X + Y )−1 ∈ GLn (S −1 R[X, Y ])
(2)
Then, G(X, Y )f (X +Y ) = f (X). Hence, G(X, 0) = I =⇒ G(X, Y ) =
I +Y H(X, Y ). Choose a c ∈ S such that cH has coefficients in R. Thus
G(X, cY ) has coefficients in R. But, determinant of M (X) is constant
11
in R. Thus implying determinant of G(X, cY ) is 1. Hence f (X) ∼
f (0).
This leads to the following generalization of the corollary,
Theorem 13. Let R be an integral domain and f an unimodular vector in R[X]n with atleast one of the leading coefficients being 1. Then
f (X) ∼ f (0) over R[X].
Proof. Let J be the set of elements c such that f (X + cY ) ∼ f (X) over
R[X, Y ]. It is easy to see that J is an ideal. Let ℘ be any prime ideal
in R. By the result of corollary 11 it is easy to see that f (X) ∼ f (0)
over R℘ [X] and hence by lemma 12 it follows that there exists c ∈ R
such that c ∈
/ ℘ satisfying f (X + cY ) ∼ f (X) over R[X, Y ]. Thus J is
not contained in any prime ideal hence is the full ring R. Thus there
exists an invertible matrix M (X, Y ) over R[X, Y ] such that f (X +Y ) =
M (X, Y )f (X). Substituting 0 for x one obtains the desired result. Finally, we state the central theorem of the lecture which was proved
by Quillen and Suslin.
Theorem 14 (Quillen-Suslin). Let k be a field and let f be a unimodular vector in k[X1 , X2 , · · · Xr ]n . Then f has uep.
Proof. The proof is obtained by inducting on the number of variables
r. If r = 1 then the ring is principal and the proof is trivial. Assume
r ≥ 2 and that the theorem is true for r − 1 variables. We view the
polynomial vector f as a vector of polynomials in the variable Xr over
the ring k[X1 , X2 , · · · Xr−1 ].
However to apply the results of theorem 13 one wants atleast one of the
polynomials to have leading coefficient 1 when viewed as a polynomial
in Xr . To do so one utilizes Nagata’s trick as in the proof of Noether’s
Normalization Theorem. That is we substitute,
Yr = Xr , Yi = Xi − Xrmi
(3)
for suitable constants mi such that the polynomial vector f (Y1 , Y2 · · · Yr )
when viewed as polynomial in Yr has one component with leading coefficient 1. Hence, there exists a matrix N (Y ) = M (X) invertible over
R[Xr ] = R[Yr ] such that g(Y1 , Y2 · · · Yr ) = N g(Y1 , Y2 · · · Yr−1 , 0). Using
the induction hypothesis g(Y1 , Y2 · · · Yr−1 , 0) is unimodular in
k[Y1 , Y2 , · · · Yr−1 ]n . Thus g(Y1 , Y2 · · · Yr ) has uep implying f has uep.
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