bài tập về Tích Tenxo
Trang 1Math Ann 299, 449 476 (1994)
9 Springer-Verlag 1994
Tensor products of modules and the rigidity of Tor
Craig Huneke, Roger Wiegand*
Department of Mathematics, Purdue University, West Lafayette, 1N 47907, USA
(E-mail: huneke@math.purdue.edu)
Received: 30 August 1993
Mathematics Subject Classifications (1991): 13C12, 13C14, 13D02, 13H10
Anyone who works with tensor products quickly learns that they tend to have torsion For ideals I and J of an integral domain R it is easy to show (1.4) that the torsion submodule of I | J is isomorphic to Tor~(R/I, R/J) In trying to understand what conditions on I and J are forced by the vanishing of TorR2(R/I,R/J) one is led rather naturally to consideration of the higher Tor's
The first systematic study of the question of torsion in tensor products was undertaken by Auslander [A] in his remarkable 1961 paper "Modules over un- ramified regular local rings" He gave a rather complete answer to the question
of when the tensor product of two modules over an unramified regular local ring
is torsion-free An important tool in his investigation was his rigidity theorem: If Tor~(M,N) = 0 for some j > 0 then T o r ~ ( M , N ) = 0 for all i>=j (Here M and N are finitely generated modules over the regular local ring R The ramified case of the theorem was proved by Lichtenbaum [L].) In the thirty years since the publication of Auslander and Lichtenbaums' papers, relatively little seems to have been done on the original question of torsion in tensor products (however, cf [Ho2]) The rigidity theorem and its related conjectures and variations, however, have played a role in the homological theory of local rings (See [Hol] for a discussion of the famous "rigid- ity conjecture" and [He] for its recent negative solution.)
Our interest in rigidity of Tor began, like Auslander's, in efforts to understand what it means for the tensor product of two finitely generated modules over a local domain to be torsion-free In the first case of interest - one-dimensional domains that are not discrete valuation rings - the problem is quite delicate Our best theorem
in this regard states, in the one-dimensional case, that if M and N are non-zero finitely generated modules over a hypersurface domain (that is, a ring of the form
S/(f) where f is a prime element of the two-dimensional regular local ring S), and
if M | N is torsion-free, then both M and N are torsion-free, and at least one of them is free
* Present address: Department of Mathematics, University of Nebraska, Lincoln, NE 68588-0323, USA (e-mail: rwiegand@unl.edu)
Trang 2450 C Hxlaeke and R Wiegand
We actually prove a natural generalization (3.1) of this result for hyper- surfaces of higher dimension, in which '~torsion-free" is replaced by "maximal Cohen-Macaulay" Also, by imposing a mild condition ("constant rank") on one
of the modules, we are able to dispense with the assumption that R be a doma'm These generalizations are forced upon us, as we need properties that are stable un- der operations such as completion and taking hyperplane sections There is a lot of dancing around among dimensions: The one-dimensional case, even for domains, uses a special case of the two-dimensional case, which in turn depends on a very special result in dimension one (i.9)
The rigidity of Tor plays a crucial rote in the proof of (3.t) We establish two rigidity theorems for hypersurfaces in Sect 2 The first one (2.4) states that
if M and N are finitely generated, if M | N has finite length, and if dim(M) + dim(N)_<dim(R), then Tor~(M,N)= 0 =~ Tor~(M,N)= 0 for all i~j (We need
to assume that the relevant regular local ring is unramified to get the full strength
of the theorem, although this assumption is not necessary in the applications.) The second rigidity ~heorem states that Tor~(M,N) = 0 for all i _~ 1 i f M | N is non-zero and reflexive
The first rigidity theorem is applied in Sect 3 to get the one-dimensional the- orem (3.7) on torsion in tensor products We need (3.7) to complete the proof of our second rigidity theorem, which is then used in Sect 3 to deduce the higher- dimensional result (3.1)
In Sect 4 we give lots of examples to show that most of our hypotheses (in the main theorems of Sect 2 and Sect 3) are essential A recurring theme is that if R
is not Gorenstein, then most of our theorems are false We show, for example, that
the ring k[[T3,T4,T5]] has a finitely generated module M, not torsion-free, whose tensor product with the ideal (T 3, T 4)/s torsion-free (in fact, reflexive!)
The last section contains some miscellaneous results related to the material in Sects 2 and 3 In particular, Theorem 5.1 proves that any two torsion-free modules over a normal hypersurface of dimension 2 satisfy the rigidity conjecture There is no assumption that one of the modules is of finite projective dimension In (5.3) we give necessary and suftieient conditions over a local 2-dimensional Cohen.Macaulay ring with canonical module for the tensor product of two finitely generated modules to
be torsion-free These conditions involve properties of certain Ext modules Finally
we raise some questions for further research Perhaps the most salient of these is the following: Are our rigidity theorems valid for complete intersections (not just hy~rsurfaees)?
Terminology, notation anti conventions All rings are assumed to be commutative and Noetherian Our standard reference for commutative algebra is Matsumura's book [Ma] Most of our rings will be local, and our main results are for hyper- surfaces, that is, rings of the form S/(f), where S is a regular local ring and f is a non-zero element of the maximal ideal of S More generally, a complete intersection
is a ring of the form S/(fl ,f,), where S is a regular local ring and (Ji fi)
is a regular sequence in the maximal ideal
If M is an R-module, the torsion submodute t(M) is the kernel of the natural map M ~ K | M, where K = {non-zero-divisors}-lR is the total quotient ring
of Ro The R-module M is torsion provided t ( M ) = M and torsion-free provided
t ( M ) = 0
Suppose (R,m, k) is a local ring and M is a finitely generated R module We let/~R(M) denote the number of elements in a minimal generating set for M If
Trang 3Tensor products of modules and the rigidity of Tor 451
dl
9 'Fn an'Fn-1 ' ' E l 'Fo a ~~ M , 0 ,
is a minimal free resolution o f M, we let syz](M), the nth syzygy o f M, de-
note the image of dn The depth of M is the least integer n = depthR(M ) for which Ext](k, M) 4: 0 I f M = 0 we put depthR(M) = - 1 We say M is a maximal Cohen-
Macaulay module provided M is finitely generated and depthR(M ) = dim(R) (the
KruU dimension of R) Thus our convention is that maximal Cohen-Macaulay mod- ules are always non-zero and finitely generated
We will also consider Serre's conditions (on a finitely generated module M over
a ring R):
(Sn) depth(Mp)>min{n, dim(Rp)} for every P E Spec(R)
Condition ($1) just says that Ass(M) consists of minimal primes of R; and i f R is Cohen-Macaulay this is equivalent to M being torsion-free Most of our local tings will be Gorenstein (in fact, complete intersections), in which case a module satisfies ($2) if and only if it is reflexive, [EG1, (3.6)] Therefore maximal Cohen-Macaulay modules over Gorenstein tings are always reflexive, hence torsion-free Conversely,
if R is Cohen-Macaulay and one-dimensional, then every non-zero finitely generated torsion-free module is maximal Cohen-Macaulay; and if R is Gorenstein and two- dimensional, every non-zero finitely generated reflexive module is maximal Cohen- Macaulay We will often use the more familiar terms "torsion-free" and "reflexive"
in contexts where they are equivalent to maximal Cohen-Macaulay
1.1 Lemma Let M and N be modules over the local ring R, and assume M | N
is torsion-free Let if4 = M/t(M) and N = NIt(N) Then ff, I |162 IV is torsion-free
I f M is free and N ~- O, then M is free
Proof Tensoring the exact sequence
with N, we get the following exact sequence:
t ( M ) | ~ , M | N ~-~-~ JQ @R N , 0 The image of ~ is torsion, and since M | N is torsion-free it follows that ~ = 0 Therefore 1~ is an isomorphism The first assertion now follows by symmetry If JQ
is free, then (1.1.1) splits, showing that M @R N has a direct summand isomorphic
to t(M) | N, which is torsion Therefore t(M) = 0, that is, M = ~r
Trang 4452 C Htmeke and R Wiegand Since we will not always work with domains we need to assume that our modules satisfy an extra condition, which amounts, in the case of reduced rings, to having constant rank at the minimal primes The following examples show why some such condition is necessary:
1.2 Examples (1) Let R = k[[X, Y]]/(XY), and put M = R/(x) Then M | M
M ~- (y), which is torsion-free, yet M is not free
(2) Let R = k[[X, Y]]/(X2),M = R/(x) Then M | M ~ (x), which is torsion- free, even though M is not free
We say that the R,module M has constant rank provided there is an integer n such that Mp is Rp-free of rank n for every P E Ass(R) (Scheja and Storch [SS, Sect 6] say simply that M has a rank.) The integer n is called the rank of M Sometimes
we will need only the weaker condition that M be generically free, that is, Me is
R?-free for each P E Ass(R) If R is reduced, then every R-module is generically free
1.3 Lemma, Let R be any ring (Noetherian, as always) with total quotient ring
K, and let M be an R-module The following conditions are equivalent:
(1) M has constant rank
(2) K | M / s K-free
(3) There is a homonwrphism ? : M ~ F, such that F is free and both Ker(~')
and Coker(y) are torsion module~:
Proof I f M has constant rook, then K | M is a projective K-module of constant rank Since K is semitoeal we have (2) Assuming (2) holds, we easily obtain an exact sequence
0 ~ t ( M ) ~ M r > R ( n ) and we want to know that 6 := ltc | ? is an isomorphism But gJ is a one-to- one map between free K-modules o f the same rank, so by McCoy's fheorem its determinant is a non-zero-divisor, that is, a unit Since (3) obviously implies (I), the proof is complete Q
1.4 Lemma Let R be a ring, and let M C F and N C_ G, where F and G are free Assume either M or N is generically free Then t(M | N) ~ Tor~(F/M, G/N) Proof We may assume M is generically free When we tensor the exact sequence
O-~ N , G ~ G/N , O
with M, we get an exact sequence
0 ~ Tor~(.~ GIN) -~ M | N > M @R G
Now M | G is torsion-free, being a direct sum of copies of M, and Tort(M, G/N)
is torsion since M is generically free It follows that Tor~(M, G/N) is isomorphic
to the torsion submodule o f M | N Next, we tensor the exact sequence
O ~ M * F -.-.~ F/M -~ O
with N The tong exact sequence o f Tor shows that TorR2(F/3.~[,G/N)~
Trang 51,5 Remark If R is a hypersurface, then Re is zero-dimensional and Gorenstein for every P E Ass(R) It follows IV, (A.I); HW, Theorem 10] that every torsion-free R-module can be embedded in a free R-module
1.6 Lemma Let (R,m,k) be a one-dimensional local ring with multiplicity e, and assume k is infinite Let M be a maximal Cohen-Macaulay R-module o f constant rank i" Then I~lr <=re
Proof We consider the multiplicity e(m,M) of M By the associativit7 formula [Ma, (14.7)] and the tact that M has constant rank r, we see that e ( m , M ) = re(m, R) = re Since k is infinite there is an element x E m which is a reduction
of m, that is, xm n-~ = m n for some n->l By [Ma, (14.13)] e ( ( x ) , M ) = e(m,M) Now x is a non-zero-divisor on M, so by [Ma, (14.10)] e((x),M) is equal to the multiplicity of M/xM, which is just its length 2(M/xM) Finally, #R(M)= 2(M/raM) =< 2(M/xM) = re []
We will make frequent use of Eisenbud's theory of matrix factorizations [E] A nice exposition of the theory can be found in Yoshino's book [Y, Chap 7] At this point we recall some of the most important consequences of the theory:
1.7 Proposition Let R be a hypersurface, and let M be a maximal Cohen- M:acaulay R-module with no non-zero free summand Then M has a minimal free resolution that is periodic o f period at most 2:
- - ~ R(m) ~ ~ R(m) ,I~ ~ R(,n) ~ ~ R (m) ~ ~ R (m) -'-* M ~ 0
In particular, we have syz~(M) ~ M and ktR(syz~(M)) =/.tR(M) Finally, syz~(M)
is also a maximal Cohen-Macaulay R-module with no n on-zero free summand
1.8 Corollary Let R be a hypersurfaee, and let M and N be maximal Cohen- Macaulay modules neither o f which has a non-zero free summand Assume M
is 9enerically free Then Tor~(?d,N) ~ Tor~(syz~(M),syz~(N)) ~ t(M | N) In particular, if M | N is torsion-free, so is syz~(M)| syz~(N)
Proof Since M and N are torsion-free, we can embed them in free modules F and G, respectively, by (1.5) We have t(M | N ) ~- Tor~(F/M,G/N) by (1.4) Tensoring the short exact sequence
O -+ M ~ F - + F / M - , O
with G/N, we have Tor~(F/M, G/N) -~ Tor~(M, G/N) by the long exact sequence of Tor Also, a s / ~ syz2(M) (by (1.7)), we have Tor~(M, G/N) ~ Tor~(M, G/N) But
Tor~(M,G/N) Tor~(M,N) (by the short exact sequence 0 -+ N , G -+ GIN -+
0) Combining these isomorphisms, we have Tor~(M,N) ~ t(M | N)
By (1.7) there is a short exact sequence 0 * N + R (n) * syzJ~(N)~ 0 This
sequence shows that Tor 2 (M,N) = Tor 3 (M, syzR(N)) Finally, Tor 3 (M, syzR(N)) =
Tor2(sYzR(M),syzR(N)) by the long exact sequence of Tor (~J
The following proposition is very special, but it plays a crucial role in the proof of our main theorem (3.1) on torsion in tensor products Actually, we will need only the case of multiplicity 2, but multiplicity 3 is not significantly harder
1.9 PrOlmsition Let R be a one-dimensional hypersurface o f multiplicity at most 3 Let M and N be R-modules o f constant rank I f M | N is torsion-free, then either M or N is free
Trang 6454 C Huneke and R Wiegand
P r o o f By (1.1) we may assume that both M and N are torsion-free and of constant ranks r and s respectively We assume, by way of contradiction, that neither M nor N is free, and that, among all such cotmterexamples, r + s is minimal Then neither M nor N has a non-zero free direct summand Let # R ( M ) = m, #R ( N ) =
n, and note that syzl(M) and syzl(N) are torsion-free modules o f constant ranks
m - r and n - s, respectively Moreover, (1.7) guarantees that neither syzygy is free Also, syzl(M) | syzl(N) is torsion-free, by (1.8) By minimality of r + s, either
m - r > r or n - s> s; we assume the first condition holds, that is,
By (1.7) we know that #R(syz~(M)) = m and pR(syzl(N)) = n Applying (1.6)
to M | and syz~(M)| syz~(N), we get mn<3rs and m n < 3 ( m - r ) ( n - s )
Adding these two inequalities and rearranging terms, we get:
We close this section with the following slight extension of a theorem due to Auslander and Buchsbaum [AB]:
1.10 Proposition Let R be a local rin9 and M an R-module o f constant rank
r If, f o r some prime ideal P, both Me and M / P M are free (over Rp and R/P, respectively), then M is free
P r o o f Since P contains an associated prime of R, Me must be free of rank r Then Me/PMe is an r-dimensional vector space over Rp/PR? It follows that r must also be the rank o f the free R/P-module M/PM In particular, #R(M/PM) = r By Nakayama's lemma, #R(M) = r Therefore (syzl(M))Q = 0 for every Q E Ass(R), and since s y z l ( M ) is torsion-free it follows that syz~(M) = 0 []
2 Rigidity
The classical theorem on the rigidity o f Tor says that if M and N are finitely generated modules over the regular local ring R, and if TorT(M,N ) = 0 for some
j > 0 , then Tor/R(M,N)= 0 for every t > j This was proved by Auslander [A] in
1961 in the unramified case In 1963 Murthy [Mu] showed that over any regular local ring the vanishing o f two consecutive Tors forces all higher Tors to vanish, and
in 1966 Lichtenbaum [L] proved rigidity for all regular local rings The celebrated rigidity conjecture asserted that the theorem should be valid over an arbitrary local ring, provided one assumes that one of the modules has finite projective dimension This was answered negatively by Heitmann [He] in 1993 An interesting question that is still open is whether rigidity holds if both modules are assumed to have finite projective dimension (We remark that rigidity can fail badly for non-finitely generated modules For example, suppose E is the injective hull of the residue field o f the complete regular local ring R of dimension d Then, for every finitely
Trang 7Tensor products of modules and the rigidity of Tor 455
generated R-module M, Tor~(M,E) - Ha-J(M) (by local duality [HK, (5.5)] and
[Rot, (9.5.1)]); and the local cohomology modules can vanish sporadically, [EG2].) The rigidity theorem arose naturally in Auslander's investigation o f torsion in the tensor product of two modules over a regular local ring Our interest in rigidity arose in the same way, but the rigidity results we obtain, for hypersurfaces, are
of independent interest9 Our results appear to have little to do with finiteness of projective dimension9 In fact, most of our modules will have infinite projective dimension9
Notation and assumptions Throughout this section all modules are finitely gen-
erated Let (&ms) be a local ring and put R = S / ( f ) , where f is a non-zero-
divisor in the maximal ideal of R Fix R-modules M and N For compactness of notation we let T~ = Tor~e(M,N) and T s = TorS(M,N) Let t/R and t s denote the lengths of T/g and T s, respectively (Since R and S have the same residue field,
it does not matter whether lengths are computed over R or over S Note also that
To R ~ T0S.) If t0 s < oo then Supp(M) A Supp(N) = {ms} It follows that all the t/~ and t s are finite9 If, further, we know that t s = 0 for n >> 0, we can define the Euler characteristic X s = zS(M,N) = ~i~_o(-1)it s and the higher Euler character-
O O istics Z s = zS(M,N) = ~-~i=j ( - 1 ) i-jts"
We begin with a well-known long exact sequence used by Murthy [Mu] and Lichtenbaum [L]
2.1 Lemma Let (&ms) be a local ring, and let f be a non-zero-divisor in ms Let M and N be modules over R := S/(f) With the notation established above,
we have the following exact sequence:
Proof There is a spectral sequence
Tor~(M, Tory'(N, R)) ==~ TorvS+q(M, N )
p
(See, for example, [Rot,(ll.64)].) It is easy to cheek that TorS(N,R)~-N and
TorS(N,R) = 0 for q>_2 Therefore the spectral sequence degenerates into the long exact sequence above9 ,_Jr'
In our applications in the next section, we need only a special ease o f one o f our main rigidity theorems (2.4) Since its proof is quite easy, we state this special case separately in (2.3) below9
Trang 8456 C Htmeke and R Wiegand
2.2 Proposition Let R = S / ( f ) be a hypersurface o f dimension d, where (S, ms, k)
is a (d + 1)-dimensional regular local ring Let M and N be R-modules, and assume
N has finite length Let i > j > d + 1 Then Tor~(M,N) and TorR(M,N) have the same length,
Proof Let fli = #s(syz~(M)), the ith Betti number of M When we tensor the min- imal resolution of M with the quotient field of S, M dies ( s i n c e f M = 0) It follows that ~d=o( 1)ifl, = 0 Since fli = dimk(TorSi(M,k)) this shows that z S ( M , k ) = O
Therefore zS(M,N) = 0, since N has finite length and xS(M, ) is additive on short exact sequences
We look at the tail end of the long exact sequence (2.1), starting with TS+l, which is 0 because j + 1 > dim(S) and S is regular The alternating sum of all the corresponding t i (for both R and S) is 0 because lengths are additive on short exact sequences Also, the alternating sum of the t s is 0 since it is just zS(M,N) Most o f the t/R cancel in pairs, leaving only the two in the upper right comer, which yield the equation t~+ 1 = t~ ~3
2.3 Corollary Let R = S / ( f ) be a hypersurface o f dimension d, where (S, ms, k) is
a (d + 1)-dimensional regular local ring Let M and N be R-modules, and assume
N has finite length I f Tor~(M,N) - 0 for some j > d + 1, then Tor~(M,N) = 0
if M | N has finite length and d i m ( M ) + dim(N) < dim(S) (His theorem is ac- tually valid over any complete intersection or isolated singularity, provided M and
N are both assumed to have finite projective dimension; but we will need only the regular case of the theorem.)
2.4 First rigidity theorem Let R = S / ( f ) be a hypersurface o f dimension d, where
S is a regular local ring o f dimension d + 1 Let U and V be R-modules such that
(1) U | V has finite length, and
j>-d + 1 we may proceed exactly as in the proof o f (2.2) Suppose now that S is unramified Then the higher Euler characteristics K s are all non-negative, [L] Since the e a s e j = 0 is trivial, we assume that j > 1 From (2.1) we get an exact
s e q u e n c e
( 2 4 1 ) 0 L ~, TS+l , T~+ 1 , T R j 1 - ~ Tj s , 0
Trang 9Tensor products of modules and the rigidity of Tor 457 Suppose j = 1 The exact sequence (2.4.1) becomes
(2.4.2) 0 ~ T 2 s , T ~ T o s ~ > T s ~ 0
In particular, tSo>_t s On the other hand, t s - t s + X2 s = Z s = 0, and since xs>=0
it follows that to s = t s (whence a is an isomorphism) and Z s = 0 It follows from [L, Theorem 2] that T s = 0 (We can harmlessly pass to the completion S, use Lichtenbaum's theorem to get Tor~(/Q,N) = 0, and then conclude by faithfully flat descent [EGA, (2.5.8)] that T o r S ( M , N ) = 0.) Another glance at (2.4.2) confirms that T2 R = 0, as desired
Now assume that j > 2 The alternating sum of the lengths o f all the modules
strictly below the top two lines displayed in (2.1) is 0 Since the tiR, i ~ j - 2 can-
cel in pairs, we are left with ] - I i=0(-1) t i + (-1}/tT_ 1 = 0 Also, z_.,i 0~X"q-lr-lxi's: i s "i + ( - l y x s = Z s = 0 , so ~-1 = Z s- Therefore, by (2.4.1), we have z s > t s But Z s =
t s - zs+l, and since ZS+l > 0 it follows that Z s = t s (whence ~ is an isomorphism) and ~(:+1 ~ 0 Again we use [L, Theorem 2] to conclude that TS+l = 0 By (2.4.1), T~+ 1 = 0, and we are done []
We wilt see in Sect 4 that hypothesis (2) cannot be omitted The hypothesis
is not as strong as it might seem, however, since the fact that M | N has finite length implies that d i m ( M ) + dim(N) <d + 1 (See, for example, [$2, Chap V, Thror~me 3].)
Our next result will be applied in Sect 3, in conjunction with our rigidity theorems, to prove our results on tensor products o f modules It will also be used
in the proof o f our second rigidity theorem (2.7), but only in order to obtain the stronger conclusion that M and N are actually reflexive, rather than just torsion-free
2.5 Proposition Let R be a complete intersection Let M and N be non-zero finitely generated R-modules such that TorR(M,N) = 0 for all i > 1 Then
depth(M) + depth(N) = dim(R) + depth(M | N )
Proof If both M and N are maximal Cohen-Macaulay modules, the inequality " > "
is automatic Therefore, in proving ">=" we will assume that M is not a maximal
Cohen-Macaulay module
Write R = A / ( ~ ,fr), where A is a regular local ring and (J~ ,fi)
is a regular sequence in A We use induction on r, starting with the case r = 0, that is, R = A A theorem due to Auslander [A,(1.2)] implies that d e p t h ( N ) = depth(M | N) + h dimn(M) But the Auslander-Buchsbaum formula lEG1, (3.1)] says that h dimR(M) = dim(R) - depth(M), and the desired equality follows
Now write R - - - S / ( f ) , where S = A / ( f l f r - 0 and f =f~ We assume, in-
ductively, that the proposition holds for S-modules Using (2.1), we see that (2.5.1) T o r S ( M , N ) ~ - M | and T o r S ( M , N ) = 0 for j > 2
Choose a short exact sequence
(2.5.2) 0 ~ E ~ S (m) , M O
Since f M = 0 , d e p t h ( M ) < d i m ( S ) = depth(S(m)) By the "depth lemma" [EG1,
(1.1)] d e p t h ( E ) = 1 + depth(M) Also, T o r S ( E , N ) = 0 for all j ~ l by (2.5.1) and (2.5.2) By our inductive assumption we have
Trang 10458 C Htmeke and R Wiegand (2.5.3) depth(E) + depth(N) = dim(S) + depth(E | N ) ,
or, equivalently,
(2.5.4) depth(M) + depth(N) = dim(R) + depth(E | N )
Therefore it will suffice to prove
Therefore it will suffice to show that depth(E | N) >_- depth(M | N), and now
we may assume that depth(M) < dim(R) If depth(E | N) < depth(M | N), it follows easily from (2.5.6) that depth(W)= depth(E | N), and then from (2.5.7) that depth(N)=depth(E | Now (2.5.4) implies that depth(M)=dim(R), contradiction []
2.6 Corollary Let R be a complete intersection, and let M and N be R-modules such that Tor~(M,N) = 0 for all i >_ 1 I f M | N satisfies Serre's property (Sn),
then so do M and N In particular, if M | N is torsion-free (respectively reflexive, respectively maximal Cohen-Macaulay) then both M and N are torsion-free (respectively reflexive, respectively maximal Cohen-Macaulay)
Proof Let P be an arbitrary prime ideal of R We have, by (2.5),
depth(Me) + depth(Ne ) =dim(Re ) + depth(Me | Ne )
~dim(Re) + min{n, dim(Re)}
Since depth(Ne)<dim(Re) this gives the inequality depth(Mp)>min{n,
dim(Re)} E1
The proof of the next Theorem 2.7 depends on (3.7), the one-dimensional case of the main result of Sect 3 We will use (2.7) in Sect 3 to implement the inductive step in proving the general (d-dimensional) result (3.1) We include (2.7) in this section because it is really a rigidity theorem The reader may rest assured that it will not be used in the proof of (3.7)!
2.7 Second rtgidity theorem Let R = S / ( f ) be a hypersurface, and let M and N
be non-zero R-modules, at least one o f which has constant rank I f M | N is reflexive, then both M and N are reflexive, and TorR(M,N) = 0 for all i~_ 1
Trang 11Tensor products of modules and the rigidity of Tor 459
Proof It is enough to prove that T o ~ ( M , N ) = 0 for all i, for then the fact that
M and N are both reflexive will follow from (2.6) Suppose we can prove the theorem under the additional assumption that M is torsion-free In the general case,
w e tensor the short exact sequence
0 ~ t ( M ) ~ M + M * 0 with N , getting an exact sequence
T o r ~ ( M , N ) , t ( M ) | ~ ' , M | [s,~I|
The image o f ~ is a torsion module, and since M | N is torsion-free,/3 is an iso- morphism Therefore ~Q @R N is reflexive B y the torsion-free case o f the theorem, Tor~(J~r,N) = 0, and it follows that t ( M ) = 0 B y symmetry, we m a y assume both
M and N are torsion-free
W e proceed b y induction on the dimension o f R There is nothing to prove
if d i m ( R ) = 0 (thanks to the constant rank hypothesis), and (3.7) takes care o f dimension one Therefore we assume dim(R) > 1
Choose a short exact sequence
where F is a free R-module Since M is torsion-free, the natural m a p M ~ M** is
y* one-to-one, and we let M1 be the cokernel o f the composition M + M** ~ F** -+
F , where the last map is the natural identification W e have an exact, commutative diagram:
0 , M** ~ F , W* ~, Ext,~(M*,R) , 0 Let X 1 = {P E Spec(R)lheight(P)< 1} I f P E X 1, then both Mp and g ~ are
maximal Cohen-Macaulay modules Since Re is Gorenstein, M~ is reflexive and Ext~e (M~,, Re) = 0 Therefore (2.7.2) becomes
(2.7.4) 0 ~ M o F o M~ - o 0 ,
(2.7.5) 0 o N - o G ~ N1 ~ 0 ,
in which F and G are free Moreover, for each P E X l we have
M e is free if and only if (Ml)e is free and
Are is free if and only if (Nl)e is free
Trang 12460 c Huneke and R Wiegand Notice that Tor~(M1,N) = Tor2R(M1,NI) = 0 by (1.4), since M | N is torsion- free Letting F = R (n), we obtain from (2.7.4) an exact sequence
0 -~M | -~N (n) ~ MI | ~ 0 Localizing this sequence at any P ~ Ass(M1 @~ N), we see that height(P)< 1
(Otherwise, since M | N is reflexive and N (n) is torsion-free, the depths of the
three localized modules would be > 2, > 1, and 0, contradicting the "depth lemma", [EGI, (1.1)].) Thus we have shown that
Composing the natural map S (n) ~ R (~) with the map R (n) ~ MI of (2.7.2), we
get an exact sequence of S-modules
0 ~ E * S (~) * MI ~ 0 When we kill f and use the fact that T o ~ ( M b R ) -~ M1 (obtained from (2.1) or directly), we get an exact sequence of R-modules
We claim that T o r ~ ( E / f E , Nl) = 0 By (2.7.9) and (2.7.10), A s s ( T o r ~ ( E / f E , Nl)) C
X 1 Therefore it will suffice to show that T o ~ e ( ( E / f E ) p , ( N l ) ? ) = 0 for every
P E X 1 To see this, 'we note that by (3.7) either Mp or Np is free If M e is free, so is (M1)p, and by (2.7.7) ( E / f E ) p is free If Np is free, then so is (N1)p In either ease, T o ~ P ( ( E / f E ) p , ( N 1 ) p ) = 0, and our claim is proved
Since f is a non-zero-divisor on E and f N 1 = 0, we have TorS(E, N1) -
T o ~ ( E / f E , NI) for all j (See, for example, [Ma, Sect 18, Lemma 2].) Therefore
by the rigidity theorem o f Auslander and Lichtenbaum [L, Corollary 1] (applied to
the regular local ring S) we have T o r ~ ( E / f E , N1) = 0 for all j > 1 It follows from
(2.7.5) that
Trang 13Tensor products of modules and the rigidity of Tor 461 (2.7.11) Tor~(E/fE, N) = 0 for all j > 1
From (2.7.7) and (2.7.8) we get a surjection Tor~(E/fE, N) * To~(M,N) But
Tor~(E/fE, N) ~- Tor~(E/fE, NO = 0 This shows that Torff(M,N) = 0
Now tensor (2.7.7) with N to get Torzn(M,N)~ Tor~(M1,N) Since at each minimal prime (in fact, at each prime in X 1) either Ml or N is free, To~(M1,N) is
a torsion module On the other hand, it embeds in the torsion-free module M | N
by (2.7.4) We conclude that Tor2R(M,N) = 0
Finally, if j> 3, we have TorT(M,N) ~" Tore_l= (M1,N) =N Tor~_ 2(M,N), by (2.7.7), (2.7.11) and (2.7.4) Since we have already shown that T o r ~ ( M , N ) = Tor2R(M,N) = 0, we are done []
We remark that (1.2) shows that the constant rank hypothesis cannot be omitted In both examples, M | is reflexive, but To~(M,M) -(x)/(x2):t:O
3 Torsion in tensor products
In this section we will apply our rigidity theorems to tensor products of modules (always finitely generated) over hypersurfaces A major goal here is (3.7), which states that a tensor product of two modules over a one-dimensional abstract hy- persurface always has torsion, provided neither module is free and at least one of them has constant rank The general (d-dimensional) version of the theorem, (3.1), follows from (3.7) and (2.7) by induction on the dimension (See (3.3).) We have, however, been unable to find a direct proof of the one-dimensional result A critical step of the proof is to pass from a one-dimensional ring to a closely related ring of dimension two and to use a weak form of the theorem in dimension two to deduce the general result in dimension one Also, our focus on modules of constant rank over general hypersurfaces, rather than on, say, torsion-free modules over integral domains, is more than an attempt to be as general as possible Rather, we need to impose conditions that are stable under operations such as completing and taking hyperplane sections
3.1 Theorem Let (R, m, k) be an abstract hypersurface, and let M and N be R- modules, at least one of which has constant rank I f M | N is a maximal Cohen- Macaulay module, then both M and N are maximal Cohen-Macaulay modules, and either M or N is free
It will suffice to prove that either M or N is free For if, say, M is free, then
N is isomorphic to a direct summand of M | N and is therefore maximal Cohen- Macaulay
The proof will proceed in several steps After some initial simplifications, we prove that the result passes from dimension d > 1 to dimension d + 1 provided M
and N are known to be maximal Cohen-Macaulay modules Next, we use (1.9) and the inductive step to prove the theorem for reflexive modules over two-dimensional rings of multiplicity 2 Then we use rigidity and a device employed by Knfrrer [K]
in the study of hypersurfaces of finite Cohen-Macaulay type, to prove the general case of the theorem in dimension one Once the one-dimensional case of the theorem
is proved, we can use (2.7) to show in all dimensions that Tor~(M,N) = 0 Then we deduce from (2.5) that M and N are maximal Cohen-Macaulay modules Another application of the inductive step completes the proof of the theorem
Trang 14462 C Huneke and R Wiegand
Preliminary reductions We assume for definiteness that M has constant rank Let
R' be the completion /~ if k is infinite Otherwise, let R ~ be the completion o f /~(Z) := J~[Z]~[z] Then R ~ is a complete local ring with infinite residue field, R ~
R ~ is faithfully fiat, and dim(R~) = dim(R) Moreover, the assumptions (in (3.1) and (3.7)) about M , N and M | N all pass to M ' := R' | M , N ' := R' | N and
Mt | Nt (Use (1.3) to see that M ' has constant rank.) If, say, M t can be shown
to be free as an R'-module, it will follow that RM is free (See, for example, [EGA, (2.5.8)].) Therefore we may harmlessly make the following assumptions:
3.2 Harmless assumptions M has constant rank, R is a hypersurface and k is infinite
The Inductive Step
3.3 1.emma Let (R, m, k) be a hypersurface o f dimension d > 2 and with k infinite Let 3: be a finite set o f maximal Cohen-Macaulay modules, each o f constant rank Then there is a non-zero-divisor # E m - m 2 such that R/(g) is a hypersurface o f dimension d - 1 with multiplicity e(R/(#)) : e(R); and such that the Jbllowing are true for every M E f :
(I) Me is Re-free for every prime P minimal over (0)
(2) M/gM is a maximal Cohen-Macaulay R/(g)-module o f constant rank
(3) I f M/oM is R/(o)-free, then M is free
Proof Fix M E ~', and choose an exact sequence
(3.3.1) 0 ~ M ~ F ~ F / M , O,
in which F is free and F / M is torsion Since depth(M) = depth(F)>2, it follows
that F / M has positive depth, that is, m ~ Ass(F/M)
Select a non-zero-divisor c E m such that cF C_ M Since R/(c) is Cohen-
Macaulay o f positive dimension, m ~ Ass(R/(c)) Consider the following set of
ideals: F(M):= Ass(R) U Ass(F/M) U Ass(R/(c)) U {m2} Now repeat the process
for each module M E ~', and let F = U M ~ : F(M) Then F is a finite collection of
ideals, each o f them properly contained in m
Write R = S / ( f ) where (S,n,k) is a regular local ring o f dimension d + 1 Let n = (xl Xd+l) Denote by G the associated graded ring of S, namely
G = S/n @ n/n 2 @ Write y * for the leading form of a non-zero element y E S
Thus y* E ni/n i+l if y E n i - n i+l Note that G ~ k[x~ ,x}+l] is a polynomial
ring o f dimension d + 1, [Ma, (14.4)] Since the associated graded ring o f R is isomorphic to G/(f*), it follows that e(R) = deg(f*), [Ma, Exercise 14.5]
Write F = {11 ls}, and let J / = n-1(I~), where n : S ~ R is the natural map Each ,/i is properly contained in n, and by Nakayama's lemma we have
(3.3.2) J i + n 2 # : n , i = 1 , s
For an element h E n, write h + n 2 = ctlx~ + + Ctd+lX}+ 1, with the ~tj E k, and put ~(h) = (~l ~td+l) Thus r : n ~ V := k (d+l) is a surjective R-homomorphism with kernel n 2 Let ~ = ~(Ji),i = 1, ,s By (3.3.2) each ~ is a proper subspace
o f V
Let dl dt be the distinct linear forms in G dividing f * Let Ui be the l-
dimensional subspace o f V spanned by the vector o f coefficients of di Since k is infinite there is an element 9 = ( o q , , a t d + ~ ) E V t/I1 U U Ws U Ux U U Ut