Self-duality of rank 2 reflexive modules

Journal of Pure and Applied @ North-Holland Publishing

Algebra 16 (1980) 275-284 Company

SELF-DUALITY

OF RANK 2 REFLEXIVE

MODULES

Matthew MILLER* Department of Mathemarics, University of Illinois ar Urbana-Champaign,

Urbana, IL 61801, USA

Communicated by H. Bass Received 31 October 1978

Every finitely generated rank 2 third syzygy module over a regular local ring is known to be self-dual. We show more generally that any finitely generated rank 2 reflexive module is self-dual, and that the isomorphism is skew symmetric. We use this ismorphism to estimate how large k may be if the module is a kth syzygy, and how closely allied rank 3 modules are related to their duals.

1. Introduction The syzygy conjecture states that if R is a regular local ring and M a kth syzygy of rank less than k, then M is free. The problem is trivial for k = 1; for k = 2 it has an affirmative solution since rank 1 reflexive modules are free over a UFD, and all second syzygies are reflexive. By the same argument any rank 1 third (or higher) syzygy is free. Thus the first interesting (and still open) case concerns the possible existence of a non-free rank 2 third syzygy. Such modules are known to be self-dual [2]. We shall prove more generally that all rank 2 second syzygies are self-dual, and that, moreover, the isomorphism is skew symmetric in a certain sense. The prototype is the skew symmetric automorphism of RZ given by the matrix

0

[ -1

1

1 0’

In brief, the idea of the proof is to represent a second syzygy M as the central term of an extension in Ext*(P, R), where P is a suitable height 2 (prime) ideal, and then to induce an isomorphism q:M-*M* by means of an endomorphism *: Ext’(P, R)-,Ext’(P, R). The paper concludes with several applications. The self-duality of a rank 2 reflexive module entails that it cannot be a very high order syzygy. Furthermore, this self-duality provides information about the relationship of rank 3 third syzygies to their duals. We relate, finally, the question of possible self-duality in the rank 3 case to a problem in the theory of complete intersections. Throughout this paper R will be a regular local ring and all modules will be finitely generated. The dual of M is M* = Hom(M, R). A module M is reflexive if the canonical map u: M-M** is an isomorphism. In [l] it is shown that M is reflexive if and only if M is a second syzygy. * Present address: Department of Mathematics, University of Tennessee, Knoxville, TN 37916, USA. 275

276

M. Miller

2. Construction

of endomorph&m

*

Lemma 1. Let Ibe an ideal of height 2. There is a set map * : Extl(l, R)*Ext’(l, induced by Hom(-,

R)

R) such that 5** = 6 for all .$ E Ext’(l, R).

Proof. Suppose 5 is represented

by a short exact sequence

O*R&Gk+O.

Apply Hom(-,

R) and obtain an exact sequence R).

04*%f*SR*:Ext’(l,

Since height I = 2, there is a regular R-sequence of length 2 contained in I, and it follows that I* = R* = R, where the second isomorphism is the usual identification. Note that S(1) = 5. Let J = ker 6. Now Ext’(1, R) =Ext’(R/I, R); so I annihilates Ext’(f, R) and 16 = 0. Then Z c J, and we may form the pull-back diagram ?j:R

--M*-J

e$“:R-N

1 I -

(1)

I

Let the lower row represent e*. A routine diagram chase verifies that this is independent of choice of representative of 5. We form [** by dualizing t*, obtaining an ideal K z I, and forming the pull back, as in diagram 2. R-N*-K

(2)

The lower row then represents 5 **. By the snake lemma applied to (l), there is an -0. After dualizing we have an exact sequence exact sequence O-,N-*M*+JjI O-,(J/I)*-+M**-+N*+Ext’(J/Z,

R).

Since grade I = 2, and I c ann(J/I), it follows that (J/I)* = Ext’(J/I, R) = 0, and hence M** = N*. When the top row 17of (1) is dualized, one would expect to obtain a new ideal containing J, just as we saw that J contained I after dualizing 5. The

Self-dualiry of rank 2 repexice modules

following further

lemma dualizing

allows us to conclude, does not produce

however,

that J is stable

277

in the sense that

a. larger ideal.

Lemma 2. Let q: R+L+Q represent a non-zero element in Ext’(Q, R), where height Q = 2. Assume L is reflexive or Q is prime. Then no pull-back is needed to form t7*. We defer the proof, and continue with the proof of Lemma 1. Since R is Gorenstein and M* is a dual, M* is reflexive. Using Lemma 2, we obtain the following

commutative

diagram.

The isomorphism J-, K is induced by idR*, so in fact J = K. The map (+ is the standard map of a module to its double dual; it is manic because M is clearly torsion-free. We may regard each of P and A4 as the submodule of M** which is the full preimage of I; thus there is a map A4 ---, P compatible with the rest of the diagram, and necessarily an isomorphism. Then 6 = t** in Ext’(Z, R), concluding the proof of Lemma 1. Proof of Lemma 2. The assertion O--,Q*-+L*-+R*~Ext’(Q,

is essentially

that Q = ker 6 in the exact sequence

R).

We already know that Q c ker 6. If Q is prime and the inclusion is proper, then height (ker 6) 2 3, and Ext’(ker 8, R) = 0. It follows that L** ss R*, and L* = L*** = (R*)* is free. Then ker 6 is a 2-generated ideal, which contradicts height (ker S) 5 3, or else ker S = R. But this is also impossible since S( 1) = n # 0. Therefore Q = ker 6. For the other case suppose L is reflexive and let Z = ker S 2 Q. Dualize the sequence O-+R+L*-+Z+O, and let Z’zZ be the kernel of the connecting diagram below in which the first two columns natural inclusion.

homomorphism. Then we obtain the are isomorphisms, and Q-Z’ is the

278

M. Miller

-0

R-L

I I cr

R -

L**-_l’

Then Q = I = I’, as desired. Note that n # 0 if I_.is reflexive, since Q is not reflexive. Remark.

It is easy to check that 0” = 0.

The next step in our program homomorphism. Lemma

is to show that the map t-3(*

is actually a

3. Let I be a height 2 ideal, a E R, and (E Ext’(I, R). Then (at)* = at*.

Proof. The extension 5 is represented, formed by the push-out diagram (:R a

say, by O+R+M+I+O.

Then a( is

-M-l

I I

aezR-M,-~

To compute the effect of *, dualize and form pull-backs. (at)*:

R -

N,

By the universal mapping property for pull-backs there is a map Ni *N, so that the diagram commutes. The top of the diagram is now precisely the pull-back which yields a&*; thus (a()* = at*.

Self-duality

of rank 2 ref7exice modules

279

One may similarly show, by explicitly manipulating the Baer sums, that * is additive, although this is really quite cumbersome. In the cases of particular interest, for instance if I = P is a height 2 prime, we will be able to give a neater, and more informative proof. The following lemma also shows how the map tc*.$* is used to relate A4 and M*. Lemma

4. Suppose

t-3

I&y)

~:Rw+ROR-L

is a (necessarily non-split) extension, where I is generated by a regular sequence (x, y). Then [* = -6. Proof. After making the usual identifications, exact rows. 5:

RV

(-3

ROR

A *. 5 * R-ROR-I

we have the following diagram with

-I

[:’ -:I

t&y)

B

I (Y.--X)

(C)

Note that A commutes, but B commutes only up to (-1). If we compute -t* we find that the diagram 5:

(-I)

R -ROR

-I

I

by)

L-Y3

-5*:

R.o_ROR-I Y

(Y.-X)

commutes, which means 5 = -[*. We are now ready to prove the main result of this section. Theorem 5. Let R be a regular local ring and P an unmixed height 2 ideal which is 2-generated at its minimal primes. The map * is an R-endomorphism of Ext’(P, R). Moreover, * is actually multiplication by -1. Proof.

Define cp:Ext’(P, R)-,Ext’(P, R) by cp(5)=5+5*. By Lemma 3 the set Im(cp) is closed under scalar multiplication. If Im(cp) Z 0, let C? be a prime minimal with respect to the property (Im(cp))o # 0. Clearly Q 1 P, so height 0 2 2. Replace R

280

M. ‘wler

by RQ, and suppose that (p(t) # 0, with .$ represented by R wM-nP. If Q is not minimal over P, that is, Qa Ass(R/P), then depth Ext’(P, R) 2 1. (Indeed any element x of Q which is a non-zero-divisor on R/P is likewise a non-zero-divisor on Ext*(R/P, R)=Ext’(P, R).) But Q is minimal in the support of R *q([)c Ext’(P, R), which means Q E Ass(Ext’(P, R)), and hence depth Ext’(P, R) = 0. We conclude Q is minimal over P. Now if M is a reflexive module over a dimension 2 regular local ring, then M is actually free, and by Lemma 4, q(t) = 0, which is again a contradiction. There remains the possibility that M is torsion free, but not free. Since P is 2-generated, R/P is Gorenstein and therefore Ext’(P, R) = R/P, with cyclic generator 11:R >-,R**P. If 5 = aq, then cp(5‘)=5+5‘*=a77+(al7)*=a77+ag*=acp(t7)=0, which is again a contradiction. endomorphism.

Therefore

Im(cp) = 0, [* = -t

for all 5, and * is an

Remark.

Bruns, Evans, and Griffith have shown [2] that if M is a rank 2 reflexive module then there is an exact sequence O+R+M-+P+O, in which P has the properties stated in the theorem. In fact, if R contains a field of characteristic 0 and dim R ~4, then P may even be chosen to be prime [6].

3. The main theorem and applications Theorem 6. LetMbe a rank 2 reflexive module. There isan isomorphism Q: M+M* such that -q* 0 u = cp. That is, after identifying M with M**, cp becomes “skew symmetric”. Proof. By the remarks above, there is an extension 6: R-M-P, for which t* = -5, by Theorem 5. By Lemma 2, f* is represented by R wM* -nP. Let 40be-the center isomorphism given by the equivalence of 5 and -[* in Ext*(P, R). 5:

-p:

R-

M-P

R-M*-P

*. The assertion of the proposition is that Consider the map T = rp +cp*u:M-+M Im(r) = 0. If Im(r) # 0, take Q minimal in its support. Then height Q 2 3 because M reflexive implies M0 is free for height Q s 2, and by Lemma 4, 0

cpo=

[ -1

1

1 = -q$ 0

Self-duality of rank 2 reflexice modules

281

for such primes. Now Im(r) is contained in M*, which is torsion free, so that (0) is the sole associated prime. Thus we must have Im(r) = 0. The next result is an immediate consequence

of the theorem.

Corollary 7. Let M be a rank 2 reflexive module with isomorphism cp: M-M* as in the theorem. Suppose {ml, . . . , rnk) is a minimal generating set for M, and let fi = cp(ml) be the corresponding generators of M*. Then (1) fi(mi)=-fj(rni) (=O if i=j). (2) If F = uF=, Rei AM with r(ei) = mi, then the composition F:M:M*-+F*

ff’

CanOn -F

is given by an alternating matrix. m EM} is generated by {f(mi)}i
The point of (3) is that rank 2 vector bundles which are not free tend to require many generators. We may also use (1) and (3) to somewhat circumscribe the possible rank 2 kth syzygies. Let us write syz(M) = k to mean that M is a kth syzygy, but no higher. Recall the Auslander-Bridger criterion [ 1, Theorem 4.251 that syz(M) = k if and only if depth Mp b min{k, depth Rr}. Hence depth M = syz(M) if M is free on the punctured spectrum. Under the additional hypothesis that 4~ R, we will show that syz(M) -pd MS 1. In effect, a rank 2 syzygy cannot appear near the back of a free resolution. We use a theorem due to Lebelt [5], which states that if 4~ R, pd M c m, and syz(M) t m + k, then M A M is a kth syzygy. Lemma 8. Suppose f~ R and M is a rank 2 reflexive module with pd M G m and syz M 2 m + 1. Then M A M and tr Mare isomorphic ideals. Proof. There is a canonical map K:M

A M-MOM

given by

K(XhJ’)=+(X@y-J&X).

The image of the evaluation map ev: MOM*R is simply tr M. If cp:M+M* is the isomorphism of Corollary 7, then it is easy to check that 4 = ev 0 (10~) 0 K maps M A M onto tr M. Since M A M is rank 1 torsion free by Lebelt’s theorem, JI is an isomorphism. Proposition 9. Suppose $E R and M is a rank 2 non-free reflexive module. Then syz M - pd M s 1. Zf M is free on the punctured spectrum of R, and dim R = n, then syz M s t(n + 1) and pd Mb&n - 1).

M. Miller

282

Proof. Let pd M = M. If syz M B m + 2, then by Lebelt’s theorem and Lemma 8, tr M z M AM is a second syzygy, and thereby reflexive. Since A4 is reflexive, it is free at height 2 primes, and therefore height (tr M) b 3. Then tr M = R, which implies M is free. To avoid this contradiction, we must have syz M c m + 1. The second assertion follows easily.

Several other proofs of this proposition are known. If the new intersection theorem holds for R, a very short proof may given by sewing a resolution F. of M to the complex Hom(F., R) along the isomorphism (p:M+M* (see [2]). For our final application, we show that there is a close relationship between rank 3 third syzygies and unmixed height 3 ideals which are generic complete intersections (i.e. 3-generated at their associated primes). Proposition 10. Let R be a regular local ring with infinite residuefield. Let Mbe a rank 3 third syzygy withoutfree summand. Then there exists an ideal I, unmixed of height 3, with Ip 3-generated for P E Ass(R/I), and which has a resolution (*)

O-+R;M:M*&+O

in which $ is skew symmetric. Conversely, if I is a height 3 unmixed generic complete intersection, there exists a third syzygy Mappearing

in a resolution of form (*).

Proof. Since M is a second syzygy of rank 3, there exists x EM such that M/Rx

=N

is a second syzygy [3]. Let CYbe defined by CY (1) = x. By dualizing the exact sequence O+R:M-+N-+O, we obtain an exact sequence O+N*-,M*+I-,O, where I = tr (x) = {f(x) 1f E M*}. Note that I # R since (Ydoes not split. Using the

isomorphism cp:N + N*, we may splice these sequences together to form an exact sequence

It is easy to see that I/, is skew symmetric; that is, 1,4= -(//* 0 a,++ Since CYsplits at height 2 primes, height I z 3. Suppose P E Ass(R/I) and height PS 4. By the Auslander-Bridger criterion depth Mp 2 3; hence depth Np = depth N*p b 3. From the reflexivity of M*, we have depth M*p 2 2, and therefore depth IP B 2. It follows that depth (R/I)p a 1, contradicting our supposition that I had associated primes of height more than three. We conclude I is unmixed of height exactly three, and a generic complete intersection since M is free at height 3 primes.

Self-duality of rank 2 reflexive modules

283

For the converse, choose a regular sequence x = (xi, ~2, ~3) c I such that xp = Ip for P E Ass(R/I). (See [7, Lemme 3.81; this is where the infinitude of the residue field is used.) Let K be the second syzygy in the Koszul resolution of R/x. This provides us with an extension A : K - R’*x. Consider the exact sequence Ext’(1, K)+Ext’(x,

K)-+Ext*(l/x,

K).

Since the last term vanishes at all PE Ass(R/I), there exists 71E Ext’(I, K) and s E R -UP, such that qp = sAp for all PE Ass(R/I). Let K w N +I represent 1). Then N is a second syzygy which is free at height 3 primes, and N* is a third syzygy. The self-duality of K enables us to construct an exact sequence of the desired form, where we set M = N*. Remarks. (1) By the same theorem of Brunssited in the proof, rank 3 third syzygies may always be obtained from third syzygies of higher rank. There is also considerable latitude in the choice of (Y. (2) If M is merely assumed to be a reflexive module, but free on the punctured spectrum, then m may become an embedded prime of I. Notice that if we regard A4 as affording a rank 3 vector bundle on Spec R -{m}, then Spec(R/I) is the zero set of the section (Y,and is locally a complete intersection on the punctured spectrum. Hartshorne and Ogus [4, Problem B] and [S] have asked whether such a variety must be a global complete intersection for dim R 2 10. By using the sequence (*), one can easily show that if R/I is Cohen-Macaulay, then syz M=syzM*, and $7 : Ext’(M*, R)-,Ext’(M, R) is onto for i = 2 and an isomorphism for i > 2. If R/I is actually a complete intersection, then it is natural to ask whether M must be free, that is, whether (*) must actually be the standard free resolution of I. (It is not unreasonable to expect at least some additional conditions on M. For motivation see Hartshorne’s discussion in [S] of the relationship between vector bundles of rank I < n on P” and their associated subvarieties of codimension r. In particular, for r = 2 the complete intersections correspond exactly to the direct sums of line bundles.) Indeed, in our case, we can show that (*) and the Koszul resolution both represent the same element in Ext*(I, R), up to multiplication by a unit u of R, provided dim R 2 5. (This unit is obtained by mapping the Koszul resolution of I to (*).) By dualizing these sequences and forming the mapping cone, one obtains a complex 0-R

(‘) -RQM+R3@M*--+R30R

(..U)

-R+O

which is readily verified to be exact. Hence MOR3 = R30M*, and, if the KrullSchmidt theorem holds (e.g. if R is complete), M = M*. Apparently the existence of a complete intersection associated to a rank 3 vector bundle would entail fairly strong conditions on the bundle itself. Even the weaker of these “self-duality” isomorphisms is sufficient to yield a nontrivial upper bound on syz A4 (see comment following Proposition 9), if M is not free.

M. Miller

284

Acknowledgement This work constitutes Illinois.

He would

discussions

part of the author’s

like to thank

and constant

Ph.D. thesis written

his adviser,

Prof.

at the University

Phil Griffith,

of

for his helpful

encouragement.

Note added in proof The module it4 of the last paragraph is indeed free. One exploits the comparison map from the Koszul resolution to (*) and the “self-duality” of (*) in the proof.

References [l] M. Auslander and M. Bridger, Stable module theory, Mem. Amer. Math. Sot. 94 (1969). [2] W. Bruns, E.G. Evans and P. Griffith, Syzygies, ideals of height two, and vector bundles, in preparation. [3] W. Bruns, “Jede” freie Aufliisung ist freie Auflosung eines von drei Elementen erzeugten Ideals, J. Algebra 39 (1976) 429-439. [4] R. Hartshorne and A. Ogus, On the factoriality of local rings of small embedding codimension, Comm. in Algebra 1 (1974) 415437. [5] K. Lebelt, Zur homologischen Dimension Lusserer Potenzen von Moduln, Arch. Math. 26 (1975) 595-601. [6] M. Miller, Ph.D. Thesis, University of Illinois, Urbana, IL, 1979. [7] C. Peskine and L. Szpiro, Liaison des varietals algebriques, Invent. Math. 26 (1974) 271-302. [8] R. Hartshorne, 1017-1032.

Varities

of small codimension

in projective

space, Bull. Amer.

Math. Sot. 80 (1974)