The Weak Lefschetz for a Graded Module

Transcription

1 1/24 The Weak Lefschetz for a Graded Module Zachary Flores

2 /24 Solomon Lefschetz

3 /24 Solomon Lefschetz 1 Lefschetz lost both his hands in an engineering accident and subsequently became a mathematician.

4 /24 Solomon Lefschetz 1 Lefschetz lost both his hands in an engineering accident and subsequently became a mathematician. 2 Lefschtez was an instructor at UNL from

5 /24 Solomon Lefschetz 1 Lefschetz lost both his hands in an engineering accident and subsequently became a mathematician. 2 Lefschtez was an instructor at UNL from Lefschetz was a professor at KU from

6 3/24 Solomon Lefschetz 1 Lefschetz once received the following letter of recommendation for John Nash.

7 3/24 Solomon Lefschetz 1 Lefschetz once received the following letter of recommendation for John Nash. 2 My roommate once held the door open for John Nash.

8 3/24 Solomon Lefschetz 1 Lefschetz once received the following letter of recommendation for John Nash. 2 My roommate once held the door open for John Nash. 3 Conclusion: I have met Solomon Lefschetz.

9 4/24 Outline 1 Basic Concepts 2 Semistable Bundles 3 Symmetric Hilbert Functions 4 Main Theorem

10 /24 The Weak Lefschetz Property Let k be an algebraically closed field and S the polynomial ring k[x 1,..., x r ].

11 /24 The Weak Lefschetz Property Let k be an algebraically closed field and S the polynomial ring k[x 1,..., x r ]. Definition Given a graded S-module N of finite length, we say that N has the Weak Lefschetz Property if for any general linear form l S 1, the map l : N t N t+1 has maximal rank for all t.

12 /24 The Weak Lefschetz Property Let k be an algebraically closed field and S the polynomial ring k[x 1,..., x r ]. Definition Given a graded S-module N of finite length, we say that N has the Weak Lefschetz Property if for any general linear form l S 1, the map l : N t N t+1 has maximal rank for all t. Question: Which graded S-modules of finite length have the Weak Lefschetz Property?

13 6/24 The Weak Lefschetz Property When N = S/I with I a homogeneous ideal of codimension r, the Weak Lefschetz Property has been studied extensively. 1 When r 2, S/I always has the Weak Lefschetz.

14 6/24 The Weak Lefschetz Property When N = S/I with I a homogeneous ideal of codimension r, the Weak Lefschetz Property has been studied extensively. 1 When r 2, S/I always has the Weak Lefschetz. 2 In [1] it is shown that if k has characteristic 0, r = 3 and I is a complete intersection generated in certain degrees, then S/I has the Weak Lefschetz. This was used to prove all complete intersections in three variables have the Weak Lefschetz.

15 6/24 The Weak Lefschetz Property When N = S/I with I a homogeneous ideal of codimension r, the Weak Lefschetz Property has been studied extensively. 1 When r 2, S/I always has the Weak Lefschetz. 2 In [1] it is shown that if k has characteristic 0, r = 3 and I is a complete intersection generated in certain degrees, then S/I has the Weak Lefschetz. This was used to prove all complete intersections in three variables have the Weak Lefschetz. What about characteristic p > 0?

16 7/24 An Example Suppose r = 3, k has characteristic 2 and I = (x 2 1, x2 2, x2 3 ). If l = ax 1 + bx 2 + cx 3 is a linear form, then a matrix for the map l : (S/I) 1 (S/I) 2 is given by b a 0 A = c 0 a 0 c b

17 7/24 An Example Suppose r = 3, k has characteristic 2 and I = (x 2 1, x2 2, x2 3 ). If l = ax 1 + bx 2 + cx 3 is a linear form, then a matrix for the map l : (S/I) 1 (S/I) 2 is given by b a 0 A = c 0 a 0 c b Then det(a) = 2abc = 0, so l is not injective. Thus S/I does not have the Weak Lefschetz.

18 7/24 An Example Suppose r = 3, k has characteristic 2 and I = (x 2 1, x2 2, x2 3 ). If l = ax 1 + bx 2 + cx 3 is a linear form, then a matrix for the map l : (S/I) 1 (S/I) 2 is given by b a 0 A = c 0 a 0 c b Then det(a) = 2abc = 0, so l is not injective. Thus S/I does not have the Weak Lefschetz. In fact, if r 3 and k has characteristic p > 0, then S/(x p 1,..., xp r) does not have the Weak Lefschetz.

19 8/24 Setup As the above results suggest, the characteristic of k plays a subtle role in determining when N has the Weak Lefschetz.

20 8/24 Setup As the above results suggest, the characteristic of k plays a subtle role in determining when N has the Weak Lefschetz. 1 k has characteristic zero (and is still algebraically closed!).

21 8/24 Setup As the above results suggest, the characteristic of k plays a subtle role in determining when N has the Weak Lefschetz. 1 k has characteristic zero (and is still algebraically closed!). 2 We set R = k[x, y, z].

22 8/24 Setup As the above results suggest, the characteristic of k plays a subtle role in determining when N has the Weak Lefschetz. 1 k has characteristic zero (and is still algebraically closed!). 2 We set R = k[x, y, z]. 3 n 1, ϕ : n+2 j=1 R( b j) n i=1 R( a i) is an R-linear map with b j b j+1 and a i a i+1.

23 8/24 Setup As the above results suggest, the characteristic of k plays a subtle role in determining when N has the Weak Lefschetz. 1 k has characteristic zero (and is still algebraically closed!). 2 We set R = k[x, y, z]. 3 n 1, ϕ : n+2 j=1 R( b j) n i=1 R( a i) is an R-linear map with b j b j+1 and a i a i+1. 4 We set M = coker(ϕ).

24 8/24 Setup As the above results suggest, the characteristic of k plays a subtle role in determining when N has the Weak Lefschetz. 1 k has characteristic zero (and is still algebraically closed!). 2 We set R = k[x, y, z]. 3 n 1, ϕ : n+2 j=1 R( b j) n i=1 R( a i) is an R-linear map with b j b j+1 and a i a i+1. 4 We set M = coker(ϕ). When M has finite length, we are interested in what numerical constraints we can place on the b j and a i so that M has the Weak Lefschetz.

25 9/24 Some Algebraic Geometry If E is a vector bundle on P r, its slope µ(e) is the rational number c 1 (E)/rank(E), where c 1 (E) is the first Chern class of E.

26 9/24 Some Algebraic Geometry If E is a vector bundle on P r, its slope µ(e) is the rational number c 1 (E)/rank(E), where c 1 (E) is the first Chern class of E. Definition We say that a vector bundle E on P r is semistable if µ(e ) µ(e) for every proper nonzero subbundle E of E.

27 9/24 Some Algebraic Geometry If E is a vector bundle on P r, its slope µ(e) is the rational number c 1 (E)/rank(E), where c 1 (E) is the first Chern class of E. Definition We say that a vector bundle E on P r is semistable if µ(e ) µ(e) for every proper nonzero subbundle E of E. For E with rank 2 and E normalized (c 1 (E) { 1, 0}), E is semistable if and only if it has no sections.

28 10/24 Some Algebraic Geometry By a theorem of Grothendieck, every vector bundle on P 1 splits as a sum of line bundles. Hence, if λ is general line in P 2 and E is a vector bundle on P 2 then E λ = O λ (e 1 ) O λ (e s )

29 10/24 Some Algebraic Geometry By a theorem of Grothendieck, every vector bundle on P 1 splits as a sum of line bundles. Hence, if λ is general line in P 2 and E is a vector bundle on P 2 then E λ = O λ (e 1 ) O λ (e s ) Where the e k are independent of λ and s = rank(e). The s-tuple (e 1,..., e s ) is called the splitting type of E.

30 11/24 The Grauert-Mülich Theorem Theorem If E is a semistable normalized vector bundle of rank two on P 2 and λ is a general line in P 2, then { (0, 0) c1 (E) = 0 (e 1, e 2 ) = (0, 1) c 1 (E) = 1

31 11/24 The Grauert-Mülich Theorem Theorem If E is a semistable normalized vector bundle of rank two on P 2 and λ is a general line in P 2, then { (0, 0) c1 (E) = 0 (e 1, e 2 ) = (0, 1) c 1 (E) = 1 We will see how this theorem plays a crucial role in determining when M has the Weak Lefschetz.

32 2/24 The Minimal Free Resolution of M M has finite length if and only if its ideal of maximal minors has codimension 3. In this case, the Buchsbaum-Rim complex gives the minimal free resolution for M:

33 2/24 The Minimal Free Resolution of M M has finite length if and only if its ideal of maximal minors has codimension 3. In this case, the Buchsbaum-Rim complex gives the minimal free resolution for M: n+2 0 F 3 F 2 R( b j ) ϕ j=1 n R( a i ) i=1 Where F 3 = n i=1 R(a i d) and F 2 = m j=1 R(b j d) with d = b j a i.

34 13/24 Bundles from Syzygies Set E = ker(ϕ), so that upon sheafification, we obtain the following exact sequence of sheaves over P 2 : 0 F 3 F 2 E 0 Thus E is a vector bundle of rank two and we want to know when E is semistable.

35 13/24 Bundles from Syzygies Set E = ker(ϕ), so that upon sheafification, we obtain the following exact sequence of sheaves over P 2 : 0 F 3 F 2 E 0 Thus E is a vector bundle of rank two and we want to know when E is semistable. Lemma Suppose n > 1. Then E is semistable when (a) d is even, i<n b i a i > 4, a n + b m + 1 < b n + b n+1 (b) d is odd, i<n b i a i > 4, a n + b m + 2 < b n + b n+1

36 14/24 Bundles from Syzygies A twist combined with the Grauert-Mülich Theorem yields Corollary If n > 1 and either condition of the above lemma is satisfied, then the splitting type (e 1, e 2 ) of E is { ( e, e) d = 2e (e 1, e 2 ) = ( e, e 1) d = 2e + 1

37 14/24 Bundles from Syzygies A twist combined with the Grauert-Mülich Theorem yields Corollary If n > 1 and either condition of the above lemma is satisfied, then the splitting type (e 1, e 2 ) of E is { ( e, e) d = 2e (e 1, e 2 ) = ( e, e 1) d = 2e + 1 With ([1], Lemma 2.1, Corollary 2.2), we have numerical conditions for all n on the a i and b j that tell us when E is semistable and, if that is the case, what its splitting type is.

38 15/24 Symmetrically Gorenstein Modules In [1], one of the key ingredients in proving that complete intersections of codimension 3 have the Weak Lefschetz Property was the symmetry of the Hilbert function. It is a well-known fact that graded Gorenstein k-algebras have a symmetric Hilbert function.

39 15/24 Symmetrically Gorenstein Modules In [1], one of the key ingredients in proving that complete intersections of codimension 3 have the Weak Lefschetz Property was the symmetry of the Hilbert function. It is a well-known fact that graded Gorenstein k-algebras have a symmetric Hilbert function. Definition We say a graded S-module N of finite length is Symmetrically Gorenstein if there is a v Z and a graded isomorphism τ : N Hom k (N, k)( v) such that τ = Hom k (τ, k)( v).

40 16/24 Kunte s Theorem When is M Symmetrically Gorenstein?

41 16/24 Kunte s Theorem When is M Symmetrically Gorenstein? We have the following special case of ([2], Theorem 1.3). Theorem Let N be graded R-module of finite length and of maximal socle degree c. Set ( ) = Hom R (, R( c 3)). Then N is Symmetrically Gorenstein if and only if its minimal graded free resolution is of the form 0 G 0 G 1 where Ψ is antisymmetric. Ψ G 1 G 0

42 16/24 Kunte s Theorem When is M Symmetrically Gorenstein? We have the following special case of ([2], Theorem 1.3). Theorem Let N be graded R-module of finite length and of maximal socle degree c. Set ( ) = Hom R (, R( c 3)). Then N is Symmetrically Gorenstein if and only if its minimal graded free resolution is of the form 0 G 0 G 1 where Ψ is antisymmetric. Ψ G 1 G 0 It is not hard to see that under mild conditions, Symmetrically Gorenstein modules have symmetric Hilbert functions.

43 7/24 Using Kunte s Theorem Lemma M has a symmetric Hilbert function if a 1 = 0 and its maximal socle degree is d 3.

44 7/24 Using Kunte s Theorem Lemma M has a symmetric Hilbert function if a 1 = 0 and its maximal socle degree is d 3. When a 1 = 0, M will have a maximal socle degree d 3 under reasonable assumptions.

45 18/24 Main Theorem Theorem If E is semistable and M has symmetric Hilbert function, then M has the Weak Lefschetz Property.

46 18/24 Main Theorem Theorem If E is semistable and M has symmetric Hilbert function, then M has the Weak Lefschetz Property. In particular, utilizing our preceding work, we have numerical conditions on the a i and b j that guarantee that M has the Weak Lefschetz.

47 18/24 Main Theorem Theorem If E is semistable and M has symmetric Hilbert function, then M has the Weak Lefschetz Property. In particular, utilizing our preceding work, we have numerical conditions on the a i and b j that guarantee that M has the Weak Lefschetz. Corollary Complete intersections in R have the Weak Lefschetz Property.

48 19/24 Sketch of proof Recall ϕ : n+2 j=1 R( b j) n i=1 R( a i) is an R-linear map with cokernel M.

49 19/24 Sketch of proof Recall ϕ : n+2 j=1 R( b j) n i=1 R( a i) is an R-linear map with cokernel M. Proof. Let l be a general linear form in R and λ be the general line defined by l in P 2. After sheafifying, some diagram chasing and an application of the Snake Lemma, we obtain following exact sequence of sheaves:

50 19/24 Sketch of proof Recall ϕ : n+2 j=1 R( b j) n i=1 R( a i) is an R-linear map with cokernel M. Proof. Let l be a general linear form in R and λ be the general line defined by l in P 2. After sheafifying, some diagram chasing and an application of the Snake Lemma, we obtain following exact sequence of sheaves: n+2 0 E λ O λ ( b j ) j=1 n O λ ( a i ) 0 i=1 ( )

51 19/24 Sketch of proof Recall ϕ : n+2 j=1 R( b j) n i=1 R( a i) is an R-linear map with cokernel M. Proof. Let l be a general linear form in R and λ be the general line defined by l in P 2. After sheafifying, some diagram chasing and an application of the Snake Lemma, we obtain following exact sequence of sheaves: n+2 0 E λ O λ ( b j ) j=1 n O λ ( a i ) 0 When d = 2e, the semistability of E gives, by our previous work, E λ = O λ ( e) 2. i=1 ( )

52 20/24 Proof. Set R = R/lR. If N = im(ϕ) n i=1 R( a i), taking global sections in ( ) yields the exact sequence of R-modules:

53 20/24 Proof. Set R = R/lR. If N = im(ϕ) n i=1 R( a i), taking global sections in ( ) yields the exact sequence of R-modules: n+2 0 R( e) 2 R( b j ) ϕ N 0 j=1 ( )

54 20/24 Proof. Set R = R/lR. If N = im(ϕ) n i=1 R( a i), taking global sections in ( ) yields the exact sequence of R-modules: n+2 0 R( e) 2 R( b j ) ϕ N 0 j=1 ( ) Since the Hilbert function is symmetric, it suffices to show that for u d 3 2 = e 2, M u l M u+1 is injective.

55 21/24 Proof. To this end, if l kills a nontrivial element of M u, there is an equation lf = ϕa with F n i=1 R( a i) and A n+2 j=1 R( b j).

56 21/24 Proof. To this end, if l kills a nontrivial element of M u, there is an equation lf = ϕa with F n i=1 R( a i) and A n+2 j=1 R( b j). Reducing modulo l, we obtain a nontrivial element A of ker(ϕ). Using ( ), it is not hard to see that this would yield a syzygy of ϕ that has degree u + 1 e, which is impossible. When d is odd, the proof is similar.

57 22/24 Example Example Let n = 2, q 3 and f 1, f 2, f 3 be a regular sequence in R q and ϕ : R( q) 4 R 2 defined by [ ] f1 f 2 f f 1 f 2 f 3 Then M = coker(ϕ) has the Weak Lefschetz and its minimal number of generators as an R-module is 2.

58 23/24 References 1 Harima, T., Migliore, J.C., Nagel, U., Watanabe, J., The Weak and Strong Lefschetz Properties for Artinian K-algebras. Journal of Algebra 262(1), pp , Kunte, M., Gorenstein Modules of Finite Length. Mathematische Nachrichten 284(7), pp , 2011.

59 4/24 Acknowledgements A big thanks to the organizers for inviting me to speak. I would like to thank Chris Peterson for suggesting this problem, his helpful comments and encouragement on this project. I would also like to thank Gioia Failla for her helpful comments in preparing the paper this talk is based on.