Multiple Structures with Arbitrarily Large Projective Dimension

Transcription

1 Multiple Structures with Arbitrarily Large Projective Dimension Jason McCullough (Joint w/craig Huneke, Paolo Mantero and Alexandra Seceleanu) October 5, 2013 AMS Sectional Meeting Special Session on Recent Advances in Commutative Algebra

2 Notation K = K an algebraically closed field

3 Notation K = K an algebraically closed field S a polynomial ring over K

4 Notation K = K an algebraically closed field S a polynomial ring over K S = i=0 S i is a standard graded ring, where S i = K-vector space of homogeneous degree i polynomials

5 Notation K = K an algebraically closed field S a polynomial ring over K S = i=0 S i is a standard graded ring, where S i = K-vector space of homogeneous degree i polynomials S( d) = rank one free module with generator in degree d (so S( d) i = S i d )

6 Notation K = K an algebraically closed field S a polynomial ring over K S = i=0 S i is a standard graded ring, where S i = K-vector space of homogeneous degree i polynomials S( d) = rank one free module with generator in degree d (so S( d) i = S i d ) I = (f 1,...,f t ) R a homogeneous ideal (i.e. each f j is in some R i )

7 Stillman s Question Question (Stillman) Is there a bound on the projective dimension of ideals in S = K[X 1,...,X n ] which are generated by N homogeneous polynomials of given degrees d 1,...,d N?

8 Stillman s Question Known cases: Easy cases: I = (f), I = (f,g), I = a monomial ideal.

9 Stillman s Question Known cases: Easy cases: I = (f), I = (f,g), I = a monomial ideal. Most general result: Yes, if char(k) 2,3 and all d i 4. (Ananyan-Hochster)

10 Stillman s Question Known cases: Easy cases: I = (f), I = (f,g), I = a monomial ideal. Most general result: Yes, if char(k) 2,3 and all d i 4. (Ananyan-Hochster) Tighter bounds:

11 Stillman s Question Known cases: Easy cases: I = (f), I = (f,g), I = a monomial ideal. Most general result: Yes, if char(k) 2,3 and all d i 4. (Ananyan-Hochster) Tighter bounds: pd(s/(3 quadrics) 4 and this bound is tight. (Eisenbud-Huneke)

12 Stillman s Question Known cases: Easy cases: I = (f), I = (f,g), I = a monomial ideal. Most general result: Yes, if char(k) 2,3 and all d i 4. (Ananyan-Hochster) Tighter bounds: pd(s/(3 quadrics) 4 and this bound is tight. (Eisenbud-Huneke) pd(s/(3 cubics) 36, likely not tight. (Engheta)

13 Stillman s Question Known cases: Easy cases: I = (f), I = (f,g), I = a monomial ideal. Most general result: Yes, if char(k) 2,3 and all d i 4. (Ananyan-Hochster) Tighter bounds: pd(s/(3 quadrics) 4 and this bound is tight. (Eisenbud-Huneke) pd(s/(3 cubics) 36, likely not tight. (Engheta) pd(s/(m quadrics of height 2)) 2m 2 and this bound is tight. (HMMS - Next!)

14 Stillman s Question Known cases: Easy cases: I = (f), I = (f,g), I = a monomial ideal. Most general result: Yes, if char(k) 2,3 and all d i 4. (Ananyan-Hochster) Tighter bounds: pd(s/(3 quadrics) 4 and this bound is tight. (Eisenbud-Huneke) pd(s/(3 cubics) 36, likely not tight. (Engheta) pd(s/(m quadrics of height 2)) 2m 2 and this bound is tight. (HMMS - Next!) pd(s/(4 quadrics)) 9.(6?). (HMMS)

15 An Approach to Stillman s Question Let I = (f,g,h).

16 An Approach to Stillman s Question Let I = (f,g,h). May assume ht(i) = ht(f,g) = 2. Get: 0 R I : h R (f,g) R I 0.

17 An Approach to Stillman s Question Let I = (f,g,h). May assume ht(i) = ht(f,g) = 2. Get: 0 R I : h R (f,g) R I Suffices to bound pd(r/(i : h)). 0.

18 An Approach to Stillman s Question Let I = (f,g,h). May assume ht(i) = ht(f,g) = 2. Get: 0 R I : h R (f,g) R I 0. Suffices to bound pd(r/(i : h)). Note: (I : h) has height 2 and multiplicity deg(f)deg(g).

19 An Approach to Stillman s Question New question: Question Is there a bound on the projective dimension of ideals in S = K[X 1,...,X n ] which are unmixed in terms of their height h and multiplicity e?

20 An Approach to Stillman s Question New question: Question Is there a bound on the projective dimension of ideals in S = K[X 1,...,X n ] which are unmixed in terms of their height h and multiplicity e? Yes, if h = 1, then pd(s/i) = 1.

21 An Approach to Stillman s Question New question: Question Is there a bound on the projective dimension of ideals in S = K[X 1,...,X n ] which are unmixed in terms of their height h and multiplicity e? Yes, if h = 1, then pd(s/i) = 1. Yes, if e = 1, then I is a linear prime and pd(s/i) = h.

22 An Approach to Stillman s Question New question: Question Is there a bound on the projective dimension of ideals in S = K[X 1,...,X n ] which are unmixed in terms of their height h and multiplicity e? Yes, if h = 1, then pd(s/i) = 1. Yes, if e = 1, then I is a linear prime and pd(s/i) = h. Yes, if h = e = 2...

23 Engheta s Classification of h = e = 2 unmixed ideals Proposition (Engheta) Let I S be a homogeneous unmixed ideal with ht(i) = e(s/i) = 2. Then pd(s/i) 3 and I is one of the following:

24 Engheta s Classification of h = e = 2 unmixed ideals Proposition (Engheta) Let I S be a homogeneous unmixed ideal with ht(i) = e(s/i) = 2. Then pd(s/i) 3 and I is one of the following: (x, q), q = irreducible quadric (prime)

25 Engheta s Classification of h = e = 2 unmixed ideals Proposition (Engheta) Let I S be a homogeneous unmixed ideal with ht(i) = e(s/i) = 2. Then pd(s/i) 3 and I is one of the following: (x, q), q = irreducible quadric (prime) (x,y) (x,z) = (x,yz)

26 Engheta s Classification of h = e = 2 unmixed ideals Proposition (Engheta) Let I S be a homogeneous unmixed ideal with ht(i) = e(s/i) = 2. Then pd(s/i) 3 and I is one of the following: (x, q), q = irreducible quadric (prime) (x,y) (x,z) = (x,yz) (w,x) (y,z) = (wy,wz,xy,xz) (intersection of linear primes)

27 Engheta s Classification of h = e = 2 unmixed ideals Proposition (Engheta) Let I S be a homogeneous unmixed ideal with ht(i) = e(s/i) = 2. Then pd(s/i) 3 and I is one of the following: (x, q), q = irreducible quadric (prime) (x,y) (x,z) = (x,yz) (w,x) (y,z) = (wy,wz,xy,xz) (intersection of linear primes) (x,y) 2 +(ax +by) such that ht(x,y,a,b) = 4

28 Engheta s Classification of h = e = 2 unmixed ideals Proposition (Engheta) Let I S be a homogeneous unmixed ideal with ht(i) = e(s/i) = 2. Then pd(s/i) 3 and I is one of the following: (x, q), q = irreducible quadric (prime) (x,y) (x,z) = (x,yz) (w,x) (y,z) = (wy,wz,xy,xz) (intersection of linear primes) (x,y) 2 +(ax +by) such that ht(x,y,a,b) = 4 (x,y 2 ) (primary)

29 Classification of Primes (minimal multiplicity) Theorem (Swinnerton-Dyer) Let p S be a homogeneous prime ideal with ht(p) = 2 and e(s/p) = 3. Then pd(s/i) = 2 and p is one of the following: (x,c), c = irreducible cubic (ux wv,uz yv,wz xy) (ux v 2,uz yv,vz xy) (wy x 2,wz xy,xz y 2 )

30 Manolache s Classification of CM Structures Theorem (Manolache) Let I S be homogeneous and primary to (x,y), Cohen-Macaulay on Spec(S) {m} with e(s/i) 4. Then I is one of the following: Multiplicity 2: (x,y 2 )

31 Manolache s Classification of CM Structures Theorem (Manolache) Let I S be homogeneous and primary to (x,y), Cohen-Macaulay on Spec(S) {m} with e(s/i) 4. Then I is one of the following: Multiplicity 2: (x,y 2 ) Multiplicity 3: (x 2,xy,y 2 ) (x,y 3 ) (x 2 +ay,y 2,xy) (ax 2 +c 2 (bx +cy),axy bc(bx +cy),ay 2 +b 2 (bx + cy),x(bx +cy),y(bx +cy),x 3,x 2 y,xy 2,y 3 )

32 Manolache s Classification of CM Structures (cont.) Theorem Multiplicity 4: (x 3,xy,y 2 ) (x 2 +y 2,xy) (x 2,y 2 ) (gx 2 fxy,hx 2 fy 2,hxy gy 2,x 3,x 2 y,xy 2,y 3 ), where ht(f,g,h) 2 (x,y 4 ) (x 3 +ay,y 2,xy) (x 2 +ay,y 2 ) (x 2 +xy +ay,y 2 ) (ax 3 +b(by+cx 2 ),axy c(by+cx 2 ),y 2,x 4,x 2 y,x(by+cx 2 )), where ht(a,b,c) 2

33 Similar Classification for Unmixed ideals? ht / e:

34 Similar Classification for Unmixed ideals? ht / e:

35 Main Result Theorem (Huneke-Mantero-M-Seceleanu) Let K be an algebraically closed field. For any integers h,e 2 with (h,e) (2,2) and for any integer p, there exists a primary (unmixed) ideal I = I h,e,p in a polynomial ring S with ht(i) = h e(s/i) = e I = (x1,x 2,...,x h ) pd(s/i) p.

36 Main Result Theorem (Huneke-Mantero-M-Seceleanu) Let K be an algebraically closed field. For any integers h,e 2 with (h,e) (2,2) and for any integer p, there exists a primary (unmixed) ideal I = I h,e,p in a polynomial ring S with ht(i) = h e(s/i) = e I = (x1,x 2,...,x h ) pd(s/i) p. Method: Direct construction of 4 unmixed ideals with fixed resolution. Use linkage to construct ideals in the main result.

37 Tools Theorem (Buchsbaum-Eisenbud) Let F be a complex free S-modules of finite rank d F : F 1 d 0 2 d F1 3 d p F2 F p 0. Set r j = p i=j ( 1)p i rankf i. Then F is a resolution of M := Coker(d 1 ) if and only if ht(i rj (d j )) j for all j = 1,...,p.

38 Tools As a corollary: Proposition Using the notation from the previous theorem, suppose F is a minimal free resolution of M. Then M satisfies Serre s condition (S k ) if and only if ht(i rj (d j )) min{dim(s),j+k} for all j = codim(m)+1,...,p.

39 Proof Sketch Proposition Suppose f,g,h S d for some d 1 such that ht(x,y,f,g,h) 4. Let L = (x,y) 3 +(y 2 f +xyg +x 2 h). Then L is (x,y)-primary, e(s/l) = 5 and S/L has the following free resolution: R d 1 R 5 d 2 R 5 d 3 R 0, where d 3 = h g f y x.

40 Proof Sketch Step 1: Note pd(ker(d 3)) = pd(ker(h,g,f, y,x)) = pd(s/(x,y,f,g,h)) 2 = BIG.

41 Proof Sketch Step 1: Note pd(ker(d 3)) = pd(ker(h,g,f, y,x)) = pd(s/(x,y,f,g,h)) 2 = BIG. Step 2: Set I = (x i,y j ) : L, where i,j 3. Then ht(i) = 2 e(s/i) = ij 5

42 Proof Sketch Step 1: Note pd(ker(d 3)) = pd(ker(h,g,f, y,x)) = pd(s/(x,y,f,g,h)) 2 = BIG. Step 2: Set I = (x i,y j ) : L, where i,j 3. Then ht(i) = 2 e(s/i) = ij 5 Step 3: 0 Im(d 2) Ker(d 3) Ext 2 S(S/L,S) 0

43 Proof Sketch Step 1: Note pd(ker(d 3)) = pd(ker(h,g,f, y,x)) = pd(s/(x,y,f,g,h)) 2 = BIG. Step 2: Set I = (x i,y j ) : L, where i,j 3. Then ht(i) = 2 e(s/i) = ij 5 Step 3: 0 Im(d 2) Ker(d 3) Ext 2 S(S/L,S) 0 pd : 1

44 Proof Sketch Step 1: Note pd(ker(d 3)) = pd(ker(h,g,f, y,x)) = pd(s/(x,y,f,g,h)) 2 = BIG. Step 2: Set I = (x i,y j ) : L, where i,j 3. Then ht(i) = 2 e(s/i) = ij 5 Step 3: 0 Im(d 2) Ker(d 3) Ext 2 S(S/L,S) 0 pd : 1 BIG

45 Proof Sketch Step 1: Note pd(ker(d 3)) = pd(ker(h,g,f, y,x)) = pd(s/(x,y,f,g,h)) 2 = BIG. Step 2: Set I = (x i,y j ) : L, where i,j 3. Then ht(i) = 2 e(s/i) = ij 5 Step 3: 0 Im(d 2) Ker(d 3) Ext 2 S(S/L,S) 0 pd : 1 BIG BIG

46 Proof Sketch Step 4: Ext 2 S(S/L,S) = I/(x i,y j )

47 Proof Sketch Step 4: Ext 2 S(S/L,S) = I/(x i,y j ) Step 5: 0 I/(x i,y j ) S/(x i,y j ) S/I 0

48 Proof Sketch Step 4: Ext 2 S(S/L,S) = I/(x i,y j ) Step 5: 0 I/(x i,y j ) S/(x i,y j ) S/I 0 pd : 2

49 Proof Sketch Step 4: Ext 2 S(S/L,S) = I/(x i,y j ) Step 5: 0 I/(x i,y j ) S/(x i,y j ) S/I 0 pd : BIG 2

50 Proof Sketch Step 4: Ext 2 S(S/L,S) = I/(x i,y j ) Step 5: 0 I/(x i,y j ) S/(x i,y j ) S/I 0 pd : BIG 2 BIG +1

51 Proof Sketch Step 6: Proposition If I,I are unmixed and in the same even linkage class (e.g. I = (x) : L,I = (x ) : L), then pd(s/i) = pd(s/i ).

52 An Example To construct an unmixed ideal I, with I = (x,y) e(s/i) = 17 pd(s/i) = 11

53 An Example To construct an unmixed ideal I, with I = (x,y) Take: e(s/i) = 17 pd(s/i) = 11 I = (x 7,x 4 ) : [ ( )] (x 5,x 3 ) : (x 3,y 3 ) : L,

54 An Example To construct an unmixed ideal I, with I = (x,y) Take: where e(s/i) = 17 pd(s/i) = 11 I = (x 7,x 4 ) : [ ( )] (x 5,x 3 ) : (x 3,y 3 ) : L, f = a 8, g = b 8, h = a 7 z 1 +a 6 bz 2 + +b 7 z 8.

55 An Example To construct an unmixed ideal I, with I = (x,y) Take: where e(s/i) = 17 pd(s/i) = 11 I = (x 7,x 4 ) : [ ( )] (x 5,x 3 ) : (x 3,y 3 ) : L, f = a 8, g = b 8, h = a 7 z 1 +a 6 bz 2 + +b 7 z 8. Note: I has several hundred generators.

56 Linkage Structure x 4,y n+3 I 2,7+4n,p L x 3,y 3 x 4,y 3 I 2,4,p L 2,8,p 1 x 5,y 3 x 4+n,y 4 I 2,8+4n,p L 2,11,p 1 x 6,y 3 x 5+n,y 4 I 2,9+4n,p L 2,14,p 1 x 6+n,y 4 I 2,10+4n,p

57 Final Questions Question Is there a finite classification for unmixed ideals of a given height h, given multiplicity e and generated in at most some degree d?

58 Final Questions Question Is there a finite classification for unmixed ideals of a given height h, given multiplicity e and generated in at most some degree d? Yes for (h,e,d) = (2,3,3) or (3,2,2).

59 Final Questions Question Is there a finite classification for unmixed ideals of a given height h, given multiplicity e and generated in at most some degree d? Yes for (h,e,d) = (2,3,3) or (3,2,2). Recall unmixed Serre s S 1 condition. So...

60 Final Questions Question Is there a finite classification for unmixed ideals of a given height h, given multiplicity e and generated in at most some degree d? Yes for (h,e,d) = (2,3,3) or (3,2,2). Recall unmixed Serre s S 1 condition. So... Question Is there a finite classification for S 2 ideals of a given height h and given multiplicity e?

61 Final Questions Question Is there a finite classification for unmixed ideals of a given height h, given multiplicity e and generated in at most some degree d? Yes for (h,e,d) = (2,3,3) or (3,2,2). Recall unmixed Serre s S 1 condition. So... Question Is there a finite classification for S 2 ideals of a given height h and given multiplicity e? Open even for (h,e) = (2,3).

62 Reference: Thank you! C Huneke, P. Mantero, J. McCullough and A. Seceleanu. Multiple structures with arbitrarily large projective dimension supported on linear subspaces. submitted. arxiv:1301:4147