Ideals of three dimensional Artin-Schelter regular algebras. Koen De Naeghel Thesis Supervisor: Michel Van den Bergh

Ideals of three dimensional Artin-Schelter regular algebras Koen De Naeghel Thesis Supervisor: Michel Van den Bergh February 17, 2006

Polynomial ring Put k = C. Commutative polynomial ring S = k[x, y, z] = k x, y, z /(f 1, f 2, f 3 ) f 1 : xy yx = 0 f 2 : yz zy = 0 f 3 : zx xz = 0 1

Polynomial ring Put k = C. Commutative polynomial ring S = k[x, y, z] = k x, y, z /(f 1, f 2, f 3 ) f 1 : xy yx = 0 f 2 : yz zy = 0 f 3 : zx xz = 0 Noncommutative polynomial rings How to define them? Pick certain properties of S. 2

Polynomial ring Put k = C. Commutative polynomial ring S = k[x, y, z] = k x, y, z /(f 1, f 2, f 3 ) f 1 : xy yx = 0 f 2 : yz zy = 0 f 3 : zx xz = 0 Noncommutative polynomial rings Artin-Schelter (1986) defined class of algebras. A is quadratic: A = k x, y, z /(g 1, g 2, g 3 ) Generic: g 1 : ayz + bzy + cx 2 = 0 g 2 : azx + bxz + cy 2 = 0 g 3 : axy + byx + cz 2 = 0 A is cubic: A = k x, y /(g 1, g 2 ) Generic: { g1 : ay 2 x + byxy + axy 2 + cx 3 = 0 g 2 : ax 2 y + bxyx + ayx 2 + cy 3 = 0 3

Projective plane Consider the projective plane P 2. Homogeneous coordinate ring S = k[x, y, z] 5

Projective plane Consider the projective plane P 2. Homogeneous coordinate ring S = k[x, y, z] What has S = k[x, y, z] to do with P 2? For any homogeneous polynomial f k[x, y, z] {p P 2 f(p) = 0} is a curve on P 2. Example: y 2 xz 6

Projective plane Consider the projective plane P 2. Homogeneous coordinate ring S = k[x, y, z] S d = {homogeneous polynomials degree d} S = k S 1 S 2... graded k-algebra 7

Projective plane Consider the projective plane P 2. Homogeneous coordinate ring S = k[x, y, z] S d = {homogeneous polynomials degree d} S = k S 1 S 2... graded k-algebra P 2 is completely determined by S. Theorem of Serre (1955) Qcoh P 2 GrMod S/Tors S What is GrMod S? An object of GrMod S is M = M 1 M 0 M 1... where M d is k-vector space action of S on M such that M i S j M i+j 9

Projective plane Consider the projective plane P 2. Homogeneous coordinate ring S = k[x, y, z] S d = {homogeneous polynomials degree d} S = k S 1 S 2... graded k-algebra P 2 is completely determined by S. Theorem of Serre (1955) Qcoh P 2 GrMod S/Tors S What is Tors S? Generated by modules M GrMod S for which m M : ms d = 0 for some d Typical: M = M 1 M 0 M 1 0 0... 10

Projective plane Consider the projective plane P 2. Homogeneous coordinate ring S = k[x, y, z] S d = {homogeneous polynomials degree d} S = k S 1 S 2... graded k-algebra P 2 is completely determined by S. Theorem of Serre (1955) Qcoh P 2 GrMod S/Tors S Noncommutative projective plane Model of noncommutative projective plane P 2 q Artin-Zhang (1994) Replace S by noncommutative k-algebra A Define Qcoh P 2 q := GrMod A/Tors A 11

Projective plane Consider the projective plane P 2. Homogeneous coordinate ring S = k[x, y, z] S d = {homogeneous polynomials degree d} S = k S 1 S 2... graded k-algebra P 2 is completely determined by S. Theorem of Serre (1955) Qcoh P 2 GrMod S/Tors S Noncommutative projective plane Model of noncommutative projective plane P 2 q Artin-Zhang (1994) Replace S by noncommutative k-algebra A Define Qcoh P 2 q := GrMod A/Tors A Arguments for taking A a quadratic Artin- Schelter algebra. 12

The points on P 2 Point p P 2 p 1 13

The points on P 2 Point p P 2 two linear forms l 1, l 2 S 1 p 1 l 1 l 2 14

The points on P 2 Point p P 2 two linear forms l 1, l 2 S 1 p 1 l 1 represented by ( l2 0 S( 2) where l 1 ) S( 1) 2 l 2 ( l1 l 2 ) S P 0 P = P 0 P 1 P 2... GrMod S P = P 0 S h P (t) := ddim k P d t d = 1 + t + t 2 +... P is called a point module. 15

The points on P 2 Correspondence is reversible point p on P 2 S-module P = P 0 S, h P (t) = 1 1 t 16

The points on P 2 Correspondence is reversible point p on P 2 S-module P = P 0 S, h P (t) = 1 1 t The points on P 2 q : by definition point p on P 2 q := right A-module P = P 0 A, h P (t) = 1 1 t 17

The points on P 2 Correspondence is reversible point p on P 2 S-module P = P 0 S, h P (t) = 1 1 t The points on P 2 q : by definition point p on P 2 q := right A-module P = P 0 A, h P (t) = 1 1 t Artin, Tate and Van den Bergh (1990): There is divisor E P 2 of deg 3 such that (closed) point p on E point on P 2 q A, P 2 q determined by the points on P 2 q Generic: E is smooth elliptic curve (a 3 + b 3 + c 3 )xyz = abc(x 3 + y 3 + z 3 ) 18

The points on P 2 versus graded S-ideals Point p P 2 two linear forms l 1, l 2 S 1 p 1 l 1 0 S( 2) ( l2 l 1 ) S( 1) 2 l 2 ( l1 l 2 ) S P 0 19

The points on P 2 versus graded S-ideals Point p P 2 two linear forms l 1, l 2 S 1 0 S( 2) ( l2 l 1 ) S( 1) 2 p 1 l 1 l 2 ( l1 l 2 ) S P 0 I I = l 1 S + l 2 S ideal polynomials vanishing at p. 20

The points on P 2 versus graded S-ideals Point p P 2 two linear forms l 1, l 2 S 1 0 S( 2) ( l2 l 1 ) S( 1) 2 p 1 l 1 l 2 ( l1 l 2 ) S P 0 I I = l 1 S + l 2 S ideal polynomials vanishing at p. In general: any graded ideal I, I S(d) is (up to Tors S) the ideal of polynomials vanishing at some points. If I S graded S-ideal: - put J = ωπi - Either J = S(d) or pd J = 1. - If pd J = 1 then J = S(d) i.e. Ext 1 S (P, J) 0 for some point module P. 22

S-ideals of projective dimension one Let I S graded ideal, pd I = 1 0 i S( i) b i M i S( i) a i I 0 23

S-ideals of projective dimension one Let I S graded ideal, pd I = 1 0 i S( i) b i M i S( i) a i I 0 Hilbert-Burch (1890) I is generated by the maximal minors of M (whose zero s determine configuration of points) Known: Given a i, b i there is such an ideal I (up to shift) if and only if deg ( S( i) b i i S( i) a i) =...... s > 0 25

A Castelnuovo polynomial is of the form s(t) = 1+2t+3t 2 + +ut u 1 +s u t u + +s v t v u s u... s v 0 for some integers u, v 0. 27

A Castelnuovo polynomial is of the form s(t) = 1+2t+3t 2 + +ut u 1 +s u t u + +s v t v u s u... s v 0 for some integers u, v 0. Visualized in form of a stair Example: 1 2 3 4 5 5 3 2 1 1 1 1 28

The points on P 2 q versus right A-ideals point on P 2 q E p 1 29

The points on P 2 q versus right A-ideals point on P 2 q two linear forms l 1, l 2 A 1 intersecting at E E l 1 p 1 l 2 30

The points on P 2 q versus right A-ideals point on P 2 q two linear forms l 1, l 2 A 1 intersecting at E E l 1 0 A( 2) ( ) w1 w 2 A( 1) 2 p 1 l 2 ( l1 l 2 ) A P 0 I 31

The points on P 2 q versus right A-ideals point on P 2 q two linear forms l 1, l 2 A 1 intersecting at E v 1 E 1 l 1 0 A( 2) ( ) w1 w 2 v 2 A( 1) 2 p 1 l 2 ( l1 l 2 ) A P 0 I If v 1, v 2 A 1 not intersecting at E 0 A( 2) ( ) v1 v 2 A( 1) 2 I 0 Then Ext 1 A (P, I ) = 0 for all point modules P. Such ideals I are called reflexive. 32

Right A-ideals of projective dimension one Let I A graded right ideal, pd I = 1 0 i A( i) b i M i A( i) a i I 0 33

Right A-ideals of projective dimension one Let I A graded right ideal, pd I = 1 0 i A( i) b i M i A( i) a i I 0 Given a i, b i there is such I (up to shift) if and only if (Theorem 6) deg ( S( i) b i i S( i) a i) = if and only if (Theorem 4) h I (t) =...... 1 (1 t) 3 s(t) 1 t for some Castelnuovo polynomial s(t). s > 0 35

Hilbert scheme of points on P 2 Classify all possible configurations of n points on P 2. Can be done by parameterspace. Formally: parameter space for subschemes of P 2 of dimension zero and degree n. moduli problem Hilb n (P 2 ) : Noeth /k Sets R Hilb n (P 2 )(R) Hilb n (P 2 )(R) ={N P 2 R N is R-flat and x Spec R N x dimension 0, degree n} 38

Hilbert scheme of points on P 2 q Initial problem: P 2 q has few zero-dimensional noncommutative subschemes 40

Hilbert scheme of points on P 2 q Initial problem: P 2 q has few zero-dimensional noncommutative subschemes Solution: consider ideal sheaves instead 41

Hilbert scheme of points on P 2 q Initial problem: P 2 q has few zero-dimensional noncommutative subschemes Solution: consider ideal sheaves instead moduli problem Hilb n (P 2 q ) : Noeth /k Sets R Hilb n (P 2 q)(r) Hilb n (P 2 q)(r) ={I coh P 2 q,r I is R-flat and x Spec R I x coh P 2 q,k(x) torsion free, pd1, normalized, rk 1}/Pic R In case A = S: agrees with Hilb n (P 2 ). Nevins and Stafford (2002) The functor Hilb n (P 2 q) is representable by projective variety Hilb n (P 2 q) = Hilb n (P 2 q)(k) - smooth - dimension 2n - connectedness proved for almost all A (using deformation theory and Hilb n (P 2 )) 43

- A graded (right) ideal I is reflexive if Ext 1 (P, I) = 0 for all point modules P - If I is reflexive then pd I 1. 44

- A graded (right) ideal I is reflexive if Ext 1 (P, I) = 0 for all point modules P - If I is reflexive then pd I 1. reflexive graded S-ideals: S(d) 45

- A graded (right) ideal I is reflexive if Ext 1 (P, I) = 0 for all point modules P - If I is reflexive then pd I 1. reflexive graded S-ideals: S(d) reflexive graded right A-ideals In case A is generic: Theorems 1,2 R(A) = {reflexive graded right A-ideals}/iso,shift n D n where D n is smooth affine variety of dim 2n points D n are given by stable representations k n X > Y > Z > k n with rank cx az by bz cy ax 2n+1 ay bx cz 46

Picture in case A is generic: Hilb n (P 2 q) D n Hilb n (P 2 q ) parameterizes {graded right A-ideals I, pd I = 1 1 h I (t) is (1 t) 3 s(t) up to shift}/iso, shift 1 t D n parameterizes {reflexive graded right A-ideals I 1 h I (t) is (1 t) 3 s(t) up to shift}/iso, shift 1 t What about the boundary? 47

For any graded right ideal J with pd J = 1 0 J J 1 P 1 (d 1 ) 0 0 J 1 J 2 P 2 (d 2 ) 0. 0 J r 1 J r P r (d r ) 0 where J r is reflexive. Note 0 r n. nhilb n (P 2 q) n D n J r J r 1... J 1 J 48

For any graded right ideal J with pd J = 1 0 J J 1 P 1 (d 1 ) 0 0 J 1 J 2 P 2 (d 2 ) 0. 0 J r 1 J r P r (d r ) 0 where J r is reflexive. Note 0 r n. nhilb n (P 2 q) n D n J r J r 1... J 1 J Let Hilb d n (P2 q ) be the J Hilb n(p 2 q ) with r d. Theorem 7 - Hilb d n (P2 q ) projective variety of dimension 2n d - boundary Hilb n (P 2 q ) \ D n has dimension 2n 1 49

For any graded right ideal J with pd J = 1 0 J J 1 P 1 (d 1 ) 0 0 J 1 J 2 P 2 (d 2 ) 0. 0 J r 1 J r P r (d r ) 0 where J r is reflexive. Note 0 r n. nhilb n (P 2 q ) n D n J r J r 1... J 1 J Let Hilb d n (P 2 q) be the J Hilb n (P 2 q) with r d. Theorem 7 - Hilb d n (P 2 q) projective variety of dimension 2n d - boundary Hilb n (P 2 q ) \ D n has dimension 2n 1 Thus the actual Hilbert scheme of points on P 2 q is Hilbn n (P2 q ), has dimension n 50

Stratification of Hilb n (P 2 q) Consider all points of Hilb n (P 2 ) parameterizing ideals of A with same Hilbert series. Hilb n (P 2 q)... 51

Stratification of Hilb n (P 2 q) Consider all points of Hilb n (P 2 ) parameterizing ideals of A with same Hilbert series. Hilb n (P 2 q) Hilb h (P 2 q)... For an appearing Hilbert series h(t) = 1 (1 t) 3 s(t) 1 t s(t) is Castelnuovo polynomial, s(1) = n put Hilb h (P 2 q) = {I Hilb n (P 2 q) h I (t) = h(t)} 52

Hilb n (P 2 q) Hilb h (P 2 q)... Chapter 3 Hilb h (P 2 q) Hilb n (P 2 q) is locally closed subvariety - smooth - connected In case A = S: Proved by Gotzmann (1988) 53

Hilb n (P 2 q) Hilb h (P 2 q)... Chapter 3 Hilb h (P 2 q ) Hilb n(p 2 q ) is locally closed subvariety - smooth - connected In case A = S: Proved by Gotzmann (1988) Formula for dimhilb h (P 2 q): constant term of (t 1 t 2 )s(t 1 )s(t)+n+1 There is an unique stratum with maximal dimension 2n 54

Hilb n (P 2 q) Hilb h (P 2 q)... Chapter 3 Hilb h (P 2 q) Hilb n (P 2 q) is locally closed subvariety - smooth - connected In case A = S: Proved by Gotzmann (1988) Formula for dimhilb h (P 2 q): constant term of (t 1 t 2 )s(t 1 )s(t)+n+1 There is an unique stratum with maximal dimension 2n Theorem 5 Hilb n (P 2 q) is connected. 55

Hilb n (P 2 ) and Hilb n (P 2 q) analogous strata Hilb n (P 2 ) Hilb n (P 2 q ) H ψ H ϕ H ψ H ϕ Incidence problem: for which ϕ, ψ do we have H ϕ H ψ? 56

Hilb n (P 2 ) and Hilb n (P 2 q) analogous strata Hilb n (P 2 ) Hilb n (P 2 q ) H ψ H ϕ H ψ H ϕ Incidence problem: for which ϕ, ψ do we have H ϕ H ψ? General incidence problem for Hilb n (P 2 ): unknown 57

Hilb n (P 2 ) and Hilb n (P 2 q) analogous strata Hilb n (P 2 ) Hilb n (P 2 q ) H ψ H ϕ H ψ H ϕ Incidence problem: for which ϕ, ψ do we have H ϕ H ψ? General incidence problem for Hilb n (P 2 ): unknown - Guerimand (2002) solved case ϕ and ψ are as close as possible under a technical condition 58

ϕ(t) = as close as possible means: writing 1 (1 t) 3 s ϕ 1 t, ψ(t) = 1 (1 t) 3 s ψ 1 t s ψ is obtained from s ϕ by minimal movement of one square to the left. 59

ϕ(t) = as close as possible means: writing 1 (1 t) 3 s ϕ 1 t, ψ(t) = 1 (1 t) 3 s ψ 1 t s ψ is obtained from s ϕ by minimal movement of one square to the left. Examples: n = 17: ϕ and ψ as close as possible s ϕ s ψ 60

If ϕ, ψ Hilbert series as close as possible : We have H ϕ H ψ if and only if I 2 1 0 C 1 C > D D 0 A 0 A < B B 1 0 C 1 C > D D 0 II III 0 1 0 2 IV 3 2 3 1 V 1 1 0 1 VI 2 3 2 4 2 Before: s ϕ, after: s ψ 63

Incidence problem: for which ϕ, ψ do we have H ϕ H ψ? Same solution for Hilb n (P 2 q ) as for Hilb n(p 2 )? 64

Incidence problem: for which ϕ, ψ do we have H ϕ H ψ? Same solution for Hilb n (P 2 q) as for Hilb n (P 2 )? No for generic A s ϕ s ψ Generic I H ϕ 0 A( 4) A( 7) A( 2) A( 3) A( 6) I 0 V = {f : I F h F = ϕ ψ} 65

Incidence problem: for which ϕ, ψ do we have H ϕ H ψ? Same solution for Hilb n (P 2 q) as for Hilb n (P 2 )? No for generic A s ϕ s ψ Generic I H ϕ 0 A( 4) A( 7) A( 2) A( 3) A( 6) I 0 V = {f : I F h F = ϕ ψ} - commutative case: V = P 2 V N : f dim k Ext 1 A (ker f,ker f) non-constant hence H ϕ H ψ - smooth elliptic case: V = three points on E V N : f dim k Ext 1 A (ker f,ker f) constant hence H ϕ H ψ 66

If ϕ, ψ Hilbert series as close as possible : We have H ϕ H ψ if and only if I 2 1 0 C 1 C > D D 0 A 0 A < B B 1 0 C 1 C > D D 0 II III 0 1 0 2 IV 3 2 3 1 V 1 1 0 1 VI 2 3 2 4 2 A = S commutative I II III IV V VI A generic I II III VI 67