QUADRATIC ALGEBRAS ASSOCIATED WITH THE UNIONOF A QUADRIC AND A LINE IN P3

M. VANCLIFF

Department of Mathematics,University of Washington,Seattle, WA, 98195, U.S.A.

E-mail : vancliff@math.washington.edu

August 29, 2003

Submitted J. Algebra

Abstract. We define a family of graded quadratic algebras Aσ (on 4 generators)depending on a fixed nonsingular quadric Q in P3, a fixed line L in P3 and an au-tomorphism σ ∈ Aut(Q ∪ L). This family contains Oq(M∈(C)), the coordinatering of quantum 2 × 2 matrices. Many of the algebraic properties of Aσ are shownto be determined by the geometric properties of Q ∪ L, σ. For instance, whenAσ = Oq(M∈(C)), then the quantum determinant is the unique (up to a scalar mul-tiple) homogeneous element of degree 2 in Oq(M∈(C)) that vanishes on the graphin P3 × P3 of σ|Q but not on the graph of σ|L. Following [ATV1, 2], we study pointand line modules over the algebras Aσ, and find that their algebraic properties areconsequences of the geometric data. In particular, the point modules are in one-to-one correspondence with the points of Q ∪ L, and the line modules are in bijectionwith the lines in P3 that either lie on Q or meet L. In the case of Oq(M∈(C)), whenq is not a root of unity, the quantum determinant annihilates all the line modulesM(`) corresponding to lines ` ⊂ Q; the determinant generates the whole annihilatorfor such ` ⊂ Q if and only if ` ∩ L = ∅.

1991 Mathematics Subject Classification. 16P40, 16W50, 14A22.Key words and phrases. Quadratic algebras, point module, line module, quantum 2 × 2 matrices,

quantum symplectic space.

Introduction

The purpose of this paper is to broaden the class of examples hitherto discussed inrelation to the non-commutative geometry developed in [ATV1, 2]. The main focus in[ATV1, 2] was on regular algebras of dimension 3, for which many algebraic proper-ties were shown to be a consequence of certain geometric data; namely, a scheme E(typically a cubic divisor in P2) and an automorphism of E. This geometric themeis continued in [SSt] and [LS] in their study of the Sklyanin algebra, a 4-dimensionalregular algebra, which is related to an elliptic curve in P3 and an automorphism in asimilar way. We assume the reader is familiar (to some extent) with [ATV1, 2] and[LS].

This paper describes a family of regular algebras of dimension 4 defined in terms ofa variety Q ∪ L in P3, where Q ⊂ P3 is a nonsingular quadric and L ⊂ P3 is a line (ingeneral position), together with an automorphism σ of Q∪L. [SSt] and [LS] show thatmany of the techniques developed in [ATV1, 2] for the 3-dimensional regular algebrascarry over in dimension 4 to the Sklyanin algebra – our goal is to show these techniquesalso carry over to the algebras associated to Q ∪ L, σ.

Our class of algebras splits into two types: those for which σ preserves the rulingson Q and those for which σ interchanges them. Examples of the first kind (presentedin §1) are Oq(M∈(C)), the coordinate ring of quantum 2× 2 matrices, and Oq(spC4),the coordinate ring of quantum symplectic 4-dimensional space. Moreover, families ofalgebras on n2 generators which contain Oq(M\(C)) are defined in [AST], [Su, §4b]and [T] respectively, and when n = 2, most of their algebras are included in our family(with σ|L = identity).

Section 1 begins with the geometric definition of the algebras A. We then describe Ain terms of generators and relations. In particular, we show they are skew-polynomialrings which are noetherian domains of global homological dimension 4 having the sameHilbert series as the polynomial ring in 4 variables. The P3 containing Q∪L is naturallyidentified with P(A∗1), so the elements of A1 are linear forms on Q ∪ L, and elementsof A1 ⊗ A1 are bilinear forms on P3 × P3. We show that any element ω ∈ A1 that isan eigenvector for σ|Q (i.e., ω σ|Q is a scalar multiple of ω) and vanishes on L is anormal element of A – there are at least two (independent) such elements. Moreover,there is a unique (up to a scalar multiple) element Ω in A2 that vanishes on the graphin P3 × P3 of σ|Q but not on the graph of σ|L – in addition to this geometric property,Ω is a normal element of A.

In §2 the point modules are shown to be in bijection with the points of Q ∪ L, andthe line modules in natural bijection with the lines in P3 that either lie on Q or meetthe line L. If M is either a point or a line module over A then M ∼= A/AW where Wis the linear subspace of A1 vanishing on the respective point or line.

The quotient algebra A/AΩ is shown in §3 to be a domain. Here the techniquesemployed from [ATV1, 2] analyse A/AΩ in terms of the geometric data Q, σ|Q,Lwhere L is the invertible sheaf of linear forms on Q. The key idea is to construct thealgebra B =

⊕n≥0 H

0(Q,L ⊗ Lσ ⊗ · · · ⊗ Lσ\−∞), where Lσ is the pullback of L byσ|Q, and to show that B ∼= A/AΩ. The algebra B (and hence A/AΩ) is analogous tothe homogeneous coordinate ring of Q, which is isomorphic to

⊕n≥0 H

0(Q,L\). In thelanguage of [AV], B is a twisted homogeneous coordinate ring. In this way, we canview Ω as playing a role similar to that of the defining equation of Q.

We investigate the annihilators of point and line modules in §4, proving that when σhas infinite orbit at “most” points of Q then AΩ is the annihilator of the line modules

1

M(`) where ` is a line on Q such that ` ∩ L = ∅. We end the section by restrictingto the algebras for which σ preserves the rulings on Q. In particular, we show thatwhen q is not a root of unity, then every line module over Oq(M∈(C)) has nonzeroannihilator. Moreover, every homogeneous prime ideal of Oq(spC4), when q is not aroot of unity, is shown to be the annihilator of some linear module.

The kernel of a surjective map from a line module to a point module is observed in§2 to be a shifted line module. We devote the last section (§5) to identifying the linecorresponding to this kernel.

Acknowledgements . The author is grateful to S. P. Smith for many helpful and inter-esting discussions.

1. Definition of the Algebras

The algebras discussed in this paper are graded algebras on four generators (of degreeone) defined by six homogeneous quadratic relations. We define the algebras in termsof a certain subvariety of P3, namely the union of a fixed quadric Q and a fixed lineL, and an automorphism σ of that variety. We prove that the algebras so obtained areskew-polynomial rings, but it is the geometric description of the algebras that will beexploited throughout the paper.

Unless otherwise stated, we will work over an algebraically closed field k, wherechar(k) 6= 2. We fix a nonsingular quadric Q ⊂ P3 and a line L ⊂ P3, L 6⊂ Q, thatmeets Q in two distinct points. Certain restrictions (see (1)-(4) below) are placed onσ ∈ Aut(Q∪L). These may seem artificial at first glance, but the remark following thedefinition of σ explains why these restrictions are natural. We define σ ∈ Aut(Q ∪ L)subject to the following conditions:

(1) σ(Q) = Q and σ(L) = L,(2) σ|Q∩L is the identity,(3) σ|Q and σ|L are restrictions of linear automorphisms of P3, and(4) σ is not the restriction of a linear automorphism of P3.

Remark . Conditions (1) and (3) are automatically satisfied by an automorphism ofQ∪L. However, if condition (4) is omitted, and σ does extend to P3, then the algebraobtained (see below) is a “twist” ([ATV2, §8]) of the commutative polynomial ring byan automorphism. Since the categories of graded modules over an algebra and its twistare equivalent ([ATV2, 8.5]), there is nothing new to say about such algebras.

If σ violates condition (2), then one obtains algebras which are not domains andtherefore are not noetherian (Auslander-)regular algebras ([ATV2], [L]). As our mainemphasis is on the techniques developed in [ATV1, 2] and [LS], this paper will notdiscuss such algebras, although they are probably straightforward to understand.

There exist choices of σ satisfying the above conditions that yield the coordinatering of quantum 2× 2 matrices and other quantum group examples (see 1.5-1.7).

The graph of σ will be denoted by Γσ; that is,

Γσ := (p, σ(p)) ∈ P3 × P3 : p ∈ Q ∪ L.Similarly, Γσ(Q) and Γσ(L) will denote the graphs of σ|Q and σ|L respectively.

2

Let V be the 4-dimensional k-vector space of linear forms on the copy of P3 thatcontains Q∪L, and let T (V ) be the tensor algebra on V . Thus, P3 = P(V ∗) and V ⊗Vacts as bilinear forms on P(V ∗)× P(V ∗). Let Iσ be the subspace of V ⊗ V defined by

Iσ := f ∈ V ⊗ V : f(Γσ) = 0.Definition . With the above notation, we define a family of algebras Aσ by

Aσ :=T (V )

〈Iσ〉.

We recall, for the benefit of the reader, certain properties of the quadric Q whichwill be used frequently.

Facts .

(a) For each x ∈ Q there are exactly two lines lying on Q which pass through x.(b) Let `1, `2 be two distinct intersecting lines on Q. Then any other line on Q meets

only one of `1, `2. This defines two families (rulings) of lines on Q.(c) Every plane meets Q in either a nondegenerate conic or in two distinct lines.(d) If a line ` 6⊂ Q meets Q in two distinct points, then there exist precisely two

planes containing ` and meeting Q in degenerate conics.

Notation . Since L meets Q at two distinct points, these facts show there are ex-actly four “special”, distinct (since L 6⊂ Q) lines lying on Q that meet L. Lete1, e2, e3, e4 ∈ Q denote the four distinct points at which these lines pairwise intersect,and set e2, e3 = Q ∩ L. Notice that e1, . . . , e4 span P3.

We define four linear forms x1, . . . , x4 on P3 by:

V(xi) = linear span ej : 1 ≤ j ≤ 4, j 6= i for 1 ≤ i ≤ 4.

Thus, x1, . . . , x4 is a basis for V . Clearly, L = V(x1, x4). We aim to show that wecan assume Q = V(x1x4 + x2x3).

Lemma 1.1. The nonsingular quadrics which contain the four special lines on Q thatmeet L are precisely the V(λ1x1x4 + λ2x2x3) where λ1, λ2 ∈ k \ 0. In particular, wecan rescale xi such that Q = V(x1x4 + x2x3).

Proof. Write U for the union of the four special lines on Q that meet L, and denotethe line passing through ei and ej by `ij.

We first consider the singular quadrics which contain U . Since `12∩`34 = ∅ = `13∩`24,U cannot lie on a rank 1 quadric (a plane). A rank 3 quadric contains a point throughwhich every line on the quadric passes, so U cannot be contained in a rank 3 quadric.A rank 2 quadric, being the union of two planes, contains U if and only if two of thefour lines lie in one of the planes and the other two in the other. Therefore, the linesdetermine the planes. It follows that V(x1x4) and V(x2x3) are the only rank 2 (andthe only singular) quadrics containing U . Set q1 = x1x4, q2 = x2x3.

Suppose there exists a nonsingular quadratic form q3 /∈ kq1⊕kq2 such that U ⊂ V(q3).Then we obtain a net N of quadrics which contain U , where

N =

qλ =

3∑

i=1

λiqi : λ = (λ1, λ2, λ3) ∈ P2

' P2.

Let Mi ∈ M4(k), resp. Mλ ∈ M4(k), denote the matrix of the bilinear form corre-sponding to qi, resp. qλ. Then Mλ =

∑λiMi. Now the vanishing of the determinant

defines a degree 4 curve λ ∈ P2 : detMλ = 0 in P2 consisting of the singular3

quadrics in N . This contradicts the fact that there exist only two singular quadricswhich contain U . The result follows.

For the rest of the paper we assume Q = V(x1x4 + x2x3) and L = V(x1, x4).

Remark 1.2. By condition (2) defining σ, the points e2 and e3 are each fixed by σ.Therefore, since σ|Q is the restriction of a linear automorphism of P3, it follows thatthe two lines on Q through e2, resp. e3, are either fixed or interchanged by σ. Hence,the set e1, e4 is stable under σ, so that only two types of automorphism σ arise.

Lemma 1.3. The automorphism σ either fixes each of the two rulings on Q or inter-changes them.

(a) If σ fixes the two rulings on Q then Aσ = k[x1, x2, x3, x4] with the six definingrelations:

x2x1 = αx1x2, x3x1 = λx1x3, x4x1 = αλx1x4,

x4x3 = αx3x4, x4x2 = λx2x4, x3x2 − βx2x3 = (αβ − λ)x1x4,

for some α, β, λ ∈ k \ 0 determined by σ, where λ 6= αβ.(b) If σ interchanges the two rulings on Q then Aσ = k[x1, x2, x3, x4] with the six

defining relations:

x3x4 = αx1x3, x2x4 = λx1x2, x24 = αλx2

1,

x4x2 = αx2x1, x4x3 = λx3x1, βx3x2 − x2x3 = (λ− αβ)x21,

for some α, β, λ ∈ k \ 0 determined by σ, where λ 6= αβ.

Proof. The first statement follows from remark 1.2. In (a) and (b), we seek a basisfor Iσ.

(a) Conditions (1), (2) and (3) defining σ show that, with respect to the basis ei,we have

σ|Q =

αλ 0 0 00 λ 0 00 0 α 00 0 0 1

and σ|L =

(β 00 1

)

for some α, β, λ ∈ k \ 0. Also, condition (4) of σ holds if and only if λ 6= αβ. Hence,σ is given by:

σ(x1, x2, x3, x4) = (αλx1, λx2, αx3, x4) on Q,

σ(0, x2, x3, 0) = (0, βx2, x3, 0) on L.

Therefore, Iσ ⊇ R where R is the linear span of the six linearly independent elements:

αx1 ⊗ x2 − x2 ⊗ x1, x4 ⊗ x2 − λx2 ⊗ x4,

αx3 ⊗ x4 − x4 ⊗ x3, x4 ⊗ x1 − αλx1 ⊗ x4,

λx1 ⊗ x3 − x3 ⊗ x1, x3 ⊗ x2 − βx2 ⊗ x3 + (λ− αβ)x1 ⊗ x4.

We will show that Iσ = R. Let f ∈ Iσ \R. Without loss of generality f =∑10i=1 αivi

is a linear combination of the elements:

v1 = x1 ⊗ x1, v2 = x2 ⊗ x2, v3 = x3 ⊗ x3, v4 = x4 ⊗ x4, v5 = x1 ⊗ x2,

v6 = x1 ⊗ x3, v7 = x2 ⊗ x4, v8 = x3 ⊗ x4, v9 = x2 ⊗ x3, v10 = x1 ⊗ x4.4

The vanishing of f at (p, σ(p)) ∈ Γσ(Q), for all points p belonging to the four speciallines on Q that meet L shows that f = α9v9 + α10v10. Then evaluating f on Γσ(L),and next at (p, σ(p)) ∈ Γσ(Q) for a point p not lying on the four special lines, showsthat f = 0. Thus, Iσ = R.

(b) With respect to the basis ei, we have

σ|Q =

0 0 0 10 α 0 00 0 λ 0αλ 0 0 0

and σ|L =

(1 00 β

),

for some α, β, λ ∈ k \ 0. Again, condition (4) of σ holds if and only if λ 6= αβ.Hence, the map σ is given by:

σ(x1, x2, x3, x4) = (x4, αx2, λx3, αλx1) on Q,

σ(0, x2, x3, 0) = (0, x2, βx3, 0) on L,

and in this case we have Iσ ⊇ R where R is the linear span of the six linearly indepen-dent elements:

αx1 ⊗ x3 − x3 ⊗ x4, x4 ⊗ x3 − λx3 ⊗ x1,

αx2 ⊗ x1 − x4 ⊗ x2, x4 ⊗ x4 − αλx1 ⊗ x1,

λx1 ⊗ x2 − x2 ⊗ x4, x2 ⊗ x3 − βx3 ⊗ x2 − (αβ − λ)x1 ⊗ x1.

Then, as in (a), evaluating any f ∈ Iσ \R at suitable points of Γσ proves that suchan f must be zero.

The algebras Aσ are defined in terms of geometric data. The next result shows that,conversely, the defining relations of the algebras determine the geometric data.

Lemma 1.4. Γσ = V(Iσ).

Proof. Since, by definition, Γσ ⊆ V(Iσ) we need only show the reverse inclusion. Letp = (p1, p2, p3, p4) and q = (q1, q2, q3, q4) ∈ P(V ∗). By 1.3 there are two cases to analyse.

(a) Suppose Aσ is given by the relations in 1.3 (a). The basis for Iσ found in proving1.3(a) shows that (p, q) ∈ V(Iσ) if and only if

x4 0 0 −αλx1

0 x4 0 −λx2

0 0 x4 −αx3

x2 −αx1 0 0x3 0 −λx1 00 x3 −βx2 (λ− αβ)x1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣p

q1

q2

q3

q4

= 0

(where the subscript |p beside the matrix means evaluation at p). Let M denote thematrix.

Suppose (p, q) ∈ V(Iσ). Then rank(M |p) < 4. We first show that p ∈ Q ∪ L. Theminor formed from the first three rows and the last row of M is (λ−αβ)x2

4(x1x4+x2x3),which must vanish at p. Therefore, if x4(p) 6= 0 then p ∈ Q. On the other hand, ifx4(p) = 0 then the 4× 4 minor:

∣∣∣∣∣∣∣∣∣

0 0 0 1x2 −αx1 0 0x3 0 −λx1 00 x3 −βx2 (λ− αβ)x1

∣∣∣∣∣∣∣∣∣= (αβ − λ)x1x2x3

5

also vanishes at p. So p ∈ V(x2, x4)∪ V(x3, x4)∪ V(x1, x4), which is a subset of Q∪L.We now claim that q = σ(p). If not, then there are two independent solutions, q and

σ(p), to the equation, implying that rank(M |p) < 3. But this is false, since for eachp ∈ Q∪L one may check that there is at least one 3× 3 minor of M |p that is nonzero.It follows that (p, q) ∈ Γσ, from which we conclude V(Iσ) = Γσ.

(b) Now suppose Aσ is given by the relations in 1.3 (b). One can check thatapplication of the above argument to the matrix

αλx1 0 0 −x4

0 λx1 0 −x2

0 0 αx1 −x3

−λx3 0 x4 0−αx2 x4 0 0

(αβ − λ)x1 βx3 −x2 0

yields the desired result.

Remarks .(a) The main reason that 1.4 holds is because of condition (4) that we imposed upon

σ; i.e., the fact that σ is not the restriction of a linear map on P3.(b) Notice that Aσ is never commutative. One explanation is as follows. If Aσ is

commutative then case 1.3(a) occurs, forcing α = λ = β = 1. But then σ is therestriction of the identity map on P3, contradicting condition (4) used in defining σ.

Example 1.5. The coordinate ring of quantum 2×2 matrices is denoted Oq(M∈(C)),where q ∈ C \ 0 and q2 6= 1 ([FRT]). It is (isomorphic to) the algebra C[x1, x2, x3, x4]with the six defining relations:

x2x1 = qx1x2, x3x1 = q−1x1x3, x4x1 = x1x4,

x4x3 = qx3x4, x4x2 = q−1x2x4, x3x2 − x2x3 = (q − q−1)x1x4.

By 1.3 and 1.4, Oq(M∈(C)) ∼= Aσ where

σ(x1, x2, x3, x4) = (x1, q−1x2, qx3, x4)

σ(0, x2, x3, 0) = (0, x2, x3, 0)

on Q,

on L.

In particular, σ preserves the rulings on Q; moreover, σ is of finite order if and only ifq is a root of unity.

Example 1.6. Families of algebras on n2 generators containing Oq(M\(C)) are de-fined in [AST], [Su, §4b] and [T, §2]. The second family contains the third and, whenn = 2, the third contains the first. Moreover, when n = 2, these families intersectwith the subfamily Aσ : σ preserves the rulings on Q and σ|L is the identity on L.However, our insistence on σ not being the restriction of a linear map on P3 excludessome of the algebras in [AST], [Su] and [T] (n = 2) from this subfamily – those omittedallow (in our notation) the possibility λ− αβ = 0.

Example 1.7. The coordinate ring of quantum symplectic 4-dimensional space is de-noted Oq(spC4), where q ∈ C\0 and q2 6= 1 ([FRT]). It is the algebra C[x1, x2, x3, x4]with the six defining relations:

x2x1 = q−1x1x2, x3x1 = q−1x1x3, x4x1 = q−2x1x4,

x4x3 = q−1x3x4, x4x2 = q−1x2x4, x3x2 − q−2x2x3 = q−2(q−1 − q)x1x4.6

By 1.3 and 1.4, Oq(spC4) ∼= Aσ where

σ(x1, x2, x3, x4) = (x1, qx2, qx3, q2x4)

σ(0, x2, x3, 0) = (0, x2, q2x3, 0)

on Q,

on L.

Here, σ preserves the rulings on Q; and σ is of finite order if and only if q is a root ofunity.

Lemma 1.8. The algebras Aσ are iterated Ore extensions, and hence skew-polynomialrings; that is, they are of the form k[X1;µ1, δ1] · · · [X4;µ4, δ4] where the µi are auto-morphisms and the δi are µi-derivations at each stage of the extension.

Proof. For the algebras in 1.3(a) the result is easy to check by adjoining the variablesin the order: x1, x2, x4, x3. For those in 1.3(b), setting

X1 =

√λ−αβ4αβλ

(√αλx1 − x4), X4 =

√λ−αβ4αβλ

(√αλx1 + x4), X2 = x2, X3 = x3

shows that each algebra of 1.3(b) is isomorphic to an algebra k[X1, X2, X3, X4] withdefining relations:

X2X4 = γX4X2, X2X1 = −γX1X2, X4X1 = −X1X4,

X4X3 = γX3X4, X1X3 = −γX3X1, X3X2 − νX2X3 = (X1 +X4)2,

where γ =√

λα

and ν = 1β. By adjoining the variables in the order: X1, X2, X4, X3,

the result is straightforward to verify.

Remarks .(a) The basis Xi in the proof of 1.8 is dual to the basis e1 +

√αλe4, e2, e3, e1 −√

αλe4. The latter set consists of eigenvectors for the linear map τ ∈ Aut(P3) whichsatisfies τ |Q = σ|Q.

(b) The algebras Aσ of 1.3(b) seem to depend on three parameters whereas the alge-bras k[X1, X2, X3, X4], in the proof of 1.8, depend on only two. This can be explainedas follows. Suppose σ1, σ2 ∈ Aut(Q ∪ L) satisfy conditions (1) to (4), used in definingσ, with σ1|L = σ2|L, and that τ1, τ2 ∈ Aut(P3) are linear maps with τi|Q = σi|Q.Then Aσ1

∼= Aσ2 if and only if there exists a linear map φ ∈ Aut(P3) such thatφ(Q) = Q, φ(L) = L and τ2 = φτ1φ

−1. (Loosely speaking, the algebras are determinedup to conjugacy of σ.)

Corollary 1.9. The algebra Aσ is a noetherian domain of global homological dimensiongldim(Aσ) = 4, with Hilbert series HAσ(t) = (1 − t)−4. Moreover, the Koszul complexis exact.

Proof. The first sentence is proved using [MR, §1.2, 7.5.3, 7.9.16]. There are a numberof ways to show exactness of the Koszul complex – one is to apply [LSV, 2.6], anotheris to exploit the PBW basis and apply [P].

Notation 1.10.

(a) We will use the symbol τ to denote the linear automorphism of P3 such thatτ |Q = σ|Q. Notice that τ(L) = L since σ fixes both e2 and e3.

(b) If u ∈ V and θ ∈ Aut(P3) is a linear map, let uθ denote the linear form u θ.(c) For 1.3(a), let ω1 = x1, ω2 = x4;

for 1.3(b), let ω1 =√αλx1 − x4, ω2 =

√αλx1 + x4.

(d) Let Ω = xτ3 ⊗ x2 − xτ2 ⊗ x3. The image of Ω in Aσ will be denoted Ω.7

Remark . If k = C and σ preserves the rulings on Q, then Ω is central if and onlyif Aσ ∼= Oq(M∈(C)) (see example 1.5). In this case, Ω is (a scalar multiple of) thequantum determinant.

Lemma 1.11.

(a) L = V(ω1, ω2).(b) Ω is the unique element in A2 (up to a scalar multiple) that vanishes on Γσ(Q)

but not identically on Γσ(L). Moreover, V(Ω) ∩ Γσ = Γσ(Q).(c) If u, v ∈ V then uτv − vτu is a scalar (possibly zero) multiple of Ω.(d) If u, v ∈ V and L ⊂ V(uv) then uτv = vτu in Aσ.(e) ω1, ω2 and Ω are normal in Aσ.

Note . If v ∈ V is an eigenvector for τ (i.e., vτ ∈ kv) and v(L) = 0 then 1.11(d) showsthat uτv ∈ kvu for all u ∈ V . Hence, v is normal in Aσ. Furthermore, the first fiverelations in both 1.3(a) and 1.3(b) are consequences of 1.11(d).

Proof. (a) follows from the definitions.

(b) By definition of τ , we have Ω(Γσ(Q)) = 0. With notation as in 1.3, recallthat λ − αβ 6= 0 due to σ not being the restriction of a linear map on P3. Takingr = (0, 1, 1, 0) ∈ L we find that Ω(r, σ(r)) is a nonzero scalar multiple of λ− αβ, and

hence Ω does not vanish identically on Γσ(L). Now Ω(p, σ(p)) = (xτ3xν2 − xτ2xν3)(p) for

all p ∈ L, where ν ∈ Aut(P3) is a linear map such that ν|L = σ|L. The polynomialxτ3x

ν2 − xτ2xν3 is nonzero of degree 2, so, by Bezout’s theorem, it can vanish at no more

than two points of L. These points are e2, e3 = Q ∩ L, which proves V(Ω) ∩ Γσ =Γσ(Q).

Furthermore, by the proof of 1.3 any element in V ⊗ V/Iσ that vanishes on Γσ(Q)must belong to kx1 ⊗ x4 ⊕ kx2 ⊗ x3 for 1.3(a) or kx1 ⊗ x1 ⊕ kx2 ⊗ x3 for 1.3(b). Since(x2⊗x3)(Γσ(Q)) 6= 0 there can only be one such element (up to a scalar multiple) thatdoes not vanish identically on Γσ(L).

(c) follows from (b) since, by definition of τ , we have uτ ⊗ v − vτ ⊗ u vanisheson Γσ(Q); and (d) holds since the extra hypothesis ensures that now uτ ⊗ v − vτ ⊗ uvanishes on Γσ(L) also.

(e) Normality of ωi follows from the previous results. (It can also be deducedfrom (d) since ωτi ∈ kωi – see previous note.) A calculation shows that Ωxi ∈ kxiΩ,1 ≤ i ≤ 4.

2. Classification of Point and Line Modules

We observe in §2.1 that the point modules are in bijection with the points on thevariety Q ∪ L; in §2.2 we show that the line modules are in bijection with those linesin P3 that either lie on Q or meet L. It is shown that both point (§2.3) and line (§2.1)modules are quotients of Aσ by a left ideal generated by an appropriate subspace of(Aσ)1.

2.1. Preliminaries. Henceforth, we assume σ is fixed and write A for Aσ. We definea grading on A by declaring the generators x1, . . . , x4 to be of degree one; that is, Ainherits the usual grading from T (V ). We will denote the homogeneous degree n partof A by An and identify V with its image, A1, in A.

The Hilbert series of a graded module M will be denoted HM(t).8

A shift of a Z-graded module M =⊕

nMn is a Z-graded module M [p] where M [p]n =Mp+n for some p ∈ Z and for all n ∈ Z. Clearly, HM [p](t) = t−pHM(t).

Definition . [ATV1, 2] Let M be a Z-graded cyclic module over A. We say M is apoint, resp. line, resp. plane, module if HM(t) = (1−t)−1, resp. (1−t)−2, resp. (1−t)−3.

Let u be a nonzero element in A1. Then, since A is a domain, we have

HA/Au(t) = HA(t)−HAu(t) = (1− t)−3.

Thus, A/Au is a plane module. Conversely, suppose M =⊕

nMn is a plane module.Since M is cyclic and dim(M1) = 3, it follows that there is a surjective homomorphismA/Au M for some nonzero element u ∈ A1. But A/Au is a plane module, whenceM ∼= A/Au. In particular, the plane modules are in bijection with the planes V(u) inP3.

Now, suppose K is a line module and M is a point module. A similar dimensionargument shows there exist surjective homomorphisms:

A

Au+ Av K and

A

Au′ + Av′ + AwM

for some nonzero elements u, v, u′, v′, w ∈ A1. In this section we will show these mapsare isomorphisms. However, for now, we can at least conclude that K defines a line,V(u, v), in P3 and M a point, V(u′, v′, w), in P3. In fact, 2.1 shows that more is true.

Proposition 2.1. The point modules for A are in bijection with the points of thevariety Q ∪ L.

Proof. Let πi : P3 × P3 → P3 denote the projection map onto the i’th coordinate fori = 1, 2. Since V(Iσ) = Γσ is the graph of σ, we have πi|V(Iσ) is injective for i = 1, 2,and π1(V(Iσ)) equals π2(V(Iσ)) – both equal Q∪L. The result now follows from [ATV1,§3].

Remark 2.2. By [ATV1, §3], the action of the algebra A on a point module M(p),corresponding to a point p ∈ Q ∪ L, can be described as follows: if x ∈ A1 and vi is abasis for M(p)i for i ≥ 0 then x · vi = x(σ−i(p))vi+1 for all i ≥ 0.

It is demonstrated in [LS] that, in some cases, it is possible for information on point,line and plane modules to be extracted from certain homological data concerning thealgebra. Following this approach, we employ terminology from [LS, §1]. Corollary 1.9and proposition 2.3 show that A satisfies a sufficient number of the hypotheses on thealgebras discussed in [LS, §2] to enable the line modules to be described in a naturalway (theorem 2.4).

Proposition 2.3. The algebra A is Auslander-regular of dimension 4 and satisfies theCohen-Macaulay property.

Proof. The elements ω1, ω2, x2, x3 form a regular normalizing sequence in A. Theresult then follows by applying [L, §5.10] and corollary 1.9.

Theorem 2.4. [LS, §2] Let A be a noetherian, graded k-algebra, generated in degree1, with Hilbert series HA(t) = (1 − t)−4, which is Auslander-regular of dimension 4and satisfies the Cohen-Macaulay property. Then the line modules are in bijection withthose lines ` in P3 such that ` = V(u, v) where u, v ∈ A1 are linearly independentelements satisfying A1u ∩ A1v 6= 0. The line module corresponding to such a line ` is(isomorphic to) A/Au+ Av.

9

Notation . The symbols M(p) and M(`) will denote the point module, resp. linemodule, corresponding to the point p or line `.

2.2. Geometric Classification of Line Modules.

Theorem 2.5. There is a bijection between the set of isomorphism classes of linemodules over A and the collection of lines in P3 that either lie on Q or meet L.

We will prove the theorem in two parts, making use of 2.4, but first we need a lemma.

Lemma 2.6. Let ` := V(u, v) be a line in P3, where u, v ∈ V . Suppose ` ∩ L = ∅and ` 6⊂ Q. Then there exists a nonzero element w ∈ ku ⊕ kv such that V(w) ∩ Q isa nondegenerate conic in the plane V(w) and V(w) ∩ L 6⊂ Q. Furthermore, there aremany such w.

Proof. Let P3 be the dual space consisting of the planes in P3, and let ˇ⊂ P3 be theline consisting of the planes containing `. Notice that L 6⊂ H for all H ∈ ˇ (otherwiseL ∩ ` 6= ∅).

Since ` 6⊂ Q we have ` ∩ Q = x, y, where possibly x = y. Let `1, `2 be the twolines on Q which pass through x. Consider a conic H ∩Q for any H ∈ ˇ. If the conic isdegenerate then it contains a line passing through x ∈ H ∩Q, and so contains either `1

or `2. Thus, if there were three distinct H ∈ ˇsuch that H∩Q is degenerate, then two ofthem, say H and H ′, must contain a common `i (i = 1 or 2). But then, `i ⊂ H∩H ′ = `,and so ` ⊂ Q which is false. Therefore, the set H ∈ ˇ: H ∩Q is nondegenerate is adense open subset of ˇ.

Now consider the subset Y := H ∈ ˇ: H ∩ L ⊂ Q of ˇ. If |Y | ≥ 3 then, since|Q ∩ L| = 2, there would exist H,H ′ ∈ Y, H 6= H ′, with H ∩ L = H ′ ∩ L. It wouldthen follow that ` ∩ L = (H ∩H ′) ∩ L = H ∩ L 6= ∅, which is a contradiction. Thus,|Y | ≤ 2 implying that H ∈ ˇ: H ∩ L 6⊂ Q is a dense open subset of ˇ.

Hence, the set H ∈ ˇ: H ∩Q is a nondegenerate conic and H ∩L 6⊂ Q is dense inˇ, which completes the proof.

Proposition 2.7. Let A/Au + Av be a line module where u, v ∈ A1. Then the line` := V(u, v) either lies on the quadric Q or meets the line L.

Proof. Comparing the Hilbert series of A with that of a line module, we see thatdim(A1u+A1v) = 7, so therefore A1u ∩ A1v 6= 0. (In fact, this is a minor part of 2.4,but is true for these elementary dimension reasons.) In particular, there exist nonzeroelements a, b ∈ A1 such that av = bu, or equivalently

(a⊗ v − b⊗ u)(Γσ) = 0. (∗)Suppose for a contradiction that ` ∩ L = ∅ and ` 6⊂ Q.

By 2.6 we can pick a plane H = V(w) with 0 6= w ∈ ku ⊕ kv, such that H ∩ Qis a nondegenerate conic in H and H ∩ L = p for some p /∈ Q. Moreover, we canassume w = u, since if w, w′ ∈ ku ⊕ kv are nonzero and linearly independent thenA1w ∩ A1w

′ 6= 0 also.Now, p and the nondegenerate conic H∩Q both lie in the plane H. Since p /∈ H∩Q

there are exactly two lines in H passing through p which are tangent to H ∩ Q. Itfollows that there are many choices of distinct points q, r, s ∈ H∩Q such that p, q, r arecollinear but with p, q, r, s not collinear. We choose such distinct points q, r, s ∈ H ∩Qsuch that q, r, s ∩ ` ∩Q is empty.

10

By construction, u vanishes at p, q, r, s whereas v is nonzero at these four points. So,applying (∗) to (σ−1(p), p), . . . , (σ−1(s), s) in turn implies that a vanishes at σ−1(p),σ−1(q), σ−1(r), σ−1(s). In particular, these four points are coplanar. We will provethat this is, in fact, not the case.

Since the map τ (defined in 1.10) is linear, it follows that τσ−1(p), q, r, s arecoplanar. But q, r, s span H, and τ(L) = L, so τσ−1(p) ∈ H ∩ L = p. Therefore,σ−1(p) = τ−1(p) so that the linear maps σ−1|L and τ−1|L agree at three distinct points,e2, e3, p, of L. It follows that σ−1|L = τ−1|L, and hence σ = τ |Q∪L, contradictingcondition (4) used in defining σ.

Proposition 2.8. Let ` be a line in P3 that either lies on Q or meets the line L. Thenthere exist elements u, v ∈ A1 such that ` = V(u, v) and A/Au+ Av is a line module.

Proof. We will show there exist nonzero elements u, v, a, b ∈ A1 such that ` = V(u, v)and av = bu, and invoke 2.4 to finish the proof.

Firstly, if ` ∩ L 6= ∅, then there exist u, v ∈ V such that ` = V(u, v) and u(L) = 0.By 1.11(d), it follows that uτv = vτu.

Now suppose `∩L = ∅ and ` ⊂ Q. Then the distinguished line L does not lie in anyplane that contains `.

Choose two distinct lines `1, `2 on Q, in the ruling on Q which does not contain`, such that `i ∩ L = ∅, for i = 1, 2. We take u ∈ V to be the plane determinedby the lines ` and `1, and similarly v ∈ V to be that determined by ` and `2. Then` = V(u, v), and u(L) 6= 0 and v(L) 6= 0.

There is exactly one point r ∈ L such that u(σ(r)) = 0 and, since V(u)∩Q = `∪`1 ⊂Q \ L, we have r /∈ Q. Put q = V(v) ∩ L. By hypothesis, q 6= σ(r).

Now u/v is a rational function on both Q and L with associated divisors

divQ

(u

v

)= `1 − `2 and divL

(u

v

)= σ(r)− q.

Define a linear form a ∈ V to be such that r ∈ V(a) and σ−1(`1) ⊂ V(a). Sinceσ−1(`1) ⊂ Q ∩ V(a), which is a conic in V(a), it follows that the divisor of zeros of aon Q is (a)0 = `3 + σ−1(`1) for some line `3 ⊂ Q. Next, define a linear form b ∈ V byrequiring that the divisor of zeros of b on Q is (b)0 = `3 + σ−1(`2).

The rational function a/b on Q has divisor

divQ

(a

b

)= σ−1(`1)− σ−1(`2).

However,

divQ

(u

v

)τ= σ−1

(divQ

(u

v

))= σ−1(`1)− σ−1(`2).

Hence, replacing a by a suitable scalar multiple we have a/b = (u/v)τ on Q; that is,

(avτ − buτ )(p) = 0 for all p ∈ Q.More precisely, we have

(a⊗ v − b⊗ u) (p, σ(p)) = 0 for all p ∈ Q.It remains to show this also holds on L. Since it is already true at p = ei for

i = 2, 3, by Bezout’s theorem we need only show it holds at some p ∈ L \ e2, e3.But, by choice of r ∈ L \ Q above, we have u(σ(r)) = 0 and a(r) = 0, implying that(a⊗ v − b⊗ u)(r, σ(r)) = 0. It follows that av = bu in A.

11

Note . Suppose u, v ∈ V satisfy the hypotheses of 2.7. Then the point (a, b) ∈ P(V×V )such that av = bu (or equivalently, such that (a ⊗ v − b ⊗ u)(Γσ) = 0) is unique (seethe proof of [LS, 4.3]).

2.3. Point Modules. By remark 2.2, if p ∈ V(u, v), where u, v are nonzero, linearlyindependent elements of V , then M(p) is a quotient of A/Au + Av. It then followsfrom proposition 2.1 and theorem 2.5 that every point module is a quotient of a linemodule. Hence, if V(u, v) ≡ ` corresponds to a line module then we have the shortexact sequence

0 −→ K −→M(`) −→M(p) −→ 0

for some A-module K ⊂ M(`). In fact, K is a shift of a line module as can be seenfrom the following result.

Proposition 2.9. Let p = V(u, v, w) ∈ Q ∪ L where u, v, w are nonzero, linearlyindependent elements of V . Then the corresponding point module M(p) is given byM(p) ∼= A/Au+ Av + Aw.

Proof. Since V(Iσ) is the graph of the automorphism σ, we can apply [LSV, 4.1.1].

The line corresponding to the shifted line module K, in the above short exact se-quence, is identified in terms of the point p and the line ` in §5.

Remarks .(a) Given an automorphism of an algebra, a procedure for twisting the algebra, and

the modules over it, by this automorphism, is given in [ATV2, §8]. The category ofgraded modules over the twisted algebra is equivalent to the category of graded modulesover the original algebra ([ATV2, 8.5]). For both kinds of σ there exists such a twist ofAσ that yields the algebra Aστ−1 which is of type 1.3(a) (where τ as defined in 1.10).Notice that σ τ−1|Q is the identity map on Q. Here the twist map depends on ω2 and,by lemma 1.11(d), it is given by τ . If M(X) = M is a linear module over Aσ, where Xis a point/line/plane, then M τ [d] corresponds to X but M [d]τ corresponds to τ−dX.In particular, we could work over Aτ

σ := Aστ−1 and “twist” the results back to Aσ.However, the work for Aτ

σ is not that much simpler, and, in fact, studying Aτσ alone is

insufficient for proving many results on annihilators of such modules M(X), since theorder of σ|Q plays a substantial role in determining annM(X). For instance, over Aτ

σ,M(p) ∼= Aτσ/annM(p) for all p ∈ Q, and the latter (as an algebra) is isomorphic to apolynomial ring in one variable ([LS, §5]) – over arbitrary Aσ this is false. For thesereasons, we consider arbitrary Aσ throughout the paper.

(b) The work in [LSV] applies to twists of Aσ, and “twisting” their results back toAσ shows that a line ` through p ∈ L ∪ (Q ∩ V(ω2)) corresponds to a line module ifand only if ` is contained in V(ω2) or a certain quadric Qp. Our results show that Qp

is independent of p and Qp = Q.

3. The Algebra A/AΩ

Recall that the normal element Ω (defined in 1.10) is the unique element of A2 (upto a scalar multiple) that vanishes on Γσ(Q) but not on Γσ(L). In this section we showthat A/AΩ is a twisted homogeneous coordinate ring ([AV]) of Q with respect to σ

12

and a suitable invertible sheaf L on Q. The general construction is taken from [ATV1]and will show that A/AΩ is a domain.

For convenience, in this section only, σ will denote σ|Q.

Notation .

(a) Let j : Q → P3 be the inclusion map and put L = |∗OP3(∞) = OQ(∞).

(b) For each n ∈ Z, write Lσ\ = (σ\)∗L.

(c) Write L′ = OQ and set L\ = L ⊗ · · · ⊗ Lσ\−∞ for n ≥ 1.(d) Write Bn = H0(Q,L\) and define B =

⊕n≥0 Bn.

We give B the structure of a graded ring by defining the multiplication as follows: ifu ∈ Bn, v ∈ Bm and vσ

ndenotes the image of v (see remark 3.1) in H0(Q,Lσ\m ) then

u · v = µn,m(u⊗ vσn) ∈ Bn+m

where µn,m : H0(Q,L\)⊗H′(Q,Lσ\m )→ H′(Q,L\+m) is the natural map.

We will show that B ∼= A/AΩ.

Remark 3.1. If U ⊂ Q is open, then by definition of Lσ, we have Lσ(U) = L(σU).Thus, if v ∈ Lσ(U) then v σ ∈ L(U). In the case that U = Q, then Lσ(Q) = L(Q)and so, for this reason, we take v ∈ L(Q) and v σ = vσ ∈ Lσ(Q). Moreover, if vi ∈ B1

then the element v1 · v2 ∈ B2 is the form on Q defined by (v1 · v2)(p) = v1(p)vσ2 (p) =v1(p)v2(σ(p)) for all p ∈ Q.

Lemma 3.2. There is a k-algebra homomorphism φ : A→ B.

Proof. Since Q is defined by one quadratic relation it follows that

B1 = H0(Q,L) ∼= H′(P3,OP3(∞)) = ‖§∞ ⊕ ‖§∈ ⊕ ‖§3 ⊕ ‖§4.So A1

∼= B1 via φ : xi 7→ xi. We claim φ induces a homomorphism from A to B. Forif f =

∑λijxixj = 0 in A then φ(f) =

∑λijxi · xj = 0 in B, because f vanishes on

Γσ(Q) ⊂ Γσ, that is, φ(f)(p) =∑λijxi(p)xj(σ(p)) = 0 for all p ∈ Q, by remark 3.1.

Lemma 3.3. B is generated by B1 as a k-algebra.

Proof. We need to prove that Bn+1 = BnB1 for all n ≥ 1. However, by the definition ofmultiplication in B, this is equivalent to showing that the cokernel of the multiplicationmap

µn,1 : H0(Q,L\)⊗H′(Q,Lσ\)→ H′(Q,L\+∞)

is zero for all n ≥ 1. But this is a consequence of [M, theorem 2] since Lσ ∼= L,L\ ∼= OQ(\) ∼= L\ and

H i(Q,L\ ⊗ (Lσ\)−〉) = H〉(Q,OQ(\ − 〉)) = ′ for all \, 〉 ≥ ∞(computed from the long exact sequence in cohomology applied to the short exactsequence 0→ OP3(\ − ∈)→ OP3(\)→ OQ(\)→ ′).Corollary 3.4. The map φ is surjective and B ∼= A/AΩ.

Proof. It follows from 3.2 and 3.3 that B = φ(A). In addition, Ω(Γσ(Q)) = 0 so Ω ∈ker(φ). Thus there is a surjective homomorphism A/AΩ B.

Now, since L ∼= Lσ, we have dimH0(Q,L\) = dimH0(Q,L\) and, moreover, the

algebra⊕H0(Q,L\) is isomorphic to the homogeneous coordinate ring of Q (since the

latter is integrally closed). Therefore B has the same Hilbert series as the homogeneous13

coordinate ring of Q which, in turn, has the same Hilbert series as A/AΩ (since A is adomain). Hence B ∼= A/AΩ as desired.

Lemma 3.5. The algebra A/AΩ is a domain.

Proof. By 3.4 we need only show that B is a domain. Suppose there exist elementsu ∈ Bn, v ∈ Bm such that 0 = u · v ∈ Bn+m. By remark 3.1, this implies that0 = uvσ

n ∈ H0(Q,L\+m). As in the proof of 3.4,⊕H0(Q,L\) is isomorphic to the

homogeneous coordinate ring of Q, which is a domain since Q is irreducible. Thereforeu = 0 or vσ

n= 0, from which the result follows.

Remark .Consider τ (defined in 1.10(a)) as an automorphism of V via the rule: vτ = v τ for

v ∈ V (see 1.10(b)). Let R denote the homogeneous coordinate ring of Q and extend τto an automorphism of R in the obvious way. Define a new multiplication ∗ on R via:a ∗ b = abτ

m, where a, b ∈ R are homogeneous elements of degrees m,n respectively.

This multiplication is “the same” as that used to define B from⊕H0(Q,L\) = R

and the “twisted” algebra, Rτ , so obtained, is isomorphic to A/AΩ. Consequently, thecategory of graded modules over A/AΩ (∼= B) is equivalent to the category of gradedmodules over R ([ATV2, 8.5]).

4. Annihilators of Point and Line Modules

The main result of this section shows that when σ has infinite orbit at “most” pointsof Q then annM(`) = AΩ where ` is any line on Q that does not meet L.

Remark 4.1. By [LS, §1, §2], point, line and plane modules are critical for GK-dimension, and hence have prime annihilators (adapt the argument of [St, 3.9] oralternatively use [ATV2, 2.30(vi)]).

For this reason, the annihilators of linear modules feature prominently in [SS] intheir search for the primitive ideals of the Sklyanin algebra, and this suggests theywarrant study here too.

Lemma 4.2. Let ω1, ω2 ∈ A1 be the normal elements defined in 1.10.

(a) The point module M(p) is annihilated by ω1 and ω2 if and only if p ∈ L.(b) The point module M(p) is annihilated by Ω if and only if p ∈ Q.

Proof. The result follows from 1.11, 2.1 and 2.2.

Note . By §2.2, M(L) ∼= A/Aω1 + Aω2 which (as an algebra) is isomorphic to aquantum plane. Hence, the points on L can be viewed as corresponding to pointmodules over a quantum plane.

Remark 4.3. Let M(`) be a line module and suppose there are infinitely many pointson the line ` that correspond to point modules. Then it follows from [S, §1] thatannM(`) =

⋂annM(p), where the intersection runs over infinitely many p ∈ ` that

correspond to point modules.

Lemma 4.4.

(a) The line module M(`) is annihilated by ω1 and ω2 if and only if ` = L.(b) The line module M(`) is annihilated by Ω if and only if ` ⊂ Q.

14

Proof. (a) follows from §2.2.(b) If ` ⊂ Q then lemma 4.2 and remark 4.3 show that Ω annihilates M(`).Conversely, if ` 6⊂ Q then, by 2.5, ` meets the line L. Therefore, we can assume

` = V(u, v) where u, v are nonzero linearly independent elements of V such thatu(L) = 0.

If Ω ∈ annM(`) then there exist elements a, b ∈ V such that

(a⊗ v − b⊗ u)(Γσ(Q)) = 0 whereas (a⊗ v − b⊗ u)(Γσ(L)) 6= 0.

In particular, the degree 2 polynomial avτ − buτ vanishes on Q (where τ as defined in1.10). Since Q is irreducible, avτ − buτ ∈ k(x1x4 + x2x3); but avτ − buτ also vanisheson the line τ−1(`) which is not contained in Q, so that avτ − buτ = 0 in the polynomialring. Thus, uτ ∈ ka or kvτ , and vτ ∈ kb or kuτ . Since u and v are linearly independentand τ is bijective we have uτ ∈ ka and vτ ∈ kb. Therefore,

a⊗ v − b⊗ u = λ1uτ ⊗ v − λ2v

τ ⊗ ufor some λ1, λ2 ∈ k. However, since uτ (L) = u(L) = 0, this expression vanisheson Γσ(L), contradicting the choice of a and b. We conclude that Ω /∈ annM(`) asdesired.

In [ATV2], a regular algebra of dimension 3 is shown to be a finite module over itscentre if and only if the associated automorphism of the scheme E (see introduction)has finite order. This same dichotomy also occurs for the Sklyanin algebra ([LS], [S])and therefore, as might be expected, to determine the annihilators more precisely, weneed to make some assumption on the order of σ. However, to assume σ has infiniteorder at every point (other than e1, . . . , e4) of Q ∪ L, excludes Oq(M∈(C)) given in1.5, for which σ|L is the identity on L. In examples 1.5 and 1.7, taking q not to be aroot of unity is equivalent to insisting σ have infinite orbit at all points of Q \ ej4

j=1

only. In fact, this is the generic case: that is, in the notation of 1.3, when σ preservesthe rulings on Q, then | 〈σ〉 · p| = ∞ for all p ∈ Q \ ej4

j=1 if and only if both α andλ are not roots of unity; when σ interchanges the rulings on Q, then | 〈σ〉 · p| =∞ forall p ∈ Q \ ej4

j=1 if and only if α/λ is not a root of unity. Thus, 4.5 and 4.6 apply toboth Oq(M∈(C)) and Oq(spC4) when q is not a root of unity.

We recall from §1 the existence of exactly four special lines on Q that meet L.

Proposition 4.5. Suppose | 〈σ〉 · p| =∞ for all p ∈ Q \ ej4j=1 and let ` be a line on

Q. If ` ∩ L = ∅, then annM(`) = AΩ.

Proof. Since Ω ∈ annM(`) ⊆ annM(p) for all p ∈ `, we have

A

AΩ A

annM(`) A

annM(p)for all p ∈ `.

Suppose for a contradiction that annM(`) 6= AΩ. Then, since AΩ is prime (by§3) and GKdim(A/AΩ) = 3, we have GKdim(A/annM(`)) = 2. However, since thepoints on ` have infinite orbit, an argument similar to that of [LS, 5.11] shows thatGKdim(A/annM(p)) ≥ 2 for all p ∈ `. But, by remark 4.1, annM(`) is prime, so

annM(`) = annM(p) for all p ∈ `. (∗)Now, ` meets one of the four special lines `′ on Q. By remark 1.2 there exists a planeV(w) such that (i) V(w) ∩ Q = two special lines, and (ii) σi(`′) ⊆ V(w) ∩ Q for all i.Therefore w ∈ annM(q) for all q ∈ `′.

15

In particular, setting r = `∩ `′ and applying (∗) we have w ∈ annM(r) = annM(`),which proves w(`) = 0. That is, ` ⊂ V(w) ∩ Q and so must be a special line on Q,contradicting our hypothesis on `.

Proposition 4.6. Suppose σ preserves the rulings on Q and that | 〈σ〉 · p| =∞ for allp ∈ Q \ ej4

j=1. Let ` := V(u, v), where u, v ∈ V , be one of the four special lines onQ. Then

(a) annM(`) = Au+ Av;(b) annM(ei) = Axj + Axm + Axn where i, j,m, n = 1, 2, 3, 4;(c) if p ∈ ` \ ej4

j=1 then annM(p) = annM(`).

Proof. (a) By 2.4, annM(`) ⊆ Au + Av so we need only show the reverse inclusion.Since σ(`) = ` we have u(σi(`)) = 0 = v(σi(`)) for all i, so applying remark 2.2we obtain Au + Av ⊆ annM(p) for all p ∈ `. It then follows from remark 4.3 thatAu+ Av ⊆ annM(`).

(b) Since σ(ej) = ej the result follows from remark 2.2.(c) We have annM(`) ⊂ annM(p), and also by (a), M(`) ∼= A/annM(`). Therefore

GKdim

(A

annM(p)

)≤ GKdim

(A

annM(`)

)= 2.

However, as in the proof of 4.5, since p has infinite orbit, GKdim(A/annM(p)) ≥ 2 –whence equals 2. Since annM(`) is prime, it follows that annM(p) = annM(`).

In reading 4.7, one should note that V(x2) ∩ V(x3) 6⊂ Q (see §1).

Proposition 4.7. Suppose σ interchanges the two rulings on Q and that | 〈σ〉 · p| =∞for all p ∈ Q \ ej4

j=1. Let ` be one of the four special lines on Q. Then

(a) for ` ⊂ V(xi) where i = 2, 3 we have annM(`) = 〈xi〉;(b) annM(e1) = annM(e4) = 〈x2, x3〉;(c) for i = 2, 3 we have annM(ei) = Axj +Axm +Axn where i, j,m, n are distinct ;(d) if p ∈ ` \ ej4

j=1 then annM(p) = annM(`).

Proof. Since ` ⊂ V(x2) ∪ V(x3), we assume ` ⊂ V(x3). Then x3 ∈ annM(p) for allp ∈ ` and so, by 4.3, x3 ∈ annM(`). If p ∈ ` has infinite orbit, then (as above) we have

2 = GKdim

(A

〈x3〉

)≥ GKdim

(A

annM(`)

)≥ GKdim

(A

annM(p)

)≥ 2.

Hence, since annM(`) is prime, we have annM(p) = annM(`), which proves (d). Toprove (a) it suffices to show 〈x3〉 is prime – we leave this for now.

Since σ interchanges e1 and e4, it follows that 〈x2, x3〉 ⊆ annM(e1) = annM(e4). ButGKdim(A/ 〈x2, x3〉) = 1, and therefore, to show equality, we need only show 〈x2, x3〉is prime. As (c) follows from the fact that σ fixes e2 and e3, the proposition would beproved if we show 〈x3〉 and 〈x2, x3〉 are prime in A. We prove these in the followinglemma.

Lemma 4.8. Suppose σ interchanges the two rulings on Q. Then the ideals 〈x2〉, 〈x3〉,〈x2, x3〉 are prime (but not completely prime) in A.

Note . When σ preserves the rulings on Q these ideals are not prime in A, since x1, x4

are normal and x1Ax4 ⊆ 〈xi〉 for i = 2, 3.16

Proof. We first show 〈x2, x3〉 is prime. By the proof of 1.8, or by 1.3(b), the ringA/ 〈x2, x3〉 is isomorphic to the ring R/ 〈x2 + y2〉 where R := k[x, y] has the defin-ing relation xy + yx = 0. Thus R is a quantum plane whose point modules areparametrized by P1 with associated automorphism ([ATV1]) σ1 : P1 → P1 given byσ1(α, β) = (α,−β). The ideal 〈x2 + y2〉 is the annihilator of the point module M(1, 1)and hence is prime in R.

Similarly, A/ 〈x3〉 is isomorphic to the ring S/ 〈x2 + z2〉 where S := k[x, y, z] hasthe three defining relations γzy = yz, yx = −γxy, xz = −zx where γ 6= 0. Thepoint modules over S are parametrized by P2 with associated automorphism ([ATV1])σ2 : P2 → P2 given by σ2(p1, p2, p3) = (−γp1, p2, γp3). We will show 〈x2 + z2〉 is theannihilator of a line module for S, and hence is a prime ideal.

Consider the line ` = V(x − z). From the above result for R, and the fact thatx2 + z2 = (x− z)2 is normal in S, it follows that

⟨x2 + z2

⟩⊆ annM(`) ( annM(1, 0, 1) =

⟨y, x2 + z2

⟩. (∗)

We claim 〈x2 + z2〉 = annM(`). To prove this it suffices, by (∗), to show that〈y〉∩annM(`) ⊂ 〈x2 + z2〉. Let f ∈ 〈y〉∩annM(`), and choose such an f to be nonzeroof minimal y degree (which is possible as S has a PBW basis). Since y is normal,there exists g ∈ S such that f = gy. However, annM(`) is prime, so g ∈ annM(`)(because y is normal). Thus, there exist g1, g2 ∈ S such that g = g1y+ g2(x2 + z2) anddegy(g1y) ≤ degy(g) < degy(f). But g1y ∈ 〈y〉 ∩ annM(`) and hence, by choice of f , itfollows that g1y = 0. Therefore, g ∈ 〈x2 + z2〉, whence f ∈ 〈x2 + z2〉, and this provesthe claim. We conclude that 〈x3〉 is prime in A.

We next consider the lines ` 6⊂ Q that meet L and the points on Q that do notlie on the four special lines. Here the behaviour varies depending on the algebra (orrather, on σ) even if we restrict to the case where σ preserves the rulings on Q. Todemonstrate this, we complete the section by contrasting examples 1.5 and 1.7.

Example 4.9. Suppose A = Oq(M∈(C)), where q is not a root of unity; or moregenerally, suppose αλ = 1 in 1.3(a) with | 〈σ〉 · p| = ∞ for all p ∈ Q \ ej4

j=1. Thenevery line module has nonzero annihilator.Note. The assumption αλ = 1 implies x1, x4 are eigenvectors for τ corresponding tothe same eigenvalue, and since they both vanish on L, any linear combination of themwill be normal in A (by 1.11(d)).Proof. Consider a line module, M(`), for some line ` in P3. If ` ⊂ Q then, by4.4, Ω ∈ annM(`). If ` 6⊂ Q, then ` meets L, so there exists (µ, ν) ∈ P1 such that` ⊂ V(µx1 − νx4). However, for all (µ, ν) ∈ P1 the element µx1 − νx4 is normal. Itfollows that A(µx1 − νx4) ⊆ annM(`).

Moreover, 〈µx1 − νx4〉 is the annihilator of the plane module A/A(µx1 − νx4), andso must be prime. If ` 6= L and ` 6⊂ Q, and if there exists p ∈ ` with | 〈σ〉 · p| = ∞then, in fact, annM(`) = 〈µx1 − νx4〉 (by a similar GKdimension argument as usedabove combined with 4.4); in addition, if p ∈ Q \ 4 special lines (i.e., if µν 6= 0)then annM(p) = 〈Ω, µx1 − νx4〉 (since a basis argument shows A/ 〈Ω, µx1 − νx4〉 is adomain).

Example 4.10. Suppose A = Oq(spC4). (Then αλ = q−2 6= 1.) If q is not a root ofunity, then annM(`) = 0 for all lines ` 6⊂ V(x1) ∪ V(x4) ∪Q.Note. Here x2, x3 are eigenvectors for τ corresponding to the same eigenvalue, butneither x2 nor x3 vanishes identically on L.

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Proof. For q not a root of unity, a “basis argument” proves that the prime ideals ofA are:

(a) 〈0〉,(b) 〈Ω〉 , 〈x1〉 , 〈x4〉,(c) 〈x2 + γx3〉 for all 0 6= γ ∈ C,〈x1, x3 + δx2x

24〉,

〈x4, x2 + δx21x3〉 for all δ ∈ C,

〈xi, xj〉 for i < j, (i, j) 6= (2, 3),(d) 〈xi, xj, xm〉 for i, j,m distinct,(e) 〈xi, xj, xl, xm − γ〉 for i, j, l,m distinct and for all γ ∈ C.

The maximal ideals are those in (e); the primitives are those in (a), (c), (e); types (b)and (d) are not primitive; and all the prime ideals are completely prime. Notice thatΩ ∈ 〈δx2 + γx3〉 for all (δ, γ) ∈ P1, and that when γ 6= 0 we have 〈x2 + γx3〉 = annM(p)for all p ∈ V(x2 + γx3) ∩ Q which satisfy | 〈σ〉 · p| = ∞. If 〈x1〉 ⊆ annM(`) for someline module M(`), then ` ⊂ V(x1) since M(`) ∼= A/Ax1 + Av for some v ∈ A1. If,in addition, ` 6= L then the list of primes shows that equality must hold, that is,〈x1〉 = annM(`). Similarly, 〈x4〉 = annM(`) for ` ⊂ V(x4), ` 6= L. This contrasts withlines ` 6⊂ V(x1) ∪ V(x4) ∪Q for which the list implies annM(`) = 0.

An observation that arises from this last example is that (when q is not a root ofunity) then every non-maximal homogeneous prime ideal of Oq(spC4) is in fact theannihilator of some point, line (or plane) module.

5. Point Modules as Quotients of Line Modules

Recall the situation in §2.3 where a point module M(p) is a quotient of a line moduleM(`), and the kernel K was shown to be a shifted line module generated by an elementof degree one; that is, we have a short exact sequence

0 −−−→ M(`′)[−1] −−−→ M(`) −−−→ M(p) −−−→ 0

where `′ is the line corresponding to K ∼= M(`′)[−1]. In this section we identify suchlines `′ in terms of p and `.

Remark 5.1. [A, corollary 2.10] Suppose that non-isomorphic point modules M(p)and M(q) are quotients of a line module M(`). Consider the diagram:

0 −−−→ M(`′)[−1] −−−→ M(`) −−−→ M(p) −−−→ 0yM(q)

If the composition M(`′)[−1] → M(q) is zero then M(q) is a quotient of M(p). Butsince both M(q) and M(p) have the same Hilbert series it follows that M(q) ∼= M(p),which is a contradiction. Therefore, the composition is a nonzero map M(`′)[−1] →M(q), with image M(q)≥1

∼= M(σ−1(q))[−1]. Hence σ−1(q) ∈ `′.Notation . The symbol `pq will denote the line through the points p and q.

Lemma 5.2. Let ` be a line in P3 that contains at least three distinct points p, q, r∈ Q ∪ L. Then there is a short exact sequence

0 −−−→ M(`′)[−1] −−−→ M(`) −−−→ M(p) −−−→ 0

where `′ = `σ−1(q) σ−1(r).18

Note . If ` in 5.2 is the distinguished line L then `′ = L. On the other hand, if ` ⊂ Qthen `′ = σ−1(`).

Proof. The result follows from remark 5.1.

Recall the definition of τ from 1.10.

Lemma 5.3. Let ` be a line in P3 that meets the distinguished line L at a point p.Then there is a short exact sequence

0 −−−→ M(τ−1(`))[−1] −−−→ M(`) −−−→ M(p) −−−→ 0.

Proof. The result for ` = L follows from 5.2. Henceforth, assume ` 6= L. Then wecan assume ` = V(u, v) such that u, v ∈ A1 with u(L) 6= 0 and v(L) 6= 0. Since p ∈ L,there exists w ∈ A1 such that p = V(u, v, w) and w(L) = 0. As w vanishes on L, itfollows from 1.11(d) that uτw−wτu = 0 = vτw−wτv. Let w denote the image of w inM(`) ∼= A/Au+Av. Then, by §2.3, Aw is the kernel of the map M(`)→M(p) and theprevious work shows that uτ w = 0 = vτ w. From linear independence of u and v, andbijectivity of τ , it follows that M(`′) ∼= A/Auτ + Avτ , which completes the proof.

Lemma 5.4. Let ` be a line in P3 that meets Q ∪ L in exactly two points, p ∈ Q \ Land q ∈ L \Q. Then there is a short exact sequence

0 −−−→ M(`′)[−1] −−−→ M(`) −−−→ M(p) −−−→ 0

where `′ = `σ−1(p) σ−1(q).

Proof. By remark 5.1, σ−1(q) ∈ `′. We will show σ−1(p) ∈ `′ also.By hypothesis ` 6⊂ Q and so, by 4.4, the normal element Ω /∈ annM(`). Therefore,

by [LS, 2.10], Ω is a nonzero divisor in M(`) which implies ΩM(`) is isomorphic to ashifted line module. As p ∈ Q we have ΩM(p) = 0, whence ΩM(`) ⊂ K where K isthe kernel of the map M(`) → M(p). Therefore, K/ΩM(`) is isomorphic to a shiftedpoint module M(r)[−1]. Since ΩM(r) = 0 and M(r) is isomorphic to a quotient ofM(`′), it follows that r ∈ Q ∩ `′. Thus `′ = `r σ−1(q).

Next, write ` = V(u, v) and p = V(u, v, w) where u, v, w ∈ A1 and u(L) = 0. Then,as in the proof of 5.3, K = Aw and uτ w = 0. Moreover, by 1.11(c), vτw−wτv ∈ kΩ, sothat vτ w = 0 in K/ΩM(`). Hence, uτ (r) = 0 = vτ (r). Therefore, τ(r) ∈ ` ∩Q = p,giving r = τ−1(p) = σ−1(p).

The only type of line ` left to consider is one like `pq in 5.4, except assuming nowthat the point q belongs to Q ∩ L. Unfortunately, the proof in 5.4 no longer works(indeed the result is different in general) as now K/ΩM(`) ∼= M(q)[−1]. The readershould compare the next two results with [ATV2, 6.28].

Lemma 5.5. Let ` := V(u, v) be a line in P3, where u, v ∈ V . If ` meets Q ∪ L inexactly two points, p ∈ Q \ L and q ∈ Q ∩ L, then there is a short exact sequence

0 −−−→ M(`′)[−1] −−−→ M(`) −−−→ M(p) −−−→ 0

where τ−1`′ = V(ΩuΩ−1, ΩvΩ−1).

Proof. As in the proof of 5.4, K/ΩM(`) ∼= M(r)[−1] for some r ∈ Q. We claim thatr = q. Assuming this holds, we have a short exact sequence

0 −−−→ ΩM(`)[1] −−−→ K[1] ∼= M(`′) −−−→ M(q) −−−→ 0

and ΩM(`)[1] ∼= M(`Ω)[−1] for some line `Ω. Applying lemma 5.3 to this sequence, itfollows that `Ω = τ−1`′. Since A is a domain and Ω is normal, there is a linear map θ ∈

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Aut(A1) such that Ωu = uθΩ and Ωv = vθΩ. In M(`) this translates to uθΩ = vθΩ = 0where Ω is the image of Ω in M(`) ∼= A/Au+ Av. Thus

M(`Ω) ∼= ΩM(`)[2] ∼= A

Auθ + Avθ.

Hence, `Ω = V(uθ, vθ) = θ−1(`) which we write as V(ΩuΩ−1, ΩvΩ−1) (cf. [ATV2,6.28]).

It remains to show that the point r found above is the given point q. Denoting thetwo lines on Q passing through p by `1, `2, and those on Q through q by `3, `4, wedefine w, a ∈ A1 to be linear forms whose divisors of zeros on Q are (w)0 = `1 + `2 and(a)0 = `3 + `4. By the hypotheses on p, q, all the lines `i are distinct, and, relabellingif necessary, `2 ∩ `3 6= ∅. Next we take v, b ∈ A1 with respective divisors of zeros on Qto be (v)0 = `2 + `3 and (b)0 = `1 + `4. Then ` = V(v, b) and w /∈ kv⊕ kb. Since ` 6= Lwe can assume L 6⊂ V(v) (by relabelling v, b if necessary) and there exists u ∈ kv⊕ kbsuch that L ⊂ V (u) (possibly u = b). In summary, we have

` = V(u, v), p = V(u, v, w), M(`) ∼= A

Au+ Av, K = Aw

where w denotes the image of w in M(`). By construction

uτw − wτu = 0 in A, vτw − wτv ∈ kΩ, (1)

and (vτ ⊗ b − aτ ⊗ w)(σ−1(s), s) = 0 for all s ∈ ∪`i. Furthermore, as in the proof of2.8, multiplying a by a suitable scalar if necessary, we can conclude the latter holds forall s ∈ Q. Therefore

vτb− aτw ∈ kΩ. (2)

Combining (1) and (2) we find uτ w = vτ w = aτ w = 0 in K/ΩM(`), proving thatr = V(uτ , vτ , aτ ) = q.

Corollary 5.6. (to previous proof).Let ` := V(u, v) be a line in P3 where u, v ∈ V . If ` 6⊂ Q then ΩM(`) ∼= M(`′)[−2]where `′ = V(ΩuΩ−1, ΩvΩ−1).

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