Modules over the 4-dimensional Sklyanin Algebra

T. Levasseur, Département de Mathématiques, Université de Bretagne Occidentale, 29287 Brest CEDEX–France. email: levasseu@ubolib.cicb.fr

S. Paul Smith, Department of Mathematics, University of Washington, Seattle, USA 98195. email: smith@math.washington.edu

RÉSUMÉ. Cet article étudie les ‘point modules’ et ‘line modules’ sur l’algèbre définie par E.K. Sklyanin dans [17]. Ces modules sont précisément les modules de Cohen-Macaulay de multiplicité 1 et dimension de Gelfand-Kirillov 1 et 2 respectivement. Il a été démontré en [21] que les ‘point modules’ sont en bijection avec les points d’une courbe elliptique E dans ℙ^{3} augmentée de 4 autres points. On prouve ici que les ‘line modules’ sont en bijection avec les droites sécantes de E. On montre que d’autres propriétés algébriques de ces modules sont conséquences et/ou analogues de propriétés géométriques de E et des 4 points. Par exemple, si deux droites non concourantes sont sur une quadrique lisse contenant E, alors les deux modules correspondant ont le même annulateur. On démontre également que l’algèbre de Sklyanin peut être définie à l’aide des formes bilinéaires s’annulant sur une certaine sous-variété de ℙ^{3} × ℙ^{3}.

ABSTRACT. This paper studies point modules and line modules over the algebra defined by E.K. Sklyanin in [17]. It was proved in [21] that the point modules are in bijection with the points of an elliptic curve E in ℙ^{3} together with 4 other points. Here it is proved that the line modules are in bijection with the lines in ℙ^{3} which are secant lines to E. The point and line modules are precisely the Cohen-Macaulay modules of multiplicity 1, and Gelfand-Kirillov dimension 1 and 2 respectively. Further algebraic properties of these modules are shown to be consequences and analogues of the geometric properties of the elliptic curve and the 4 points. For example, if two lines lie on a smooth quadric containing E, and they do not intersect, then the two corresponding line modules have the same annihilator. It is also shown that the Sklyanin algebra may be defined in terms of the bilinear forms vanishing on a certain subvariety of ℙ^{3} × ℙ^{3}.

Key words and phrases: Point modules, line modules, Cohen-Macaulay modules, Auslander-regular rings, elliptic curve.

AMS Mathematics Subject Classification (1980): 16A33, 16A62, 14K07.

Introduction.

The 4-dimensional Sklyanin algebra is the graded algebra A = ℂ[x_{0}, x_{1}, x_{2}, x_{3}] defined by the six relations:

\begin{aligned}x_{0}x_{i} − x_{i}x_{0} & = α_{i}(x_{j}x_{k} + x_{k}x_{j}) \\ x_{0}x_{i} + x_{i}x_{0} & = x_{j}x_{k} − x_{k}x_{j} \end{aligned}

where (i, j, k) is a cyclic permutation of (1, 2, 3), and (α_{1}, α_{2}, α_{3}) ∈ ℂ^{3} lies on the surface α_{1} + α_{2} + α_{3} + α_{1}α_{2}α_{3} = 0. (One also excludes a certain finite set of points on this surface – see §1 for details). This two parameter family of algebras was defined and first studied by E.K. Sklyanin in 1982 [17]. Although the above succinct description of A suffices for the purposes of this introduction, an alternative ”better” definition (which explains the restriction on the α_{i}) is given at the start of Section 1. There A will be defined in terms of an elliptic curve ℂ/Λ, and a point τ ∈ ℂ/Λ which is not of order 4.

Sklyanin’s initial study of A showed (among other things) that A has various finite dimensional representations in spaces of theta functions, and that A has two homogeneous central elements of degree two [17], [18]. Further algebraic properties of A were obtained in [21]. It was proved there that A is a noetherian domain, and has the same Hilbert series as the polynomial ring in 4 variables, namely (1−t)^{−4}. Furthermore, A is a Koszul algebra of global homological dimension 4, and is regular in the sense of Artin and Schelter [1]. The methods in [21] follow closely those in [2] and [3]. Further homological properties of A were established in [11]; in particular A is Auslander-regular.

The purpose of this paper is to begin a study of the representation theory of A. All the results mentioned in the previous paragraph will play a key role in our analysis. Following the ideas in [2] and [3], we study three classes of A-modules: point modules, line modules and plane modules. These are defined to be cyclic modules with Hilbert series (1 − t)^{−n} where n = 1, 2, 3 respectively. Thus the Hilbert series of a line module is the same as that of the homogeneous coordinate ring of the projective line ℙ^{1}. The point modules were classified in [21]. They are in bijection with the points of a subvariety E ∪ S of ℙ^{3}, where E is a smooth elliptic curve, and S consists of 4 more points. If p ∈ E ∪ S, we write M(p) for the corresponding point module. It is rather easy to see that plane modules are in bijection with the hyperplanes in ℙ^{3}. Thus our main interest is in line modules.

We will prove that the line modules are in bijection with the set of lines in ℙ^{3} which are secant lines of E (we note that E has no trisecants). If p, q ∈ E, then M(p, q) denotes the corresponding line module. Both M(p) and M(q) are quotients of M(p, q), and the kernel of each of these surjections is again a line module. The line modules can also be characterized by their homological properties. They are precisely the Cohen-Macaulay modules of projective dimension 2, and multiplicity 1 (see §1 for definitions).

As in [21], the proof of these results is closely related to the geometry of E. The algebra A determines not just E ∪ S (as the space parametrizing the point modules), but also an automorphism σ of E ∪ S. It is shown in [21] that σ|_{S} = Id and σ|_{E} is translation p ↦ p + τ by a certain point τ ∈ E.

For this paragraph, suppose that τ is of infinite order. Then the center of A is a polynomial ring ℂ[Ω_{1}, Ω_{2}] in the two central elements found by Sklyanin. The annihilator of M(p, q) is generated by a non-zero element Ω ∈ ℂΩ_{1} ⊕ ℂΩ_{2}. Moreover, up to scalar multiples, Ω depends only on p + q, so we write Ann M(p, q) = ⟨Ω(p + q)⟩. Furthermore, if r, s ∈ E, then ℂΩ(r) = ℂΩ(s) if and only if either r = s or r+s = −2τ. As an indication of the parallels between the algebraic properties of A and the geometric properties of E ⊂ ℙ^{3} we will prove that, if z ∈ E is fixed, then all the secant lines through p and z − p lie on a common quadric containing E, and all the line modules M(p, z − p) have a common annihilator.

The results are presented as follows. Section 1 begins with a definition of the Sklyanin algebra, and shows that the defining relations have a succinct geometric description. This way of viewing the relations, and the degree two central elements of A, will be extremely useful for us. Section 1 also contains background material on homological properties (e.g. the Auslander condition) and Hilbert series of graded algebras. Section 2 gives a homological classification of point, line and plane modules: they are precisely the Cohen-Macaulay modules of multiplicity 1. Section 3 examines the geometry of the secant lines of E, and the quadrics on which they lie. Section 4 proves that the line modules are in bijection with the secant lines of E. Section 5 examines point modules, their finite dimensional simple quotients, and their relationship to line modules. For example, if p, q ∈ E then there is a short exact sequence 0 → M(p + τ, q − τ) → M(p, q) → M(q) → 0. Section 6 describes the annihilators of the line modules.

Acknowledgements. The first author would like to thank the University of Washington for its hospitality while part of this work was carried out. The second author was supported in part by NSF grant DMS-8901890, and by a US-Israel Binational Science Foundation grant 88-00130. He thanks both organisations, and the Weizmann Insitute of Science and l’Université de Bretagne Occidentale for their hospitality while part of this work was carried out.

§1. Preliminaries.

1.1. The defining equations of the Sklyanin Algebra.

We now define the Sklyanin algebra in a way which will be more useful than that given in the introduction. From our point of view (1.2b) below is the best way to define the Sklyanin algebra.

Fix, once and for all η ∈ ℂ with Im(η) > 0, and write Λ = ℤ⊕ℤη. Let θ_{00}, θ_{01}, θ_{10}, θ_{11} be Jacobi’s four theta functions associated to Λ, as defined in Weber’s book [24, p.71]. In particular,

θ_{ab}(z + 1) = (−1)^{a}θ_{ab}(z), θ_{ab}(z + η) = exp(−πiη − 2πiz − πib)θ_{ab}(z)

and the zeroes of θ_{ab} are at the points (\frac{1+b}{2}) + (\frac{1+a}{2})η + Λ. Furthermore θ_{11} is an odd function, and the other θ_{ab} are even functions.

Fix, once and for all τ ∈ ℂ, such that τ is not of order 4 in ℂ/Λ. Whenever {ab, ij, kl} = {00, 01, 10}, define

α_{ab} = (−1)^{a+b} [\frac{θ_{11}(τ)θ_{ab}(τ)}{θ_{ij}(τ)θ_{kℓ}(τ)}]^{2} ,

and set α_{1} = α_{00}, α_{2} = α_{01}, α_{3} = α_{10}. It is not difficult to show that these satisfy α_{1} + α_{2} + α_{3} + α_{1}α_{2}α_{3} = 0, and {α_{1}, α_{2}, α_{3}} ∩ {−1, 0, 1} = ∅.

Let V be a 4-dimensional vector space with basis x_{0}, x_{1}, x_{2}, x_{3}. Define A to be the quotient of the tensor algebra T (V), with defining relations as in the introduction. Thus A = T (V)/I where I is the graded ideal generated by its six dimensional subspace of degree two elements, namely I_{2} ⊂ V ⊗V. There is another ideal in T (V) which is important for us. In [17] Sklyanin found two central elements in A_{2}; see also [21, §3.9]. It will be convenient for us to take these central elements to be any two of the following:

\begin{aligned}Ω_{0} & = (1 + α_{3})x_{1}^{2} + (1 + α_{1}α_{3})x_{2}^{2} + (1 − α_{1})x_{3}^{2} \\ Ω_{1} & = (1 + α_{3})x_{0}^{2} + (α_{1}α_{3} − α_{3})x_{2}^{2} − (α_{1} + α_{3})x_{3}^{2} \\ Ω_{2} & = (1 + α_{1}α_{3})x_{0}^{2} + (α_{3} − α_{1}α_{3})x_{1}^{2} − (α_{1} + α_{1}α_{3})x_{3}^{2} \\ Ω_{3} & = (1 − α_{1})x_{0}^{2} + (α_{1} + α_{3})x_{1}^{2} + (α_{1} + α_{1}α_{3})x_{2}^{2}. \end{aligned}

(The hypothesis that τ ∉ E_{4} ensures that none of the coefficients of the x_{i}^{2} is zero.) We will write Z_{2} for the two dimensional space spanned by these elements, and define B := A/⟨Z_{2}⟩ to be the algebra obtained by quotienting out these elements. Define J to be the kernel of the map T (V) → B. Thus J is generated by its 8-dimensional subspace J_{2} ⊃ I_{2}.

For each ab ∈ {00, 01, 10, 11} define

g_{ab}(z) = γ_{ab}θ_{ab}(τ)θ_{ab}(2z) where γ_{ab} = \{ \begin{matrix}i = \sqrt{−1} & ab = 00, 11 \\ 1 & ab = 01, 10 \end{matrix}

Define j_{τ} : ℂ/Λ → ℙ(V^{∗}) = ℙ^{3} by

j_{τ} (z) = (g_{11}(z), g_{00}(z), g_{01}(z), g_{10}(z))

with respect to the homogeneous coordinates x_{0}, x_{1}, x_{2}, x_{3}. Write E = j_{τ} (ℂ/Λ). It follows from [21, §2.4] that E is defined by any two of the following quadratic forms:

\begin{aligned}g_{0} & = (1 + α_{2}α_{3})x_{1}^{2} + (1 + α_{3})x_{2}^{2} + (1 − α_{2})x_{3}^{2} \\ g_{1} & = (1 + α_{2}α_{3})x_{0}^{2} + (α_{2}α_{3} − α_{3})x_{2}^{2} + (α_{2}α_{3} + α_{2})x_{3}^{2} \\ g_{2} & = (1 + α_{3})x_{0}^{2} + (α_{3} − α_{2}α_{3})x_{1}^{2} + (α_{2} + α_{3})x_{3}^{2} \\ g_{3} & = (α_{2} − 1)x_{0}^{2} + (α_{2} + α_{2}α_{3})x_{1}^{2} + (α_{2} + α_{3})x_{2}^{2}. \end{aligned}

(The hypothesis that τ ∉ E_{4} ensures that none of the coefficients of the x_{i}^{2} is zero.)

We will show that the defining relations of A have a succinct description in terms of the geometry of E. Consider V as linear forms on ℙ(V^{∗}), and V ⊗ V as bi-homogeneous forms on ℙ(V^{∗}) × ℙ(V^{∗}). We define the following subvarieties:

e_{0} = (1, 0, 0, 0), e_{1} = (0, 1, 0, 0), e_{2} = (0, 0, 1, 0), e_{3} = (0, 0, 0, 1)

\begin{aligned}S & = \{e_{i} | 0 ≤ i ≤ 3\} \\ Δ_{S} & = \{(e_{i}, e_{i}) | 0 ≤ i ≤ 3\} \\ Δ_{τ} & = \{(p, p + τ) | p ∈ E\} \\ Γ & = Δ_{S} ∪ Δ_{τ} . \end{aligned}

Thus Γ is the graph of the automorphism σ of E ∪ S which was introduced in [21, §2.8]. Recall that σ(p) = p + τ for p ∈ E, and σ(e_{i}) = e_{i} for i = 0, 1, 2, 3. The following is one of the main results in [21, §§2,3].

Theorem 1.1. The subvarieties of ℙ(V^{∗})×ℙ(V^{∗}) defined by I_{2} and J_{2} are V(I_{2}) = Γ, and V(J_{2}) = Δ_{τ}.

Theorem 1.2.

(a) I_{2} is the subspace of J_{2} consisting of those forms which vanish on Δ_{S}. Thus I_{2} is the subspace of V ⊗ V consisting of those forms which vanish on Γ = Δ_{τ} ∪ Δ_{S}.

(b) If f ∈ J_{2} vanishes at two points of Δ_{S} then f ∈ I_{2}. Proof. (a) Since I_{2} vanishes on Γ, we need to show that if f ∈ J_{2} vanishes on Δ_{S},

then f ∈ I_{2}. Let Ω̃_{1} and Ω̃_{2} be preimages of the central elements Ω_{1} and Ω_{2} in V ⊗ V. Thus J_{2} = I_{2} ⊕ ℂΩ̃_{1} ⊕ ℂΩ̃_{2}. Since Ω̃_{1}, Ω̃_{2} ∈ ⨁_{0≤i≤3} ℂ(x_{i} ⊗ x_{i}) never vanish at all the points e_{i}, if f ∈ J_{2} and f(Δ_{S}) = 0, then f ∈ I_{2}.

(b) The result follows from the fact that a non-zero element in ℂΩ̃_{1} ⊕ ℂΩ̃_{2} cannot be a linear combination of just two of the x_{i} ⊗ x_{i}. ■

Remarks. 1. In Section 3 it will be proved that the points e_{i} ∈ S are the only points in ℙ^{3} which lie on infinitely many secant lines of E. Thus S is determined by E, and hence Γ ⊂ ℙ^{3} × ℙ^{3} is determined by E and τ. Therefore (1.2a) gives a succinct geometric description of the defining relations of A.

2. We shall use (1.2) as a way of recognizing non-zero central elements in A_{2}. If f ∈ V ⊗ V vanishes on Δ_{τ} but not on Δ_{S}, then the image of f in A is a non-zero central element. Furthermore if two elements of J_{2} vanish at a common point (e_{i}, e_{i}) ∈ Δ_{S}, then the elements are scalar multiples of one another.

1.2. Homological properties of the Sklyanin Algebra.

Let k be a field. For the sake of simplicity the algebras considered in this section will be noetherian graded k-algebras of the form A = k ⊕ A_{1} ⊕ A_{2} ⊕ . . .. The dimension of a k-vector space E is denoted by dimE.

We denote by Mod(A), respectively Mod^{g}(A), the category of left or right A-modules, respectively ℤ-graded A-modules. The subcategories of finitely generated A-modules will be denoted by Mod_{f}(A) and Mod_{f}^{g}(A), respectively. Let M = ⨁_{m} M_{m} be in Mod^{g}(A) and p ∈ ℤ. The shifted module M[p] is defined by setting M[p]_{m} := M_{p+m}. If M and N are in Mod^{g}(A), then HOM_{A}(M, N) denotes the ℤ-graded group such that HOM_{A}(M, N)_{p} = {ϕ : M → N | ϕ is A-linear and ϕ(M_{m}) ⊂ M_{m+p} for all m}. It is well know that if M ∈ Mod_{f}^{g}(A), then HOM_{A}(M, N) coincides with the usual Hom_{A}(M, N). In that case the derived functors of HOM_{A}(M, −) and Hom_{A}(M, −) are the same, namely Ext_{A}^{q}(M, −).

When not otherwise specified a map M →^{ϕ} N between modules of Mod^{g}(A) will be an element of HOM_{A}(M, N)_{0}, i.e. ϕ(M_{m}) ⊂ N_{m} for all m.

The projective dimension of M ∈ Mod(A) is denoted by pd(M). The algebra A is said to have finite global homological dimension if d = sup{pd(M) | M ∈ Mod(A)} is finite. In this case we write gldim(A) = d. We say that A has finite injective dimension if the left and the right A-module A both have finite injective dimension. They are then equal (because A is noetherian) and we set μ = injdim(A).

Let _{A}M be in Mod_{f}^{g}(A) with p = pd(M) < ∞. There exists a minimal graded free resolution of length p of M. That is: P_{•} → M → 0, P_{j} = 0 if j > p, P_{j} = ⨁_{i} A[−i]^{a_{ij}}, and ∂_{j} : P_{j+1} → P_{j} is given by a matrix where the non-zero entries are elements of A of positive degree. We call P_{•} →^{∂} M → 0 the minimal resolution of M.

The dual of M is an element M^{D} of the derived category D_{r}^{b}(A) of bounded complexes of finitely generated graded right modules. It is defined by RHom_{A}(M, A) and is represented by 0 ← P_{•}^{∨} ← …, where P_{•}^{∨} = Hom_{A}(P_{•}, A). The cohomology of this complex gives the groups Ext_{A}^{q}(M, A) ∈ Mod_{f}^{g}(A). The Hilbert series of M is defined to be h_{M}(t) := ∑_{m}(dim M_{m})t^{m} and is then equal to ∑_{j}(−1)^{j}h_{P_{j}}(t) = ∑_{j}(−1)^{j}[∑_{i} a_{i,j}t^{i}h_{A}(t)].

Let M = ⨁_{m} M_{m} be in Mod_{f}^{g}(A). Define a function f_{M} : ℤ → ℕ by f_{M}(n) := ∑_{m≤n} dim M_{m}. From now on we shall assume that, for all M ∈ Mod_{f}^{g}(A), the growth of f_{M}, as defined in [9, Chapter 1], is a polynomial of degree s. The Gelfand-Kirillov dimension of M, GKdim(M), is defined to be s = \overline{lim}_{n} \frac{log f(n)}{log n}. A module of Gelfand-Kirillov dimension s is said to be s-homogeneous if every non-zero submodule is also of Gelfand-Kirillov dimension s. A module is s-critical if it is of Gelfand-Kirillov dimension s and all its proper quotients have strictly smaller Gelfand-Kirillov dimension.

We shall be interested in the case where h_{M}(t) = g_{M}(t)(1 − t)^{−d} for some fixed d ∈ ℕ and g_{M}(t) ∈ ℤ[t, t^{−1}]. As in [3, 2.21] one can see that f_{M} has polynomial growth of degree s which is the order of the pole at t = 1 of h_{M}(t). Hence h_{M}(t) = g_{M}(t)(1 − t)^{−s}, where g_{M}(t) ∈ ℤ[t, t^{−1}], and the multiplicity of M is defined to be e(M) := g_{M}(1) = [(1 − t)^{s}h_{M}(t)]_{t=1} ∈ ℕ.

Let M be in Mod_{f}(A). The grade of M is the element j(M) ∈ ℕ ∪ {+∞} defined by j(M) := inf{i | Ext_{A}^{i}(M, A) ≠ 0}. When injdim(A) < ∞ and M ≠ 0 we have j(M) < ∞, (see [10, Theorem 3.1] for instance). We say that M is pure if j(N) = j(M) for all nonzero submodules N of M. When n = j(M) we say that M is n-pure. We abbreviate Ext_{A}^{q}(M, A) by E_{A}^{q}(M) or E^{q}(M).

Definition. The algebra A is Auslander-Gorenstein, respectively Auslander-regular, of dimension μ if

(a) inj.dim(A) = μ < ∞, respectively gldim(A) = μ < ∞, and

(b) for all M ∈ Mod_{f}(A), and for all q ≥ 0, j(N) ≥ q for every A-submodule N of E^{q}(M).

Remarks. 1. Condition (b) is called the Auslander condition. It implies that E^{p}(E^{q}(M)) = 0 for all p < q, which may be taken as the definition of a Gorenstein ring in the commutative case. This condition is discussed in detail in [6].

2. By [8, Theorem 0.1] one can replace Mod_{f}(A) by Mod_{f}^{g}(A) in the definition.

3. Assume A is Auslander-Gorenstein. Then the grade number j(M) is exact on short exact sequences: that is, if 0 → M′ → M → M′′ → 0 is exact then j(M) = inf{j(M′), j(M′′)}, (see [6,§1.8]).

Proposition 1.3. [11] Let A be Auslander-regular. Then A is a domain and is regular in the sense of Artin-Schelter (cf. [1] and [2, 2.12]).

The following summarizes results in [5, Chapter 2], [6, §1], [7], [10, §4], and [12].

Theorem 1.4. Let A be an Auslander-Gorenstein algebra of dimension μ, and let M be a non-zero finitely generated A-module.

(a) There exists a convergent spectral sequence in Mod_{f}(A):

E_{2}^{p,−q} := E_{A}^{p}(E_{A}^{q}(M)) ⟹ ℍ^{n}

where ℍ^{n} = 0 if n ≠ 0 and ℍ^{0} = M. The resulting filtration on M has the form:

0 = F^{μ+1}M ⊂ F^{μ}M ⊂ … ⊂ F^{1}M ⊂ F^{0}M = M.

(b) There is an exact sequence

0 → \frac{F^{p}M}{F^{p+1}M} → E^{p}(E^{p}(M)) → Q(p) → 0

where Q(p) is a subquotient of ⨁_{i≥1} E^{p+i+1}(E^{p+i}(M)) and satisfies j(Q(p)) ≥ p + 2.

(c) F^{p}M is the largest submodule X of M such that j(X) ≥ p. In particular j(M) = max{p | F^{p}M = M}.

(d) E^{j(M)}(M) is pure and E^{p}(E^{p}(M)) is either 0 or p-pure.

Remark. When M ∈ Mod_{f}^{g}(A) the spectral sequence in (1.4a) takes place in Mod_{f}^{g}(A). Hence the filtration F^{•}M consists of finitely generated graded submodules of M.

Definition. Let μ be an integer. We say that A satisfies the Cohen-Macaulay property (CM-property for short) if GKdimM + j(M) = μ for all 0 ≠ M ∈ Mod_{f}(A).

Remarks. 1. If A satisfies the Cohen-Macaulay property, then GKdimA = μ.

2. If (A, 𝔐) is a commutative noetherian local ring, then A is Cohen-Macaulay if and only if Kdim(M) + j(M) = Kdim(A) for all 0 ≠ M ∈ Mod_{f}(A) (here Kdim is the Krull dimension). This explains the terminology in the definition.

Corollary 1.5. Suppose that A is Auslander-Gorenstein of dimension μ. Let 0 ≠ M ∈ Mod_{f}(A) and suppose that GKdim(A) = μ. If A satisfies the Cohen-Macaulay property, then:

(a) M is n-pure if and only if M is (μ − n)-homogeneous

(b) The submodule F^{p}M is the largest submodule of M of GK-dimension ≤ μ − p.

(c) If N is a submodule of E_{A}^{q}(M), then GKdim(N) ≤ μ − q.

Proof. The assertions follow easily from the definitions and Theorem 1.4. ■

There is an example of an Auslander-Gorenstein algebra which is of particular importance for us. For simplicity assume that k is algebraically closed.

Let E be a smooth elliptic curve, L an invertible sheaf of degree ≥ 3 on E, σ a k-automorphism of E. As in [2], [4], or [21] one can construct a graded algebra B := B(E, σ, L) = k ⊕ B_{1} ⊕ B_{2} ⊕ …, which for σ = Id_{E} is the homogeneous coordinate ring of the projective embedding E ↪ ℙ(H^{0}(E, L)^{∗}). By [2] and [4] it is known that B is a noetherian graded algebra generated by B_{1}, GKdim(B) = 2, and B is a domain.

The next two results are proved in [11].

Proposition 1.6. [11] Let B = B(E, σ, L) be as above. Then B is Auslander-Gorenstein of dimension 2 and satisfies the Cohen-Macaulay property.

Remarks. 1. This result (which is well known in the commutative case viz. σ = Id_{E}) is based on results of A. Yekutieli [25].

2. It is easily seen that if M ∈ Mod_{f}^{g}(B) then the Hilbert series h_{M}(t) is of the form q_{M}(t)(1 − t)^{−2} for some q_{M}(t) ∈ ℤ[t, t^{−1}].

Definition. An element Ω ∈ A, is said to be normal if ΩA = AΩ. Recall also that a sequence {Ω_{1}, …, Ω_{ℓ}} ⊂ A is a regular normalizing (respectively centralizing) sequence if Ω_{i+1} is a regular (i.e. non zero divisor) normal (respectively central) element in the ring A/⟨Ω_{1}, …, Ω_{i}⟩, for all 0 ≤ i ≤ ℓ − 1.

Theorem 1.7. [11] Suppose that {Ω_{1}, …, Ω_{ℓ}} is a regular normalizing sequence of homogeneous elements of positive degree in A. If B = A/⟨Ω_{1}, …, Ω_{ℓ}⟩ is Auslander-Gorenstein of dimension ν and satisfies the Cohen-Macaulay property, then A is Auslander-Gorenstein of dimension μ = ν + ℓ and satisfies the Cohen-Macaulay property.

Corollary 1.8. Suppose that A contains a regular normalizing sequence {Ω_{1}, …, Ω_{ℓ}} of homogeneous elements of positive degree, such that B := A/⟨Ω_{1}, …, Ω_{ℓ}⟩ ≅ B(E, σ, L) for some triple (E, σ, L) as above. If gldim(A) < ∞, then A is Auslander-regular of dimension ℓ + 2 and satisfies the Cohen-Macaulay property: GKdim(M) + j(M) = ℓ + 2 for all 0 ≠ M ∈ Mod_{f}(A).

By [21, 5.4], Corollary 1.8 applies to the Sklyanin algebra. Since we are mainly interested in modules over this algebra we isolate:

Corollary 1.9. The 4-dimensional Sklyanin algebra is an Auslander-regular ring of dimension 4 which satisfies the Cohen-Macaulay property.

Remark. There are some other cases where (1.8) can be applied. In [2,§2] some “regular” algebras are introduced which have the property A/⟨Ω⟩ ≅ B(E, σ, L) for a central element Ω ∈ A_{3} and a triple (E, σ, L) with degL = 3. In [23], J.T. Stafford modifies the construction of the Sklyanin algebra to construct some families of “regular” algebras which satisfy A/⟨Ω_{1}, Ω_{2}⟩ ≅ B(E, σ, L), where {Ω_{1}, Ω_{2}} is a regular normalizing (not centralizing in general) sequence of quadratic elements and degL = 4.

For the rest of Section 1 we make the following assumptions on A: (a) the Hilbert series of A is h_{A}(t) = (1 − t)^{−μ},

(b) A is Auslander-regular of dimension μ,

(c) A satisfies the Cohen-Macaulay property.

The 4-dimensional Sklyanin algebra satisfies these conditions.

As noticed in (1.3), A is then regular in the sense of Artin-Schelter and by [3, §2.18] every M ∈ Mod_{f}^{g}(A) satisfies h_{M}(t) = q_{M}(t)(1 − t)^{−s}, for some q_{M}(t) ∈ ℤ[t, t^{−1}] such that q_{M}(1) ≠ 0. Furthermore in this situation s = GKdim(M) = μ − j(M) whenever M ≠ 0.

If 0 ≠ M ∈ Mod_{f}^{g}(A) with n = j(M), then we define the dual of M to be M^{∨} := E^{n}(M). When A is Auslander-Gorenstein M^{∨} is an n-pure graded right module, so (1.4b) gives a natural map 0 → M/F^{n+1}M → M^{∨∨}. This map generalizes the well known map from M modulo its torsion to the bidual of M.

Definition. A non-zero module M ∈ Mod_{f}^{g}(A) is called Cohen-Macaulay (CM for short) if pd(M) = j(M), i.e. E_{A}^{i}(M) = 0 if i ≠ n = j(M).

Proposition 1.10. The duality M → M^{∨} gives a bijection between left and right Cohen-Macaulay-modules of projective dimension n. In particular M ≅ M^{∨∨}, h_{M^{∨}}(t) = (−1)^{μ−n}t^{−μ}h_{M}(t^{−1}), and e(M^{∨}) = e(M).

Proof. We first show that M^{∨} is CM. We have E^{i}(E^{n}(M)) = 0 if i < n by the Auslander condition and E^{n}(E^{n}(M)) ≠ 0 by (1.4). Thus j(M^{∨}) = n. We must prove E^{j}(M^{∨}) = E^{j}(E^{n}(M)) = 0 if j > n. Recall the spectral sequence of (1.4):

E_{2}^{p,−q} = E^{p}(E^{q}(M)) ⟹ ℍ^{p−q} = \{ \begin{matrix}0  p ≠ q \\ M  p = q \end{matrix}

It follows that E_{∞}^{p,−q} = 0 if p > q. If r ≥ 2, the differential d_{r} in the spectral sequence gives:

E_{r}^{j−r,−n+(r−1)} ⟶^{d_{r}} E_{r}^{j,−n} ⟶^{d_{r}} E_{r}^{j+r,−n−(r−1)}.

Since n ∓ (r − 1) ≠ n, E^{n∓(r−1)}(M) = 0. Hence E_{r}^{j+r,−n±(r−1)} = 0 for all r ≥ 2 and it follows that E_{2}^{j,−n} = E_{3}^{j,−n} = … = E_{∞}^{j,−n} = 0 for all j > n. Thus E^{j}(E^{n}(M)) = E_{2}^{j,−n} = 0 if j > n.

Because E^{p}(M) = 0 when p ≠ n, the filtration on M has the form: M = F^{0}M = … = F^{n}M⫌F^{n+1}M = … = F^{μ}M = 0. By (1.4b) M = F^{n}M ≅ M^{∨∨} = E^{n}(E^{n}(M)). Since M is CM we have h_{M^{D}}(t) = (−1)^{n}h_{M^{∨}}(t). From [3, §2.35,§2.36] we deduce that h_{M^{D}}(t) = (−1)^{μ}t^{−μ}h_{M}(t^{−1}), and e(M^{D}) = (−1)^{n}e(M) (recall μ = n + GKdimM) so the proof is complete. See also the remark after (1.12). ■

Corollary 1.11. Let M ∈ Mod_{f}^{g}(A) be a Cohen-Macaulay module. Set n = j(M), and m = GKdimM. Then M is n-pure, (equivalently m-homogeneous). Furthermore, if e(M) = 1, then M is critical for GK-dimension.

Proof. By (1.9) M = (M^{∨})^{∨} = E^{n}(M^{∨}), so the first assertion follows from (1.4d) and the CM-property. The second assertion is then obvious by additivity of the multiplicity.■

Lemma 1.12. Let 0 → M′ → M → N → 0 be an exact sequence in Mod_{f}^{g}(A). Suppose that j(M) = n, and j(N) = n + 1.

(a) If M and N are Cohen-Macaulay then so is M′

(b) If M and M′ are Cohen-Macaulay then so is N.

Proof. Use the definition and the long exact sequence in cohomology associated to 0 → M′ → M → N → 0. ■

Remark. Let P_{•} → M → 0 be the minimal resolution of a Cohen-Macaulay module M. Then P_{•}^{∨} → M^{∨} → 0 is the minimal resolution of M^{∨}, where P_{•}^{∨} = Hom_{A}(P_{•}, A). This follows from the following two observations. Firstly, if ∂_{j} : P_{j+1} = ⨁_{i} A[−i]^{a_{i,j+1}} → P_{j} = ⨁_{i} A[−i]^{a_{i,j}} is the differential in P_{•}, its dual ∂_{j}^{∨} : P_{j}^{∨} = ⨁_{i} A[i]^{a_{i,j}} → P_{j+1}^{∨} = ⨁_{i} A[i]^{a_{i,j+1}} is also given by a matrix where entries have positive degree. Secondly, by definition of a CM-module, the complex (P_{•}^{∨}, ∂^{∨}) has cohomology only in the top degree n = j(M) and this cohomology is M^{∨}.

¿From this remark and the fact that h_{A}(t) = (1 − t)^{−μ}, one obtains the equalities h_{M^{∨}}(t) = (−1)^{μ−n}t^{−μ}h_{M}(t^{−1}) and e(M^{∨}) = e(M) of (1.10).

We shall be interested in Cohen-Macaulay modules of multiplicity 1 over the Sklyanin algebra. It may be useful to recall the situation for a commutative polynomial ring in μ variables (which satisfies the assumptions made after (1.9) which are in force for the remainder of this section). Because there seems to be no suitable reference we include a proof of the following:

Theorem 1.13. Let A = k[X_{1}, …, X_{μ}] be a polynomial ring in μ variables, graded by setting deg X_{i} = 1. Let M be in Mod_{f}^{g}(A). Then the following are equivalent:

(a) M is cyclic with Hilbert series t^{p}(1 − t)^{−d} for some p ∈ ℤ.

(b) there exists p ∈ ℤ such that M[p] ≅ A/⟨Y_{1}, …, Y_{μ−d}⟩ where Y_{1}, …, Y_{μ−d} are linearly independent elements of A_{1}.

(c) M is a Cohen-Macaulay module, GKdim(M) = d and e(M) = 1. Proof. Replacing M by a suitable shift M[p] if necessary, we can assume that M =

⊕_{m≥0}M_{m} with M_{0} ≠ 0.

(a) ⟹ (b) Write M ≅ A/I for some graded ideal I. Since h_{M}(t) = (1 − t)^{−d} we have dim I_{1} = μ − d. If {Y_{1}, …, Y_{μ−d}} is a basis of I_{1} we get a natural graded surjective map: A/⟨Y_{1}, …, Y_{μ−d}⟩ →^{ϕ} M. Since these two modules have the same Hilbert series, ϕ is an isomorphism.

(b) ⟹ (c) This is easy and well known (for instance use (1.12b)).

(c) ⟹ (a) Notice that (a) and (c) remain true under base extension. If k ⊂ L is a field extension, set A_{L} = L ⊗_{k} A = L[X_{1}, …, X_{μ}], and M_{L} = L ⊗_{k} M = ⨁_{m=0}^{∞}(L ⊗_{k} M)_{m} ∈ Mod_{f}^{g}(A_{L}). We make three useful observations. Firstly M is CM if and only if M_{L} is CM (because A ↪ A_{L} is faithfully flat). Secondly h_{M}(t) = h_{M_{L}}(t), so GKdimM_{L} = GKdimM and e(M_{L}) = e(M). Thirdly M_{L}/⟨X_{1}, …, X_{μ}⟩M_{L} ≅ L ⊗_{k} M/⟨X_{1}, …, X_{μ}⟩M so M is cyclic if and only if M_{L} is cyclic.

Thus to prove the implication we may assume that k is an uncountable field.

1. If m = (m_{1}, …, m_{μ}) ∈ ℕ^{μ} with m_{1} < m_{2} < … < m_{μ} we put f_{m}(X) = det[X_{i}^{jm_{i}}]_{1≤i,j≤μ} = ∏_{1≤i≤μ} X_{i}^{m_{i}} ∏_{1≤i<j≤μ}(X_{j}^{m_{j}} − X_{i}^{m_{i}}). Since k is uncountable we can find 0 ≠ λ ∈ k which is not a root of unity. If we set λ = (λ, λ, …, λ) ∈ ℕ^{μ} then f_{m}(λ) ≠ 0 for all m as above.

2. Let 𝔐_{1}, …, 𝔐_{t} be proper ideals of A and assume A_{1} ⊂ ⋃_{j=1}^{t} 𝔐_{j}. Choose λ as above and notice that the set S = {λ^{m}X_{1} + λ^{2m}X_{2} + … + λ^{μm}X_{μ} | m ∈ ℕ} is infinite. Since S ⊂ A_{1} ⊂ ⋃_{j=1}^{t} 𝔐_{j} there exists ℓ ∈ {1, …, t} such that S ∩ 𝔐_{ℓ} is infinite. Choose m = (m_{1} < m_{2} < … < m_{μ}) ∈ ℕ^{μ} such that ∑_{i=1}^{μ} λ^{im_{j}}X_{i} ∈ 𝔐_{ℓ} for all j ∈ {1, …, μ}. Since f_{m}(λ) ≠ 0 we deduce that X_{i} ∈ 𝔐_{ℓ} for all i, that is 𝔐_{ℓ} = ⟨X_{1}, …, X_{μ}⟩.

3. We now prove the implication by induction on d = GKdimM. If d = 0, then dim M = e(M) = 1, hence M = M_{0} = A/⟨X_{1}, …, X_{μ}⟩. If d ≥ 1, denote by 𝔐_{1}, …, 𝔐_{t} the associated prime ideals of M. Recall that ⋃_{j=1}^{t} 𝔐_{j} is the set of zero-divisors in M. Suppose A_{1} ⊂ ⋃_{j=1}^{t} 𝔐_{j}. By part 2 we get 𝔐_{ℓ} = ⟨X_{1}, …, X_{μ}⟩ for some ℓ. It follows thatA/⟨X_{1}, …, X_{μ}⟩ ↪ M which implies pd(M) = μ, i.e. d = 0, which is a contradiction. Therefore we can find a ∈ A_{1} which is a non-zero divisor in M. The graded module \overline{M} = M/aM is CM with Hilbert series h\overline{M}(t) = (1 − t)h_{M}(t), and \overline{M}/⟨X_{1}, …, X_{μ}⟩\overline{M} = M/⟨X_{1}, …, X_{μ}⟩M. By induction \overline{M} is cyclic, h\overline{M}(t) = (1 − t)^{1−d}, and from above we conclude that M is cyclic and h_{M}(t) = (1 − t)^{−d}. ■

§2. Homological Characterization of Point, Line and Plane Modules.

Throughout this section we will assume that A is a ℂ−algebra of finite global dimension containing a regular normalizing sequence {Ω_{1}, Ω_{2}}, where Ω_{1}, Ω_{2} ∈ A_{2} are such that B := A/⟨Ω_{1}, Ω_{2}⟩ ≅ B(E, σ, L) (graded isomorphism) for some (E, σ, L) with L of degree 4. As we noticed after (1.8) the Skylyanin algebra is an example of such a ring; also see the remark after (1.9).

Since h_{B}(t) = (1 + t)^{2}(1 − t)^{−2} = (1 − t^{2})^{2}h_{A}(t) we have h_{A}(t) = (1 − t)^{−4}. By (1.8) A is Auslander regular of dimension 4 and satisfies the Cohen-Macaulay property. Furthermore, since B is generated by B_{1}, the algebra A is generated by A_{1}. Hence if 𝔐 = ⨁_{m≥1} A_{m} is the augmentation ideal we have: 𝔐 = A_{1}A = AA_{1}.

The following proposition provides a version of [3, Theorem 4.1, Corollary 4.2].

Proposition 2.1 Let 0 ≠ M ∈ Mod_{f}^{g}(A), set n = j(M), and m = GKdimM. Then (a) n + m = 4;

(b) M^{∨} = E^{n}(M) is m-homogeneous and e(M^{∨}) = e(M);

(c) GKdimE^{q}(M) ≤ 4 − q and the following are equivalent: (i) GKdimE^{q}(M) = 4 − q, (ii) E^{q}(E^{q}(M)) ≠ 0, (iii) GKdimN = 4 − q for some submodule 0 ≠ N ⊆ M;

(d) There is a canonical map M →^{ϕ} M^{∨∨} and an exact sequence 0 → M/kerϕ → M^{∨∨} → Q → 0 with Kerϕ equal to the maximal submodule of M of GKdim < m, and GKdimQ ≤ m − 2;

(e) The following are equivalent:

(i) pd(M) < 4, (ii) the socle of M is Soc(M) = Hom_{A}(A/𝔐, M) = 0, (iii) E^{4}(M) = 0, (iv) the torsion submodule of M is T (M) := {x ∈ M | 𝔐^{i}x = 0 for some i} = 0; furthermore E^{4}(M) ≅ E^{4}(T (M)) is a finite dimensional vector space of the same dimension as T (M) and E^{i}(M) ≅ E^{i}(M/T (M)) for all i < 4;

(f) If m ∈ {0, 1, 2}, M^{∨} is Cohen-Macaulay. If m = 3, E^{j}(M^{∨}) = 0 for j = 0, 3, 4 and E^{2}(M^{∨}) ⊂ E^{4}(E^{2}(M)) is a finite dimensional vector space. Proof. (a) This is the Cohen-Macaulay property.

(b) This follows from (1.4) and [3,§2.8] using the inequality GKdimE^{q}(M) ≤ 4 − q of (c).

(c) The inequality is (1.5c). By §1.2, GKdimE^{q}(M) = 4 − q ⇔ j(E^{q}(M)) = q ⇔ E^{q}(E^{q}(M)) is q-pure ⇔ F^{q}M/F^{q+1}M ≠ 0 ⇔ M contains a submodule of GK-dimension 4 − q.

(d) The map ϕ is the composition of M/F^{n+1}M → M^{∨∨} with the natural projection M → M/F^{n+1}M. Hence Kerϕ = F^{n+1}M, so (d) is a consequence of (1.4) and (1.5).

(e) This is proved in [3,§2.46].

(f) We shall use the notation in the proof of (1.10). Recall that the spectral sequence E_{2}^{p,−q} ⇒ ℍ^{p−q} implies E_{∞}^{p,−q} = 0 if p − q ≠ 0. Furthermore E_{r+1}^{p,−q} = Kerd_{r}/Imd_{r}. If r ≥ 2, then n + 1 − r < n so E_{2}^{p−r,−n−1+r} = 0 and E_{r+1}^{p,−n} = Ker(d_{r}). By the Auslander condition E^{p}(M^{∨}) = E^{p}(E^{n}(M)) = 0 if p < n = 4 − m.

Suppose that p > n, and that m ∈ {0, 1, 2} or equivalently n ≥ 2. If r ≥ 2, then p + r > 4, so E_{2}^{p−r,−n−1+r} = 0 because gldim(A) = 4. It follows that E_{2}^{p,−n} = E_{3}^{p,−n} = ⋯ = E_{∞}^{p,−n} = 0.

Suppose that m = 3, or equivalently that n = 1. When p ≥ 3 and r ≥ 2 the previous argument shows that E_{2}^{p,−n} = ⋯ = E_{∞}^{p,−n} = 0 because p + r ≥ 5. Assume p = 2. If r ≥ 3

then 2 + r ≥ 5 so 0 = E_{2}^{2+r,−1}, whence 0 = E_{∞}^{2,−1} = ⋯ = E_{3}^{2,−1} = Ker(E_{2}^{2,−1}⟶^{d_{2}}E_{2}^{4_{,}−2}).

Thus E_{2}^{2,−1} = E^{2}(M^{∨}) is a submodule of E^{4}(E^{2}(M)) = E_{2}^{4,−2}, hence is finite dimensional by (e). ■

We want to study Cohen-Macaulay modules with multiplicity 1. There are two easy cases. If M is Cohen-Macaulay and GKdimM = 0 then M ≅ k[p], for some p ∈ ℤ. If M is Cohen-Macaulay, and GKdimM = 4 then M ≅ A[p], for some p ∈ ℤ. Therefore we shall only consider the case 1 ≤ GKdimM ≤ 3. If A is a commutative polynomial ring in 4 variables, Theorem 1.13 shows that these modules are (up to a shift) in bijection with the linear subvarieties of ℙ(A_{1}^{∗}) ≅ ℙ^{3}. On the other hand [3] describes such modules over a regular algebra A of dimension 3 with A/⟨Ω⟩ ≅ B(E, σ, L) where Ω is a central cubic element and L has degree 3. They can be classified (up to a shift) as follows:

–If GKdimM = 1, M is a “point module”, i.e. M is cyclic, and h_{M}(t) = (1 − t)^{−1}. Point modules are parametrized by the points of E ⊂ ℙ^{2} ≅ ℙ(A_{1}^{∗}). See [3,§6.17].

–If GKdimN = 2, M is a “line module”, i.e. M is cyclic, h_{M}(t) = (1 − t)^{−2}. Line modules are parametrized by the lines in ℙ^{2} ≅ ℙ(A_{1}^{∗}). See [3,§6.1].

In view of these remarks we make the following definition:

Definition. Let M be in Mod_{f}^{g}(A). We say that M is a plane module if M is cyclic and h_{M}(t) = (1 − t)^{−3}, M is a line module if M is cyclic and h_{M}(t) = (1 − t)^{−2}, M is a point module if M is cyclic and h_{M}(t) = (1 − t)^{−1}.

A shift of a plane, respectively line or point, module is a cyclic module with Hilbert series t^{p}(1 − t)^{−i}, p ∈ ℤ, i = 3, 2, 1 respectively. The main result in this section is the following.

Theorem 2.2. The module M ∈ Mod_{f}^{g}(A) is a shift of a plane, respectively line, respectively point, module if and only if M is a Cohen-Macaulay module, e(M) = 1 and GKdimM = 3, respectively 2, respectively 1.

The proof of Theorem 2.2 occupies the rest of §2. We shall examine separately the different types of modules. Because of the following consequence of (1.10) and (1.11), we will work with left A−modules unless otherwise specified.

Corollary 2.3. The duality M → M^{∨} gives a bijection between left and right shifted plane, respectively line or point, modules. These modules are critical.

We first investigate the case of plane modules. Proposition 2.5 below is a particular case of [3,§2.43]. For the convenience of the reader we include a proof of it in our setting. We first begin with a lemma taken from [3, §2.41].

Lemma 2.4. Let f : _{⨁i}A[−i]^{b_{i}} ⟶ ⨁_{i} A[−i]^{a_{i}} be an injective linear map of degree 0 between finitely generated graded free modules. Assume that the non-zero matrix entries of f all have positive degree. Then b_{j} ≤ ∑_{i<j}(a_{i} − b_{i}) for all j.

Proof. The assumptions on f imply that f(⨁_{i≤j} A[−i]^{b_{i}}) ⊂ ⨁_{i<j} A[−i]^{a_{i}}. If X is the quotient of these two modules, then h_{X}(t) = (1 − t)^{−4}(∑_{i<j} a_{i}t^{i} − ∑_{i≤j} b_{i}t^{i}). Since [(1 − t)^{GKdimX}h_{X}(t)]_{t=1} = e(X) ≥ 0, we have 0 ≤ [(1 − t)^{4}h_{X}(t)]_{t=1} = ∑_{i<j} a_{i} − ∑_{i≤j} b_{i}. The result follows. ■

Proposition 2.5. If M ∈ Mod_{f}^{g}(A) the following are equivalent:

(a) M is a shift of a plane module;

(b) M ≅ (A/Aa)[p], for some p ∈ ℤ, where 0 ≠ a ∈ A_{1}.

(c) M is Cohen-Macaulay, GKdimM = 3 and e(M) = 1. Proof. We may assume that M = M_{0} ⊕ M_{1} ⊕ M_{2} ⊕ ⋯, and M_{0} ≠ 0.

(a) ⇒ (b) We have dimM_{0} = 1, dimM_{1} = 3, and dimA_{1} = 4. Hence there exists 0 ≠ a ∈ A_{1} such that aM_{0} = 0. It follows that M = AM_{0} is a quotient of A/Aa. But A is a domain, so h_{A/Aa}(t) = (1 − t)^{−3} = h_{M}(t). Thus M = A/Aa.

(b) ⇒ (c). The short exact sequence 0 → A[−1] ⟶^{×a} A ⟶ M → 0 is the minimal resolution of M and h_{M}(t) = (1 − t)^{−3}. This shows that GKdimM = 3, e(M) = 1, and pd(M) = 1.

(c) ⇒ (a). Let 0 → ⨁_{i} A[−i]^{b_{i}} →^{∂} ⨁_{i} A[−i]^{a_{i}} → M → 0 be the minimal resolution of M. As noticed in §1, 0 → ⨁_{i} A[i]^{a_{i}} →^{f=∂^{∨}} ⨁_{i} A[i]^{b_{i}} → M^{∨} → 0 is the minimal resolution of M^{∨}. By Lemma 2.4 applied to f we have a_{j} ≤ ∑_{i>j}(b_{i} − a_{i}). Therefore

(∗) ∑_{i≥1} i(b_{i} − a_{i}) = ∑_{i≥1}(b_{i} − a_{i}) + ∑_{i≥2}(b_{i} − a_{i}) + ∑_{i≥3}(b_{i} − a_{i}) + ⋯ ≥ ∑_{j≥0} a_{j} ≥ 1.

It follows from the presentation of M that

h_{M}(t) = \frac{1}{(1 − t)^{4}} ∑_{i≥0}(a_{i} − b_{i})t^{i} = \frac{1}{(1 − t)^{4}} ∑_{j≥0}(−1)^{j} (∑_{i≥j} ({i \atop j})(a_{i} − b_{i})) (1 − t)^{j}.

Since GKdimM = 3 and e(M) = 1, this implies that ∑_{i≥0}(a_{i} − b_{i}) = 0 and ∑_{i≥1} i(b_{i} − a_{i}) = 1. By (∗) it follows that ∑_{j≥0} a_{j} = 1. Thus ∑_{j≥0} b_{j} = 1.

But a_{0} ≠ 0 because M_{0} ≠ 0. Hence a_{0} = 1 and a_{i} = 0 if i ≥ 1. Thus 1 = ∑_{i≥1} i(b_{i} − a_{i}) = ∑_{i≥1} ib_{i}. But ∑_{j≥0} b_{j} = 1 so b_{1} = 1, and b_{i} = 0 if i ≠ 1. Hence the minimal resolution takes the form 0 → A[−1] → A → M → 0, which proves (a). ■

We now consider the point modules.

Proposition 2.6. Let M be a Cohen-Macaulay module, with GKdimM = 1 and e = e(M). Then

(a) h_{M}(t) = t^{p} (∑_{0≤i≤d−1} f_{i}t^{i} + et^{d}(1 − t)^{−1}) for some p ∈ ℤ, d ∈ ℕ, and f_{i} ∈ ℕ such that 1 ≤ f_{i} ≤ e − 1 for all i ∈ {0, ⋯, d − 1}.

(b) If e = 1, then h_{M}(t) = t^{p}(1 − t)^{−1}, p ∈ ℤ.

(c) If e = 2, then h_{M}(t) = t^{p}(1 + t^{d})(1 − t)^{−1}, for some p ∈ ℤ, d ∈ ℕ. Proof. By shifting the grading we may assume that M = ⨁_{n≥0} M_{n}, and that M_{0} ≠ 0.

(a) Since GKdimM = 1 we have h_{M}(t) = g(t)(1 − t)^{−1}, where g(t) ∈ ℤ[t]. Write g(t) = e + (1 − t)p(t), where p(t) ∈ ℤ[t] is of degree d − 1. It follows that h_{M}(t) = f_{0} + f_{1}t + ⋯ + f_{d−1}t^{d−1} + et^{d}(1 − t)^{−1}, with f_{i} ∈ ℕ , and f_{0} ≠ 0. We can choose d as small as possible with respect to this property. For this choice we obviously must have f_{d−1} ≠ e.

Note that f_{j} ≠ 0 for all j ∈ {0, ⋯, d − 1}. If this doesn’t hold choose j minimal such that f_{j−1} ≠ 0, and f_{j} = 0. Then A_{1}M_{j−1} ⊂ M_{j} = 0 implies M_{j−1} ⊂ Soc(M); but M is Cohen-Macaulay so its socle is zero by (2.5e). This contradiction shows that f_{j} ≠ 0.

Recall that M^{∨} := E^{3}(M) is a CM-module whose Hilbert series is −t^{−4}h_{M}(t^{−1}) by (1.10). Hence h_{M^{∨}}(t) = (e−f_{d−1})t^{−d−3}+(e−f_{d−2})t^{−d−2}+⋯+(e−f_{0})t^{−4}+et^{−3}(1−t)^{−1}. In particular e ≥ f_{j} for all j. Since e > f_{d−1} and Soc(M^{∨}) = 0 we can conclude as above that e > f_{j} for all j. Hence 1 ≤ f_{j} ≤ e − 1 for all j and (a) is proved.

(b) If e = 1 the condition 1 ≤ f_{j} ≤ 0 forces d = 0, i.e., h_{M}(t) = (1 − t)^{−1}.

(c) If e = 2 it follows from 1 ≤ f_{j} < 2 that h_{M}(t) = 1 + t + ⋯ + t^{d−1} + 2t^{d}(1 − t)^{−1} = (1 + t^{d})(1 − t)^{−1}. ■

Proposition 2.7. Let M be in Mod_{f}^{g}(A). Then the following are equivalent:

(a) M is a shift of a point module;

(b) M is Cohen-Macaulay, and GKdim(M) = e(M) = 1. Proof. (a) ⇒ (b) As usual we may assume that M is a point module. Say M = Av_{0}

where v_{0} ∈ M_{0} and h_{M}(t) = (1 − t)^{−1}. By (2.1e) it suffices to show that Soc(M) = 0. Suppose that 0 ≠ x ∈ Soc(M)_{p}. Then A_{1}x = 0 and M_{p} = ℂx = A_{1}^{p}v_{0}, so A_{1}^{p+1}.v_{0} = 0 which is a contradiction.

(b) ⇒ (a) By (2.6b) h_{M}(t) = t^{p}(1 − t)^{−1}. We must show that M is cyclic. Pick 0 ≠ x ∈ M_{p}. Since T (M) = 0 we know that A_{n}.x = A_{1}^{n}.x ≠ 0 for all n ≥ 0. Thus A_{n}.x = M_{n+p} and M = A.x. ■

To prove Theorem 2.2 it remains to investigate the line modules. We begin with a characterization of line modules which furnishes half of the desired result: a shift of a line module is Cohen-Macaulay of GK-dimension 2 and multiplicity 1.

Proposition 2.8.

(a) If u, v ∈ A_{1} are linearly independent and A_{1}u ∩ A_{1}v ≠ 0 then L = A/Au + Av is a line module

(b) If L is a line module, then there exist linearly independent u, v ∈ A_{1} such that A_{1}u ∩ A_{1}v ≠ 0 and L ≅ A/Au + Av.

(c) If L is a line module then L is Cohen-Macaulay, GKdimL = 2, and e(L) = 1. Proof. (a) Let 0 ≠ a ∈ A_{1} satisfy au ∈ A_{1}v. Write M = A/Av and \overline{u} ∈ M_{1}

for the image of u. There is a surjective map ψ : A/Aa → A\overline{u}[1]. Since u and v are linearly independent A\overline{u} is a non-zero submodule of the plane module M. Since M is critical and GKdimM = 3 it follows that GKdim(A\overline{u}) = 3. But A/Aa is also a plane module, so critical of GKdim3. Hence ψ is an isomorphism and h_{L}(t) = h_{M}(t) − h_{A\overline{u}}(t) = (1 − t)^{−3} − t(1 − t)^{−3} = (1 − t)^{−2}. Thus L is a line module.

(b) If L is a line module then dimL_{1} = 2 so there exist linearly independent u, v ∈ A_{1} such that uL_{0} = vL_{0} = 0. Hence there is a surjective map ϕ : A/Au + Av → L. However dimL_{2} = 3, dimA_{2} = 10, dimA_{1}u = dimA_{1}v = 4, so A_{1}u ∩ A_{1}v ≠ 0. By (1) h_{A/Au+Av}(t) = h_{L}(t) = (1 − t)^{−2} so ϕ is an isomorphism.

(c) Write L = A/Au + Av as in (b). We must show that L is CM. As in the proof of (1) write M = A/Av and \overline{u} for the image of u. Then M′ = A\overline{u} is a shift of a plane module and we have an exact sequence 0 → M′ → M → L → 0 where M′ and M are CM, j(M) = 1, and j(L) = 2. By (1.12b) L is Cohen-Macaulay. ■

Corollary 2.9.

(a) A line module L has a minimal resolution of the form

0 → A[−2] ⟶^{∂_{1}} A[−1] ⊕ A[−1] ⟶^{∂_{0}} A → L → 0.

(b) The line modules are in bijection with those lines ℓ in ℙ^{3} := ℙ(A_{1}^{∗}) such that ℓ = V(u, v) where u, v ∈ A_{1} are linearly independent elements satisfying A_{1}u ∩ A_{1}v ≠ 0. The line module corresponding to such an ℓ is M(ℓ) := A/Au + Av. Proof. (a) Write L ≅ A/Au + Av as in (2.8) and denote by β a non-zero element

in A_{1}u ∩ A_{1}v. Define ∂_{0} by the matrix [{u \atop v}] and ∂_{1} by [β]. It is clear that this gives the minimal resolution of L.

(b) If L and L′ are isomorphic line modules then Ann_{A_{1}}(L_{0}) = Ann_{A_{1}}(L′_{0}) so the elements u, v ∈ A_{1} and u′, v′ ∈ A_{1} guaranteed by (2.8) span the same subspace of A_{1} and hence define the same line in ℙ^{3}. ■

Remember that throughout this section we are assuming that A_{2} contains a regular normalizing sequence {Ω_{1}, Ω_{2}} such that A/⟨Ω_{1}, Ω_{2}⟩ := B ≅ B(E, σ, L).

Our next goal is to prove a converse of (2.8c). Two preliminary results are necessary before this is done in (2.12).

Lemma 2.10. Let 0 ≠ M ∈ Mod_{f}(A) be critical and let 0 ≠ Ω be a normal element of A. Then Ω is a non-zero divisor in M if and only if ΩM ≠ 0.

Proof. Since A is a domain we can define an automorphism α of A by the formula Ωα(a) = aΩ, for all a ∈ A. Denote by ^{α}M the abelian group M with A acting by a · x = α(a)x, for a ∈ A, and x ∈ M. Then ^{α}M is a critical A-module and multiplication by Ω defines an A-module map ^{α}M ⟶^{Ω} M with image ΩM and kernel K = {x ∈ M | Ωx = 0}. The isomorphism ^{α}M/K →^{∼} ΩM shows that GKdim(M/K) = GKdim(ΩM) < GKdim(M) if K ≠ 0. Since GKdim(ΩM) = GKdim(M) if ΩM ≠ 0, the result follows.■

Proposition 2.11.

(a) Suppose M ∈ Mod_{f}^{g}(A) is 2–homogeneous with e(M) = 1. Then M is critical. Furthermore if Ω_{1}M = 0, then Ω_{2} is a non-zero divisor in M.

(b) Let M be a Cohen-Macaulay module, with GKdim(M) = 2 and e(M) = 1. Then h_{M}(t) = t^{p}(1 + t^{d})(1 − t)^{−1}(1 − t^{2})^{−1} where p ∈ ℤ and d ∈ ℕ is odd. Proof. (a) The first assertion is obvious. Suppose Ω_{1}M = 0 and that 0 ≠ y ∈ M is

such that Ω_{2}y = 0. We may assume that y is homogeneous of degree q. Then N = Ay is a B-module and N = By ≅ \frac{B}{L} [−q] for some graded left ideal L of B. Since B is a domain and GKdim B = GKdim N = 2 we must have L = 0. Thus h_{N}(t) = t^{q} \frac{(1+t)^{2}}{(1−t)^{2}} which implies e(N) = 4 which contradicts e(N) ≤ e(M) = 1. Thus Ω_{2}M ≠ 0 and by (2.10) Ω_{2} is a non-zero divisor in M.

(b) Without loss of generality we may assume that M = ⨁_{n≥0} M_{n}, and M_{0} ≠ 0. By (a) and (2.10) one of the elements Ω_{1}, Ω_{2} is a non-zero divisor in M, say Ω_{2}. Set \overline{M} = M/Ω_{2}M and notice that h\overline{M}(t) = (1 − t^{2})h_{M}(t) so GKdim\overline{M} = 1, and e(\overline{M}) = 2. Denote by α the automorphism of A defined by Ω_{2}α(a) = aΩ_{2} for a ∈ A, and define αM ∈ Mod_{f}^{g}(A) as in the proof of (2.10). It is clear that ^{α}M is Cohen-Macaulay and that there is an exact sequence 0 ⟶ ^{α}M ⟶^{×Ω_{2}} M[2] ⟶ \overline{M}[2] ⟶ 0. Thus, by (1.12b), \overline{M}[2] is Cohen-Macaulay. From (2.6c) we deduce h\overline{M}(t) = (1 + t^{d})(1 − t)^{−1}, for some d ∈ ℕ. Therefore h_{M}(t) = (1+t^{d})(1−t)^{−1}(1−t^{2})^{−1}. But h_{M}(t) must have the form f(t)(1−t)^{−4} for some f(t) ∈ ℤ[t], whence 1 + t divides 1 + t^{d}. Thus d is odd. ■

Proposition 2.12. Let M be a Cohen-Macaulay module, with d(M) = 2, and e(M) = 1. Then M is a shift of a line module.

Proof. As usual we shift the grading and assume that M = M_{0} ⊕ M_{1} ⊕ ⋯, and M_{0} ≠ 0. Write the minimal resolution of M in the form

0 ⟶ ⨁_{i} A[−i]^{b_{i}} ⟶^{∂_{1}} ⨁_{i} A[−i]^{a_{i}} ⟶^{∂_{0}} ⨁_{i} A[−i]^{c_{i}} ⟶^{φ}M ⟶ 0.

By the Remark after (1.12), the minimal resolution of M^{∨} is

0 ⟶ ⨁_{i} A[i]^{c_{i}} ⟶^{∂_{0}^{∨}} ⨁_{i} A[i]^{a_{i}} ⟶^{∂_{1}^{∨}} ⨁_{i} A[i]^{b_{i}} ⟶^{ε^{∨}}M^{∨} ⟶ 0

Notice that c_{i}, resp. b_{i}, is the number of elements of degree i, resp. −i, in a minimal set of homogeneous generators for M, resp. M^{∨}. Since M = ⨁_{n≥0} M_{n} it is easily seen that a_{0} = b_{0} = b_{1} = 0, and a_{i} = b_{i} = c_{i} for all i < 0. Thus h_{M}(t) == (1 − t)^{−4} ∑_{i≥0}(b_{i} − a_{i} + c_{i})t^{i}.

By (2.11b), there is an odd integer d ∈ ℕ, such that

h_{M}(t) = \frac{1 + t^{d}}{(1 − t)^{2}(1 + t)}
= 1 + t + 2t^{2} + 2t^{3} + ⋯
+ (\frac{d−1}{2})t^{d−3} + (\frac{d−1}{2})t^{d−2} + (\frac{d+1}{2})t^{d−1} + (\frac{d+3}{2})t^{d} + · · · .

We first have to show that d = 1, or equivalently that h_{M}(t) = (1 − t)^{−2}. Suppose to the contrary that d ≥ 3. Then dimM_{0} = 1, and dimM_{2} = 2. Since A_{2}M_{0} ≠ 0 it follows that c_{2} ≤ 1. We will show in the next paragraph that b_{2} ≤ 1, but assume for the moment that this is true. By comparing the coefficient of t^{2} in the two expressions for h_{M}(t) above, it follows that b_{2} − a_{2} + c_{2} = 4. Therefore 4 ≤ a_{2} + 4 = b_{2} + c_{2} ≤ 2 which is absurd. Thus d = 1.

Now we prove that b_{2} ≤ 1. First we have

h_{M^{∨}}(t) = t^{−4}h_{M}(t^{−1})
= t^{−(d+1)}h_{M}(t)
= t^{−(d+1)} + t^{−d} + 2t^{−(d−1)} + ⋯
+ (\frac{d−1}{2})t^{−4} + (\frac{d−1}{2})t^{−3} + (\frac{d+1}{2})t^{−2} + · · · .

By (1.10) M^{∨} is Cohen-Macaulay, GKdimM^{∨} = 2 and e(M^{∨}) = 1. By (2.11a) applied to M^{∨}, either Ω_{1} or Ω_{2} is a non-zero divisor in M^{∨}. Let Ω be this element. Let w_{1}, ⋯, w_{b_{2}} be elements of M_{−2}^{∨} which are part of a minimal set of homogeneous generators for M^{∨}. Then the sum ΩM_{−4}^{∨} + ∑_{1≤i≤b_{2}} ℂw_{i} must be direct and dimM_{−4}^{∨} = dimΩM_{−4}^{∨}. Hence b_{2} + dimM_{−4}^{∨} ≤ dimM_{−2}^{∨}. Thus b_{2} ≤ dimM_{−2}^{∨} − dimM_{−4}^{∨} = \frac{d+1}{2} − \frac{d−1}{2} = 1.

It remains to prove that M is cyclic. Since M ≅ (M^{∨})^{∨}, it is enough to show that M^{∨} is cyclic. Since h_{M^{∨}}(t) = t^{−2} + 2t^{−1} + ⋯, it follows that b_{2} = 1, and b_{i} = 0 if i ≥ 3. But b_{0} = b_{1} = 0, so M^{∨} is cyclic generated by an element of degree 2 (cf (2.9a)). ■

Proposition 2.13. Let M be a point module. Then M ≅ A/Au + Av + Aw for some u, v, w ∈ A_{1}. In fact a minimal projective resolution of M has the form

0 ⟶ A[−3]⟶^{∂_{2}}A[−2]^{3}⟶^{∂_{1}}A[−1]^{3}⟶^{∂_{0}}A[0]⟶^{ε}M ⟶ 0.

Proof. Take a minimal resolution for M, say

0 ⟶ ⨁_{i} A[−i]^{d_{i}} ⟶^{∂_{2}} ⨁_{i} A[−i]^{c_{i}} ⟶^{∂_{1}} ⨁_{i} A[−i]^{b_{i}} ⟶^{∂_{0}} ⨁_{i} A[−i]^{a_{i}} ⟶^{ε}M ⟶ 0.

The dual complex is a minimal resolution of M^{∨} by the remark after (1.12). Recall that M^{∨} is a shifted point module, generated in degree −3. Because M and M^{∨} are cyclic, a_{0} = d_{3} = 1 and all other a_{i} and d_{i} are zero. As usual we have H_{M}(t) = H_{A}(t). ∑_{i}(a_{i} − b_{i} + c_{i} − d_{i})t^{i} and thus ∑_{i}(−b_{i} +c_{i})t^{i} = −3t+3t^{2}. Because dim(M_{1}) = dim(A_{1})−3 and the resolution is minimal it follows that b_{1} = 3; the same argument applied to M^{∨} shows that c_{2} = 3. The minimality of the resolution ensures that b_{i} = 0 for i ≤ 0 and c_{i} = 0 for i ≤ 1. Therefore ∑_{i≥2} b_{i}t^{i} = ∑_{i≥3} c_{i}t^{i}. In particular b_{2} = 0 so it remains to show that for all i ≥ 3 we have b_{i} = c_{i} = 0. Suppose this is not the case.

Because A is noetherian, and M is finitely generated we can set k := max{i | b_{i} ≠ 0}. Thus k ≥ 3. Write

F′ := ⨁_{i<k} A[−i]^{c_{i}}, G′ := A[−k]^{c_{k}}, F := ⨁_{i<k} A[−i]^{b_{i}}, G := A[−k]^{b_{k}}.

Thus ∂_{1} : F′ ⊕ G′ → F ⊕ G. Since deg(∂_{1}) = 0 and G_{m} = 0 for m < k it follows that ∂_{1}(F′) ⊂ F. Since the resolution is minimal, ∂_{1}(G′) ⊂ 𝔐F ⊕ 𝔐G. But G′ is generated in degree k and 𝔐G is zero in degree k so ∂_{1}(G′) ⊂ F. Thus ∂_{1}(F′ ⊕ G′) ⊂ F.

On the other hand ⨁_{i} A[−i]^{a_{i}} ≅ A[0] so ∂_{0}(F) and ∂_{0}(G) are left ideals of A. The minimality of the resolution ensures they are both non-zero. Since A is a noetherian domain it follows that ∂_{0}(F) ∩ ∂_{0}(G) ≠ 0. Hence there exists 0 ≠ f ∈ F and 0 ≠ g ∈ G such that ∂_{0}(f) = ∂_{0}(g). It follows that f − g ∈ ker(∂_{0}) = Im(∂_{1}) but f − g ∉ F. This contradicts the conclusion of the previous paragraph, so the result follows. ■

§3. Quadrics and secant lines.

This section gives proofs of the following properties of the quadrics which contain E, and the secant lines of E. There is a pencil of quadric hypersurfaces containing E, and each point of ℙ^{3}\E belongs to a unique quadric in this pencil. Each line on one of these quadrics is a secant line of E, and every secant line lies on some (in fact, a unique) quadric in the pencil. There are exactly 4 singular quadrics in the pencil, say Q_{j} (0 ≤ j ≤ 3). Each Q_{j} may be characterized as the unique quadric which contains E and e_{j} where e_{j} is as defined in §1.1. Furthermore, Q_{j} is defined by a rank 3 quadratic form so has a unique singular point, and that point is e_{j}. If p∈Q_{j} then the line through p and e_{j} is contained in Q_{j}, so e_{j} lies on infinitely many secant lines of E, and Q_{j} is the union of all the secant lines of E which pass through e_{j}. In contrast any point in ℙ^{3}\(E ∪ S) lies on at most two secant lines of E. Thus the points e_{j} are in very special position relative to E: they are the only points in ℙ^{3}\E lying on infinitely many secant lines of E.

The map ℂ/Λ → E described in the introduction fixes the group law on E, and the identity element 0∈E. The group law is related to the geometry of the secant lines. Since E is a degree 4 curve, being an intersection of two quadrics, any hyperplane in ℙ^{3} meets E at 4 points (counted with multiplicity). The sum of these 4 points of E is zero, and conversely, if 4 points of E sum to zero then they are precisely the points of intersection (counted with multiplicity) of some hyperplane with E. Since E is not contained in any hyperplane, E has no trisecants, so a secant line meets E at two points (counted with multiplicity, where p∈ℓ ∩ E has multiplicity two if ℓ is tangent to E at p). If p, q ∈ ℙ^{3} we will write ℓ_{pq} for the line through p and q. The subgroup E_{2} of E, of points of order 2, contains 4 elements. These elements may be labeled as ω_{j} (0 ≤ j ≤ 3) in such a way that if p, q ∈ E, then ℓ_{pq} passes through e_{j} ⟺ p + q = ω_{j}. In fact, Q_{j} is the union of all the secant lines ℓ_{pq} such that p + q = ω_{j}. Now suppose that Q is a smooth quadric in the pencil. Hence Q ≅ ℙ^{1} × ℙ^{1} and there are two families of lines on Q (all of which are secant lines of E). There is a point z ∈ E (determined by Q up to sign) such that if p, q ∈ E, then ℓ_{pq} lies on Q ⟺ p + q = ±z. Moreover, all those ℓ_{pq} such that p + q = z do not intersect, and ℓ_{pq} intersects ℓ_{p′q′} ⟺ p + q = −(p′ + q′).

All the results in this section are straightforward and rather elementary. In fact they are all well-known to the average 19th century geometer; see for example [16, Art. 347]. Hence the point of this section is to state these facts, and to provide proofs for the convenience of some readers.

If 0 ≠ x ∈ A_{1}, then E meets the hyperplane x = 0 at 4 points, counted with multiplicity, and the corresponding degree 4 divisor on E will be denoted by (x)_{0} = p_{1} +p_{2} +p_{3} +p_{4}. There will be potential for confusion, since the sum of points of E, in the group law on E, will also be denoted by p_{1} + p_{2} + p_{3} + p_{4}. The context should make it clear whether + denotes addition in E or in DivE.

Lemma 3.1. Let p_{1}, p_{2}, p_{3}, p_{4} ∈ E. Then there exists 0 ≠ u ∈ A_{1} such that (u)_{0} = p_{1} + p_{2} + p_{3} + p_{4} if and only if the sum p_{1} + p_{2} + p_{3} + p_{4} in (E, +) is zero.

Proof. By Abel’s theorem if p_{1}, …, p_{r}, q_{1}, …, q_{r} ∈ E, then O(p_{1} + … + p_{r}) ≅ O(q_{1} + … + q_{r}) ⟺ p_{1} + … + p_{r} = q_{1} + … + q_{r} (on the LHS + is in DivE, on the RHS + is in E). The 4 points of E which lie on the plane x_{0} = 0 are the four z ∈ ℂ/Λ, such that θ_{11}(z) = 0, namely 0, \frac{1}{2}, \frac{1}{2}η, \frac{1}{2} + \frac{1}{2}η (the points of E_{2}). Hence, if j : E → ℙ^{3} is the inclusion, then j^{∗}O_{ℙ^{3}}(1) ≅ O(0 + (\frac{1}{2}) + (\frac{1}{2}η) + (\frac{1}{2} + \frac{1}{2}η)). Therefore, if p_{1}, p_{2}, p_{3}, p_{4} ∈ E, then they are the points of intersection of a hyperplane and E ⟺ O(p_{1} + p_{2} + p_{3} + p_{4}) ≅ j^{∗}O_{ℙ^{3}}(1) ⟺ p_{1} + p_{2} + p_{3} + p_{4} = 0 + \frac{1}{2} + \frac{1}{2}η + (\frac{1}{2} + \frac{1}{2}η) = 0. ■

Remark. It is easy to see that E has no trisecants. Suppose to the contrary that ℓ is a trisecant. Since E = X ∩ Y is an intersection of quadrics, the defining equations of X and Y restricted to ℓ are quadratic forms so either have two zeroes or vanish identically. However, by hypothesis they must have at least 3 zeroes, so we are forced to conclude that ℓ ⊆ X ∩ Y = E. This is absurd.

Let g_{1} and g_{2} be any two (linearly independent) quadratic forms which define E. To be definite we can take g_{1} and g_{2} to be the functions so labelled in §1.1. For each λ = (λ_{1}, λ_{2}) ∈ ℙ^{1}, define Q_{λ} := V(λ_{1}g_{1} + λ_{2}g_{2}). This gives a pencil of quadrics, each of which contains E.

Proposition 3.2.

(a) If Q is a quadric containing E, then Q = Q_{λ} for some λ ∈ ℙ^{1}.

(b) If p ∈ ℙ^{3}\E, then there is a unique λ ∈ ℙ^{1} such that p ∈ Q_{λ}.

(c) If λ ≠ μ, then Q_{λ} ∩ Q_{μ} = E.

(d) If ℓ is a secant line of E, then there is a unique Q_{λ} such that ℓ ⊂ Q_{λ}.

(e) If ℓ is a line lying on a quadric Q containing E then ℓ is a secant line of E. Proof. (a) By [21, 2.5], the polynomial ring modulo ⟨g_{1}, g_{2}⟩ is reduced. In particular,

if g is a quadratic form vanishing on E, then g∈\sqrt{⟨g_{1}, g_{2}⟩} = ⟨g_{1}, g_{2}⟩, so g is a linear combination of g_{1} and g_{2}.

(b) Since p ∉ E, either g_{1}(p) ≠ 0 or g_{2}(p) ≠ 0. Hence there is a unique λ = (λ_{1}, λ_{2}) ∈ ℙ^{1} such that (λ_{1}g_{1} + λ_{2}g_{2})(p) = 0. Of course (c) is an immediate consequence of (b).

(d) The uniqueness of such a Q_{λ} is guaranteed by (b). Set {p, q} = ℓ ∩ E (counted with multiplicity), and note that each g_{λ} vanishes at both p and q. Now fix r ∈ ℓ\E, and note that either g_{1}(r) ≠ 0 or g_{2}(r) ≠ 0. So there exists λ = (λ_{1}, λ_{2}) ∈ ℙ^{1} such that λ_{1}g_{1}(r) + λ_{2}g_{2}(r) = 0. But then g_{λ} = λ_{1}g_{1} + λ_{2}g_{2} vanishes at 3 points of ℓ, and hence on all of ℓ since deg(g_{λ}) = 2.

(e) Let g be a quadratic form vanishing on E but not on Q. Then ℓ ∩ E is given by the zeroes of the restriction of g to ℓ. Since there are two such zeroes, ℓ meets E with multiplicity 2. ■

Lemma 3.3. Let p, q, r, s ∈ E be such that ℓ_{pq} and ℓ_{rs} are distinct. Then ℓ_{pq} ∩ℓ_{rs} ≠ ∅ if and only if p + q = −(r + s).

Proof. There is a plane containing both ℓ_{pq} and ℓ_{rs} if and only if ℓ_{pq} ∩ ℓ_{rs} ≠ ∅. This happens if and only if p, q, r, s are coplanar. The result now follows from (3.1). ■

Proposition 3.4.

(a) For each j = 0, 1, 2, 3 there is a unique quadric, Q_{j} say, which contains e_{j} and E.

(b) Each Q_{j} is singular, of rank 3, and these are the only singular quadrics in the pencil of quadrics which contain E.

(c) The only singular point on Q_{j} is e_{j}.

(d) If p ∈ Q_{j} then the line ℓ_{pe_{j}} lies on Q_{j}. Furthermore, every line on Q_{j} passes through e_{j}, and Q_{j} is the union of the lines it contains.

(e) For each j, Q_{j} is the union of all those secant lines of E which pass through e_{j}. Proof. (a) This is a special case of (3.2b).

(b) Since E is not contained in any hyperplane, E can not be contained in any rank 2 quadric. Thus the defining equation of Q_{j} is the unique (up to scalar multiple) linear combination of g_{1} and g_{2} in which the coefficient of x_{j}^{2} is zero. This is the equation denoted by g_{j} in §1.1.

A quadratic form q on ℙ^{n}, defines a singular variety ⇔ rank(q) < n + 1. Hence a pencil λ_{1}q_{1} + λ_{2}q_{2} generated by two quadratic forms, contains n + 1 singular forms (given by the zeroes of the determinant of an (n + 1) × (n + 1) matrix whose entries are linear in λ_{1}, λ_{2}). Hence in our situation there are 4 such, and these must be the Q_{j}.

(c) Since rank(q_{j}) = 3 it has a unique singular point. Indeed, the singular point of V(y_{1}^{2} + y_{2}^{2} + y_{3}^{2}) in coordinates y_{0}, y_{1}, y_{2}, y_{3} is (1, 0, 0, 0).

(d) Let Q be a rank 3 quadric in ℙ^{3} defined by y_{1}^{2} + y_{2}^{2} + y_{3}^{2} in suitable coordinates y_{0}, y_{1}, y_{2}, y_{3}. Then e=(1, 0, 0, 0) is the unique singular point of Q. Let p = (p_{0}, p_{1}, p_{2}, p_{3}) and q = (q_{0}, q_{1}, q_{2}, q_{3}) be distinct points of Q such that ℓ_{pq}, the line through p and q, lies on Q. Then for each (s, t) ∈ ℙ^{1}, sp+tq ∈ Q_{j}. In particular, this gives st(p_{1}q_{1}+p_{2}q_{2}+p_{3}q_{3}) = 0

because p, q ∈ Q_{j}. Thus p_{1}q_{1} + p_{2}q_{2} + p_{3}q_{3} = 0. Hence \{ \begin{matrix}p_{1} & p_{2} & p_{3} \\ q_{1} & q_{2} & q_{3} \end{matrix} \} \{ \begin{matrix}p_{1} & q_{1} \\ p_{2} & q_{2} \\ p_{3} & q_{3} \end{matrix} \} = 0.

If rank ( \begin{matrix}p_{1} & p_{2} & p_{3} \\ q_{1} & q_{2} & q_{3} \end{matrix} ) = 2, then its transpose is also of rank 2, and it is impossible for

such a product to be zero. Hence the rank is ≤1. So we can choose s, t ∈ ℙ^{1} such that sp_{1} + tq_{1} = sp_{2} + tq_{2} = sp_{3} + tq_{3} = 0. Therefore sp + tq = (1, 0, 0, 0), so e lies on ℓ_{pq}.

To see that Q is the union of the lines it contains, suppose that p = (p_{0}, p_{1}, p_{2}, p_{3}) ∈ Q and p ≠ e. Then a typical point of the line ℓ_{ep} through e and p is (sp_{0} + t, sp_{1}, sp_{2}, sp_{3}) for (s, t) ∈ ℙ^{1}. It is clear that y_{1}^{2} + y_{2}^{2} + y_{3}^{2} vanishes at this point, so ℓ_{ep} ⊆ Q. In particular, p lies on a line contained in Q.

(e) If p, q ∈ E and ℓ_{pq} passes through e_{j}, then the unique quadric which contains ℓ_{pq} (such a quadric exists by (3.2d)) also contains e_{j}. But the only quadric containing e_{j} is Q_{j}, so ℓ_{pq} ⊆ Q_{j}. ■

Corollary 3.5 If p ∈ E, then the line ℓ_{pe_{j}} is a secant line of E. In particular, each e_{j} lies on infinitely many secant lines of E.

Notation. The points of E_{2} are labelled as follows:

ω_{0} = 0, ω_{1} = \frac{1}{2} + \frac{1}{2}η, ω_{2} = \frac{1}{2}η, ω_{3} = \frac{1}{2}.

Proposition 3.6. Set ζ_{0} = 0, ζ_{1} = \frac{1}{4} + \frac{1}{4}η, ζ_{2} = \frac{1}{4}η, ζ_{3} = \frac{1}{4}. Then

(a) the cosets of E_{2} in E_{4} are ζ_{j} + E_{2} (j = 0, 1, 2, 3);

(b) if p ∈ E, then ℓ_{pe_{j}} is tangent to E at p ⟺ p ∈ ζ_{j} + E_{2} ⟺ 2p = ω_{j}. Proof. The zeroes of g_{ab} occur at \frac{1+a}{4}η + \frac{1−b}{4} + \{0, \frac{1}{2}, \frac{1}{2}η, \frac{1}{2} + \frac{1}{2}η\} + Λ. Hence ζ_{0} + E_{2}

is the zero set of g_{11}, ζ_{1} +E_{2} is the zero set of g_{00}, ζ_{2} +E_{2} is the zero set of g_{01}, and ζ_{3} +E_{2} is the zero set of g_{10}. It is clear that (a) is true.

Let p = (p_{0}, p_{1}, p_{2}, p_{3}) ∈ E_{2}. The line ℓ_{pe_{0}} is tangent to E at p ⇔ ℓ_{pe_{0}} meets E at p alone. Since e_{0} = (1, 0, 0, 0), ℓ_{pe_{0}} consists of the points {(sp_{0} +t, sp_{1}, sp_{2}, sp_{3}) | (s, t) ∈ ℙ^{1}}. It follows from the defining equations of E, that such a point of ℓ_{pe_{0}} will lie on E if and only if 2stp_{0} + t^{2} = 0. Hence ℓ_{pe_{0}} is tangent to E at p ⇔ the only solution (s, t) ∈ ℙ^{1} to the equation 2stp_{0} + t^{2} = 0 is (s, t) = (1, 0) ⇔ p_{0} = 0 ⇔ p = j(z) where z is a zero of g_{11} ⇔ p ∈ ζ_{0} + E_{2} = E_{2}.

This proves (b) for j = 0. The proof for j = 1, 2, 3 is similar. Finally, since 2ζ_{j} = ω_{j}, it is clear that p ∈ ζ_{j} + E_{2} ⇔ 2p = ω_{j}. ■

Proposition 3.7. Let p, q ∈ E. The line ℓ_{pq} passes through e_{j} if and only if p+q = ω_{j}; that is, ℓ_{pq} ⊂ Q_{j} ⇔ p + q = ω_{j}.

Proof. (⇒) Suppose that e_{j} ∈ ℓ_{pq}. If p = ζ_{j}, then (3.6b) implies that q = ζ_{j} also, whence p + q = 2ζ_{j} = ω_{j}. Suppose that p ≠ ζ_{j}. Then the lines ℓ_{pq} and ℓ_{ζ_{j}e_{j}} are distinct and intersect at e_{j}. Hence they are contained in a single plane. Therefore, by (3.1), p + q + ζ_{j} + ζ_{j} = 0, whence p + q = ω_{j}.

(⇐) Suppose that p + q = ω_{j}. If p = ζ_{j}, then q = ζ_{j} too, so ℓ_{pq} is tangent to E at ζ_{j}, and (3.6b) shows this line passes through e_{j}. Suppose that p ≠ ζ_{j}. The three points p, q, ζ_{j} lie on a common plane, H say. By (3.1) H meets E at a fourth point, namely −(p + q + ζ_{j}) = ζ_{j}. Hence H meets E at ζ_{j} with multiplicity 2, so H contains the tangent line to E at ζ_{j}. But this tangent line passes through e_{j}, so e_{j} ∈ H, whence ℓ_{pe_{j}} ⊂ H. But ℓ_{pe_{j}} is a secant line, so meets E at another point of E ∩ H. Since p ≠ ζ_{j}, that other point must be q. Hence ℓ_{pe_{j}} = ℓ_{pq}, and e_{j} ∈ ℓ_{pq}. ■

Corollary 3.8. A line through two distinct points of S is not a secant line of E.

Let Q ⊆ ℙ^{3} be a smooth quadric. Then Q ≅ ℙ^{1} × ℙ^{1} (via the Segre embedding), so P ic(Q) ≅ ℤ ⨁ ℤ, where the hyperplane section is (1,1) and the intersection pairing is given by (m, n) · (m′, n′) = mn′ + nm′. There are two families of lines on Q, corresponding under the isomorphism to those of the form {p}×ℙ^{1} and ℙ^{1} ×{q}. Lines in the first family correspond to (1,0) in P ic(Q), and those in the second to (0, 1). Each point of Q lies on exactly two lines, one from each family. If Q′ is any other rank 4 quadric in ℙ^{3}, then Q∩Q′ is a curve of class (2,2) in P ic(Q). Hence any line on Q meets Q ∩ Q′ at two points.

Lemma 3.9. If p ∈ ℙ^{3}\(E ∪ S) then p lies on at most two secant lines of E. In fact if p lies on a smooth quadric containing E then p lies on two secant lines, whereas if p lies on a singular quadric then p lies on one secant line.

Proof. Let Q be the quadric containing p and E. If Q is singular then it follows from (3.4) that p lies on a unique secant line. If Q is smooth, then by the above comments, p lies on two lines contained in Q, say ℓ and ℓ′. Since E = (2, 2) in P ic(Q), both ℓ and ℓ′ are secant lines of E. If ℓ′′ is any secant line through p, then the quadric containing ℓ′′ also contains p, so by uniqueness must be Q. Thus ℓ′′ ⊂ Q, so ℓ′′ is either ℓ or ℓ′. ■

Theorem 3.10. Fix r, s ∈ E such that r + s ∉ E_{2}. Let Q be the unique quadric containing E and ℓ_{rs}. Then

(a) Q is smooth;

(b) every line on Q is a secant line of E;

(c) if ℓ is a line on Q with ℓ ∩ E = {p, q}, then (i) ℓ ∩ ℓ_{rs} = ∅ ⇔ p + q = r + s, (ii) ℓ ∩ ℓ_{rs} ≠ ∅ ⇔ p + q = −(r + s);

(d) if p, q ∈ E, then ℓ_{pq} ⊂ Q ⇔ either p + q = r + s, or p + q = −(r + s). Proof. (a) Since r + s ∈ E_{2}, ℓ_{rs} is not contained in any of the four singular quadrics

Q_{j} by (3.7b). Hence Q is smooth.

(b) This is contained in the remarks prior to (3.9).

(c)(ii) This is (3.3).

(c)(i) (⇐). Since r + s ∉ E_{2}, p + q = r + s ⇒ p + q ≠ −(r + s) so by (ii), ℓ_{pq} ∩ ℓ_{rs} = ∅. (c)(i) (⇒). Since ℓ ∩ ℓ_{rs} = ∅, ℓ and ℓ_{rs} give the same element of P ic(Q). Hence there

is a line ℓ′ on Q such that ℓ′ · ℓ = ℓ′ · ℓ_{rs} = 1. Hence ℓ and ℓ′ are coplanar; let a = 0 be the plane containing ℓ and ℓ′. Similarly, there is a plane b = 0 containing ℓ_{rs} and ℓ′. If ℓ′ ∩ E = {z, w}, it follows that the divisors on E consisting of the zeroes of a and b are (a)_{0} = p + q + z + w, and (b)_{0} = r + s + z + w. Hence the divisor of the rational function \frac{a}{b} ∈ ℂ(E) is div(\frac{a}{b}) = p + q − r − s. By Abel’s Theorem p + q = r + s in E.

(d) (⇒) This follows at once from (c).

(d) (⇐). Since p ∈ E ⊂ Q, and Q is the union of the lines it contains, there is a line on Q passing through p. In fact there are two such lines, ℓ′ and ℓ′′ say, and they satisfy ℓ′ · ℓ_{rs} = 1 and ℓ′′ · ℓ_{rs} = 0.

Suppose that p + q = r + s. Since ℓ′′ ∩ ℓ_{rs} = ∅, if ℓ′′ ∩ E = {p, t} then p + t = r + s by (c)(i). Therefore t = q, so ℓ′′ = ℓ_{pq}, and ℓ_{pq} ⊂ Q.

Suppose that p+q = −(r+s). Since ℓ′ ∩ℓ_{rs} = ∅, if ℓ′ ∩E = {p, t} then p+t = −(r+s) by (c)(ii). Therefore t = q, so ℓ′ = ℓ_{pq}, and ℓ_{pq} ⊂ Q. ■

Corollary 3.11. Given any quadric Q_{λ}, λ ∈ ℙ^{1}, in the pencil of quadrics containing E, there exists z_{λ} ∈ E such that, if p, q ∈ E then ℓ_{pq} ⊂ Q_{λ} ⇔ p + q = ±z_{λ}.

Proof. Combine (3.4), (3.7) and (3.10). ■

§4. Geometric Classification of Line Modules.

The main result in this section is that the line modules are in bijection with the secant lines of E.

Throughout this section u and v will denote linearly independent elements of A_{1}, and a and b will denote non-zero elements of A_{1}. We shall make frequent use of (1.1) and (1.2).

Proposition 4.1. If A/Au + Av is a line module, then V(u, v) is a secant line of E. Proof. By (2.8), A_{1}u ∩ A_{1}v ≠ 0, so there exist non-zero a, b ∈ A_{1} such that a ⊗ v −

b ⊗ u ∈ I_{2}. Since the proposition is only concerned with the subspace of A_{1} spanned by u and v, we may replace u and v by any linear combinations which are themselves linearly independent. In particular, we can assume that there exist distinct e_{j}, e_{k} ∈ S such that u(e_{j}) = v(e_{k}) = 0 and u(e_{k}) ≠ 0, v(e_{j}) ≠ 0. By (1.1) a⊗v − b⊗u vanishes at (e_{j}, e_{j}) and (e_{k}, e_{k}), so a(e_{j}) = b(e_{k}) = 0.

Let p∈E. Since u(e_{j}) = 0, if u(p) = 0 then so too is u(ω_{j} − p) = 0 by (3.9). Hence there exist p_{1}, p_{2}, q_{1}, q_{2} ∈ E such that the zeroes of u and v on E are given by

(∗) \{ \begin{matrix}(u)_{0} = (p_{1}) + (ω_{j} − p_{1}) + (p_{2}) + (ω_{j} − p_{2}) \\ (v)_{0} = (q_{1}) + (ω_{k} − q_{1}) + (q_{2}) + (ω_{k} − q_{2}). \end{matrix}

Suppose that u and v have no common zero on E. Since a(p)v(p+τ)−b(p)u(p+τ) = 0 for all p∈E, taking p to be each of the zeroes of u in turn, it follows that (a)_{0} = (p_{1} − τ) + (ω_{j} − p_{1} − τ) + (p_{2} − τ) + (ω_{j} − p_{2} − τ). Since these 4 points are coplanar, their sum is zero. Hence 4τ = 0. But this contradicts the fact that τ ∉ E_{4}. Hence u and v have at least one common zero.

Suppose that u and v have exactly one common zero on E. With a change of notation if necessary, we may assume that p_{1} = q_{1}. Therefore {ω_{j} − p_{1}, p_{2}, ω_{j} − p_{2}} ∩ {ω_{k} − p_{1}, q_{2}, ω_{k} − q_{2}} = ∅. It will be useful to rewrite this as

(†) {ω_{j} − p_{1} − τ, p_{2} − τ, ω_{j} − p_{2} − τ} ∩ {ω_{k} − p_{1} − τ, q_{2} − τ, ω_{k} − q_{2} − τ} = ∅.

Since a(p−τ)v(p)−b(p−τ)u(p) = 0 for all p ∈ E, evaluating this at the points p which are zeroes of only one of u and v, it follows that a vanishes at ω_{j} − p_{1} − τ, p_{2} − τ, ω_{j} − p_{2} − τ, and b vanishes at ω_{k} − p_{1} − τ, q_{2} − τ, ω_{k} − q_{2} − τ. Both a and b have a fourth zero on E which can be determined from the other three zeroes by the fact that four points are coplanar if and only if their sum is zero. Hence

\begin{aligned}(a)_{0} & = (ω_{j} − p_{1} − τ) + (p_{2} − τ) + (ω_{j} − p_{2} − τ) + (p_{1} + 3τ) \\ (b)_{0} & = (ω_{k} − p_{1} − τ) + (q_{2} − τ) + (ω_{k} − q_{2} − τ) + (p_{1} + 3τ). \end{aligned}

Since a(e_{j}) = 0, it follows that a(p) = 0 ⇔ a(ω_{j} − p) = 0, for p ∈ E. Hence

(1) {ω_{j} − p_{1} − τ, p_{2} − τ, ω_{j} − p_{2} − τ, p_{1} + 3τ} = {p_{1} + τ, ω_{j} − p_{2} + τ, p_{2} + τ, ω_{j} − p_{1} − 3τ}.

Similarly, since b(e_{k}) = 0, it follows that

(2) {ω_{k} − p_{1} − τ, q_{2} − τ, ω_{k} − q_{2} − τ, p_{1} + 3τ} = {p_{1} + τ, ω_{k} − q_{2} + τ, q_{2} + τ, ω_{k} − p_{1} − 3τ}.

Since p_{1}+τ belongs to the right hand side of both (1) and (2), it belongs to the intersection of the left hand side of (1) with the left hand side of (2). However, by (†) this intersection is {p_{1} +3τ}. Hence p_{1} +τ = p_{1} +3τ, whence 2τ = 0. This contradicts the fact that τ ∉ E_{4}.

Hence u and v have at least two common zeroes on E. ■

Lemma 4.2. Let u, v ∈ A_{1} define a secant line ℓ = V(u, v) of E. Suppose that the divisor of the rational function \frac{u}{v} on E is div(\frac{u}{v}) = (x + τ) + (y + τ) − (w + τ) − (z + τ).

(a) Define a symmetric and transitive relation ∼ on E, by s ∼ t if s + t = −(x + y). If s ∼ t, then there exists a unique plane a = 0, such that (a)_{0} = x + y + s + t.

(b) If 0 ≠ a ∈ A_{1} is such that (a)_{0} = x + y + s + t, then there exists a unique b ∈ A_{1} such that a ⊗ v − b ⊗ u vanishes on Δ_{τ}; furthermore (b)_{0} = w + z + s + t.

(c) Let ≈ denote the equivalence relation generated by ∼ i.e. s ≈ t if either s = t or x + y + s + t = 0; then there is a bijective map

φ : E/≈ ⟶ {(a, b) ∈ ℙ(A_{1} × A_{1}) | (a ⊗ v − b ⊗ u)(Δ_{τ}) = 0},

given by φ(s) = (a, b) where a ∈ A_{1} satisfies (a)_{0} = x + y + s + (−x − y − s) and b is determined as in (b). Proof. (a) If s ∼ t, then x + y + s + t = 0, so there certainly exists 0 ≠ a ∈ A_{1}

(unique up to non-zero scalar multiple) such that (a)_{0} = x + y + s + t.

(b) By Abel’s Theorem x + y = w + z, so w + z + s + t = 0. Hence there exists b ∈ A_{1} such that (b)_{0} = w + z + s + t. Notice that b is only determined up to a (non-zero) scalar multiple. Both a ⊗ v and b ⊗ u vanish at the six points (x, x + τ), (y, y + τ), (s, s + τ), (t, t + τ), (w, w + τ), and (z, z + τ).

By [21, 2.8], there are cubic forms f and g on E such that (\frac{u}{v})^{σ} = \frac{f}{g}; that is, (\frac{f}{g})(p) = (\frac{u}{v}) (p + τ) for all p ∈ E. Since E is a degree 4 curve in ℙ^{3}, and f and g are forms of degree 3, both f and g have 12 zeroes on E. However, since div (\frac{u}{v}) is as above, div (\frac{f}{g} ) = x + y − z − w, so f and g must have 10 common zeroes on E; let these common zeroes be p_{1}, ⋯, p_{10}. Hence both ag and bf vanish at the 16 points x, y, s, t, w, z, p_{1}, ⋯, p_{10} of E. Replace b by a suitable scalar multiple of itself, such that ag − bf vanishes at some seventeenth point of E. Since the degree 4 form ag − bf now has 17 zeroes on E, it is identically zero on E. Hence (a ⊗ v − b ⊗ u)(Δ_{τ}) = 0.

If some other b′ ∈ A_{1} also satisfies (a⊗v −b′ ⊗u)(Δ_{τ}) = 0, then ((b−b′)⊗u)(Δ_{τ}) = 0. If b ≠ b′, then Δ_{τ} ⊆ ℙ^{3} × V(u), which is absurd since E is not contained in a hyperplane.

(c) There certainly is such a map φ. The injectivity of φ is clear: if φ(s) = φ(s′), then (a)_{0} = (a′)_{0}, whence {s, t} = {s′, t′}, so the equivalence classes of s and s′ coincide. To see that φ is surjective, suppose that a, b∈A_{1} are such that (a⊗v −b⊗u)(Δ_{τ}) = 0. Evaluating a ⊗ v − b ⊗ u at the points (x, x + τ) and (y, y + τ) gives a(x)v(x + τ) = a(y)v(y + τ) = 0, whence a(x) = a(y) = 0. So (a)_{0} = x + y + s + t for some s, t. ■

Remarks. 1. By (1.2), the element av − bu determined by (4.2) is in the center of A, so annihilates the line module A/Av + Au.

2. We may restate part of (4.2) as follows. If V(u, v) = ℓ_{pq} and a, b ∈ A_{1} satisfy (a ⊗ v − b ⊗ u)(Δ_{τ}) = 0, then V(a, b) is a secant line ℓ_{rs} for some r, s such that r + s = p + q + 2τ, and every such r and s may occur.

Lemma 4.3. Let p, q ∈ E. Let u, v ∈ A_{1} be linearly independent. Then there exists at most one point (a, b) ∈ ℙ(A_{1} × A_{1}) such that (a ⊗ v − b ⊗ u)(Γ) = 0.

Proof. By (1.2) (a ⊗ v − b ⊗ u)(Γ) = 0 ⇔ av = bu. Hence we must show that there is at most one b ∈ ℙ(A_{1}) such that bu ∈ Av (such b then uniquely determines a, and hence determines (a, b)∈ ℙ(A_{1} × A_{1})).

Consider \overline{u} ∈ A/Av. Since A/Av is a plane module, it is 3-critical. If bu ∈ Av, there is a surjective map A/Ab ⟶ A\overline{u}. Since A/Ab is also 3-critical, this map must be an isomorphism; hence b is unique up to (non-zero) scalar multiples. ■

Proposition 4.4. Let p, q ∈ E. Suppose that u, v ∈ A_{1} are such that V(u, v) = ℓ_{p+τ,q+τ} and div(\frac{u}{v}) = (x + τ) + (y + τ) − (w + τ) − (z + τ). If s ∈ E, and φ(s) = (a, b) ∈ ℙ(A_{1} × A_{1}) where φ is as in (4.2), then av − bu ∈ A_{2} is a central element. Furthermore, av − bu = 0 if and only if (a)_{0} = x + y + (p + 2τ) + (q + 2τ). In that case (b)_{0} = w + z + (p + 2τ) + (q + 2τ).

Proof. By (4.2) and (1.2) av − bu is central. By (4.3) there is at most one pair (a, b) ∈ ℙ(A_{1} × A_{1}) such that av = bu. Hence it is enough to show that (a, b) = φ(p + 2τ) satisfies (a ⊗ v − b ⊗ u)(Γ) = 0. By (1.2) and (4.2) it therefore suffices to prove that (a ⊗ u − b ⊗ v)(e_{i}, e_{i}) = 0 for two e_{i} ∈ S.

There are at least two i such that p+2τ ≠ ω_{i} −p−2τ and p+τ ≠ ω_{i} −p−τ. For such an i, e_{i} lies on the secant lines ℓ_{p+2τ,ω_{i}−p−2τ} and ℓ_{p+τ,ω_{i}−p−τ}. Since p + 2τ ≠ ω_{i} − p − 2τ and p + τ ≠ ω_{i} − p − τ there are scalars λ, ξ, μ, ρ such that

e_{i} = λ(p + 2τ) + μ(ω_{i} − p − 2τ) = ξ(p + τ) + ρ(ω_{i} − p − τ).

All these scalars are non-zero, since e_{i} ∉ E. Now

a(e_{i})v(e_{i}) − b(e_{i})u(e_{i}) =
(λa(p + 2τ) + μa(ω_{i} − p − 2τ)).(ξv(p + τ) + ρv(ω_{i} − p − τ))
− (λb(p + 2τ) + μb(ω_{i} − p − 2τ)).(ξu(p + τ) + ρu(ω_{i} − p − τ))

= μρ(a(ω_{i} − p − 2τ)v(ω_{i} − p − τ) − b(ω_{i} − p − 2τ)u(ω_{i} − p − τ)).

But (a ⊗ v − b ⊗ u)(Δ_{τ}) = 0, so this expression is zero, and the result follows. ■

Theorem 4.5. The isomorphism classes of line modules are in bijection with the secant lines of E.

Proof. By (2.8) and (4.1), if M is a line module, then M ≅ A/Au + Av where u, v ∈ A_{1} are such that ℓ=V(u, v) is a secant line of E. Conversely, if u, v ∈ A_{1} are such ℓ:=V(u, v) is a secant line of E, then by (4.4) there exist 0≠a, b∈A_{1} such that av = bu ≠ 0, so A_{1}u ∩ A_{1}v ≠ 0. By (2.8) A/Au + Av is a line module.

To show that distinct secant lines give non-isomorphic line modules, let V(u, v) and V(u′, v′) be distinct secant lines. If φ : A/Au + Av ⟶ A/Au′ + Av′ is a (graded) isomorphism of the corresponding line modules, then φ maps the degree 0 part to the degree 0 part; say φ(\overline{1}) = e ∈ (A/Au′ + Av′)_{0}. But Ann(\overline{1})_{1} = ℂu + ℂv and Ann(e)_{1} = ℂu′ + ℂv′. However, since the lines are distinct ℂu + ℂv ≠ ℂu′ + ℂv′, so the line modules cannot be isomorphic. ■

Proposition 4.6. Let M(p, q, r, s) be a plane module. Then there is a short exact sequence 0 → M(p + τ, q + τ, r − τ, s − τ)[−1] → M(p, q, r, s) → M(p, q) → 0.

Proof. Let u, v ∈ A_{1}, be such that (u)_{0} = p + q + r + s, and V(u, v) = ℓ_{pq}. Thus M(p, q) ≅ A/Au + Av, giving an exact sequence 0 → A.\overline{v} → A/Au → A/Au + Av → 0. By (4.3) and (4.4) there is a unique a ∈ ℙ(A_{1}) such that av ∈ A_{1}u. Since plane modules are 3-critical, the map A/Aa → A\overline{v} is an isomorphism. By (4.2) and (4.4), (a)_{0} = (p + τ) + (q + τ) + (r − τ) + (s − τ). This proves the result. ■

Remark. Iterating this proposition, M(p, q, r, s) contains a submodule isomorphic to M(p + n_{1}τ, q + n_{2}τ, r + n_{3}τ, s + n_{4}τ) if (n_{1}, n_{2}, n_{3}, n_{4}) ∈ ℤ^{4} satisfies n_{1} + n_{2} + n_{3} + n_{4} = 0.

§5. Point modules.

The first result in this section, namely (5.2), shows that if p ∈ E ∪ S, then the corresponding point module is M(p) ≅ A/Au + Av + Aw where u, v, w ∈ A_{1} are linear forms such that V(u, v, w) = {p}. It is obvious from the definition of M(p) that there is a surjective map A/Au + Av + Aw → M(p), since M(p) = ⨁_{j=0}^{∞} ℂe_{j} and x ∈ A_{1} acts via x.e_{j} = x(p − jτ)e_{j+1}. The main result in this section is that if p, q ∈ E, then there is a short exact sequence 0 → M(p + τ, q − τ) → M(p, q) → M(p) → 0.

Lemma 5.1. Let p ∈ E, and let W ⊂ A_{1} be the subspace vanishing at p. Then J ⊂ AW whence A/AW ≅ B/BW.

Proof. Consider A_{1} ⊗A_{1} as bihomogeneous forms on ℙ^{3} ×ℙ^{3}. Then V(I_{2} +A_{1} ⊗W) = V(I_{2}) ∩ V(A_{1} ⊗ W) = Γ ∩ (ℙ^{3} × {p}) = (p^{σ}, p). Since I_{2} + A_{1} ⊗ W vanishes at a unique point of ℙ^{3} × ℙ^{3}, it is of codimension ≤ 1 in A_{1} ⊗ A_{1}. Hence any function vanishing at (p^{σ}, p) lies in I_{2} + A_{1} ⊗ W. In particular, as J_{2}(Δ_{τ}) = 0, J_{2} ⊂ I_{2} + A_{1} ⊗ W. ■

Remark. The lemma does not apply to e_{i} ∈ S, because J_{2} does not vanish at (e_{i}, e_{i}). In fact, if e_{i} ∈ S, then M(e_{i}) is not a B-module. Hence there are two cases in the proof of the following proposition, depending on whether or not p ∈ E.

Proposition 5.2. Let p ∈ E ∪ S. Then the point module corresponding to p is M(p) = A/AW where W ⊂ A_{1} is the subspace vanishing at p.

Proof. Let p ∈ E. There is a surjective map A/AW → M(p), so it suffices to show that dim(A_{n+1}/A_{n}W) = 1 for all n ≥ 1 (we did the case n = 1 in (5.1)). Since A/AW ≅ B/BW it is enough to show that dim(B_{n+1}/B_{n}W) = 1 for all n ≥ 1.

Let i : E → ℙ^{3} be the inclusion, and let L = i^{∗}O(1). Define L_{n} = L ⊗ σ^{∗}(L) ⊗ … ⊗ (σ^{n−1})^{∗}(L). By [21,§3], it follows that B_{n+1}/B_{n}W is the cokernel of the map H^{0}(E, L_{n}) ⊗ W → H^{0}(E, L_{n+1}) which is the restriction of the map

H^{0}(E, L_{n}) ⊗ H^{0}(E, L) → H^{0}(E, L_{n}) ⊗ H^{0}(E, L^{σ^{n}}) → H^{0}(E, L_{n+1}).

Note that W = H^{0}(E, L(−p)). Hence we wish to show that the cokernel of the map

H^{0}(L_{n}) ⊗ H^{0}(L(−p)) → H^{0}(L_{n}) ⊗ H^{0}(L(−p)^{σ^{n}}) → H^{0}(L_{n} ⊗ L(−p)^{σ^{n}}) → H^{0}(E, L_{n+1})

is of dimension 1.

Apply L_{n} ⊗ − to the sequence 0 → L(−p)^{σ^{n}} → L^{σ^{n}} → L^{σ^{n}}/L(−p)^{σ^{n}} → 0 and take cohomology. Since deg(L_{n} ⊗ L(−p)^{σ^{n}}) > 0 (actually = 4n + 3), H^{1}(L_{n} ⊗ L(−p)^{σ^{n}}) = 0 and we have an exact sequence

0 → H^{0}(L_{n} ⊗ L(−p)^{σ^{n}}) → H^{0}(L_{n+1}) → H^{0}(L_{n} ⊗ L^{σ^{n}}/L(−p)^{σ^{n}}) → 0.

These spaces have dimensions 4n + 3, 4n + 4, and 1 respectively. Hence it suffices to prove that the cokernel of the map H^{0}(L_{n}) ⊗ H^{0}(L(−p)^{σ^{n}}) → H^{0}(L_{n} ⊗ L(−p)^{σ^{n}}) is zero. Apply [13, Theorem 2(b)] with F = L_{n}, L = L(−p)^{σ^{n}} and i = 0, to conclude that this cokernel is indeed zero (the hypothesis of Mumford’s Theorem only needs to be checked for i = 1, and by degree arguments it is satisfied).

Now suppose that p = e_{i} ∈ S. Let W_{i} := ℂx_{0} ⊕ ⋯ ⊕ ℂ̂x_{i} ⊕ ⋯ ⊕ ℂx_{3} be the subspace of A_{1} vanishing at e_{i}. A careful examination of the defining relations of A shows that AW_{i} is a 2-sided ideal, and that A/AW_{i} ≅ ℂ[x_{i}]. In particular, there is a surjective map ℂ[x_{i}] ⟶ M(e_{i}), and because the Hilbert series are the same this map is an isomorphism. Thus, A/AW_{i} ≅ M(e_{i}) as required. ■

A more elementary proof of (5.2) follows from (2.13).

Lemma 5.3. Let p, q ∈ E. Then there is a short exact sequence

0 ⟶ K ⟶ M(p, q) ⟶ M(p) ⟶ 0

where the kernel K is a shifted line module.

Proof. Let u, v, w ∈ A_{1} be such that ℓ_{pq} = V(u, v), and V(u, v, w) = {p}. Thus M(p) ≅ A/Au + Av + Aw, and M(p, q) ≅ A/Au + Av. This shows the existence of a surjective map M(p, q) → M(p) with kernel K ≅ A\overline{w}, the submodule of A/Au + Av generated by the image of w. Its Hilbert series is (1 − t)^{−2} − (1 − t)^{−1} = t(1 − t)^{−2}, so A.\overline{w} is a shifted line module. ■

Remarks. 1. If p + q ∉ E_{2}, then ℓ_{pq} meets E ∪ S only at p and q, so the only point modules which are quotients of M(p, q) are M(p) and M(q). However, if p + q = ω_{i} ∈ E_{2}, then ℓ_{pq} passes through the point e_{i} ∈ S, so M(e_{i}) is a quotient of M(p, q). This describes all the ways in which a point module can arise as a quotient of a line module.

2. Our notation obscures the fact that there is only one surjection (up to scalar multiples) from M(p, p) to M(p). This is implicit in the proof of (5.3): p determines ℂu + ℂv + ℂw, the subspace of A_{1} vanishing at p, so if ℓ_{pp} = V(u, v), then the kernel of any map M(p, p) → M(p) must contain ℂ\overline{w} = (ℂu + ℂv + ℂw)/(ℂu + ℂv).

3. Next we determine the line module K in (5.3). This is done in (5.5), and (5.7) describes the kernel of the map M(p, ω_{i} − p) → M(e_{i}).

Lemma 5.4. Let p, q ∈ E. The elements of E_{2} may be labelled E_{2} = {ω_{i}, ω_{j}, ω_{k}, ω_{ℓ}} in such a way that

(a) p + q ∉ {ω_{i}, ω_{j}}, ω_{i} + ω_{j} ∉ {p − q, p − q − 2τ} and

(b) there exists 0 ≠ u ∈ A_{1} such that (u)_{0} = p + q + (ω_{i} − p) + (ω_{i} − q) and u(e_{j}) ≠ 0. Proof. If p + q ∉ E_{2} then (a) is easily satisfied. On the other hand if p + q = ω_{ℓ}

then {ω_{i} + ω_{j}, ω_{i} + ω_{k}, ω_{j} + ω_{k}} consists of 3 distinct elements, so one of these is not in {p − q, p − q − 2τ}. Again (a) holds.

Now pick 0 ≠ u ∈ A_{1} with the zero locus as in the statement of the lemma. Now u(e_{j}) = 0 if and only if either p + q = ω_{j} or p + (ω_{i} − q) = ω_{j}. Since these possibilities are excluded by (a), (b) is also true. ■

Theorem 5.5. Let p, q ∈ E. Then there is a short exact sequence

0 ⟶ M(p + τ, q − τ)[−1] ⟶ M(p, q) ⟶ M(p) ⟶ 0.

Proof. Choose ω_{i}, ω_{j} ∈ E_{2} and u ∈ A_{1} as in (5.4). Thus (u)_{0} = (p) + (q) + (ω_{i} − p) + (ω_{i} − q). Define 0 ≠ v, w ∈ A_{1} by (v)_{0} = (p) + (q) + (ω_{j} − p) + (ω_{j} − q) and (w)_{0} = (p) + (ω_{i} − p) + (ω_{j} − p) + (ω_{i} + ω_{j} + p). The careful choice of ω_{i} and ω_{j} ensures that u, v, w are linearly independent. Therefore V(u, v, w) = {p}, and the secant lines V(u, v) = ℓ_{pq}, V(u, w) = ℓ_{p,ω_{i}−p} and V(v, w) = ℓ_{p,ω_{j}−p} are pairwise distinct. By (5.3), A\overline{w} ⊂ A/Au + Av is the line module to be determined. By (4.4) there exist non-zero elements x, y, b, b′ ∈ A_{1} such that xw − bu = yw − b′v = 0. Furthermore, since div(\frac{u}{w}) = (q) + (ω_{i} − q) − (ω_{j} − p) − (ω_{i} + ω_{j} + p) we must have (x)_{0} = (q − τ) + (ω_{i} − q − τ) + (p + τ) + (ω_{i} − p + τ). Similarly (y)_{0} = (q−τ)+(ω_{j}−q−τ)+(p+τ)+(ω_{j}−p+τ). Notice that x and y are linearly independent: if not then (x)_{0} = (y)_{0} so (ω_{i} − q − τ) + (ω_{i} − p + τ) = (ω_{j} − q − τ) + (ω_{j} − p + τ), whence ω_{i} − q − τ = ω_{j} − p + τ contradicting our choice of ω_{i} and ω_{j}. Therefore A\overline{w} ≅ A/Ax + Ay and V(x, y) = ℓ_{p+τ,q−τ} . ■

The next result should be compared with [14, Theorem 3.2].

Proposition 5.6. Suppose that p − q ∉ ℤ · 2τ. Then M(p, q) has a basis {e_{ij} | (i, j) ∈ ℕ^{2}} with the property that

(a) M(p, q)_{n} has basis {e_{ij} | i + j = n};

(b) A.e_{ij} ≅ M(p + (j − i)τ, q + (i − j)τ);

(c) A.e_{ij} has basis {e_{i+k,j+ℓ} | k, ℓ ≥ 0};

(d) if x ∈ A_{1}, then x.e_{ij} ∈ ℂe_{i+1,j} ⊕ ℂe_{i,j+1};

(e) if x ∈ A_{1}, then x.e_{ij} ∈ ℂe_{i+1,j} if and only if x(q + (i − j)τ) = 0, and x.e_{ij} ∈ ℂe_{i,j+1} if and only if x(p + (j − i)τ) = 0. Proof. Write M = M(p, q), and pick 0 ≠ e_{00} ∈ M_{0}. Clearly (a) is true for n = 0,

and (b) is true for i = j = 0. The truth of (c) and (d) will follow from the way in which the basis is constructed

Since p ≠ q, there are short sequences 0 → K(p) → M(p, q) → M(p) → 0 and 0 → K(q) → M(p, q) → M(q) → 0 where, by (5.4), K(p) = Ae_{01} ≅ M(p + τ, q − τ) and K(q) = Ae_{10} ≅ M(p − τ, q + τ), for some e_{01}, e_{10} ∈ M_{1}. The elements e_{01} and e_{10} are linearly independent, because if not, then M(p) ≅ M/Ae_{01} = M/Ae_{10} ≅ M(q), whence p = q (which contradicts the hypothesis on p and q). Since dim(M_{1}) = 2, it follows that {e_{01}, e_{10}} is a basis for M_{1}. This proves (a) for n = 1, and (b) for i + j = 1.

Since p − q ∉ ℤ.2τ, p + τ ≠ q − τ, so the previous paragraph may be applied to Ae_{01} ≅ M(p + τ, q − τ). Hence there exist e_{02}, e_{11} ∈ M_{2} such that A_{1}.e_{01} = ℂe_{02} ⊕ ℂe_{11}, and Ae_{02} ≅ M(p + 2τ, q − 2τ) and Ae_{11} ≅ M(p, q). Similarly, there exist e′_{11}, e_{20} ∈ M_{2} such that A_{1}.e_{10} = ℂe_{20} ⊕ ℂe′_{11}, and Ae_{20} ≅ M(p − 2τ, q + 2τ) and Ae′_{11} ≅ M(p, q).

We will now show that ℂe_{11} = ℂe′_{11}. By (6.1) there exists a central element Ω ∈ A_{2} such that Ω.M(p, q) ≠ 0. Then 0 ≠ Ω.e_{00} ∈ M_{2}. Since a point module associated to a point of E is a B-module, Ω.M(p) = Ω.M(q) = 0. Hence Ω.e_{00} ∈ Ae_{01} ∩ Ae_{10}. Let u, v ∈ A_{1} be such that ℓ_{pq} = V(u, v). Then u.e_{00} = v.e_{00} = 0. Since Ae_{11} ≅ M(p, q), we also have v.e_{11} = u.e_{11} = 0. On the other hand, if u.e_{02} = ve_{02} = 0, then ℓ_{p+2τ,q−2τ} = V(u, v) = ℓ_{pq}; this forces {p + 2τ, q − 2τ} = {p, q} which is impossible, since τ ∉ E_{2} and p − q ∉ ℤ.2τ. Hence ℂe_{11} is the unique 1-dimensional subspace of A_{1}.e_{01} which is annihilated by u and v. But, since Ω is central, Ω.e_{00} is also killed by u and v, so ℂe_{11} = ℂΩe_{00}. Similarly, ℂe′_{11} = ℂΩe_{00}. Thus ℂe_{11} = ℂe′_{11}, and we can take e′_{11} = e_{11} in the previous paragraph.

Now M_{2} = A_{2}.e_{00} = A_{1}e_{01} + A_{1}e_{10} = ℂe_{02} + ℂe_{11} + ℂe_{20}. Since dim(M_{2}) = 3, {e_{02}, e_{11}, e_{20}} is a basis for M_{2}. Hence (a) is true for n = 2, and (b) is true for i + j = 2.

We proceed by induction. Suppose we have obtained e_{ij} for all i + j ≤ n, that (a) is true for all m ≤ n, and that (b) is true for all i + j ≤ n. We apply the earlier arguments to A_{1}.e_{i,n−i}. This gives elements e_{i+1,n−i} and e′_{i,n−i+1} in M_{n+1} such that A_{1}.e_{i,n−i} = ℂe_{i+1,n−i} +ℂe′_{i,n−i+1}. The previous argument applied to A_{1}e_{i,n−i} and A_{1}e_{i−1,n−i+1} shows that we can take e_{i,n−i+1} = e′_{i,n−i+1}. Hence we have elements e_{i,n−i+1}(0 ≤ i ≤ n+1), such that A.e_{i,n−i+1} ≅ M(p+(n+1−2i)τ, q−(n+1−2i)τ) and A_{1}.e_{i,n−i} = ℂe_{i+1,n−i}+ℂe_{i,n−i+1}. (Notice that we needed to use the fact that p−q ∉ ℤ·2τ). Since M_{n+1} = A_{1}.M_{n}, it follows that {e_{0,n+1}, e_{1,n} ⋯, e_{n,1}, e_{n+1,0}} spans M_{n+1}, and since dim(M_{n+1}) = n+2, it is actually a basis. This proves (a) for n + 1, and (b) for i + j = n + 1. Hence (a) and (b) follow by induction. ■

Remark. If p−q ∈ ℤ·2τ, then M(p, q) contains a submodule isomorphic to M(p′, p′) for some p′, and the arguments we have just used fail - indeed there is only one submodule of M(p′, p′) such that the quotient is isomorphic to M(p′), so we are unable to obtain e_{01} and e_{10} as in the proof of the proposition.

The basis has the property that if Ω∈A_{2} is central, then Ω.e_{ij} ∈ ℂe_{i+1,j+1} for all i, j (since Ω annihilates every point module M(p) with p ∈ E).

Theorem 5.7. Let p, q ∈ E be such that p + q = ω_{i} ∈ E_{2}. Then there is a short exact sequence

0 → M(p − τ, q − τ)[−1] → M(p, q) → M(e_{i}) → 0.

Proof. Let u, v, w ∈ A_{1} be such that V(u, v) = ℓ_{pq} and V(u, v, w) = {e_{i}}. By (5.3), the kernel of the surjection M(p, q) → M(e_{i}) is K ≅ A.\overline{w} contained in A/Au + Av. Thus the kernel is isomorphic to M(ℓ) where ℓ is the line V(x, y), and x, y ∈ A_{1} satisfy x.w, y.w ∈ A_{1}u + A_{1}v.

We can choose u, v, w such that V(u, v), V(u, w) and V(v, w) are three distinct secant lines of E. For example, choose u, v, w such that (u)_{0} = p + q + r + s, (v)_{0} = p + q + r′ + s′ and (w)_{0} = r + s + r′ + s′ with r, s, r′, s′ in general position subject to r + s = r′ + s′ = ω_{i}. Now apply (4.4) to these three secant lines. This gives x, y, b, b′ ∈ A_{1} such that xw − bu = yw − b′v = 0. Since w is not zero at p or q but both u and v are, it follows that (x)_{0} and (y)_{0} both contain (p − τ) + (q − τ). Thus V(x, y) = ℓ_{p−τ,q−τ} as required. ■

We now study the finite dimensional simple quotients of point modules. First consider a graded algebra A = ℂ ⊕ A_{1} ⊕ A_{2} ⊕ ⋯ generated by A_{1}, with dim(A_{1}) < ∞.

By [2] the point modules for A are parametrised by a space Γ which is an inverse limit of projective varieties. Furthermore, there is a map σ:Γ → Γ such that if p ∈ Γ, and M(p) is the corresponding point module, then M(p^{−σ}) is isomorphic to the submodule M(p)^{≥1} = ⨁_{j≥1} M(p)_{j} of M(p). Thus M(p)^{≥k} ≅ M(p^{σ^{−k}}).

Fix a basis x_{0}, …, x_{n−1} for A_{1}, and consider these as a system of homogeneous coordinates on ℙ(A_{1}^{∗}). Suppose (as is the case for the Sklyanin algebra, and the 3-generated 3-dimensional regular algebras of [1]) that there is an embedding Γ → ℙ(A_{1}^{∗}), such that σ is an automorphism of Γ, and for each p ∈ Γ, M(p)_{0} and M(p)_{1} have bases e_{0} and e_{1} with the property that x_{i}.e_{0} = x_{i}(p)e_{1} for all i. Then M(p) has a basis e_{0}, e_{1}, … such that x.e_{j} = x(p^{σ^{−j}})e_{j+1} for all x ∈ A_{1}, and all j. This assumption applies in (5.8) and (5.9).

The point modules for the Sklyanin algebra are parametrised by E ∪ S [21, §2], and the automorphism is given by σ(p) = p + τ if p ∈ E, and σ(e_{i}) = e_{i}. Hence if p ∈ E then M(p) has basis e_{j} such that x.e_{j} = x(p − jτ)e_{j+1} for all x ∈ A_{1}.

Lemma 5.8. Let M = M(p) = Ae_{0} be a point module as above. Let a ∈ ℕ.

(a) Let N be a submodule of M which does not contain any e_{i}. If dim(M/N) = a, then p^{σ^{b}} = p for some b ∈ {1, …, a}; in particular |⟨σ⟩ · p| divides b.

(b) Suppose that |⟨σ⟩ · p| = a, and N is a submodule of M such that dim(M/N) = a. Suppose that N′⫌N is a submodule of M. If N′ ≠ M then N′ contains some e_{j}.

(c) Suppose that |⟨σ⟩ · p| = a. For each 0≠λ∈ℂ, N=A(e_{0} − λe_{a}) is a proper submodule of M with basis {e_{i} − λe_{i+a} | i ≥ 0}. Furthermore, M/N is simple with basis \{\overline{e}_{0}, \overline{e}_{1}, …, \overline{e}_{a−1}\}.

(d) If |⟨σ⟩ · p| = ∞, then the only simple quotient of M(p) is the trivial module. Proof. (a) Choose b minimal such that N contains an element of the form e =

∑_{j=0}^{b} λ_{j}e_{i+j} with λ_{0}λ_{b} ≠ 0. Clearly 1 ≤ b ≤ a. Let x_{0}, x_{1}, …, x_{n−1} be a basis for A_{1}. For each k, N contains x_{k} · e = ∑_{0≤j≤b} λ_{j}x_{k}(p^{σ^{i+j}})e_{i+j+1}. Consider the 2 × n matrix, the rows of which are the homogeneous coordinates of the points p^{σ^{i}} and p^{σ^{i+b}}, namely

X = (x_{k}(p^{σ^{i+j}}))_{j=0,b;0≤k≤n−1}.

If rank(X) = 2, then x_{k}(p^{σ^{i}})x_{ℓ}(p^{σ^{i+b}}) − x_{ℓ}(p^{σ^{i}})x_{k}(p^{σ^{i+b}}) ≠ 0 for some k and ℓ. Hence N contains the element (x_{ℓ}(p^{σ^{i+b}})x_{k} − x_{k}(p^{σ^{i+b}})x_{ℓ}) · e which is a linear combination of e_{i+1}, …, e_{i+b+1} with the coefficient of e_{i+1} non-zero, and the coefficient of e_{i+b+1} equal to zero. This contradicts the minimality of b, so we conclude that rank(X) = 1. Hence the rows are scalar multiples of each other. Therefore p^{σ^{i}} = p^{σ^{i+b}}, whence p = p^{σ^{b}}.

(b) Suppose that N′ does not contain any e_{j}. Then dim(M/N′) = d < a and (a) applies to M/N′. That is, a = |⟨σ⟩ · p| divides b for some b ∈ {1, ..., d}. This is impossible.

(c) Define φ : M(p) → M(p)^{≥a} by φ(e_{i}) = e_{i+a}. Because p^{σ^{a}} = p, φ is A-linear. Thus N = Im(1 − λφ) has basis {(1 − λφ)(e_{i}) | i ≥ 0} = {e_{i} − λe_{i+a} | i ≥ 0}, and M/N has basis as claimed.

Suppose that N′⫌N is a submodule. If N′ contains some e_{j}, choose the minimal such j. If j ≥ a, then e_{j−a} − λe_{j} ∈ N ⊆ N′ implies e_{j−a} ∈ N′. Hence j < a. But then e_{a} ∈ N′, whence e_{0} ∈ N′, so N′ = M. Hence by (b) M/N is simple.

(d) Suppose that M/N is a simple quotient of M. If N contains e_{j}, then M/N is trivial, since M/Ae_{j} = M/M^{≥j} has a composition series all of whose factors are the trivial module. Hence if M/N is a non-trivial simple quotient, then (a) applies, and this contradicts the fact that |⟨σ⟩ · p| = ∞. ■

Proposition 5.9. Let M(p) be a point module as above. The following are equivalent:

(a) p^{σ} = p;

(b) M(p) has at least one non-trivial 1-dimensional quotient module;

(c) for all λ ∈ ℂ^{×}, M(p)/A(e_{0} − λe_{1}) is a 1-dimensional A-module;

(d) Ann(e_{0}) is a 2-sided ideal, and A/Ann(e_{0}) is a polynomial ring in 1 variable. Proof. (c) ⟹ (b) is obvious, and (a) ⟺ (c) by (5.8c).

(b) ⟺ (a) Let M(p)/N be a non-trivial 1-dimensional module. Since every composition factor of M(p)/Ae_{i} is the trivial module, e_{i} ∉ N. Hence p = p^{σ} by (5.8a).

(d) ⟹ (b) The Hilbert series of M(p) ≅ A/Ann(e_{0}) is (1 − t)^{−1}, so M(p) ≅ ℂ[X] has plenty of 1-dimensional quotients.

(a) ⟹ (d) We have M(p) = M(p^{σ^{k}}) ≅ M(p)^{≥k}, so Ann(e_{0}) annihilates every e_{k}. Thus Ann(e_{0}) = AnnM(p) is a 2-sided ideal, and A/Ann(e_{0}) ≅ ℂ[X], the only ring with Hilbert series (1 − t)^{−1}. ■

We now return to the Sklyanin algebra, and observe that (5.9) applies to the 4 points e_{i}, since e_{i}^{σ} = e_{i}.

Proposition 5.10. Let {i, j, k, ℓ} = {0, 1, 2, 3}. For each i, define I_{i} := Ax_{j} + Ax_{k} + Ax_{ℓ}. Then M(e_{i}) ≅ A/I_{i}, and I_{i} = AnnM(e_{i}).

Proof. It follows at once from the defining relations that I_{i} is a two-sided ideal, and that A/I_{i} ≅ ℂ[x_{i}]. It is obvious that I_{i} ⊂ Ann(M(e_{i})_{0}), so the result follows. ■

Theorem 5.11. If τ is of infinite order, then AnnM(p) = ⟨Ω_{1}, Ω_{2}⟩ for all p ∈ E.

Proof. Set J = AnnM(p). Since M(p) is critical, J is a prime ideal (this well known result can be proved by combining [9, Prop. 5.6] with the argument in [22, Prop. 3.9]). By the remarks after [21, Corollary 2.8], J ⊃ ⟨Ω_{1}, Ω_{2}⟩. Suppose the result is false.

Then A/J is prime of GK-dimension 1, so by [19], it is a finite module over its center. By Goldie’s Theorem A/J has a simple artinian ring of fractions, Q say. Since M(p) is a faithful A/J-module, Q ⊗_{A/J} M(p) ≠ 0, whence Q embeds in a direct sum of copies of this. Therefore A/J embeds in a (finite) direct sum of copies of M(p). In particular, a finite dimensional simple A/J-module is also a quotient of M(p). However, by (5.8d), the only such module is the trivial module. Hence A/J has a unique finite dimensional simple module. This is absurd since dim(A/J) = ∞, and A/J is a finite module over its center (for example, it contradicts [20]). ■

§6. Annihilators of Line Modules.

Write Z_{2} = ℂΩ_{1} ⊕ ℂΩ_{2} for the linear span of the two central elements in A_{2} described in [21, §3.9], and in Section 1 above.

We summarize the results in this section. Theorem 6.3 proves that if p, q ∈ E, then there exists a unique (up to scalar multiple) non-zero element of Z_{2} which annihilates M(p, q). We write Ω(p, q) for this element. It is best to think of Ω(p, q) as an element of the projective space ℙ(Z_{2}) ≅ ℙ^{1}, and we shall usually do this. Next (6.5) and (6.6) show that Ω(p, q) depends only on p+q, so we prefer to write Ω(p, q) = Ω(p+q). It is then proved that Ω(r) = Ω(r′) if and only if either r = r′ or r + r′ = −2τ. Finally (6.13) proves that every element of ℙ(Z_{2}) is of the form Ω(r); that is every element of Z_{2} annihilates some line module. It is also shown that if τ is of infinite order then AnnM(p, q) is generated by Ω(p + q).

Lemma 6.1. If M is a line module, then Ann_{Z_{2}}M is a 1-dimensional subspace of Z_{2}. That is, M is annihilated by a unique (up to scalar multiples) non-zero homogeneous central element of degree 2.

Proof. Let M = A/Au + Av. Since V(u, v) is a secant line, (4.4) shows that there exists 0 ≠ a, b ∈ A, such that av − bu is a non-zero central element of A annihilating M.

Now we show that M cannot be annihilated by both Ω_{1} and Ω_{2}. If it were, then M would be a B-module. However, B is a domain of GK-dimension 2, so any cyclic B-module of GK-dimension 2 must be isomorphic to B itself. But B and M have different Hilbert series, so M≇B. ■

Proposition 6.2. Let 0 ≠ Ω ∈ Z_{2}. Then \overline{A} = A/⟨Ω⟩ is a domain.

Proof. Suppose not. Then there are non-zero homogeneous elements a, b ∈ A such that ab ∈ ⟨Ω⟩, but a ∉ ⟨Ω⟩ and b ∉ ⟨Ω⟩. Choose such a and b, such that deg(a) + deg(b) is minimal. Choose Ω′ such that ℂΩ ⊕ ℂΩ′ = ℂΩ_{1} ⊕ ℂΩ_{2}.

\underline{Claim:} The image of Ω′ in \overline{A} is regular. \underline{Proof:} If not then Ω′x = Ωy for some x, y with x ∉ ⟨Ω⟩. Write Ω′ = λ_{1}Ω_{1} + λ_{2}Ω_{2}, and Ω = μ_{1}Ω_{1} + μ_{2}Ω_{2}. Thus (λ_{1}Ω_{1} + λ_{2}Ω_{2})x = (μ_{1}Ω_{1} + μ_{2}Ω_{2})y ⟹ (λ_{1}x − μ_{1}y)Ω_{1} = (μ_{2}y − λ_{2}x)Ω_{2} ⟹ μ_{2}y − λ_{2}x = Ω_{1}z, and λ_{1}x − μ_{1}y = Ω_{2}z for some z ∈ A, since (Ω_{1}, Ω_{2}) is a regular sequence in A. In particular Ω′z = (λ_{1}Ω_{1} + λ_{2}Ω_{2})z = λ_{1}(μ_{2}y − λ_{2}x) + λ_{2}(λ_{1}x − μ_{1}y) = (λ_{1}μ_{2} − μ_{1}λ_{2})y. Since Ω and Ω′

are linearly independent, ( \begin{matrix}λ_{1} & λ_{2} \\ μ_{1} & μ_{2} \end{matrix} ) is invertible, whence y = λΩ′z for some 0 ≠ λ ∈ ℂ.

Hence Ω′x = λΩΩ′z so x = λΩz ∈ ⟨Ω⟩. This contradiction proves the claim.

Since A/⟨Ω, Ω′⟩ = A/⟨Ω_{1}, Ω_{2}⟩ is a domain, either a ∈ ⟨Ω, Ω′⟩ or b ∈ ⟨Ω, Ω′⟩. Suppose that a = Ωc + Ω′d. By cancelling off high order terms, we can assume that deg(c) = deg(d) = deg(a) − 2. Now Ωcb + Ω′db = ab ∈ ⟨Ω⟩, so Ω′db ∈ ⟨Ω⟩. By the previous paragraph, db ∈ ⟨Ω⟩. But deg(d) + deg(b) < deg(a) + deg(b), and b ∉ ⟨Ω⟩, so by choice of a, b it follows that d ∈ ⟨Ω⟩. Therefore, a (= Ωc + Ω′d) is also in ⟨Ω⟩. This again is a contradiction, so a ∉ ⟨Ω, Ω′⟩. Therefore b ∈ ⟨Ω, Ω′⟩. A similar argument shows that b ∈ ⟨Ω⟩. This too is a contradiction, and we are forced to conclude that no such a and b exist. Thus \overline{A} is a domain. ■

Remark. We will prove in (6.12) that the center of A/⟨Ω⟩ is a polynomial ring in one variable when τ is of infinite order.

Theorem 6.3. Suppose that τ is of infinite order. Then the annihilator of a line module is generated by a non-zero homogeneous central element of degree 2.

Proof. Let p, q ∈ E, and set I = AnnM(p, q). Since M(p) is a quotient of M(p, q), and AnnM(p) = ⟨Ω_{1}, Ω_{2}⟩ by (5.9) we have I ⊂ ⟨Ω_{1}, Ω_{2}⟩. By (6.1), there exists 0 ≠ Ω ∈ A_{2} central, such that Ω ∈ I. By the uniqueness of Ω, we have I ≠ ⟨Ω_{1}, Ω_{2}⟩. However, A/⟨Ω⟩ is prime of GK-dimension 3, and since I is also prime (because M(p, q) is critical), we conclude that ⟨Ω⟩ = I. ■

Write Ω(p, q) for “the” element determined by (6.3). Thus when τ is of infinite order AnnM(p, q) = ⟨Ω(p, q)⟩. The rest of this section is devoted to getting more precise information about Ω(p, q). In particular, it will be shown that it depends only on p + q, so will be denoted by Ω(p + q) once that has been done.

Define non-zero central elements Ω(ω_{i}), i = 0, 1, 2, 3, in A_{2} as follows:

\begin{aligned}Ω(ω_{0}) & = (1 + γ)x_{1}^{2} + (1 + αγ)x_{2}^{2} + (1 − α)x_{3}^{2} \\ Ω(ω_{1}) & = (1 + γ)x_{0}^{2} + (αγ − γ)x_{2}^{2} + (α + γ)x_{3}^{2} \\ Ω(ω_{2}) & = (1 + αγ)x_{0}^{2} + (γ − αγ)x_{1}^{2} − (α + αγ)x_{3}^{2} \\ Ω(ω_{3}) & = (1 − α)x_{0}^{2} + (α + γ)x_{1}^{2} + (α + αγ)x_{2}^{2}. \end{aligned}

These are the only degree two central elements of “rank 3” i.e. which are linear combinations of only three of the x_{j}^{2}.

Proposition 6.4. If ℓ is a secant line passing through e_{i}, then Ω(ω_{i}) annihilates the line module M(ℓ). In particular, all line modules corresponding to lines on the singular quadric Q(w_{i}), are annihilated by the central element Ω(ω_{i}).

Proof. Write ℓ = V(u, v), and let av − bu ∈ A_{2} be a non-zero central element annihilating M(ℓ). Since av−bu is central it is a linear combination of x_{j}^{2} with j = 0, 1, 2, 3. Say av−bu = ∑ μ_{j}x_{j}^{2}. Then a⊗v−b⊗u−∑ μ_{j}x_{j} ⊗x_{j} ∈ I_{2}. Since e_{i} ∈ ℓ, u(e_{i}) = v(e_{i}) = 0. Therefore, a ⊗ v and b ⊗ u both vanish at (e_{i}, e_{i}), as do all the elements of I_{2}. Therefore (∑ μ_{j}x_{j} ⊗ x_{j})(e_{i}, e_{i}) = 0, so μ_{i} = 0. But there is a unique (up to scalar multiple) non-zero central element in A_{2}, whose coefficient of x_{i}^{2} is zero, namely Ω(ω_{i}). Hence av−bu = Ω(ω_{i}), up to a scalar multiple. ■

After (6.4), one can say, somewhat inaccurately, that the singular quadrics correspond to the singular central elements.

Proposition 6.5. Let ℓ and ℓ′ be lines lying on a common quadric containing E, and suppose that ℓ ∩ ℓ′ = ∅. Then there exists a non-zero central element Ω ∈ A_{2} which annihilates both M(ℓ) and M(ℓ′).

Proof. Write ℓ ∩ E = {p, q} and ℓ′ ∩ E = {p′, q′}. By (3.4d), Q is smooth, so by (3.10c) p + q = p′ + q′. Choose any x, y, z, w ∈ E such that {x, y} ≠ {z, w} and x + y + p + q = z + w + p + q = 0. Then there exist linear forms u, v, u′, v′ ∈ A_{1} such that

\begin{aligned}(u)_{0} & = (x) + (y) + (p) + (q) \\ (v)_{0} & = (z) + (w) + (p) + (q) \\ (u′)_{0} & = (x) + (y) + (p′) + (q′) \\ (v′)_{0} & = (z) + (w) + (p′) + (q′) \end{aligned}

Thus ℓ = ℓ_{pq} = V(u, v) and ℓ′ = ℓ_{p′q′} = V(u′, v′).

Choose r ∈ E, such that r ∉ {p, q} + E_{2}, and such that r is not a zero of u, v, u′ or v′. Replace u′ and v′ by non-zero scalar multiples of themselves such that (u − u′)(r) = (v − v′)(r) = 0. Then (u − u′)_{0} = (x) + (y) + (r) + (r̂) and (v − v′)_{0} = (z) + (w) + (r) + (r̂) for some r̂ ∈ E. In particular, V(u − u′, v − v′) = ℓ_{rr̂} is secant line of E, and since r + r̂ = −(x + y) = p + q, this line also lies on Q by (3.10d).

Apply (4.4) to ℓ_{rr̂}. There exist 0 ≠ a, b ∈ A_{1} such that (a ⊗ (v − v′) − b ⊗ (u − u′)) (Γ) = 0. In fact, we must have (a)_{0} = (x − τ) + (y − τ) + (r + τ) + (r̂ + τ), and (b)_{0} = (z − τ) + (w − τ) + (r + τ) + (r̂ + τ). In particular, av − bu = av′ − bu′ in A.

Now apply (4.2) to V(u, v) = ℓ_{pq}. There exists λ ∈ ℂ such that (a ⊗ v − λb ⊗ u)(Δ_{τ}) = 0. Similarly there exists λ′ ∈ ℂ such that (a ⊗ v′ − λ′b ⊗ u′)(Δ_{τ}) = 0. Hence b ⊗ ((λ − 1)u − (λ′ − 1)u′) vanishes on Δ_{τ}. Thus (λ − 1)u − (λ′ − 1)u′ vanishes on E, and since E spans ℙ^{3}, (λ − 1)u = (λ′ − 1)u′. However, (u)_{0} ≠ (u′)_{0} since ℓ ≠ ℓ′. Therefore λ = λ′ = 1.

Therefore (a ⊗ v − b ⊗ u)(Δ_{τ}) = (a ⊗ v′ − b ⊗ u′)(Δ_{τ}) = 0, so av − bu is central in A. Since V(u, v) = ℓ_{pq} and r + τ ∉ {p + τ, q + τ} it follows from (4.4) that av − bu ≠ 0. Therefore av − bu is a non-zero central element of A. Hence the result. ■

Corollary 6.6. Let z ∈ E. Then there exists a non-zero element, Ω(z) say, in Z_{2}, such that Ω(z).M(p, q) = 0 whenever p+q = z. In particular, Ω(p+q) annihilates M(p, q).

Proof. Suppose that z = p + q = r + s. Set ℓ = ℓ_{pq} and ℓ′ = ℓ_{rs}. If z = ω_{i}, then ℓ and ℓ′ both pass through e_{i}, so (6.4) shows that there exists 0 ≠ Ω ∈ Z_{2} such that Ω.M(ℓ) = Ω.M(ℓ′) = 0. If z ∈ E_{2}, then ℓ and ℓ′ lie on the smooth quadric Q(z), and by (3.10) do not intersect. Hence by (6.5), there exists 0 ≠ Ω ∈ Z_{2} such that Ω.M(ℓ) = Ω.M(ℓ′) = 0. The notational definitions give Ω(p + q).M(p, q) = 0. ■

We now turn to the problem of deciding exactly when two line modules are annihilated by the same homogeneous degree two central element. The result we prove in (6.9d) is that M(p, q) and M(p′, q′) have the same homogeneous degree two central annihilator if and only if either p + q = p′ + q′ or (p + q) + (p′ + q′) = −2τ. The proof begins with a lemma which contains most of the technicalities.

Lemma 6.7. Let r, r′ ∈ E. Pick ω_{i} ∈ E_{2} such that r ≠ ω_{i} and r′ ≠ ω_{i}. Suppose that r + r′ ≠ 0. Let p ∈ E be such that 2p ∉ {2r + r′, r + 2r′, r + r′ + ω_{i}, r − 2τ, r′ − 2τ}. Define

\begin{aligned}q = x′ & = r − p \\ x = q′ & = r′ − p \\ y = y′ & = p − (r + r′) \\ w = w′ & = ω_{i} − p \\ z & = p + ω_{i} − r \\ z′ & = p + ω_{i} − r′. \end{aligned}

Then (i) w ∉ {x, y, q} (ii) y ∉ {p, q, x, z} (iii) z′ ∉ {q, y} (iv) q − τ ∉ {p + τ, q + τ}

(v) q′ − τ ∉ {p + τ, q′ + τ} (vi) 0 = p + q + x + y = p + q′ + x′ + y′ = p + q + w + z = p + q′ + w′ + z′. Proof. All the following conclusions contradict the hypotheses

(i) w = x ⟹ r′ = ω_{i}. w = y ⟹ 2p = r + r′ + ω_{i}. w = q ⟹ r = ω_{i}.

(ii) y = p ⟹ r + r′ = 0. y = q ⟹ 2p = 2r + r′. y = z ⟹ r′ = ω_{i}. y = x ⟹ 2p = 2r′ + r.

(iii) z′ = q ⟹ 2p = r + r′ + ω_{i}.

z′ = y ⟹ r = w_{i}. (iv) q − τ = p + τ ⟹ 2p = r − 2τ.

q − τ = q + τ ⟹ 2τ = 0.

(v) q′ − τ = p + τ ⟹ 2p = r′ − 2τ. q′ − τ = q′ + τ ⟹ 2τ = 0.

(vi) This is straightforward. ■

Proposition 6.8. Let r, r′ ∈ E. Suppose that r ≠ r′ and r + r′ ≠ 0. Then Ω(r) = Ω(r′) (up to scalar multiple) ⟺ r + r′ = −2τ.

Proof. Choose ω_{i}, p, q, x, y, w, z, q′, x′, y′, w′, z′ ∈ E as in (6.7). Thus Ω(r) = Ω(p + q) and Ω(r′) = Ω(p + q′).

By (6.7(vi)) we can pick u = u′, v, v′ ∈ A_{1} such that

\begin{aligned}(u)_{0} & = p + q + x + y \\ (v)_{0} & = p + q + w + z \\ (u′)_{0} & = p + q′ + x′ + y′ \\ (v′)_{0} & = p + x + w′ + z′. \end{aligned}

Each of these linear forms is determined up to a scalar multiple. By (6.7(i)), the planes u = 0 and v = 0 are distinct. Thus ℓ = V(u, v) and ℓ′ = V(u′, v′) so Ω(r) ∈ A_{1}v + A_{1}u and Ω(r′) ∈ A_{1}v′ + A_{1}u′.

By (6.7(ii)), v(y) ≠ 0 and by (6.7(i),(ii),(iii)), v′(y) ≠ 0. Hence we can, and do, pick v and v′ such that (v − v′)(y) = 0. Therefore u and v − v′ have two (distinct) common zeroes on E, namely y and p. However v(e_{i}) = v′(e_{i}) = 0 since p + w = p′ + w′ = ω_{i}. Therefore (v − v′)(e_{i}) = 0. In contrast u(e_{i}) ≠ 0 because w = ω_{i} − p ∉ {x, y, q}. Therefore V(u, v − v′) = ℓ_{py} is a secant line of E.

Let a = a′ ∈ A_{1} be chosen such that (a)_{0} = (a′)_{0} = (x−τ)+(y−τ)+(q−τ)+(p+3τ). By (6.7) q − τ ∉ {p + τ, q + τ} so by (4.2) and (4.4) applied to V(u, v) = ℓ_{pq}, there exists b ∈ A_{1} such that Ω(r) = av−bu. Similarly there exists b′ ∈ A_{1} such that Ω(r′) = a′v′−b′u′.

Therefore Ω(r) − Ω(r′) = a(v − v′) − (b − b′)u, since u = u′, so Ω(r) = Ω(r′) if and only if (a ⊗ (v − v′) − (b − b′) ⊗ u) (Γ) = 0. However (a ⊗ (v − v′) − (b − b′) ⊗ u)(Δ_{τ}) = 0, since av − bu and a′v′ − b′u′ are both central. Consideration of (a)_{0}, together with (4.4) applied to V(u, v) = ℓ_{py}, shows that (a ⊗ (v − v′) − (b − b′) ⊗ u)(Γ) = 0 if and only if y − τ ∈ {p + τ, y + τ}. Since τ ∉ E_{2}, it follows that av − bu = a′v′ − b′u′ if and only if −2τ = p − y. Equivalently Ω(r) = Ω(r′) if and only if r + r′ = −2τ. ■

Corollary 6.9. Let r, r′ ∈ E.

(a) If r + r′ = −2τ, then Ω(r) = Ω(r′).

(b) If r ∈ E\E_{2}, then Ω(r) ≠ Ω(−r).

(c) If Ω(r) = Ω(r′), then either r = r′, or r + r′ = −2τ.

(d) Ω(r) = Ω(r′) if and only if either r = r′ or r + r′ = −2τ. Proof. (a) The result is true if r = r′, so suppose that r ≠ r′. Since τ ∉ E_{2}, r+r′ ≠ 0,

so (6.8) applies. Hence Ω(r) = Ω(r′).

(b) Set s = −2τ − r, and s′ = −2τ + r. Thus s + s′ = −4τ ≠ 0, and s ≠ s′ because τ ∉ E_{2}, so by (6.8) Ω(s) ≠ Ω(s′). However, by (a) Ω(s) = Ω(r) and Ω(s′) = Ω(−r), hence the result.

(c) Suppose that Ω(r) = Ω(r′) and r ≠ r′. If r + r′ ≠ 0, then (6.8) applies, proving that r + r′ = −2τ. On the other hand, if r + r′ = 0, then r ∈ E_{2} by (b). This in turn implies that r = r′, a contradiction.

(d) This follows from (a) and (c). ■

Theorem 6.10. Let p, q, p′, q′ ∈ E. If τ is of infinite order then AnnM(p, q) = AnnM(p′, q′) if and only if either p + q = p′ + q′ or (p + q) + (p′ + q′) = −2τ.

Proof. This is a consequence of (6.3) and (6.9d). ■

For completeness we also describe the annihilator of a plane module when τ is of infinite order.

Proposition 6.11. Suppose that τ is of infinite order. Let M(p, q, r, s) be a plane module. Then AnnM(p, q, r, s) = 0.

Proof. Set M = M(p, q, r, s), and J = Ann(M). Let I = ⟨Ω(p + q)⟩. Then J ⊆ I, since M(p, q) is a quotient of M. Since A/J is prime, and GKdim(A/I) = 3, either J = 0 or J = I. If J = 0 we are finished, so suppose that J = I. By (6.2) A/J is a domain. But M is a quotient of A/J of GK-dimension 3, so A/J ≅ M. However, A/J = A/ΩA with Ω ∈ A_{2}, so H_{A/J}(t) = (1 − t^{2}). H_{A}(t) ≠ (1 − t)^{−3} = H_{M}(t). This contradiction shows that J = 0. ■

The next result answers a question of Sklyanin [17, p.269].

Proposition 6.12. Suppose that τ is of infinite order. Then the center of A is Z(A) = ℂ[Ω_{1}, Ω_{2}].

Proof. Consider B = A/⟨Ω_{1}, Ω_{2}⟩. By [21], B is a noetherian domain of GK-dimension 2. Hence if the center of B is strictly larger than ℂ, then B satisfies a polynomial identity by [20]. Hence by [9,§10.3] GKdim(A/AnnM(p)) = GKdimM(p) = 1, which contradicts (5.11). Hence the center of A is contained in ℂ + ⟨Ω_{1}, Ω_{2}⟩.

First we show that the center of \overline{A} = A/⟨Ω_{1}⟩ is ℂ[Ω_{2}]. Suppose not. Pick a homogeneous w ∈ \overline{A} of minimal degree such that w is central, but w ∉ ℂ[Ω_{2}]. By the previous paragraph the center of \overline{A} is contained in ℂ + ⟨Ω_{2}⟩, so we can write w = Ω_{2}w′. But \overline{A} is a domain by (6.2), so w′ is also central in \overline{A}. However, deg(w′) < deg(w), so w′ ∈ ℂ[Ω_{2}]. Hence w ∈ ℂ[Ω_{2}] also.

Suppose the result is false. Pick a homogeneous central element z of minimal degree such that z ∉ ℂ[Ω_{1}, Ω_{2}]. Since z ∈ ⟨Ω_{1}, Ω_{2}⟩ we can write z = Ω_{1}a_{1} + Ω_{2}a_{2}. Moreover, since (Ω_{1}, Ω_{2}) is a regular sequence on A, we may choose a_{1}, a_{2} such that deg(z) = max{deg(a_{1}), deg(a_{2})}. The image of z is central in A/⟨Ω_{1}⟩, so a_{2} ∈ ℂ[Ω_{2}]+⟨Ω_{1}⟩. Hence we can write a_{2} = b_{2}+Ω_{1}c_{2} where b_{2} ∈ ℂ[Ω_{2}]. Similarly, we can write a_{1} = b_{1}+Ω_{2}c_{1} where b_{1} ∈ ℂ[Ω_{1}]. Therefore z = Ω_{1}b_{1} + Ω_{2}b_{2} + Ω_{1}Ω_{2}(c_{1} + c_{2}). But Ω_{i}b_{i} ∈ ℂ[Ω_{i}] is central, and since A is a domain, it follows that c_{1} + c_{2} is central. However, deg(a_{i}) > deg(c_{i}), so deg(c_{1} + c_{2}) < deg(z). Therefore c_{1} + c_{2} ∈ ℂ[Ω_{1}, Ω_{2}]. Hence z ∈ ℂ[Ω_{1}, Ω_{2}] also, which contradicts our choice of z. ■

Our final result, which completes the circle of ideas in this section, shows that every non-zero element of Z_{2} annihilates some line module. The proof is rather unsatisfactory, because it does not show the real reason this happens. It would be good to find another proof. One possibility is to describe the action of A on a general M(p, q) in an explicit way using the basis in (5.6). That would then allow one to explicitly calculate the action of the central elements Ω_{i} on the generator e_{00}, and thus determine the annihilator Ω(p + q).

The key technical points in the proof of (6.13) are as follows. Firstly, the Grassmannian of d-planes in the vector space W is denoted by G(d, W). The map G(d, W) → G(dimW − d, W^{∗}) given by U ↦ U^{⊥} is an isomorphism of varieties. If m ∈ ℕ, and V is a fixed subspace of W, and Y is the subset of G(d, W) consisting of those subspaces U such that dim(U ∩ V) = m, then the map Y → G(m, V) given by U ↦ U ∩ V is continuous. If m ∈ ℕ, and Y is the subspace of G(d, W) × G(d, W) consisting of pairs (U_{1}, U_{2}) such that dim(U_{1} + U_{2}) = m, then the map Y → G(m, W) given by (U_{1}, U_{2}) ↦ U_{1} + U_{2} is continuous.

We will denote the linear span of points p, q, r ∈ ℙ^{3} by \overline{pqr}.

Theorem 6.13. Every element of Z_{2} annihilates some line module.

Proof. We must show that the map E → ℙ(Z_{2}) ≅ ℙ^{1} given by z ↦ Ω(z) is surjective. Define an equivalence relation on E by z∼z′ if either z = z′ or z+z′ = −2τ. The projection from e_{0} to a general hyperplane in ℙ^{3} sends the singular quadric Q_{0}, and hence E, to a smooth conic. By (3.4) and (3.6), the fibres are ±z. Identifying the conic with ℙ^{1}, gives a morphism g : E → ℙ^{1}. Hence the map z ↦ g(z + τ) is a morphism whose fibres are precisely the equivalence classes. Thus E/∼ ≅ ℙ^{1}.

By (6.10) there is an injective map f : E/∼ → ℙ(Z_{2}) given by f(z) = Ω(z). Since we are working over ℂ, ℙ^{1} is homeomorphic to the 2-sphere S^{2}, so we have an injective map f : S^{2} → S^{2} which we wish to show is surjective. If f is continuous, then f must be surjective. To see this, first recall that a continuous bijective map from a compact space to a Hausdorff space is a homeomorphism. If f is not surjective, then its image is contained in a disc S^{2} −{p} so there would be a copy of S^{2} inside the disc. This is impossible. Hence the rest of the proof is devoted to showing that f : E/∼ → ℙ(Z_{2}) is continuous.

Let z, z′ ∈ E. Our goal is to show that if z′ is ‘close to z’ then Ω(z′) is ‘close’ to Ω(z). Fix some p, q, r, s ∈ E in general position, such that p + q = z. Fix u, v ∈ A_{1} such that ℂu = \overline{pqr}^{⊥}, ℂv = \overline{pqs}^{⊥} and V(u, v) = ℓ_{pq}. Since the addition law on E is continuous, there exist p′, q′ ∈ E which are close to p and q respectively, and p′ + q′ = z′. It follows that the lines ℂu′ = \overline{p′q′r}^{⊥} and ℂv′ = \overline{p′q′s}^{⊥} are close to ℂu and ℂv respectively in ℙ(A_{1}).

Since V(u, v) and V(u′, v′) are secant lines dim(A_{1}u + A_{1}v) = dim(A_{1}u′ + A_{1}v′) = 7. Therefore A_{1}u + A_{1}v and A_{1}u′ + A_{1}v′ are close to each other in G(7, A_{2}). The fact that these are secant lines also implies Z_{2}∩(A_{1}u+A_{1}v) and Z_{2}∩(A_{1}u′+A_{1}v′) are lines in ℙ(Z_{2}) which are close to each other. But these lines are precisely Ω(z) and Ω(z′) respectively. Hence the result. ■

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