Correspondence Between Cubic Algebras and Twisted Cubic Forms

THESIS FOR THE DEGREE OF MASTER

CORRESPONDENCE BETWEEN CUBIC ALGEBRAS AND TWISTED CUBIC FORMS

Zongbin CHEN

Supervisor: Dr. Lenny Taelman

ii iii iv Preface

The correspondence between cubic algebras and twisted binary cubic forms was discovered by Gan, Gross and Savin in [GGS]. It states that:

Proposition 0.0.1. There is a bijection between the set of GL2(Z)-orbits on the space of twisted binary cubic forms with integer coefﬁcients and the set of isomorphism classes of cubic algebras over Z.

Deligne explained in the letters [Del1] and [Del2] how to generalize this correspondence to any base schemes. In fact, he established equivalences between the following three kinds of categories, with morphisms being isomorphisms:

1. Twisted cubic forms, i.e. pairs (V,p), with V a vector bundle of rank 2 over S and p ∈ Γ(S, Sym3(V ) ⊗ (∧2V )−1).

2. Geometric cubic forms, i.e. triples (P, O(1), a), in which π : P → S is a family of genus 0 curves, O(1) is an invertible sheaf on P of relative degree 1 over S, and ∗ 2 −1 a ∈ Γ(P, O(3) ⊗ π (∧ π∗O(1)) ).

3. Cubic algebras, i.e. vector bundles over S of rank 3 endowed with a commutative algebra structure.

In this thesis, we give a detailed proof of the above equivalences following Deligne’s ideas sketched in [Del1] and [Del2].

v vi Acknowledgements

I want to thank deeply Dr. Lenny Taelman and Prof. Bas Edixhoven for their helps and encouragements.

vii viii Contents

Preface v

Acknowledgements vii

1 Main result 1 1.1 Basicdeﬁnitions...... 1 1.2 Statementofmainresults ...... 3 1.3 Sketchoftheproof ...... 4

2 The algebraic stack of cubic algebras 7 2.1 Classiﬁcation over algebraically closed ﬁelds ...... 7 2.2 Smoothnessanddimension ...... 11 2.3 Gorensteincubicalgebras...... 13

3 The algebraic stack of twisted cubic forms 17 3.1 Classiﬁcation over algebraically closed ﬁelds ...... 17 3.2 Smoothnessanddimension ...... 20 3.3 Primitivegeometriccubicforms ...... 22

4 Proof of the main result 23 4.1 Fromgeometriccubicformstocubicalgebras ...... 23 4.2 Fromcubicalgebrastogeometriccubicforms ...... 28 4.3 Thecorrespondenceisbijective...... 34

ix Bibliography 37

x Chapter 1

Main result

In our treatise, we use the language of algebraic stacks freely. The standard reference to algebraic stacks is [L-M], but one can also consult [Vis].

1.1 Basic deﬁnitions

Let S be the category of all schemes, we will endow it with Zariski topology.

Deﬁnition 1.1.1. Suppose that S is an object of S. A twisted cubic form over S is a pair 3 (V,p), where V is a locally free sheaf of OS-modules of rank 2 and p ∈ Γ(S, Sym (V ) ⊗ (∧2V )−1).

Deﬁnition 1.1.2. We deﬁne the category F of twisted cubic forms as follows. The objects of F are twisted cubic forms (V,p) over objects S in S. A morphism from a twisted cubic form (V1,p1) over S1 to another twisted cubic form (V2,p2) over S2 consists of a morphism ∗ ∗ ∗ f : S1 → S2 and an isomorphism g : V1 → f V2 such that g (f (p2)) = p1. The functor q : F → S taking a twisted cubic form (V,p) over S to S makes F into a category over S.

Deﬁnition 1.1.3. Suppose that S is an object of S. By a family of genus 0 curves over S we mean a proper, smooth morphism π : P → S whose ﬁbers over the geometric points are isomorphic to projective lines.

1 2 CHAPTER 1. MAIN RESULT

Deﬁnition 1.1.4. Suppose that S is an object of S.A geometric cubic form over S is a triple

(P, OP (1), a), where π : P → S is a family of genus 0 curves, OP (1) is an invertible sheaf ∗ 2 −1 on P of relative degree 1 over S, and a ∈ Γ(P, OP (3) ⊗ π (∧ π∗(OP (1))) ).

Remark 1.1.5. The existence of the invertible sheaf OP (1) assures that over every closed −1 1 m point x ∈ S, π (x) is isomorphic to Pk(x), where k(x)= OS,x/ x.

Deﬁnition 1.1.6. We deﬁne the category G of geometric cubic forms as follows. The objects of G are geometric cubic forms (P, OP (1), a) over objects S in S. A morphism from a geometric cubic form (P1, OP1 (1), a1) over S1 to a geometric cubic form (P2, OP2 (1), a2) over S2 consists of two morphisms f : S1 → S2, g : P1 → P2 and an isomorphism ∗ h : OP1 (1) → g OP2 (1) such that the diagram

g P1 −−−→ P2

    y f y S1 −−−→ S2

∗ ∗ is Cartesian, and a1 = h (g (a2)). The functor q : G → S taking a geometric cubic form

(P, OP (1), a) over S to S makes G into a category over S.

Deﬁnition 1.1.7. A cubic algebra A over S is a sheaf over S of commutative unital OS- algebras, locally free of rank 3 as an OS-module.

Deﬁnition 1.1.8. We deﬁne the category A of cubic algebras as follows. The objects of

A are cubic algebras A over objects S in S. A morphism from a cubic algebra A1 over S1 to a cubic algebra A2 over S2 consists of a morphism f : S1 → S2 and an isomorphism ∗ g : A1 → f A2 as cubic algebras over S1. The functor q : A → S taking a cubic algebra A over S to S makes A into a category over S.

Proposition 1.1.9. The categories F , G and A are all categories ﬁbered in groupoids over S.

Proof. The main reason for this result is the existence of a well defined pull back. As an illustration, we verify for the category F the two axioms defining a category fibered in groupoids over S. (For the two axioms, the reader is referred to chapter 2 in [L-M].) 1.2. STATEMENT OF MAIN RESULTS 3

(1) Suppose that f : S1 → S2 is a morphism in S and (V2,p2) is a twisted cubic form ∗ ∗ over S2. Deﬁne V1 = f V2, p1 = f p2, then (V1,p1) is a twisted cubic form over S1. ∗ The pair of morphisms consisting of f and the identity map id : V1 → f V2 deﬁnes

a morphism ϕ from the twisted cubic form (V1,p1) over S1 to (V2,p2) over S2 such that q(ϕ)= f.

(2) Suppose that (Vi,pi) are twisted cubic forms over Si, i = 1, 2, 3. Suppose that for ∗ i = 1, 2, we are given morphisms fi : Si → S3, isomorphisms hi : Vi → fi V3

defining a morphism from the twisted cubic form (Vi,pi) over Si to (V3,p3) over ∗ S3. Suppose that f : S1 → S2 is a morphism satisfying f1 = f2 ◦ f, then f1 V = ∗ ∗ ∗ ∗ (f2 ◦ f) (V ) = f (f2 (V )). Pulling back the isomorphism h2 : V2 → f2 V by f to ′ ∗ ∗ ∗ ∗ S1, we get an isomorphism h2 : f V2 → f (f2 (V )) = f1 V . Define the isomorphism ′−1 ∗ h = h2 ◦ h1 : V1 → f V2, then h and f defines the desired morphism from the

twisted cubic form (V1,p1) over S1 to (V2,p2) over S2.

Proposition 1.1.10. The categories ﬁbered in groupoids F , G and A are stacks over S.

Proof. Since we work with Zariski topology, the two axioms of stacks holds automatically by the deﬁnition of sheaf.

Remark 1.1.11. In fact, it will be proved in §2.2 and §3.2 that A and F are both smooth Artin stacks.

1.2 Statement of main results

Theorem 1.2.1. The stacks F and G are equivalent.

Proof. Suppose that (V,p) is a twisted cubic form over S, i.e. V is a locally free sheaf of 3 2 −1 OS-modules of rank 2 and p is a section of Sym (V ) ⊗OS (∧ V ) . Deﬁne P = P(V )= Proj( ∞ Symn V ), then π : P = P(V ) → S is a family of genus 0 curves. Let O(1) Ln=0 be Serre’s twisting sheaf of P , obviously deg(O(1)) = 1. 4 CHAPTER 1. MAIN RESULT

By [Hart], chap. III, theorem 5.1(a) and the deﬁnition of Serre’s twisting sheaf, there is a natural identiﬁcation π∗O(1) = V . As a consequence,

∗ 2 −1 3 2 −1 Γ(P, O(3) ⊗ π (∧ π∗(O(1))) )=Γ(S, Sym (V ) ⊗ (∧ V ) ).

∗ 2 −1 Denote by a the element in Γ(P, O(3) ⊗ π (∧ π∗(O(1)) ) that corresponds to p. The triple (P, O(1), a) deﬁnes a geometric cubic form over S, which is canonically associated to the twisted cubic form (V,p). Conversely, given a geometric cubic form (P, O(1), a) over S, where π : P → S is a family of genus 0 curves, O(1) is an invertible sheaf of relative degree 1, a ∈ Γ(P, O(3) ⊗ ∗ 2 −1 π (∧ π∗(O(1)) ). Let V = π∗(O(1)), then V is a locally free sheaf of rank 2 over ∗ 2 −1 3 2 −1 S. As above, we have Γ(P, O(3) ⊗ π (∧ π∗(O(1)) ) = Γ(S, Sym (V ) ⊗ (∧ V ) ). Let p be the element in Γ(S, Sym3(V ) ⊗ (∧2V )−1) corresponding to a ∈ Γ(P, O(3) ⊗ ∗ 2 −1 π (∧ π∗(O(1)) ). Then (V,p) deﬁnes a twisted cubic form on S. It is obvious that the above two constructions are inverse to each other, so we get an equivalence of stacks F and G .

The main result in this thesis is

Main Theorem 1. The stacks A and G are equivalent.

As a corollary of theorem 1.2.1 and the main theorem 1, we have

Corollary 1.2.2. The stacks A and F are equivalent.

1.3 Sketch of the proof

To prove the main theorem, we need to construct two 1-morphisms F1 : G → A and

F2 : A → G inverse to each other. Suppose that (P, O(1), a) is a geometric cubic form over S. Let π : P → S be the ∗ 2 structural morphism. Denote J := O(−3) ⊗ π (∧ π∗(O(1))). Since a ∈ Γ(P, O(3) ⊗ ∗ 2 −1 a π (∧ π∗(O(1))) ), we can deﬁne a complex J −→O. Let

0 a A := R π∗(J −→O), 1.3. SKETCH OF THE PROOF 5

0 where R π∗ denotes the 0-th hypercohomology group. (For the deﬁnition of hypercohomology group, the reader is referred to [EGA 0] §12.4.1.) It can be proved that A is a locally free sheaf of OS-modules of rank 3. What’s more, A can be endowed with a product structure. It comes from a product on the complex J −→Oa , deduced from the O-module structure of J . Calculating with Cechˇ cocycle, we can prove that A becomes a cubic algebra over S with this product structure. The details of this construction can be found in §4.1.

This construction commutes with arbitrary base change, so we get a 1-morphism F1 : G → A .

The construction of the 1-morphism F2 is more complicated. First of all, we restrict ourselves to the sub-category fibered in groupoids V of Gorenstein cubic algebras. Let A be a Gorenstein cubic algebra over S. Then we can associate to A a family of genus 0 curves π : P → S together with a closed immersion φ : Spec(A) → P . Let D be the image of φ. It is an effective relative divisor of P over S of degree 3 and is isomorphic to 1 Spec(A). Define O(1) := O(D) ⊗ ΩP/S. The section 1 of O(D) defines a section a of 1 ∨ ∼ ∗ 2 −1 O(1) ⊗ (ΩP/S) = O(3) ⊗ π (∧ (π∗O(1))) . In this way we obtain a geometric cubic form (P, O(1), a) over S from a Gorenstein cubic algebra A over S. This construction ′ commutes with arbitrary base change, so we get a 1-morphism of algebraic stacks F2 : V → G . In the general case, we use the fact that V is an open sub algebraic stack of A A ′ A G and its complement has codimension 4 in to extend F2 to a 1-morphism F2 : → . The details of this construction can be found in §4.2. In §4.3, we prove that the two 1-morphisms are inverse to each other. The idea is to restrict to the "nice" open sub algebraic stacks of primitive geometric cubic forms and Gorenstein cubic algebras. In these two cases, we have concrete and simple descriptions of the functors F1 and F2 respectively. 6 CHAPTER 1. MAIN RESULT Chapter 2

The algebraic stack of cubic algebras

2.1 Classiﬁcation over algebraically closed ﬁelds

In this section we suppose that k is an algebraically closed ﬁeld.

Lemma 2.1.1. Suppose that A is a cubic algebra over a commutative ring R, free as an R-module. Suppose further that A/R · 1 is free. There exist α, β ∈ A such that (1, α, β) is a basis of A over R and αβ ∈ R.

Proof. Take x, y ∈ A such that (1, x, y) forms a basis of A. Suppose that xy = ax+by+c, a, b, c ∈ R, then (x − b)(y − a)= ab + c ∈ R. Obviously (1, x − b, y − a) is again a basis of A over R, so we can take α = x − b, β = y − a.

We call a basis (1,α,β) a good basis of A over R if αβ ∈ R.

Theorem 2.1.2. A cubic algebra over k is isomorphic to one of the following:

(1) A1 = k × k × k.

2 (2) A2 = k × (k[ε]/(ε )).

3 (3) A3 = k[ε]/(ε ).

2 2 (4) A4 = k[α, β]/(α ,αβ,β ).

7 8 CHAPTER 2. THE ALGEBRAIC STACK OF CUBIC ALGEBRAS

Proof. Suppose that A is a cubic algebra over k. For any x ∈ A, let P (T ) be the minimal polynomial of x.

(i) Suppose that there exists x ∈ A such that deg(P (T )) = 3, then A = k[x] =∼ k[T ]/(P (T )) by dimension reason. Since k is algebraically closed, P (T )=(T −

a1)(T − a2)(T − a3), for some a1, a2, a3 ∈ k. ∼ If all the three roots of P (T ) are distinct, we get A = k[T ]/((T − a1)(T − a2)(T − ∼ a3)) = k × k × k.

If there is one root of P (T ) appearing with multiplicity 2, for example a1 = a2 =6 a3, ∼ 2 ∼ 2 we get A = k[T ]/((T − a1) (T − a3)) = k × (k(ε)/(ε ))

∼ 3 ∼ 3 If a1 = a2 = a3, A = k[T ]/(T − a1) = k[ε]/(ε ).

(ii) Suppose that for any x ∈ A, deg(P (T )) ≤ 2. By lemma 2.1.1, we can ﬁnd a basis (1, α, β) of A over k such that x := αβ ∈ k. By assumption, there are a, b, c, d ∈ k such that α2 − aα − c =0 and β2 − bβ − d =0.

Consider the homomorphism of algebras ρ : A → Endk(A) deﬁned by ρ(m)(n) = mn, ∀m, n ∈ A. Under the basis (1, α, β), we ﬁnd that

0 1  0 0 1 ρ(α)= c a , ρ(β)= x 0 ,         x 0 d b

and ρ(1) = id. By calculation,

 x   x  ρ(αβ)= ρ(α)ρ(β)= ax 0 c , ρ(βα)= ρ(β)ρ(α)= x .          x bx d 0

By assumption, ρ(αβ) = ρ(βα) = ρ(x) = diag(x, x, x). Compare these identities we get x = c = d =0. So αβ =0, α2 = aα, β2 = bβ.

We claim that a = b =0. In fact, for any r, s ∈ k, by assumption the element rα+sβ 2.1. CLASSIFICATION OVER ALGEBRAICALLY CLOSED FIELDS 9

satisﬁes an equation

(rα + sβ)2 + m(rα + sβ)+ n =0,

for some m, n ∈ k which depend on r and s. Expanding the left hand side, we get equalities 0= r2a + mr = s2b + ms = n,

for any r, s ∈ k, so we must have a = b = m =0. Now we have α2 = β2 = αβ =0, so A = k[α, β]/(α2, β2, αβ). For any r, s, t ∈ k, (r + sα + tβ)2 = −r2 +2r(r + sα + tβ), so the minimal polynomial of any element of A will have degree at most 2.

Remark 2.1.3. Observe that we don’t use the hypothesis that k is algebraically closed in step (ii) of the above proof. So if A is a cubic algebra over an arbitrary ﬁeld k such that every element in A has minimal polynomialof degree at most 2, then A ∼= k[α, β]/(α2,αβ,β2).

Theorem 2.1.4. We have the automorphism groups of the cubic algebras:

∼ (1) Autk(k × k × k) = S3.

2 ∼ × (2) Autk(k × (k(ε)/(ε ))) = k .

3 b c × (3) Aut (k(ε)/(ε )) ∼= 2 | b ∈ k , c ∈ k . k n b o 2 2 ∼ (4) Autk(k[α, β]/(α ,αβ,β )) = GL2(k).

Proof. (1) For A = k × k × k, take e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). Then

2 ei = ei, eiej =0, ∀i, j =1, 2, 3; i =6 j,

i.e. ei are idemptents. So f ∈ Autk(A) if and only if f(ei), i = 1, 2, 3 are also

idemptent. It is obvious that for any σ ∈ S3, the homomorphism deﬁned by σ(ei)=

eσ(i), i = 1, 2, 3, is an automorphism of A over k. Conversely, suppose that x = 10 CHAPTER 2. THE ALGEBRAIC STACK OF CUBIC ALGEBRAS

2 x1e1 + x2e2 + x3e3, x1, x2, x3 ∈ k satisﬁes x = x, a simple calculation tells us that 2 xi = xi, so xi =0 or 1, i =1, 2, 3. Taking into account that f(ei) =06 , f(ei)f(ej)=

0, ∀i, j = 1, 2, 3; i =6 j, we get that f(ei) = eσ(i), i = 1, 2, 3, for some σ ∈ S3. So ∼ Autk(k × k × k) = S3.

(2) For A = k × (k[ε]/(ε2)), 1 = (1, 1). Take α = (1, 0), β = (0,ε), then (1,α,β) 2 2 forms a basis of A over k and α = α, β =0, αβ =0. So f ∈ Autk(A) if and only if f(1) = 1, f(α)2 = f(α), f(β)2 =0, f(α)f(β)=0.

Suppose that f(α) = aα + bβ + c, f(β) = mα + nβ + p, a,b,c,m,n,p ∈ k, then f(β)2 =0 is equivalent to m = p = 0. And n =6 0 since f is an automorphism. By f(α)f(β)=0, we get c = 0. Finally by f(α)2 = f(α) we get a2 = a, b = 0, so a = 0 or 1. But if a = 0, then f(A) ⊂ k · 1 ⊕ kβ, contradiction to the assumption that f is an automorphism. So a = 1. In conclusion, f(1) = 1, f(α) = α, f(β) = nβ, n ∈ k×. It is easy to check that any such f deﬁnes an automorphism of A over 2 ∼ × k, so Autk(k × (k[ε]/(ε ))) = k .

3 (3) For A = k[ε]/(ε ). Let f ∈ Autk(A), then f(1) = 1. Suppose that f(ε)= a + bε + cε2, a, b, c ∈ k. By f(ε)3 = f(ε3)=0, we get a =0. By f(ε2)= f(ε)2 = b2ε2, we get b =06 since f is an automorphism. In conclusion, under the base (1,ε,ε2), f will be of the form 1  b c , b ∈ k×, c ∈ k.    2  b  It is easy to check that any such f deﬁnes an automorphism of A over k. So

3 b c × Aut (k(ε)/(ε )) =∼ 2 | b ∈ k , c ∈ k . k n b o

2 2 (4) For A = k[α, β]/(α ,αβ,β ). Let f ∈ Autk(A), then f(1) = 1. Suppose f(α) = a + bα + cβ, f(β)= m + nα + pβ, a,b,c,m,n,p ∈ k. By f(α)2 =0, we get a =0. By f(β)2 = 0, we get m = 0. Because f is an automorphism, the matrix b c is n p 2.2. SMOOTHNESS AND DIMENSION 11

invertible. So under the basis (1,α,β), f will be of the form

1  b c ∈ GL .   3    n p

It is easy to see that any f of this form deﬁnes an automorphism of A over k. So ∼ Autk(A) = GL2(k).

2.2 Smoothness and dimension

Deﬁnition 2.2.1. Let S be an arbitrary scheme, a based cubic algebra over S is a pair ⊕3 (A, φ), where A is a cubic algebra over S, φ : OS → A is an isomorphism of OS-modules such that φ(1, 0, 0)=1.

Deﬁnition 2.2.2. Let S be an arbitrary scheme, a good based cubic algebra over S is a pair ⊕3 (A, φ), where A is a cubic algebra over S, φ : OS → A is an isomorphism of OS-modules such that φ(1, 0, 0)=1, φ(0, 1, 0)φ(0, 0, 1) ∈ OS(S).

Deﬁnition 2.2.3. Let B1 be the category of based cubic algebras. Its objects are the based cubic algebras over S. A morphism from a based cubic algebra (A1,φ1) over S1 to (A2,φ2) ∗ over S2 consists of a morphism f : S1 → S2, an isomorphism h : A1 → f A2 such that −1 ∗ φ1 = h ◦ (f (φ2)).

Deﬁnition 2.2.4. Let B1,g be the category of good based cubic algebras. Its objects are the good based cubic algebras over S. A morphism from a good based cubic algebra (A1,φ1) over S1 to (A2,φ2) over S2 consists of a morphism f : S1 → S2, and an isomorphism ∗ −1 ∗ h : A1 → f A2 such that φ1 = h ◦ (f (φ2)).

Proposition 2.2.5. The categories B1 and B1,g are both categories ﬁbered in groupoids over S.

Proof. This is a simple consequence of the existence of a well deﬁned pull-back. 12 CHAPTER 2. THE ALGEBRAIC STACK OF CUBIC ALGEBRAS

1,g 4 1 Proposition 2.2.6. The category ﬁbered in groupoids B is representable by AZ and B 6 is representable by AZ.

Proof. We reproduce the proof in [Poonen], proposition 5.1. In [GGS], it was shown that if (1,α,β) is a good basis of a cubic algebra A over a ring R, then the conditions of associativity and commutativity show that

α2 = −ac + bα − aβ β2 = −bd + dα − cβ (2.1) αβ = −ad, for some a,b,c,d ∈ R, and conversely by the multiplication table any a,b,c,d ∈ R deﬁnes a good based cubic algebra. This correspondence is obviously bijective, from which we 1,g 4 deduce B = AZ. G2 B1 G2 There is a left action of a on by which (m, n) ∈ a(S) maps an algebra A over S with basis (1,α,β) to A with basis (1, α + m, β + n). This action gives an isomorphism:

G2 B1,g B1 a × → , (2.2)

1 6 so B = AZ.

Denote by H the closed subgroup scheme of GL3 stabilizing the element (1, 0, 0). H acts naturally on the category ﬁbered in groupoids B1. For (A, φ) a based cubic algebra ⊕3 ⊕3 over S, let hφ : OS → A be deﬁned as hφ(x) = φ(hx), ∀x ∈ OS , h ∈ H. Then h(A, φ):=(A, hφ).

Theorem 2.2.7. The category ﬁbered in groupoids A is a smooth Artin stack.

Proof. Since every cubic algebra is locally free, it admits a basis locally. So we have 1 1 6 A = [B /H] as stack over S. By [L-M], example 4.6.1, [B /H] = [AZ/H] is an Artin 6 1 stack. Since AZ is smooth and H is a smooth group scheme over Z, [B /H] is a smooth Artin stack. So A is a smooth Artin stack.

Proposition 2.2.8. The dimension of A is 0. 2.3. GORENSTEIN CUBIC ALGEBRAS 13

6 6 6 Proof. Since dim(AZ) = dim(H)=6 and A = [AZ/H], we get dim(A ) = dim(AZ) − dim(H)=0.

2.3 Gorenstein cubic algebras

Let A be a cubic algebra over S, i.e. A is a sheaf of commutative unital OS-algebras, locally free of rank 3 as an OS-module. Let X = Spec(A), let f : X → S be the structure morphism. Then f is a ﬁnite ﬂat morphism of degree 3 and A = f∗OX .

Since f is a ﬁnite ﬂat morphism, the sheaf HomOS (f∗OX , OS) is a coherent f∗OX - module and there is a coherent sheaf ω on X such that f∗ω = HomOS (f∗OX , OS).

Deﬁnition 2.3.1. The cubic algebra A is called Gorenstein over S if ω is an invertible sheaf on X.

Example 2.3.1. Suppose that k is a ﬁeld, let A be a cubic algebra over k.

(1) If there exists an element x ∈ A whose minimal polynomial is of degree 3, A = k[x] by dimension reason. Deﬁne f ∈ A∨ by f(1) = 0, f(x)=0, f(x2)=1. A simple calculation shows that (f, xf, x2f) is a basis of A∨ over k. So A∨ is a free A-module with generator f and A is Gorenstein.

(2) If every element in A has minimal polynomial of degree at most 2, remark 2.1.3 shows that A = k[α, β]/(α2,αβ,β2). Deﬁne h ∈ A∨ to be h(a · 1+ bα + cβ)= a. It is easy to check that for any x ∈ A, f ∈ A∨, xf lies in the k-linear span of h in A∨. So A∨ can not be a free A-module of rank 1 and hence is not Gorenstein.

In particular, in the classiﬁcation result of theorem 2.1.2, the cubic algebras A1, A2, A3 are

Gorenstein, while A4 is not Gorenstein.

Since the pull-back of a Gorenstein cubic algebra is always Gorenstein, we can deﬁne 4 the sub stack V of Gorenstein cubic algebras of A . Let O the origin of AZ, let V be the G2 4 6 image of a × (AZ\O) in AZ under the isomorphism (2.2).

Theorem 2.3.2. We have V = [V /H] as stack. In particular, V is an open sub algebraic stack of A whose complement is of codimension 4. 14 CHAPTER 2. THE ALGEBRAIC STACK OF CUBIC ALGEBRAS

First of all, we prove the following lemma, which is a generalization of [GGS], proposition 5.2.

Lemma 2.3.3. Suppose that S is a locally noetherian scheme and A is a cubic algebra over S. Then A is Gorenstein over S if and only if for every closed point x of S, the cubic algebra A ⊗OS k(x) is Gorenstein over Spec(k(x)).

Proof. The necessity is obvious. For the sufﬁciency, we need to prove that for any closed ∨ point x, Ax is a projective Ax-module. So we can assume that S = Spec(R), R a noetherian local ring with residue ﬁeld k. Let M = HomR(A, R), we need to prove that M is a projective A-module. Since A is a free R-module, M is a free R-module. By assumption,

A ⊗R k is Gorenstein. By the result in example 2.3.1, M ⊗R k is a free A ⊗R k-module of rank 1. Take m ∈ M such that its image in M ⊗R k is a generator of M ⊗R k as an

A ⊗R k-module. Deﬁne ϕ : A → M by ϕ(a) = am, ∀a ∈ A. By Nakayama’s lemma, ϕ is surjective. Denote its kernel by N. We have the exact sequence

0 → N → A → M → 0.

Applying the functor • ⊗R k we get the exact sequence

R ϕ⊗idk Tor1 (M,k)=0 → N ⊗R k → A ⊗R k −−−→ M ⊗R k → 0,

the ﬁrst equality is because M is a free R-module. Since ϕ ⊗ idk is an isomorphism,

N ⊗R k =0. By Nakayama’s lemma, N =0. So M is a free A-module.

6 Proof of theorem 2.3.2. Since A = [AZ/H], we can suppose that the base scheme S is noetherian without any loss of generality. Suppose that A is a cubic algebra over S with good basis (1,α,β), α, β ∈ OS. Then A together with this good basis corresponds uniquely 4 to an S-point Q =(a,b,c,d) of AZ as described in formula (2.1) in the proof of proposition 2.2.6. By lemma 2.3.3 and example 2.3.1, A is Gorenstein over S if and only if the image 4 of Q doesn’t intersect with the closed subscheme O of AZ. So Q factors through the open 4 subscheme AZ\O and the category ﬁbered in groupoids of Gorenstein cubic algebras with 4 4 good basis is represented by the open subscheme AZ\O of AZ. Similar to the proof of the V G2 4 second conclusion in proposition 2.2.6, we have = [ a × (AZ\O)/H] = [V /H]. 2.3. GORENSTEIN CUBIC ALGEBRAS 15

V A G2 The complement of in is [ a × O/H], which is of dimension 2 − 6 = −4, so it is of codimension 4 in A . 16 CHAPTER 2. THE ALGEBRAIC STACK OF CUBIC ALGEBRAS Chapter 3

The algebraic stack of twisted cubic forms

3.1 Classiﬁcation over algebraically closed ﬁelds

Suppose that k is an algebraically closed ﬁeld. A twisted cubic form over k is a pair 3 2 −1 (V, p), where V is a vector space of dimension 2 over k and p ∈ Sym (V ) ⊗k (∧ V ) .

Two twisted cubic forms (V1, p1) and (V2, p2) are said to be isomorphic if there exists an ∗ isomorphism of vector spaces f : V1 → V2 such that p1 = f p2. Because all the 2-dimensional vector spaces over k are isomorphic, we only need to classify the cubic forms (V, p) for a ﬁxed vector space V . Fix a basis (v1, v2) of V .

Theorem 3.1.1. A twisted cubic form (V, p) over k is isomorphic to one of the following:

−1 (1) p1 = v1v2(v1 − v2) ⊗k (v1 ∧ v2) .

2 −1 (2) p2 = v1v2 ⊗k (v1 ∧ v2) .

3 −1 (3) p3 = v1 ⊗k (v1 ∧ v2) .

(4) p4 =0.

3 2 −1 Proof. Obviously (V, 0) is a twisted cubic form. Suppose that p ∈ Sym (V ) ⊗k (∧ V ) , 3 2 2 3 −1 p =06 . Let p =(av1 + bv1 v2 + cv1v2 + dv2) ⊗k (v1 ∧ v2) , a,b,c,d ∈ k. By assumption,

17 18 CHAPTER 3. THE ALGEBRAIC STACK OF TWISTED CUBIC FORMS

k is algebraically closed, so p can be written as

−1 p =(x1v1 − y1v2)(x2v1 − y2v2)(x3v1 − y3v2) ⊗k (v1 ∧ v2) ,

1 for some xi,yi ∈ k, i =1, 2, 3. Let ai = [xi : yi], i =1, 2, 3 be three points on Pk.

(1) Suppose that ai, i = 1, 2, 3 are distinct to each other. It is well known that there exists an element γ = a b ∈ PGL (k) such that c d 2

γ(a1)=[1:0], γ(a2)=[0:1], γ(a3)=[1:1],

az+b 1 ′ where γ(z) = cz+d , ∀z ∈ Pk. For any γ ∈ GL2(k) whose image in PGL2(k) is γ, we have ′∗ −1 ′ γ (v1v2(v1 − v2) ⊗k (v1 ∧ v2) )= m(γ )p,

for some m(γ′) ∈ k×. Since k is a ﬁeld, we can choose appropriately an element

γ˜ ∈ GL2(k) whose image in PGL2(k) is γ, such that

∗ −1 γ˜ (v1v2(v1 − v2) ⊗k (v1 ∧ v2) )= p.

(2) Suppose that a1 = a2 =6 a3. As in the above, there exists one element γ ∈ PGL2(k) such that

γ(a1)= γ(a2)=[1:0], γ(a3)=[0:1].

Similar to the case (1), we can choose an element γ˜ ∈ GL2(k) whose image in

PGL2(k) is γ, such that

∗ 2 −1 γ˜ (v1 v2 ⊗k (v1 ∧ v2) )= p.

The cases a1 = a3 =6 a2 and a2 = a3 =6 a1 can be treated in the same way.

(3) Suppose that a1 = a2 = a3. There exists one element γ ∈ PGL2(k) such that

γ(a1)= γ(a2)= γ(a3)=[1:0]. 3.1. CLASSIFICATION OVER ALGEBRAICALLY CLOSED FIELDS 19

Similar to case (1), we can choose an element γ˜ ∈ GL2(k) whose image in PGL2(k) is γ, such that ∗ 3 −1 γ˜ (v1 ⊗k (v1 ∧ v2) )= p.

Theorem 3.1.2. The automorphism groups of the above twisted cubic forms are

∼ (1) Autk((V,p1)) = S3.

∼ × (2) Autk((V,p2)) = k .

a b × (3) Aut ((V,p )) ∼= 2 ∈ GL (k)| a ∈ k , b ∈ k . k 3 a 2

(4) Autk((V,p4)) = GL2(k).

Proof. Let γ = a b ∈ GL (k), then γ(v )= av + cv , γ(v )= bv + dv . c d 2 1 1 2 2 1 2

(1) Suppose that γ ∈ Autk((V, p1)). Denote by X the zero section of the cubic form 1 v1v2(v1 − v2) over P(V ) = Pk. Let x1 = [0 : 1], x2 = [1 : 0], x3 = [1 : 1], then 1 ∗ X = x1 + x2 + x3 as a divisor on Pk. By assumption γ p1 = p1, so γ induces ∼ an automorphism of X. Since Aut(X) = S3, we obtain a homomorphism j :

Autk((V, p1)) → S3.

We claim that j is injective. In fact, for γ ∈ Autk((V, p1)) such that j(γ)=1, because

−1 −1 (ad − bc) (av1 + cv2)(bv1 + dv2)[(a − b)v1 +(c − d)v2] ⊗k (v1 ∧ v2) −1 = v1v2(v1 − v2) ⊗k (v1 ∧ v2) ,

and γ(xi)= xi, i =1, 2, 3, we get equalities

c =0, b =0

av1 − dv2 = v1 − v2,

i.e. γ = 1 . So j is injective. 1 20 CHAPTER 3. THE ALGEBRAIC STACK OF TWISTED CUBIC FORMS

Conversely, for any permutation σ ∈ S3, there exists a unique aσ ∈ PGL2(k) = 1 Autk(Pk), such that aσ(xi) = xσ(i). Then for any A ∈ GL2(k) whose image in ∗ × PGL2(k) is aσ, we have A (p1)= m(A)p1, for some m(A) ∈ k . Since k is a ﬁeld, ∗ we can ﬁnd an A0 ∈ GL2(k) whose image in PGL2 is aσ such that A0(p1)= p1. So

j is surjective, hence an isomorphism between Autk((V, p)) and S3.

(2) Suppose that γ ∈ Autk((V, p2)). Denote by X the zero section of the cubic form 2 1 v1v2 over P(V )= Pk. Let x1 = [0 : 1], x2 = [1 : 0], then X =2x1 + x2 as a divisor 1 ∗ on Pk. Since γ (p2) = p2, we get γ(x1) = x1 and γ(x2) = x2. Combining with the equality

−1 2 −1 (ad − bc) (av1 + cv2) (bv1 + dv2) ⊗k (v1 ∧ v2) 2 −1 = v1v2 ⊗k (v1 ∧ v2) .

We get a = 1, b = c = 0, i.e. γ = 1 with d ∈ k×. It is easy to check that any d ∼ × such γ deﬁnes an automorphism of (V, p2). So Autk((V, p2)) = k .

(3) γ ∈ Autk((V, p3) if and only if

−1 3 −1 3 −1 (ad − bc) (av1 + cv2) ⊗k (v1 ∧ v2) = v1 ⊗k (v1 ∧ v2) .

Comparing the two sides of the equality, we get c =0, a2 = d. So

a b × Aut ((V, p )) = 2 | a ∈ k , b ∈ k . k 3 n a o

(4) This case is obvious.

3.2 Smoothness and dimension

Deﬁnition 3.2.1. Let S be an arbitrary scheme, a based twisted cubic form over S is a pair ⊕2 (V,p,ϕ), where (V,p) is a twisted cubic form over S, φ : OS → V is an isomorphism of

OS-modules. 3.2. SMOOTHNESS AND DIMENSION 21

Deﬁnition 3.2.2. Let F ′ be the category of based twisted cubic forms. Its objects are the based twisted cubic forms. A morphism from a based twisted cubic form (V1,p1,ϕ1) over S1 to (V2,p2,ϕ2) over S2 consists of a morphism f : S1 → S2, an isomorphism ∗ ∗ ∗ −1 ∗ h : V1 → f V2 such that p1 = h (f (p2)) and ϕ1 = h ◦ (f (ϕ2)).

Proposition 3.2.3. The category F ′ is a category ﬁbered in groupoids over S, it can be 4 represented by AZ.

Proof. The first conclusion is a simple consequence of the existence of a well defined pull- back. For the second conclusion, since any based twisted cubic form over S has trivial isomorphism group, we only need to give the 1-morphism over the affine base scheme. Suppose that (V,p,ϕ) is a based twisted cubic form over S = Spec(R), R a commutative ring, such that V is a free rank 2 sheaf of OS-module. Let e1 = ϕ((1, 0)), e2 = ϕ((0, 1)), 3 2 2 3 −1 then p can be written uniquely as (ae1 + be1e2 + ce1e2 + de2) ⊗ (e1 ∧ e2) , for some a,b,c,d ∈ R, from which the proposition is easily deduced.

′ The linear algebraic group GL2,Z acts naturally on F . Suppose that (V,p,ϕ) is a ⊕2 based twisted cubic form on S. Let g ∈ GL2(OS), define gϕ : OS → V to be gϕ(x) = ⊕2 ∗ ϕ(gx), ∀x ∈ OS . Define g(V,ϕ,p)=(V,gϕ,g (p)), it is easily verified that this defines ′ an action of GL2 on F .

Theorem 3.2.4. The stack F is a smooth Artin stack.

′ 4 4 Proof. Obviously we have F = [F / GL2] = [A / GL2]. Since both A and GL2 are 4 smooth, [A / GL2] is a smooth Artin stack. So F is also a smooth Artin stack.

Proposition 3.2.5. The dimension of F is 0.

′ 4 4 Proof. Since F = [F / GL2] = [A / GL2] and dim(A ) = dim(GL2)=4, we have dim(F )=0.

Corollary 3.2.6. The stack G of geometric cubic forms is a smooth Artin stack of dimension 0.

Proof. By theorem 1.2.1, G ∼= F as stack, so the corollary results from the above two propositions. 22 CHAPTER 3. THE ALGEBRAIC STACK OF TWISTED CUBIC FORMS

3.3 Primitive geometric cubic forms

4 4 Let O the origin of AZ, V = AZ\O. Obviously [V / GL2] is an open sub algebraic stack 4 of F = [AZ/ GL2], with complement of codimension 4. The objects of [V / GL2] will be called primitive twisted cubic forms. By theorem 1.2.1, F ∼= G . Denote by W the open sub algebraic stack of G that corresponds to [V / GL2] in F . Obviously the complement of W is of codimension 4 in G . The objects of W will be called primitive geometric cubic forms.

Proposition 3.3.1. Suppose that (P, O(1), a) is a primitive geometric cubic form over S, ∗ 2 −1 a let J := O(3) ⊗ π (∧ π∗O(1)) . Then the morphism of sheaves J −→O is injective.

Proof. By assumption, a is non zero at every point, so the morphism of sheaves J −→Oa is injective. Chapter 4

Proof of the main result

4.1 From geometric cubic forms to cubic algebras

In this section, we will construct a 1-morphism of algebraic stacks F1 : G → A . Suppose that (P, O(1), a) is a geometric cubic form, i.e. π : P → S is a family of ∗ 2 −1 ∗ 2 genus 0 curves, a ∈ Γ(P, O(3) ⊗ π (∧ π∗O(1)) ). Let J := O(−3) ⊗ π (∧ π∗O(1)), −1 ∗ 2 −1 −1 then we have J = O(3) ⊗ π (∧ π∗O(1)) . So a ∈ Γ(P, J ) = Γ(P, Hom(J , O)) deﬁnes a morphism of sheaves of modules

a : J → O, which will be regarded as a complex of coherent O-modules with deg(J ) = −1. Deﬁne 0 a A := R π∗(J −→O), the 0-th hypercohomology of the complex J → O. From the exact sequence of complexes

0 →O→ (J −→Oa ) →J [1] → 0,

we obtain the following long exact sequence by applying π∗,

i i i a i i+1 ···→ R π∗O = R π∗O→ R π∗(J −→O) → R π∗(J [1]) = R π∗(J ) →··· . (4.1)

23 24 CHAPTER 4. PROOF OF THE MAIN RESULT

Lemma 4.1.1. We have

π∗O = OS, i R π∗O =0, if i =06 .

1 Proof. Firstly, we suppose that S is afﬁne and P = PS. Let S = Spec(R), R a commu- i i 1 ∼ tative ring. By [Hart], chapter III, proposition 8.5, R π∗O = H (PR, O) . So by [Hart], chapter III, theorem 5.1, 1 ∼ π∗O = Γ(PR, O) = OS, and for i =06 , i i 1 ∼ R π∗O = H (PR, O) =0.

In general case we can cover S by small enough afﬁne open subsets U such that −1 1 π (U) = PU . Since the above equalities are canonical, we get the same equalities in general case.

Lemma 4.1.2. We have

i R π∗(J )=0, if i =16 , 1 ∨ R π∗(J )=(π∗O(1)) .

Proof. First of all, observe that

∗ 2 1 J = O(−3) ⊗ π (∧ π∗O(1)) = O(−1) ⊗ ΩP/S.

1 Suppose that S is afﬁne and P = PS. Let S = Spec(R), R a commutative ring. By i i 1 ∼ [Hart], chapter III, proposition 8.5, R π∗J = H (PR, J ) . So by [Hart], chapter III, theorem 5.1,

2 0 1 ∼ 2 π∗J = π∗O(−3) ⊗ (∧ π∗O(1)) = H (PR, O(−3)) ⊗ (∧ π∗O(1)) = 0.

By Serre’s duality,

1 1 1 1 ∼ 0 1 ∼ ∨ R π∗J = H (PR, O(−1) ⊗ ΩP/S ) = HomR(H (PR, O(1)), R) =(π∗O(1)) . 4.1. FROM GEOMETRIC CUBIC FORMS TO CUBIC ALGEBRAS 25

i For i =06 , 1, R π∗J =0 since the relative dimension of P over S is 1. In the general case we can cover S by small enough afﬁne open subsets U such that −1 1 π (U) = PU . Since the above equalities and isomorphisms are canonical, we get the same equalities as claimed in the lemma.

Proposition 4.1.3. We have the exact sequence

0 ∨ 0 → OS → R π∗(J → O) → (π∗O(1)) → 0, and i R π∗(J → O)=0, if i =06 .

Proof. This is the result of the long exact sequence 4.1 and lemma 4.1.1, 4.1.2.

0 Corollary 4.1.4. A = R π∗(J → O) is a locally free OS-module of rank 3.

Proof. This is because of proposition 4.1.3 and the fact that π∗O(1) is a locally free OS - module of rank 2,

Now we want to deﬁne a commutative unital OS-algebra structure on A. First of all, by deﬁnition

a a (J −→O) ⊗O (J −→O) id ⊗a⊕(−a)⊗id ⊗ ⊗ = J ⊗J −−−−−−−−→ (J ⊗ O) ⊕ (O⊗J ) −−−−−−−→Oa id+id a ⊗ O (4.2) id ⊗a⊕(−a)⊗id = J ⊗J −−−−−−−−→J ⊕ J −−→Oa+a .

Deﬁne r : J⊕J →J by r((i, j)) = i + j, ∀i, j ∈ J . It is easy to check that the following diagram is commutative

id ⊗a⊕(−a)⊗id J ⊗J −−−−−−−−→ J ⊕ J −−−→a+a O

0 r   y0 −−−→ Jy −−−→a O , 26 CHAPTER 4. PROOF OF THE MAIN RESULT

in particular it deﬁnes a morphism θ1 of the two horizontal complexes. By identity (4.2),

θ1 deﬁnes a homomorphism of complexes

a a a θ1 :(J −→O) ⊗O (J −→O) → (J −→O), which induces a morphism

0 a a 0 a ϑ1 : R π∗[(J −→O) ⊗O (J −→O)] → R π∗(J −→O).

Consider the ﬁber product π ×S π : P ×S P → S, and the complex of sheaves

⊠ ∗ ∗ (J → O) (J → O) := pr1(J → O) ⊗OP ×S P pr2(J → O)

on P ×S P .

Proposition 4.1.5. There is a natural isomorphism

0 0 ∼ 0 ⊠ ϑ2 : R π∗(J → O) ⊗OS R π∗(J → O) = R (π ×S π)∗[(J → O) (J → O)].

Proof. Since both J and O are invertible sheaves on P and the morphism π : P → S i is flat, J and O are both flat over S. By proposition 4.1.3, R π∗(J → O) are locally free OS-modules for all i ∈ Z, in particular they are flat OS-modules. So the hypothesis of [EGA III], theorem 6.7.8 holds, and we have the isomorphism stated in the proposition.

Let ∆ : P → P ×S P be the diagonal morphism, it is a closed immersion. Because

∗ ∆ ((J → O) ⊠ (J → O))=(J → O) ⊗O (J → O),

∆ induces a homomorphism

∗ 0 0 ∆ : R (π ×S π)∗[(J → O) ⊠ (J → O)] → R π∗[(J → O) ⊗O (J → O)].

∗ 0 0 0 Deﬁne Θ= ϑ1 ◦ ∆ ◦ ϑ2 : R π∗(J → O) ⊗OS R π∗(J → O) → R π∗(J → O), it ∗ is obvious that this is a bilinear homomorphism. Because ϑ1, ∆ , ϑ2 are symmetric in the 4.1. FROM GEOMETRIC CUBIC FORMS TO CUBIC ALGEBRAS 27

two variables, Θ is also symmetric. 0 Now we will describe Θ more concretely. Because A = R π∗(J → O) is a coherent sheaf on S, we only need to consider the case when S is afﬁne, i.e. S = Spec(R) for some commutative ring R. By [EGA 0] §12.4.3,

0 0 ∼ R π∗(J → O)= H (P, J → O) , where Hi denotes the i-th hypercohomology group. U 1 ˇ Suppose that = {U0, U1} is the standard covering of P = PR. The Cech complex C•(U , J → O) is a J (U01) −−−→ O(U01)

xd xd    a⊕a  J (U0) ⊕J (U1) −−−→ O(U0) ⊕ O(U1), where the degree of O(U01) is (0, 1). By [EGA 0] §12.4.7, H0(P, J → O) = H0(C•(U , J → O)), i.e. it is the 0-th cohomology group of the following complex, in which O(U01) is of degree 1,

(−d)⊕(a⊕a) a⊕d J (U0) ⊕J (U1) −−−−−−−→J (U01) ⊕ O(U0) ⊕ O(U1) −−→O(U01).

0 Let c1,c2 ∈ H (P, J → O), suppose that c1,c2 have representatives (i, f0, f1), (j, g0,g1) ∈

J (U01) ⊕ O(U0) ⊕ O(U1) respectively. Then Θ(c1, c2) has representative (f0j + g0i + aij, f0g0, f1g1) ∈J (U01)) ⊕ O(U0) ⊕ O(U1). Calculating by Cechˇ cocycle, it is easy to see that the injective morphism of sheaves 0 OS ֒→ R π∗(P, J → O) in the exact sequence of proposition 4.1.3 deﬁnes the unital structure of A.

0 Proposition 4.1.6. Let c1, c2, c3 ∈ H (P, J → O), then

Θ((c1, c2), c3)=Θ(c1, (c2, c3)).

0 Proof. Suppose that ci ∈ H (P, J → O) have representatives by Cechˇ cocyles (ji, fi,gi) ∈

J (U01)⊕O(U0)⊕O(U1) for i =1, 2, 3. A direct computationshows that both Θ((c1, c2), c3) 28 CHAPTER 4. PROOF OF THE MAIN RESULT

and Θ(c1, (c2,c3)) can be represented by the cocycle

(f1f2j3 + f2f3j1 + f1f3j2 + af1j2j3 + af2j1j3 + af3j1j2 2 +a j1j2j3, f1f2f3, g1g2g3).

So Θ((c1, c2), c3)=Θ(c1, (c2, c3)).

0 In conclusion Θ deﬁnes a commutative unital OS-algebra structure on A = R π∗(J → O) and A is thus a cubic algebra over S.

∗ 2 −1 Example 4.1.1. Suppose that a ∈ Γ(P, O(3)⊗π (∧ π∗O(1)) ) is not a zero divisor, then we have the exact sequence

0 →J −→Oa → O/J ′ → 0,

′ 0 a where J is the image of J in O under the morphism a. So A = R π∗(J −→O) = ′ π∗(O/J ), and the ring structure on A deﬁned by Θ comes down to the one induced by the ring structure of O/J ′.

Example 4.1.2. Suppose that a = 0, then the complex J → O equals O⊕J [1]. So 0 1 ∨ ∨ A = R π∗(P, O⊕J [1]) = π∗O ⊕ R π∗J = OS ⊕ (π∗O(1)) . Denote m =(π∗O(1)) . Calculating by Cechˇ cocycle, it is easy to ﬁnd that m2 = Θ(m, m)=0.

In this way, we have constructed a cubic algebra A over S from a geometric cubic form (P, O(1), a) over S, the construction commutes with arbitrary base change by corollary 4.1.4 and the expressions with Cechˇ cocycles. So in fact we have constructed a 1-morphism of algebraic stacks F1 : G → A .

4.2 From cubic algebras to geometric cubic forms

In this section, we will construct a 1-morphism F2 : A → G of algebraic stacks. In the following, we will always assume the base scheme S to be noetherian. In fact, since 6 A = [AZ/H], we will lose no generality with this assumption. 4.2. FROM CUBIC ALGEBRAS TO GEOMETRIC CUBIC FORMS 29

Suppose that A is a cubic algebra over S, i.e. A is a sheaf of OS-algebras, locally free of rank 3 as OS -module. Let X = Spec(A), let f : X → S be the structure morphism.

Then f is a ﬁnite ﬂat morphism of degree 3 and A = f∗OX .

Proposition 4.2.1. The sheaf f∗OX /OS is locally free of rank 2 as a sheaf of OS -modules.

Proof. Applying HomOS (•, OS) to the morphism OS ֒→ f∗OX , we get a morphism

HomOS (f∗OX , OS) → OS . We claim that it is surjective. This is a local question, we can assume that S = Spec(R) for a noetherian local ring R. Suppose that x is the only closed point of S. Because the morphism R ֒→ A comes from the unity of A as an R-algebra, we have the exact sequence

0 → R ⊗R k(x)= k(x) → A ⊗R k(x).

So the homomorphism

Homk(x)(A ⊗R k(x),k(x)) → k(x) is surjective. Because A is a free R-module,

Homk(x)(A ⊗R k(x),k(x)) = HomR(A, R) ⊗R k(x).

By Nakayama’s lemma, the homomorphism HomR(A, R) → R is surjective.

Now that HomOS (f∗OX , OS) → OS is surjective, the exact sequence

0 → OS → f∗OX → f∗OX /OS → 0

is locally split. So f∗OX /OS is a locally free OS-module. Obviously it is of rank 2.

∨ Denote V = (f∗OX /OS) , deﬁne P = P(V ). Let π : P → S be the structure morphism. Then P is a family of genus 0 curves on S.

Now we begin the construction of the 1-morphism F2 : A → G . First of all, we restrict ourselves to the case that A is Gorenstein over S. Recall that in §2.2, we have deﬁned a coherent sheaf ω on X such that f∗ω = HomOS (f∗OX , OS), X is said to be Gorenstein if ω is an invertible sheaf on X. 30 CHAPTER 4. PROOF OF THE MAIN RESULT

By the exact sequence

0 → OS → f∗OX → f∗OX /OS → 0, (4.3)

∨ ∨ ∗ we get a morphism of sheaves ψ : V = (f∗OX /OS) ֒→ (f∗OX ) = f∗ω. Because f is ∗ the left adjoint of f∗, we get a morphism of sheaves ϕ : f V → ω.

Proposition 4.2.2. The morphism of sheaves ϕ : f ∗V → ω is surjective.

Proof. Because the problem is local, we can assume that S = Spec(R), R a noetherian local ring. Then A is a R-algebra, free of rank 3 as R-module. We need to prove that the following morphism is surjective:

θ : HomR(A/R, R) ⊗R A → HomR(A, R),

with θ(f ⊗ a)(b) = f(ab), ∀a, b ∈ A, f ∈ HomR(A/R, R), where ab denotes the image of ab in A/R under the projection. Let y ∈ S be the unique closed point of S. By Nakayama’s lemma, we only need to prove that the morphism

θ ⊗ idk(y) : (HomR(A/R, R) ⊗R A) ⊗R k(y) → HomR(A, R) ⊗R k(y) is surjective. Since A is a free R-module, we have

(HomR(A/R, R) ⊗R A) ⊗R k(y) = Homk(y)((A ⊗R k(y))/k(y), k(y)) and

HomR(A, R) ⊗R k(y) = Homk(y)(A ⊗R k(y), k(y)).

So we are reduced to the case that R is a ﬁeld. In fact we can even assume that R is algebraically closed since this is a homomorphism of vector spaces. By the classiﬁcation result of theorem 2.1.2 and example 2.3.1, we need to check the surjectivity of θ for A =

A1, A2 and A3.

(1) For A1 = R × R × R, e1 = (1, 1, 1) is the unity of A1. Let e2 = (0, 1, 0), e3 = 4.2. FROM CUBIC ALGEBRAS TO GEOMETRIC CUBIC FORMS 31

(0, 0, 1), V = Re2 ⊕ Re3. We identify V with A/Re1. Let (f1, f2, f3) be the basis ∨ of A dual to (e1, e2, e3). It is obvious that θ(f2|V ) = f2, θ(f3|V ) = f3, where fi|V

denotes the restriction of fi to V , for i =2, 3. By calculation,

θ(f2 ⊗ (e2 − e1))(e1)= f2(e2 − e1)=1,

θ(f2 ⊗ (e2 − e1))(e2)= f2(e2 − e2)=0,

θ(f2 ⊗ (e2 − e1))(e3)= f2(0) = 0.

So θ(f2 ⊗ (e2 − e1)) = f1 and θ is hence surjective.

2 (2) For A2 = R×(R[ǫ]/(ǫ )), e1 = (1, 1) is the unity of A1. Let e2 = (1, 0), e3 = (0, ǫ), ∨ V = Re2 ⊕ Re3. We identify V with A/Re1. Let (f1, f2, f3) be the basis of A dual

to (e1, e2, e3). It is obvious that θ(f2|V )= f2, θ(f3|V )= f3. A simple calculation as

above tells us that θ(f2 ⊗ (e2 − e1)) = f1, so θ is surjective.

3 2 (3) For A3 = R[ǫ]/(ǫ ), e1 =1 is the unity of A1. Let e2 = ǫ, e3 = ǫ , V = Re2 ⊕ Re3. ∨ We identify V with A/Re1. Let (f1, f2, f3) be the basis of A dual to (e1, e2, e3). It

is obvious that θ(f2|V ) = f2, θ(f3|V ) = f3. A simple calculation as above tells us

that θ(f2 ⊗ e2)= f1, so θ is surjective.

By [Hart], Chap II, proposition 7.12, the surjective morphism of sheaves ϕ : f ∗V → ω determines a morphism φ : X → P = P(V ) of S-schemes.

Lemma 4.2.3. Let h : Y → Z be a ﬁnite morphism of schemes, then h is a closed im- ♯ mersion if and only if the induced morphism of structure sheaves h : OZ → h∗OY is surjective.

Proof. The necessity is obvious. For the sufficiency, since this is a local question, we can assume that Z is affine. Since h is finite, Y is also affine, and replacing Z by the scheme- theoretic image of h, we can assume that h is surjective. Let Y = Spec(R), Z = Spec(T ). ♯ Since h is surjective, the induced morphism of structure sheaves h : T → h∗R is injective. By assumption, it is also surjective, so h♯ is an isomorphism. In particular,e eh♯(Z) : T =

T (Z) → h∗R(Z)= S is an isomorphism, so h is an isomorphism. e e 32 CHAPTER 4. PROOF OF THE MAIN RESULT

Proposition 4.2.4. The S-morphism φ : X → P = P(V ) is a closed immersion.

Proof. The morphism π ◦ φ = f : X → S is finite, in particular it is proper. Since π is separated, φ must be proper by [Hart], chap II, corollary 4.8(e). In fact, φ is a finite morphism since it is obviously quasi-finite. So to prove that φ is a closed immersion, we ♯ only need to prove that the induced morphism of sheaves φ : OP → φ∗OX is surjective by lemma 4.2.3. For this purpose, it is enough to check it for every geometric point on S by Nakayama’s lemma. So we can assume that S = Spec(k), k an algebraically closed field. By the classification result of theorem 2.1.2 and example 2.3.1, we need to check the cases

A = A1, A2, A3.

(1) For A1 = k × k × k, take e1 = (1, 1, 1), e2 = (0, 1, 0), e3 = (0, 0, 1). Let (f1, f2, f3) ∨ ∨ be the basis of A1 dual to (e1, e2, e3). Deﬁne h ∈ A1 by h(ae1 +be2 +ce3)= a+b+

c, ∀a, b, c ∈ k. A simple calculation shows that (−e1 + e2 + e3)h = f1, (e1 − e3)h =

f2, (e1 − e2)h = f3. Let xi = fi/h, i =2, 3, we need to prove that Ax2 = k[x3/x2]

and Ax3 = k[x2/x3]. By calculation, x2 = e1 − e3, x3 = e1 − e2 and Ax2 = e1 e1−e2 e3 e1−e2 But in the ring , e3 since . k[ e1−e3 , e1−e3 ]= k[ e1−e3 , e1−e3 ]. Ax2 e1−e3 =0 e3(e1 −e3)=0

So Ax2 = k[x3/x2]. The same proof gives Ax3 = k[x2/x3].

2 (2) For A = A2 = k × (k[ǫ]/(ǫ )), take e1 = (1, 1), e2 = (0, 1), e3 = (0, ǫ). Let ∨ ∨ (f1, f2, f3) be the basis of A2 dual to (e1, e2, e3). Deﬁne h ∈ A2 by h(ae1 + be2 +

ce3)=2a + b + c, ∀a, b, c ∈ k. A simple calculation shows that (e1 − e2)h =

f1, (−e1 +e2 +e3)h = f2, (e2 −e3)h = f3. Let xi = fi/h, i =2, 3, we need to prove

that Ax2 = k[x3/x2] and Ax3 = k[x2/x3]. By calculation, x2 = −e1 + e2 + e3, x3 = and e1 e2−e3 e3 e2−e3 . But in the ring e2 − e3 Ax2 = k[ −e1+e2+e3 , −e1+e2+e3 ]= k[ −e1+e2+e3 , −e1+e2+e3 ] , e3 since . So . Ax2 −e1+e2+e3 =0 e3(−e1 + e2 + e3)=0 Ax2 = k[x3/x2] Similarly, e1 −e1+e2+e3 e3 −e1+e2+e3 . But in the ring , Ax3 = k[ e2−e3 , e2−e3 ] = k[ e2−e3 , e2−e3 ] Ax3 e3 since . So . e2−e3 =0 e3(e2 − e3)=0 Ax3 = k[x2/x3]

3 2 (3) For A = A3 = k[ǫ]/(ǫ ), (1, ǫ, ǫ ) is a basis of A3. Let (f1, f2, f3) be the basis of ∨ 2 ∨ 2 A3 dual to (1, ǫ, ǫ ). Deﬁne h ∈ A3 by h(a + bǫ + cǫ ) = a + b + c, ∀a, b, c ∈ k. 2 2 A simple calculation shows that ǫ h = f1, (ǫ − ǫ )h = f2, (1 − ǫ)h = f3. Let

xi = fi/h, i = 2, 3, we need to prove that Ax2 = k[x3/x2] and Ax3 = k[x2/x3]. 4.2. FROM CUBIC ALGEBRAS TO GEOMETRIC CUBIC FORMS 33

2 1 1−ǫ By calculation, x2 = ǫ − ǫ , x3 = 1 − ǫ and Ax2 = k[ ǫ−ǫ2 , ǫ−ǫ2 ]=0= k[x3/x2].

Similarly, Ax3 = A = k[x2/x3] since x3 is invertible in A3.

Let D the image of φ : X → P . Since φ is a closed immersion, D is isomorphic to X. Since X is flat over S, D is an effective relative divisor of P over S of degree 3. Define 1 1 ∨ ∗ 2 −1 O(1) = O(D) ⊗ ΩP/S. Then O(D) = O(1) ⊗ (ΩP/S) = O(3) ⊗ π (∧ π∗O(1)) . So ∗ 2 −1 the element 1 ∈ O(D) defines a global section a ∈ Γ(P, O(3) ⊗ π (∧ π∗O(1)) ). In this way we have defined a geometric cubic form (P, O(1), a) over S from a Gorenstein cubic algebra A over S. It is obvious that this construction commutes with arbitrary base change, ′ V G so it defines a 1-morphism of algebraic stacks F2 : → . Generally, consider the category fibered in groupoids B1 of based cubic algebras. By 6 proposition 2.2.6, it is representable by AZ. So there exists a universal based cubic algebra 6 A over AZ. By theorem 2.3.2, the category of based Gorenstein cubic algebra is represented e G2 4 6 by the open subscheme V = a × (AZ\O) of AZ. The restriction A1 of A on V is a ′ e Gorenstein cubic algebra. Let F2(A1)=(P1, OP1 (1), a1), which is a geometric cubic form over V .

Proposition 4.2.5. The geometric cubic form (P1, OP1 (1), a1) over V extends uniquely to 6 a geometric cubic form (P, O(1), a) over AZ.

∨ 6 6 Proof. Deﬁne P = P((A/OAZ · 1) ). Let π : P → AZ be the structural morphism, then e 6 ∨ P is a family of genus 0 curves over AZ. Recall that P1 = P((A1/OV · 1) ), so P is an 6 extension of P1 to AZ. Let π1 : P1 → V be the structural morphism. 6 6 Since the complement of V in AZ is of codimension 4 and AZ is smooth, the following lemma shows that OP1 (1) can be extended uniquely to an invertible sheaf O(1) of P over 6 ∗ 2 −1 AZ of relative degree 1, and the section a1 ∈ Γ(P1, OP1 (3) ⊗ π1 (∧ π1,∗(OP1 (1))) ) can ∗ 2 −1 be extended uniquely to the section a ∈ Γ(P, OP (3) ⊗ π (∧ π∗(OP (1))) ). In this way we obtain a geometric cubic form (P, O(1), a).

Lemma 4.2.6. Suppose that Y is a noetherian, integral, regular, separated scheme. Let U be an open subscheme of Y with a complement of codimension at least 2. Let L be an invertible sheaf on U and s ∈ Γ(U, L), then both L and s can be extended uniquely to Y . 34 CHAPTER 4. PROOF OF THE MAIN RESULT

Proof. Since Y is noetherian, integral, regular and separated, by [Hart], chapter II, proposition 6.11 and proposition 6.15, Pic(Y ) = Cl(Y ) and Pic(U) = Cl(U). Since the complement of U has codimension at least 2, by [Hart], chapter II, proposition 6.5(b), Cl(U) = Cl(Y ). So Pic(U) = Pic(Y ), i.e. every invertible sheaf on U can be extended uniquely to an invertible sheaf on Y . Let L′ be the extension of L to Y . Now that every invertible sheaf on U and Y can be embedded into the constant sheaf K(U) = K(Y ), every section s ∈ Γ(U, L) can be extended uniquely to a section s′ ∈ Γ(Y, L′).

6 Proposition 4.2.7. The geometric cubic form (P, O(1), a) over AZ is H-equivariant.

6 Proof. For any point x on AZ, denote by (Px, OPx (1), ax) the ﬁber of (P, O(1), a) at x.

We need to prove that the geometric cubic form (Phx, OPhx (1), ahx) on Spec(k(hx)) is isomorphic to the geometric cubic form (Px, OPx (1), ax) on Spec(k(x)) for any h ∈ H. ′ ∗ 6 Let h ∈ H, A := h A is a cubic algebra over AZ. We constructed as above a geometric ′ e e′ 6 ′ ′ ′ cubic form (P , OP ′ (1), a ) on AZ corresponding to A . Then the ﬁber of (P , OP ′ (1), a ) e at x is just (Phx, OPhx (1), ahx). ′ Recall that the construction of F2 doesn’t depend on the choices of basis, so the restric- ′ ′ ′ tion of (P , OP (1), a ) on V and (P1, OP1 (1), a1) are the same. Since (P, O(1), a) is the ′ ′ 6 ′ unique extension of (P1, OP1 (1), a1) to AZ, we have (P, O(1), a)=(P , OP (1), a ), in particular their ﬁbers at x will be the same.

6 The geometric cubic form (P, O(1), a) over AZ corresponds to a 1-morphism of alge- 1 6 braic stacks F : B = AZ → G . By proposition 4.2.7, F factors through the quotient 1 A = [B /H] and deﬁnes a 1-morphism F2 : A → G .

4.3 The correspondence is bijective

In this section we prove that the 1-morphisms F1 : G → A and F2 : A → G are inverse to each other. As a corollary, we get the main theorem 1.

Proposition 4.3.1. The 1-morphism F1 : G → A is left inverse to F2 : A → G . 4.3. THE CORRESPONDENCE IS BIJECTIVE 35

Proof. By theorem 2.2.7 and 2.3.2, V is an open sub algebraic stack of the smooth Artin stack A with a complement of codimension 4. Since F2 is constructed as the unique ′ V extension of F2, we only need to prove that F1 ◦ F2 is isomorphic to the identity on .

Suppose that A is a Gorenstein cubic algebra over S and F1(A)=(P, O(1), a). We use ∼ the notation of section 4.2. Since φ is a closed immersion, A = π∗OD. By deﬁnition of ∗ 2 −1 ∗ 2 O(1), we have O(D)= O(3) ⊗ π (∧ π∗O(1)) ). Let J := O(−3) ⊗ π (∧ π∗O(1)) = ∗ 2 −1 O(−D). By deﬁnition, the section a ∈ Γ(P, O(3) ⊗ π (∧ π∗O(1)) ) correspond to 1 ∈ Γ(P, O(D)). So we have the commutative diagram of exact sequences

0 −−−→ J −−−→a O −−−→ O/J ′ −−−→ 0

1 0 −−−→ O(−D) −−−→ O −−−→ OD −−−→ 0, where J ′ is the image of J in O under a. Since 1 is obviously not a zero divisor of ′ Γ(P, O(D)), a is not a zero divisor. So by example 4.1.1, F2((P, O(1), a)) = π∗(O/J )= ∼ π∗OD = A and F1 ◦ F2 is identity on V .

Proposition 4.3.2. The 1-morphism F1 : G → A is right inverse to F2 : A → G .

Proof. Since W is an open sub algebraic stack of the smooth Artin stack G with complement of codimension 4, we only need to prove that F2 ◦ F1 is identity on W . Suppose that (P, OP (1), a) is a primitive geometric cubic form over S. Let J := OP (−3) ⊗ ∗ 2 1 π (∧ π∗OP (1)) = OP (−1) ⊗ ΩP/S. By proposition 3.3.1, the morphism of sheaves a ′ ′ J −→OP is injective. Let J be the image of J in OP under the morphism a. J is an ideal sheaf in OP , it deﬁnes an effective relative divisor D of P over S of degree 3. In other words, We have the commutative diagram of exact sequences

a ′ 0 −−−→ J −−−→ OP −−−→ OP /J −−−→ 0

  y 1 0 −−−→ OP (−D) −−−→ OP −−−→ OD −−−→ 0, in which the ﬁrst downward arrow is an isomorphism.

Let A := F1((P, OP (1), a)). Since a is not a zero divisor, by example 4.1.1, we have 36 CHAPTER 4. PROOF OF THE MAIN RESULT

′ ′ ′ A = π∗(O/J )= π∗OD. Let F2(A)=(P , OP ′ (1), a ). By proposition 4.1.3, we have the exact sequence 1 ∨ 0 → OS −→ A → (π∗OP (1)) → 0.

′ ∨ By construction in §4.2, P = P((A/OS · 1) ) = P(π∗OP (1)) = P by the above exact .sequence. Let φ : Spec(A) ֒→ P ′ = P be the closed immersion constructed in §4.2 ′ ∼ Since AutS(P ) acts strictly three times transitive on P , we have D := φ(Spec(A)) = ′ ′ ′ 1 ∼ D after applying an automorphism of P . So OP (1) = OP (D ) ⊗ ΩP ′/S = OP (D) ⊗ 1 ∼ ∗ 2 −1 1 ΩP/S = (OP (3) ⊗ π (∧ π∗OP (1)) ) ⊗ ΩP/S = OP (1). Denote the composition of these ∗ 2 ′ isomorphisms by η. Let J1 = OP ′ (−3) ⊗ π (∧ (π∗OP ′ (1))). By the construction of a , we have the following commutative diagram of exact sequences

′ a ′ ′ ′ 0 −−−→ J1 −−−→ OP −−−→ OP /J1 −−−→ 0

′ 1 0 −−−→ OP ′ (−D ) −−−→ OP ′ −−−→ OD′ −−−→ 0,

′ ′ ′ where J1 is the image of J1 in OP under the morphism a . Combining the above two com- ∼ mutative diagram of exact sequences, we ﬁnd that the isomorphism η : OP ′ (1) = OP (1) ∗ 2 −1 ∗ 2 −1 induces an isomorphism between OP ′ (3)⊗π (∧ π∗OP ′ (1)) and O(3)⊗π (∧ π∗O(1)) ′ sending a to a. So F2(A)=(P, OP (1), a) in G and F2 ◦ F1 is identity on W . Bibliography

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