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 Mathematisch Instituut, Universiteit Leiden June 2008

c Copyright by Zongbin CHEN 2008 All Rights Reserved ii

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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 GL 2 (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, Sym 3 (V ) ( 2 V ) 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 a Γ(P, O(3) π ( 2 π O(1)) 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

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

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Contents Preface Acknowledgements v vii 1 Main result 1 1.1 Basic definitions................................ 1 1.2 Statement of main results........................... 3 1.3 Sketch of the proof.............................. 4 2 The algebraic stack of cubic algebras 7 2.1 Classification over algebraically closed fields................ 7 2.2 Smoothness and dimension.......................... 11 2.3 Gorenstein cubic algebras........................... 13 3 The algebraic stack of twisted cubic forms 17 3.1 Classification over algebraically closed fields................ 17 3.2 Smoothness and dimension.......................... 20 3.3 Primitive geometric cubic forms....................... 22 4 Proof of the main result 23 4.1 From geometric cubic forms to cubic algebras................ 23 4.2 From cubic algebras to geometric cubic forms................ 28 4.3 The correspondence is bijective........................ 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 (V, p), where V is a locally free sheaf of O S -modules of rank 2 and p Γ(S, Sym 3 (V ) ( 2 V ) 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 (V 1, p 1 ) over S 1 to another twisted cubic form (V 2, p 2 ) over S 2 consists of a morphism f : S 1 S 2 and an isomorphism g : V 1 f V 2 such that g (f (p 2 )) = p 1. 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, O P (1), a), where π : P S is a family of genus 0 curves, O P (1) is an invertible sheaf on P of relative degree 1 over S, and a Γ(P, O P (3) π ( 2 π (O P (1))) 1 ). Remark 1.1.5. The existence of the invertible sheaf O P (1) assures that over every closed point x S, π 1 (x) is isomorphic to P 1 k(x), where k(x) = O S,x/m 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, O P (1), a) over objects S in S. A morphism from a geometric cubic form (P 1, O P1 (1), a 1 ) over S 1 to a geometric cubic form (P 2, O P2 (1), a 2 ) over S 2 consists of two morphisms f : S 1 S 2, g : P 1 P 2 and an isomorphism h : O P1 (1) g O P2 (1) such that the diagram P 1 S 1 g P 2 f S 2 is Cartesian, and a 1 = h (g (a 2 )). The functor q : G S taking a geometric cubic form (P, O P (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 O S - algebras, locally free of rank 3 as an O S -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 A 1 over S 1 to a cubic algebra A 2 over S 2 consists of a morphism f : S 1 S 2 and an isomorphism g : A 1 f A 2 as cubic algebras over S 1. 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 : S 1 S 2 is a morphism in S and (V 2, p 2 ) is a twisted cubic form over S 2. Define V 1 = f V 2, p 1 = f p 2, then (V 1, p 1 ) is a twisted cubic form over S 1. The pair of morphisms consisting of f and the identity map id : V 1 f V 2 defines a morphism ϕ from the twisted cubic form (V 1, p 1 ) over S 1 to (V 2, p 2 ) over S 2 such that q(ϕ) = f. (2) Suppose that (V i, p i ) are twisted cubic forms over S i, i = 1, 2, 3. Suppose that for i = 1, 2, we are given morphisms f i : S i S 3, isomorphisms h i : V i f i V 3 defining a morphism from the twisted cubic form (V i, p i ) over S i to (V 3, p 3 ) over S 3. Suppose that f : S 1 S 2 is a morphism satisfying f 1 = f 2 f, then f 1V = (f 2 f) (V ) = f (f2 (V )). Pulling back the isomorphism h 2 : V 2 f2 V by f to S 1, we get an isomorphism h 2 : f V 2 f (f2(v )) = f1v. Define the isomorphism h = h 1 2 h 1 : V 1 f V 2, then h and f defines the desired morphism from the twisted cubic form (V 1, p 1 ) over S 1 to (V 2, p 2 ) over S 2. 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 O S -modules of rank 2 and p is a section of Sym 3 (V ) OS ( 2 V ) 1. Define P = P(V ) = Proj( n=0 Symn V ), then π : P = P(V ) S is a family of genus 0 curves. Let O(1) 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, Γ(P, O(3) π ( 2 π (O(1))) 1 ) = Γ(S, Sym 3 (V ) ( 2 V ) 1 ). Denote by a the element in Γ(P, O(3) π ( 2 π (O(1)) 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) π ( 2 π (O(1)) 1 ). Let V = π (O(1)), then V is a locally free sheaf of rank 2 over S. As above, we have Γ(P, O(3) π ( 2 π (O(1)) 1 ) = Γ(S, Sym 3 (V ) ( 2 V ) 1 ). Let p be the element in Γ(S, Sym 3 (V ) ( 2 V ) 1 ) corresponding to a Γ(P, O(3) π ( 2 π (O(1)) 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 F 1 : G A and F 2 : 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 structural morphism. Denote J := O( 3) π ( 2 π (O(1))). Since a Γ(P, O(3) π ( 2 π (O(1))) 1 ), we can define a complex J a O. Let A := R 0 π (J a O),

1.3. SKETCH OF THE PROOF 5 where R 0 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 O S -modules of rank 3. What s more, A can be endowed with a product structure. It comes from a product on the complex J a O, deduced from the O-module structure of J. Calculating with Čech 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 F 1 : G A. The construction of the 1-morphism F 2 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 Spec(A). Define O(1) := O(D) Ω 1 P/S. The section 1 of O(D) defines a section a of O(1) (Ω 1 P/S ) = O(3) π ( 2 (π O(1))) 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 F 2 : V G. In the general case, we use the fact that V is an open sub algebraic stack of A and its complement has codimension 4 in A to extend F 2 to a 1-morphism F 2 : A G. 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 F 1 and F 2 respectively.

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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) A 1 = k k k. (2) A 2 = k (k[ε]/(ε 2 )). (3) A 3 = k[ε]/(ε 3 ). (4) A 4 = k[α, β]/(α 2, αβ, β 2 ). 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 a 1 )(T a 2 )(T a 3 ), for some a 1, a 2, a 3 k. If all the three roots of P(T) are distinct, we get A = k[t]/((t a 1 )(T a 2 )(T a 3 )) = k k k. If there is one root of P(T) appearing with multiplicity 2, for example a 1 = a 2 a 3, we get A = k[t]/((t a 1 ) 2 (T a 3 )) = k (k(ε)/(ε 2 )) If a 1 = a 2 = a 3, A = k[t]/(t a 1 ) 3 = k[ε]/(ε 3 ). (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 α 2 aα c = 0 and β 2 bβ d = 0. Consider the homomorphism of algebras ρ : A End k (A) defined by ρ(m)(n) = mn, m, n A. Under the basis (1, α, β), we find that and ρ(1) = id. By calculation, 0 1 0 0 1 ρ(α) = c a, ρ(β) = x 0, x 0 d b 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, α 2 = aα, β 2 = 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 = r 2 a + mr = s 2 b + ms = n, for any r, s k, so we must have a = b = m = 0. Now we have α 2 = β 2 = αβ = 0, so A = k[α, β]/(α 2, β 2, αβ). For any r, s, t k, (r +sα +tβ) 2 = r 2 +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 polynomial of degree at most 2, then A = k[α, β]/(α 2, αβ, β 2 ). Theorem 2.1.4. We have the automorphism groups of the cubic algebras: (1) Aut k (k k k) = S 3. (2) Aut k (k (k(ε)/(ε 2 ))) = k. (3) Aut k (k(ε)/(ε 3 )) { ( = b c ) } b b k, c k. 2 (4) Aut k (k[α, β]/(α 2, αβ, β 2 )) = GL 2 (k). Proof. (1) For A = k k k, take e 1 = (1, 0, 0), e 2 = (0, 1, 0), e 3 = (0, 0, 1). Then e 2 i = e i, e i e j = 0, i, j = 1, 2, 3; i j, i.e. e i are idemptents. So f Aut k (A) if and only if f(e i ), i = 1, 2, 3 are also idemptent. It is obvious that for any σ S 3, the homomorphism defined by σ(e i ) = e σ(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 x 1 e 1 + x 2 e 2 + x 3 e 3, x 1, x 2, x 3 k satisfies x 2 = x, a simple calculation tells us that x 2 i = x i, so x i = 0 or 1, i = 1, 2, 3. Taking into account that f(e i ) 0, f(e i )f(e j ) = 0, i, j = 1, 2, 3; i j, we get that f(e i ) = e σ(i), i = 1, 2, 3, for some σ S 3. So Aut k (k k k) = S 3. (2) For A = k (k[ε]/(ε 2 )), 1 = (1, 1). Take α = (1, 0), β = (0, ε), then (1, α, β) forms a basis of A over k and α 2 = α, β 2 = 0, αβ = 0. So f Aut k (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 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 a 2 = 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 k, so Aut k (k (k[ε]/(ε 2 ))) = k. (3) For A = k[ε]/(ε 3 ). Let f Aut k (A), then f(1) = 1. Suppose that f(ε) = a + bε + cε 2, a, b, c k. By f(ε) 3 = f(ε 3 ) = 0, we get a = 0. By f(ε 2 ) = f(ε) 2 = b 2 ε 2, we get b 0 since f is an automorphism. In conclusion, under the base (1, ε, ε 2 ), f will be of the form 1 b c, b k, c k. b 2 It is easy to check that any such f defines an automorphism of A over k. So Aut k (k(ε)/(ε 3 )) { ( = b c ) } b b k, c k. 2 (4) For A = k[α, β]/(α 2, αβ, β 2 ). Let f Aut k (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 n p) is

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 Aut k (A) = GL 2 (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 (A, φ), where A is a cubic algebra over S, φ : O 3 S such that φ(1, 0, 0) = 1. A is an isomorphism of O S-modules Definition 2.2.2. Let S be an arbitrary scheme, a good based cubic algebra over S is a pair (A, φ), where A is a cubic algebra over S, φ : O 3 S A is an isomorphism of O S-modules such that φ(1, 0, 0) = 1, φ(0, 1, 0)φ(0, 0, 1) O S (S). Definition 2.2.3. Let B 1 be the category of based cubic algebras. Its objects are the based cubic algebras over S. A morphism from a based cubic algebra (A 1, φ 1 ) over S 1 to (A 2, φ 2 ) over S 2 consists of a morphism f : S 1 S 2, an isomorphism h : A 1 f A 2 such that φ 1 = h 1 (f (φ 2 )). Definition 2.2.4. Let B 1,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 (A 1, φ 1 ) over S 1 to (A 2, φ 2 ) over S 2 consists of a morphism f : S 1 S 2, and an isomorphism h : A 1 f A 2 such that φ 1 = h 1 (f (φ 2 )). Proposition 2.2.5. The categories B 1 and B 1,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 Proposition 2.2.6. The category fibered in groupoids B 1,g is representable by A 4 Z and B1 is representable by A 6 Z. 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 α 2 = ac + bα aβ β 2 = bd + dα cβ αβ = ad, (2.1) 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 deduce B 1,g = A 4 Z. There is a left action of G 2 a on B1 by which (m, n) G 2 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: G 2 a B 1,g B 1, (2.2) so B 1 = A 6 Z. Denote by H the closed subgroup scheme of GL 3 stabilizing the element (1, 0, 0). H acts naturally on the category fibered in groupoids B 1. For (A, φ) a based cubic algebra over S, let hφ : O 3 S h(a, φ) := (A, hφ). A be defined as hφ(x) = φ(hx), x O 3 S, h H. Then 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 A = [B 1 /H] as stack over S. By [L-M], example 4.6.1, [B 1 /H] = [A 6 Z /H] is an Artin stack. Since A 6 Z is smooth and H is a smooth group scheme over Z, [B1 /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 Proof. Since dim(a 6 Z ) = dim(h) = 6 and A = [A6 Z /H], we get dim(a ) = dim(a6 Z ) 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 O S -algebras, locally free of rank 3 as an O S -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 O X. Since f is a finite flat morphism, the sheaf Hom OS (f O X, O S ) is a coherent f O X - module and there is a coherent sheaf ω on X such that f ω = Hom OS (f O X, O S ). 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(x 2 ) = 1. A simple calculation shows that (f, xf, x 2 f) 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]/(α 2, αβ, β 2 ). 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 A 1, A 2, A 3 are Gorenstein, while A 4 is not Gorenstein. Since the pull-back of a Gorenstein cubic algebra is always Gorenstein, we can define the sub stack V of Gorenstein cubic algebras of A. Let O the origin of A 4 Z, let V be the image of G 2 a (A4 Z \O) in A6 Z 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, A x is a projective A x -module. So we can assume that S = Spec(R), R a noetherian local ring with residue field k. Let M = Hom R (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 Tor R 1 (M, k) = 0 N R k A R k ϕ id k M R k 0, the first equality is because M is a free R-module. Since ϕ id k is an isomorphism, N R k = 0. By Nakayama s lemma, N = 0. So M is a free A-module. Proof of theorem 2.3.2. Since A = [A 6 Z /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 O S. Then A together with this good basis corresponds uniquely to an S-point Q = (a, b, c, d) of A 4 Z 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 of Q doesn t intersect with the closed subscheme O of A 4 Z. So Q factors through the open subscheme A 4 Z \O and the category fibered in groupoids of Gorenstein cubic algebras with good basis is represented by the open subscheme A 4 Z \O of A4 Z. Similar to the proof of the second conclusion in proposition 2.2.6, we have V = [G 2 a (A 4 Z \O)/H] = [V /H].

2.3. GORENSTEIN CUBIC ALGEBRAS 15 The complement of V in A is [G 2 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 (V, p), where V is a vector space of dimension 2 over k and p Sym 3 (V ) k ( 2 V ) 1. Two twisted cubic forms (V 1, p 1 ) and (V 2, p 2 ) are said to be isomorphic if there exists an isomorphism of vector spaces f : V 1 V 2 such that p 1 = f p 2. 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 (v 1, v 2 ) of V. Theorem 3.1.1. A twisted cubic form (V, p) over k is isomorphic to one of the following: (1) p 1 = v 1 v 2 (v 1 v 2 ) k (v 1 v 2 ) 1. (2) p 2 = v1 2v 2 k (v 1 v 2 ) 1. (3) p 3 = v1 3 k (v 1 v 2 ) 1. (4) p 4 = 0. Proof. Obviously (V, 0) is a twisted cubic form. Suppose that p Sym 3 (V ) k ( 2 V ) 1, p 0. Let p = (av1 3 + bv1v 2 2 + cv 1 v2 2 + dv2) 3 k (v 1 v 2 ) 1, 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 p = (x 1 v 1 y 1 v 2 )(x 2 v 1 y 2 v 2 )(x 3 v 1 y 3 v 2 ) k (v 1 v 2 ) 1, for some x i, y i k, i = 1, 2, 3. Let a i = [x i : y i ], i = 1, 2, 3 be three points on P 1 k. (1) Suppose that a i, i = 1, 2, 3 are distinct to each other. It is well known that there exists an element γ = ( a b c d ) PGL2 (k) such that γ(a 1 ) = [1 : 0], γ(a 2 ) = [0 : 1], γ(a 3 ) = [1 : 1], where γ(z) = az+b cz+d, z P1 k. For any γ GL 2 (k) whose image in PGL 2 (k) is γ, we have γ (v 1 v 2 (v 1 v 2 ) k (v 1 v 2 ) 1 ) = m(γ )p, for some m(γ ) k. Since k is a field, we can choose appropriately an element γ GL 2 (k) whose image in PGL 2 (k) is γ, such that γ (v 1 v 2 (v 1 v 2 ) k (v 1 v 2 ) 1 ) = p. (2) Suppose that a 1 = a 2 a 3. As in the above, there exists one element γ PGL 2 (k) such that γ(a 1 ) = γ(a 2 ) = [1 : 0], γ(a 3 ) = [0 : 1]. Similar to the case (1), we can choose an element γ GL 2 (k) whose image in PGL 2 (k) is γ, such that γ (v 2 1 v 2 k (v 1 v 2 ) 1 ) = p. The cases a 1 = a 3 a 2 and a 2 = a 3 a 1 can be treated in the same way. (3) Suppose that a 1 = a 2 = a 3. There exists one element γ PGL 2 (k) such that γ(a 1 ) = γ(a 2 ) = γ(a 3 ) = [1 : 0].

3.1. CLASSIFICATION OVER ALGEBRAICALLY CLOSED FIELDS 19 Similar to case (1), we can choose an element γ GL 2 (k) whose image in PGL 2 (k) is γ, such that γ (v 3 1 k (v 1 v 2 ) 1 ) = p. Theorem 3.1.2. The automorphism groups of the above twisted cubic forms are (1) Aut k ((V, p 1 )) = S 3. (2) Aut k ((V, p 2 )) = k. (3) Aut k ((V, p 3 )) = {( a b a 2 ) GL2 (k) a k, b k }. (4) Aut k ((V, p 4 )) = GL 2 (k). Proof. Let γ = ( a b c d ) GL2 (k), then γ(v 1 ) = av 1 + cv 2, γ(v 2 ) = bv 1 + dv 2. (1) Suppose that γ Aut k ((V, p 1 )). Denote by X the zero section of the cubic form v 1 v 2 (v 1 v 2 ) over P(V ) = P 1 k. Let x 1 = [0 : 1], x 2 = [1 : 0], x 3 = [1 : 1], then X = x 1 + x 2 + x 3 as a divisor on P 1 k. By assumption γ p 1 = p 1, so γ induces an automorphism of X. Since Aut(X) = S 3, we obtain a homomorphism j : Aut k ((V, p 1 )) S 3. We claim that j is injective. In fact, for γ Aut k ((V, p 1 )) such that j(γ) = 1, because (ad bc) 1 (av 1 + cv 2 )(bv 1 + dv 2 )[(a b)v 1 + (c d)v 2 ] k (v 1 v 2 ) 1 and γ(x i ) = x i, i = 1, 2, 3, we get equalities = v 1 v 2 (v 1 v 2 ) k (v 1 v 2 ) 1, c = 0, b = 0 av 1 dv 2 = v 1 v 2, i.e. γ = ( 1 1 ). So j is injective.

20 CHAPTER 3. THE ALGEBRAIC STACK OF TWISTED CUBIC FORMS Conversely, for any permutation σ S 3, there exists a unique a σ PGL 2 (k) = Aut k (P 1 k ), such that a σ(x i ) = x σ(i). Then for any A GL 2 (k) whose image in PGL 2 (k) is a σ, we have A (p 1 ) = m(a)p 1, for some m(a) k. Since k is a field, we can find an A 0 GL 2 (k) whose image in PGL 2 is a σ such that A 0(p 1 ) = p 1. So j is surjective, hence an isomorphism between Aut k ((V, p)) and S 3. (2) Suppose that γ Aut k ((V, p 2 )). Denote by X the zero section of the cubic form v1 2v 2 over P(V ) = P 1 k. Let x 1 = [0 : 1], x 2 = [1 : 0], then X = 2x 1 + x 2 as a divisor on P 1 k. Since γ (p 2 ) = p 2, we get γ(x 1 ) = x 1 and γ(x 2 ) = x 2. Combining with the equality (ad bc) 1 (av 1 + cv 2 ) 2 (bv 1 + dv 2 ) k (v 1 v 2 ) 1 = v1 2 v 2 k (v 1 v 2 ) 1. We get a = 1, b = c = 0, i.e. γ = ( ) 1 d with d k. It is easy to check that any such γ defines an automorphism of (V, p 2 ). So Aut k ((V, p 2 )) = k. (3) γ Aut k ((V, p 3 ) if and only if (ad bc) 1 (av 1 + cv 2 ) 3 k (v 1 v 2 ) 1 = v 3 1 k (v 1 v 2 ) 1. Comparing the two sides of the equality, we get c = 0, a 2 = d. So { ( Aut k ((V, p 3 )) = a b ) } a a k, b k. 2 (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 (V, p, ϕ), where (V, p) is a twisted cubic form over S, φ : O 2 S O S -modules. V is an isomorphism of

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 (V 1, p 1, ϕ 1 ) over S 1 to (V 2, p 2, ϕ 2 ) over S 2 consists of a morphism f : S 1 S 2, an isomorphism h : V 1 f V 2 such that p 1 = h (f (p 2 )) and ϕ 1 = h 1 (f (ϕ 2 )). Proposition 3.2.3. The category F is a category fibered in groupoids over S, it can be represented by A 4 Z. Proof. The first conclusion is a simple consequence of the existence of a well defined pullback. 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 O S -module. Let e 1 = ϕ((1, 0)), e 2 = ϕ((0, 1)), then p can be written uniquely as (ae 3 1 + be2 1 e 2 + ce 1 e 2 2 + de3 2 ) (e 1 e 2 ) 1, for some a, b, c, d R, from which the proposition is easily deduced. The linear algebraic group GL 2,Z acts naturally on F. Suppose that (V, p, ϕ) is a based twisted cubic form on S. Let g GL 2 (O S ), define gϕ : O 2 S V to be gϕ(x) = ϕ(gx), x O 2 S. Define g(v, ϕ, p) = (V, gϕ, g (p)), it is easily verified that this defines an action of GL 2 on F. Theorem 3.2.4. The stack F is a smooth Artin stack. Proof. Obviously we have F = [F / GL 2 ] = [A 4 / GL 2 ]. Since both A 4 and GL 2 are smooth, [A 4 / GL 2 ] is a smooth Artin stack. So F is also a smooth Artin stack. Proposition 3.2.5. The dimension of F is 0. Proof. Since F = [F / GL 2 ] = [A 4 / GL 2 ] and dim(a 4 ) = dim(gl 2 ) = 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 Let O the origin of A 4 Z, V = A4 Z \O. Obviously [V / GL 2] is an open sub algebraic stack of F = [A 4 Z / GL 2], with complement of codimension 4. The objects of [V / GL 2 ] 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 / GL 2 ] 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, let J := O(3) π ( 2 π O(1)) 1. Then the morphism of sheaves J a O is injective. Proof. By assumption, a is non zero at every point, so the morphism of sheaves J a O 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 F 1 : G A. Suppose that (P, O(1), a) is a geometric cubic form, i.e. π : P S is a family of genus 0 curves, a Γ(P, O(3) π ( 2 π O(1)) 1 ). Let J := O( 3) π ( 2 π O(1)), then we have J 1 = O(3) π ( 2 π O(1)) 1. So a Γ(P, J 1 ) = 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 A := R 0 π (J a O), the 0-th hypercohomology of the complex J O. From the exact sequence of complexes 0 O (J a O) J [1] 0, we obtain the following long exact sequence by applying π, R i π O = R i π O R i π (J a O) R i π (J [1]) = R i+1 π (J ). (4.1) 23

24 CHAPTER 4. PROOF OF THE MAIN RESULT Lemma 4.1.1. We have π O = O S, R i π O = 0, if i 0. Proof. Firstly, we suppose that S is affine and P = P 1 S. Let S = Spec(R), R a commutative ring. By [Hart], chapter III, proposition 8.5, R i π O = H i (P 1 R, O). So by [Hart], chapter III, theorem 5.1, π O = Γ(P 1 R, O) = O S, and for i 0, R i π O = H i (P 1 R, O) = 0. In general case we can cover S by small enough affine open subsets U such that π 1 (U) = P 1 U. Since the above equalities are canonical, we get the same equalities in general case. Lemma 4.1.2. We have Proof. First of all, observe that R i π (J ) = 0, if i 1, R 1 π (J ) = (π O(1)). J = O( 3) π ( 2 π O(1)) = O( 1) Ω 1 P/S. Suppose that S is affine and P = P 1 S. Let S = Spec(R), R a commutative ring. By [Hart], chapter III, proposition 8.5, R i π J = H i (P 1 R, J ). So by [Hart], chapter III, theorem 5.1, π J = π O( 3) ( 2 π O(1)) = H 0 (P 1 R, O( 3)) ( 2 π O(1)) = 0. By Serre s duality, R 1 π J = H 1 (P 1 R, O( 1) Ω 1 P/S) = Hom R (H 0 (P 1 R, O(1)), R) = (π O(1)).

4.1. FROM GEOMETRIC CUBIC FORMS TO CUBIC ALGEBRAS 25 For i 0, 1, R i 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 π 1 (U) = P 1 U. 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 O S R 0 π (J O) (π O(1)) 0, and R i π (J O) = 0, if i 0. Proof. This is the result of the long exact sequence 4.1 and lemma 4.1.1, 4.1.2. Corollary 4.1.4. A = R 0 π (J O) is a locally free O S -module of rank 3. Proof. This is because of proposition 4.1.3 and the fact that π O(1) is a locally free O S - module of rank 2, Now we want to define a commutative unital O S -algebra structure on A. First of all, by definition (J a O) O (J a O) = J J = J J id a ( a) id a id + id a (J O) (O J ) O O id a ( a) id J J a+a O. (4.2) 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 J J 0 id a ( a) id J J r 0 J a+a O a O,

26 CHAPTER 4. PROOF OF THE MAIN RESULT in particular it defines a morphism θ 1 of the two horizontal complexes. By identity (4.2), θ 1 defines a homomorphism of complexes θ 1 : (J a O) O (J a O) (J a O), which induces a morphism ϑ 1 : R 0 π [(J a O) O (J a O)] R 0 π (J a O). Consider the fiber product π S π : P S P S, and the complex of sheaves (J O) (J O) := pr 1 (J O) O P S P pr 2 (J O) on P S P. Proposition 4.1.5. There is a natural isomorphism ϑ 2 : R 0 π (J O) OS R 0 π (J O) = R 0 (π S π) [(J O) (J O)]. Proof. Since both J and O are invertible sheaves on P and the morphism π : P S is flat, J and O are both flat over S. By proposition 4.1.3, R i π (J O) are locally free O S -modules for all i Z, in particular they are flat O S -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 : R 0 (π S π) [(J O) (J O)] R 0 π [(J O) O (J O)]. Define Θ = ϑ 1 ϑ 2 : R 0 π (J O) OS R 0 π (J O) R 0 π (J O), it is obvious that this is a bilinear homomorphism. Because ϑ 1,, ϑ 2 are symmetric in the

4.1. FROM GEOMETRIC CUBIC FORMS TO CUBIC ALGEBRAS 27 two variables, Θ is also symmetric. Now we will describe Θ more concretely. Because A = R 0 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, R 0 π (J O) = H 0 (P, J O), where H i denotes the i-th hypercohomology group. Suppose that U = {U 0, U 1 } is the standard covering of P = P 1 R. The Čech complex C (U, J O) is J (U 01 ) d J (U 0 ) J (U 1 ) where the degree of O(U 01 ) is (0, 1). a O(U 01 ) d a a O(U 0 ) O(U 1 ), By [EGA 0] 12.4.7, H 0 (P, J O) = H 0 (C (U, J O)), i.e. it is the 0-th cohomology group of the following complex, in which O(U 01 ) is of degree 1, J (U 0 ) J (U 1 ) ( d) (a a) J (U 01 ) O(U 0 ) O(U 1 ) a d O(U 01 ). Let c 1, c 2 H 0 (P, J O), suppose that c 1, c 2 have representatives (i, f 0, f 1 ), (j, g 0, g 1 ) J (U 01 ) O(U 0 ) O(U 1 ) respectively. Then Θ(c 1, c 2 ) has representative (f 0 j + g 0 i + aij, f 0 g 0, f 1 g 1 ) J (U 01 )) O(U 0 ) O(U 1 ). Calculating by Čech cocycle, it is easy to see that the injective morphism of sheaves O S R 0 π (P, J O) in the exact sequence of proposition 4.1.3 defines the unital structure of A. Proposition 4.1.6. Let c 1, c 2, c 3 H 0 (P, J O), then Θ((c 1, c 2 ), c 3 ) = Θ(c 1, (c 2, c 3 )). Proof. Suppose that c i H 0 (P, J O) have representatives by Čech cocyles (j i, f i, g i ) J (U 01 ) O(U 0 ) O(U 1 ) for i = 1, 2, 3. A direct computation shows that both Θ((c 1, c 2 ), c 3 )

28 CHAPTER 4. PROOF OF THE MAIN RESULT and Θ(c 1, (c 2, c 3 )) can be represented by the cocycle (f 1 f 2 j 3 + f 2 f 3 j 1 + f 1 f 3 j 2 + af 1 j 2 j 3 + af 2 j 1 j 3 + af 3 j 1 j 2 So Θ((c 1, c 2 ), c 3 ) = Θ(c 1, (c 2, c 3 )). +a 2 j 1 j 2 j 3, f 1 f 2 f 3, g 1 g 2 g 3 ). In conclusion Θ defines a commutative unital O S -algebra structure on A = R 0 π (J O) and A is thus a cubic algebra over S. Example 4.1.1. Suppose that a Γ(P, O(3) π ( 2 π O(1)) 1 ) is not a zero divisor, then we have the exact sequence 0 J a O O/J 0, where J is the image of J in O under the morphism a. So A = R 0 π (J a 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 A = R 0 π (P, O J [1]) = π O R 1 π J = O S (π O(1)). Denote m = (π O(1)). Calculating by Čech cocycle, it is easy to find that m 2 = 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 Čech cocycles. So in fact we have constructed a 1-morphism of algebraic stacks F 1 : G A. 4.2 From cubic algebras to geometric cubic forms In this section, we will construct a 1-morphism F 2 : A G of algebraic stacks. In the following, we will always assume the base scheme S to be noetherian. In fact, since A = [A 6 Z /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 O S -algebras, locally free of rank 3 as O S -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 O X. Proposition 4.2.1. The sheaf f O X /O S is locally free of rank 2 as a sheaf of O S -modules. Proof. Applying Hom OS (, O S ) to the morphism O S f O X, we get a morphism Hom OS (f O X, O S ) O S. 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 Hom k(x) (A R k(x), k(x)) k(x) is surjective. Because A is a free R-module, Hom k(x) (A R k(x), k(x)) = Hom R (A, R) R k(x). By Nakayama s lemma, the homomorphism Hom R (A, R) R is surjective. Now that Hom OS (f O X, O S ) O S is surjective, the exact sequence 0 O S f O X f O X /O S 0 is locally split. So f O X /O S is a locally free O S -module. Obviously it is of rank 2. Denote V = (f O X /O S ), 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 F 2 : 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 = Hom OS (f O X, O S ), 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 O S f O X f O X /O S 0, (4.3) we get a morphism of sheaves ψ : V = (f O X /O S ) (f O X ) = 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 : Hom R (A/R, R) R A Hom R (A, R), with θ(f a)(b) = f(ab), a, b A, f Hom R (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 id k(y) : (Hom R (A/R, R) R A) R k(y) Hom R (A, R) R k(y) is surjective. Since A is a free R-module, we have (Hom R (A/R, R) R A) R k(y) = Hom k(y) ((A R k(y))/k(y), k(y)) and Hom R (A, R) R k(y) = Hom k(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 = A 1, A 2 and A 3. (1) For A 1 = R R R, e 1 = (1, 1, 1) is the unity of A 1. Let e 2 = (0, 1, 0), e 3 =

4.2. FROM CUBIC ALGEBRAS TO GEOMETRIC CUBIC FORMS 31 (0, 0, 1), V = Re 2 Re 3. We identify V with A/Re 1. Let (f 1, f 2, f 3 ) be the basis of A dual to (e 1, e 2, e 3 ). It is obvious that θ(f 2 V ) = f 2, θ(f 3 V ) = f 3, where f i V denotes the restriction of f i to V, for i = 2, 3. By calculation, θ(f 2 (e 2 e 1 ))(e 1 ) = f 2 (e 2 e 1 ) = 1, θ(f 2 (e 2 e 1 ))(e 2 ) = f 2 (e 2 e 2 ) = 0, θ(f 2 (e 2 e 1 ))(e 3 ) = f 2 (0) = 0. So θ(f 2 (e 2 e 1 )) = f 1 and θ is hence surjective. (2) For A 2 = R (R[ǫ]/(ǫ 2 )), e 1 = (1, 1) is the unity of A 1. Let e 2 = (1, 0), e 3 = (0, ǫ), V = Re 2 Re 3. We identify V with A/Re 1. Let (f 1, f 2, f 3 ) be the basis of A dual to (e 1, e 2, e 3 ). It is obvious that θ(f 2 V ) = f 2, θ(f 3 V ) = f 3. A simple calculation as above tells us that θ(f 2 (e 2 e 1 )) = f 1, so θ is surjective. (3) For A 3 = R[ǫ]/(ǫ 3 ), e 1 = 1 is the unity of A 1. Let e 2 = ǫ, e 3 = ǫ 2, V = Re 2 Re 3. We identify V with A/Re 1. Let (f 1, f 2, f 3 ) be the basis of A dual to (e 1, e 2, e 3 ). It is obvious that θ(f 2 V ) = f 2, θ(f 3 V ) = f 3. A simple calculation as above tells us that θ(f 2 e 2 ) = f 1, 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 immersion if and only if the induced morphism of structure sheaves h : O Z h O Y 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 schemetheoretic 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, h (Z) : T = T(Z) h R(Z) = S is an isomorphism, so h is an isomorphism.

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 : O P φ O X 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 = A 1, A 2, A 3. (1) For A 1 = k k k, take e 1 = (1, 1, 1), e 2 = (0, 1, 0), e 3 = (0, 0, 1). Let (f 1, f 2, f 3 ) be the basis of A 1 dual to (e 1, e 2, e 3 ). Define h A 1 by h(ae 1+be 2 +ce 3 ) = a+b+ c, a, b, c k. A simple calculation shows that ( e 1 +e 2 +e 3 )h = f 1, (e 1 e 3 )h = f 2, (e 1 e 2 )h = f 3. Let x i = f i /h, i = 2, 3, we need to prove that A x2 = k[x 3 /x 2 ] and A x3 = k[x 2 /x 3 ]. By calculation, x 2 = e 1 e 3, x 3 = e 1 e 2 and A x2 = e 1 e 1 e 3, e 1 e 2 e 3 k[ e 1 e 3 ] = k[ e 1 e 3, e 1 e 2 e 1 e 3 ]. But in the ring A x2, So A x2 = k[x 3 /x 2 ]. The same proof gives A x3 = k[x 2 /x 3 ]. e 3 e 1 e 3 = 0 since e 3 (e 1 e 3 ) = 0. (2) For A = A 2 = k (k[ǫ]/(ǫ 2 )), take e 1 = (1, 1), e 2 = (0, 1), e 3 = (0, ǫ). Let (f 1, f 2, f 3 ) be the basis of A 2 dual to (e 1, e 2, e 3 ). Define h A 2 by h(ae 1 + be 2 + ce 3 ) = 2a + b + c, a, b, c k. A simple calculation shows that (e 1 e 2 )h = f 1, ( e 1 +e 2 +e 3 )h = f 2, (e 2 e 3 )h = f 3. Let x i = f i /h, i = 2, 3, we need to prove that A x2 = k[x 3 /x 2 ] and A x3 = k[x 2 /x 3 ]. By calculation, x 2 = e 1 + e 2 + e 3, x 3 = e 2 e 3 and A x2 = k[ A x2, e 1 e 1 +e 2 +e 3, e 2 e 3 e 3 e 1 +e 2 +e 3, e 1 +e 2 +e 3 ] = k[ e 3 e 1 +e 2 +e 3 = 0 since e 3 ( e 1 + e 2 + e 3 ) = 0. So A x2 = k[x 3 /x 2 ]. e 1 e 2 e 3, e 1+e 2 +e 3 e 3 e 2 e 3, e 1+e 2 +e 3 Similarly, A x3 = k[ e 2 e 3 ] = k[ e 3 e 2 e 3 = 0 since e 3 (e 2 e 3 ) = 0. So A x3 = k[x 2 /x 3 ]. e 2 e 3 e 1 +e 2 +e 3 ]. But in the ring e 2 e 3 ]. But in the ring A x3, (3) For A = A 3 = k[ǫ]/(ǫ 3 ), (1, ǫ, ǫ 2 ) is a basis of A 3. Let (f 1, f 2, f 3 ) be the basis of A 3 dual to (1, ǫ, ǫ 2 ). Define h A 3 by h(a + bǫ + cǫ 2 ) = a + b + c, a, b, c k. A simple calculation shows that ǫ 2 h = f 1, (ǫ ǫ 2 )h = f 2, (1 ǫ)h = f 3. Let x i = f i /h, i = 2, 3, we need to prove that A x2 = k[x 3 /x 2 ] and A x3 = k[x 2 /x 3 ].

4.2. FROM CUBIC ALGEBRAS TO GEOMETRIC CUBIC FORMS 33 By calculation, x 2 = ǫ ǫ 2, x 3 = 1 ǫ and A x2 = k[ 1 ǫ ǫ 2, 1 ǫ ǫ ǫ 2 ] = 0 = k[x 3 /x 2 ]. Similarly, A x3 = A = k[x 2 /x 3 ] since x 3 is invertible in A 3. 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 O(1) = O(D) Ω 1 P/S. Then O(D) = O(1) (Ω1 P/S ) = O(3) π ( 2 π O(1)) 1. So the element 1 O(D) defines a global section a Γ(P, O(3) π ( 2 π O(1)) 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, so it defines a 1-morphism of algebraic stacks F 2 : V G. Generally, consider the category fibered in groupoids B 1 of based cubic algebras. By proposition 2.2.6, it is representable by A 6 Z. So there exists a universal based cubic algebra à overa 6 Z. By theorem 2.3.2, the category of based Gorenstein cubic algebra is represented by the open subscheme V = G 2 a (A 4 Z \O) of A6 Z. The restriction A 1 of à on V is a Gorenstein cubic algebra. Let F 2 (A 1) = (P 1, O P1 (1), a 1 ), which is a geometric cubic form over V. Proposition 4.2.5. The geometric cubic form (P 1, O P1 (1), a 1 ) over V extends uniquely to a geometric cubic form (P, O(1), a) over A 6 Z. Proof. Define P = P((Ã/O A 6 1) ). Let π : P A 6 Z Z be the structural morphism, then P is a family of genus 0 curves over A 6 Z. Recall that P 1 = P((A 1 /O V 1) ), so P is an extension of P 1 to A 6 Z. Let π 1 : P 1 V be the structural morphism. Since the complement of V in A 6 Z is of codimension 4 and A6 Z is smooth, the following lemma shows that O P1 (1) can be extended uniquely to an invertible sheaf O(1) of P over A 6 Z of relative degree 1, and the section a 1 Γ(P 1, O P1 (3) π 1( 2 π 1, (O P1 (1))) 1 ) can be extended uniquely to the section a Γ(P, O P (3) π ( 2 π (O P (1))) 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 ). Proposition 4.2.7. The geometric cubic form (P, O(1), a) over A 6 Z is H-equivariant. Proof. For any point x on A 6 Z, denote by (P x, O Px (1), a x ) the fiber of (P, O(1), a) at x. We need to prove that the geometric cubic form (P hx, O Phx (1), a hx ) on Spec(k(hx)) is isomorphic to the geometric cubic form (P x, O Px (1), a x ) on Spec(k(x)) for any h H. Let h H, à := h à is a cubic algebra over A 6 Z. We constructed as above a geometric cubic form (P, O P (1), a ) on A 6 Z corresponding to Ã. Then the fiber of (P, O P (1), a ) at x is just (P hx, O Phx (1), a hx ). Recall that the construction of F 2 doesn t depend on the choices of basis, so the restriction of (P, O P (1), a ) on V and (P 1, O P1 (1), a 1 ) are the same. Since (P, O(1), a) is the unique extension of (P 1, O P1 (1), a 1 ) to A 6 Z, we have (P, O(1), a) = (P, O P (1), a ), in particular their fibers at x will be the same. The geometric cubic form (P, O(1), a) over A 6 Z corresponds to a 1-morphism of algebraic stacks F : B 1 = A 6 Z G. By proposition 4.2.7, F factors through the quotient A = [B 1 /H] and defines a 1-morphism F 2 : A G. 4.3 The correspondence is bijective In this section we prove that the 1-morphisms F 1 : G A and F 2 : A G are inverse to each other. As a corollary, we get the main theorem 1. Proposition 4.3.1. The 1-morphism F 1 : G A is left inverse to F 2 : 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 F 2 is constructed as the unique extension of F 2, we only need to prove that F 1 F 2 is isomorphic to the identity on V. Suppose that A is a Gorenstein cubic algebra over S and F 1 (A) = (P, O(1), a). We use the notation of section 4.2. Since φ is a closed immersion, A = π O D. By definition of O(1), we have O(D) = O(3) π ( 2 π O(1)) 1 ). Let J := O( 3) π ( 2 π O(1)) = O( D). By definition, the section a Γ(P, O(3) π ( 2 π O(1)) 1 ) correspond to 1 Γ(P, O(D)). So we have the commutative diagram of exact sequences 0 J 0 O( D) a O O/J 0 1 O O D 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, F 2 ((P, O(1), a)) = π (O/J ) = π O D = A and F1 F 2 is identity on V. Proposition 4.3.2. The 1-morphism F 1 : G A is right inverse to F 2 : 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 F 2 F 1 is identity on W. Suppose that (P, O P (1), a) is a primitive geometric cubic form over S. Let J := O P ( 3) π ( 2 π O P (1)) = O P ( 1) Ω 1 P/S. By proposition 3.3.1, the morphism of sheaves J a O P is injective. Let J be the image of J in O P under the morphism a. J is an ideal sheaf in O P, 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 0 J 0 O P ( D) a O P O P /J 0 1 O P O D 0, in which the first downward arrow is an isomorphism. Let A := F 1 ((P, O P (1), a)). Since a is not a zero divisor, by example 4.1.1, we have