UNIVERSIDAD NACIONAL AUTÓNOMA DE MÉXICO PROGRAMA DE MAESTRÍA Y DOCTORADO EN CIENCIAS MATEMÁTICAS Y DE LA ESPECIALIZACIÓN EN ESTADÍSTICA APLICADA TWO CONJECTURES ON HIGHER NASH BLOWUPS OF TORIC VARIETIES TESIS QUE PARA OPTAR POR EL GRADO DE: DOCTOR (A) EN CIENCIAS PRESENTA: ENRIQUE CHÁVEZ MARTÍNEZ DIRECTOR DE LA TESIS: DR. MARK SPIVAKOVSKY INSTITUT DE MATHÉMATIQUES DE TOULOUSE UNIVERSITÉ PAUL SABATIER CODIRECTOR DE LA TESIS: DR. ANDRÉS DANIEL DUARTE CONACYT-UNIVERSIDAD AUTÓNOMA DE ZACATECAS UNIDAD ACADÉMICA DE MATEMÁTICAS MIEMBROS DEL COMITÉ TUTOR DR. MARK SPIVAKOVSKY, INSTITUT DE MATHÉMATIQUES DE TOULOUSE, UNIVERSITÉ PAUL SABATIER DR. ANDRÉS DANIEL DUARTE, CONACYT-UNIVERSIDAD AUTÓNOMA DE ZACATECAS, UNIDAD ACADÉMICA DE MATEMÁTICAS DR. JAWAD SNOUSSI, UNIDAD CUERNAVACA DEL INSTITUTO DE MATEMÁTICAS, UNIVERSIDAD NACIONAL AUTÓNOMA DE MÉXICO CIUDAD DE MÉXICO, NOVIEMBRE 2021. UNAM – Dirección General de Bibliotecas Tesis Digitales Restricciones de uso DERECHOS RESERVADOS © PROHIBIDA SU REPRODUCCIÓN TOTAL O PARCIAL Todo el material contenido en esta tesis esta protegido por la Ley Federal del Derecho de Autor (LFDA) de los Estados Unidos Mexicanos (México). El uso de imágenes, fragmentos de videos, y demás material que sea objeto de protección de los derechos de autor, será exclusivamente para fines educativos e informativos y deberá citar la fuente donde la obtuvo mencionando el autor o autores. Cualquier uso distinto como el lucro, reproducción, edición o modificación, será perseguido y sancionado por el respectivo titular de los Derechos de Autor. Agradecimientos En primer lugar quiero agradecer a mi familia, por todo el apoyo que me han dado desde que comencé este largo viaje, que me han educado y han sido un ejemplo de vida en todo momento para mi llevándome por un buen camino y ayudándome a corregir mis errores. Y más que nada por darme la oportunidad llegar hasta aqúı. Me encuentro particularmente agradecido con mi asesor de tesis el doc- tor Daniel Duarte por su incondicional ayuda y apoyo en todo momento durante la elaboración de este trabajo, su gran paciencia, dedicación y todo el tiempo que invirtió en este proyecto, que a pesar de haber sido una de- cisi´on de ultima hora, decidió tomarme bajo su tutela. Por sus grandes consejos y apoyo tanto académicos como personales, que me ayudaron du- rante todo este recorrido. Además, gracias a su gran carisma y pasión por las matemáticas, logro que todo este trabajo se volviera una experiencia sumamente divertida y enriquecedora. Al mi otro director de tesis, el doctor Mark Spivakovsky que fue la persona con la que se inicio mi trabajo doctoral y que me tuvo la confianza para este proyecto, quien siempre lograba hacer un espacio en su apretada agenda para atenderme. Además, agradezco sus valiosas aportaciones en el trabajo, mostrandome detalles que no hubiera visto en su momento. A todos mis compañeros y maestros del IMATE-Cuernavaca quienes me acompañaron a lo largo de todos mis estudios de doctorado, que con sus enseñanzas logre superarme acadẽmicamente y con su buena compañ́ıa y largas pláticas, lograron que los estudios se volvieran algo que disfrute mucho. A toda la gente de la Unidad Académica de Matemáticas de la UAZ, tanto profesores como estudiantes que me recibieron con los brazos abiertos durante mi estancia de elaboración de tesis, que gracias a su gran hospitali- dad, lograron que todo este proceso se volviera algo de disfrute de principio afin. Con su amistad logre salir de la rutina y olvidarme por momentos del estrés causado por este trabajo, haciéndolo una experiencia que nunca 1 2 olvidaré. A la UNAM y a la UAZ las dos instituciones que me abrieron sus puertas y me extendieron su mano durante todo mi proceso del doctorado. Expreso mi sincero agradecimiento al Consejo Nacional de Ciencia y Tecnoloǵıa ya que sin el apoyo económico proporcionado a lo largo de estos cuatro años el mero hecho de estudiar un programa de posgrado hubiera sido simplemente impensable. Y por ultimo a la DGAPA-UNAM por ofrecerme apoyo económico du- rante la primera parte de mi estancia para la culminación de este trabajo a través del proyecto PAPIIT IN108320. Introduction The Nash blowup of an algebraic variety is a modification that replaces singular points by limits of tangent spaces at non-singular points. It was proposed to resolve the singularities by iterating this process [19, 23]. This question has been treated in [19, 21, 13, 14, 17, 24, 1, 26, 7]. The particular case of toric varieties is treated in [13, 15, 16, 8, 11] using their combinatorial structure. There is a generalization of Nash blowups, called higher Nash blowups or Nash blowups of order n, that was proposed by Takehiko Yasuda. This mod- ification replaces singular points by limits of infinitesimal neighborhoods of certain order at non-singular points. In particular, the higher Nash blowup looks for resolution of singularities in one step [27]. There are several papers that deal with higher Nash blowups in the special case of toric varieties. The usual strategy for this special case is to translate the original geometric problem into a combinatorial one and then try to solve the latter. So far, the combinatorial description of higher Nash blowups of toric varieties has been obtained using Gröebner fans or higher-order Jacobian matrices. The usage of Gröebner fans for higher Nash blowups of toric varieties was initiated in [9], which in turn was inspired by [29]. Later, this tool was further developed in [26] to show that the Nash blowup of order n of the toric surface singularity A3 is singular for any n > 0, over the complex numbers. This result was later revisited to show that it is also holds in prime characteristic [12]. In recent years several authors have introduced higher-order versions of the Jacobian matrix. In [10], a higher-order Jacobian matrix is studied in relation with the higher Nash blowup of a hypersurface. More recently, in [2, 3], a similar matrix is introduced for any finitely generated algebra. In these articles the matrices are used to study singularities in arbitrary characteristic or to study algebraic properties of the module of Kähler differentials of high order. In another but related direction, article [7] describes a matrix 3 4 associated to a relative compactification of the induced map on the main components of jet schemes of a projective birational morphism. In chapter 1 we introduce a matrix that represents a higher-order tan- gent map of a morphism. This matrix involves higher-order derivatives, making it more suitable for some computations related to jet-spaces. Our main application of this matrix is the solution of two conjectures of T. Ya- suda. The first is related to the combinatorial shape of the higher Nash blowup of formal curves and the second is related to the factorization of the normalization of the Nash blow-up of order n of the toric surface An by the minimal resolution. We also develop a combinatorial description for the higher Nash blowup of toric varieties. This result is based and inspired in the analogous de- scription of the usual Nash blowup of toric varieties given in [15, 16]. Our results depend strongly on the general framework developed in [15] for not necessarily normal toric varieties. In a subsequent paper [28], T. Yasuda gave a conjectural explicit de- scription of the semigroup of the higher Nash blowup of formal curves. In chapter 2 we prove this conjecture for toric curves using the higher order matrix defined in chapter 1 and other combinatorial tools. We also present a family of non-monomial curves showing that Yasuda’s conjecture fails in general. By combining the results we obtained for mono- mial morphisms and the general construction of the matrix representing the higher-order tangent map, we are able to describe a particular element of the semigroup of the higher Nash blowup of this family of curves which does not belong to the semigroup suggested by Yasuda. The results of chapters 1 and 2 are published in [5]. The techniques from [26] can be used to compute the Gröebner fan of the normalization of higher Nash blowup of An for some n’s. Those com- putations suggest that the essential divisors of the minimal resolution of An appear in the normalization of the Nash blow-up of order n of An for some n’s. T. Yasuda conjectures that this happens for all n. In particular, this implies that the normalization of the Nash blowup of order n of An factors through its minimal resolution. Chapter 3 is devoted to proving the second conjecture of T. Yasuda previously mentioned. The results of chapter 1 and the results of [15] told us that the normalization of the higher Nash blowup of An is a toric variety associated to a fan that subdivides the cone determining An. An explicit description of this fan could be obtained by effectively computing all minors of the corresponding higher order Jacobian matrix. This is a difficult task given the complexity of the matrix for large n. However, for the problem we 5 are interested in, we do not require an explicit description of the entire fan. The rays that subdivide the cone of An to obtain its minimal resolution can be explicitly specified. Thus, in order to show that these rays appear in the fan associated to the normalization of the higher Nash blowup we need to be able to control only certain minors of the matrix. A great deal of this chapter is devoted to construct combinatorial tools that allow us to accomplish that goal. The results of chapter 3 are contained in the preprint [4]. 6 Contents 1 Higher Nash blowup of toric varieties 9 1.1 A higher-order Jacobian matrix of a morphism . . . . . . . . 9 1.1.1 A higher-order Jacobian matrix of a morphism of affine spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.1.2 Higher-order Jacobian matrices and birational mor- phisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.2 Higher order Jacobian matrices for monomial morphisms . . . 14 1.3 Higher Nash blowup of toric varieties . . . . . . . . . . . . . . 17 1.3.1 Higher Nash blowup . . . . . . . . . . . . . . . . . . . 17 1.3.2 An explicit open cover of the higher Nash blowup of a toric variety by affine toric varieties . . . . . . . . . 18 1.3.3 The logarithmic Jacobian ideal of order n is indepen- dent of the generators of Γ . . . . . . . . . . . . . . . 22 2 Higher Nash blowup of toric curves 27 2.1 Higher Nash blowup of toric curves . . . . . . . . . . . . . . . 27 2.1.1 A partial description of Nashn(XΓ) . . . . . . . . . . 30 2.1.2 Proof of conjecture 2.1.2 for toric curves and some consequences . . . . . . . . . . . . . . . . . . . . . . . 33 2.2 Counterexample to the conjecture . . . . . . . . . . . . . . . 37 3 Factorization by minimal resolution 41 3.1 The main result . . . . . . . . . . . . . . . . . . . . . . . . . . 41 3.2 A particular basis for the vector space Cλ2,n . . . . . . . . . . 43 3.2.1 Linear independence of Tη . . . . . . . . . . . . . . . . 46 3.2.2 Moving Tj,η along a diagonal preserves linear indepen- dence . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 3.3 Proof of Theorem 3.1.5 . . . . . . . . . . . . . . . . . . . . . . 58 3.3.1 A distinguished element of SAn . . . . . . . . . . . . . 61 7 8 CONTENTS 3.3.2 Jηk ∈ SAn is minimal with respect to fk . . . . . . . . 69 References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 Chapter 1 Higher order matrices and higher Nash blowup in toric varieties 1.1 A higher-order Jacobian matrix of a morphism 1.1.1 A higher-order Jacobian matrix of a morphism of affine spaces In this section we study a higher-order derivative of a morphism between affine varieties and find a matrix representation of this linear map. Notation 1.1.1. The following notation will be constantly used in this paper. • The entries of vectors α ∈ Nt are denoted as α = (α(1), . . . , α(t)). • α ≤ β ⇔ α(i) ≤ β(i) ∀i ∈ {1, . . . , t}. In particular, α < β if and only if α(i) ≤ β(i) ∀i ∈ {1, . . . , t} and α(i) < β(i) for some i ∈ {1, . . . , t}. • |α| = α(1) + · · ·+ α(t). • α! = α(1)!α(2)! · · ·α(t)! • ∂α = ∂α(1)∂α(2) · · · ∂α(t). • For t, n ∈ N, Λt,n := {γ ∈ Nt|1 ≤ |γ| ≤ n}. In addition, we denote λt,n := |Λt,n| = ( n+t t ) − 1. 9 10 CHAPTER 1. HIGHER NASH BLOWUP OF TORIC VARIETIES Let K an algebraic closed field. Consider a morphism ϕ : Kd → Ks, x = (x1, . . . , xd) 7→ (g1(x), . . . , gs(x)). Assume that ϕ is regular at some x ∈ Kd and let y = ϕ(x) ∈ Ks. Let m ⊂ C[X1, . . . , Xd] and n ⊂ C[Y1, . . . , Ys] be the maximal ideals correspond- ing to x and y, and mx, ny the maximal ideals in (C[X1, . . . , Xd])m and (C[Y1, . . . , Ys])n, respectively. Let ϕ∗ : (C[Y1, . . . , Ys])n → (C[X1, . . . , Xd])m be the induced homomor- phism on local rings, where ϕ∗(ny) ⊂ mx. In particular, there is a homo- morphism of K−vector spaces for each n ∈ N: (ϕ̄∗)n : ny/n n+1 y → mx/m n+1 x . the elements Ax = {(X−x)α := (X1−x1) α(1) · · · (Xd−xd) α(d)|α ∈ Λd,n} form a basis ofmx/m n+1 x as aK−vector space. Similarly, By = {(Y−y)β |β ∈ Λs,n} forms a basis of ny/n n+1 y . The dual bases of Ax and By are, respectively, A∨ x = { 1 α! ∂α ∂Xα ∣ ∣ ∣ x |α ∈ Λd,n } , B∨ y = { 1 β! ∂β ∂Y β ∣ ∣ ∣ y |β ∈ Λs,n } . Since (ϕ̄∗)n((Y − y)β) = (g1− g1(x)) β(1) · · · (gs− gs(x)) β(s) = (ϕ−ϕ(x))β , it follows that the dual morphism (ϕ̄∗)∨n : (mx/m n+1 x )∨ → (ny/n n+1 y )∨ satisfies (ϕ̄∗)∨n ( 1 α! ∂α ∂Xα ∣ ∣ ∣ x ) = 1 α! ∂α ∂Xα ∣ ∣ ∣ x ◦ (ϕ̄∗)n : ny/n n+1 y → K, (1.1) (Y − y)β 7→ 1 α! ∂α(ϕ− ϕ(x))β ∂Xα ∣ ∣ ∣ x . It follows that the matrix representation of (ϕ̄∗)∨n in these bases is: [ (ϕ̄∗)∨n ]B∨ y A∨ x = ( 1 α! ∂α(ϕ− ϕ(x))β ∂Xα ∣ ∣ ∣ x ) β∈Λs,n,α∈Λd,n . (1.2) Definition 1.1.2. Let ϕ : Kd → Ks be as before, where ϕ(x) = y. We call the linear map (ϕ̄∗)∨n the derivative of order n of ϕ at x. In addition, let Dn x(ϕ) := ( 1 α! ∂α(ϕ− ϕ(x))β ∂Xα |x ) β∈Λs,n,α∈Λd,n . 1.1. A HIGHER-ORDER JACOBIAN MATRIX OF A MORPHISM 11 We call Dn x(ϕ) the Jacobian matrix of order n of ϕ at x or the higher-order Jacobian matrix of ϕ at x. Notice that Dn x(ϕ) is a (λs,n × λd,n)-matrix. We order the rows and columns of this matrix increasingly using graded lexicographical order on Λs,n and Λd,n. This order is denoted . Remark 1.1.3. Notice that, for each β ∈ Λs,n, the β row of Dn x(ϕ) cor- responds precisely to the coefficients of the truncated Taylor expansion of order n of (ϕ− ϕ(x))β centered at x. Remark 1.1.4. A similar higher-order Jacobian matrix of a single poly- nomial F was defined in [10] and is denoted Jacn(F ). See also [2, 3] for a further development of this matrix. Example 1.1.5. Let ϕ : K → K2, t 7→ (t, t2). The usual matrix represen- tation of the derivative of ϕ at 0 ∈ K is given by the Jacobian matrix: D0(ϕ) = ( dt dt |0 dt2 dt |0 ) = ( 1 2t ) ∣ ∣ 0 . Following the construction of the higher-order Jacobian matrix given previ- ously, in the case n = 2, we obtain: D2 0(ϕ) =         dt dt |0 1 2! d2t dt2 |0 dt2 dt |0 1 2! d2t2 dt2 |0 d(t)2 dt |0 1 2! d2(t)2 dt2 |0 d(t·t2) dt |0 1 2! d2(t·t3) dt2 |0 d(t2)2 dt |0 1 2! d2(t2)2 dt2 |0         =       1 0 2t 1 0 1 0 2t 0 4t2       ∣ ∣ 0 . The higher-order Jacobian matrix satisfies the following basic properties. Lemma 1.1.6. Let ϕ : Kd → Ks be as before. (i) If β ∈ Λs,n is such that |β| = 1 then ∂α(ϕ−ϕ(x))β ∂Xα = ∂αϕβ ∂Xα for every α ∈ Λd,n. (ii) Let α ∈ Λd,n, β ∈ Λs,n be such that |α| < |β|. Then ∂α(ϕ−ϕ(x))β ∂Xα |x = 0. (iii) D1 x(ϕ) is the usual Jacobian of ϕ evaluated at x. (iv) If ϕ : Kd → Kd is the identity then Dn x(ϕ) is the identity matrix. (v) Let ψ : Ks → Kr be another morphism. Then Dn x(ψ◦ϕ) = Dn y (ψ)D n x(ϕ). 12 CHAPTER 1. HIGHER NASH BLOWUP OF TORIC VARIETIES Proof. (i) If β ∈ Λs,n is such that |β| = 1 then (ϕ − ϕ(x))β = gi − gi(x) for some i ∈ {1, . . . , s}. Since gi(x) is a constant, the result follows. (ii) The hypothesis on α and β means that in ∂α(ϕ−ϕ(x))β ∂Xα the order of the derivative is less than the number of factors in (ϕ− ϕ(x))β . This implies that in every summand of ∂α(ϕ−ϕ(x))β ∂Xα there is a factor gi−gi(x). Thus ∂α(ϕ−ϕ(x))β ∂Xα |x = 0. (iii) This follows from the definition of Dn x(ϕ) and (i). (iv) If ϕ is the identity then (ϕ̄∗)n : mx/m n+1 x → mx/m n+1 x is also the identity. With respect to the common basis chose for both vector spaces, we conclude that Dn x(ϕ) is the identity matrix. (v) We know that (ψ ◦ ϕ)∗ = ϕ∗ ◦ ψ∗. Thus, ((ψ ◦ ϕ)∗)n = (ϕ̄∗)n ◦ (ψ̄∗)n. Taking duals ((ψ ◦ ϕ)∗)∨n = ((ϕ̄∗)n ◦ (ψ̄∗)n) ∨ = (ψ̄∗)∨n ◦ (ϕ̄∗)∨n . The result follows. Now suppose that X ⊂ Kd and Y ⊂ Ks are affine varieties and let ϕ : X → Y be a morphism which is regular at x ∈ X and let y = ϕ(x). Denote by m̄x and n̄y the maximal ideals of the corresponding local rings. Since ϕ is the restriction of a morphism ϕ : Kd → Ks, the diagram X ϕ //  _ i  Y  _ i  Kd ϕ // Ks induces the diagram (m̄x/m̄ n+1 x )∨ //  _  (n̄y/n̄ n+1 y )∨  _  (mx/m n+1 x )∨ // (ny/n n+1 y )∨ Taking bases as before we identify (mx/m n+1 x )∨ ∼= Kλd,n and (ny/n n+1 y )∨ ∼= Kλs,n . The commutativity of the diagram (m̄x/m̄ n+1 x )∨ //  _  (n̄y/n̄ n+1 y )∨  _  Kλd,n Dn x (ϕ) // Kλs,n 1.1. A HIGHER-ORDER JACOBIAN MATRIX OF A MORPHISM 13 allows us to define a higher-order tangent map of ϕ : X → Y at x ∈ X as the restriction Dn x(ϕ) : (m̄x/m̄ n+1 x )∨ → (n̄y/n̄ n+1 y )∨. 1.1.2 Higher-order Jacobian matrices and birational mor- phisms Let Y ⊂ Ks be an irreducible algebraic variety and y ∈ Y . In this subsection, we use the higher-order Jacobian matrix to explicitly compute the space (n̄y/n̄ n+1 y )∨ in some cases. Lemma 1.1.7. Let X and Y be irreducible varieties and let ϕ : X 99K Y be a birational morphism. Let U ⊂ X and V ⊂ Y be open subsets isomorphic to each other. Let x ∈ U and y = ϕ(x) ∈ V . Then ϕ induces an isomorphism n̄y/n̄ n+1 y ∼= m̄x/m̄ n+1 x . Proof. Since ϕ|U : U → V is an isomorphism, there is an induced isomor- phism on local rings OY,y ∼= OX,x. In particular, ϕ∗(n̄y) = m̄x. The result follows. Proposition 1.1.8. Let ϕ : Kd 99K Y ⊂ Ks be a birational morphism, U ⊂ Kd and V ⊂ Y open subsets isomorphic to each other, and y = ϕ(x) for some x ∈ U . Then the vector space (n̄y/n̄ n+1 y )∨ is isomorphic to the image of the linear map defined by Dn x(ϕ). In particular, rank(Dn x(ϕ)) = λd,n. Proof. We have the following commutative diagram Kd i◦ϕ // ϕ !! Ks Y ? i OO This diagram induces in turn the following commutative diagram (mx/m n+1 x )∨ (i◦ϕ)∗ ∨ // (ϕ̄∗)∨ ∼= '' (ny/n n+1 y )∨ (n̄y/n̄ n+1 y )∨, ? (ī∗)∨ OO where the isomorphism in the diagonal arrow comes from lemma 1.1.7. Fix- ing bases for mx/m n+1 x and ny/n n+1 y as in the previous section, we identify 14 CHAPTER 1. HIGHER NASH BLOWUP OF TORIC VARIETIES (mx/m n+1 x )∨ ∼= Kλd,n and (ny/n n+1 y )∨ ∼= Kλs,n . In addition, from (1.2), it fol- lows that (i ◦ ϕ)∗ ∨ is the linear map defined by the matrixDn x(i◦ϕ) = Dn x(ϕ). We thus obtain the diagram Kλd,n Dn x (ϕ) // && Kλs,n (n̄y/n̄ n+1 y )∨. ? OO The commutativity of this diagram proves the proposition. Remark 1.1.9. Notice that the proofs in this section consider local rings of points of a variety. Therefore, these results are also valid in the analytic case. In particular, we can define a higher-order Jacobian matrix for germs of analytic maps ϕ : (X,x) → (Y, y). Example 1.1.10. Let ϕ : K → C = V(y − x2) ⊂ K2, t 7→ (t, t2). We computed D2 0(ϕ) : K 2 → K5 in the previous section. Let n̄0 be the maximal ideal of (0, 0) ∈ C. Using proposition 1.1.8 we obtain (n̄0/n̄ 3 0) ∨ = Im(D2 0(ϕ)) = Im       1 0 2t 1 0 1 0 2t 0 4t2       ∣ ∣ 0 ⊂ K5. 1.2 Higher order Jacobian matrices for monomial morphisms Let a1, . . . , as ∈ Zd. We assume that d ≤ s. In this section we study the higher-order Jacobian matrix of the monomial morphism ϕ : (K \ {0})d → Ks (1.3) x = (x1, . . . , xd) 7→ (xa1 , . . . , xas), where xai := x ai(1) 1 · · ·x ai(d) d . Notation 1.2.1. The following notation will be used constantly. • A denotes the (d×s)-matrix whose columns are the vectors a1, . . . , as. By abuse of notation, the set {a1, . . . , as} is also denoted as A. 1.2. HIGHERORDER JACOBIANMATRICES FORMONOMIALMORPHISMS15 • Ai := (a1(i), . . . , as(i)), i = 1, . . . , d, denote the rows of A. In parti- cular, for γ ∈ Ns, XAγ = XA1·γ 1 · · ·XAd·γ d , where Aγ is a product of matrices and Ai ·γ is the usual inner product in Rs. • For β ∈ Ns, we denote (XA − xA)β := (Xa1 − xa1)β(1) · · · (Xas − xas)β(s). • For λ, τ ∈ Nt, denote ( λ τ ) := (λ(1) τ(1) ) · · · (λ(t) τ(t) ) . With this notation, the higher-order Jacobian of ϕ at a point x ∈ (K \ {0})d is given by: Dn x(ϕ) = ( 1 α! ∂α(XA − xA)β ∂Xα |x ) β∈Λs,n,α∈Λd,n . We are interested in computing the maximal minors of this matrix. This will be done in several steps. Lemma 1.2.2. Let γ ∈ Ns and α ∈ Nd. Then 1 α! ∂α(XAγ) ∂Xα = ( Aγ α ) XAγ−α. Proof. This is a direct computation. Lemma 1.2.3. Let β ∈ Λs,n, α ∈ Λd,n and x ∈ (K \ {0})d. Then 1 α! ∂α(XA − xA)β |x = cβ,αx Aβ−α, where cβ,α := ∑ γ≤β,γ 6=0(−1)|β−γ| ( β γ )( Aγ α ) . Proof. From the binomial theorem we obtain, for each i ∈ {1, . . . , s}: (Xai − xai)β(i) = β(i) ∑ γ(i)=0 (−1)β(i)−γ(i) ( β(i) γ(i) ) (Xai)γ(i)(xai)β(i)−γ(i). 16 CHAPTER 1. HIGHER NASH BLOWUP OF TORIC VARIETIES Thus, letting γ := (γ(1), . . . , γ(s)), (XA − xA)β = β(1) ∑ γ(1)=0 · · · β(s) ∑ γ(s)=0 (−1)|β−γ| ( β γ ) s ∏ i=1 (Xai)γ(i)(xai)β(i)−γ(i) = ∑ γ≤β (−1)|β−γ| ( β γ ) (X ∑ γ(i)ai)(x ∑ β(i)ai− ∑ γ(i)ai) = ∑ γ≤β (−1)|β−γ| ( β γ ) (XAγ)(xAβ−Aγ). With this formula and the previous lemma now it is easy to compute the derivative evaluated at x: 1 α! ∂α(XA − xA)β |x = ∑ γ≤β,γ 6=0 (−1)|β−γ| ( β γ ) (xAβ−Aγ) 1 α! ∂α(XAγ)|x = ∑ γ≤β,γ 6=0 (−1)|β−γ| ( β γ ) (xAβ−Aγ) ( Aγ α ) XAγ−α|x = ∑ γ≤β,γ 6=0 (−1)|β−γ| ( β γ )( Aγ α ) xAβ−α = [ ∑ γ≤β,γ 6=0 (−1)|β−γ| ( β γ )( Aγ α ) ] xAβ−α. Using this lemma it follows that the higher-order Jacobian of ϕ at each x ∈ (K \ {0})d has the following shape: Dn x(ϕ) = ( cβ,αx Aβ−α ) β∈Λs,n,α∈Λd,n . (1.4) Proposition 1.2.4. Let J = {β1, . . . , βλd,n } ⊂ Λs,n, where β1 ≺ . . . ≺ βλd,n (see definition 1.1.2 for the notation ≺). Let LJ denote the submatrix of Dn x(ϕ) formed by the rows β1, . . . , βλd,n and all of its columns α1, . . . , αλd,n . Then, if x ∈ (K \ {0})d, det(LJ) = x Aβ1+···+Aβλd,n x α1+···+αλd,n det(Lc J), where Lc J := (cβi,αj )i,j. 1.3. HIGHER NASH BLOWUP OF TORIC VARIETIES 17 Proof. The matrix whose determinant we want to compute is the following: LJ =        cβ1,α1x Aβ1−α1 · · · cβ1,αλd,n x Aβ1−αλd,n cβ2,α1x Aβ2−α1 · · · cβ2,αλd,n x Aβ2−αλd,n ... · · · ... cβλd,n ,α1x Aβλd,n −α1 · · · cβλd,n ,αλd,n x Aβλd,n −αλd,n        . Multiply the αjth column by xαj . Then multiply the βith row by x−Aβi . Let Lc J = (cβi,αj )i,j . Then det(LJ) = x Aβ1+···+Aβλd,n x α1+···+αλd,n det(Lc J). (1.5) Remark 1.2.5. If n = 1 then λd,n = d. Letting βik = eik ∈ Ns for k = 1, . . . , d and J = {βi1 , . . . , βid} ⊂ Λs,1, it follows that L c J is the (d×d)-matrix whose rows are ai1 , . . . , aid . In particular, in view of (1.5), det(LJ) 6= 0 if and only if ai1 , . . . , aid are linearly independent. This remark allows a comparison between the so-called logarithmic Jacobian ideal of a toric variety and an ideal whose blowup defines the Nash blowup of the variety [13, 19]. This, in turn, gives place to the fact that the Nash blowup of a toric variety can be obtained as the blowup of its logarithmic Jacobian ideal (see [13, 18, 15]). As a result, there is an explicit combinatorial description of the Nash blowup in this context [13, 15, 16]. 1.3 Higher Nash blowup of toric varieties In this section we exhibit an open cover for the higher Nash blowup of a toric variety. We start by recalling the definition of the higher Nash blowup of an algebraic variety. Subsection 1.3.2 is based on the general theory of (not necessarily normal) toric varieties developed in [15] and also uses some ideas appearing in [16]. 1.3.1 Higher Nash blowup Notation 1.3.1. Given an irreducible algebraic variety X ⊂ Ks of dimen- sion d and a point x ∈ X, we denote Tn xX := (m̄x/m̄ n+1 x )∨. This is a vector space of dimension λd,n, whenever x is a non-singular point. 18 CHAPTER 1. HIGHER NASH BLOWUP OF TORIC VARIETIES Notice that X ⊂ Ks implies Tn xX ⊂ Tn x K s ∼= Kλs,n . Thus, if x is a non-singular point, we can see Tn xX as an element of the Grassmanian Gr(λd,n,K λs,n). Definition 1.3.2. [19, 20, 27] Let X ⊂ Ks be an irreducible algebraic variety of dimension d. Consider the Gauss map of order n: Gn : X \ Sing(X) → Gr(λd,n,K λs,n) x 7→ Tn xX, where Sing(X) denotes the set of singular points ofX. Denote byNashn(X) the Zariski closure of the graph of Gn. Call πn the restriction to Nashn(X) of the projection of X ×Gr(λd,n,K λs,n) to X. The pair (Nashn(X), πn) is called the higher Nash blowup of X or the Nash blowup of X of order n. It was proposed by T. Yasuda ([27]) to resolve the singularities of X by applying once the higher Nash blowup for n sufficiently large. Yasuda himself proved that his method works for curves ([27, Corollary 3.7]). More- over, Yasuda suggested in [29, Remark 1.5] that the A3-singularity might be a counterexample to his conjecture on the one-step resolution. R. Toh-Yama recently proved in [26] that Nashn(A3) is singular for every n ≥ 1. 1.3.2 An explicit open cover of the higher Nash blowup of a toric variety by affine toric varieties Let us recall the definition of an affine toric variety (see, for instance, [6, Section 1.1] or [25, Chapter 4]). Definition 1.3.3. Let A = {a1, . . . , as} ⊂ Zd. Let Γ := NA denote the semigroup generated by A, i.e., Γ = { ∑ i λiai|λi ∈ N}. In addition, as- sume that ZA = { ∑ i λiai|λi ∈ Z} = Zd. Consider the following monomial morphism: ϕΓ : (K∗)d → Ks (1.6) x = (x1, . . . , xd) 7→ (xa1 , . . . , xas), where K∗ = K \ {0}. Let XΓ denote the Zariski closure of the image of ϕΓ. We call XΓ the affine toric variety defined by Γ. It is well known that XΓ is an irreducible variety of dimension d, contains a dense open set isomorphic to (K∗)d and such that the natural action of (K∗)d on itself extends to an action on the variety. In addition, XΓ does not depend on the generating set A (see [6, Theorem 1.1.17] for various equivalent characterizations of affine toric varieties). 1.3. HIGHER NASH BLOWUP OF TORIC VARIETIES 19 Proposition 1.3.4. [6, Prop. 1.2.12],[15, Prop. 15] Let XΓ ⊂ Ks be an affine toric variety, σ∨ := R≥0Γ ⊂ Rd the cone generated by Γ, and σ its dual cone. The following statements are equivalent: (a) 0 ∈ XΓ. (b) XΓ has a 0-dimensional orbit. (c) The cone σ is of dimension d. (d) The cone σ∨ is strongly convex. We want to show that the higher Nash blowup of a toric variety having a 0-dimensional orbit, has a finite open cover given by affine toric varieties with the same property. The proof of this fact is based on the following combinatorial construction of blowing ups of monomials ideals in toric vari- eties (see [15, Section 2.6]). Combinatorial description of the blowup of a monomial ideal. Let XΓ ⊂ Ks be an affine toric variety having a 0-dimensional orbit and σ∨ = R≥0Γ ⊂ Rd (which is strongly convex, by the previous proposition). (i) Let I = 〈xm|m ∈ B〉 ⊂ K[XΓ] be a monomial ideal. (ii) Let N (I) be the Newton polyhedron of I, i.e., the convex hull in Rd of the set {m+ σ∨|m ∈ B}. (iii) Let m′ ∈ B. Denote Γm′ := Γ + N({m−m′|m ∈ B}). (iv) Given m′,m′′ ∈ B, the affine toric varieties XΓm′ and XΓm′′ can be glued together along the principal open subsets XΓm′ \ V(xm ′′−m′ ) and XΓm′′ \V(xm ′−m′′ ). There is an isomorphism between these open subsets which is induced by localizations of coordinate rings: K[XΓm′ ] xm ′′ xm ′ ∼= K[XΓm′′ ] xm ′ xm ′′ . (v) The variety resulting from the previous glueing is the blowup of XΓ along I (see [15, Proposition 32]). We denote it as BlIXΓ. (vi) Finally, let B′ = {m′ ∈ B|m′ is a vertex of N (I)}. Then BlIXΓ = ⊔ m′∈B′ XΓm′ / ∼ 20 CHAPTER 1. HIGHER NASH BLOWUP OF TORIC VARIETIES (see the proof of Proposition 32, [15]). By proposition 1.3.4, for m′ ∈ B′, XΓm′ has a 0-dimensional orbit. In particular, BlIXΓ has an open cover by affine toric varieties having a 0-dimensional orbit. Remark 1.3.5. The variety resulting from the previous construction is an example of an abstract toric variety having a good action (see [15, Section 2.8]). These varieties are characterized by the fact that they can be described in combinatorial terms by families of semigroups labeled by fans (see [15, Theorem 44]). In order to use the previous construction and compare it to the higher Nash blowup of a toric variety, we need to introduce some monomial ideal. In addition, we use the Plücker embedding of Gr(λd,n,K λs,n) into the projective space P (λ2,n λd,n )−1 . First, some notation. Notation 1.3.6. Let A = {a1, . . . , as} ⊂ Zd and Γ = NA a semigroup defining a toric variety XΓ ⊂ Ks. • Given J = {β1, . . . , βλd,n } ⊂ Λs,n such that β1 ≺ · · · ≺ βλd,n , we denote by UJ the affine chart of P( M D)−1 where the J-coordinate is non-zero (see definition 1.1.2 for the notation ≺). • Let SA := {J = {β1, . . . , βλd,n } ⊂ Λs,n|β1 ≺ · · · ≺ βλd,n , det(Lc J) 6= 0}. Notice that SA 6= ∅ by propositions 1.1.8 and 1.2.4. • For each J = {β1, . . . , βλd,n } ⊂ Λs,n, denote mJ := Aβ1+ · · ·+Aβλd,n . Definition 1.3.7. Let In := 〈XmJ |J ∈ SA〉 ⊂ K[XΓ]. Following the usual terminology, we call In the logarithmic Jacobian ideal of order n of XΓ. Remark 1.3.8. In the following subsection we show that In does not de- pend on the set of generators of Γ. We want to apply the combinatorial description of the blowup of a mono- mial ideal to In. To that end, we simplify a little the notation coming from that description. For XmJ ∈ In, instead of using ΓmJ as in (iii), we simply write ΓJ . Now we are ready to prove the main theorem of this chapter. Theorem 1.3.9. Let XΓ ⊂ Ks be an affine toric variety having a 0-dimensional orbit. Then Nashn(XΓ) is isomorphic to the blowup of the logarithmic Ja- cobian ideal of order n of XΓ. In particular, Nashn(XΓ) has a finite open covering given by affine toric varieties having a 0-dimensional orbit. 1.3. HIGHER NASH BLOWUP OF TORIC VARIETIES 21 Proof. We divide this proof into two steps: the first one describes locally Nashn(XΓ) and the second one is a glueing argument. Step I: According to proposition 1.1.8 and (1.4), for a point p := ϕΓ(x) ∈ XΓ, for some x ∈ (K∗)d, we have Tn p XΓ = Im(Dn x(ϕΓ)) = Im ( cβ,αx Aβ−α ) β∈Λs,n,α∈Λd,n . Thus, the Plücker coordinates of Tn p XΓ ∈ Gr(λd,n,K λs,n) →֒ P( M D)−1 are given by the maximal minors of ( cβ,αx Aβ−α ) β,α . According to (1.5), for a choice J = {β1, . . . , βλd,n } ⊂ Λs,n, where β1 ≺ . . . ≺ βλd,n , the corresponding minor is: det(Lc J) x Aβ1+···+Aβλd,n x α1+···+αλd,n . Fix J0 ∈ SA. It follows that: 1. If J ∈ SA we can make a change of coordinates in UJ0 ∼= K s+(λs,nλd,n )−1 to turn the non-zero constant det(Lc J ) det(Lc J0 ) into 1. Thus, we can assume that the J-coordinate of Nashn(XΓ) ∩ UJ0 is equal to 1 for every J ∈ SA. 2. If J /∈ SA the J-coordinate of Nashn(XΓ) is zero. This implies that we can embed Nashn(XΓ) ∩ UJ0 in Ks+|SA|−1. These two remarks imply that Nashn(XΓ) ∩ UJ0 ∼= { ( ϕΓ(x), x ∑ βi∈J Aβi x ∑ β0 i ∈J0 Aβ0 i ) |J ∈ SA \ {J0}, x ∈ (K∗)d} = {(ϕΓ(x), x mJ−mJ0 )|J ∈ SA \ {J0}, x ∈ (K∗)d} (1.7) = Im(ϕΓJ0 ) ⊂ Ks+|SA|−1. In particular, this affine chart of Nashn(XΓ) is an affine toric variety. Step II: By Step I, for each J ∈ SA, XΓJ ∼= Nashn(XΓ) ∩ UJ . Since both BlInXΓ and Nashn(XΓ) are obtained by glueing XΓJ and Nashn(XΓ)∩UJ , respectively, we only need to check that the glueing is the same. The glueing in Nashn(XΓ) ⊂ XΓ × P( M D)−1 is given by the usual glueing in P( M D)−1, i.e., the one induced by the following isomorphisms of localizations of coordinate 22 CHAPTER 1. HIGHER NASH BLOWUP OF TORIC VARIETIES rings for each couple J1, J2 ∈ SA: K[xa1 , . . . , xas , xmJ−mJ1 |J ∈ SA /∈ {J1}]xmJ2 x mJ1 ∼= K[xa1 , . . . , xas , xmJ−mJ2 |J ∈ SA /∈ {J2}]xmJ1 x mJ2 . This is exactly the glueing described in the combinatorial description of the blowup of a monomial ideal. Remark 1.3.10. For n = 1, the previous theorem was proved in [13, 18, 15]. Remark 1.3.11. The previous theorem and its proof show that Nashn(XΓ) can be covered by open affine varieties which are invariant under the action of a torus. This statement could be obtained directly using results of [15, 27]. Indeed, by [27, Section 2.2], the higher Nash blowup of a toric variety is an equivariant morphism; in particular, it is the blowup of some monomial ideal. Then [15, Corollary 34] implies the statement. We want to emphasize that the contribution of this section is that one can take the logarithmic Jacobian ideal of order n as such monomial ideal. In addition, we describe an explicit method to construct this ideal. 1.3.3 The logarithmic Jacobian ideal of order n is indepen- dent of the generators of Γ In this subsection we show that the ideal In does not depend on the set of generators A of Γ. To that end, we need to modify temporarily the notation In. We denote as IC the logarithmic Jacobian ideal of order n, where C is an arbitrary set of generators of Γ. Theorem 1.3.12. Let A = {a1, . . . , as} ⊂ Zd and B = {b1, . . . , bt} ⊂ Zd be such that Γ = NA = NB. Then IA = IA∪B = IB. In particular, the logarithmic Jacobian ideal of order n of XΓ does not depend on the generators of Γ. Proof. It is enough to show IA = IA∪B. Lemma 1.3.13 states that IA ⊂ IA∪B. Applying repeatedly lemma 1.3.14 we obtain the other inclusion. Lemma 1.3.13. With the notation of theorem 1.3.12, IA ⊂ IA∪B. Proof. For J ∈ SA, define J̄ := {(β, 0, . . . , 0) ∈ Ns+t|β ∈ J}. The submatrix of Dn x(ϕA∪B) defined by J̄ is the same as the submatrix of Dn x(ϕΓ) defined by J . Therefore J̄ ∈ SA∪B. Thus, X mJ = XmJ̄ ∈ IA∪B. 1.3. HIGHER NASH BLOWUP OF TORIC VARIETIES 23 Lemma 1.3.14. Let A be as in theorem 1.3.12 and b ∈ NA. Let A′ = A ∪ {b}. Then IA′ ⊂ IA. Proof. Consider the following partition of SA′ : S1 := {J̄ ∈ SA′ |β(s+ 1) = 0 for all β ∈ J̄}, S2 := {J̄ ∈ SA′ |β(s+ 1) > 0 for some β ∈ J̄}. By definition, IA′ = 〈{XmJ̄ |J̄ ∈ S1} ∪ {XmJ̄ |J̄ ∈ S2}〉. As in the proof of lemma 1.3.13, {XmJ̄ |J̄ ∈ S1} ⊂ IA. We claim that {XmJ̄ |J̄ ∈ S2} ⊂ 〈{XmJ̄ |J̄ ∈ S1}〉, implying the lemma. Now, to prove the claim we show that for J̄ ∈ S2 there exists J ∈ S1 and γ̄ ∈ Γ such that mJ̄ = mJ + γ̄. First, we need some notation. • For γ ≤ βi, let ǫγ := (−1)|βi−γ| ( βi γ ) . Then, by definition, cβi,αj = ∑ γ≤βi,γ 6=0 ǫγ ( A′γ αj ) (see lemma 1.2.3). • cβi := ( ∑ γ≤βi,γ 6=0 ǫγ ( A′γ αj ) ) 1≤j≤λd,n (cβi is the βith row of Lc J̄ ). • vγ := ( ( A′γ αj ) ) 1≤j≤λd,n . Notice that by remark 2.1.3, cβi = ∑ γ≤βi,γ 6=0 ǫγvγ (that remark is stated for toric curves but it also holds for toric vari- eties of any dimension). Let J̄ = {β1, . . . , βλd,n } ∈ S2. Then det(Lc J̄ ) 6= 0 and we can assume that β1(s+ 1) > 0. Then the following holds: 1. There exists γ′ ≤ β1 such that the matrix obtained by replacing the β1th row of Lc J̄ by vγ′ has non-zero determinant. 2. There exists δ0 ∈ Ns+1 such that δ0(s+ 1) = 0 and A′δ0 = A′γ′. 3. There exists δ ∈ Ns+1 such that δ ≤ δ0, δ(s + 1) = 0, and the matrix having as rows cδ, cβ2 , . . . , cβλd,n has non-zero determinant. 4. Let J1 := J̄ \{β1}∪{δ}. Then J1 ∈ SA′ and mJ̄ equals mJ1 plus some element in Γ. Notice that by applying 1 - 4 to any element of J̄ whose (s+ 1)-entry is greater than zero, we obtain J ∈ S1 and γ̄ ∈ Γ with the desired properties. Now we prove the previous statements. 24 CHAPTER 1. HIGHER NASH BLOWUP OF TORIC VARIETIES 1. It follows immediately from: 0 6= det(Lc J̄ ) = det      cβ1 cβ2 ... cβλd,n      = det      ∑ γ≤β1,γ 6=0 ǫγvγ cβ2 ... cβλd,n      = ∑ γ≤β1,γ 6=0 ǫγ det      vγ cβ2 ... cβλd,n      . 2. If γ′(s + 1) = 0, let δ0 := γ′. Now suppose that γ′(s + 1) = k > 0. Since b ∈ NA, b = ∑s l=1 λlal. Let δ0(l) := γ′(l) + kλl for l < s+1 and δ0(s+ 1) = 0. Then A′δ0 = s ∑ l=1 δ0(l)al = s ∑ l=1 (γ′(l) + kλl)al = s ∑ l=1 γ′(l)al + kb = A′γ′. 3. Let M denote the matrix whose rows are cδ0 , cβ2 , . . . , cβλd,n , in this order. If det(M) 6= 0 let δ := δ0. Suppose that det(M) = 0. Then 0 = det(M) = ∑ γ<δ0,γ 6=0 ǫγ det      vγ cβ2 ... cβλd,n      + det      vδ0 cβ2 ... cβλd,n      . On the other hand, A′δ0 = A′γ′ implies vγ′ = vδ0 and so 0 6= det      vγ′ cβ2 ... cβλd,n      = det      vδ0 cβ2 ... cβλd,n      . Therefore 0 6= ∑ γ<δ0,γ 6=0 ǫγ det      vγ cβ2 ... cβλd,n      . 1.3. HIGHER NASH BLOWUP OF TORIC VARIETIES 25 Thus there exists δ1 < δ0 such that det(vδ1 cβ2 · · · cβλd,n ) 6= 0. If det(cδ1 cβ2 · · · cβλd,n ) 6= 0, let δ := δ1. Otherwise repeat the previous process. This leads to a sequence δ0 > δ1 > · · · . Since this sequence cannot decrease infinitely many times, we conclude that there exists k ≥ 0 such that δ0 > δ1 > · · · > δk =: δ and det(cδ cβ2 · · · cβλd,n ) 6= 0. In addition, since δ ≤ δ0 and δ0(s+ 1) = 0, we have δ(s+ 1) = 0. 4. To show that J1 ∈ SA′ we only need to show that |δ| ≤ n because we already know that det(Lc J1 ) 6= 0. If |δ| > n then, by lemma 1.1.6 (ii), cδ = 0, which contradicts that det(Lc J1 ) 6= 0. On the other hand, we know that δ ≤ δ0 and γ′ ≤ β1. Let δ0 = δ+ δ′ and β1 = γ′+ γ′′. Then A′β1 = A′γ′ +A′γ′′ = A′δ0 +A′γ′′ = A′δ +A′δ′ +A′γ′′. This implies that mJ̄ equals mJ1 plus an element from Γ. 26 CHAPTER 1. HIGHER NASH BLOWUP OF TORIC VARIETIES Chapter 2 Combinatorial structure of higher Nash blowup of toric curves 2.1 Higher Nash blowup of toric curves In this chapter we study in detail the higher Nash blowup of toric curves. In this chapter we use the following notation: A = {a1, . . . , as} ⊂ N, where 0 < a1 < . . . < as and gcd(a1, . . . , as) = 1. Let Γ = NA ⊂ N. We assume that A is the minimal generating set of Γ. LetXΓ ⊂ Ks be the corresponding toric curve. According to theorem 1.3.9, Nashn(XΓ) is isomorphic to the blowup of the ideal In. Since Γ ⊂ R≥0, it follows that the Newton polyhedron of In has only one vertex mJ0 = min{mJ |J ∈ SA}. In particular, Nashn(XΓ) is determined by a single semigroup. We denote it as: Nashn(Γ) := Γ + N({mJ −mJ0 |J ∈ SA \ {J0}}). Let us show how this semigroup looks like for n = 1. In this case, S = {e1, . . . , es}, where the e′is denote the canonical basis of Ns, mei = ai, and so mini{mei} = a1. Therefore Nash1(Γ) = Γ + N({ak − a1|k > 1}). Remark 2.1.1. The previous description is a particular case of the com- binatorial description of the Nash blowup of toric varieties given in [15, 16] (see also [11], where the Nash blowup of toric curves is studied in detail). 27 28 CHAPTER 2. HIGHER NASH BLOWUP OF TORIC CURVES We may ask the question: is there an explicit description for Nashn(Γ) as in n = 1? T. Yasuda made the following conjecture in a more general context. Conjecture 2.1.2. [28, Conjecture 5.6] Let X be a formal curve with asso- ciated semigroup Γ = {0 = s0 < s1 < · · · }. Let Nashn(Γ) be the associated semigroup of Nashn(X). Let Γ(n) be the semigroup generated by sm − sl, where l ≤ n < m. Then Nashn(Γ) = Γ(n). In what follows we prove that this conjecture is true for toric curves. However, in the final section we show that it fails in general. In order to prove the conjecture in the toric case, first we need to study carefully some maximal minors of the higher-order Jacobian matrix. In sec- tion 1.2 we defined, for J = {β1, . . . , βn} ⊂ Λs,n, the matrix Lc J = (cβi,j)i,j , where cβi,j = ∑ γ≤βi,γ 6=0 (−1)|βi−γ| ( βi γ )( A · γ j ) . Notice that in this case A is a vector in Ns and A ·γ is the usual dot product. Remark 2.1.3. (a) For a fixed i, every entry of the i-th row of Lc J has the same amount of summands and the same coefficients (−1)|βi−γ| ( βi γ ) . In other words, for a fixed row of Lc J , the amount of summands and coefficients of its entries do not depend on j. (b) Fix i ∈ {1, . . . , n}. We rewrite the sums cβi,j as follows: cβi,j = ( si,1 j ) + ti,2 ( si,2 j ) + · · ·+ ti,ki ( si,ki j ) , where si,1 := A · βi, si,l ∈ {A · γ|γ ≤ βi, 0 6= γ 6= βi} for 1 < l ≤ ki, and ti,l ∈ Z. Assume that si,1 > si,2 > . . . > si,ki > 0. By (a), {si,l}l, {ti,l}l and ki do not depend on j. Therefore, the i-th row of Lc J can be written as: ( ( si,1 1 ) + ti,2 ( si,2 1 ) + · · ·+ ti,ki ( si,ki 1 ) , . . . , ( si,1 n ) + ti,2 ( si,2 n ) + · · ·+ ti,ki ( si,ki n ) ) . Now we define some elementary operations on a matrix having the same shape as Lc J . Given ki ∈ N for i ∈ {1, . . . , n}, and si,l ∈ N\{0}, ti,l ∈ Q\{0} for l ∈ {1, . . . , ki}, consider a matrix D = ( ti,1 ( si,1 j ) + ti,2 ( si,2 j ) + · · ·+ ti,ki (si,ki j ) ) 1≤i≤n 1≤j≤n . 2.1. HIGHER NASH BLOWUP OF TORIC CURVES 29 Notice that if we fix i, the terms ki, si,l, ti,l do not depend on j. Assume that si,1 > si,2 > · · · > si,ki for all i ∈ {1, . . . , n}. Finally, let Ri denote the i-th row of D. Definition 2.1.4. We say that D satisfies (⋆) if there exist i, i′ ∈ {1, . . . , n} such that si,1 = si′,1. Using the following algorithm we show that, under some assumptions, we can perform elementary operations on the rows of D to obtain a matrix that does not satisfy the property (⋆). Algorithm 2.1.5. Assume detD 6= 0 and that D satisfies (⋆). 1. Replace the row Ri by ti′,1Ri − ti,1Ri′ . 2. Since detD 6= 0 the new row cannot be the vector 0̄. Write this new vector as: R′ i := ( t′i,1 (s′i,1 j ) + ti,2 (s′i,2 j ) + · · ·+ t′ i,k′i (s′ i,k′ i j ) ) 1≤j≤n , where t′i,l 6= 0 for all l ∈ {1, . . . , k′i} and s′i,1 > · · · > s′ i,k′i . Notice that si,1 > s′i,1 > 0. 3. Let D′ :=         R1 ... R′ i ... Rn         . (i) If there exists i′′ ∈ {1, . . . , n} \ {i} such that s′i,1 = si′′,1, then apply step 1 to R′ i. As before, we obtain a new element s′′i,1 ∈ N such that si,1 > s′i,1 > s′′i,1 > 0. (ii) If there is no i′′ ∈ {1, . . . , n} \ {i} such that s′i,1 = si′′,1 then stop. Because of the decreasing sequence si,1 > s′i,1 > s′′i,1 > · · · , this algorithm must stop and it produces a new row that looks like ( ui,1 ( ri,1 j ) + ui,2 ( ri,2 j ) + · · ·+ ui,mi (ri,mi j ) ) 1≤j≤n , where ui,l 6= 0 for all l ∈ {1, . . . ,mi}, ri,1 > · · · > ri,mi > 0 and ri,1 6= sl,1 for all l ∈ {1, . . . , n} \ {i}. 30 CHAPTER 2. HIGHER NASH BLOWUP OF TORIC CURVES Applying this process every time that the new matrix satisfies (⋆), we finally get a matrix D D = ( ui,1 ( ri,1 j ) + ui,2 ( ri,2 j ) + · · ·+ ui,mi (ri,mi j ) ) 1≤i≤n 1≤j≤n , such that ri,1 > · · · > ri,mi for each i and ri,1 6= ri′,1 for all i 6= i′. Example 2.1.6. Consider the following matrix: D =   ( 2 1 ) ( 2 2 ) ( 2 3 ) ( 6 1 ) − 2 ( 3 1 ) ( 6 2 ) − 2 ( 3 2 ) ( 6 3 ) − 2 ( 3 3 ) ( 6 1 ) − 3 ( 4 1 ) + 3 ( 2 1 ) ( 6 2 ) − 3 ( 4 2 ) + 3 ( 2 2 ) ( 6 3 ) − 3 ( 4 3 ) + 3 ( 2 3 )   . Notice that D satisfies (⋆). Applying algorithm 2.1.5 to the third row we obtain the matrix: D =   ( 2 1 ) ( 2 2 ) ( 2 3 ) ( 6 1 ) − 2 ( 3 1 ) ( 6 2 ) − 2 ( 3 2 ) ( 6 3 ) − 2 ( 3 3 ) −3 ( 4 1 ) + 2 ( 3 1 ) + 3 ( 2 1 ) −3 ( 4 2 ) + 2 ( 3 2 ) + 3 ( 2 2 ) −3 ( 4 3 ) + 2 ( 3 3 ) + 3 ( 2 3 )   . 2.1.1 A partial description of Nashn(XΓ) The first step towards proving conjecture 2.1.2 for toric curves is to deter- mine minJ∈SA {mJ}. Recall that for J = {β1, . . . , βn} ⊂ Λs,n, we defined mJ = A · β1 + · · · + A · βn. On the other hand, A · βi ∈ Γ since A is the vector formed by the generators of Γ. Therefore, it is natural to expect that minJ∈SA {mJ} = ∑n i=1 si. The goal of this subsection is to prove that this is indeed the case. Proposition 2.1.7. Let J ⊂ Λs,n, |J | = n, then minJ∈SA {mJ} = ∑n i=1 si. Proof. This is proved in lemmas 2.1.9 and 2.1.13. This proposition gives the following preliminary description ofNashn(Γ). Corollary 2.1.8. Nashn(Γ) = Γ + N({mJ − ∑n i=1 si|J ∈ SA}). Lemma 2.1.9. Let J ∈ SA. Then mJ ≥ s1 + · · · + sn. In particular, minJ∈SA {mJ} ≥ ∑n i=1 si. Proof. Let J = {β1, β2, . . . , βn}. Using (b) of remark 2.1.3 we have Lc J = ( ( si,1 j ) + ti,2 ( si,2 j ) + · · ·+ ti,ki (si,ki j ) ) 1≤i≤n 1≤j≤n , 2.1. HIGHER NASH BLOWUP OF TORIC CURVES 31 where si,l ∈ Γ for each l, si,1 = A · βi, and si,1 > si,2 > · · · > si,ki . If si,1 6= si′,1 for all 1 ≤ i 6= i′ ≤ n then the statement follows. Now suppose that there exist βi, βi′ ∈ J , i 6= i′, such that si,1 = si′,1, i.e., Lc J satisfies (⋆). Since J ∈ SA, i.e., det(L c J) 6= 0, we can apply algorithm 2.1.5 to obtain some elements r1,1, . . . , rn,1 ∈ Γ satisfying ri,1 6= ri′,1 for all i 6= i′ and si,1 > ri,1 for some i ∈ {1, . . . , n}. Under these conditions we have mJ = n ∑ i=1 A · βi = n ∑ i=1 si,1 > n ∑ i=1 ri,1 ≥ n ∑ i=1 si. To show that minJ∈SA {mJ} = ∑n i=1 si we need to show that, if J = {β1, . . . , βn} ⊂ Λs,n is such that A ·βi = si, then J ∈ SA. In other words, we need to study J ’s such that det(Lc J) 6= 0. We do not have a characterization of such J ’s. However, in the following definition and lemma we give sufficient conditions for J to be in SA. Definition 2.1.10. Let J ⊂ Ns be a finite subset. We say that J satisfies (∗) if the following conditions hold: 1) For all β, β′ ∈ J such that β 6= β′, it holds that A · β 6= A · β′. 2) For all β ∈ J and 0 6= γ < β, there exists β′ ∈ J such that A·γ = A·β′. Example 2.1.11. (i) Let J = {ej , 2ej , . . . , nej}, where ej is a basic vec- tor. Then J satisfies (∗). Indeed, 1) follows by definition and 2) by the definition of <. (ii) Let J = {β1, β2, . . . , βn} be such that A · βi = si. Then J satisfies (∗). Indeed, 1) follows by definition and 2) follows from the fact that γ < β implies A · γ < A · β. Remark 2.1.12. Let β ∈ Ns be such that |β| ≥ n+1. Then A · β > na1 ≥ sn. This implies that for J = {β1, . . . , βn} ⊂ Ns such that A · βi = si for each i, we have J ⊂ Λs,n. Lemma 2.1.13. Let J ⊂ Λs,n, |J | = n. If J satisfies (∗) then J ∈ SA. In particular, minJ∈SA {mJ} ≤ ∑n i=1 si. Proof. The last statement follows from (ii) of example 2.1.11 and remark 2.1.12. Take a set J = {β1, . . . , βn} satisfies (∗). In particular, A ·βi 6= A ·βi′ 32 CHAPTER 2. HIGHER NASH BLOWUP OF TORIC CURVES for all i 6= i′. Assume A · β1 < · · · < A · βn. By (b) of remark 2.1.3, we can rewrite the matrix Lc J as ( cβi,j ) 1≤i≤n 1≤j≤n = ( ( si,1 j ) + ti,2 ( si,2 j ) + · · ·+ ti,ki (si,ki j ) ) 1≤i≤n 1≤j≤n , where si,1 = A · βi, si,1 > si,2 > · · · , and si,l = A · γ for some γ ≤ βi. Now we do some elementary operations on the n-th row of Lc J : ( ( sn,1 j ) + tn,2 ( sn,2 j ) + · · ·+ tn,kn ( sn,kn j ) ) 1≤j≤n . We know that sn,2 = A ·γ for some γ < βn. Since J satisfies (∗), there exists βj0 ∈ J such that A · βj0 = A · γ = sn,2. Then we subtract tn,2-times the row j0 from the row n, thus obtaining ( (s′n,1 j ) + t′n,2 (s′n,2 j ) + · · ·+ t′n,k′n (s′ n,k′n j ) ) 1≤j≤n , where s′n,1 = sn,1, s ′ n,2 > s′n,3 > · · · , and sn,2 > s′n,2. Now we have s′n,2 = A·γ′ for some γ′ < βn or some γ′ < βj0 . Once again, by (∗) we can repeat the previous process to obtain a new element s′′n,2 such that sn,2 > s′n,2 > s′′n,2. Because of this decreasing sequence of natural numbers, the iteration of this process must stop turning the n-th row into ( ( sn,1 j ) ) 1≤j≤n . Applying this process to the other rows of Lc J in an ascending way we obtain the matrix ( ( si,1 j ) ) 1≤i≤n 1≤j≤n . Notice that si,1 = A · βi 6= A · βi′ = si′,1 for all i 6= i′. The following lemma shows that this matrix has non-zero determinant, thus concluding that J ∈ S. Lemma 2.1.14. Let 0 < c1 < c2 < · · · < cn be natural numbers. Consider the matrix L = ( ( ci j ) ) 1≤i≤n 1≤j≤n . Then detL 6= 0. Proof. For j ∈ {1, . . . , n}, consider the polynomial bj(x) = x(x−1)···(x−j+1) j! . Notice that if x ∈ N, bj(x) = ( x j ) and deg bj(x) = j. Thus L = ( ( ci j ) ) 1≤i≤n 1≤j≤n = ( bj(ci) ) 1≤i≤n 1≤j≤n . 2.1. HIGHER NASH BLOWUP OF TORIC CURVES 33 Now we show that the columns of this matrix are linearly independent. Let α1, . . . , αn ∈ R be such that ∑n j=1 αjbj(ci) = 0, for each i ∈ {1, . . . , n}. Let f(x) = ∑n j=1 αjbj(x). Then {c1, . . . , cn} are roots of f(x). But we also have f(0) = 0. Since deg f(x) ≤ n it follows that f(x) = 0. As deg bj(x) = j for each j, we conclude that αj = 0 for all j. In particular, detL 6= 0. 2.1.2 Proof of conjecture 2.1.2 for toric curves and some con- sequences Now we are ready to prove the main theorem of this chapter. Recall that by definition and corollary 2.1.8: Γ(n) = N({sm − sl|m > n, l ≤ n}), Nashn(Γ) = Γ + N({mJ − n ∑ l=1 sl|J ∈ SA}). Theorem 2.1.15. Γ(n) = Nashn(Γ). Proof. This is proved in propositions 2.1.16 and 2.1.18. Proposition 2.1.16. Nashn(Γ) ⊂ Γ(n). Proof. By corollary 2.1.8, it is enough to show that ai ∈ Γ(n) for each i ∈ {1, . . . , s} and mJ − ∑n l=1 sl ∈ Γ(n) for each J ∈ SA. We first prove ai ∈ Γ(n). For ai ≤ sn there exists m ∈ N such that mai ≤ sn < (m + 1)ai. Then ai = (m + 1)ai −mai ∈ Γ(n). If ai ≥ sn then ai + a1 > sn, and ai = (ai + a1)− a1 ∈ Γ(n). Now we prove that mJ − ∑n l=1 sl ∈ Γ(n) for each J ∈ SA. Consider J = {β1, β2, . . . , βn} ∈ SA, let si,1 := A · βi and assume s1,1 ≤ · · · ≤ sn,1. Let k := max{l ∈ {1, . . . , n}|sl,1 ≤ sn}. Case I: Suppose that s1,1 < s2,1 < · · · < sk,1 ≤ sn. Let ψ = {s1, . . . , sn} \ {s1,1, · · · , sk,1}. Write ψ = {rk+1, . . . , rn}. By definition of k and ψ we obtain: mJ − n ∑ l=1 sl = n ∑ l=1 sl,1 − n ∑ l=1 sl = n ∑ l=k+1 sl,1 − n ∑ l=k+1 rl ∈ Γ(n). Case II: Suppose that there exist i, i′ ≤ k such that si,1 = si′,1. We claim that for all j ≤ k there exist rj,1 ∈ Γ such that sj,1 − rj,1 ∈ Γ(n) and rj,1 6= rj′,1 for all j 6= j′. Assume this claim for the moment. For the elements sm,1 with m > k, we have that sm,1 > sn and so sm,1−sl ∈ Γ(n) for 34 CHAPTER 2. HIGHER NASH BLOWUP OF TORIC CURVES any l ≤ n. Let ψ = {s1, . . . , sn} \ {r1,1, . . . , rk,1}. Write ψ = {rk+1, . . . , rn}. As in the previous case, we conclude that mJ − n ∑ l=1 sl = ( k ∑ l=1 sl,1 − k ∑ l=1 rl,1 ) + ( n ∑ l=k+1 sl,1 − n ∑ l=k+1 rl ) ∈ Γ(n). Now we prove the claim. Since J ∈ SA, we can apply algorithm 2.1.5 to any pair of rows of Lc J , i, i ′ ≤ k such that si,1 = si′,1, to get a matrix D = ( ui,1 ( ri,1 j ) + ui,2 ( ri,2 j ) + · · ·+ ui,mi (ri,mi j ) ) 1≤i≤n 1≤j≤n , where ri,1 6= ri′,1 for all i, i′ ≤ k. Let us show that si,1 − ri,1 ∈ Γ(n) for all i ∈ {1, . . . , k}. We can assume that si,1 6= ri,1. In the first run of the algorithm we obtain an element s′i,1 ∈ Γ such that si,1 > s′i,1 ≥ ri,1 and s′i,1 = A · γ for some γ < βi or some γ < βi′ . This implies that si,1 − s′i,1 ∈ Γ. On the other hand, we know that s′i,1 ≥ ri,1 and ri,1 ∈ Γ. Therefore si,1 − s′i,1 + ri,1 ∈ Γ and si,1 − s′i,1 + ri,1 ≤ sn. Consider the following set φi := {sl ∈ Γ\{0}|sl+ri,1 ≤ sn}. This set is not empty since si,1−s ′ i,1 ∈ φi. Let st := maxφi. If si,1 + st ≤ sn, we have that (si,1 − s′i,1 + st) + ri,1 = si,1+st−(s′i,1−ri,1) ≤ si,1+st ≤ sn and si,1−s ′ i,1+st > st, which contradicts the maximality of st. Thus si,1 + st > sn and si,1 − ri,1 = (si,1 + st)− (st + ri,1) ∈ Γ(n). We need the following lemma to prove the remaining inclusion in theorem 2.1.15. Lemma 2.1.17. Let sm, si ∈ Γ be such that m > n ≥ i and sm − si /∈ Γ. Let βm ∈ Ns be such that A · βm = sm. Then there exists β0 ≤ βm such that A · β0 > sn and |β0| ≤ n. Proof. If |βm| ≤ n then βm satisfies the conditions of β0. Assume that |βm| > n. Suppose first that a2 ≤ sn. The set {a1, 2a1, . . . , (n − 1)a1, a2} has n different elements of Γ implying na1 > sn. Let β be such that |β| = n. Then A · β ≥ ns1 = na1 > sn. In particular, any β ≤ βm such that |β| = n satisfies the conditions of the Lemma. 2.1. HIGHER NASH BLOWUP OF TORIC CURVES 35 Now suppose sn < a2. Then sk = ks1 for all k ≤ n. Notice that if βm(j) = 0 for all j > 1, then sm − si = |βm|a1 − ia1 ∈ Γ which contradicts the hypothesis. Thus there exists j > 1 such that βm(j) 6= 0. Consider β0 = ej , then A · β0 = A · ej = aj ≥ a2 > sn and |β0| = 1 ≤ n. Proposition 2.1.18. Γ(n) ⊂ Nashn(Γ). Proof. Throughtout this proof we fix m, i ∈ N such that m > n ≥ i. Let sm − si ∈ Γ(n). Case I: Suppose that sm − si ∈ Γ. Then sm − si ∈ Nashn(Γ) by definition. Case II: Suppose that sm − si /∈ Γ. Fix βm ∈ Ns such that A · βm = sm. We claim that there exist β0 ∈ Ns, J0 = {β1, . . . , βn} ⊂ Ns, and i ≤ l ≤ n such that: (1) β0 ≤ βm. (2) A · βj = sj for each j ∈ {1, . . . , n}. (3) sl − si ∈ Γ. (4) J := (J0 \ {βl}) ∪ {β0} satisfies (∗). In particular, A · β0 − sl = mJ − n ∑ i=1 si ∈ Nashn(Γ). Assume this claim for the moment. Let δ ∈ Ns be such that βm = β0+δ. In particular, sm = A · βm = A · β0 +A · δ. Then, since A · δ ∈ Γ, we conclude sm − si = (A · β0 − sl) +A · δ + (sl − si) ∈ Nashn(Γ). Now we prove the claim. We first show that there is a β0 ∈ Ns satisfying (1) and some extra conditions needed for the proof of (4). Let T := {γ ∈ Ns|γ ≤ βm}. We write this set as T = T≤ ⊔ T>, where T≤ := {γ ∈ T |A · γ ≤ sn}, T> := {γ ∈ T |A · γ > sn}. Notice that βm ∈ T>. Let β0 ≤ βm be a minimal element in T> such that β0 ∈ Λs,n (such an element exists by lemma 2.1.17). By construction, β0 has the following properties: a) For all γ < β0, γ ∈ T≤. 36 CHAPTER 2. HIGHER NASH BLOWUP OF TORIC CURVES b) For all β̄i such that A · β̄i = si it holds β0 ≯ β̄i (this is true because βm ≥ β0 and sm − si /∈ Γ implies that βm ≯ β̄i). Now we prove the existence of J0 = {β1, . . . , βn} ⊂ Ns, and i ≤ l ≤ n satisfying (2) and (3) and some extra conditions needed for the proof of (4). Define the set (recall that i is fixed): Ω := {sj ∈ {s1, . . . , sn}|∀β ′ such that A·β′ = sj , ∃γ ≤ β′ such that A·γ = si}. If sj ∈ {s1, . . . , sn} \ Ω, consider βj ∈ Ns such that A · βj = sj and for all γ ≤ βj , A · γ 6= si. If sj ∈ Ω, consider a βj ∈ Ns such that A · βj = sj . Let J0 := {β1, . . . , βn} ⊂ Ns. By remark 2.1.12, we have that J0 ⊂ Λs,n. Notice that Ω 6= ∅, since si ∈ Ω. Let sl := max{Ω}. In particular, sl ∈ Ω and so sl − si ∈ Γ. It remains to prove that J := (J0\{βl})∪{β0} satisfies (∗) (see definition 2.1.10). By construction and since A · β0 > sn, we have 1) in the definition of (∗). Now let βk ∈ J0 \ {βl}. We want to show that if γ < βk then there exists β′ ∈ J such that A · β′ = A · γ. If k < l this condition is satisfied (see example 2.1.11). Suppose k > l (in particular, sk /∈ Ω). If γ < βk is such that A · γ = sj 6= sl then by making β′ = βj the condition is satisfied. Suppose that A·γ = sl. Since sl ∈ Ω there exists γ′ ≤ γ such that A·γ′ = si. Since γ < βk, it follows that γ′ < βk. This is a contradiction since βk was chosen so that for all δ < βk we have A · δ 6= si. Therefore, for all γ ≤ βk, A · γ 6= sl. This shows that every element of J0 \ {βl} satisfies 2) in the definition of (∗). Now consider γ < β0. By property a) above, we have that A · γ ≤ sn. If γ < β0 is such that A · γ = sj 6= sl then, as before, by making β′ = βj the condition is satisfied. Suppose that A ·γ = sl. As before, there exists γ ′ ≤ γ such that A · γ′ = si. Since γ < β0, it follows that γ ′ < β0. This contradicts property b) above. We conclude that J satisfies (∗). Theorem 2.1.15 has two immediate consequences. The first one is about resolving toric curves by applying once the higher Nash blowup for n suf- ficiently large. This gives a combinatorial proof of Yasuda’s theorem on one-step resolution of curves by higher Nash blowups in the case of toric curves. The second result is the analogue of Nobile’s theorem for the higher Nash blowup of toric curves. Corollary 2.1.19. Nashn(XΓ) is non-singular if and only if sn + 1 ∈ Γ. In particular, Nashn(XΓ) is non-singular for n≫ 0. 2.2. COUNTEREXAMPLE TO THE CONJECTURE 37 Proof. Notice that for all m > n and i ≤ n, we have that sn+1 + si ≤ sm + sn, then sn+1 − sn ≤ sm − si. Thus, sn+1 − sn = min{Γ(n) \ {0}} = min{Nashn(Γ) \ {0}}. Then Nashn(XΓ) is non-singular if and only if Nashn(Γ) = N({1}) if and only if 1 = sn+1 − sn if and only if sn + 1 = sn+1 ∈ Γ. Corollary 2.1.20. Nashn(XΓ) ∼= XΓ if and only if XΓ is non-singular. Proof. Suppose that XΓ is singular, i.e., 1 < a1. We are going to show that Γ ( Nashn(Γ), which implies Nashn(XΓ) 6∼= XΓ. Let a2 = qa1 + r, where 0 < r < a1 and q ≥ 1. With this notation we have s1 = a1, . . . , sq = qa1, sq+1 = a2. If n ≤ q then sn ≤ sq = qa1 < a2 and so a2−a1 ∈ Γ(n) = Nashn(Γ). But we also have a2−a1 = (q−1)a1+r /∈ Γ. Suppose that n > q. Consider the following subset of Γ: {sq+1 = qa1 + r, (q + 1)a1, (q + 1)a1 + r, (q + 2)a1, (q + 2)a1 + r, . . .}. The elements on this subset are not necessarily consecutive elements in Γ. Therefore, for p > q it follows sp+1−sp ≤ max{a1−r, r} < a1. In particular, sn+1 − sn < a1. Thus, sn+1 − sn ∈ Nashn(Γ) but sn+1 − sn /∈ Γ. 2.2 Counterexample to the conjecture In section 2.1 we stated and proved a conjecture by T. Yasuda for toric curves. In this section we exhibit a family of non-monomial curves showing that the conjecture is false in general. Example 2.2.1. Consider the plane curve C ⊂ C2 parametrized by t 7→ (t4, t6 + t7). The associated semigroup of C is Γ = {0, 4, 6, 8, 10, 12, 13, 14,m|m ≥ 16}. Yasuda’s conjecture states that the semigroup of Nash1(C) is Γ (1) = N(2, 9). However, the Nash blowup of order 1 of C is parametrized by t 7→ ( t4, t6 + t7, 6 4 t2 + 7 4 t3 ) . Using the first and third terms of the parametrization we obtainNash1(Γ) = N(2, 5). We conclude that Nash1(Γ) 6= Γ(1). 38 CHAPTER 2. HIGHER NASH BLOWUP OF TORIC CURVES We may still ask whether the conjecture holds for n≫ 0. In what follows we construct a family of plane curves {Cn}n≥1, with numerical semigroup Γn, such that Nashn(Γn) 6= (Γn) (n). Fix n ≥ 1. Consider the plane curve Cn with parametrization ϕ(t) = (t4, t4n+2 + t4n+3). Let Γn be the corresponding semigroup. Notice that the first n non-zero terms of Γn is the set {4, 8, . . . , 4n}. In addition, the first odd number that appears in Γn is 8n+5 (it appears as the order of (t4n+2+t4n+3)2−(t4)2n+1). In particular, the first odd number that appears in (Γn) (n) is 8n+ 5− 4n = 4n+ 5. We claim that 5 ∈ Nashn(Γn) implying that Nashn(Γn) 6= (Γn) (n). To prove the claim we need to compute some maximal minors of the matrix ( 1 α! ∂α(ϕ− ϕ(t))β ∂Tα |t ) β∈Λ2,n,α∈Λ1,n Let J1 = {e1, 2e1, . . . , ne1} and J2 = {e1, 2e1 . . . , (n − 1)e1, e2}. We first show that the minors of the submatrices defined by J1 and J2 are not zero. Let LJ1 be the submatrix defined by J1. Notice that the rows of LJ1 only involve the first term of ϕ(t), which is a monomial. Therefore, by example 2.1.11 and lemma 2.1.13, detLJ1 6= 0. In addition, by proposition 1.2.4, detLJ1 = c · t ∑n k=1 4k−k = c · t 3n(n+1) 2 , with c a non-zero constant. Now, for J2, notice that the first n − 1 rows of LJ2 only involve the monomial term of ϕ(t). Using lemma 1.2.3 we obtain that the (i, j)-entry of LJ2 is cie1,jt 4i−j , for 1 ≤ i < n and 1 ≤ j ≤ n. On the other hand, the nth row of LJ2 can be described as follows. Since |e2| = 1, by lemma 1.1.6 we obtain 1 j! ∂j(ϕ− ϕ(t))e2 ∂T j |t = ( 4n+ 2 j ) t4n+2−j + ( 4n+ 3 j ) t4n+3−j . Summarizing, the matrix LJ2 is:      ce1,1t 4−1 · · · ce1,nt 4−n ... ... c(n−1)e1,1t 4(n−1)−1 · · · c(n−1)e1,nt 4(n−1)−n ( 4n+2 1 ) t4n+2−1 + ( 4n+3 1 ) t4n+3−1 · · · ( 4n+2 n ) t4n+2−n + ( 4n+3 n ) t4n+3−n      . Multiply the jth column by tj . Then, for 1 ≤ i < n multiply the ith row by 2.2. COUNTEREXAMPLE TO THE CONJECTURE 39 t−4i. Finally, multiply the nth row by t−4n−2 to obtain detLJ2 = ( t 3n(n+1) 2 +2 ) det      ce1,1 · · · ce1,n ... ... c(n−1)e1,1 · · · c(n−1)e1,n ( 4n+2 1 ) + ( 4n+3 1 ) t · · · ( 4n+2 n ) + ( 4n+3 n ) t      . Applying the method of proof of Proposition 2.1.13 to the first n − 1 rows and using basic properties of determinants, we get that detLJ2 = t 3n(n+1) 2 +2 det      ( 4 1 ) · · · ( 4 n ) ... ... ( 4(n−1) 1 ) · · · ( 4(n−1) n ) ( 4n+2 1 ) · · · ( 4n+2 n )      + t 3n(n+1) 2 +3 det      ( 4 1 ) · · · ( 4 n ) ... ... ( 4(n−1) 1 ) · · · ( 4(n−1) n ) ( 4n+3 1 ) · · · ( 4n+3 n )      . The determinants appearing in the sum are non-zero by lemma 2.1.14. Therefore detLJ2 6= 0. Now we need to prove that detLJ1 has the minimum order over all non- zero minors of the higher-order Jacobian matrix of ϕ. Consider β = (a1, a2) ∈ N2 and suppose that a2 > 0. Notice that if the polynomial 1 m! ∂m(T 4 − t4)a1(T 4n+2 + T 4n+3 − t4n+2 − t4n+3)a2 ∂Tm ∣ ∣ t is non-zero, then its order is greater or equal than 4n + 2 − m. Let J = {β1, . . . , βn} ⊂ Λ2,n be such that J 6= {e1, . . . , ne1}. In particular, the second entry of βi is non-zero, for some i. Reorder J in such a way that βi(2) 6= 0 for 1 ≤ i ≤ k and βj(2) = 0 for k < j ≤ n. Then, if j > k, βj = mje1 with 1 ≤ mj ≤ n and if j > i > k, mj 6= mi. Let us show that if detLJ 6= 0 then ord(det(LJ)) > 3n(n+1) 2 . To begin with, detLJ = ∑ σ∈Sn sgn(σ) n ∏ i=1 ai,σ(i) = ∑ σ∈Sn sgn(σ) n ∏ i=1 1 σ(i)! ∂σ(i)(ϕ− ϕ(t))βi ∂T σ(i) |t, 40 CHAPTER 2. HIGHER NASH BLOWUP OF TORIC CURVES where Sn is the symmetric group. Let Aσ = ∏n i=1 1 σ(i)! ∂σ(i)(ϕ−ϕ(t))βi ∂Tσ(i) |t. The claim follows if we can prove that, for all σ such that Aσ 6= 0, ord(Aσ) > 3n(n+1) 2 . But this is true since: ord(Aσ) = n ∑ i=1 ord (∂σ(i)(ϕ− ϕ(t))βi ∂T σ(i) |t ) ≥ k ∑ i=1 (4n+ 2− σ(i)) + n ∑ j=k+1 (4mj − σ(j)) = 2k + 4(nk + n ∑ j=k+1 mj)− n(n+ 1) 2 ≥ 2k + 4 n ∑ j=1 j − n(n+ 1) 2 = 2k + 3n(n+ 1) 2 > 3n(n+ 1) 2 . Using all previous claims we see that the Plücker coordinates of Tn ϕ(t)Cn, for t 6= 0, look like: (· · · : ct 3n(n+1) 2 : c1t 3n(n+1) 2 +2 + c2t 3n(n+1) 2 +3 : · · · ), with c, c1, c2 non-zero constants. Since the coordinate defined by J1 has the minimum order, Nashn(Cn) ⊂ UJ1 , i.e., the affine chart obtained from dividing over ct 3n(n+1) 2 . In particular, the parametrization of Nashn(Cn) has the term c1 c t2 + c2 c t3. Now proceed as in example 2.2.1 to show that 5 ∈ Nashn(Γn). Chapter 3 Factorization of the normalization of the Nash blow-up of order n of An by the minimal resolution 3.1 The main result In this section we state the main result of this chapter. First, we introduce some notation that will be constantly used throughout this chapter and recall the notation of the previous chapters. Notation 3.1.1. Let γ, β ∈ Nt and v ∈ N2. 1) We denote by πi(β) the projection to the i-th coordinate of β. 2) γ ≤ β if and only if πi(γ) ≤ πi(β) for all i ∈ {1, . . . , t}. In particular, γ < β if and only if γ ≤ β and πi(γ) < πi(β) for some i ∈ {1, . . . , t}. 3) ( β γ ) := ∏t i=1 (πi(β) πi(γ) ) . 4) |β| = ∑t i=1 πi(β). 5) Λt,n := {β ∈ Nt | 1 ≤ |β| ≤ n}. In addition, λt,n := |Λt,n| = ( n+t n ) − 1. 6) v̄ := ( ( v α ) ) α∈Λ2,n ∈ Nλ2,n . 41 42 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION 7) Let An := ( 1 1 n 0 1 n+ 1 ) . 8) Given J ⊂ Λ3,n, let mJ := ∑ β∈J Anβ ∈ N2. This entire chapter is devoted to studying some aspects of the higher Nash blowup of the An singularity (recall definition 1.3.2). Let us recall its definition and the notation we will use. Definition 3.1.2. Consider the cone σn = R≥0{(0, 1), (n+1,−n)} ⊂ (R2)∨. We denote by An the normal toric surface corresponding to σn, i.e., An = V (xz − yn+1). The following definition is the particular case of the minor of the higher order matrix describe in chapter 1 applied in the case of the surface An. Definition 3.1.3. Let J ⊂ Λ3,n be such that |J | = λ2,n. We define the matrix Lc J := ( cβ ) β∈J , where cβ = ∑ γ≤β(−1)|β−γ| ( β γ ) Anγ ∈ Nλ2,n . In addition, we denote SAn := {J ⊂ Λ3,n | |J | = λ2,n and detLc J 6= 0}. Let In = {mJ ∈ R2 | J ∈ SAn}. The set In defines an order function: ordIn : σn →R v 7→ min mJ∈In 〈v,mJ〉. This function induces the following cones: σmJ := {v ∈ σn | ordIn(v) = 〈v,mJ〉}. These cones form a fan Σ(In) := ⋃ mJ∈In σmJ . This fan is a refinement of σ. This construction is a particular case of the construction given in section 5 of [15] applied to our context. Proposition 3.1.4. With the previous notation, we have: Nashn(An) ∼= XΣ(In), where Nashn(An) is the normalization of the Nash blow-up of An of order n and XΣ(In) is the normal variety corresponding to Σ(In). 3.2. A PARTICULAR BASIS FOR THE VECTOR SPACE Cλ2,N 43 Proof. By proposition 1.3.9 we have that Nashn(An) is a monomial blowing up of the ideal In = 〈xmJ | J ∈ SAn〉. The result follows from proposition 5.1 and remark 4.6 of [15]. The goal of this chpater is to prove the following result about the shape of the fan Σ(In). Theorem 3.1.5. For each k ∈ {1, . . . , n}, there exist J, J ′ ∈ SAn such that (k, 1−k) ∈ σmJ ∩σmJ′ . In particular, the rays generated by (k, 1−k) appear in the fan Σ(In). Corollary 3.1.6. Let A′ n be the minimal resolution of An and let Nashn(An) be the normalization of the higher Nash blow-up of An of order n. Then there exists a proper birational morphism φ : Nashn(An) → A′ n such that the following diagram commutes Nashn(An) φ // %% A′ n  An. Proof. It is well-known that A′ n is obtained by subdividing σn with the rays generated by the vectors (k, 1− k), for k ∈ {1, . . . , n}. The result follows by Theorem 3.1.5. 3.2 A particular basis for the vector space Cλ2,n As stated in Theorem 3.1.5, we need to find some subsets J ⊂ Λ3,n such that the determinant of Lc J is non-zero. This will be achieved by reducing the matrix Lc J to another matrix given by vectors formed by certain bino- mial coefficients. In this section, we prove that those vectors are linearly independent. We will see that this is equivalent to finding some basis of the vector space Cλ2,n . Definition 3.2.1. Consider a sequence η = (z, d0, d1, d2, . . . , dr), where z ∈ Z2, d0 = 0, {di} r i=1 ⊂ N \ {0} and ∑r i=0 di = n. We denote by Ω the set of all such sequences. With this set let us define a subset of vectors of N2. 44 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION Definition 3.2.2. Let η = (z, d0, d1, . . . , dr) ∈ Ω. We construct a set of vectors {vj,η} n j=1 ⊂ N2 as follows. For each j ∈ {1, . . . , n}, there exists a unique t ∈ {1, . . . , r} such that ∑t−1 i=0 di < j ≤ ∑t i=0 di. This implies that j = ∑t−1 i=0 di + c, where 0 < c ≤ dt. Then we define vj,η =                          ( ∑ i odd i n−d+,r or π1(v) > n−d−,r, we obtain that av̄=0 in (3.1). Assume this claim for the moment. For each 0 ≤ s ≤ d+,r, define the set Es = {v ∈ Tη | π1(v) = s and π2(v) ≤ d−,r}. Notice that |Es| ≤ d−,r + 1 = n− d+,r + 1 ≤ n− s+ 1. 3.2. A PARTICULAR BASIS FOR THE VECTOR SPACE Cλ2,N 49 Using the claim and taking s = d+,r we obtain the conditions of lemma 3.2.7. Thus av̄ = 0 for all v ∈ Ed+,r . Now we can repeat the same argument for s = d+,r − 1. Applying this process in a decreasing way for each s ∈ {0, . . . , dr,+} we obtain that av̄ = 0 for all v ∈ ∪ d+,r s=0Es. Then for v ∈ Tη, we have three possibilities: v ∈ ∪ d+,r s=0Es, π1(v) > d+,r, or π2(v) > d−,r. In any case, we obtain that av̄ = 0 by the previous argument or the claim. This implies that Tη is linearly independent. Now we proceed to prove the claim. For each 1 ≤ l ≤ r, define d+,l = ∑ i≤l i odd di, d−,l = ∑ i≤l i even di. We prove the claim by induction on l. By definition, we have that d+,1 = d1 and d−,1 = 0. Therefore we only have to prove that if π2(v) > n − d1, then av̄ = 0. We claim that for all v ∈ Tη such that π2(v) > n − d1, we have that π1(v) ≥ π2(v). We proceed to prove this claim by contrapositive. Let v ∈ Tη be such that π2(v) > π1(v). This implies that v = (0, π2(v) − π1(v)) + π1(v)(1, 1) = vj,η + π1(v)(1, 1) for some j ≤ n, where π1(v) ≤ n− j by definition 3.2.2. By 5) of lemma 3.2.5, there exists i < j such that vi,η = (0, 1). Moreover, by definition 3.2.2, i = d1 + 1. By lemma 3.2.8, we obtain π2(v) = π2(vj,η + π1(v)(1, 1)) ≤ π2(vj,η + (n− j)(1, 1)) ≤ π2(vd1+1,η + (n− d1 − 1)(1, 1)) = n− d1, as we claim. For each s ∈ {n − d1 + 1, . . . , n}, we define the set E(s) = {v ∈ Tη|π2(v) = s}. By 4) of lemma 3.2.5 and the previous claim, we have that for each s ∈ {n − d1 + 1, . . . , n} we have |E(s)| ≤ n − s + 1. Now we are in the conditions of lemma 3.2.7. Applying the lemma for each s in a descendant way, we obtain the result. Now suppose that the claim is true for l, i.e., for all v ∈ Tη such that π2(v) > n−d+,l or π1(v) > n−d−,l for some l ≥ 1, we have that av̄ = 0 and we prove the claim for l + 1. We have two cases: l odd or l even. We prove the case l odd, the other case is analogous. Since l is odd, we obtain that d+,l = d+,l+1 and d−,l+dl+1 = d−,l+1. Then, by the induction hypothesis, we only need to check that for all v ∈ Tη such that n−d−,l+1 < π1(v) ≤ n−d−,l and π2(v) ≤ n − d+,l, we have av̄ = 0. For this, we are going to apply lemma 3.2.8 in an iterative way. By definition, v∑l i=0 di,η = (d+,l, 0). We 50 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION claim that for all v ∈ Tη such that π1(v) ≥ n − d−,l+1 + 1, we have that π2(v) > π1(v)− d+,l − 1. We proceed to prove this claim by contrapositive. Let v ∈ Tη be such that π2(v) ≤ π1(v) − d+,l − 1. This implies that v = (π1(v) − π2(v), 0) + π2(v)(1, 1) = vj,η + π2(v)(1, 1), where π2(v) ≤ n − j by definition 3.2.2. Since π1(v) − π2(v) ≥ d+,l + 1, by 5) of lemma 3.2.5, there exist i < j such that vi,η = (d+,l +1, 0). Moreover, by definition 3.2.2, i = ∑l+1 i=0 di + 1. By lemma 3.2.8, we obtain π1(v) = π1(vj,η + π2(v)(1, 1)) ≤ π1(vj,η + (n− j)(1, 1)) ≤ π1(v∑l+1 i=0 di+1,η − (n− l+1 ∑ i=0 di − 1)(1, 1)) = d+,l + 1 + n− l+1 ∑ i=0 di − 1 < n− d−,l+1 + 1 as we claim. For each s ∈ {n − d−,l+1 + 1, . . . , n − d−,l}, we define the set E(s) = {v ∈ Tη|π1(v) = s and π2(v) ≤ n − d+,l}. Notice that, by the previous claim, we have that for each s ∈ {n − d−,l+1 + 1, . . . , n − d−,l}, |E(s)| ≤ (n− d+,l)− (s− d+,l − 1) = n− s+1. By the induction hypothesis we are in the conditions of lemma 3.2.7 for s = n−d−,l. Applying the lemma for each s in a descendant way, we obtain the result. In the case z = 0 the claim becomes: for each v ∈ Tη such that π2(v) > n−d−,r or π1(v) > n−d+,r then av̄ = 0. The proof of this case is analogous. 3.2.2 Moving Tj,η along a diagonal preserves linear indepen- dence Proposition 3.2.9 shows that Tη is a basis of C λ2,n for all η ∈ Ω. Our following goal is to show that we can move the set Tj,η along a diagonal without losing the linear independence for all j ∈ {1, . . . , n}. First we need the following combinatorial identities. Lemma 3.2.10. [22, Chapter 1] Given n,m, p ∈ N, we have the following identities: 1) ( n m )( m p ) = ( n p )( n−p m−p ) . 2) ∑ j(−1)j ( n−j m )( p j ) = ( n−p m−p ) = ( n−p n−m ) . 3.2. A PARTICULAR BASIS FOR THE VECTOR SPACE Cλ2,N 51 3) ∑ j ( n m−j )( p j ) = ( n+p m ) . 4) ∑ j ( n−p m−j )( p j ) = ( n m ) . Lemma 3.2.11. For allm ∈ N, we have that (m,m) ∈ spanC{(1, 1), . . . , (n, n)}. Proof. Recalling notation 3.1.1, consider the vector vj = j ∑ i=1 (−1)j−i ( j i ) (i, i), for each j ∈ {1, . . . , n}. Notice that for all j ∈ {1, . . . , n}, we have vj ∈ spanC{(1, 1), . . . , (n, n)}. We claim that (m,m) = ∑n j=1 ( m j ) vj . We have to prove the identity: ( m p− q )( m q ) = n ∑ j=1 j ∑ i=1 (−1)j−i ( m j )( j i )( i q )( i p− q ) , for all 1 ≤ p ≤ n and 0 ≤ q ≤ p. By 1) of lemma 3.2.10, we obtain the identities: n ∑ j=1 j ∑ i=1 (−1)j−i ( m j )( j i )( i q )( i p− q ) = n ∑ j=1 j ∑ i=1 (−1)j−i ( m j )( j q )( j − q i− q )( i p− q ) = n ∑ j=1 j ∑ i=1 (−1)j−i ( m q )( m− q j − q )( j − q i− q )( i p− q ) = ( m q ) n ∑ j=1 j ∑ i=1 (−1)j−i ( m− q j − q )( j − q i− q )( i p− q ) . With this, the claim is reduced to proving that ( m p− q ) = n ∑ j=1 j ∑ i=1 (−1)j−i ( m− q j − q )( j − q i− q )( i p− q ) . Now we have the following identities, where the second identity comes from the rearrangement of the coefficients and the fourth identity by 2) of lemma 52 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION 3.2.10. n ∑ j=1 j ∑ i=1 (−1)j−i ( m− q j − q )( j − q i− q )( i p− q ) = n ∑ j=1 (−1)j ( m− q j − q ) ( j ∑ i=1 (−1)i ( j − q i− q )( i p− q ) ) = n ∑ j=1 (−1)j ( m− q j − q ) ( ∑ i (−1)j−i ( j − i p− q )( j − q i ) ) = n ∑ j=1 (−1)j ( m− q j − q ) (−1)j ( ∑ i (−1)i ( j − i p− q )( j − q i ) ) = n ∑ j=1 ( m− q j − q )( q p− j ) . (3.2) Finally, we have the following identities, where the first identity comes from replace j by j + q and the second by 3) of lemma 3.2.10, n ∑ j=1 ( m− q j − q )( q p− j ) = n ∑ j=1 ( m− q j )( q (p− q)− j ) = ( m p− q ) , proving the claim. Lemma 3.2.12. For all a, r ∈ N and l ≤ n, we have that l ∑ i=0 (−1)i ( l i ) n−l+1 ∑ j=0 (−1)n−l+1+j ( n− l + 1 j ) (a+ r + j, r + i+ j) = 0̄. Proof. The proof is by induction on r. First, consider r = 0, we need to show that for all (p− q, q) ∈ Λ2,n, l ∑ i=0 (−1)i ( l i ) n−l+1 ∑ j=0 (−1)n−l+1+j ( n− l + 1 j )( a+ j p− q )( i+ j q ) = 0. 3.2. A PARTICULAR BASIS FOR THE VECTOR SPACE Cλ2,N 53 First, notice l ∑ i=0 (−1)i ( l i ) n−l+1 ∑ j=0 (−1)n−l+1+j ( n− l + 1 j )( a+ j p− q )( i+ j q ) = l ∑ i=0 n−l+1 ∑ j=0 (−1)n−l+1+j+i ( n− l + 1 j )( a+ j p− q )( i+ j q )( l i ) = n−l+1 ∑ j=1 (−1)n−l+1+j ( n− l + 1 j )( a+ j p− q ) ( l ∑ i=0 (−1)i ( i+ j q )( l i ) ) . (3.3) Now we have the following identity, where the first identity comes from the rearrangement of the sum and the second comes from 2) of lemma 3.2.10, l ∑ i=0 (−1)i ( i+ j q )( l i ) = (−1)l l ∑ i=0 (−1)i ( l i )( l + j − i q ) = (−1)l ( j q − l ) , Replacing this identity in the sum (3.3) and using 1) of lemma 3.2.10, we obtain n−l+1 ∑ j=0 (−1)n+1+j ( n− l + 1 j )( a+ j p− q )( j q − l ) = n−l+1 ∑ j=0 (−1)n+1+j ( n− l + 1 q − l )( n− q + 1 j − q + l )( a+ j p− q ) = (−1)n+1 ( n− l + 1 q − l ) n+1+j ∑ j=0 (−1)j ( n− q + 1 (n− l + 1)− j )( a+ j p− q ) . (3.4) Replacing i by n − l + 1 − j on the sum 3.4 and using 2) of lemma 3.2.10, we obtain that n+1+j ∑ j=0 (−1)j ( n− q + 1 (n− l + 1)− j )( a+ j p− q ) = n+1+j ∑ j=0 (−1)n−l+1−j ( n− q + 1 j )( a+ (n− l + 1)− j p− q ) = ( a+ q − l p− n− 1 ) . 54 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION Since p ≤ n, the claim is true for r = 0. Now suppose that is true for r − 1, i.e., l ∑ i=0 (−1)i ( l i ) n−l+1+j ∑ j=0 ( n− l + 1 j )( a+ (r − 1) + j p− q )( (r − 1) + i+ j q ) = 0, and we have to show that l ∑ i=0 (−1)i ( l i ) n−l+1+j ∑ j=0 ( n− l + 1 j )( a+ r + j p− q )( r + i+ j q ) = 0, for all (p− q, q) ∈ Λ2,n. Using basic properties of binomial coefficients, we have ( a+ r + j p− q )( r + i+ j q ) = ( a+ (r − 1) + j p− q )( (r − 1) + i+ j q ) + ( a+ (r − 1) + j p− q )( (r − 1) + i+ j q − 1 ) + ( a+ (r − 1) + j p− q − 1 )( (r − 1) + i+ j q ) + ( a+ (r − 1) + j p− q − 1 )( (r − 1) + i+ j q − 1 ) . Then l ∑ i=0 (−1)i ( l i ) n−l+1+j ∑ j=0 ( n− l + 1 j )( a+ r + j p− q )( r + i+ j q ) = l ∑ i=0 (−1)i ( l i ) n−l+1+j ∑ j=0 ( n− l + 1 j )( a+ (r − 1) + j p− q )( (r − 1) + i+ j q ) + l ∑ i=0 (−1)i ( l i ) n−l+1+j ∑ j=0 ( n− l + 1 j )( a+ (r − 1) + j p− q )( (r − 1) + i+ j q − 1 ) + l ∑ i=0 (−1)i ( l i ) n−l+1+j ∑ j=0 ( n− l + 1 j )( a+ (r − 1) + j p− q − 1 )( (r − 1) + i+ j q ) + l ∑ i=0 (−1)i ( l i ) n−l+1+j ∑ j=0 ( n− l + 1 j )( a+ (r − 1) + j p− q − 1 )( (r − 1) + i+ j q − 1 ) = 0. Notice that each element of {(p−l, l), (p−l, l−1), (p−l−1, l), (p−l−1, l−1)} belongs to Λ2,n or has a negative entry. In any case, by induction hypothesis, each of the four sums are zero, obtaining the result. 3.2. A PARTICULAR BASIS FOR THE VECTOR SPACE Cλ2,N 55 Corollary 3.2.13. For all a, r ∈ N and l ≤ n, we have that l ∑ i=0 (−1)i ( l i ) n−l+1 ∑ j=0 (−1)n−l+1+j ( n− l + 1 j ) (r + i+ j, a+ r + j) = 0̄. Proof. We need to prove that l ∑ i=0 (−1)i ( l i ) n−l+1+j ∑ j=0 ( n− l + 1 j )( r + i+ j p− q )( a+ r + j q ) = 0, for all (p− q, q) ∈ Λ2,n. Notice that by definition of Λ2,n, if (p− q, q) ∈ Λ2,n then (q, p − q) ∈ Λ2,n. With this and the previous lemma we obtain the result. Now we are ready to show the other important result of this section. As we mentioned before, the goal is to show that we can move the sets Tj,η along a diagonal without losing the linear independence. We are going to prove this with some additional properties. Proposition 3.2.14. Let η ∈ Ω and l ∈ {1, . . . , n}. Let (r1, . . . , rl) ∈ Nl and Ti,η + ri := {v + (ri, ri) | v ∈ Ti,η}. Then, we have spanC{v̄ ∈ Cλ2,n | v ∈ T0,η ⋃ (∪l i=1Ti,η + ri)} = spanC{v̄ ∈ Cλ2,n | v ∈ ∪l i=0Ti,η}. In particular, vl,η + (r, r) ∈ spanC{v̄ ∈ Cλ2,n | v ∈ ∪l i=0Ti,η}, for all r ∈ N. Proof. Let η ∈ Ω. The proof is by induction on l. Consider l = 1. There are two cases, v1,η = (1, 0) or v1,η = (0, 1). Suppose that v1,η = (1, 0). Consider the sums f0,r = n ∑ j=0 (−1)n+j ( n j ) (1 + r + j, r + j), f1,r = n ∑ j=0 (−1)n+j ( n j ) (1 + r + j, 1 + r + j). Applying lemma 3.2.12 for a = 1 and l = 1, we obtain that f1,r − f0,r = 0̄, (3.5) for all r ∈ N. 56 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION By lemma 3.2.11, we have that {(1 + r + j, 1 + r + j)}nj=0 ⊂ spanC{v̄ ∈ Cλ2,n | v ∈ T0,η}, for all r, j ∈ N. In particular f1,r ∈ spanC{v̄ ∈ Cλ2,n | v ∈ T0,η}. Moreover, since v1,η = (1, 0), for r = 0, we have that f0,0 − (1 + n, n) ∈ spanC{v̄ ∈ Cλ2,n | v ∈ T1,η}. Then (1 + n, n) = f1,0 − f0,0 + (1 + n, n) ∈ spanC{v̄ ∈ Cλ2,n | v ∈ T0η ∪ T1,η}. Notice that the coefficient of (1, 0) is not zero. By elementary results from linear algebra, we have that spanC{v̄ ∈ Cλ2,n | v ∈ T0η ∪ T1,η} = ( spanC{v̄ ∈ Cλ2,n | v ∈ T0η ∪ T1,η} \ {(1, 0)} ) ∪ {(1 + n, n)} = spanC{v̄ ∈ Cλ2,n | v ∈ T0η ∪ ( T1,η + 1 ) }. Applying the same argument for r = 1 in (3.5), we obtain that spanC{v̄ ∈ Cλ2,n | v ∈ T0η ∪ T1,η + 1} = ( spanC{v̄ ∈ Cλ2,n | v ∈ T0η ) ∪ T1,η + 1} \ {(2, 1)} ∪ {(2 + n, 1 + n)} = spanC{v̄ ∈ Cλ2,n | v ∈ T0η ∪ ( T1,η + 2 ) }. Repeating the argument r1 times for each r and putting together all the identities, we obtain that spanC{v̄ ∈ Cλ2,n | v ∈ T0η∪T1,η} = spanC{v̄ ∈ Cλ2,n | v ∈ T0η∪ ( T1,η+r1 ) }. This finish the proof for l = 1 and v1,η = (1, 0). For v1,η = (0, 1) the proof is analogous using the corollary 3.2.13. Now suppose that the statement is true for l−1 and let (r1, . . . , rl) ∈ Nl. We claim that spanC{v̄ ∈ Cλ2,n | v ∈ ∪l−1 i=0Ti,η ∪ ( Tl,η + rl ) } = spanC{v̄ ∈ Cλ2,n | v ∈ ∪l i=0Ti,η}. Assume this claim for the moment. By induction hypothesis, we have that spanC{v̄ ∈ Cλ2,n | v ∈ T0,η ⋃ (∪l−1 i=1Ti,η + ri)} = spanC{v̄ ∈ Cλ2,n | v ∈ ∪l−1 i=0Ti,η}. 3.2. A PARTICULAR BASIS FOR THE VECTOR SPACE Cλ2,N 57 This implies that spanC{v̄ ∈ Cλ2,n | v ∈ T0,η ⋃ (∪l i=1Ti,η + ri)} = spanC{v̄ ∈ Cλ2,n | v ∈ ∪l i=0Ti,η}. Now we proceed to prove the claim. There are two cases, vl,η = (a, 0) or vl,η = (0, a), where 0 < a ≤ l by 3) of lemma 3.2.5. Suppose that vl,η = (a, 0). For each i ∈ {0, . . . , l} and r ∈ N, consider the sum fi,r = n−l+1 ∑ j=0 (−1)n−l+1+j ( n− l + 1 j ) (a+ r + j, i+ r + j). Applying lemma 3.2.12 for l and a, we have that l ∑ i=0 (−1)i ( l i ) fi,r = 0̄. (3.6) By 6) of lemma 3.2.5 we have that {(a− 1, 0), . . . , (1, 0), (0, 1), . . . , (0, l − a)} = {vi,η} l−1 i=1. Notice that if i = a, then (a+ r + j)(1, 1) ∈ spanC{v̄ ⊂ Cλ2,n | v ∈ T0,η} by lemma 3.2.11. If 1 ≤ i < a, then (a− i, 0) + (i+ r + j, i+ r + j) = (a+ r + j, i+ r + j), and if a < i ≤ l, then (0, i− a) + (a− i, a− i) + (i+ r + j, i+ r + j) = (a+ r + j, i+ r + j). By the induction hypothesis, we obtain that {(a+ r + j, i+ r + j)}n−l+1 j=0 ⊂ spanC{v̄ ∈ Cλ2,n | v ∈ ∪l−1 i=0Ti,η}, for all i ∈ {1, . . . , l}, r ∈ N. In particular fi,r ∈ spanC{v̄ ∈ Cλ2,n | v ∈ ∪l−1 i=0Ti,η} for all i ∈ {1, . . . , l}. Moreover, since vl,η = (a, 0), for r = 0, we have that f0,0 − (a+ n− l + 1, n− l + 1) ∈ spanC{v̄ ∈ Cλ2,n | v ∈ Tl,η}. Then (a+ n− l + 1, n− l + 1) = − ( l ∑ i=0 (−1)i ( l i ) fi,0 ) +(a+ n− l + 1, n− l + 1) ∈ spanC{v̄ ∈ Cλ2,n | v ∈ ∪l i=0Ti,η}. 58 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION Applying the same argument that the case l = 1, we obtain spanC{v̄ ∈ Cλ2,n | v ∈ ∪l i=0Ti,η} = spanC{v̄ ∈ Cλ2,n | v ∈ ∪l−1 i=0Ti,η ∪ Tl,η + 1} = spanC{v̄ ∈ Cλ2,n | v ∈ ∪l−1 i=0Ti,η ∪ Tl,η + 2} = ... spanC{v̄ ∈ Cλ2,n | v ∈ ∪l−1 i=0Ti,η ∪ Tl,η + rl}. Now suppose that vl,η = (0, a). In this case we have that {(0, a− 1), . . . , (0, 1), (1, 1), (1, 0), . . . , (l − a, 0)} = {vi,η} l−1 i=0. Obtaining {(i+ r + j, a+ r + j)}n−l+1 j=0 ∈ spanC{v̄ ∈ Cλ2,n | v ∈ ∪l−1 i=0Ti,η}. The proof is analogous using corollary 3.2.13. 3.3 Proof of Theorem 3.1.5 In this section we give a proof of the main theorem of this chapter. We first associate to each η ∈ Ω a unique Jη ∈ SAn with certain properties. Secondly, we construct a distinguished element Jηk for each k ∈ {1, . . . , n} and prove that there exists another element Jη ∈ SAn with the same value with respect to an order function. Finally, we prove that Jηk is minimal in SAn with respect to the previous function. Definition 3.3.1. Let η ∈ Ω, {vi,η} n i=1 and {Ti,η} n i=1 ⊂ N2 as in definition 3.2.2. Consider ri,η := n · π2(vi,η) for all i ∈ {1, . . . , n}. We define T ′ η := T0,η ∪ (∪n i=1Ti,η + ri,η), where Ti,η + ri,η := {v + (ri,η, ri,η) | v ∈ Ti,η}. Example 3.3.2. Let n = 6, r = 5 and η = (1, 0, 1, 1, 1, 1, 2). By definition 3.2.2, we have that v1,η = (1, 0), v2,η = (0, 1), v3,η = (2, 0), v4,η = (0, 2), v5,η = (3, 0) and v6,η = (4, 0). By definition, we obtain that r1,η = 0, r2,η = 6, r3,η = 0, r4,η = 12, r5,η = 0 and r6,η = 0. Thus T ′ η = T0,η ∪ T1,η ∪ (T2η + 6) ∪ T3,η ∪ (T4,η + 12) ∪ T5,η ∪ T6,η, where T2,η + 6 = {(6, 7), (7, 8), (8, 9), (9, 10), (10, 11)}, T4,η + 12 = {(12, 14), (13, 15), (14, 16)}. 3.3. PROOF OF THEOREM 3.1.5 59 Figure 3.2: Example of T ′ η, with η = (1, 0, 1, 1, 1, 1, 2) Remark 3.3.3. Recall notation 3.1.1. Let β, β′ ∈ Λ3,n be such that β 6= β′. Then Anβ 6= Anβ ′. Proposition 3.3.4. For each η ∈ Ω, there exists a unique Jη ⊂ Λ3,n such that An · Jη := {An · β ∈ N2 | β ∈ Jη} = T ′ η. Moreover, Jη ∈ SAn. Proof. We need to show that for each v ∈ T ′ η, there exists a unique element β ∈ Λ3,n such that Anβ = v. The uniqueness comes from remark 3.3.3. Now, let v ∈ T ′ η. Then v ∈ Tη,0 or v ∈ ∪n i=1Tη,i + rη,i. For the first case we have that v0 = (t, t) with t ≤ n. In this case we take β = (0, t, 0). For the second case we have v = vi,η + (s, s) + ri,η(1, 1) = vη,i + (s, s) + nπ2(vi,η)(1, 1), where s ≤ n − i. By definition 3.2.2, vi,η = (q, 0) or vi,η = (0, q), where q ≤ i. Then v = (q + s, s) or v = (nq + s, (n+ 1)q + s). For these we take β = (q, s, 0) and β = (0, s, q) respectively. Using the previous inequalities, we obtain that β ∈ Λ3,n. Now we have to see that Jη ∈ SAn . Since λ2,n = |Tη| = |T ′ η| = |Jη|, we only have to see that detLc Jη 6= 0. Let {β1, β2, . . . , βλ2,n} = Jη be such that β1 ≺ β2 ≺ · · · ≺ βλ2,n , where ≺ denotes the lexicographic order. Notice that 60 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION if β′ < β (see notation 3.1.1), then β′ ≺ β. By definition of Lc Jη , we need to check that det ( ∑ γ≤βi (−1)|βi−γ| ( βi γ ) Anγ ) 1≤i≤λ2,n 6= 0. For this, first we turn the previous matrix into ( Aβi ) 1≤i≤λ2,n using el- ementary row operations. This implies the result since AnJη = T ′ η and {v ∈ Zλ2,n |v ∈ T ′ η} is linearly independent by proposition 3.2.14 and propo- sition 3.2.9. Fix the λ2,n-row. Consider βλ2,n ≻ γ1,λ2,n ≻ · · · ≻ γrλ2,n ,λ2,n , where {γi,λ2,n} λ2,n i=1 = {γ ∈ Λ3,n|γ < βλ2,n}. We can write this row as the sum Anβλ2,n + (−1) |βλ2,n −γ1,λ2,n | ( βλ2,n γ1,λ2,n ) Anγ1,λ2,n+ · · ·+ (−1) |βλ2,n −γrλ2,n ,λ2,n | ( βλ2,n γrλ2,n ,λ2,n ) Anγrλ2,n ,λ2,n , Since Anβλ2,n ∈ T ′ η we have that βλ2,n have the shape (q, s, 0) or (0, s, q) with s+ q ≤ n and Anβλ2,n equals one of (q+ s, s) or (nq+ s, (n+ 1)q+ s). Since γ1,λ2,n < βλ2,n , we obtain that γ1,λ2,n have the shape (q′, s′, 0) or (0, s′, q′) with s′ < s or q′ < q. Thus, Anγ1,λ2,n have the shape (q′ + s′, s′) or (nq′ + s′, (n+ 1)q′ + s′). In any case, we have that Anγ1,λ2,n ∈ T ′ η. By the first part of the proposition, we have that γ1,λ2,n = βi, for some i < λ2,n. Then we subtract (−1) |βλ2,n −γ1,λ2,n | ( βλ2,n γ1,λ2,n ) -times the row i to the row λ2,n in the matrix Lc Jη . Notice that if γ < βi, we have γ < βλ2,n . Thus we obtain that Anβλ2,n + c2Anγ2,λ2,n + · · ·+ crλ2,nAnγrλ2,n ,λ2,n is the new λ2,n-row, for some constants {c2, . . . , crλ2,n} ⊂ Z. Applying the same argument for each γi,λ2,n in a increasing way, we turn the λ2,n-th row into Anβλ2,n . Applying this process to the other rows of Lc J in an ascending way we obtain the matrix ( Aβi ) 1≤i≤λ2,n . 3.3. PROOF OF THEOREM 3.1.5 61 3.3.1 A distinguished element of SAn Let k ∈ {1, . . . , n}. Consider the function fk :N2 → Z v 7→ 〈(k, 1− k), v〉. Definition 3.3.5. Let n ∈ N \ {0}, 1 ≤ k ≤ n and dk,0 = 0. If fk((1, 0)) ≤ fk((n, n+ 1)), we take zk = 1. If fk((n, n+ 1)) < fk((1, 0)), we take zk = 0. Now, we define dk,l for l > 0 in an iterative way. Let dk,l = min{n− l−1 ∑ j=0 dk,j , tl − sl}, where tl =                                      max{m ∈ N | m · fk((1, 0)) ≤ fk(( ∑l−1 j even dk,j + 1)(n, n+ 1))} if zk = 1 and l odd. max{m ∈ N | m · fk((n, n+ 1)) ≤ fk(( ∑l−1 j odd dk,j + 1)(1, 0))} if zk = 1 and l even, max{m ∈ N | m · fk((n, n+ 1)) ≤ fk(( ∑l−1 j even dk,j + 1)(1, 0))} if zk = 0 and l odd, max{m ∈ N | m · fk((1, 0)) ≤ fk(( ∑l−1 j odd dk,j + 1)(n, n+ 1))} if zk = 0 and l even, and sl =          0 if l = 1, ∑l−1 j odd dk,j if l odd and l > 1, ∑l−1 j even dk,j if l even. If ∑l j=1 dk,j < n, we define dk,l+1. Otherwise, we finish the process and we define ηk = (zk, dk,0, . . . , dk,r). Example 3.3.6. Let n = 6 and k = 3. We have that d3,0 = 0. On the other hand, we have f3((1, 0)) = 3 < 4 = f3((6, 7)). Then z3 = 1. For l = 1, we have that t1 = max{m ∈ N | m · 3 = m · f3((1, 0)) ≤ f3((6, 7)) = 4} = 1, 62 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION and s1 = 0. Then d3,1 = min{6, 1− 0} = 1. Now we computed d3,2. By definition t2 = max{m ∈ N | m · 4 = m · f3((6, 7)) ≤ f5(2(1, 0)) = 6} = 1, and s2 = 0. This implies that d3,2 = min{6− 2 = 4, 1− 0} = 1. In an analogous way we obtain that d3,3 = 1 and d3,4 = 1. Now we computed d3,5. We have that t5 = max{m ∈ N | m · 3 = m · f3((1, 0)) ≤ f5(3(6, 7)) = 12} = 4, and s3 = d3,1 + d3,3 = 2. Then d3,5 = min{6− 1− 1− 1− 1 = 2, 4− 2 = 2} = 2. Since n − ∑3 j=0 dk,j = 6 − 2 − 1 − 3 = 0 we finish the process. Thus η5 = (1, 0, 1, 1, 1, 1, 2). Lemma 3.3.7. Let 1 ≤ k ≤ n and ηk be as in definition 3.3.5. Then we have the following properties: 1) dk,l > 0 for all l ∈ {1, . . . , r}. In particular, ηk ∈ Ω. 2) For each i ∈ {1, . . . , n}, let l ∈ {1, . . . , r} be the unique element such that ∑l−1 j=0 dk,j < i ≤ ∑l j=1 dk,j. Then, we have the following inequali- ties: fk(vi,ηk ) + ri,ηk ≤ fk(( ∑l j even dk,j + 1)(n, n+ 1)) if zk = 1 and l odd, fk(vi,ηk ) + ri,ηk ≤ fk(( ∑l j odd dk,j + 1)(1, 0)) if zk = 1 and l even, fk(vi,ηk ) + ri,ηk ≤ fk(( ∑l j even dk,j + 1)(1, 0)) if zk = 0 and l odd, fk(vi,ηk ) + ri,ηk ≤ fk(( ∑l j odd dk,j + 1)(n, n+ 1)) if zk = 0 and l even. 3) Let i′, i ∈ N \ {0} be such that ∑l−1 j=0 dk,j < i < i′ ≤ ∑l j=0 dk,j , for some l ∈ {1, . . . , r}. Then fk(vi,ηk) + ri,ηk ≤ fk(vi′,ηk) + ri′,ηk . 3.3. PROOF OF THEOREM 3.1.5 63 4) For all 1 ≤ i < i′ ≤ n, we have that fk(vi,ηk + ri,ηk(1, 1)) ≤ fk(vi′,ηk + ri′,ηk(1, 1)). 5) If l > 2 and fk(vl,ηk + rl,ηk(1, 1)) = fk(vl−1,ηk + rl−1,ηk(1, 1)), then fk(vl,ηk + rl,ηk(1, 1)) ≥ fk(vl−2,ηk + rl−2,ηk(1, 1)) + 2 Proof. 1) By construction n − ∑l−1 j=0 dk,j > 0. Then, by definition 3.3.5, we only have to check that tl−sl > 0. Notice that by definition, t1 > 0 and s1 = 0. This implies that is true for l = 1. Now suppose that l > 1. We have four cases: zk = 1 and l odd; zk = 1 and l even; zk = 0 and l odd; zk = 1 and l even. Consider zk = 1 and l odd. By definition of tl−1 fk((tl−1 + 1)(n, n+ 1)) > fk(( l−2 ∑ j odd dk,j + 1)(1, 0)). Since l is odd, ∑l−2 j odd dk,j = ∑l−1 j odd dk,j . It follows that fk(( l−2 ∑ j odd dk,j + 1)(1, 0)) = fk(( l−1 ∑ j odd dk,j + 1)(1, 0)) = fk((sl + 1)(1, 0)). On the other hand, notice that if dk,l−1 = n − ∑l−2 j=0 dk.j , then n = ∑l−1 j=0 dk,j and so there is no dk,l, which is a contradiction. This implies that dk,l−1 = tl−1 − sl−1. Thus fk((tl−1 + 1)(n, n+ 1)) = fk((dk,l−1 + sl−1 + 1)(n, n+ 1)). Since l−1 is even and sl−1 = ∑l−2 j even dk,j , we have that dk,l−1+sl−1 = ∑l−1 j even dk,j . Then fk(( l−1 ∑ j even dk,j + 1)(n, n+ 1)) > fk((sl + 1)(1, 0)). By definition of tl, we obtain that tl ≥ sl + 1 and so tl − sl > 0. The other three cases are analogous. 64 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION 2) Let i ∈ {1, . . . , n} and l ∈ {1, . . . , r}. We have four cases: zk = 1 and l odd; zk = 1 and l even; zk = 0 and l odd; zk = 1 and l even. Suppose that zk = 1 and l odd. In this case, by definition, vi,ηk = ( ∑l−1 j odd dk,j + c)(1, 0) with c ≤ dk,l, ri,ηk = 0 and sl = ∑l−1 j odd dk,j . Then l−1 ∑ j odd dk,j + c ≤ l−1 ∑ j odd dk,j + dk,l = sl + dk,l ≤ tl. By definition of tl, we have that fk(vi,ηk)+ri,ηk = fk(( l−1 ∑ j odd dk,j+c)(1, 0)) ≤ fk(( l−1 ∑ j even dk,j+1)(n, n+1)). Since l is odd, we have that ∑l−1 j even dk,j = ∑l j even dk,j . This implies the inequality that we need. Now suppose that zk = 1 and l even. In this case, by definition, vi,ηk = ( ∑l−1 j even dk,j + c)(0, 1) with c ≤ dk,l, ri,η = n( ∑l−1 j even dk,j + c) and sl = ∑l−1 j even dk,j . Using the above and the linearity of fk we obtain fk(vi,ηk) + ri,ηk = fk(vi,ηk) + fk(ri,ηk(1, 1)) = fk(vi,ηk + ri,ηk(1, 1)) = fk(( l−1 ∑ j even dk,j + c)(n, n+ 1)). Since l−1 ∑ j even dk,j + c ≤ l−1 ∑ j even dk,j + dk,l ≤ tl, by definition of tl, we obtain the inequality fk(vi,ηk)+ri,ηk = fk(( l−1 ∑ j even dk,j+c)(n, n+1)) ≤ fk(( l−1 ∑ j odd dk,j+1)(1, 0)). Since l is even, we have that ∑l−1 j odd dk,j = ∑l j odd dk,j , obtaining the result. The other two cases are analogous. 3.3. PROOF OF THEOREM 3.1.5 65 3) The hypothesis implies that i = ∑l−1 j=0 dk,j+ci and i ′ = ∑l−1 j=0 dk,j+ci′ , where 0 < ci < ci′ ≤ dk,l. We have four cases: zk = 1 and l odd; zk = 1 and l even; zk = 0 and l odd; zk = 1 and l even. Consider zk = 1 and l even. By definition 3.2.2, vi,ηk = (0, ∑l−1 j even dk,j + ci) and vi′,ηk = (0, ∑l−1 j even dk,j + ci′). Then fk(vi,ηk) + ri,ηk = (1− k)( l−1 ∑ j even dk,j + ci) + n( l−1 ∑ j even dk,j + ci) = (n− k + 1)( l−1 ∑ j even dk,j) + (n− k + 1)ci ≤ (n− k + 1)( l−1 ∑ j even dk,j) + (n− k + 1)ci′ = (1− k)( l−1 ∑ j even dk,j + ci′) + n( l−1 ∑ j even dk,j + ci′) = fk(vi′,ηk) + ri′,ηk . The other three cases are analogous. 4) Let 1 ≤ i < i′ ≤ n. Let 1 ≤ l ≤ l′ ≤ r be such that i = ∑l−1 j=0 dk,j + ci and i′ = ∑l′−1 j=0 dk,j + ci′ . By hypothesis, we have that l ≤ l′. If l = l′, the result follows from 3). Suppose that l < l′, this implies that l′ = l + c with c > 0. We have four cases (zk = 1 and l odd; zk = 1 and l even; zk = 0 and l odd; zk = 1 and l even). Consider zk = 0 and l even. By definition 3.2.2, we have that vi,ηk = ( ∑l−1 j even dk,j + ci, 0). By 2), we have that fk(vi,ηk) + ri,ηk ≤ fk(( l ∑ j odd dk,j + 1)(n, n+ 1)). On the other hand, by definition 3.2.2, we obtain that v∑l j=0 dk,j+1,ηk = (0, ∑l j odd dk,j + 1). Then 66 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION fk(v∑l j=0 dk,j+1,ηk ) + r∑l j=0 dk,j+1,ηk = fk((0, l ∑ j odd dk,j + 1) + n · ( l ∑ j odd dk,j + 1)(1, 1)) = fk(( l ∑ j odd dk,j + 1)(n, n+ 1)) ≥ fk(vi,ηk) + ri,ηk . Now, if c > 1, by 2) and knowing that l + 1 is odd, we obtain that fk(v∑l j=0 dk,j+1,ηk ) + r∑l j=0 dk,j+1,ηk ≤ fk(( l+1 ∑ j even dk,j + 1)(1, 0)). In addition, using definition 3.2.2, we have the vector v∑l+1 j=0 dk,j+1,ηk = ( ∑l+1 j even dk,j + 1, 0). Then fk(v∑l j=0 dk,j+1,ηk ) + r∑l j=0 dk,j+1,ηk ≤ fk(( l+1 ∑ j even dk,j + 1)(1, 0)) = fk(v∑l+1 j=0 dk,j+1,ηk ) = fk(v∑l+1 j=0 dk,j+1,ηk ) + r∑l+1 j=0 dk,j+1,ηk . Repeating this argument c times, we obtain fk(vi,ηk) + ri,ηk ≤ fk(v∑l j=0 dk,j+1,ηk ) + r∑l j=0 dk,j+1,ηk ≤ fk(v∑l+1 j=0 dk,j+1,ηk ) + r∑l+1 j=0 dk,j+1,ηk ... ≤ fk(v∑l′−1 j=0 dk,j+1ηk ) + r∑l′−1 j=0 dk,j+1ηk ≤ fk(vi′,ηk) + ri′,ηk , where the last inequality comes from 2). The other cases are analogous. 3.3. PROOF OF THEOREM 3.1.5 67 5) Notice that if k = n, fn(t(n, n + 1)) = t for all t ∈ {1, . . . , n} and fn((1, 0)) = n. Then by definition 3.3.5, ηn = (0, 0, n), i.e., vj,ηn = (0, j) for all j ∈ {1, . . . , n}. In particular, fn(vi,ηn + ri,ηn) < fn(vj,ηn + rj,ηn) if 1 ≤ i < j ≤ n,. Then we cannot have the conditions of lemma. Analogous, if k = 1, η1 = (1, 0, n), and f1(vi,η1 + ri,η1) < f1(vj,η1 + rj,η1), for all 1 ≤ i < j ≤ n. This implies that if there exists l ∈ {1, . . . , n} such that fk(vl,ηk +rl,ηk) = fk(vl−1,ηk +rl−1,ηk), we have that k ∈ {2, . . . , n− 1}. Now, suppose that there exist l ∈ {1, . . . , n} such that fk(vl,ηk+rl,ηk) = fk(vl−1,ηk + rl−1,ηk). If vl,ηk = (0, s) and vl−1,ηk = (0, s− 1), then fk(vl,ηk + rl,ηk ) = s(n− k + 1) > (s− 1)(n− k + 1) = fk(vl−1,ηk + rl−1,ηk ). In an analogous way, obtain a contradiction if vl,ηk = (t, 0) and vl−1,ηk = (t − 1, 0). This implies that vl,ηk = (t, 0) and vl−1,ηk = (0, s) or vl,ηk = (0, s) and vl−1,ηk = (t, 0). Consider the first case, the other case is analogous. By definition fk(vl,ηk + rl,ηk) = fk((t, 0)) = fk((0, s) + (ns, ns)), (3.7) By 5) of lemma 3.2.5, we deduce that vl−2,ηk = (0, s− 1) or vl−2,ηk = (t− 1, 0). Suppose that vl−2,ηk = (0, s− 1). Then we have fk(vl,ηk + rl,ηk) =fk((0, s) + (ns, ns)) =fk(s(n, n+ 1)) =s(n− k + 1) =(s− 1)(n− k + 1) + n− k + 1 =fk((s− 1)(n, n+ 1)) + n− k + 1 ≥fk(vl−2,ηk + rl−2,ηk) + 2, where the first equality comes from equation (3.7) and the last in- equality comes from k ≤ n− 1. Now suppose that vl−2,ηk = (t− 1, 0). In an analogous way, we obtain that fk(vl,ηk + rl,ηk) =fk((t, 0)) =k(t− 1) + k ≥fk(vl−2,ηk + rl−2,ηk) + 2, where the first equality comes from equation (3.7) and the last in- equality comes from k ≤ 2. Obtaining the result. 68 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION The previous lemma will be constantly used in the rest of the section. Proposition 3.3.8. Let k ∈ {1, . . . , n}. Let ηk ∈ Ω be as in Definition 3.3.5. Then vn,ηk = (0, k) or vn,ηk = (n− k + 1, 0). Proof. Assume the statement is false, aiming for contradiction. By defini- tion, we have that ∑r j=0 dk,j = n. Using lemma 3.2.5 and since vn,ηk is not (n − k + 1, 0) or (0, k), we have that there exist m < n such that vm,ηk = (n − k + 1, 0) or vm,ηk = (0, k). Let l ≤ r be such that m = ∑l−1 j=0 dk,j + c and 0 < c ≤ dk,l. We have four cases: zk = 1 and l odd; zk = 1 and l even; zk = 0 and l odd; zk = 1 and l even. Consider z = 1 and l odd. In this case, by definition 3.2.2, vm,ηk = ( ∑l−1 j odd dk,j + c, 0) = (n − k + 1, 0). Hence ∑l−1 j odd dk,j + c = n− k + 1. Then n− k + 1 = m− l−1 ∑ j even dk,j < n− l−1 ∑ j even dk,j . This implies that ∑l−1 j even dk,j + 1 < k. Thus fk(( l−1 ∑ j even dk,j + 1)(n, n+ 1)) < fk(k(n, n+ 1)) = k(n− k + 1) = fk((n− k + 1, 0)) = fk(vm,ηk) + rm,ηk . This is a contradiction to lemma 3.3.7 2). Now, suppose that zk = 1 and l even. For this case, by definition 3.2.2, we have that vm,ηk = (0, k) and k = ∑l−1 j even dk,j + c. Then k = l−1 ∑ j even dk,j + c = m− l−1 ∑ j odd dk,j < n− l−1 ∑ j odd dk,j . 3.3. PROOF OF THEOREM 3.1.5 69 This implies that ∑l−1 j odd dk,j + 1 < n− k + 1. Thus fk(( l−1 ∑ j odd dk,j + 1)(1, 0)) < fk((n− k + 1)(1, 0)) = k(n− k + 1) = fk(k(n, n+ 1)) = fk((0, k) + k · n(1, 1)) = fk(vm,ηk) + rm,ηk . This is a contradiction to lemma 3.3.7 2). The other two cases are analogous. Recall that for J ∈ SAn , we denote mJ = ∑ β∈J Anβ. Corollary 3.3.9. Let n ∈ N \ {0} and 1 ≤ k ≤ n. Then, there exists η ∈ Ω such that η 6= ηk and fk(mJη) = fk(mJηk ). Proof. By the previous Proposition, we have that vn,ηk = (n − k + 1, 0) or vn,ηk = (0, k). By lemma 6) of 3.2.5, we obtain that {vi,ηk} n−1 i=1 = {(t, 0)}n−k t=1 ∪ {(0, s)}k−1 s=1 . Moreover, we can deduce that vn−1,ηk is (n− k, 0) or (0, k − 1). Suppose that vn−1,ηk = (n−k, 0). Since fk(k(n, n+1)) = fk((n−k+1, 0)) and by construction of ηk, we have that vn,ηk = (n− k + 1, 0). If vn−1,ηk = (0, k−1), we obtain that vn,ηk = (0, k). In any case, we obtain that dk,r ≥ 2. Then we define η = (z′, d′0, d ′ 1, . . . , d ′ r, d ′ r+1), where z ′ = zk, d ′ i = dk,i for all i < r, d′r = dk,r − 1 and d′r+1 = 1. By construction ∑n j=0 d ′ j = n and d′j > 0 for all j ∈ {1, . . . , n}. This implies that η ∈ Ω. On the other hand, we have that vj,ηk = vj,η for all j ≤ n − 1 and vn,η = (n − k + 1, 0) if vn,ηk = (0, k) or vn,η = (0, k) if vn,ηk = (n− k + 1, 0). Since fk(k(n, n+ 1)) = fk((n− k + 1, 0)), we obtain that fk(mJη) = fk(mJηk ). 3.3.2 Jηk ∈ SAn is minimal with respect to fk Lemma 3.3.10. Let β′, β ∈ N3 be such that β′ ≤ β (recall notation 3.1.1). Then fk(Anβ ′) ≤ fk(Anβ). Proof. This is a straightforward computation. 70 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION Lemma 3.3.11. Let β ∈ N3 be such that Anβ 6= v + q(1, 1) for all v ∈ T ′ ηk and q ∈ N. Then fk(Anβ) ≥ fk(v) for all v ∈ T ′ ηk . Proof. We claim that fk(v) ≤ k(n − k + 1) ≤ fk(Anβ) for all v ∈ T ′ ηk and for all β ∈ N3 satisfying the hypotheses of the Lemma. We are going to prove the first inequality of the claim. By definition 3.3.1, we have that T ′ ηk = T0,ηk ⋃ ∪n j=1Tj,ηk +rj,ηk , where T0,ηk = {(q, q)}nq=1, rj,ηk = n · π2(vj,ηk) and Tj,ηk + rj,ηk = {vj,ηk + (p + rj,ηk)(1, 1)} n−j p=0 . By proposition 3.3.8 we have that vn,ηk = (0, k) or vn,ηk = (n − k + 1, 0). Moreover, by 5) of lemma 3.2.5, we have that {vj,ηk} n−1 j=1 = {(t, 0)}n−k t=1 ∪ {(0, s)}k−1 s=1 . By definition, Tn,ηk +rn,ηk = {vn,ηk +rn,ηk(1, 1)}. Since we know the two possibilities for vn,ηk , we obtain that fk(vn,ηk + rn,ηk(1, 1)) = k(n − k + 1). On the other hand, if v ∈ T0,ηk , we have that v = (q, q) with q ≤ n. Since 1 ≤ k ≤ n, obtaining that fk(v) = q ≤ n ≤ k(n− k+1). With this, we only have to check the desired inequality for v ∈ ∪n−1 j=1 {vj,ηk +(p+rj,ηk)(1, 1)} n−j p=0 . This implies that v = vj,ηk + (p+ rj,ηk)(1, 1), for 1 ≤ j ≤ n− 1 and 0 ≤ p ≤ n− j. Suppose that vj,ηk = (t, 0) for some t ≤ j and recall that, t ≤ n − k. Then fk(v) =fk(vj,ηk + (p+ rj,ηk)(1, 1)) =fk((t, 0)) + p+ rj,ηk ≤kt+ n− j ≤kt+ n− t =(k − 1)t+ n ≤(k − 1)(n− k) + n =nk − k2 + k. Now suppose that vj,ηk = (0, s) for some s ≤ j and recall that s < k. 3.3. PROOF OF THEOREM 3.1.5 71 Then fk(v) =fk(vj,ηk + (p+ rj,ηk)(1, 1)) =fk((0, s) + p(1, 1) + n · s(1, 1)) =fk(s(n, n+ 1)) + p ≤s(n− k + 1) + (n− j) ≤s(n− k + 1) + (n− s) ≤s(n− k + 1) + (n− k) + (k − s) ≤s(n− k + 1) + (k − s)(n− k) + (k − s) =nk − k2 + k. This proves the first inequality of the claim. For the second inequality notice that Anβ =π1(β)(1, 0) + π2(β)(1, 1) + π3(β)(n, n+ 1) =(π1(β), 0) + (π2(β), π2(β)) + (nπ3(β), nπ3(β) + π3(β)) =(π1(β), π3(β)) + (π2(β) + nπ3(β))(1, 1) =(π1(β)− π3(β), 0) + (π2(β) + (n+ 1)π3(β))(1, 1). Similarly, we obtain the expression Anβ = (0, π3(β)− π1(β)) + (π1(β) + π2(β) + nπ3(β))(1, 1)). Working with the first expression of Anβ and applying fk to this vector, we obtain that fk(Anβ) =fk((π1(β)− π3(β), 0) + (π2(β) + (n+ 1)π3(β))(1, 1)) =fk((π1(β)− π3(β), 0)) + π2(β) + (n+ 1)π3(β) =k(π1(β)− π3(β)) + π2(β) + (n+ 1)π3(β). By the hypothesis over β and recalling that {vηk,j} n−1 j=1 = {(t, 0)}n−k t=1 ∪ {(0, s)}k−1 s=1 , we obtain that π1(β) − π3(β) ≥ n − k + 1. Using the second expression of Anβ, we obtain π3(β) − π1(β) ≥ k. Suppose that π1(β) − π3(β) ≥ n− k + 1. Then fk(Anβ) =k(π1(β)− π3(β)) + π2(β) + (n+ 1)π3(β) ≥k(n− k + 1) + π2(β) + (n+ 1)π3(β) ≥k(n− k + 1). 72 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION Now suppose that π3(β)− π1(β) ≥ k. In particular, π3(β) ≥ k. Then fk(Anβ) =k(π1(β)− π3(β)) + π2(β) + (n+ 1)π3(β) =(n+ 1)π3(β)− kπ3(β) + kπ1(β) + π2(β) =(n− k + 1)π3(β) + kπ1(β) + π2(β) ≥(n− k + 1)k + kπ1(β) + π2(β) ≥nk − k2 + k. In any case, we obtain that fk(Anβ) ≥ k(n− k+1) for all β ∈ N3 satisfying the hypotheses of the Lemma as we claim. Lemma 3.3.12. Let v = vl,ηk + q(1, 1) ∈ N2 be with l ≤ n and q ≥ n− l + 1 + rl,ηk . Then fk(v) ≥ fk(u) for all u ∈ T0,ηk ⋃ ∪l j=1Tj,ηk + rj,ηk . Proof. We proceed by induction on l. Consider l = 1. Then v = v1,ηk + q(1, 1), with q ≥ n + r1,ηk and we need to prove that fk(v) ≥ fk(u) for all u ∈ T0,ηk ∪T1,ηk + r1,ηk . If u ∈ T1,ηk + r1,ηk , then u = v1,ηk +(p+ r1,ηk)(1, 1), with p ≤ n− 1. It follows that fk(v) =fk(v1,ηk + q(1, 1)) =fk(v1,ηk) + q ≥fk(v1,ηk) + n+ r1,ηk ≥fk(v1,ηk) + p+ r1,ηk =fk(v1,ηk + p+ r1,ηk(1, 1)) =fk(u). If u ∈ T0,ηk , then u = p(1, 1), with p ≤ n. By definition 3.2.2, v1,ηk = (1, 0) or v1,ηk = (0, 1). Thus fk(v) = k + q ≥ k + n or fk(v) = (1 − k) + q ≥ n+ (n− k + 1). Since k ∈ {1, . . . , n}, in any case we have that fk(v) > n ≥ p = fk(u). (3.8) We conclude that the Lemma is true for l = 1. Next, assume that the Lemma is true for all l′ < l, i.e., fk(vl′,ηk + q′(1, 1)) ≥ fk(u) for all u ∈ T0,ηk ⋃ ∪l′ j=1Tj,ηk+rj,ηk and q′ ≥ n−l′+1+rl′,ηk . Let v = vl,ηk + q(1, 1) with q ≥ n− l+ 1+ rl,ηk . If u ∈ Tl,ηk + rl,ηk , we have 3.3. PROOF OF THEOREM 3.1.5 73 that u = vl,ηk + (p+ rl,ηk)(1, 1), with p ≤ n− l. Then fk(v) =fk(vl,ηk) + q ≥fk(vl,ηk) + n− l + 1 + rl,ηk ≥fk(vl,ηk) + p+ rl,ηk =fk(vl,ηk + (p+ rl,ηk)(1, 1)) =fk(u). Thus fk(v) ≥ fk(u) for all u ∈ Tl,ηk + rl,ηk . Consider u ∈ Tl−1,ηk + rηk,l−1. By definition, u = vl−1,ηk + (p+ rl−1,ηk)(1, 1) with p ≤ n− l+ 1. By lemma 3.3.7 4), fk(vl,ηk + rl,ηk) ≥ fk(vl−1,ηk + rl−1,ηk). Then fk(v) =fk(vl,ηk) + q ≥fk(vl,ηk) + n− l + 1 + rl,ηk =fk(vl,ηk + rl,ηk(1, 1)) + n− l + 1 ≥fk(vl−1,ηk + rl−1,ηk(1, 1)) + p =fk(vl−1,ηk + (p+ rl−1,ηk)(1, 1)) =fk(u). Obtaining that the statement is true for all u ∈ Tl−1,ηk + rl−1,ηk . Suppose fk(vl,ηk + rl,ηk) ≥ fk(vl−1,ηk + rl−1,ηk) + 1. Obtaining that fk(v) ≥fk(vl,ηk + rl,ηk(1, 1)) + n− l + 1 ≥fk(vl−1,ηk + rl−1,ηk(1, 1)) + n− l + 2. Then, by the induction hypothesis for l − 1, fk(v) ≥ fk(u) for all u ∈ T0,ηk ⋃ ∪l−1 j=1Tj,ηk + rj,ηk , obtaining the result. Now suppose that fk(vl,ηk + rl,ηk) = fk(vl−1,ηk + rl−1,ηk). For this, we have two cases; l = 2 or l > 2. If l = 2, by (3.8), we have that fk(v) =fk(v2,ηk) + q ≥fk(v2,ηk) + n− 1 + r2,ηk =fk(v2,ηk + r2,ηk(1, 1)) + n− 1 =fk(v1,ηk + r1,ηk(1, 1)) + n− 1 =fk(v1,ηk + (n− 1 + r1,ηk)(1, 1)) ≥n ≥fk(u), 74 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION for all u ∈ T0,ηk . If l > 2, then for all u ∈ T0,ηk ⋃ ∪l−2 j=1Tj,ηk + rj,ηk , we have that fk(v) ≥fk(vl,ηk + rl,ηk(1, 1)) + n− l + 1 ≥fk(vl−2,ηk + rl−2,ηk(1, 1)) + n− l + 3 ≥fk(u), where the second inequality comes from 5) of lemma 3.3.7 and the last inequality comes from the induction hypothesis over l − 2. Obtaining the result. Now we are ready to prove the other important result of this section. Proposition 3.3.13. Let ηk ∈ Ω and let Jηk be the element of SAn as- sociated to ηk by Proposition 3.3.4. Then for all J ∈ SAn we have that fk(mJηk ) ≤ fk(mJ). Proof. Let J = {β1, . . . , βλ2,n} ∈ SAn . By definition of SAn we have that 0 6= det ( cβi ) 1≤i≤λ2,n , where cβi := ∑ γ≤βi (−1)|βi−γ| ( βi γ ) Anγ (recall notation 3.1.1). Fixing the β1th row of this matrix and using basic properties of deter- minants, we obtain that 0 6= det ( cβi ) 1≤i≤λ2,n = ∑ γ≤β1 (−1)|βi−γ| ( βi γ ) det     Anγ cβ2 . . . cβλ2,n     . Since the determinant is not zero, there exists β′1 ≤ β1 such that det     Anβ′1 cβ2 . . . cβλ2,n     6= 0. Applying this process for each row, we obtain the set of vectors B = {β′i} λ2,n i=1 ⊂ Λ3,n such that β′i ≤ βi for all i ∈ {1, . . . , λ2,n} and with the property det ( Anβ′i ) 1≤i≤λ2,n 6= 0. 3.3. PROOF OF THEOREM 3.1.5 75 The goal is to construct a bijective correspondence ϕ : B → T ′ ηk , such that fk(Anβ ′ i) ≥ fk(ϕ(β ′ i)). Consider the set vj,ηk +L := {vjηk +p(1, 1) | p ∈ N}. Now, consider the following partition of B: B0 = {β′i ∈ B | Anβ ′ i ∈ T ′ ηk }, B1 = {β′i ∈ B | Anβ ′ i ∈ (vj,ηk + L) \ Tj,ηk + rj,ηk for some j ∈ {1, . . . , n}}, B2 = {β′i ∈ B | Anβ ′ i = q(1, 1) for some q > n}, B3 = {β′i ∈ B | Anβ ′ i /∈ (vj,ηk + L) for all j ∈ {0, . . . , n}}, For all β′i ∈ B0, we define ϕ(β′i) = Anβ ′ i. Since det ( Anβ′i ) 1≤i≤λ2,n 6= 0, we have that ϕ(β′i) 6= ϕ(β′j) for all β ′ i, β ′ j ∈ B0. Now, if B1 6= ∅, we rearrange B in such a way that {β′1, β ′ 2, . . . , β ′ m} = B1. Consider β′1 ∈ B1. By construction of B1, there exist l ≤ n and q ∈ N such that Anβ ′ 1 = vl,ηk + q(1, 1). By proposition 3.2.14 we have that Anβ′1 ∈ spanC{v ∈ Cλ2,n | v ∈ l ⋃ j=0 Tj,ηk} = spanC{v ∈ Cλ2,n | v ∈ T0,ηk ⋃ ∪l j=1Tj,ηk + rj,ηk}. This implies that Anβ′1 = ∑ v∈T0,ηk ⋃ ∪l j=1Tj,ηk +rj,ηk avv, for some constants av ∈ C. Using again basic properties of the determinant, we obtain that there exists uβ′ 1 ∈ T0,ηk ⋃ ∪l j=1Tj,ηk + rj,ηk such that det       uβ′ 1 Anβ′2 ... Anβ′λ2,n       6= 0. Applying this process for each element of B1, we obtain the vectors {uβ′ j }mj=1. We define ϕ(β′j) = uβ′ j for all j ∈ {1, . . . ,m}. Now, we need to check that ϕ is injective on B0 ∪B1 and fk(Anβ ′ i) ≥ fk(ϕ(β ′ i)). 76 CHAPTER 3. FACTORIZATION BY MINIMAL RESOLUTION Notice that, by construction det            uβ′ 1 ... uβ′ m Anβ′m+1 ... Anβ′λ2,n            6= 0. This implies that uβ′ i 6= uβ′ j for all 1 ≤ i < j ≤ m. In particular we have that uβ′ i 6= uβ′ j . Moreover, using the same argument, we have that uβ′ i 6= Anβ ′ j = ϕ(β′j) for all β ′ j ∈ B0. Thus, ϕ is injective on B0 ∪B1. On the other hand, Anβ ′ i = vl,ηk + q(1, 1) /∈ Tl,ηk + rl,ηk for some l ≤ n. This implies that q ≥ n− l + 1 + rl,ηk . Then fk(Anβ ′ i) =fk(vl,ηk + q(1, 1)) =fk(vl,ηk) + q ≥fk(vl,ηk) + n− l + 1 + rl,ηk =fk(vl,ηk + (n− l + 1 + rl,ηk)(1, 1)) ≥fk(uβ′ i ). where the last inequality comes from lemma 3.3.12, obtaining the inequality we are looking for. For all β′i ∈ B2, we have that Anβ ′ i = q(1, 1), for some q > n. By lemma 3.2.11, we have that Anβ′i = ∑ v∈T0,ηk avv. Applying the same method that for the elements of B1, we define ϕ with the properties that we need. Now, since | T ′ ηk |=| B |= λ2,n, we have that | B3 |=| Tηk \ {ϕ(β ′ j) | β ′ j ∈ B0 ∪ B1 ∪ B2} |. Then we take ϕ(β′i) = v, with v ∈ Tηk \ {ϕ(β′j) | β′j ∈ B0 ∪ B1 ∪ B2} in such a way that ϕ(β′i) 6= ϕ(β′j) for all β′i, β ′ j ∈ B3 and β′i 6= β′j . By construction we obtain that ϕ is a bijective correspondence and by definition of B3 and lemma 3.3.11 we have that fk(Anβ ′ i) ≥ fk(ϕ(β ′ i)) for all 3.3. PROOF OF THEOREM 3.1.5 77 β′i ∈ B3. Then fk(mJ) = ∑ βi∈J fk(Anβi) ≥ ∑ β′ i∈B fk(Anβ ′ i) ≥ ∑ β′ i∈B fk(ϕ(β ′ i)) = ∑ b∈T ′ ηk fk(v) =fk(mJηk ), where the first inequality comes from lemma 3.3.10 and the second comes from the construction of ϕ. Now we are ready to prove Theorem 3.1.5. Proof. By proposition 3.3.4, Jηk ∈ SAn . 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