aa r X i v : . [ m a t h . A C ] O c t EQUATIONS DEFINING CERTAIN GRAPHS
YOUNGSU KIM AND VIVEK MUKUNDAN
Dedicated to Professor Bernd Ulrich on the occasion of his 65th birthday A BSTRACT . Consider the rational map φ : P n − k [ f : ··· : f n ] −−−→ P n k defined by homogeneous polynomials f ,... , f n of the samedegree d in a polynomial ring R = k [ x ,... , x n ] over a field k . Suppose I = ( f ,... , f n ) is a height two perfect ideal satisfying µ ( I p ) ≤ dim R p for p ∈ Spec ( R ) \ V ( x ,... , x n ) . We study the equations defining the graph of φ whose coordinate ring isthe Rees algebra R [ It ] . We provide new methods to construct these equations using work of Buchsbaum and Eisenbud.Furthermore, for certain classes of ideals satisfying the conditions above, our methods lead to explicit equations definingRees algebras of the ideals in these classes. These classes of examples are interesting, in that, there are no known methods tocompute the defining ideal of the Rees algebra of such ideals. These new methods also give rise to effective criteria to checkthat φ is birational onto its image.
1. I
NTRODUCTION
Our primary goal in this paper is to understand certain rational maps from the projective n − n space. In particular, we will provide an explicit description of the equations defining the image and graph of certainrational maps. Fix a field k . Let X = P n − k and W a linear system in H ( X , O X ( d )) for some positive integer d . Ingeneral, W does not induce a morphism. However, it defines a rational map from X to Y : = P dim W − k with the baselocus defined by W . We denote such a rational map by φ : X W Y . The image of φ may not be closed in Y . To studythe algebraic properties of the image, we take the closure of the image φ ( X ) in Y . The questions we are interested inare the following: Question 1. ( a ) What are the equations defining φ ( X ) in Y . More generally, let Γ φ denotes the closure of thegraph of φ . What are the equations defining Γ φ in X × Y ? ( b ) Is the rational map φ birational onto its image?These questions have a tight connection with commutative algebra as their coordinate rings are well-studied ringsamong algebraists. In the coordinate ring R = k [ x , . . . , x n ] of X = P n − k , let I be the ideal generated by the elements in W . The ideal I is a homogeneous ideal generated in degree d . The Rees algebraR [ It ] : = R ⊕ It ⊕ I t ⊕ · · · and the special fiber ring F ( I ) : = R [ It ] ⊗ k ∼ = R / m ∼ = R [ It ] / m R [ It ] , where m is the maximal ideal ( x , . . . , x n ) are the coordinate rings of the projective varieties Γ φ in X × Y and φ ( X ) in Y , respectively. Hence, the answer to Question 1(a) is nothing but the equations in the defining ideals of these rings(see Section 3 for details). In the case where ideal I is nice , for instance, if W = O X ( d ) the d-th Veronese embeddingor I is a complete intersection ideal, the defining ideal of the Rees algebra of I is well-understood, cf. [HS06, Sec.5.5]. However, determining defining ideals of Rees algebras is a challenging task, and there is a series of work on thistopic.Our main case concerns perfect ideals of codimension (equivalently height) two. Such ideals satisfy the struc-ture theorem of Burch, and a number of papers has been devoted to understanding this class of ideals; for instance[HSV83, Mor96, MU96, T`02, HSV08, CHW08, Bus09, KPU11, CBtD13, CBtD14, Lan14, Mad15, BM16, KPU17]. Date : October 31, 2019.2010
Mathematics Subject Classification. primary 13A30; secondary 13D02,13H15,14A10,14E05.
Key words and phrases. blowup algebra, degree of a variety, morphism, multiplicity, rational map, Rees ring, special fiber ring. he novelty in our approach is to incorporate the work of Buchsbaum and Eisenbud [BE74], in particular, Buchsbaum-Eisenbud multipliers.We work on Question 1 in the following setup. Let W ⊂ H ( P n − , O P n − ( d )) be a subsystem. Assume that dim k W = n + φ : P n − k W P n k ), codim Fitt i ( I ) ≥ i + ≤ i ≤ n −
1, and Proj R / I is arithmetically Cohen-Macaulay ofcodimension 2. Here, Fitt i ( I ) denotes the i th Fitting ideal of I . The condition that Proj R / I is arithmetically Cohen-Macaulay of codimension 2 is equivalent to the condition of the base locus being defined by a height 2 perfect ideal I .The defining ideal of the Rees algebra R [ It ] is the kernel of the surjective R -algebra homomorphism B : = R ⊗ k S = R [ T , . . . , T n ] → R [ It ] , (1)where S = k [ T , . . . , T n ] . This kernel a bigraded ideal in B . We often study the defining ideal of R [ It ] in Sym R ( I ) ,the symmetric algebra of I , since the presentation in eq. (1) factors through Sym R ( I ) , and the kernel of this inducedpresentation is well-understood, see Section 3. By abuse of notation, we often call the kernel of the induced mapSym R ( I ) → R [ It ] the defining ideal of the Rees algebra R [ It ] .Under this setup, Sym R ( I ) is a complete intersection ring, and the defining ideal of the Rees algebra R [ It ] in Sym R ( I ) is equal to A : = H m ( Sym R ( I )) , the zeroth local cohomology module supported in the maximal ideal m = ( x , . . . , x n ) R .These facts allow us to take the advantage of the bigraded structure of A and the duality theorem (Theorem 2). In thesequel, for a bigraded module M = ⊕ i , j ∈ Z M i , j in B , we set M i : = ⊕ j ∈ Z M i , j . Theorem 2 (Grothendieck, [Jou96, Section 3.6], [KPU17, Theorem 2.4] ) . For each i ∈ { , . . . , d − n } , there exists anisomorphism of finitely generated graded S-modules A i ∼ = Hom S ( Sym ( I ) d − n − i , S ( − n )) . In [Jou96], Jouanolou constructed the isomorphisms in Theorem 2 explicitly. Hence to understand A i it suffices tostudy its dual Hom S ( Sym ( I ) d − n − i , S ( − n )) . Our first result concerns this Hom module. Fix i ∈ { , . . . , d − n } and let F α → F → [ Sym R ( I )] d − n − i be a graded presentation of [ Sym R ( I )] d − n − i . Then Hom S ( Sym ( I ) d − n − i , S ( − n )) corresponds to the kernel of α ∗ . InTheorem 13, we show that there exists a complex which induces elements in ker α ∗ . Theorem A (Theorem 13) . We have a complex of graded free S -modules ∧ r − F ⊗ S ( − s ) ∂ → F ∗ α ∗ −→ F ∗ , (2)where r = rank [ Sym R ( I )] d − n − i , i.e., Im ( ∂ )( − n ) ⊂ ker α ∗ ( − n ) ∼ = A i .We note that the above is a complex for any degree of A i (equivalently, [ Sym R ( I )] d − n − i ). It is natural to ask un-der what conditions the complex in eq. (2) is exact. We show that for r ≤
2, this complex is exact provided that [ Sym R ( I )] d − n − i satisfies Serre’s condition ( S r ) (Theorems 15 and 16), and we ask that whether this holds true in gen-eral (Question 18). Our theorems provide the differential map ∂ and the shift s in eq. (2) explicitly. To do this we usea structure theorem of Buchsbaum-Eisenbud [BE74], and their main theorem and lemmas are our main technical toolsin this paper.Our next theorem concerns Question 1(b). It is a well-known fact that the closed image φ ( X ) is defined by a singleequation, for instance, see [UV93, Proposition 2.4]. Hence, it suffices to study A ∼ = Hom S ( Sym ( I ) d − n , S ( − n )) = ker α ∗ ( − n ) , where α ∗ as in eq. (2). We provide equivalent conditions that the complex in Theorem A is exact. The equivalence ( ) and ( ) in Theorem B below was first established in [KPU16, Corollary 3.7]. Theorem B.
Assume k is a field of characteristic zero and n ≥
3. Then with s as in eq. (2), the generating degree of A is at most s + n . Furthermore, the following statements are equivalent: ( ) The rational map φ is birational to its image. ( ) A is generated in degree s + n . ( ) The greatest common divisor of the entries of ker α ∗ is 1. ) e ( F ( I )) = e ( R / ( g , . . . , g n − ) : I ) , where g , . . . , g n − are general k -linear combinations of the f ′ i s (see Sec-tion 2.3 for the definition of the term general). Here, e ( − ) denotes the Hilbert-Samuel multiplicity.The explicit isomorphism of Jouanolou provides an explicit form of the defining equation from ker α ∗ , and there area few other ways to treat this case. In [BCJ09], Bus´e, Chardin, and Jouanolou achieved this by analyzing [ Sym R ( I )] q for q ≫ [ Sym R ( I )] q . We note that as q becomes larger, the size ofthe presentation matrix of [ Sym R ( I )] q grows in the binomial order of d = dim R . With our approach, one only needs toanalyze the presentation matrix of [ Sym R ( I )] d − n . As far as item (1) in the above theorem is concerned, one may applya theorem of Doria, Hassanzadeh, and Simis [DHS12, Section 2.3]. Their approach is more general, but it requiresunderstanding an additional graded piece A of the defining ideal and its Jacobian dual. Lastly, we compare ours witha result by Boswell and Mukundan [BM16], where the authors use an iterative Jacobian dual. This iteration involvescomputing large size matrices, and it is computationally not as efficient as our approach. However, their statementprovides a closed formula for A in terms of colon ideals.One of the advantages of our approach is that every condition we impose is general, for instance see [BJ03, p. 316]and Theorem 26. By the semi-continuity theorem [Eis95, Thm 14.8b], one can see that the exactness of the complexin eq. (2) is a general condition. That is, there exists an open subset U , which may be empty, in a parameter spacesuch that the fiber of each point of U satisfies this condition. The difficult part is to show that such an open subset U is non-empty. In other words, one needs to exhibit an example whose corresponding point belongs to U . This turnedout to be the most challenging part of our paper. We believe that for d , . . . , d n ∈ N and a presentation matrix ϕ = x d x d · · · x d n x d x d · · · x d n ... ... . . . ... x d n x d n n · · · x d n n g g · · · g n , where g i ’s are symmetric polynomials of degree d i , the ideal generated by the n × n minors of ϕ would provide afamily of examples. As an evidence, we show that a variant of this with d = , d = , d ≥ Theorem C (Theorem 26) . Let R = k [ x , y , z ] , where k is a field of characteristic 0. Let M be a 4 × R and let I = I ( M ) . If M is general of type ( , , q ) , where q > R [ It ] is minimally generated bybidegree number of elements l ( , ) l ( , ) l ( q , ) A ( , q + ) A q − i ( q − i , i + ) (cid:0) + i (cid:1) A q ( q , ) ≤ i < q −
1. In particular, the defining ideal is minimally generated by (cid:0) q + (cid:1) + l , l , l denotethe equations defining Sym R ( I ) in B , see Section 3.The paper is organized as follows. In Section 2, we set up the notation and provide preliminaries. In Section 3,we explain our setup mentioned in the introduction, the work of Buchsbaum and Eisenbud, and the duality theorem indetail. In Section 4, we prove Theorem A and related statements. Section 5 is devoted to the equivalence in TheoremB. In Section 6, we present Theorem C and the promised example. Acknowledgment:
We would like to thank professors Bernd Ulrich and Craig Huneke for their helpful comments andsuggestions. Also, we owe a lot to the anonymous referee for his/her invaluable comments. The earlier version of thepaper had an erroneous definition of the notion of general elements which was kindly pointed out to us by the referee. . P RELIMINARIES
In this section, we will setup the notation and review some background materials. We refer the reader to [Eis95,Har77] for basic definitions and notations for algebraic geometry and commutative algebra.2.1.
Morphisms between projective spaces and Rees algebras.
Let k be a field, k [ x , . . . , x n ] , the coordinate ringfor P n − k , and f , . . . , f n homogeneous polynomials of degree d in k [ x , . . . , x n ] . Then we have a rational map betweenprojective spaces φ : P n − k [ f : ··· : f n ] −−−→ P n k (3)defined by the polynomials f , . . . , f n . This map is defined on P n − k \ V ( I ) (equivalently, the base locus is V ( I ) ). Theimage Im φ is not a closed subscheme of P n k in general. In this article, we study the closed subscheme Im φ in P n k . Let k [ y , . . . , y n ] be the coordinate ring of P n k . Then Im φ = Proj k [ f , . . . , f n ] , and the (rational) maps between projectiveschemes P n − k φ Im φ ⊂ P n k corresponds to the maps between k -algebras k [ y , . . . , y n ] ։ k [ f , . . . , f n ] ⊂ k [ x , . . . , x n ] , where the first map is defined by y i f i for i = , . . . , n . Observe that Im φ = Proj k [ f , . . . , f n ] = V ( J ) for somehomogeneous ideal J of k [ y , . . . , y n ] .Let R = k [ x , . . . , x n ] , m = ( x , . . . , x n ) R , and I = ( f , . . . , f n ) . The Rees algebra of I is the graded ring R [ It ] = R ⊕ It ⊕ I t ⊕ · · · ⊂ R [ t ] , and the special fiber ring of I is the graded ring F ( I ) = R [ It ] ⊗ R R / m = R [ It ] / m R [ It ] . The Rees algebra R [ It ] is the coordinate ring of the closure of the graph of the rational map in eq. (3). Here, the graph of a rational map φ : X Y with base locus W between projective schemes is Γ φ : = { ( x , y ) ∈ X × Y | y = φ ( x ) , x W } . In addition, if deg f i = d for i = , . . . , n , then the special fiber ring F ( I ) is an integral domain and isisomorphic to the subring k [ f , . . . , f n ] of k [ x , . . . , x n ] .Let Quot ( A ) denote the total ring of fractions of a ring A . In the case where A is an integral domain, Quot ( A ) is thefield of fractions. For fields F ⊂ K , let [ K : F ] denote the field extension degree. Remark 3 (cf. [DHS12, Prop. 2.11]) . The rational map in eq. (3) is birational to the image if and only if k [ f , . . . , f n ] and k [ x , . . . , x n ] have the same field of fractions.2.2. Free resolutions and minors of matrices.
Let R be a Noetherian ring. In this subsection, we review two the-orems of Buchsbaum-Eisenbud on finite free complexes. One provides a characterization of the acyclicity of a finitefree (graded) R -complex, and the other one provides a structure theorem for a finite free (graded) acyclic R -resolution.Let ϕ : F → G be a map between finite free R -modules of rank f and g , respectively. Once we fix ordered bases for F and G , we obtain a matrix representation M of ϕ , which is an g × f matrix with entries in R . For an m × n matrix N with entries in R , let I t ( N ) be the ideal generated by t × t minors of N if 1 ≤ t ≤ min { m , n } , and we set I ( N ) = R and I t ( N ) = t > min { m , n } . By abuse of notation, let I t ( ϕ ) denote I t ( M ) , where M is a matrix representation for ϕ . Matrix representations depend on the choice of bases. However, the ideal I t ( M ) does not depend on the choice ofbases. The rank of ϕ , denoted rk ϕ , is the number t where I t + ( ϕ ) =
0, but I t ( ϕ ) =
0, and we set I ( ϕ ) = I rk ϕ ( ϕ ) . For aproper ideal I of R , the grade of I , denoted grade I (equivalently, the depth of I in R denoted by depth I R ), is the lengthof a maximal R regular sequence contained in I . It is well-known that the maximal length is independent of regularsequences, cf. [BH93, Def. 1.2.11]. Such construction holds for arbitrary set of homogeneous ideals of the same degree in S , but this is the set up we will work in this paper. heorem 4 ([BE73, Theorem]) . Let C • be a finite complex of free R-modules of finite rank → F n ϕ n → · · · ϕ → F ϕ → F . Then C • is acyclic if and only if for k = , . . . , n, ( ) rk F k = rk ϕ k + + rk ϕ k and ( ) grade I ( ϕ k ) ≥ k or I ( ϕ k ) = R. The map ϕ : F → G induces maps between exterior powers ∧ k ϕ : ∧ k F → ∧ k G for any k . We also denote the imageof ∧ k ϕ by I k ( ϕ ) . (Once we fix bases for F , G and a matrix representation M for ϕ , the matrix representation of ∧ k ϕ is the k -minors of M (up to sign). Therefore, the image of ∧ k ϕ in R is I k ( ϕ ) .) For F a free module of rank f , anisomorphism η : ∧ f F → R is called an orientation of F . We say a finite free module is oriented if it is equipped withan orientation. For an oriented finite free module F , we have ∧ k F ⊗ ∧ f − k F → ∧ f F η → R . Hence, we identify ( ∧ k F ) ∗ with ∧ f − k F for oriented free modules. Here, for any R -module L , L ∗ : = Hom R ( L , R ) denotes the R -dual of L . Theorem 5 ([BE74, Theorem 3.1]) . Consider a finite free acyclic R-complex → F n ϕ n → · · · ϕ → F ϕ → F . Write r i = rk ϕ i . For k = , . . . , n, there exists unique R-homomorphism a k : R → ∧ r k F k − such that ( ) a n : = ∧ r n ϕ n : R = ∧ r n F n → ∧ r n F n − , and ( ) for k < n, the diagram ∧ r k F k ∧ rk ϕ k / / a ∗ k + ! ! ❉❉❉❉❉❉❉❉ ∧ r k F k − R a k ; ; ①①①①①①①①① commutes. ( ) For all k > , p I ( a k ) = p I ( ϕ k ) . General property.
In this subsection, we recall the notion of a general property. We will follow the section“general object” in [Har92]. Let X be a variety (or a scheme) parametrized by (closed) points in an (irreducible andreduced) variety Y and P a property on X . We say that P is general or a general property with respect the pair X and Y , if the set { p ∈ Y | object parametrized by p satisfies P } ⊂ Y is a dense open subset. If P is a general propertyand y ∈ Y satisfies P , then y or the object of X parameterized by y is called general or a general member with respectto P . In the sequel, whenever we use the phrase an object G is general or a general member, it is understood that itrefers to a general property for a pair X and Y , and G a general member with respect to this general property.We will use this notion of a general property in the following setup. Let k be a field of characteristic zero, R = k [ x , . . . , x n ] a polynomial ring, and A another polynomial ring over k . Further, let B = A ⊗ k R ∼ = A [ x , . . . , x n ] and J a homogeneous ideal of B which does not contain any element of A other than zero, i.e., J ∩ A =
0. Here, weset X = Proj Y B / J , where Y = Spec A . Our parameter space Y will be always affine space over a field of characteristiczero, and we will consider only closed points of Y . By abuse of terminology, we say an R -ideal I is general or ageneral member with respect to some general property P if Proj k R / I is a general member with respect to P for thepair X and Y . We also say a sequence of elements g , . . . , g s of R (or a matrix M whose entries are in R ) is general or a general member with respect to some general property P if the ideal ( g , . . . , g s ) (or I s ( M ) for some fixed s ) isa general member with respect to P . For the sake of completeness, we list the setups of two cases in detail. We willuse Case 1 in Section 5 and Case 2 for Theorem 26 and in its proof. Case 1:
Let R = k [ x , . . . , x n ] be a polynomial ring over a field of characteristic zero k and f , . . . , f l homogeneouspolynomials of degree d ≥ R . Fix a positive integer s . Consider the parameter space A : = k [ u i j | ≤ i ≤ l , ≤ j ≤ s ] , and for 1 ≤ j ≤ s , set F j : = u j f + u j f + · · · + u l j f l , in B : = A ⊗ k R ∼ = A [ x , . . . , x n ] . Write J = ( F , . . . , F s ) ⊂ B . Here Y = Spec A and X = Proj Y B / J . A general membersatisfying some property with respect to the pair X and Y is also often called as s-general elements or s-general k -linear combination of f , . . . , f l . ase 2: R = k [ x , . . . , x n ] be a polynomial ring over a field of characteristic zero k . Fix positive integers m , l and d , . . . , d l . For 1 ≤ j ≤ l , let h j denote the number of monomials of R of degree d j , and define h : = h + · · · + h l . Ourparameter space is A = ⊗ k A i j , where A i j = k [ u i j , k | ≤ k ≤ h ] for 1 ≤ i ≤ m , ≤ j ≤ l . In B : = A ⊗ k R = A [ x , . . . , x n ] ,consider the m by l matrix M ∼ whose ( i , j ) -entry is u i j , m d j , + u i j , m d j , + · · · + u i j , h j m d j , h j , where { m d j , , . . . , m d j , h j } is a fixed monomial basis of R of degree d j . Let J = I s ( M ∼ ) , where s is an integer. Here, X = Proj Y B / J with Y = Spec A . We say that an m × l matrix M with entries in R is general of type ( d , . . . , d l ) if forsome s , I s ( M ) is a general member for some general property with respect to this pair X and Y .We apply the general property in Case 2 to the height of I s ( M ) for some integer s . The open subset defining sucha general property corresponds to the complement of the closed subset defined by the ideal I e in Theorem 6(b) (in ournotation, S = B / J ), provided that such complement is non-empty. We will use this correspondence in the proof of themain theorem (Theorem 26). Theorem 6.
Suppose that A is a Noetherian ring and S = S ⊕ S ⊕ · · · is a positively graded ring which is finitelygenerated over A = S . Then we have the following statements. ( a ) [Eis95, Theorem 14.8(b)] For any integer e, there exists an A-ideal I e (depending on e) such that for any maximalideal m of A, dim A / m ⊗ A S ≥ e if and only if m ⊃ I e . ( b ) Write S = A [ x , . . . , x n ] / J. For any integer e, there exists an A-ideal I e (depending on e) such that for any maximalideal m of A, ht J ( A / m )[ x , . . . , x n ] ≤ n − e if and only if m ⊃ I e . Proof.
We show part (b). Fix a maximal ideal m of A and write T = A / m ⊗ A A [ x , . . . , x n ] = ( A / m )[ x , . . . , x n ] . Noticethat T / JT = A / m ⊗ A S . Now, part (b) follows from the first part and the following identity ([Har77, Theorem 1.8A])ht JT + dim T / JT = dim T (= n ) . (cid:3) We note that [Eis95, Theorem 14.8(b)] is stated for prime ideals of A . In this paper, we only consider closed pointsof a parameter space, so we stated part (a) for maximal ideals. Remark 7. ( a ) We list examples of general properties. • For an ideal I = ( f , . . . , f l ) of grade ≥ s , the property of the grade of the ideal generated by s -generallinear combinations of f , . . . , f l being at least s is a general property. • For an ideal I = ( f , . . . , f l ) minimally generated by the f i ’s and s ≤ l , the condition that s -general linearcombinations of f , . . . , f l are part of minimal generating set is a general property. ( b ) An advantage of having a general property is that one may ask finitely many general properties simultaneouslysince a finite intersection of non-empty dense open subsets remains non-empty dense open. ( c ) For those who are familiar with algebraic geometry, generic freeness and generic smoothness of C -varietiesare examples of general properties [Eis95, Theorem 14.4].3. D EFINING IDEALS OF R EES ALGEBRAS AND SPECIAL FIBER RINGS
Let R = k [ x , . . . , x n ] be a polynomial ring in n variables over a field k , m = ( x , . . . , x n ) R the homogeneous maximalideal, f , . . . , f n homogeneous polynomials of degree d in R , and I = ( f , . . . , f n ) . Consider a homogeneous surjective R -linear map π π : R [ T , . . . , T n ] → R [ It ] , where T i f i t for 0 ≤ i ≤ n . Since F ( I ) = R [ It ] / m R [ It ] , the map π induces a surjective k -linear map π for F ( I ) π : k [ T , . . . , T n ] → F ( I ) . The kernel of π is called the defining ideal of the Rees algebra of I , and the kernel of π is called the defining ideal of the special fiber ring F ( I ) . Recall that f , . . . , f n define a rational map φ : Proj ( R ) = P n − k [ f : ··· : f n ] −−−→ P n k , (4) ( ker π ) defines the closure of the graph of φ in P n − k × P n k , and V ( ker π ) defines the closure of the image of φ in P n k .The study of ker π can be simplified via the symmetric algebra of I . For an ideal I , the symmetric algebra of I isthe graded ring Sym R ( I ) : = R ⊕ I ⊕ Sym R ( I ) ⊕ · · · , where Sym kR ( I ) denotes the k th symmetric power of I . By the universal property of Sym R ( I ) , the homogeneouspresentation π factors through Sym R ( I ) R [ T , . . . , T n ] π / / π ′ & & ◆◆◆◆◆◆◆◆◆◆ R [ It ] Sym R ( I ) . π ′′ : : ✉✉✉✉✉✉✉✉✉ (5)The kernel of π ′ is easy to describe from the graded presentation matrix of I : Consider a homogeneous presentation ϕ of the ideal I ⊕ m R ( − d − d i ) ϕ → R ( − d ) n + → I → , (6)where d i are positive integers. Notice that ϕ is a n + × m matrix. Then ker π ′ = I ([ T . . . T n ] ϕ ) . Hence by abuse ofnotation, we call A = ( ker π ) Sym R ( I ) the defining ideal of the Rees algebra.Since R is a graded ring, R [ T , . . . , T n ] has a natural bi-graded structure; we set deg x i = ( , ) and deg T i = ( , ) .Hence ker π is a bi-graded ideal of R [ T , . . . , T n ] , and ker π is a homogeneous ideal of k [ T , . . . , T n ] . Write B = R [ T , . . . , T n ] and S = k [ T , . . . , T n ] . It is often convenient to view B as B ∼ = R ⊗ k S = k [ x , . . . , x n ] ⊗ k k [ T , . . . , T n ] .Hence B is free as an S -module. For a graded module M , M ( a ) denotes the grade shift by a . That is [ M ( a )] i = M a + i for any i ∈ Z . Similarly, for a bigraded module M , M ( a , b ) means [ M ( a , b )] ( i , j ) = M ( a + i , b + j ) for any i , j ∈ Z .Our main theorems are stated in the same hypothesis of the following proposition. The statement (as well as itsproof) is well-known. We present this proposition to fix the notation, and we will use it as a quick reference for oursetup. Proposition 8.
Let R = k [ x , . . . , x n ] be a polynomial ring in n variables over a field k , and I an R-ideal, and m =( x , . . . , x n ) R. Assume that I is codimension perfect, that the degrees of the entries of the columns of a presentationmatrix ϕ of I in eq. (6) are d ≤ d ≤ · · · ≤ d n , and that µ ( I p ) ≤ dim R p for all p ∈ Spec ( R ) \ { m } . Then we have thefollowing: ( ) The complex → ⊕ n R ( − d i ) ϕ → R n + → I ( d ) → is exact. ( ) I is generated in degree d = Σ n d i . ( ) Sym I ( R ) is a complete intersection. ( ) A = H m ( Sym I ( R )) , where H m ( Sym I ( R )) denotes the 0-th local cohomology module of Sym I ( R ) with supportin m .Proof. ( ) and ( ) follow from the Hilbert-Burch theorem [Eis95, Theorem 20.15] since I is a codimension two perfectgraded ideal, ( ) follows from the fact that dim Sym R ( I ) = n + π ′ in eq. (5) is generated by n elements, and ( ) follows from [HSV82, Theorem 2.6]. (cid:3) Remark 9.
For a ring and an ideal I , we say that I satisfies the condition ( G s ) if µ ( I p ) ≤ dim R p for all p ∈ Spec ( R ) such that dim R p (cid:12) s . Hence the ideal I in Proposition 8 satisfies ( G dim R ) condition. There are plenty of ideals satis-fying this condition, e.g., complete intersection ideals, (homogeneous) ideals primary to the homogeneous maximalideal, and this condition can be checked with the Fitting ideals of I (that is I satisfies ( G s ) ⇐⇒ ht F i ( I ) ≥ i + ≤ i ≤ s − Remark 10.
With the notation and hypothesis of proposition 8, let l , . . . , l n be in B = R [ T , . . . , T n ] such that [ l . . . l n ] =[ T . . . T n ] ϕ ; hence Sym R ( I ) = B / ( l , . . . , l n ) , and l , . . . , l n form a bi-homogeneous regular sequence in B . Then theKoszul complex K : = K ( l i ; B ) is a bi-graded B -resolution for Sym R ( I ) ; K : 0 → B ( − Σ n d i , − n ) → · · · → ⊕ n B ( − d i , − ) → B . ince B is a free graded S = k [ T , . . . , T n ] -module, K is a graded free S -resolution of Sym R ( I ) . Each component of K is not of finite rank as an S -module. However, for each k ∈ Z , K k : = K ( k , ∗ ) is a finite free graded S -resolution for [ Sym R ( I )] ( k , ∗ ) , and each component of K k is of finite rank as S -module; K k : 0 → F n → · · · → F → F , where F i = L ≤ j ≤ j ≤···≤ j i ≤ n S ( − i )( k − ( dj + ··· + dji )+ n − n − ) . Furthermore, these S -resolutions are linear resolutions, i.e., thenon-zero entries of the differential maps are of degree 1.Recall the notation that for a bigraded module M = ⊕ i , j ∈ Z M i , j in B , M i : = ⊕ j ∈ Z M i , j . Hence Sym ( I ) k = [ Sym R ( I )] ( k , ∗ ) and A k = [ A ] ( k , ∗ ) . In the sequel, when we use a single grading for B = R ⊗ S , we will always follow this convention. Theorem 11 (Grothendieck, [Jou96, Section 3.6], [KPU17, Theorem 2.4] ) . For each i ∈ { , . . . , δ } , there exists anisomorphism of finitely generated graded S-modules A i ∼ = Hom S ( Sym ( I ) δ − i , S ( − n )) , where δ = d − n = d + · · · + d n − n. In [Jou96, Section 3.6], Jouanolou describes the isomorphism in terms of Morley forms (for instance, see [KPU17,Chapter 4]). Therefore, an explicit computation of Hom S ( Sym ( I ) δ − i , S ( − n )) leads to a generating set, not only theirbidegrees, of A i . In the following two sections, we study A i and give a generalized method to compute a generatingset for A i . 4. T HE DUAL GENERATORS OF THE DEFINING IDEAL
We will adapt the notation in Section 3 and the notation and hypothesis of Proposition 8. The graded free S -resolution of Sym ( I ) δ − i in Remark 10 is K δ − i : 0 → F m α m → · · · → F α → F , (7)where F t = L ≤ j ≤ j ≤···≤ j i ≤ n S ( − t )( δ − i − ( dj + ··· + dji )+ n − n − ) for t = , . . . , m . Let r t = rk α t and f t = rk F t . By Theorem 4(1), f t = r t + r t + for t = , . . . , m , and by Theorem 11, A i ∼ = Hom S ( Sym ( I ) δ − i , S ( − n )) = ker α ∗ i ( − n ) . For the followingconstruction, it is worth mentioning that the above minimal graded resolution is linear. Remark 12.
In order to apply the theorems in [BE74] to our set up, we need to specify shifts. It is not a hard taskto do, but for the convenience of the reader and to set up the notation, we review their construction below. We alsonote that their constructions are for projective modules, but in our paper we only need their theorems for (graded) freemodules. Hence our review is written for graded free modules of their work. ( ) First for 0 ≤ t ≤ m , fix a basis for F t . This enables us to have a matrix representation of α t and ∧ r t α t for1 ≤ t ≤ m , respectively. Henceforth whenever we talk about these maps, we use their matrix representations.We use Theorem 5 on eq. (7) to construct maps a t for t = , . . . , m . Since α m : F m → F m − is an injective map,the entries of a m = ∧ r m α m = ∧ f m α m are of degree r m . Thus a m : S ( − r m m )) → ∧ r m F m − . (Recall that sinceeq. (7) is a linear resolution, the shift of F t is − t for all t = , . . . , m .) Using Theorem 5, we have have thecommutative diagram ∧ r m − F m − ∧ rm − α m − / / a ∗ m ( ( ❘❘❘❘❘❘❘❘❘❘❘❘❘ ∧ r m − F m − S ( − r m − ( m − ) + r m ) a m − ❧❧❧❧❧❧❧❧❧❧❧❧❧ otice that a ∗ m is a row matrix whereas a m − is a column matrix; a ∗ m = [ a ∗ m , · · · a ∗ m , ( fm − rm )] a m − = a m − , ... a m − , ( fm − rm − ) . Then one has the factorization ∧ r m − α m − = a m − ◦ a ∗ m . Not all of the entries of a ∗ m is zero as a ∗ m = ( ∧ r m α m ) ∗ and rk α m = r m . Thus there exists an entry, say a ∗ m , u = a m − , consider the u -th column of ∧ r m − α m − and divide its entries by a ∗ m , u . From this wededuce that the entries of a m − are of degree r m − − r m and that a m − : S ( − r m − ( m − ) + r m ) → ∧ r m − F m − .Iteratively, one can see that a : S ( − ∑ mp = ( − ) p − r p ) → ∧ r F . ( ) Let s t = r t · t − ∑ mp = t + ( − ) p − t − r p . Then a t : S ( − s t ) → ∧ r t F t − for t = , . . . , m . In particular, s = r − r + · · · + ( − ) m − r m = ∑ mp = ( − ) p − r p . ( ) The map α : F → F induces the map F ∗ ⊗ F → S . By dualizing it, we obtain the map f α : S → F ⊗ F ∗ .Consider the map ∂ : ∧ f − F → ∧ f F ⊗ F ∗ , (8)which is the composition of the following maps ∧ f − F ∼ = ∧ f − F ⊗ S id ⊗ f α −−−→ ∧ f − F ⊗ F ⊗ F ∗ m ∧ ⊗ id −−−−→ ∧ f F ⊗ F ∗ , where m ∧ : ∧ f − F ⊗ F → ∧ f F is the usual multiplication in the exterior algebra ∧ F . Fix an orientation η for F and let a : S ( − s ) → ∧ r F be the map in Theorem 5(2). This map ∂ is used in the proof of Theorem 13. Theorem 13.
With the hypothesis of Proposition 8, for any i ∈ { , . . . , δ } and α in the graded free S-resolution of Sym R ( I ) δ − i in eq. (7) , we have the following statements. ( ) The following is a complex of graded free S-modules ∧ f − r − F ⊗ S ( − s ) η ◦ m ∧ ◦ ( id ⊗ a ) −−−−−−−−→ F ∗ α ∗ −→ F ∗ , where a is the map in Theorem 5. In particular, if f = r + , then the image of a is in ker α ∗ . ( ) In addition, if rk Sym ( I ) δ − i > , then ∧ f − r − F ⊗ S ( − s ) ∧ f − r − α ⊗ id −−−−−−−−−→ ∧ f − r − F ⊗ S ( − s ) η ◦ m ∧ ◦ ( id ⊗ a ) −−−−−−−−→ F ∗ α ∗ −→ F ∗ is a complex of graded free S-modules.Proof. (1): [BE74, Lemma 3.2(a)] (the map ∂ in eq. (8) is equal to d α f − r − ) implies that the composition ∧ f − r − F ⊗ ∧ r F m ∧ ◦ ( id ⊗∧ r α ) −−−−−−−−−→ ∧ f − F ∂ −→ ∧ f F ⊗ F ∗ (9)is zero.We identify ∧ f − F ∼ = F ∗ and ∧ f F ∼ = S following the fixed orientation η of F . Lemma 3.2(c) in [BE74] impliesthat the following diagram commutes up to sign: ∧ f − F ∂ / / η (cid:15) (cid:15) ∧ f F ⊗ F ∗ η ⊗ id (cid:15) (cid:15) ∧ F ∗ ( m ∧ ◦ ( id ⊗ α )) ∗ / / ∧ F ∗ ⊗ F ∗ . (10)Notice that in the above diagram, the vertical maps are isomorphisms and the bottom map, i.e., ( m ∧ ◦ ( ⊗ α )) ∗ is acomposition of the following maps ∧ F ∗ m ∗∧ −−→ ∧ F ∗ ⊗ ∧ F ∗ ( id ⊗ α ) ∗ −−−−−→ ∧ F ∗ ⊗ ∧ F ∗ . ince ∧ F ∗ ∼ = S , the map ( m ∧ ◦ ( ⊗ α )) ∗ can be identified with the map F ∗ α ∗ −→ F ∗ . (11)From (9), (10), and (11), we conclude that ∧ f − r − F ⊗ ∧ r F η ◦ m ∧ ◦ ( id ⊗∧ r α ) −−−−−−−−−−−→ F ∗ α ∗ −→ F ∗ (12)is a complex. Thus Im ( η ◦ m ∧ ◦ ( id ⊗ ∧ r α )) ⊆ ker α ∗ .With the commutative diagram in Theorem 5(a) ∧ r F a ∗ $ $ ■■■■■■■■■ ∧ r α / / ∧ r F S ( − s ) , a : : ✉✉✉✉✉✉✉✉✉ the first morphism in eq. (12) factors as follows ∧ f − r − F ⊗ ∧ r F η ◦ m ∧ ◦ ( id ⊗∧ r α ) / / id ⊗ a ∗ (cid:15) (cid:15) F ∗ α ∗ / / F ∗ ∧ f − r − F ⊗ S ( − s ) id ⊗ a / / ∧ f − r − F ⊗ ∧ r F m ∧ / / ∧ f − F . η O O (13)Now let J = Im a ∗ be an S -ideal (with a shift). Then Im id ⊗ a ∗ = ∧ f − r − F ⊗ J . By Theorem 5(c), p I ( α ) = p I ( a ) .So, the ideal J has a positive grade. Since the composition of maps in the top row is zero and the above diagramcommutes, we have 0 = ( α ∗ ◦ η ◦ m ∧ ◦ ( id ⊗ ∧ r α ))( ∧ f − r − F ⊗ ∧ r F ) (14) = ( α ∗ ◦ η ◦ m ∧ ◦ ( id ⊗ a ) ◦ ( id ⊗ a ∗ ))( ∧ f − r − F ⊗ ∧ r F )= ( α ∗ ◦ η ◦ m ∧ ◦ ( id ⊗ a )) ◦ ( id ⊗ a ∗ )( ∧ f − r − F ⊗ ∧ r F )= ( α ∗ ◦ η ◦ m ∧ ◦ ( id ⊗ a ))( ∧ f − r − F ⊗ J ( − s ))= J ( α ∗ ◦ η ◦ m ∧ ◦ ( id ⊗ a ))( ∧ f − r − F ⊗ S ( − s )) . The last equality holds as the maps are S -module homomorphisms and tensor products are over S . Since the lastequality holds in the free module F ∗ , and J has positive grade, α ∗ ◦ η ◦ m ∧ ◦ ( id ⊗ a ) =
0. Thus, ∧ f − r − F ⊗ S ( − s ) η ◦ m ∧ ◦ ( id ⊗ a ) −−−−−−−−→ F ∗ α ∗ −→ F ∗ is a complex of graded free S -modules.(2): By item (1), it suffices to show that [ η ◦ m ∧ ◦ ( id ⊗ a )] ◦ [ ∧ f − r − α ⊗ id S ] =
0. We first extend the complex inthe top row of eq. (13) to ∧ f − r − F ⊗ ∧ r F ∧ f − r − α ⊗ id −−−−−−−−−→ ∧ f − r − F ⊗ ∧ r F η ◦ m ∧ ◦ ( id ⊗∧ r α ) −−−−−−−−−−−→ F ∗ α ∗ −→ F ∗ . (15)Since eq. (7) is a free S -resolution for Sym ( I ) δ − i , we have rk Sym ( I ) δ − i = rk F − rk α = f − r > f − > r . Hence from η ◦ m ∧ ◦ ( id ⊗ ∧ r α ) ◦ ( ∧ f − r − α ⊗ id ) = η ◦ m ∧ ◦ ( ∧ f − r − α ⊗ ∧ r α ) = η ◦ ∧ f − α = , we conclude that eq. (15) is a complex. ith the factorization ∧ r α = a ◦ a ∗ , we obtain the following a commutative diagram ∧ f − r − F ⊗ ∧ r F ∧ f − r − α ⊗ id / / id ⊗ a ∗ (cid:15) (cid:15) ∧ f − r − F ⊗ ∧ r F η ◦ m ◦ ( id ⊗∧ r α ) / / F ∗ α ∗ / / F ∗ . ∧ f − r − F ⊗ S ( − s ) ∧ f − r − α ⊗ id S / / ∧ f − r − F ⊗ S ( − s ) η ◦ m ∧ ◦ ( id ⊗ a ) ✐✐✐✐✐✐✐✐✐✐✐✐✐✐✐✐✐✐✐✐✐ Let J = Im a ∗ be an S -ideal. Then the argument used to prove Item (1) (from eq. (14)) shows that η ◦ m ∧ ◦ ( id ⊗ a ) ◦∧ f − r − α ⊗ id S =
0. This complete the proof of the theorem. (cid:3)
We present two examples demonstrating Theorem 13(b). It is worth mentioning that in the first example, thecorresponding complex is exact whereas in the second example, it is not.
Example 14. ( a ) (cf. Theorem 26) Let R = k [ x , y , z ] and I = I ( ϕ ) , where ϕ = x y z y z yz z x y xy + yz + xz . It is easy to verify that this example satisfies the hypothesis of Proposition 8. Consider B = R [ T , T , T , T ] and let K denote the Koszul complex on l , l , l where [ l l l ] = [ T T T T ] · ϕ . Since K is a graded freeresolution of Sym ( I ) , we can extract the S -resolution of Sym ( I ) → S ( − ) α −→ S → Sym ( I ) → , where α = (cid:2) T T T (cid:3) t . Here, we have f = , f = , s = r =
1. Theorem 13(b) shows that ∧ F ⊗ S ( − s ) ∧ α ⊗ id −−−−−→ ∧ F ⊗ S ( − s ) m ∧ ◦ ( id ⊗ a ) −−−−−−−→ ∧ F ∼ = η F ∗ α ∗ −→ F ∗ is a complex. For this example, we will verify that α ∗ ◦ η ◦ m ∧ ◦ ( id ⊗ a ) = η ◦ m ∧ ◦ ( id ⊗ a ) = ◦ ∧ α ◦ id = F , F as { e , e , e } , { g } respectively, we can fix a basis { e ∧ e , e ∧ e , e ∧ e } , { g } for ∧ F and ∧ F , respectively. A basis { e ⊗ g , e ⊗ g , e ⊗ g } of ∧ F ⊗ ∧ F gives us a following matrixrepresentation of η ◦ m ∧ ◦ ( id ⊗ a ) T − T − T T T − T . As ∧ α ⊗ id ∼ = α and α ∗ is the transpose α , we have ∧ F ⊗ S ( − s ) T T T −−−→ ∧ F ⊗ S ( − s ) T − T − T T T − T −−−−−−−−−−−−−−→ F ∗ h T T T i −−−−−−−−−→ F ∗ . Then it is easy to see that this is a complex. In fact, the above complex is a graded minimal free S -resolutionof coker α ∗ → ∧ F ⊗ S ( − s ) → ∧ F ⊗ S ( − s ) → F ∗ → F ∗ → coker α ∗ → . ( b ) Let R = k [ x , y , z ] and I = I ( ϕ ) , where ϕ = x x y x y z y z z x z . It is easy to verify that this example satisfies the hypothesis of Proposition 8. We will use the same notation asin part (a). The number δ is d − n = − =
5. The graded free S -resolution of Sym ( I ) δ is0 → S ( − ) α −→ S ( − ) α −→ S → Sym ( I ) → . ere, we have f = , f = , f = , rk Sym ( I ) = r = , r =
3. Therefore, f − r − = − − = s = r − r =
17. Theorem 13(a) implies that ∧ F ⊗ S ( − s ) m ∧ ◦ ( id ⊗ a ) −−−−−−−→ ∧ F ∼ = η F ∗ α ∗ −→ F ∗ is a complex. However, this complex does not extend to a free resolution of coker α ∗ (cf. Theorems 15 and 20).As the examples above demonstrate the complexes in Theorem 13 are not exact in general. It is natural to ask whichconditions guarantee the exactness of these complexes. A positive answer to this question provides part of minimalgenerating equations for the defining ideal of the corresponding Rees algebra by Jouanolou. In the following theorems,we provide sufficient conditions for the exactness in the case where Sym ( I ) k is of rank 1 or 2, respectively. Theorem 15.
For a fixed integer i, consider the complex ∧ f − r − F ⊗ S ( − s ) η ◦ m ∧ ◦ ( id ⊗ a ) −−−−−−−−→ F ∗ α ∗ −→ F ∗ , in Theorem 13. Assume that rk Sym ( I ) δ − i = . Then the complex is exact if and only if grade I ( a ) ≥ .Proof. Since rk Sym ( I ) δ − i =
1, we have f − r − =
0. Therefore, the complex is isomorphic to the complex S ( − s ) η ◦ a −−−→ F ∗ α ∗ −→ F ∗ . Notice that rk α ∗ ≥ I ( η ( a )) = grade I ( a ) . Hence by Theorem 4, the complex is acyclic if and only ifgrade I ( a ) ≥ (cid:3) Theorem 16.
For a fixed integer i, consider the complex ∧ f − r − F ⊗ S ( − s ) η ◦ m ∧ ◦ ( id ⊗ a ) −−−−−−−−→ F ∗ α ∗ −→ F ∗ , in Theorem 13. Assume that rk Sym ( I ) δ − i = . Consider the following statements: ( a ) Sym ( I ) δ − i satisfies Serre’s condition ( S ) . ( b ) grade I ( α t ) ≥ t + for t = , . . . , m. ( c ) The complex above is exact.Then we have ( a ) ⇒ ( b ) ⇒ ( c ) . Before proving the theorem, we present a lemma which explains the relationship between the conditions ( a ) and ( b ) . Lemma 17.
With the setup of Theorem 16, we have a complex of graded free S-modules → F m ⊗ S ( − s ) α m ⊗ id −−−−→ · · · → F ⊗ S ( − s ) α ⊗ id −−−→ F ⊗ S ( − s ) η ◦ m ∧ ◦ ( id ⊗ a ) −−−−−−−−→ F ∗ α ∗ −→ F ∗ . (16) Furthermore, this complex is acyclic if
Sym ( I ) δ − i satisfies Serre’s condition ( S ) .Proof. Recall that 0 → F m α m −→ F m − → · · · → F α −→ F α −→ F is a graded minimal free S -resolution of Sym ( I ) δ − i , and so is its shift. Hence for the first part of the statement, itsuffices to show that F ⊗ S ( − s ) α ⊗ id −−−→ F ⊗ S ( − s ) η ◦ m ∧ ◦ ( id ⊗ a ) −−−−−−−−→ F ∗ α ∗ −→ F ∗ is a complex. But this follows from Theorem 13(b) with f − r − = rk Sym ( I ) δ − i − = η ◦ m ∧ ◦ ( id ⊗ a ) : F ⊗ S ( − s ) → F ∗ . By choosing a basis { e , . . . , e f } of F , thiscan be described as a square matrix. As a : S ( − s ) → ∧ r F ∼ = ∧ f − F ∗ , by abuse of notation, we list the basis for ∧ r F as e ∗ i ∧ e ∗ j and a = ⊕ c i , j e ∗ i ∧ e ∗ j , where c i , j ∈ S . With this notation, one sees that η ◦ m ∧ ◦ ( id ⊗ a ) a skewsymmetric matrix whose i , j th entry for i < j is c i , j up to sign. Hence we have p I ( η ◦ m ∧ ◦ ( id ⊗ a )) ⊃ I ( a ) andgrade I ( η ◦ m ∧ ◦ ( id ⊗ a )) ≥ grade I ( a ) ≥ grade I ( α ) . Furthermore, as grade α ∗ = grade α , by Theorem 4, the com-plex in eq. (16) is exact if and only if grade α m ≥ m + i = , . . . , m . t is a well-known theorem of Auslander and Bridger [AB69, Theorem 4.25] that a module M is k th syzygy ifand only if M satisfies Serre’s condition ( S k ) . Hence Sym ( I ) δ − i satisfies ( S ) if and only if for some free S -modules F − , F − , the following complex is acyclic0 → F m α m −→ F m − → · · · → F α −→ F α −→ F → F − → F − . By Theorem 4, this condition implies that grade α m ≥ m + i = , . . . , m . (cid:3) Proof of Theorem 16.
Now the proof follows from Lemma 17 and its proof. (cid:3)
A similar argument as in the proof of Lemma 17, one can show that the exactness of the complex in Theorem 15follows from the condition that rk Sym ( I ) δ − i = ( I ) δ − i satisfies Serre’s condition ( S ) . Hence it is natural toask the following question. Question 18.
Consider the complex ∧ f − r − F ⊗ S ( − s ) η ◦ m ∧ ◦ ( id ⊗ a ) −−−−−−−−→ F ∗ α ∗ −→ F ∗ in Theorem 13. Assume that rk Sym ( I ) δ − i = r >
0. Then is this complex exact if Sym ( I ) δ − i satisfies Serre’s condition ( S r ) ? Furthermore, in the case where this complex is exact, is there a natural way to construct the minimal freeresolution of coker α ∗ ?This question has an affirmative answer when r ≤ EGREES OF I MPLICIT EQUATION
In this section, we provide a concrete description of the bi-degree of the equation defining the special fiber ring F ( I ) . With the help of Morley forms, the results in this section can be used to find the equation defining F ( I ) . Weassume the notation and set up of Proposition 8, but for the convenience of the reader we recall the key notation andsetup. By d , . . . , d n we denote the column degrees of the graded presentation matrix ϕ of I , n is the dimension of R , I = I n ( ϕ ) , and δ = d + · · · + d n − n = d − n . Consider the diagram, where the rows are exact0 / / A / / (cid:15) (cid:15) Sym ( I ) / / (cid:15) (cid:15) R [ It ] / / (cid:15) (cid:15) / / ker π / / S = k [ T , . . . , T n ] π / / F ( I ) / / . The defining ideal of F ( I ) is ker π ∼ = A ⊗ k ∼ = A S . By Theorem 11, to find the defining ideal of F ( I ) , it is enoughto compute Hom S ( Sym R ( I ) δ , S ( − n )) .We begin with the resolution of Sym ( I ) δ , which is induced from the resolution in Remark 10.0 → F m α m −→ F m − → · · · → F α −→ F α −→ F → Sym ( I ) δ → , (17)where m ≤ n − δ + n = d = d + · · · + d n , we may write F t = M ≤ j ≤ j ≤···≤ j t ≤ n S ( − t )( δ − ( dj + ··· + djt )+ n − n − ) = M ≤ k ≤ k ≤···≤ k n − t ≤ n S ( − t )( dk + ··· + dkn − t − n − ) , where { j , , . . . , j t , k , . . . , k n − t } = { , . . ., n } . As in the previous section, we let r t = rk α t , f t = rk F t and A ∼ = Hom ( Sym ( I ) δ , S ( − n )) = ker α ∗ ( − n ) . Remark 19.
The length of the Koszul complex in Remark 10 is n whereas the length of the resolution of Sym ( I ) δ ineq. (17) can be strictly less than n −
1. In particular, F t = m + ≤ t ≤ n −
1. In other words, we have (cid:18) d k + · · · + d k n − t − n − (cid:19) = m + ≤ t . y [UV93, Proposition 2.4], ker π is a principal prime ideal and hence, so is ker α ∗ . By Theorem 13, we have acomplex ∧ f − r − F ⊗ S ( − s ) η ◦ m ∧ ◦ ( id ⊗ a ) −−−−−−−−→ F ∗ α ∗ −→ F ∗ , where η is a fixed orientation of F and a : S ( − s ) → ∧ r F is the map obtained by applying Theorem 5 to theresolution eq. (17). We refer to Remark 12(2) for the description of the shifts s in the preceding complex. Sincerk Sym ( I ) δ =
1, we have f − r − =
0, and the above complex simplifies to S ( − s ) η ( a ) −−−→ F ∗ α ∗ −→ F ∗ . Thus η ( a ( S ( − s ) )) ∈ ker α ∗ . Henceforth, we will often denote the basis element 1 S ( − s ) of S ( − s ) simply by 1 if noconfusion arises.Our main result of this section is the following theorem. Theorem 20.
Assume k is a field of characteristic zero and the notation and hypothesis of Proposition 8. LetR = k [ x , . . . , x n ] with n ≥ , and let φ : Proj R = P n − k [ f : ··· : f n ] −−−→ P n k be the rational map defined in eq. (4) . WriteI = ( f , . . . , f n ) . Then the following statements are equivalent: ( ) The rational map φ is birational onto its image. ( ) The ker α ∗ is generated by ( η ◦ a )( ) . ( ) grade I ( η ( a )) ≥ . ( ) The greatest common divisor of the entries of η ◦ a (this is a column matrix) is 1. ( ) e ( F ( I )) = e ( R / ( g , . . . , g n − ) : I ) , where g , . . . , g n − are general k -linear combinations of the f ′ i s. Here, e ( − ) denotes, the Hilbert-Samuel multiplicity. The Hilbert-Samuel multiplicity e ( F ( I )) is also called the degree of the projective variety Proj F ( I ) = Φ ( P n − ) .We use the next two lemmas to prove Theorem 20. Lemma 21 ([KPU16, Corollary 3.7]) . Let k be a field of characteristic zero, R = k [ x , . . . , x n ] with n ≥ . Further,let I be a homogeneous ideal generated by forms of the same degree d and k [ R d ] the d-th Veronese subring of R. Ifg , . . . , g n − are general k -linear combinations of the generators f , . . . , f n of I, then the following equality holds.e ( F ( I )) = [ k [ R d ] : k [ f , . . . , f n ]] · e (cid:18) R ( g , . . . , g n − ) : I ∞ (cid:19) . In the above theorem [ k [ R d ] : k [ f , . . . , f n ]] denotes the field extension degree [ Quot ( k [ R d ]) : Quot ( k [ f , . . . , f n ])] .Using Remark 3, we have that the rational map φ (as defined in eq. (4)) is birational onto its image if and only if [ k [ R d ] : k [ f , . . . , f n ]] = Definition 22.
For any matrix A with entries of the same degree, we define deg A to be the degree of the entries of A . Lemma 23.
Assume the notation and hypothesis of Theorem 20. For general k -linear combinations g , . . . , g n − ofthe generators f , . . . , f n of I, we have deg Im ( η ◦ a ) + dim R = e (cid:18) R ( g , . . . , g n − ) : I (cid:19) = e (cid:18) R ( g , . . . , g n − ) : I ∞ (cid:19) . Proof.
First we show that for general linear combinations g , . . . , g n − , e ( R / ( g , . . . , g n − ) : I ∞ ) = e ( R / ( g , . . . , g n − ) : I ) . For each λ = ( λ , · · · , λ n ) ∈ ( A n + k ) n + , where λ i = ( λ i : · · · : λ in ) ∈ A n + k we set g i = ∑ nj = λ i j f j ∈ R for 0 ≤ i ≤ n .Let U ′ be an open subset of ( A n + k ) n + such that det ( a i j ) ∈ k \{ } . This shows that { g , . . . , g n } is a minimal generat-ing set of I if λ ∈ U ′ . Further, let U ′ = U ′ ∩ ( A n + k ) n − (the first n − ( A n + k ) n + ).Notice that the ideal I satisfies µ ( I p ) ≤ dim R p for every p ∈ Spec ( R ) \{ m } . By [Ulr94, Corollary 1.6, Proposi-tion 1.7] (see also [PX13, Lemma 3.1]), there exists a dense open set U ′′ ⊆ ( A n + k ) n − such that for every λ ∈ U ′′ , ( g , . . . , g n − ) : I is a geometric ( n − ) -residual intersection. (We note that in [PX13], the non-empty open subset U ′′ corresponds to general R -linear combinations. Thus, the same U ′′ can be used as the desired non-empty open subsetfor k -linear combinations.) t follows that ( g , . . . , g n − ) P = I P for P ∈ V ( I ) such that ht P ≤ n −
1. Since I is a codimension two perfect ideal, I is a strongly Cohen-Macaulay ideal. Now we apply [Hun83, Theorem 3.1] to conclude that ( g , . . . , g n − ) : I ∞ =( g , . . . , g n − ) : I . The open set U = U ′ ∩ U ′′ provides the desired non-empty open subset.As g , . . . , g n form a minimal homogeneous generators of I , we may choose a presentation matrix ϕ ′ for ( g , . . . , g n ) with the same column degrees d , . . . , d n of ϕ . This follows from the uniqueness of the graded Betti numbers of I .Furthermore, we assume that g i are signed minors of the ( n + ) × n matrix ϕ ′ .Let M = I / ( g , . . . , g n − ) . Notice that the last two rows of ϕ ′ is a presentation matrix for M , and we call this presen-tation matrix by ϕ ′′ . We note that ϕ ′′ is a 2 × n matrix. Furthermore, I ( ϕ ′′ ) = Fitt ( M ) = ann M = ( g , . . . , g n − ) : I (cf.[BE77, Theorem 3.1(2)]) has the maximum possible height n −
1. Thus we reduce to the case of computing the multi-plicity e ( R / I ( ϕ ′′ )) . Since the grade I ( ϕ ′′ ) is the maximum possible number, n −
1, the Eagon-Northcott complex of ϕ ′′ , EN ( ϕ ′′ ) , forms a graded (minimal) free R -resolution for R / I ( ϕ ′′ ) . Let N = R / I ( ϕ ′′ ) . Clearly, e ( N ) = e ( N ( )) ,and the shifted Eagon-Northcott complex of ϕ ′′ is the resolution of N ( ) .EN ( ϕ ′′ )( ) :0 → ( Sym ( R ) n − ) ∗ ⊗ R n − ( − ( d + · · · d n − )) → ( Sym ( R ) n − ) ∗ ⊗ M ≤ j ≤···≤ j n − ≤ n R n − ( − ( d j + · · · + d j n − − )) → · · · → ( Sym ( R ) ) ∗ ⊗ M ≤ j ≤ j ≤ n R ( − ( d j + d j − )) → R ( ) → . Furthermore, the Hilbert Series of N ( ) is n − ∑ t = " ( − ) t · t ∑ ≤ j ≤···≤ j t + ≤ n z ( d j + ··· + d jt + − ) + z − ( − z ) n . Let p ( z ) = n − ∑ t = " ( − ) t · t ∑ ≤ j ≤···≤ j t + ≤ n z ( d j + ··· + d jt + − ) + z − . Then e ( N ( )) = ( − ) n − p ( n − ) ( )( n − ) ! = ( − ) n − " n − ∑ t = ( − ) t · t · ∑ ≤ j ≤···≤ j t + ≤ n (cid:18) d j + · · · + d j t + − n − (cid:19) + ( − ) n − = n − ∑ t = " ( − ) n − + t · t · ∑ ≤ j ≤···≤ j t + ≤ n (cid:18) d j + · · · + d j t + − n − (cid:19) + , where p ( n − ) ( z ) denotes the ( n − ) st derivative of p ( z ) . We compare the terms with the ranks in eq. (17). By Re-mark 19, we have the following equality. ∑ ≤ j ≤···≤ j t + ≤ n (cid:18) d j + · · · + d j t + − n − (cid:19) = ( rk F n − t − = f n − t − t ≥ n − m − , t < n − m − . Under this identification, we may rewrite e ( N ( )) as follows e ( N ( )) = n − ∑ t = n − m − (cid:2) ( − ) n − − t · t · f n − t − (cid:3) + = ( − ) m ( n − m − ) f m + ( − ) m − ( n − m ) f m − + ( − ) m − ( n − m + ) f m − + · · · + ( n − ) f + . e note that since rk Sym ( I ) δ = ∑ mj = ( − ) j f j =
1. Then one has n − m − = ( n − m − ) m ∑ j = ( − ) j f j = ( − ) m ( n − m − ) f m + ( − ) m − ( n − m − ) f m − + ( − ) m − ( n − m − ) f m − + · · · + ( n − m − ) f = ( − ) m ( n − m − ) f m + ( − ) m − ( n − m ) f m − + ( − ) m − ( n − m + ) f m − + · · · + ( n − ) f − [( − ) m − ( ) f m − + ( − ) m − ( ) f m − + · · · + ( m ) f ]= [ e ( N ( )) − ] − [( − ) m − ( ) f m − + ( − ) m − ( ) f m − + · · · + ( m ) f ] . Hence e ( N ( )) = ( − ) m − ( ) f m − + ( − ) m − ( ) f m − + · · · + ( m ) f + n − m = m ∑ k = ( − ) m − k · k · f m − k + n − m . Thus, we have shown that e ( R / ( g , . . . , g n − ) : I ∞ ) = m ∑ k = ( − ) m − k · k · f m − k + ( n − m ) . We will complete the proof by showing that the right hand side of the equation is equal to deg Im ( η ◦ a ) + n . Firstobserve that deg Im ( η ◦ a ) + n = deg ( a ) + n since η is an isomorphism of degree 0. In Remark 12, we showed thatthe degree of Im a is s = ∑ mt = ( − ) t − r t , and r t = ∑ mj = t ( − ) j − t f j . Note that since ∑ mj = ( − ) j f j =
1, we have1 = f − f + · · · + ( − ) t − f t − + ( − ) t f t + ( − ) t + f t + + · · · + ( − ) m f m | {z } ( − ) t r t . Therefore, r t = ( − ) t + ( f − f + · · · ( − ) t − f t − ) + ( − ) t = t − ∑ j = ( − ) t + j + f j + ( − ) t . Finally, we havedeg Im ( η ◦ a ) + n = deg ( a ) + n = m ∑ t = ( − ) t − r t + n = m ∑ t = ( − ) t − t − ∑ j = ( − ) t + j + f j + ( − ) t ! + n = m ∑ t = t − ∑ j = ( − ) j f j − ! + n = m ∑ t = t − ∑ j = ( − ) j f j + m ∑ t = ( − ) + n = m ∑ t = t − ∑ j = ( − ) j f j + m ( − ) + n = f + ( f − f ) + ( f − f + f ) + · · · + m − ∑ j = ( − ) j f j !! + ( n − m )= m ∑ k = ( − ) m − k · k · f m − k + ( n − m ) . This completes the proof. (cid:3) roof of Theorem 20. We note that the rational map φ is birational to the image if and only if [ k [ R d ] : k [ f , . . . , f n ]] = ( ) ⇔ ( ) : This follows from Lemmas 21 and 23. ( ) ⇔ ( ) : As F ( I ) is a hypersurface, e ( F ( I )) is the degree of a generating element of ker π . The equivalence followsfrom Lemmas 21 and 23 which say e ( F ( I )) = deg Im ( η ◦ a ) + n [ k [ R d ] : k [ f , . . . , f n ]] . ( ) ⇔ ( ) : This is a consequence of Theorem 15. ( ) ⇔ ( ) : This equivalence is clear as the ambient ring is a UFD. (cid:3) Example 14(b) is an example where I satisfies the conditions in Proposition 8, but the rational map is not birationalto its image. In this example deg a + n =
20, but e ( F ( I )) =
10, so that the extension degree is 2.6. T HE M AIN RESULT AND APPLICATION
Let R = k [ x , y , z ] be a polynomial ring over a characteristic zero field k and M a 4 by 3 matrix with entries in R .Our main result states that if M is general of type ( , , q ) for some q > I = I ( M ) satisfies the equivalence conditions in Theorems 15 and 16. Therefore, we are able to statethe bi-degrees and the minimal number of equations in each bi-degree of the defining ideal of the Rees algebra R [ It ] ,explicitly (hence for F ( I ) as well). Furthermore, by the duality of Jouanolou (Theorem 11), one may recover theexplicit equations in Sym R ( I ) . Remark 24.
Let B = S [ x , . . . , x d ] , where S is a ring, and f ( x , . . . , x d ) a non-zero homogeneous polynomial of degree q in B . The graded homomorphism Φ : B ( − q ) · f → B induces a homomorphism in each degree i . We call the inducedmap Φ fi : [ B ( − q )] i → B i . This map can be described explicitly once we fix a basis of B i . In this note, we will alwaysuse the Lex (monomial) order on x , . . . , x d . For instance, when d = q = f = A x + A x x + A x , where A i ∈ S ,then Φ f has a matrix representation A A A
00 0 A A
00 0 A A A
00 0 A basis z }| { x x x x x x x x x x x x x x x x x x . Here each column is the presentation of x i f by the monomials of degree 3.To prove our main theorem, we need to compute the height of the maximal minors of Φ fi . The following lemmageneralizes [Eis05, Theorem A.2.60] in which there is no gap between nonzero entries. Lemma 25.
Let R be a ring and g , . . . , g t are elements of R. Let M = ( m i j ) be a r × s matrix (r ≥ s) satsifying ( ) each entry of M is either zero or g , . . . , g t , ( ) the i-th non zero entry of each column is g i , ( ) the last non-zero entry of each column is g t , and ( ) if m i j = g q , then m k j + = g q for some k > i.Then I s ( M ) = ( g , . . . , g t ) s .Proof. It suffices to show the statement that if R ′ = Z [ y · · · y t ] and M = ( m i j ) a r × s matrix ( r ≥ s ) satisfying(1) each entry of M is either zero or y , . . . , y t ,(2) the i -th non zero entry of each column is y i ,(3) the last non-zero entry of each column is y t , and(4) if m i j = y q , then m k j + = y q for some k > i , hen I s ( M ) = ( y , . . . , y t ) s .Clearly, I s ( M ) ⊆ ( y , . . . , y t ) s . We now show the other inclusion. We prove by induction on the number of variables t and the number of columns s of the matrix. Suppose t =
1. The case of s = I t ( M ) = ( y ) . In fact, if s >
1, then it is easy to see from item (4) that I s ( M ) = ( y ) s settling the base case of induction.Now suppose by induction hypothesis, the result is true for all matrices M following the pattern in the hypothesis, withentries in the ring T = Z [ y , . . . , y q ] where q ≤ t − s .If s =
1, in which case M has only one column, then clearly I s ( M ) = ( y , . . . , y t ) . By induction hypothesis on s assume that the result is true for all matrices M , satisfying the pattern in the hypothesis, with the number of columnsbeing strictly less than s . Now let M be a matrix with s columns and j , . . . , j s be indices such that m j = m j = · · · = m s j s = y . If M ′ denotes the matrix obtained by deleting the j -th row and the first column of M , then by induction hy-pothesis on s , it is clear that I s − ( M ′ ) = ( y , . . . , y t ) s − . Since the j -th row of M has y appearing in the first entry andzero otherwise, we have f = y · g where g ∈ I s − ( M ′ ) = ( y , · · · , y t ) s − . Thus one has y ( y , . . . , y t ) s − ⊆ I s ( M ) . Weshow similarly that y u ( y , . . . , y t ) s − u ⊆ I s ( M ) . Let M u be the matrix obtained by removing the rows j , . . . , j u and thecolumns 1 , . . . , u from M . Then by induction hypothesis I s − u ( M u ) = ( y , . . . , y t ) s − u . Now consider M and an s -minor f ′ obtained by choosing the rows j , . . . , j u , i , . . . , i s − u where 1 ≤ i q ≤ r , i q
6∈ { j , . . . , j u } for 1 ≤ q ≤ s − u , of M . Thenwe have that f ′ = y u g ′ where g ′ ∈ I s − u ( M u ) = ( y , . . . , y t ) s − u . Thus we have y u ( y , . . . , y t ) s − u ⊆ I s ( M ) for 1 ≤ u ≤ s .Now, consider the ring T = T / ( y ) and the matrix M obtained by extending the matrix M to T . M follows thepattern in the hypothesis and hence by induction hypothesis on t , we have I s ( M ) = ( y , · · · , y t ) s . Thus we have I s ( M ) + ( y ) ⊇ ( y , . . . , y t ) s . It suffices to show that for each monominal generator m for ( y , . . . , y t ) s is in I s ( M ) . Thisfollows immediately as y i are variables. Thus, we have ( y , . . . , y t ) s ⊆ I s ( M ) . (cid:3) Theorem 26.
Let R = k [ x , y , z ] , where k is a field of characteristic . Let M be a × matrix whose entries are in Rand let I = I ( M ) . If M is general of type ( , , q ) , where q > (see Case 2 in Section 2.3), then the defining ideal ofthe Rees algebra R [ It ] is minimally generated bybidegree number of elementsl ( , ) l ( , ) l ( q , ) A ( , q + ) A q − i ( q − i , i + ) (cid:0) + i (cid:1) A q ( q , ) for ≤ i ≤ q − . In particular, the defining ideal is minimally generated by (cid:0) q + (cid:1) + elements.Proof. The general property P needs to satisfy the following conditions.(1) I ( M ) is of height 2, and I satisfies ( G ) , i.e, ht I ( M ) ≥ I ( M ) ≥ ≤ i ≤ q , rk Sym i = I ( α j ) , where α j are the differentials in the complex by Theorem 4.)Notice that we have finitely many conditions. If each condition is general corresponding to a non-empty open subset U k , then the desired non-empty subset for P is the intersection of the open subsets U k . To show that each conditionis general, first we apply Theorem 6(b) to show the existence of such open subsets. To finish the proof, we need toshow that each open subset is not empty. This can be done by demonstrating an example satisfying the conditions in(1) and (2). We do this in Example 27. The generating degrees follow from Theorems 15 and 16. This completes theproof. (cid:3) Example 27.
Let R = k [ x , y , z ] , where k is a field of characteristic 0, m = ( x , y , z ) R , and M the matrix M = x y γ z q y z yz q − z x y q xy + yz + xz , here γ ∈ k \ { } . We will show that there exists a nonzero γ such that the ideal I = I ( M ) satisfies the desired prop-erties required in the proof of Theorem 26. Claim 1: I is a height two perfect ideal satisfying the G condition. Proof of Claim 1.
By Krull’s principal ideal theorem, it suffices to show that ( I , x ) is of height three. (We will use thistechnique multiple times in the proof.) Equivalently, we show that height I ( R / xR ) is two. By abuse of notation, weuse the same symbols for y , z for their images in R : = R / xR . Let M = y γ z q y z yz q − z y q yz . Notice that from the determinant after deleting the second row of M is γ yz q + , and the determinant after deleting thelast row of M is y q + − y z q + γ z q + . Hence any prime containing I ( M ) contains ( y , z ) . Since I ( M ) R = I ( M ) , thisshows that ht I ≥
2. Then by the Hilbert-Burch theorem [Eis95, Theorem 20.15], ht I = G condition, it suffices to show that ht F ( M ) = I ( M ) ≥ I ( M ) is of height 3 (indeed, it will turn out that ( x , y , z ) is the only prime containing I ( M ) ). Let p be aprime ideal containing I ( M ) . From the minor (of rows 2,3 and columns 1,3 of M ) (cid:12)(cid:12)(cid:12)(cid:12) y yz q − z y q (cid:12)(cid:12)(cid:12)(cid:12) = y q + − yz q = y q ( y − z ) ,we see that y or y − z is in p . If y ∈ p , then from the minors yx − z , x − zy (from the first two columns of M ), we seethat x , z are also in p . So, p = ( x , y , z ) . If y − z ∈ p , then from the minor y − xz , we have y − xy = y ( x − y ) is in p . If y is in p , then we are done by the previous argument. So, we assume that x − y ∈ p . Then from the minor x ( xy + yz + xz ) ,we see that x ( x ) = x is in p as x − y , y − z ∈ p . Hence x , y − x , z − x is in p , and this implies p = ( x , y , z ) . Thus, wehave ht I ( M ) =
3, and this completes the proof of claim 1. (cid:3)
Claim 2:
Let S = k [ T , T , T , T ] , B = R ⊗ k S = S [ x , y , z ] , and δ = + + q − = q . Then we have the following. ( a ) [ Sym R ( I )] is S -free of rank 1, and for 1 ≤ i ≤
2, the graded S -resolutions of [ Sym R ( I )] i are0 → S ( − ) α → S → [ Sym R ( I )] → → S ( − ) ⊕ S ( − ) α → S → [ Sym R ( I )] → , (18)respectively, and ht I ( α ) = ( b ) For 3 ≤ i ≤ δ −
1, the graded S -resolution of [ Sym R ( I )] i is0 → S ( + i − )( − ) α → S ( + i − )( − ) ⊕ S ( + i − )( − ) α → S ( + i ) → [ Sym R ( I )] i → , (19)and ht I ( α ) = I ( α ) = ( c ) The graded S -resolution of [ Sym R ( I )] δ = [ Sym R ( I )] q is0 → S ( + q − )( − ) α → S ( + q − )( − ) ⊕ S ( + q − )( − ) ⊕ S ( − ) α → S ( + q ) → [ Sym R ( I )] q → , and ht I ( α ) = Proof of claim 2.
The shape of the complexes follows from Remark 10. The rank of each free module in thesecomplexes can be calculated easily as they are the number of monomials in ( x , y , z ) of degree i , and the rank of [ Sym R ( I )] i = i = , , δ and 2 otherwise. By Theorems 15 and 16, to check the exactness of these complexes, itsuffices to verify the claimed heights of I ( α j ) (Theorem 4).We will use the notation [ l l l ] = [ T . . . T ] M . That is l = xT + yT + zT , l = y T + z T + x T + ( xy + yz + zx ) T , l = γ z q T + yz q − T + y q T . irst, we show that I ( α ) ≥ α is Φ − l i Φ l i . Each column consists of exactly 7 non-zero entries including T , T , T , T , and it satisfies the conditions of Lemma 25.Thus, we conclude that ht I ( α ) = i = α involves only l , and the map α = Φ l is the column matrix consists of T , T , T . Hence ht I ( α ) =
3. Notice that for 2 ≤ i ≤ δ − S -resolution in eqs. (18) and (19) involves only the elements l , l . To showthat ht I ( α ) ≥
3, by Krull’s principal ideal theorem, it suffices to show that ht I ( α ) + ( T ) ≥
4. Equivalently, we showthat ht I ( α )( S / ( T )) ≥
3. In fact, [ Sym R ( I )] i = [ Sym R ( J )] i for J = I ( M ′ ) , where M ′ = x y y z z x . So, we may assume that S = k [ T , T , T ] , M = M ′ , and I = J . We show that ht p I ( α ) =
3. Since [ l l ] =[ T T T ] M , and the shape of M , the matrices Φ l i , Φ l i is stable under the action by the cycle ( T T T ) . In other words,if f ( T , T , T ) ∈ I ( α ) , then f ( T , T , T ) , f ( T , T , T ) are also in I ( α ) . Thus, to show that ht p I ( α ) =
3, if sufficesto show that T ∈ p I ( α ) .Assume to the contrary T is not in p I ( α ) . Then p I ( α ) S T is a proper ideal of S T . Consider g = l g = A y + A yz + A z , where A = T + T T , A = T T , A = T T + T , and K ′ : = K ( g , g ; B ) . Even though ( g , g ) ( ( l , l ) in B ,these two ideals agree in B T . (In the localization at T , l = xT + yT + zT ⇐⇒ x = l − yT / T − zT / T , so we may re-place x by a combination of l , y , z , T , T , T .) In other words, the localization of [ K ′ ] i at T is also a free S T -resolutionfor ([ Sym R ( I )] i ) T . By Lemma 28 below, we conclude that ( A , A , A ) S T ∈ p I ( α ′ ) S T . Notice that ( A , A , A ) S T isa unit ideal in S T . Hence the Fitting lemma [Eis95, Cor-Def 20.4, Cor. 20.5] imples that I ( α ) B T = S T as well. Thisis a contradiction and completes the proof of ht ( α ) = I ( α ) = r = rank α . If T r ∈ I ( α ) , then modding out by ( T , T ) ,we see that the image of T r is in the image of I ( α ) in S / ( T , T ) . Hence we are done by Krull’s principal idealtheorem . Now, suppose that T p I ( α ) . We will show that ht p I ( α ) S T =
2. By the assumption, p I ( α ) S T = S T .As in the previous case, we will use g = l , g = A y + A yz + A z , g = l , where A = T + T T , A = T T , A = T T + T , and K ′ : = K ( g , g , g ; B ) . Then the Koszul complex K ′ indegree q is a free S T -resolution of [ Sym R ( I )] q . If we name α ′ the first differential of [ K ′ ] q , then I ( α ) S T = I ( α ′ ) S T since they are both the Fitting ideal of the same module. Therefore, we will show that ht I ( α ′ ) S T =
2. Notice that α ′ Computations using Macaulay2 [GS] for small q shows that this case does not happen, but this is necessary for our proof. s (cid:20) Φ g q Φ g q Φ g q (cid:21) = Φ g q z }| { T · · · T T · · · ∗ ∗ · · · T ∗ ∗ · · · ∗ ... ... . . . ...... ... . . . ...... ... . . . ...... ... . . . ... ∗ ∗ · · · ∗ | {z } ( + q − q − ) columns Φ g q z }| { ∗ · · · ∗∗ · · · ∗ ... ... ...... ... ...... ... ...... ... ...... ... ...... ... ...... ... ... ∗ · · · ∗ · · · · · · ...... ... · · · ...... ... · · · ... A · · · A A · · · A A . . . 0... ... ... ...0 0 · · · A · · · A | {z } ( + q − q − ) columns Φ g q z }| { T T γ T basis z }| { x q x q − y ... xz q − y q y q − z ...... yz q − z q . Since T is invertible in S T , I ( N ) ⊂ I ( α ′ ) S T , where N is the matrix obtained from the lower right corner of α ′ . N = A · · · A A · · · A A . . . 0 0... ... ... ... ...0 0 · · · A A · · · A T T γ T . We will show that after modifying N , ht I ( N ) ≥
2, so ht I ( α ′ ) S T = N andits variations N ′ , N ′′ , N T below will be matrices of rank j and size ( j + ) × j (with different j ).) Since T is invertiblein S T , we may multiply the last row by − T / γ T and add it to the first row. Also, we multiply the last row by − T / γ T and add it to the second last row. We call this new matrix N ′ N ′ = A · · · − A T / γ T A A · · · A A . . . 0 0 0... ... ... ... ... ...0 0 · · · A A − A T / γ T
00 0 · · · A γ T : = (cid:20) N ′′ ∗ γ T (cid:21) . Here N ′′ is the matrix obtained by deleting the last row and column of N ′ . It is straightforward to check that I ( N ′ ) = I ( N ′′ ) as T is a unit in S T . Consider the following matrix N T in S T [ T ] (here, T is a new variable) N T = A T · · · − A T / ( γ T ) A A · · · A A . . . 0 0... ... ... ... ...0 0 · · · A A · · · A A − A T / γ T nd let J T : = I ( N T ) . Since J ⊆ I ( α ′ ) S T [ T ] , ht ( J T ) < ∞ . We claim that ht J T ≥
2. We will prove this by showing that p ( J T , T ) = ( T , T , T ) is of height 3. Once we have proven the claim, then by Theorem 6, there exists a non-emptyopen subset U of Spec k [ T ] such that for any (closed) point γ ′ in U , ht J γ ′ ≥ ( J T , T ) ⊂ ( A , A , T ) since the ideal generated by A , A , T contains all the entries of the firstcolumn. Furthermore, if a prime p contains ( A , A , T ) = ( T T , T T + T , T ) , then T , T , T are in p . (Recall thatwe are in S T , so T is a unit.) Thus, we conclude that p ( A , A , T ) = ( T , T , T ) . Consider the minor obtained bydeleting the second row of N T and then putting the first row last (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A A · · · A A A · · · · · · A A A · · · A A − A T / γ T A T · · · − A T / ( γ T ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (20)From this minor, we conclude that A rank N T T / ( γ T ) is in ( J T , T ) . Therefore, if p is a prime ideal containing ( J T , T ) ,then it contains A or T . If A ∈ p , then from the minor obtained by deleting the first row of N T , we have A ∈ p . Inthis case, we have p ( J T , T ) = p ( A , A , T ) = ( T , T , T ) ⊂ p , so ( T , T , T ) = p since ( T , T , T ) is a prime of height3. We show that we have the same conclusion in the case where T ∈ p . Assume T ∈ p , then since A = T T , A is in p . Hence from the minor obtained by deleting the first row of N T and from the fact that T , T , A ∈ p , the determinant (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A · · · A A · · · A · · · · · · A A · · · A − A T / γ T (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) is in p . By Lemma 29, A q / A q / is in p if q is even, and A ( q − ) / A ( q − ) / ( − A T / γ T ) is in p if q is odd. Since T ∈ ( A , T ) ⊂ ( A , p ) is a unit ideal, A p . Therefore, in both cases we have A or T is in p . Since T is in p and T is a unit, A = T T + T is in p if and only if T is in p . Hence in both cases we have A ∈ p . We already showed that if A ∈ p , then p = ( T , T , T ) (in the paragraph before eq. (20)). Hence we have p ( J T , T ) = p ( A , A , T ) = ( T , T , T ) ,and this proves the claim.By Theorem 6, there exists a non zero γ ′ in k such that ht N γ ′ =
2. But the maximal minors of the following matrices N γ ′ = A γ ′ · · · − A T / ( γ T ) A A · · · A A . . . 0 0... ... ... ... ...0 0 · · · A A , N ′′ = A · · · − A T / ( γγ ′ T ) A A · · · A A . . . 0 0... ... ... ... ...0 0 · · · A A generate the same ideal. We replace γ by γγ ′ . This does not change our argument since the previous calculation andproof depended only on the fact that γ was not zero. Therefore, with new γ , we obtain the desired height for I ( N ) , andthis completes the proof. (cid:3) Lemma 28.
Let S = k [ T , T , T ] , where k is a field, B = S [ x , y , z ] , m = ( x , y , z ) B , g = T x + T y + T z , g = A y + A yz + A z with A i ∈ S. If g , g form an B-regular sequence, then for any i ≥ , the ith graded component (with espect to x , y , z degree) of B / ( g , g ) has a representation α of the form (cid:20) Φ g i Φ g i (cid:21) = Φ g i z }| { T · · · T T · · · ... ... . . . ∗ ∗ · · · T ∗ ∗ · · · ∗ ... ... . . . ...... ... . . . ...... ... . . . ... ∗ ∗ · · · ∗ | {z } ( + i − i − ) columns Φ g i z }| { ∗ · · · ∗∗ · · · ∗ ... ... ...... ... ...... ... ...... ... ...... ... ...... ... ... ∗ · · · ∗ · · · ... ... · · · ...... ... · · · ...... ... · · · ...A · · · A A · · · A A . . . ... ... ... ... · · · A | {z } ( + i − i − ) columns basis z }| { x i x i − y...xz i − y i y i − z......z i Furthermore, we have rk α = (cid:0) + i − (cid:1) + i − and T ( + i − ) ( A , A , A ) i − ⊂ I ( α ) .Proof. We use the polynomial grading on B and write M : = B / ( g , g ) . Since g , g form a regular sequence, theKoszul complex K : = K ( g , g ; B ) is a grade B -resolution of M . By Remark 10, K i is a free S -resolution of M i . Thatis 0 → S ( + i − ) α → S ( + i − ) ⊕ S ( + i − ) α → S ( + i ) → M i → M i = (cid:18) + i (cid:19) + (cid:18) + i − (cid:19) − (cid:18) + i − (cid:19) − (cid:18) + i − (cid:19) = (cid:20)(cid:18) + i (cid:19) − (cid:18) + i − (cid:19)(cid:21) − (cid:20)(cid:18) + i − (cid:19) − (cid:18) + i − (cid:19)(cid:21) = (cid:18) + i − (cid:19) − (cid:18) + i − (cid:19) = , and rk α = (cid:0) + i (cid:1) −
2. The presentation follows by using the Lex order on the monomials (in terms of x , y , z ) of B .Since (cid:18) + i − (cid:19) + dim k ( y , z ) i − = ( i + ) i / + ( i − ) = ( i + i ) / − = ( i + i + ) / − = ( i + )( i + ) / − = (cid:18) + i (cid:19) − = rank α , we see that the T ( + i − ) I ( N ) ⊂ I ( α ) , where N = A · · · A A · · · A A · · · · · · A . By Lemma 25, I ( N ) = ( A , A , A ) i − , and this proves the statement. (cid:3) emma 29. Let x , y , z be variables over Z . Then we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x · · · y x · · · y x · · · ... ... ... ... ... ... · · · y x · · · y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( ( − ) n x n y n , i f n is even , i f n is odd and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x · · · y x · · · y x · · · ... ... ... ... ... ... · · · y x · · · y z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( ( − ) n x n y n , i f n is even ( − ) n − x n − y n − z , i f n is odd . Proof.
For the first determinant, one can use induction on the size of the matrix, and for the second determinant, onecan use the Laplace expansion along the last row (or column) and use the result of the first determinant. (cid:3)
Remark 30.
The non-empty open subset for the conditions in Theorem 26 is a proper subset of the parameter space. The followingmatrix provides an example where I : = I ( M ) satisfies condition (1) in the proof of Theorem 26, but not condition (2). M = x x y x y z z . It is interesting that the rational map induced by I is birational to its image.We will end the paper with a couple of questions. Question 31. (1) Let U be the non-empty open subset corresponds to the general condition in Theorem 26. Canwe determine the closed set which is the complement of U ? In the case where R = k [ x , x ] , Kustin, Polini,Ulrich were able to connect this condition to a geometric condition (See [KPU17, Lemma 2.10]).(2) In the general case where R = k [ x , . . . , x n ] , does the following presentation matrix (or its variations) of I = I n ( ϕ ) ϕ = x d x d · · · x d n x d x d · · · x d n ... ... . . . ... x d n x d n n · · · x d n n g · · · · · · g n , where g i are symmetric polynomials of degree d i , provide an example of defining a similar general propertydescribed in Theorem 26?Note that the corresponding presentation matrices for the graded pieces of Sym R ( I ) are invariant underthe action of the cyclic group generated by the cycle ( T , . . . , T n − ) . We hope an expert can utilize this factand provide an alternative way to verify Example 27. Recall that the more general ϕ becomes, the morechallenging to verify conditions. R EFERENCES[AB69] Maurice Auslander and Mark Bridger.
Stable module theory . Memoirs of the American Mathematical Society, No. 94. American Math-ematical Society, Providence, R.I., 1969.[BCJ09] Laurent Bus´e, Marc Chardin, and Jean-Pierre Jouanolou. Torsion of the symmetric algebra and implicitization.
Proc. Amer. Math. Soc. ,137(6):1855–1865, 2009.[BE73] David A. Buchsbaum and David Eisenbud. What makes a complex exact?
J. Algebra , 25:259–268, 1973.[BE74] David A. Buchsbaum and David Eisenbud. Some structure theorems for finite free resolutions.
Advances in Math. , 12:84–139, 1974.[BE77] David A. Buchsbaum and David Eisenbud. What annihilates a module?
J. Algebra , 47(2):231–243, 1977.[BH93] Winfried Bruns and J¨urgen Herzog.
Cohen-Macaulay rings , volume 39 of
Cambridge Studies in Advanced Mathematics . CambridgeUniversity Press, Cambridge, 1993.[BJ03] Laurent Bus´e and Jean-Pierre Jouanolou. On the closed image of a rational map and the implicitization problem.
J. Algebra , 265(1):312–357, 2003.[BM16] Jacob A. Boswell and Vivek Mukundan. Rees algebras and almost linearly presented ideals.
J. Algebra , 460:102–127, 2016. Bus09] Laurent Bus´e. On the equations of the moving curve ideal of a rational algebraic plane curve.
J. Algebra , 321(8):2317–2344, 2009.[CBtD13] Teresa Cortadellas Ben´ı tez and Carlos D’Andrea. Rational plane curves parameterizable by conics.
J. Algebra , 373:453–480, 2013.[CBtD14] Teresa Cortadellas Ben´ı tez and Carlos D’Andrea. Minimal generators of the defining ideal of the Rees algebra associated with a rationalplane parametrization with µ = Canad. J. Math. , 66(6):1225–1249, 2014.[CHW08] David Cox, J. William Hoffman, and Haohao Wang. Syzygies and the Rees algebra.
J. Pure Appl. Algebra , 212(7):1787–1796, 2008.[DHS12] A.V. Doria, S.H. Hassanzadeh, and A. Simis. A characteristic-free criterion of birationality.
Advances in Mathematics , 230(1):390 – 413,2012.[Eis95] David Eisenbud.
Commutative algebra , volume 150 of
Graduate Texts in Mathematics . Springer-Verlag, New York, 1995. With a viewtoward algebraic geometry.[Eis05] David Eisenbud.
The geometry of syzygies , volume 229 of
Graduate Texts in Mathematics . Springer-Verlag, New York, 2005. A secondcourse in commutative algebra and algebraic geometry.[GS] Daniel R. Grayson and Michael E. Stillman. Macaulay2, a software system for research in algebraic geometry. Available at https://faculty.math.illinois.edu/Macaulay2/ .[Har77] Robin Hartshorne.
Algebraic geometry . Springer-Verlag, New York-Heidelberg, 1977. Graduate Texts in Mathematics, No. 52.[Har92] Joe Harris.
Algebraic geometry , volume 133 of
Graduate Texts in Mathematics . Springer-Verlag, New York, 1992. A first course.[HS06] Craig Huneke and Irena Swanson.
Integral closure of ideals, rings, and modules , volume 336 of
London Mathematical Society LectureNote Series . Cambridge University Press, Cambridge, 2006.[HSV82] J. Herzog, A. Simis, and W. V. Vasconcelos. Approximation complexes of blowing-up rings.
J. Algebra , 74(2):466–493, 1982.[HSV83] J. Herzog, A. Simis, and W. V. Vasconcelos. Koszul homology and blowing-up rings. In
Commutative algebra (Trento, 1981) , volume 84of
Lecture Notes in Pure and Appl. Math. , pages 79–169. Dekker, New York, 1983.[HSV08] Jooyoun Hong, Aron Simis, and Wolmer V. Vasconcelos. On the homology of two-dimensional elimination.
J. Symbolic Comput. ,43(4):275–292, 2008.[Hun83] Craig Huneke. Strongly Cohen-Macaulay schemes and residual intersections.
Trans. Amer. Math. Soc. , 277(2):739–763, 1983.[Jou96] Jean-Pierre Jouanolou. R´esultant anisotrope, compl´ements et applications.
Electron. J. Combin. , 3(2):Research Paper 2, approx. 91 pp.(electronic), 1996. The Foata Festschrift.[KPU11] Andrew R. Kustin, Claudia Polini, and Bernd Ulrich. Rational normal scrolls and the defining equations of Rees algebras.
J. ReineAngew. Math. , 650:23–65, 2011.[KPU16] Andrew R. Kustin, Claudia Polini, and Bernd Ulrich. Blowups and fibers of morphisms.
Nagoya Math. J. , 224(1):168–201, 2016.[KPU17] Andrew Kustin, Claudia Polini, and Bernd Ulrich. The bi-graded structure of symmetric algebras with applications to Rees rings.
J.Algebra , 469:188–250, 2017.[Lan14] Nguyen P. H. Lan. On Rees algebras of linearly presented ideals.
J. Algebra , 420:186–200, 2014.[Mad15] Jeff Madsen. Equations of rees algebras of ideals in two variables, 2015.[Mor96] Susan Morey. Equations of blowups of ideals of codimension two and three.
J. Pure Appl. Algebra , 109(2):197–211, 1996.[MU96] Susan Morey and Bernd Ulrich. Rees algebras of ideals with low codimension.
Proc. Amer. Math. Soc. , 124(12):3653–3661, 1996.[PX13] Claudia Polini and Yu Xie. j -multiplicity and depth of associated graded modules. J. Algebra , 379:31–49, 2013.[T`02] H`a Huy T`ai. On the Rees algebra of certain codimension two perfect ideals.
Manuscripta Math. , 107(4):479–501, 2002.[Ulr94] Bernd Ulrich. Artin-Nagata properties and reductions of ideals. In
Commutative algebra: syzygies, multiplicities, and birational algebra(South Hadley, MA, 1992) , volume 159 of
Contemp. Math. , pages 373–400. Amer. Math. Soc., Providence, RI, 1994.[UV93] Bernd Ulrich and Wolmer V. Vasconcelos. The equations of Rees algebras of ideals with linear presentation.
Math. Z. , 214(1):79–92,1993.D
EPARTMENT OF M ATHEMATICAL S CIENCES , U
NIVERSITY OF A RKANSAS , F
AYETTEVILLE , A
RKANSAS
E-mail address : [email protected] D EPARTMENT OF M ATHEMATICS , U
NIVERSITY OF V IRGINIA , C
HARLOTTESVILLE , V
IRGINIA
E-mail address : [email protected]@virginia.edu