On the Jacobian ideal of central arrangements
aa r X i v : . [ m a t h . A C ] J a n ON THE JACOBIAN IDEAL OF CENTRAL ARRANGEMENTS
RICARDO BURITY, ARON SIMIS AND S¸TEFAN O. TOH ˇANEANUA
BSTRACT . Let A denote a central hyperplane arrangement of rank n in affine space K n over aninfinite field K and let l , . . . , l m ∈ R := K [ x , . . . , x n ] denote the linear forms defining the corre-sponding hyperplanes, along with the corresponding defining polynomial f := l · · · l m ∈ R . Let J f denote the ideal generated by the partial derivatives of f and let I designate the ideal generated by the ( m − -fold products of l , . . . , l m . This paper is centered on the relationship between the two ideals J f , I ⊂ R , their properties and two conjectures related to them. Some parallel results are obtained inthe case of forms of higher degrees provided they fulfill a certain transversality requirement. I NTRODUCTION
Let A denote a central hyperplane arrangement of rank n in affine space K n over an infinite field K and let l , . . . , l m ∈ R := K [ x , . . . , x n ] denote the linear forms defining the correspondinghyperplanes. Set f := l · · · l m ∈ R , the defining polynomial of A . The module of logarithmicderivations associated to f is defined as Derlog( A ) := { θ ∈ Der( R ) | θ ( f ) ⊂ h f i} . Its R -modulestructure is well known if char( K ) does not divide m = deg( f ) : Derlog( A ) = Syz( J f ) ⊕ Rθ E . Here
Syz( J f ) ⊂ R n is up to the identification Der( R ) = R n the syzygy module of the partialderivatives of the f and θ E = P ni =1 x i ∂∂x i is the Euler derivation.In the case where the arrangement is generic, – meaning that every subset of { l , . . . , l m } with n elements is K -linearly independent – and that m ≥ n + 1 , Rose-Terao ([9]) and Yuzvinsky ([16])have established that the homological dimension of Derlog( A ) is n − . The statement is equivalentto having depth( R/J f ) = 0 , or still that the irrelevant maximal ideal m := h x , . . . , x n i is anassociated prime of R/J f . In their beautiful paper, Rose and Terao took the approach of establishingexplicit free resolutions of the modules Ω q ( A ) of logarithmic q -forms, for ≤ q ≤ n − , in thespirit of Lebelt’s work ([5]). Then, drawing on the isomorphism Derlog( A ) ≃ Ω n − ( A ) , theyderived the free resolution of the module of logarithmic derivations, and hence of R/J f as well.Motivated by this, the present authors consider the problem under a more intrinsic perspective,trying to envisage what can be said, avoiding details of free resolutions, by drawing upon a few wellestablished facts coming from commutative algebra. As such, it seemed natural to ask if there is anyresult of this sort in the case of forms f , . . . , f m ∈ R, m ≥ , of degrees deg( f i ) = d i ≥ , i =1 , . . . , m . Accordingly, the paper is focused on two major aspects: first, the nature of J f from theideal theoretic and homological point of view, for a central generic hyperplane arrangement; second,the nature of J f for arrangements of forms that are isolated singularities. Mathematics Subject Classification.
Primary 13A30; Secondary 13C15, 14N20, 52C35.
Key words and phrases.
Jacobian ideal, arrangement of hypersurfaces, depth, reduction of an ideal.The first author was partially supported by CAPES (Brazil) (grant: PVEX - 88881.336678/2019-01).The second author was partially supported by a grant from CNPq (Brazil) (302298/2014-2).Burity’s address: Departamento de Matem´atica, Universidade Federal da Paraiba, J. Pessoa, Paraiba, 58051-900,Brazil, Email: [email protected]’ address: Departamento de Matem´atica, Universidade Federal da Pernambuco, Recife, Pernambuco, 50740-560, Brazil, Email: [email protected]’s address: Department of Mathematics, University of Idaho, Moscow, Idaho 83844-1103, USA, Email:[email protected].
A full grasp of the algebraic properties of a central hyperplane arrangement does not seem to beavailable. As is well-known to experts, even in the plane case a question as to how the combinato-rial side affects the homological properties of the Jacobian ideal of the defining polynomial of thearrangement is not entirely settled. The best example is Terao’s conjecture which states that, over afield of characteristic zero, freeness of
Derlog( A ) depends only on the combinatorics.For a central hyperplane arrangement A of rank n over an infinite field K one introduces the ideal I ⊂ R generated by the ( m − − fold products of the defining linear forms of A . While workingon the above problem, we came up with the following conjecture, previously unknown at least tous:C ONJECTURE
1. ( char( K ) ∤ m ) Assume that A is a central hyperplane arrangement of rank n and size m . Let I ⊂ R = K [ x , . . . , x n ] denote the ideal generated by the ( m − -fold products of l , . . . , l m . Then the ideal J f ⊂ R generated by the partial derivatives of f is a minimal reductionof I .Since A has rank n , one has m ≥ n . The case where m = n is clear because J f = I since theyhave the same number of minimal generators, and the analytic spread of I is n .We answer affirmatively this conjecture assuming that every subset of { l , . . . , l m } with n − elements is K -linearly independent – this includes the generic case and the arbitrary n = 3 case.While focusing on this conjecture, we felt naturally compelled to ask about further behaviorpattern by the Jacobian ideal of a central hyperplane arrangement. We thus came up with a secondquestion:C ONJECTURE
2. ( char( K ) ∤ m ) Let f ∈ R = K [ x , . . . , x n ] denote the defining polynomial ofa central hyperplane arrangement of size m ≥ n . Then the Jacobian ideal J f is of linear type.The conjecture trivially holds for n = 2 . In this work we show that this conjecture holds incomplete generality for n = 3 . In the generic case and arbitrary n ≥ , we show that J f at leastsatisfies condion G ∞ (also denoted F ), which is the so to say easier half of the linear type property.We now describe the main contents of the paper.Under the assumption that A is generic and m ≥ n + 1 , the first section deals with the idealtheoretic and homological properties of the Jacobian ideal J f . We take a kind of ad hoc approach,by closely establishing the algebraic properties of J f and some of the invariants attached to it. Fromthese, we derive that depth R/J f = 0 and that its minimal free resolution is -step linear withsyzygy module equigenerated in degree m − n + 1 . Among those properties, we give a priori theexact value of the initial degree of Syz ( J f ) , the regularity and the saturation exponent of J f . When n ≥ we show that the saturation of J f with respect to m is the ideal I generated by the ( m − -foldproducts of f .The second section looks more closely at the inclusion J f ⊂ I . We already know that these idealsshare the same radical, but wish to go further to obtain that this inclusion is a minimal reductionsituation, which is the content of the first of the above conjectures. By drawing upon a couple ofmethods, we verify this conjecture in a first few cases, including the case where n = 2 and the casewhere n = 3 with a coloop. We also describe an effective partition method.Unfortunately, none of this is directly applicable to the arbitrary generic case, which requiresa separate argument. This is then the main result of the section. The proof is a kind of descentargument involving the various types of natural generators of the critical power I n . The reason oneaims at this power is because n − is an upper bound for the reduction number of I ([3, Corollary 2.6(d)]), the later being universally well-defined since the special fiber of I (the Orlik–Terao algebra)is Cohen–Macaulay by [7]).The section ends with the announced results on the linear type property of the Jacobian ideal J f that include an affirmative answer to the second of the above conjectures when n = 3 . As aconsequence we obtain a characterization of freeness in rank in terms of the Rees algebra of thecorresponding Jacobian ideal. N THE JACOBIAN IDEAL OF CENTRAL ARRANGEMENTS 3
The third section is largely experimental as it stands. As it aims at arrangement of forms ofdegrees ≥ , in whatever analogy exists to the hyperplane case, we introduce conditions on theseforms that mimic some of the properties of the latter. We have chosen to work with a couple of con-ditions, under the designation of near transversality . This includes assuming that every individualform f i is such that Proj ( R/ ( f i )) is smooth, a sort of extravagant requirement to mimic the fact thata straight line is smooth. It also includes the requirement that the forms be mutually non-tangent,a condition that we chose to encode via a Jacobian matrix in relation to certain primes. Since thedegrees of the forms are quite arbitrary, there is no apparent obstruction to assume only that m ≥ .The case of two forms is already sufficiently appealing, so we prove a main result under the neartransversality assumption (or close to this). For m arbitrary, we also state a lower bound for theinitial degree of the Jacobian ideal of the product of the forms. Finally, under a special condition onthe degrees of the forms we are able to obtain the analog of Rose–Terao–Yuzvinsky theorem.1. T HE J ACOBIAN IDEAL OF A HYPERPLANE ARRANGEMENT
The initial degree of the syzygies.
Let A denote a central hyperplane arrangement in affinespace K n over an infinite field K . Although the notions make sense in positive characteristic, theuse of the Euler relation in the main results require that char ( K ) does not divide the degree of thedefining polynomial of the arrangement, a hypothesis that we take for granted throughout refrainingfrom stating it on spot.For our purpose it will be convenient to think of A as being given by mutually independent linearforms l , . . . , l m in the polynomial ring R := K [ x , . . . , x n ] . We assume throughout that n ≥ and m ≥ n + 1 .The arrangement is said to be generic if any n of the defining linear forms are K -linearly in-dependent. Clearly, an arrangement for which the coefficients of its forms are randomly picked isautomatically generic and this conveys a fast way of producing generic arrangements.Let f := l · · · l m ∈ R be the defining polynomial of the arrangement. We will closely focuson the Jacobian (or gradient) ideal J f ⊂ R of f , i.e., the ideal generated by the partial derivatives f x i := ∂f /∂x i (1 ≤ i ≤ n ) of f . By abuse, we let Syz( J f ) denote the module of first syzygies of { f x , . . . , f x n } .Fixing a linear form l = l i of the arrangement, one has the well-known notions of deletion A \ { H } and restriction A H of A with respect to a given hyperplane H = V ( l i ) . In algebraic terms,deletion is just the arrangement of the forms l , . . . , l i − , b l i , l i +1 , . . . , l m in the same polynomial ring R . Restriction is the arrangement of the residual forms {h l i , l j i / h l i i | j = i } in the residual ring R/ h l i i , upon identification of the latter with a polynomial ring over K in n − variables.By this description, it is immediate to verify that if A is a generic arrangement, then so are therespective deletion and restriction arrangements with respect to any of the defining linear forms.Borrowing from the notation of [1], we set r ( A ) := indeg(Syz( J f )) for the corresponding initialdegree, i.e., r ( A ) := min r ∈ Z { r | (Syz( J f )) r = 0 } . We will make use of the following theorem:
Theorem 1.1. ([1, Theorem 2.14])
Let l = l i be a linear form defining the hyperplane H of acentral arrangement A . If r ( A \ { H } ) < r ( A H ) , where H = V ( l ) , then r ( A ) = r ( A \ { H } ) + 1 . Though [1] assumes char( K ) = 0 throughout, it is rather clear that the arguments only requiresthat char( K ) does not divide m = deg( f ) .It is proved in [9, Corollary 4.4.3] that, in the generic case with m ≥ n + 1 , the syzygy module of J f is equigenerated in degree m − n + 1 . Of course, this comes out as a conflagration of introducingvarious chain complexes, none of which is trivial. As it turns, one can walk a good mile just by RICARDO BURITY, ARON SIMIS AND S¸TEFAN O. TOH ˇANEANU knowing that the initial degree of Syz ( J f ) is m − n + 1 . For this reason, and also for the sake ofcompleteness, we state and prove the following: Proposition 1.2.
Let A denote a generic central hyperplane arrangement of m linear forms inaffine space K n , satisfying m ≥ n + 1 . Then r ( A ) = m − n + 1 . Proof.
By the discussions in the Appendix, one can assume that the arrangement is now { x , . . . , x n , l , . . . , l m − n } ⊂ R := K [ x , . . . , x n ] , for suitable linear forms l , . . . , l m − n . In par-ticular, the updated defining polynomial is f := x · · · x n l · · · l m − n . Starting Case: m = n + 1 . Here, say, f = x · · · x n l, with l = α x + · · · + α n x n , where α , . . . , α n ∈ K \ { } since the arrangement is generic. As easily seen, one has(1) f x i = x · · · b x i · · · x n ( α i x i + l ) . Therefore, two indices i = j yield an obvious reduced Koszul relation of degree . Thus, it sufficesto show that the partial derivatives have no syzygy of degree ≤ .C LAIM { f x , . . . , f x n } is linearly independent over K .This is a far more general result, but in the present case it follows simply because if one hasscalars λ i ∈ K , not all zero, such that λ ( x · · · x n l + x · · · x n α ) + · · · + λ n ( x · · · x n − l + x · · · x n α n ) = 0 then a scalar λ ∈ K comes out such that λ x · · · x n = X i λ i x · · · b x i · · · x n ! l. Now, either λ = 0 , in which case the parenthetical content would have to vanish; since some λ i = 0 by assumption, this would say that the partial derivatives of x · · · x n are linearly dependent, anabsurd since, as is well-known, they are even algebraically independent over K . We are left with theoption that λ = 0 , and hence x i would have to be a factor of l , because λ i = 0 so x i cannot dividethe parenthetical summation.So much for this claim.C LAIM { f x , . . . , f x n } admits no linear syzygy.Suppose that L , . . . , L n ∈ R are linear forms, not all zero, such that L f x + · · · + L n f x n = 0 . Then,(2) ( L x · · · x n + · · · + L n x · · · x n − ) l = − x · · · x n ( α L + · · · + α n L n ) . There are two cases to have in mind:
First case : α L + · · · + α n L n = 0 .This implies that ( L , . . . , L n ) t is a linear syzygy of the Boole arrangement of equation x · · · x n .But the syzygies of the latter are well-known to be given by the Hilbert–Burch matrix x · · ·− x x · · · − x x · · · − x · · · ... ... ... · · · Writing up this relation yields L = β x , L = − ( β + β ) x , . . . , L i = ± ( β i − + β i ) x i , . . . , L n = − β n − x n , N THE JACOBIAN IDEAL OF CENTRAL ARRANGEMENTS 5 which, when plugged back into α L + · · · + α n L n = 0 , implies a relation α β x − α ( β + β ) x + · · · + α n − ( β n − + β n − ) x n − − α n β n − x n = 0 . Since some L i = 0 , the corresponding β -like coefficient is nonzero. But α i = 0 for any i . Thiswould give a K -linear dependence relation for the variables. Second case : L := α L + · · · + α n L n = 0 .In this case, L must divide one of the factors on the left side of the relation (2). If L divides theparenthetical summation, after canceling, one gets an equality P l = x · · · x n , for some polynomial P ∈ R . Since no x i divides l by the generic shape of l , this is an absurd because of the UFD property.Therefore, we must have, say, L = βl , for some nonzero β ∈ K . Moreover, upon substitution in (2and cancellation, one obtains L x · · · x n + · · · + L n x · · · x n − = − βx · · · x n . This means that L , . . . , L n are the coordinates of a derivation of R/ h x · · · x n i , hence L i = β i x i ,for certain β i ∈ K . Canceling x · · · x n yields β + · · · + β n = − β , which in turns yields the relation α β x + · · · + α n β n x n = − L = − βl and hence, a syzygy α ( β + β ) x + · · · α n ( β + β n ) x n = 0 with scalar coefficients. Since no α i vanishes, every term β + β i must. Together with the relation β + · · · + β n = − β this easily leads to β i = 0 for all i , hence also L i = 0 for all i , as was to beshown. Inductive Step : m ≥ n + 2 .We will induct on m . Recall that both the deletion and the restriction are still generic ar-rangements, and m drops on either. Take both with respect to l = x . Thus, r ( A \ { H } ) =( m − − n + 1 = m − n by the inductive hypothesis. Likewise, the defining polynomialassociated to the restriction A H is e f = x · · · x n e l · · · ] l m − n , with e l i = l i | x =0 , and one has r ( A H ) = ( m − − ( n −
1) + 1 = m − n + 1 by the inductive hypothesis. Consequently, r ( A \ { H } ) < r ( A H ) . Then, by Theorem 1.1 we have r ( A ) = r ( A \ { H } ) + 1 = m − n + 1 . (cid:3) The Rose–Terao–Yuzvinsky theorem under an algebraic angle.
We keep the notation of theprevious section, under the assumption that m does not divide char( K ) , a proviso we will refrainfrom repeating throughout.Given the defining polynomial f = l · · · l m ∈ R of a generic central arrangement A , undera slight danger of confusion, we set f i := f /l i (i.e., f i is not the partial derivative f x i as mightsometimes be the chosen notation). Consider the ideal I = I ( A ) := h f , . . . , f m i = h l · · · l m , . . . , l · · · l m − i ⊂ R, generated by these ( m − -fold products of the linear forms l , . . . , l m . Note that this is a minimalset of generators of I (see, e.g., [3, Lemma 3.1 (a)]). Theorem 1.3.
Let f = l l · · · l m as above denote the defining polynomial of a generic centralarrangement in R = K [ x , . . . , x n ] with m ≥ n + 1 . Then: (i) [ I ] m − n − = [ J f ] m − n − . (ii) If n ≥ then ( J f ) sat = I . In particular, J f and I coincide locally everywhere on thepunctured spectrum Spec ( R ) \ { m } . (iii) reg( J f ) = sat( J f ) = 2 m − n − , where sat( I ) denotes the saturation exponent of an ideal I ⊂ R ..Proof. Let I := h l · · · l m , . . . , l · · · l m − i be the ideal generated by all ( m − -fold products ofthe linear forms l , . . . , l m . Clearly, I has height and, as one easily verifies, J f ⊂ I . Therefore, J f has height at most and, since f is a reduced form, the height is exactly (actually, it can be seenthat J f and I have the same minimal primes, all of codimension ). RICARDO BURITY, ARON SIMIS AND S¸TEFAN O. TOH ˇANEANU (i) One inclusion being obvious, it suffices to show that [ I ] m − n − ⊂ [ J f ] m − n − . Now, let P ∈ I , with deg( P ) = 2 m − n − . Write(3) P = ( l · · · l m ) P + · · · + ( l · · · l m − ) P m , for certain forms P i ∈ [ m m − n ] m − n , where m = ( x , . . . , x n ) .C LAIM
1. ( A generic) Let I m − n ( A ) denote the ideal of R generated by all ( m − n ) -fold productsof the given linear forms l , . . . , l m . Then I m − n ( A ) = m m − n .The proof is by induction on m . It is quite obvious for m = n + 1 . Thus, assume that m ≥ n + 2 and, for every ≤ i ≤ m , consider the deletion A i := A \ l i . By the inductive hypothesis, I m − − n ( A i ) = m m − − n , hence l i m m − − n = l i I m − − n ( A i ) ⊂ I m − n ( A ) , for all i . But since { l , . . . , l n } generates m then m m − n = h l , . . . , l n i m m − − n , so we are through..(A more conceptual argument follows by noting that, since the linear forms are generic, the linearcode dual to l , . . . , l m has minimum distance m − n + 1 , hence the assertion follows from [15,Theorem 3.1]).Thus, for each i = 1 , . . . , m , write P i = X ≤ j < ··· 2. The (2 m − n − -form ∆ := ( l · · · l m − n ) l m − n +1 · · · l m − belongs to J f .The proof is again by induction on m ≥ n + 1 .If m = n + 1 , ∆ = l l · · · l n ∈ J f . As in the proof of Proposition 1.2, up to a change ofcoordinates we can assume that l i = x i for i = 1 , . . . , n and l n +1 = ℓ = α x + · · · + α n x n , with α i ∈ K \ { } . Thus, f = x · · · x n ℓ and f x = x · · · x n ℓ + x x · · · x n α by (1). Clearly then, x x · · · x n = 1 α ( x f x − f ) ∈ J f . Let now m ≥ n + 2 . Write ∆ = ( l ) ( l · · · l m − n ) l m − n +1 · · · l m − | {z } ∆ ′ . By the inductive hypothesis, ∆ ′ ∈ J f ′ , where f ′ = l · · · l m . Write ∆ ′ = Q f ′ x + · · · + Q n f ′ x n , forcertain Q i ∈ R . But, for i = 1 , . . . , n , one has f x i = ( l ) x i f ′ + l f ′ x i , and therefore ( l ) f ′ x i ∈ J f for i = 1 , . . . , n . But then, ∆ = l ∆ ′ ∈ J f , as was to be shown.(ii) This follows from (i) since any form in [ m m − n ] m − n multiplies any f /l i into [ I ] m − n − , thusgiving I ⊂ ( J f ) sat . The reverse inclusion is obvious since I is saturated, being a codimension twoperfect ideal and n ≥ .(iii) As mentioned, sat( I ) denotes the saturation exponent (or the satiety) of an ideal I ⊂ R .C LAIM . reg( J f ) = sat( J f ) . It is well known that reg( J f ) = max { sat( J f ) , reg( J sat f ) } . N THE JACOBIAN IDEAL OF CENTRAL ARRANGEMENTS 7 By Proposition 1.2, at least reg( J f ) ≥ m − n − > m − , since m > n . On the other hand,assume that n ≥ . Then reg( J sat f ) = reg( I ) by (ii). But, since I is a linearly presented codimensiontwo perfect ideal, reg( I ) = m − . This implies the claim if n ≥ . If n = 2 then J f is m -primary,hence J sat f = (1) , whose regularity is zero by convention. Thus, we are home in this case too.Suppose sat( J f ) ≥ m − n . Then, by [2, Proposition 2.1], there exists a minimal generator h ∈ J f : m \ J f , of degree d := sat( J f ) − . Thus, d ≥ m − n − .But ( J f : m ) ⊂ ( J f ) sat = I , so h = ( l · · · l m ) P + ... + ( l · · · l m − ) P m , for some P j ∈ R ofdegree d − m + 1 ≥ m − n . Thus, each P j is a polynomial combination of monomials of degree m − n . An analogous argument to the proof of Claim 2 in the item (i) yields h ∈ J f , contradictingour choice of h . Therefore, sat( J f ) ≤ m − n − , hence reg( J f ) = 2 m − n − . (cid:3) Let f := l · · · l m ∈ R := K [ x , . . . , x n ] be the defining polynomial of a central hyper-plane arrangement A in K n . The module of logarithmic derivations associated to f is definedas Derlog( A ) := { θ ∈ Der( R ) | θ ( f ) ⊂ h f i} . If char( K ) does not divide m = deg( f ) , upon theidentification Der( R ) = R n this module has a splitting structure(4) Derlog( A ) = Syz( J f ) ⊕ Rθ E , where J f ⊂ R denotes the Jacobian ideal of f and θ E is the Euler derivation. Here, as discussedearlier, Syz( J f ) ⊂ R n denotes the module of first syzygies of the partial derivatives of f .As a consequence of Theorem 1.3, we now retrieve part of the findings of Rose–Terao andYuzvinsky. Theorem 1.4. Let f = l l · · · l m as above denote the defining polynomial of a generic centralarrangement in R = K [ x , . . . , x n ] with m ≥ n + 1 . Then (i) depth( R/J f ) = 0 . (ii) The minimal graded free resolution of J f has the following shifts → R b n ( − (2 m − → R b n − ( − (2 m − → · · · → R b ( − (2 m − n )) → R n ( − ( m − . Proof. (i) The assertion is trivial for n = 2 since J f is m -primary. Thus, assume that n ≥ andrecall that J f has height two.Take a reduced primary decomposition J = J un ∩ T i N i , where J un is the unmixed part of J and each N i is a primary component of codimension at least .If we assume that depth ( R/J ) > , then ≤ codim N i < n , for every i . Therefore, byTheorem 1.3 (ii), I = J : m ∞ = ( J un : m ∞ ) ∩ T i ( N i : m ∞ ) = J un ∩ T i N i . Therefore, J = I ,which is nonsense since the minimal number of generators of I is at leat n + 1 .(ii) By definition, reg( J f ) is the maximum value β k − k , where β k is the maximum shift at step k in the minimal graded free resolution of J f . By Proposition 1.2, the minimum shift at step k = 1 is r ( A ) + ( m − 1) = 2 m − n , which equals reg( J f ) + 1 by Theorem 1.3 (iii). But the maximum shiftat each step must increase by at least one and the homological dimension of R/J f over R is n byitem (i). This implies the claimed result by a standard argument (see, e.g., [6, Corollary 9 (2)]). (cid:3) Remark 1.5. (1) Proposition 1.2, Theorem 1.3 and Theorem 1.4 depend strongly on the hypothe-ses that m ≥ n + 1 and that the arrangement be generic. The failure is pretty bad if m ≤ n ,while genericity cannot be replaced by assuming, e.g., that all subsets of n linear forms, with oneexception, are linearly independent. For that, consider the following simple example in K [ x, y, z ] : A : { x, y, z, x + y } , a free arrangement (i.e., J f is a perfect ideal); here, the initial degree of Syz ( J f ) is < m − n + 1 .(2) In [9] the ground field K is assumed to be of characteristic zero throughout, but the authorsremark at the end (Section 4.6) that the result is valid provided char( K ) does not divide m = RICARDO BURITY, ARON SIMIS AND S¸TEFAN O. TOH ˇANEANU deg( f ) . As to our approach, the vast majority of the above results crumbles down without thisproviso. A simplest example is as follows: let R = K [ x, y, z ] , where char( K ) = 2 and let A = { x, y, z, x + y + z } . Clearly, A is generic.Calculation with [4] yields: • f J f . • J f is perfect; in particular, J sat f = J f = I . • The free resolution of J f has the shape → R ( − ⊕ R ( − → R ( − → J f → ; inparticular, J f has a syzygy of degree < m − n + 1 – in fact, is the degree of aminimal syzygy of maximal degree, which happens to be given by a reduced Koszul syzygy. • Derlog ( A ) does not split as in (4); in fact, it admits many generators of degree . • Derlog ( A ) has homological dimension , while Syz ( J f ) is a free module. • J f is not a reduction of I since J f : I ∞ = (1) .We note that characteristic is not the issue, as similar examples are easily handled in any positivecharacteristic. 2. T HE J ACOBIAN IDEAL AS A MINIMAL REDUCTION Approach via the Orlik–Terao algebra. Consider the Orlik-Terao algebra of A , which asknown is graded-isomorphic to the special fiber F ( I ) of I (see [3]). Let F ( I ) ≃ K [ T , . . . , T m ] /Q denote a presentation.2.1.1. Rank plus one. The following lemma is an adaptation to the ideal J f of a well-known fibercriterion for reductions (see, e.g., [14]). Let a i,j := ( l j ) x i := ∂l j /∂x i be the x i -coefficient of l j ,and let T , . . . , T m be presentation variables as above. Lemma 2.1. The following are equivalent: (1) J f is a minimal reduction of I . (2) The set { P mj =1 a ,j T j , . . . , P mj =1 a n,j T j } is a regular sequence on F ( I ) . (3) The ideal h P mj =1 a ,j T j , . . . , P mj =1 a n,j T j , Q i ⊂ K [ T , . . . , T m ] is h T , . . . , T m i -primary. With this at hand, one can easily handle the case where m = n + 1 . Lemma 2.2. Let A = { l , . . . , l n +1 } be a central arrangement of rank n . Let f = l · · · l n +1 and I be as before. Then J f is a minimal reduction of I .Proof. As done once and over, since { l , . . . , l m } span [ R ] , up to a change of variables we canassume that l i = x i for ≤ i ≤ n and l n +1 = α x + · · · + α n x n ,Thus, f = x · · · x n ( α x + · · · + α n x n ) . Letting f i = f /l i , for i = 1 , . . . , n + 1 , one has f x i = f i + α i f n +1 , for i = 1 , . . . , n +1 . Note that l n +1 = α l + · · · + α n l n is the dependency that gives theonly minimal generator of the Orlik-Terao ideal. Suppose α j +1 = · · · = α n = 0 for some j ≥ and α , . . . , α j = 0 . That generator is Q = T · · · T j − ( α T · · · T j T n +1 + · · · + α j T · · · T j − T n +1 ) . So, the ideal in the third part of Lemma 2.1 is h T + α T n +1 , · · · , T j + α j T n +1 , T j +1 , · · · , T n , T · · · T j − ( α T · · · T j T n +1 + · · · + α j T · · · T j − T n +1 ) i = h T + α T n +1 , · · · , T j + α j T n +1 , T j +1 , · · · , T n , (1 − ( − j − j ) α · · · α j T jn +1 i which is indeed h T . . . , T n i - primary. By Lemma 2.1, J f is a minimal reduction of I . (cid:3) Rank . Proposition 2.3. Let f ∈ R = K [ x, y ] be the defining polynomial of an arrangement of rank andsize m ≥ . Then J f is a reduction of I with reduction number one. N THE JACOBIAN IDEAL OF CENTRAL ARRANGEMENTS 9 Proof. It suffices to prove that µ ( I ) = µ ( J f I ) since J f I ⊂ I and both ideals are equigenerated inthe same degree. Now, since J f is generated by a regular sequence (of two forms), then µ ( J f I ) =2 µ ( I ) − (cid:0) (cid:1) = 2 m − (see, e.g., [8, Proposition 2.22]).On the other hand, one has µ ( I ) = (cid:0) m +12 (cid:1) − µ ( Q ) , where Q is as above the defining ideal ofthe Orlik–Terao algebra. It is known that Q is generated by a minimal set of generators among thecircuits corresponding to the dependencies – all of size in the rank case, hence the generatorsare quadrics. This minimal set has (cid:0) m − (cid:1) elements. Therefore, one gets µ ( I ) = (cid:18) m + 12 (cid:19) − (cid:18) m − (cid:19) = 2 m − , as was to be shown. (cid:3) Partition method. Let R = K [ x , . . . , x r ; y , . . . , y s ] denote a polynomial ring in r + s vari-ables. Consider a central arrangement A of K r + s = Spec ( R ) of size m . Assume that A = B ∪ C isa -partition, where B (respectively, C ) is an arrangement of K r = Spec ( K [ x , . . . , x r ]) of size m x (respectively, is an arrangement of K s = Spec ( K [ y , . . . , y s ]) of size m y ), so that m = m x + m y .One assumes that m x ≥ r, m y ≥ s .Let f and g denote, respectively, the polynomials of B and C . Thus, F := f g is the polynomialof the total arrangement A . Let I f ⊂ K [ x , . . . , x r ] denote the ideal of ( m x − -fold products ofthe first arrangement; similarly, define I g ⊂ K [ y , . . . , y s ] and I F ⊂ R .Finally, let J f ⊂ K [ x , . . . , x r ] denote the Jacobian ideal of f ; similarly, take J g ⊂ K [ y , . . . , y s ] and J F ⊂ R . Lemma 2.4. One has I F = h f I g , g I f i and J F = h f J g , gJ f i as ideals in R . Proof. This is immediate from the data. (cid:3) Proposition 2.5. Suppose that J f (respectively, J g ) is a reduction of I f of reduction number ≤ r − (respectively, of I g of reduction number ≤ s − ). Then J F is a reduction of I F of reduction number ≤ r + s − . Proof. The assumption is that I rf = J f I r − f and I sg = J g I s − g and we wish to prove that I r + s − F = J F I r + s − F . Quite generally, J F ⊂ I F , hence it suffices to show the inclusion I r + s − F ⊂ J F I r + s − F .By Lemma 2.4 and the ‘binomial expansion’, I r + s − F is spanned by the ideals of the shape a t := f r + s − − t I r + s − − tg g t I tf , for ≤ t ≤ r + s − .As usual, if t < r then r + s − − t ≥ s . Thus, it will suffice to assume that r + s − − t ≥ s .In this case, one has I r + s − − tg = J g I r + s − − tg by the assumption, hence a t = f r + s − − t I r + s − − tg g t I tf = ( f J g )( f I g ) r + s − − t ( g t I tf ) ⊂ J F I r + s − − tF I tF = J F I r + s − F , as was to be shown. (cid:3) In order to derive the consequence below one emphasizes that the above proposition applies toany -partition of A , not necessarily one given by components. Corollary 2.6. In the above notation, if J f is a reduction of I f ⊂ R for any non-decomposable cen-tral arrangement of size m ≥ dim R , then the same holds for any central arrangement whatsoeverand the reduction number is at most the size of the arrangement less the number of components.. Proof. Given a full decomposition A = A ∪ A · · · ∪ A t of a central arrangement into indecom-posable arrangements, apply the proposition to A ∪ A , then to ( A ∪ A ) ∪ A , and so on soforth. (cid:3) Given a central arrangement A = { l , . . . , l m } of rank n , a form l i is a coloop if the correspond-ing deletion has rank n − . Corollary 2.7. Let R = K [ x, y, z ] . If A admits a coloop then J f is a minimal reduction of I . Proof. By a change of coordinates, we may assume that the coloop is z . Therefore, one has apartition of A following the separation R = K [ x, y ; z ] . The case of rank one is trivial (or vacuous),while the case of rank two follows from Proposition 2.3. (cid:3) Main theorem. Once again, let I ⊂ R := K [ x , . . . , x n ] stand for the ideal generated by all ( m − -fold products of the linear forms l , . . . , l m , m ≥ n + 1 . We assume throughout that n ≥ as before. Setting f = l · · · l m , we prove that the Jacobian ideal J f is a minimal reduction of I ,when A satisfies some genericity conditions.Recall that the arrangement defined by these linear forms is called generic if any n of them is K -linearly independent. This is equivalent to requiring that the Jacobian matrix of these formshas no vanishing n -minor. In particular, the set { l , . . . , l n } has only one component, where thenumber of components of the arrangement is defined with respect to the variables x , . . . , x n . By [3,Proposition 2.7 (d)], the reduction number of I is n − u , where u is the number of such components.Another important number measures the amount of genericity of the arrangement. In the arbitrarycase, one talks about the arrangement being d -generic if any d among the defining linear forms are K -linearly independent.Our standing assumption in this part is that the arrangement A is ( n − -generic. This includesthe arbitrary rank case since any central arrangement is -generic by definition.Recall the notation Notation 2.8. R = K [ x , . . . , x n ] . • I ⊂ R is generated by L , . . . , L m , where L i := l · · · b l i · · · l m , i = 1 , . . . , m . • With f = l · · · l m , For each j = 1 , . . . , n , one has f x j = a ,j l · · · l m + · · · + a i,j l · · · b l i · · · l m + · · · + a m,j l · · · l m − , where a i,j := ( l i ) x j , for i = 1 , . . . , m . • m = h x , . . . , x n i . A key lemma. In the above notation, one has: Lemma 2.9. ( char( K ) ∤ m ) If A is ( n − -generic then L L · · · L n ∈ J f I n − .Proof. Set P := L L · · · L n = ( l l · · · l n ) n − ( l n +1 · · · l m ) n and consider the two cases as towhen { l , l , . . . , l n } has rank either n or n − .R ANK n . Note that P ⊂ h f i and set P = f Q . We contend that m Q ⊂ I n − .For this, since m = h l , . . . , l n i in this case, it suffices to show that for each j ∈ { , . . . , n } , onehas l j Q ⊂ I n − . But, indeed one easily sees that l j Q = L · · · c L j · · · L n ∈ I n − .To conclude, apply the Euler relation of f , thus getting mP = mf Q ∈ m J f I n − .R ANK n − . Up to reordering, l n is a K -linear combination of { l , . . . , l n − } . Then, up to a changeof variables we can, and will, suppose that l i = x i for ≤ i ≤ n − , so P = ( x · · · x n − l n ) n − ( l n +1 · · · l p l p +1 · · · l m ) n , where p ≥ n is such that l n , . . . , l p ∈ K [ x , . . . , x n − ] and rank( l , . . . , l n − , l k ) = n for p + 1 ≤ k ≤ m .Now, for j = 1 , . . . , n − , set ∆ j := x · · · b x j · · · x n − l n l n +1 · · · l m = f x j − x · · · x j · · · x n − ( a n,j l n +1 · · · l m ) − x · · · x j · · · x n − m X s = n +1 a s,j l n l n +1 · · · b l s · · · l m ) ! N THE JACOBIAN IDEAL OF CENTRAL ARRANGEMENTS 11 and introduce it in the expression of P to enhance its dependence on x j : P = x n − j ( x · · · b x j · · · x n − l n ) n − ( l n +1 · · · l m ) n − ∆ j = x n − j ( x · · · b x j · · · x n − l n ) n − ( l n +1 · · · l m ) n − f x j − ( x · · · x j · · · x n − ) n − x j ( a n,j l n − n ( l n +1 · · · l m ) n ) − ( x · · · x j · · · x n − l n ) n − x j p X s = n +1 a s,j l n − s ( l n +1 · · · b l s · · · l m ) n ! − ( x · · · b x j · · · x n − ) n − x nj m X s = p +1 a s,j l n − s ( l n +1 · · · b l s · · · l m ) n . Rearranging: P = x n − j ( x · · · b x j · · · x n − l n ) n − ( l n +1 · · · l m ) n − f x j − ( x · · · x j · · · x n − ) n − l n − n ( l n +1 · · · l m ) n ( a n,j x j ) − ( x · · · x j · · · x n − l n ) n − ( l n +1 · · · l m ) n − p X s = n +1 a s,j x j l n +1 · · · b l s · · · l m ! − ( x · · · b x j · · · x n − l n ) n − x nj m X s = p +1 a s,j l n − s ( l n +1 · · · b l s · · · l m ) n Summing up for j = 1 , . . . , n − gives ( n − P = n − X j =1 x n − j ( x · · · b x j · · · x n − l n ) n − ( l n +1 · · · l m ) n − f x j − ( x · · · x n − ) n − l n − n ( l n +1 · · · l m ) n n − X j =1 a n,j x j − ( x · · · x n − l n ) n − ( l n +1 · · · l m ) n − n − X j =1 p X s = n +1 a s,j x j l n +1 · · · b l s · · · l m − ( x · · · b x j · · · x n − l n ) n − n − X j =1 x nj m X s = p +1 a s,j l n − s ( l n +1 · · · b l s · · · l m ) n . Setting P n − j =1 a s,j x j = l s , for every n ≤ s ≤ p , the above can be rewritten as ( n − P = n − X j =1 x n − j ( x · · · b x j · · · x n − l n ) n − ( l n +1 · · · l m ) n − f x j − ( p − n + 1) P − ( x · · · b x j · · · x n − l n ) n − d X j =1 m X s = p +1 a s,j l n − s ( x j l n +1 · · · b l s · · · l m ) n , which finally affords pP = n − X j =1 x n − j ( x · · · b x j · · · x n − l n ) n − ( l n +1 · · · l m ) n − f x j − ( x · · · b x j · · · x n − l n ) n − n − X j =1 m X s = p +1 a s,j l n − s ( x j l n +1 · · · b l s · · · l m ) n = n − X j =1 L · · · c L j · · · L n f x j − ( x · · · b x j · · · x n − l n ) n − n − X j =1 m X s = p +1 a s,j l n − s ( x j l n +1 · · · b l s · · · l m ) n Thus, we just have to take care of the summands in the second term: they are of the form ( x · · · b x j · · · x n − l n l s ) n − ( x j l n +1 · · · b l s · · · l m ) n , with ≤ j ≤ n − and p + 1 ≤ s ≤ m , thus obtained by trading x j and l s in X := { x , . . . , x n − , l n } to get X ′ := { x , . . . , b x j , . . . , x n − , l n , l s } .Now, { x , . . . , b x j , . . . , x n − , l n } is linearly independent otherwise one would have a linear de-pendence of n − forms. Next, since l s involves effectively the variable x n it is linearly independentof { x , . . . , b x j , . . . , x n − , l n } . Therefore, rank( X ′ ) = n . a case taken care of before. (cid:3) The main theorem. Theorem 2.10. ( char( K ) ∤ m ) For an ( n − -generic central hyperplane arrangement, the Jaco-bian ideal J f is a minimal reduction of I with reduction number at most n − .Proof. Since I has analytic spread n and J f is minimally generated by at most n elements, if it is areduction then it is minimal. Thus, it suffices to prove that I n ⊂ J f I n − . As in Notation 2.8, I = h L , . . . , L m i , where L i = l · · · b l i · · · l m , i = 1 , . . . , m. The standard generators of I n have the form P := L r i L r i · · · L r u i u , where ≤ i < · · · < i u ≤ m , r + r + · · · + r u = n and r j ≥ , for all j = 1 , . . . , u . The set ofthe above indices { i , . . . , i u } will be called the support of the standard generator P . For any suchgenerator the cardinality of its support is ≤ u ≤ n .The method of the proof that P belongs to J f I n − is reverse induction on u . Thus, set v := n − u ,so ≤ v ≤ n − .I NITIAL STEP . v = 0 . Then u = n , and hence P = L i L i · · · L i n . Therefore, the result followsfrom Lemma 2.9 by an obvious symmetry of the indices.I NDUCTIVE STEP . Let v ≥ , and suppose that the result is true for v − , for any ( n − -genericarrangement), meaning that any standard generator thereof whose support has cardinality u + 1 isin J f I n − .Since v ≥ , then u ≤ n − , and at least one of the exponents r j is greater than or equal to 2.Fix an index k ∈ { , . . . , u } such that r k ≥ .Since the arrangement is ( n − -generic and u ≤ n − , the linear forms l i , . . . , l i u are K -linearlyindependent. Thus, there is a change of variables y ↔ l i , . . . , y u ↔ l i u , y u +1 ↔ x u +1 , . . . , y n ↔ x n , N THE JACOBIAN IDEAL OF CENTRAL ARRANGEMENTS 13 where y , . . . , y n are new variables over K . As per the Appendix, the change preserves the hypoth-esis and the sought conclusion; in particular, the transformed arrangement remains ( n − -generic.By this change, f transforms into ¯ f := y · · · y u ℓ u +1 · · · ℓ m , where ℓ u +1 , . . . , ℓ m are linear formsin S := K [ y , . . . , y n ] . Setting ¯ L i := ¯ f /ℓ i , i = 1 , . . . , m , where ℓ j := y j , for j = 1 , . . . , u , thetransform of P is ¯ P := ¯ L r i ¯ L r i · · · ¯ L r u i u = y n − r y n − r · · · y n − r u u ( ℓ u +1 · · · ℓ m ) n , while the transform of the ideal I is the ideal ¯ I := h ¯ L , . . . , ¯ L m i ⊂ S. Set ¯ P = ¯ L i k ¯ P ′ , where ¯ P ′ := ¯ L r i ¯ L r i · · · ¯ L r k − i k · · · ¯ L r u i u ∈ ¯ I n − . Coming back to our fixed index k ∈ { , . . . , u } above, one has ¯ f y k = ¯ L i k + X u +1 ≤ j ≤ m b k,j ¯ L j , where b k,j is the coefficient of y k in the expression of ℓ j .That is, ¯ L i k = ¯ f y k − X j / ∈{ i ,...,i u } b k,j ¯ L j . Plugging into ¯ P = ¯ L i k ¯ P ′ yields ¯ P = ¯ f y k ¯ P ′ − X j / ∈{ i ,...,i u } b k,j ¯ P ′ ¯ L j Clearly, ¯ f y k ¯ P ′ ∈ J ¯ f ¯ I n − . At the other end, for each j / ∈ { i , . . . , i u } , ¯ P ′ ¯ L j is a standard generatorwith support X := { i , . . . , i k , . . . , i u , j } since r k − ≥ . Thus, | X | = u + 1 .By the inductive hypothesis as applied to the transformed arrangement, each such ¯ P ′ ¯ L j is anelement of J ¯ f ¯ I n − , and hence ¯ P ∈ J ¯ f ¯ I n − too. Reversing back to the original variables x , . . . , x n ,as per the Appendix, we obtain that P ∈ J f I n − , as was to be shown. (cid:3) Remark 2.11. The exact reduction number above is a function of the number of components ofthe arrangement – a proof is given in [3, Proposition 2.7]. Alternatively, if one can prove that forany indecomposable hyperplane arrangement the reduction number is exactly n , then Corollary 2.6allows to deduce that for any hyperplane arrangement the reduction number is exactly n − c , where c is the number of components. Note that this corollary is not applicable within the restricted classof ( n − -generic arrangements since a component of rank r might fail to be ( r − -generic. As aremedy, we believe that Theorem 2.10 holds regardless of generic like restrictions.2.4. On the linear type property. Closing this part, we state the following Conjecture 2.12. ( char( K ) ∤ m ) Let f ∈ R = K [ x , . . . , x n ] denote the defining polynomial of acentral hyperplane arrangement of size m ≥ n . Then the Jacobian ideal J f is of linear type. Recall that an ideal I ⊂ R is said to be of linear type if the natural surjection from its symmetricalgeba to its Rees algebra is an isomorphism.We continue to assume that char( K ) ∤ m throughout.Note the following result, where an ideal a ⊂ R is said to have property G ∞ if µ ( a ℘ ) ≤ ht ℘ ,for every prime ideal ℘ ⊂ R . Lemma 2.13. Let f ∈ R = K [ x , . . . , x n ] denote the defining polynomial of a central hyperplanearrangement of size m ≥ n . If either n ≤ or else the arrangement is generic then J f satisfiesproperty G ∞ . Proof. Since J f is globally generated by n = ht m elements, it suffices to consider a prime ℘ = m .If n = 2 , J f is a complete intersection. If n = 3 , the ideal J f is generically a complete intersection. Indeed, let ℘ be a minimal prime of J f , necessarily of height . Write l i = a i x + b i x + c i x , i =1 , . . . , m . Then f x = a l · · · l m + · · · + a m l · · · l m − f x = b l · · · l m + · · · + b m l · · · l m − f x = c l · · · l m + · · · + c m l · · · l m − all belong to ℘ . In particular, the Euler relation implies that f = l l · · · l m ∈ ℘ , hence there exists j ∈ { , . . . , m } with l j ∈ ℘ . But then, a j ( f /l j ) , b j ( f /l j ) , c j ( f /l j ) also belong to ℘ , and since a j , b j , c j cannot all be zero we have f /l j ∈ ℘ . Hence there exists k = j , with l k ∈ ℘ . Since l j and l k are not proportional, then ℘ = h l j , l k i . By a change of variables, we may assume that ℘ = h x, y i ,corresponding to the point [0 : 0 : 1] of P = Proj( K [ x, y, z ]) . Therefore, the result follows from[10, Lemma 2.4].Finally, if the arrangement is generic, then the result follows from Theorem 1.3 (ii) and the factthat I satisfies G n in the generic case ([3, The proof of Proposition 4.1]) (cid:3) Proposition 2.14. Conjecture 2.12 holds for n = 3 . Proof. It follows from Lemma 2.13 since in this case J f is generated by three elements (see [12,Proposition 3.7]). (cid:3) Recall that the arrangement A is free if and only if Derlog( A ) is a free R -module or, equivalently,if and only if Syz( J f ) is a free R -module. In particular, a rank central hyperplane arrangement A is free if and only if the presentation ideal of the symmetric algebra of J f is minimally generatedby biforms. Corollary 2.15. Let A be a rank central hyperplane arrangement, with defining polynomial f ∈ R = K [ x, y, z ] . Then A is free if and only if the Rees algebra R R ( J f ) is a complete intersection. Proof. If A is free then its syzygy matrix is of size × , thus implying that the symmetric algebraof J f is a complete intersection. By Proposition 2.14, R R ( J f ) is a complete intersection.Conversely, since µ ( J f ) = 3 , the presentation ideal of R R ( J f ) has codimension . Since it isa complete intersection, then it is generated by a regular sequence of two minimal bihomogeneousgenerators. By Proposition 2.14, these are a set of generators of the presentation ideal of the sym-metric algebra of J f . Then the syzygy matrix of the -generated ideal J f must have size × , i.e., A is free. (cid:3) Remark 2.16. Even for n ≥ and m = n + 1 in the generic case the question is not triviallyhandled. By [9, Corollary 4.5.4], the minimal free resolution of R/J f has the same Betti number asthose of a complete intersection of n -forms of length n . In particular, Syz ( J f ) is generated by the (cid:0) n (cid:1) degree m − n + 1 = 2 reduced Koszul syzygies of the partial derivatives. It should be possibleto show that I in this situation is unmixed – note that the argument may eventually profit from theknowledge that the Fitting ideals of J f have the expected codimension bounds. At the other end,one of the available ways to prove the linear type property, namely, proving instead that the Koszulhomology modules of J f satisfy the sliding-depth conditions, is not an obvious alternative as in noorder the partial derivatives seem to be a d -sequence.3. F ORMS OF HIGHER DEGREES Let K be an infinite field, of characteristic zero or of characteristic not dividing certain criticalintegers. Let f , . . . , f m ∈ R := K [ x , . . . , x n ] , m ≥ , be forms of degrees deg( f i ) = d i ≥ , i = 1 , . . . , m . Let c := min { m, n } . We assume throughout that n ≥ – otherwise most of theassumptions become fuzzy or else the present objective becomes irrelevant. Definition 3.1. The forms f , . . . , f m will be called nearly transversal if they satisfy the followingconditions: N THE JACOBIAN IDEAL OF CENTRAL ARRANGEMENTS 15 (a) For every ≤ i ≤ m , the hypersurface Proj ( R/ ( f i )) is smooth – in particular, f i is irre-ducible and depends on all variables.(b) The Jacobian matrix of any subset S c ⊂ { f , . . . , f m } with c elements has maximal rankmodulo primes of height n − containing S c .By abuse, when a form satisfies condition (a) it will be said to be smooth.Let F = f f · · · f m , and let J F ⊂ R be the Jacobian ideal of F , i.e. J F = h F x , . . . , F x n i ,where F x i := ∂F∂x i , i = 1 , . . . , n . Condition (a) implies that, for any ≤ i ≤ m , the Jacobian ideal J f i is a complete intersection of codimension n .Note that J F is contained in the ideal generated by the ( m − -fold products of { f , . . . , f m } .In particular, J F has codimension at most . But, since F is reduced, its codimension is exactly .Finally, observe that, though the given forms may have different degrees d , . . . , d m , the ideal J F is equigenerated in degree d + · · · + d m − .The goal is to show that depth( R/J F ) = 0 , which is equivalent to showing that J F is non-saturated with respect to m := h x , . . . , x n i . Such a result would be a rough analog of Rose–Terao–Yuzvinsky theorem of Subsection 1.4. The overall expectation is to induct on the number m ≥ offorms. Thus, the case where m = 2 sticks up as the first to look at.3.1. The two forms case. Recall that Syz ( J F ) denotes the syzygy module of the partial derivativesof F (Subsection 1.1.) Proposition 3.2. Let f, g ∈ R be forms of respective degrees ≤ d ≤ e . Setting F := f g , one has: (i) If f is smooth then g ∈ J sat F , where F = f g . (ii) Assume that char( K ) does not divide either e or d + e . If f and g have no proper commonfactors, then indeg( Syz ( J F )) ≥ indeg( Syz ( J g )) . In particular, if g is smooth then indeg( Syz ( J F )) ≥ e − ≥ (cid:22) d + e (cid:23) − . (iii) Assume that char( K ) does not divide any of d, e, d + e and suppose that the followingconditions are satisfied: (a) f is smooth. (b) The Jacobian matrix of h f, g i has maximal rank over R/℘ , for every prime ideal ℘ ⊃h f, g i of height n − . (c) The ideal h f, J g i is m -primary.Then depth R/J F = 0 . Proof. (i) By Leibniz,(5) F x i = f x i · g + f · g x i , for every ≤ i ≤ n . Multiplying through by g yields f x i g ∈ J F for every ≤ i ≤ n . Since f is asmooth form of degree d ≥ , J f is m -primary, and therefore g ∈ ( J F ) sat .(ii) Let ( P · · · P n ) t ∈ Syz ( J F ) be a nonzero syzygy. By (5), one has f ( P g x + · · · + P n g x n ) + g ( P f x + · · · + P n f x n ) = 0 . Since gcd( f, g ) = 1 , then(6) P g x + · · · + P n g x n = H · g, for a suitable form H of degree deg P i − . Thus, ( P · · · P n ) t ∈ Derlog( g ) = Syz( J g ) ⊕ Rθ E ,where θ E denotes the Euler vector of g (see (4)). Therefore, ( P − e Hx · · · P n − e Hx n ) t ∈ Syz( J g ) . Now, if some P i − e Hx i does not vanish, then deg( P i ) = deg P i − e Hx i for all i , andhence deg( P i ) ≥ indeg( Syz ( J g )) . Else, P i = e Hx i for all i . Since H = 0 in this situation, this means that ( x · · · x n ) t ∈ Syz ( J F ) . But since d + e = 0 by assumption, then the Euler relationimplies that F = 0 – an absurd.The supplementary assertion is clear since for a complete intersection the syzygies are the Koszulrelations. The appended inequality is obvious since e = (cid:4) e (cid:5) ≥ (cid:22) d + e (cid:23) .(iii) Let Θ denote the Jacobian matrix of h f, g i .C LAIM 1. The homogeneous ideal J := h f, I (Θ) i is m -primary.Suppose otherwise and pick a prime ideal J ⊂ ℘ ⊂ R properly contained in m . Since f ∈ ℘ isassumed to be smooth, there is a partial derivative, say, f x not belonging to ℘ . At the other end, g x i f x − g x f x i ∈ J ⊂ ℘ , for every i . Using the Euler relation of g , we can write egf x = x f x g x + x f x g x + · · · + x n f x g x n = x f x g x + [ x ( f x g x − f x g x ) + x f x g x ] + · · · + [ x n ( f x g x n − f x n g x ) + x n f x n g x ]= g x ( x f x + · · · + x n f x n ) + x ( f x g x − f x g x ) + · · · + x n ( f x g x n − f x n g x )= dg x f + x ( f x g x − f x g x ) + · · · + x n ( f x g x n − f x n g x ) . Clearly, the last expression belongs to J ⊂ ℘. Since f x / ∈ ℘ then g ∈ ℘ . We thus conclude that h f, g i ⊂ ℘ .Next, we take care of the dichotomy as to whether g x ∈ ℘ or g x / ∈ ℘ . If g x ∈ ℘ is the case,then g x i ∈ ℘ for every i , since g x i f x − g x f x i ∈ J ⊂ ℘ , and f x / ∈ ℘ . Therefore, if g x ∈ ℘ , then h f, J g i ⊂ ℘ as well, contradicting assumption (c).Thus, we must have g x / ∈ ℘ . We may assume that ℘ has height n − . We then contend thatthis contradicts condition (b) in the hypothesis. And indeed, one has the following nonzero relationbetween the two rows of the Jacobian matrix of h f, g i : ( g x , g x , . . . , g x n ) f x − ( f x , f x , . . . , f x n ) g x = (0 , f x g x − f x g x , . . . , f x g x n − f x n g x ) , which vanishes modulo ℘ . Since neither f x nor g x vanishes modulo ℘ , we have a contradiction.This proves Claim 1.C LAIM g J ⊂ J F .From (5) one easily deduces the relations g x i F x j − g x j F x i = ( g x i f x j − g x j f x i ) g, for ≤ i < j ≤ n . Therefore, g I (Θ) ⊂ J F . At the other end, gf = F ∈ J F by the Euler relationof F . Together they imply the claim.Collecting the assertions of the two claims, one gets g ∈ ( J F ) sat . Clearly, g / ∈ J F for degreereasons, since indeg( J F ) = e + d − . Therefore, m is an associated prime of R/J F , as was to beshown. (cid:3) Remark 3.3. In the light of the proposition, perhaps it should be noted that finding reduced homo-geneous free divisors looks in a sense like a sparse deal. Accordingly, the above result tells us thatif the data are in some sort of good position, the chances are dimmed. Note that the result generatesno conflict with [11, Theorem 2.12] because in loc. cit. the form g – that plays the role of f here –is only smooth when d ≤ . Question 3.4. Note that the hypotheses of Proposition 3.2 (iii) imply that h f, g i is an isolatedsingularity, i.e., Proj( R/ h f, g i ) is smooth. Is the converse true?3.2. Arbitrary number of forms. We now go back to the case of forms f , . . . , f m of respectivedegrees d , . . . , d m . As before, set F = f · · · f m . Proposition 3.5. indeg( Syz ( J F )) ≥ (cid:22) d + · · · + d m m (cid:23) − . N THE JACOBIAN IDEAL OF CENTRAL ARRANGEMENTS 17 Proof. We prove the result by induction on m ≥ . For m = 2 , this is Proposition 3.2 (ii).For the inductive step, suppose m ≥ . Set δ := (cid:22) d + · · · + d m m (cid:23) . Say, d is least amongthe degrees. Set f := f , and G := F/f . Let δ ′ := (cid:22) d + · · · + d m m − (cid:23) . Clearly, δ ′ ≥ δ since ( m − d + · · · + d m ) ≤ m ( d + · · · + d m ) .Suppose to the contrary, that J F has a syzygy ( P , . . . , P n ) of degree δ − . By the same tokenas in the proof of Proposition 3.2 (ii), since gcd( f, G ) = 1 (because gcd( f i , f j ) = 1 , i = j ), wehave the following syzygies of degree δ − P − ( H/d ) x ) G x + · · · + ( P n − ( H/d ) x n ) G x n = 0( P + ( H/d ) x ) f x + · · · + ( P n + ( H/d ) x n ) f x n = 0 , where d := deg( G ) . By the inductive hypothesis, J G doesn’t have any syzygy of degree ≤ δ ′ − .Then, from the first equation we have P i = ( H/d ) x i , for all i = 1 , . . . , n . This leads to x f x + · · · + x n f x n = 0 , contradicting the Euler relation of f = 0 . (cid:3) Proposition 3.6. Suppose there exist j ∈ { , . . . , m } such that ( d + · · · + d j − + d j +1 + · · · + d m ) − d j ≤ (cid:22) d + · · · + d j − + d j +1 + · · · + d m m − (cid:23) − . Then depth( R/J F ) = 0 .Proof. Let f := f j and G := F/f . By Proposition 3.2 (i), we have G ∈ ( J F ) sat . We will shownow that G / ∈ J F .Let e := d j = deg( f ) , and d := d + · · · + d j − + d j +1 + · · · + d m = deg( G ) . Suppose G ∈ J F . Then there exists L , . . . , L n ∈ R d − e +1 , such that G = ( L f x + · · · + L n f x n ) G + f ( L G x + · · · + L n G x n ) . By grouping, we have G [ G − ( L f x + · · · + L n f x n )] = ( L G x + · · · + L n G x n ) f. Clearly, gcd( f, G ) = 1 since gcd( f i , f j ) = 1 for every i = j . therefore, G − ( L f x + · · · + L n f x n ) = B · fL G x + · · · + L n G x n = B · G, for some B ∈ R d − e .Euler’s relation says that dG = x G x + · · · + x n G x n , which plugged into the second equationleads to the syzygy of degree d − e + 1 . ( L − ( B/d ) x ) G x + · · · + ( L n − ( B/d ) x n ) G x n = 0 . Since from hypotheses d − e ≤ ⌊ d/ ( m − ⌋− , from Proposition 3.5, we have that L i = ( B/d ) x i for all i = 1 , . . . , n . This plugged in the first equation leads to G = Bf ( e/d − a contradiction. (cid:3) 4. A PPENDIX Calculus on Jacobian ideals. In order to ease the amount of calculations needed in our proofs,we often appeal to change of variables/coordinates technique to simplify the shape of the definingpolynomial of our hyperplane arrangement. The chain rule of partial differentiation and linearalgebra will safely allow us to do this without altering the homological properties of the Jacobianideal we are studying throughout these notes.Concretely, let F = l · · · l n l n +1 · · · l m ∈ R := K [ x , . . . , x n ] , where l i , . . . , l i n are linearlyindependent linear forms in R , for some distinct indices i , . . . , i n ∈ { , . . . , m } . After a re-labeling, we can assume that i = 1 , . . . , i n = n . Then there is a unique change of variables l ↔ y , . . . , l n ↔ y n , given by [ x , . . . , x n ] · M = [ y , . . . , y n ] , where M is the n × n invertible matrix whose columns are the coefficients of the linear forms l , . . . , l n , respectively.Then, we have [ x , . . . , x n ] = [ y , . . . , y n ] · M − , and hence x = L , . . . , x n = L n , where L , . . . , L n are linear forms in S := K [ y , . . . , y n ] . Then we consider G := F ( L , . . . , L n ) = y · · · y n ( l ′ n +1 · · · l ′ m ) ∈ S = K [ y , . . . , y n ] . The chain rule says that for each i = 1 , . . . , n , in K [ y , . . . , y n ] we have ∂G∂y i = ∂F∂x ( L , . . . , L n ) · ∂L ∂y i + · · · + ∂F∂x n ( L , . . . , L n ) · ∂L n ∂y i , or, as vectors with entries elements of S , we can write [ G y , . . . , G y n ] = [ F x ( L , . . . , L n ) , . . . , F x n ( L , . . . , L n )] · ( M − ) τ , where τ denotes the transposition of the matrix. Since the matrix ( M − ) τ is invertible, in terms ofideals of S = K [ y , . . . , y n ] , we have h G y , . . . , G y n i = h F x ( L , . . . , L n ) , . . . , F x n ( L , . . . , L n ) i . Step-by-step, this is the path we are following:(1) We start with J F = h F x , . . . , F x n i in R = K [ x , . . . , x n ] .(2) We make the change of variables (from x i ’s to y j ’s) x ↔ L , . . . , x n ↔ L n , to obtain anideal ¯ J in S = K [ y , . . . , y n ] that is generated by F x ( L , . . . , L n ) , . . . , F x n ( L , . . . , L n ) .(3) The ideal ¯ J of S obtained at (2) has the same homological properties as the ideal J F of R .(4) But, as we observed, ¯ J = J G ⊂ S . Therefore, in the shape of F , we can assume that l = x , . . . , l n = x n , as we did throughout this paper.This procedure works also when we want to show inclusions of the form I k ⊆ J F I k − , for some k . Denote ¯ I to be the ideal of S = K [ y , . . . , y n ] generated by all ( m − -fold products of linearforms y , . . . , y n , l ′ n +1 , . . . , l ′ m . Suppose we showed ¯ I k ⊆ J G ¯ I k − in S = K [ y , . . . , y n ] .If we go back to variables x , . . . , x n , via the change of variables y ↔ l , . . . , y n ↔ l n , since J G will transform into J F (by making this substitution in the chain rule above, and because M isinvertible), and since ¯ I will transform back into I , we will get also that I k ⊆ J F I k − . N THE JACOBIAN IDEAL OF CENTRAL ARRANGEMENTS 19 Example 4.1. Suppose n = 2 , and say, F = ( x + x ) | {z } l (2 x + x ) | {z } l ( x − x ) | {z } l ( x + 3 x ) | {z } l ∈ R := K [ x , x ] . Then we have: [ x , x ] (cid:20) (cid:21)| {z } M = [ y , y ] . Then, [ x , x ] = [ y , y ] (cid:20) − − (cid:21)| {z } M − = [ − y + y | {z } L , y − y | {z } L ] , and so G = y y ( − y + 2 y )(5 y − y ) ∈ S = K [ y , y ] . 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