Freeness versus maximal degree of the singular subscheme for surfaces in P 3
aa r X i v : . [ m a t h . AG ] S e p FREENESS VERSUS MAXIMAL DEGREE OF THE SINGULARSUBSCHEME FOR SURFACES IN P ALEXANDRU DIMCA Abstract.
We show that a free surface in P is characterized by the maximalityof the degree of its singular subscheme, in the presence of an additional tamenesscondition. This is similar to the characterization of free plane curves by the maxi-mality of their global Tjurina number given by A. A. du Plessis and C.T.C. Wall.Simple characterizations of the nearly free tame surfaces are also given. Introduction
Let D : f = 0 be a reduced hypersurface in the projective complex space P n ,defined by a homogeneous polynomial f ∈ S = C [ x , ..., x n ] of degree d . Let Σ bethe singular subscheme of D , defined by the Jacobian ideal J f , which is the ideal in S spanned by the first order partial derivatives f , ..., f n of f with respect to x , ..., x n .If dim Σ = m , then the Hilbert polynomial P ( M ( f )) of the Milnor (a.k.a. Jacobian)algebra M ( f ) = S/J f has degree m , and the degree of Σ is by definition m ! · a , where a is the leading coefficient of P ( M ( f )). The minimal degree of a Jacobian relationfor f is the integer mdr ( f ) defined to be the smallest integer q ≥ X j =0 ,n a j f j = 0among the partial derivatives f j ’s of f with coefficients a j in S q , the vector space ofhomogeneous polynomials of degree q . In this paper we assume mdr ( f ) >
0, whichis equivalent to asking D not to be a cone over a hypersurface in P n − .When n = 2, then D is a plane curve with isolated singularities and the degree ofits singular subscheme Σ is the global Tjurina number τ ( D ) = X p ∈ Σ τ ( D, p ) , where τ ( D, p ) is the Tjurina number of the isolated plane curve singularity (
D, p ),see for instance [4]. A. A. du Plessis and C.T.C. Wall have proved the following, see[7].
Mathematics Subject Classification.
Primary 14J70; Secondary 14B05, 13D02.
Key words and phrases.
Jacobian ideal, singular subscheme, Hilbert polynomial, free surface,nearly free surface. Partially supported by Institut Universitaire de France.
Theorem 1.1.
In the class of reduced plane curves with a fixed degree e = mdr ( f ) satisfying e ≤ d − , the free curve D (if it exists) has a singular locus Σ of maximaldegree. More precisely, for a curve D with fixed degree e , one has deg Σ ≤ s − s , where e = d − − e , and s = e + e , s = e e are the elementary symmetricfunctions in e , e . Moreover, the equality holds if and only if the curve D is free. This result is also discussed in [5], where a similar fact is proved for the nearly freeplane curves.In this note we investigate to what extent such results hold for n = 3, i. e. forsurfaces in P . The basic properties of a free surface D are reviewed in the nextsection, for now we just say that freeness of D is the same as asking the 3-foldsingularity given by the cone over D to be a free divisor germ in ( C ,
0) in thesense of K. Saito, who introduced this important notion of free divisor in [9]. Sincefree surfaces have a 1-dimensional singular locus, we restrict our attention to suchsurfaces, i.e. from now on we supposedim Σ = 1 . Instead of looking only at the minimal degree e of a Jacobian relation, we haveto consider a pair of degrees e ≤ e , corresponding to the lowest degrees of twoindependent generators ρ and ρ of the module AR ( f ) of all the Jacobian syzygiesof f . The behaviour of these two generators (i.e. how independent they are) is usedto define a class of surfaces, called tame surfaces, see Definition 2.3. In particular,the free surfaces and most (if not all, see Question 2.6) of the nearly free surfacesdefined in [6] are shown to be tame, see Proposition 2.5. The main results of thisnote are the following. Theorem 1.2.
Let D : f = 0 be a reduced surface in P , tame with respect to thepair of syzygies ρ , ρ of degrees e = deg ρ ≤ e = deg ρ . Set e = d − − e − e and assume e ≥ e . Then one has the following. (1) For any k < e + d − , one has dim M ( f ) k = (cid:18) k + 33 (cid:19) − (cid:18) k − d + 43 (cid:19) + X j =1 , (cid:18) k − d − e j + 43 (cid:19) . In particular, these dimensions are independent of f once the degrees e and e are fixed. (2) For any k ≥ e + d − , one has dim M ( f ) k ≤ (cid:18) k + 33 (cid:19) − (cid:18) k − d + 43 (cid:19) + X j =1 , (cid:18) k − d − e j + 43 (cid:19) , and the equality holds for any k ≥ e + d − if and only if it holds for k = e + d − . Moreover, these equalities hold exactly when D : f = 0 is afree surface with exponents e , e , e . REENESS VERSUS MAXIMAL DEGREE OF SINGULAR SUBSCHEME FOR SURFACES 3
Theorem 1.3.
In the class of tame surfaces D with fixed degrees e , e satisfying e + 2 e ≤ d − , the free surface (if it exists) has a singular locus Σ of maximaldegree. More precisely, for such a tame surface D with fixed degrees e , e , one has deg Σ ≤ s − s , where e = d − − e − e , and s = P j =1 , e j , s = P i Free, nearly free and tame surfaces in P Let f be a homogeneous polynomial of degree d in the polynomial ring S = C [ x, y, z, w ] and denote by f x , f y , f z , f w the corresponding partial derivatives. Let D be the surface in P defined by f = 0 and assume that D is reduced and not a coneover a plane curve. We denote by J f the Jacobian ideal of f , i.e. the homogeneousideal in S spanned by f x , f y , f z , f w , and by M ( f ) = S/J f the corresponding gradedring, called the Jacobian (or Milnor) algebra of f .Consider the graded S − submodule AR ( f ) ⊂ S of all relations involving thederivatives of f , namely ρ = ( α, β, γ, δ ) ∈ AR ( f ) q if and only if αf x + βf y + γf z + δf w = 0 and α, β, γ, δ are in S q . We set ar ( f ) k =dim AR ( f ) k and m ( f ) k = dim M ( f ) k for any integer k . Definition 2.1. The surface D : f = 0 is a free divisor if the following equivalentconditions hold.(1) M ( f ) is a Cohen-Macaulay S -module, i.e. depth M ( f ) = dim M ( f ) = 2.(2) H m ( M ( f )) = H m ( M ( f )) = 0 , with m = ( x, y, z, w ) the maximal homoge-neous ideal in S = C [ x, y, z, w ].(3) The minimal resolution of the Milnor algebra M ( f ) has the following form0 → ⊕ j =1 , S ( − d j − d + 1) → S ( − d + 1) ( f x ,f y ,f z ,f w ) −−−−−−−→ S for some positive integers d ≤ d ≤ d .(4) The graded S -module AR ( f ) is free of rank 3, i.e. there is an isomorphism AR ( f ) = S ( − d ) ⊕ S ( − d ) ⊕ S ( − d )for some positive integers d ≤ d ≤ d . ALEXANDRU DIMCA (5) The coherent sheaf F on P associated to the graded S -module AR ( f ) splitsas a direct sum of line bundles, i.e. F = O ( − d ) ⊕ O ( − d ) ⊕ O ( − d )for some positive integers d ≤ d ≤ d .When D is a free divisor, the integers d ≤ d ≤ d are called the exponents of D . They satisfy the relation d + d + d = d − P ( M ( f ))( k ) = ak + b of the Milnor algebra M ( f ) are given by(2.1) a = s − s and b = 2 a − s + 32 s s − s , where s = P j =1 , d j , s = P i The surface D : f = 0 is a nearly free divisor if the followingequivalent conditions hold.(1) The Milnor algebra M ( f ) has a minimal resolution of the form0 → S ( − d − d ) → ⊕ j =1 , S ( − d − d j + 1) → S ( − d + 1) ( f x ,f y ,f z ,f w ) −−−−−−−→ S for some integers 1 ≤ d ≤ d ≤ d = d , called the exponents of D .(2) There are 4 syzygies ρ , ρ , ρ , ρ of degrees d ≤ d ≤ d = d = d − ( d + d )which form a minimal system of generators for the first syzygies module AR ( f ).In down-to-earth terms, this definition says that the module AR ( f ) is not free ofrank 3, but it has 4 generators ρ , ρ , ρ and ρ of degree respectively d , d , d and d and the second order syzygy module is spanned by a unique relation(2.2) R : a ρ + a ρ + a ρ + a ρ = 0 , where a , a , a , a are homogeneous polynomials in S of degrees d − d + 1 , d − d + 1 , , D : f = 0 is nearly free, then the exponents d ≤ d ≤ d ( d is omittedsince it coincides to d ) determine the Hilbert polynomial P ( M ( f ))( k ) = ak + b as follows. Define d ′ = d , d ′ = d , d ′ = d − a ′ and b ′ be computed using the formulas in (2.1), i.e. as if a ′ , b ′ were the coefficients of theHilbert polynomial corresponding to a free surface D ′ with exponents d ′ , d ′ , d ′ . Thenone has the formulas(2.3) a = a ′ − b = b ′ + d + d − , see [6]. For both a free and a nearly free surface D : f = 0, it is clear that mdr ( f ) = d .To an element ρ = ( α, β, γ, δ ) ∈ S , we can associate the differential 1-form ω ( ρ ) = αdx + βdy + γdz + δdw ∈ Ω , and consider the Koszul complex of α, β, γ, δ in S given by K ∗ ( α, β, γ, δ ) : 0 → Ω → Ω → Ω → Ω → Ω → , REENESS VERSUS MAXIMAL DEGREE OF SINGULAR SUBSCHEME FOR SURFACES 5 where Ω k denotes the S -module of global algebraic differential k -forms on C and themorphisms are given by the wedge product by ω ( ρ ). If we assume that α, β, γ, δ do nothave any common factor of degree > 0, then the grade of the ideal I = ( α, β, γ, δ )(i.e. the maximal length of a regular sequence contained in I ), which is equal tothe codimension of I , is clearly at least 2. This implies the vanishing of the firstcohomology of the Koszul complex K ∗ ( α, β, γ, δ ), see for instance Thm. A.2.48 in[8], i.e. the following sequence(2.4) 0 → Ω ω ( ρ ) −−→ Ω ω ( ρ ) −−→ Ω is exact. This applies in particular for a surface D : f = 0 if we choose ρ = ρ to be anonzero syzygy of minimal degree, say e , in AR ( f ). Let now ρ be a homogeneousrepresentative in AR ( f ) of a nonzero homogeneous element of minimal degree, say e with e ≥ e , in the quotient module AR ( f ) / ( Sρ ). Note that both ρ and ρ are primitive syzygies , i.e. they are not nonconstant multiples of lower degree syzygies.Let Ω = { ω ( ρ ) : ρ ∈ AR ( f ) } , and consider the sequence of graded S -modules(2.5) S ( − e − ⊕ S ( − e − u =( ω ( ρ ) ,ω ( ρ )) −−−−−−−−−→ Ω v = ω ( ρ ) ∧ ω ( ρ ) −−−−−−−−→ Ω ( e + e + 2) , where the first morphism is u : ( a, b ) aω ( ρ ) + bω ( ρ ) and the second one v isinduced by the wedge product by ω ( ρ ) ∧ ω ( ρ ). Definition 2.3. The surface D is tame with respect to the syzygies ρ and ρ if thesequence (2.5) is exact. The surface D is tame if there is a pair of syzygies ρ and ρ as above, such that the surface is tame with respect to ρ and ρ .Note that a 1-form ω ( ρ ) is in the kernel of v if and only if the matrix M ( ρ , ρ , ρ )with 3 rows, corresponding to the 4 components of ρ , ρ and ρ has rank 2 over thefield of fractions K = C ( x, y, z, w ). The exactness of the sequence (2.4) shows thatthis happens if and only if(2.6) c · ρ = c · ρ + c · ρ , for some polynomials c, c , c in S with c = 0. This implies the following. Lemma 2.4. (i) The surface D : f = 0 is not tame with respect to ρ and ρ exactlywhen there is a primitive syzygy ρ satisfying the equality (2.6) with deg c > and c and c relatively prime polynomials.(ii) Suppose there is a closed Zariski subset B ⊂ C such that the matrix M ( ρ , ρ ) with 2 rows, corresponding to the 4 components of ρ and ρ has rank 2 for any point p = ( x, y, z, w ) / ∈ B . Then the corresponding sequence (2.5) is exact. The assumption in (ii) is equivalent to asking the six 2 × M ( ρ , ρ ) not to have a common divisor in S . Proof. The first claim (i) is clear. To prove (ii), assume that the sequence (2.5) isnot exact. Note that the equality (2.6) implies that the rank of the matrix M ( ρ , ρ )is 1 on the zero set of c , which has codimension one since c is not a constant, hencewe get a contradiction. (cid:3) ALEXANDRU DIMCA Proposition 2.5. (i) Any free surface is tame.(ii) A nearly free surface is tame if and only if the linear forms a and a whichoccur in the second order syzygy (2.2) are linearly independent.Proof. (i) Indeed, if we choose ρ and ρ as the first two elements in the basis of AR ( f ) described in Definition 2.1 (4), it is clear that a syzygy ρ satisfies (2.6) if andonly if it belongs to the image of u , i.e. it is a linear combination of ρ and ρ withcoefficients in S .(ii) If the linear forms a and a which occur in the second order syzygy (2.2) arelinearly dependent, we may assume a = 0 by choosing ρ and ρ appropriatedly.Then ρ = ρ satisfies a ρ = − a ρ − a ρ , which shows that D : f = 0 is not tamewith respect to ρ and ρ . If d > d , then the choice for ρ and ρ is essentialyunique, and we are done. If d = d > d or d = d = d , then the proof can beeasily adapted by the reader, since the choices for ρ and ρ can be listed.Suppose now that the linear forms a and a are linearly independent. Choose ρ and ρ as the first two elements in the system of minimal generators of ρ j ’s,with j = 1 , ..., AR ( f ) described in Definition 2.2 (2) and write ρ , an elementin the kernel of v , as ρ = P i =1 , b i ρ i . Then Lemma 2.4 (ii) implies the existence ofpolynomials c, c , c , a such that c , c are relatively prime and(2.7) aa = cb − c , aa = cb − c , aa = cb and aa = cb . The first two equalities imply that a and c are relatively prime (since c , c arerelatively prime). Then the last two equalities imply that c divides both a and a ,which is possible only if c is a constant. Hence D : f = 0 is tame with respect to ρ and ρ . (cid:3) All the examples of nearly free surfaces described in [6] are tame, because thelinear forms a and a which occur in the second order syzygy (2.2) are linearlyindependent. This can be seen also by applying to each example Lemma 2.4 (ii). Itis natural to ask the following. Question 2.6. Does there exist a nearly free surface in P which is not tame?Note that for a nearly free plane curve, the corresponding two linear forms occuringin the second syzygy similar to (2.2) are always linearly independent, see Remark5.2 in [6]. Example 2.7. (i) We give now an example of a tame surface which is neither freenor nearly free, and for which the choice of the syzygies ρ and ρ is more subtle.Let D : f = ( xw + y ) + y z = 0. Then one has the following generating syzygies:syz[1] : ( x ) f x + (0) f y + (0) f z + ( − w ) f w = 0 , syz[2] : (2 y ) f x + ( − yw ) f y + ( zw ) f z + (0) f w = 0 , syz[3] : (0) f x + ( − yz ) f y + (2 y + z + 2 xw ) f z + (0) f w = 0 , syz[4] : (0) f x + ( xy ) f y + ( − xz ) f z + ( − y ) f w = 0 REENESS VERSUS MAXIMAL DEGREE OF SINGULAR SUBSCHEME FOR SURFACES 7 and some higher degrees ones. If we choose ρ = syz[1] and ρ = syz[2], the sequence(2.5) is not exact. Indeed, one has the following relation wρ = 2 y ρ − xρ , where ρ = syz[4], compare with the equality (2.6). On the other hand, if we choose ρ = syz[1] and ρ = syz[3], then it is easy to see that the rank of matrix M ( ρ , ρ ) is2 outside a subset of codimension 2. Then the corresponding sequence (2.5) is exactby Lemma 2.4. In conclusion, the surface D is tame.(ii) Here we give an example of a surface which is not tame. Let D : f = ( x + y + zw ) + y z = 0. This surface satisfies H m ( M ( f )) = 0 and has the followinggenerators for AR ( f )syz[1] : ( − z ) f x + (0) f y + (0) f z + (2 x ) f w = 0 , syz[2] : (0) f x + ( − yz ) f y + ( z ) f z + (2 y − zw ) f w = 0 , syz[3] : ( − y + zw ) f x + (2 xy ) f y + ( − xz ) f z + (0) f w = 0 , plus some higher degree ones. The relation(2 y − zw ) syz[1] − x syz[2] = z syz[3]shows that the only possible choices, i.e. ρ = syz[1] and ρ a linear combination ofsyz[2] and syz[3], cannot produce an exact sequence due to an equality as in (2.6).(iii) Finally we describe a plane arrangement which is not tame. Consider the ar-rangement D : f = w ( x + y )( y + z )( x + z )( y − z )( x + 2 y + 3 z )(11 x + 7 y + 5 z + 3 w ) . Then AR ( f ) has as generating syzygies one syzygy ρ of degree 2, two syzygies ρ and ρ of degree 3, and some other higher degree generators. The matrix M ( ρ , ρ , ρ )has rank 2 over the field K = C ( x, y, z, w ), and this implies that D is not tame.A natural question is the following. Question 2.8. Can the tameness of a plane arrangement in P be characterizedcombinatorially?A positive answer to this question would have implications for the Terao’s conjec-ture (see [14] for a discussion of this conjecture), similar to those for line arrangementsdiscussed in [5]. This comes from the fact that deg Σ is known to be determined bythe combinatorics, see [10] and [13].3. Bourbaki ideal of the syzygy module For a reduced surface D : f = 0 in P , we choose ρ = ρ to be a nonzero syzygy ofminimal degree, say e , and ρ a homogeneous representative in AR ( f ) of a nonzerohomogeneous element of minimal degree, say e with e ≥ e , in the quotient module AR ( f ) / ( Sρ ). ALEXANDRU DIMCA Let X = ∇ f be gradient vector field of f on C and denote by ι X : Ω k → Ω k − the interior product given by the contraction of a differential fom by the vector field X . The (homology) complex(3.1) 0 → Ω ι X −→ Ω ι X −→ Ω ι X −→ Ω ι X −→ Ω → , is nothing else but the Koszul complex of the partial derivatives f x , f y , f z , f w of f .Since D is reduced, it has a singular set of codimension at most 2 in P , and we get,exactly as we have obtained the exact sequence 2.4, an exact sequence(3.2) 0 → Ω ι X −→ Ω ι X −→ Ω . Now note that a 1-form ω is in Ω , i.e. it comes from a syzygy in AR ( f ), if and onlyif ι X ( ω ) = 0. Since ι X is an anti-derivation, i.e. it satisfies a graded Leibnitz rule, itfollows that the image of the morphism v from (2.5) is contained inΩ = { ω ∈ Ω : ι X ( ω ) = 0 } . And the exact sequence (3.2) gives an isomorphism S ∼ = Ω (4) ∼ = Ω ( d + 2) . The above proves the following result. Theorem 3.1. Let D : f = 0 be a reduced surface in P , tame with respect to thepair ρ , ρ having degrees e = deg ρ ≤ e = deg ρ . Set e = d − − e − e . Thenone has the following exact sequence of graded S -modules → S ( − e ) ⊕ S ( − e ) u ′ −→ AR ( f ) v ′ −→ S ( − e ) , where u ′ ( a, b ) = aρ + bρ and v ′ ( ρ ) = h , with h the unique polynomial such that ω ( ρ ) ∧ ω ( ρ ) ∧ ω ( ρ ) = hι X ( dx ∧ dy ∧ dz ∧ dw ) . If we denote by B ( f ) ⊂ S the image of the morphism v ′ , it follows that B ( f ) isa Bourbaki ideal for the syzygy module AR ( f ), see [3], Chapitre 7, § 4, Thm. 6, aswell as section 3 in [12]. Corollary 3.2. With the above notation, the minimal number of generators of theBourbaki ideal B ( f ) is the same as the minimal number of syzygies that one mustadd to ρ and ρ in order to get a generating set of the syzygy module AR ( f ) . Remark 3.3. If we denote by M ( E, ρ , ρ , ρ ) the 4 × x, y, z, w ), and as the next rowsthe components of ρ , ρ and ρ respectively, then the equality v ′ ( ρ ) = h is equivalentto det M ( E, ρ , ρ , ρ ) = h · f. By the choice of the syzygies ρ , ρ , it is clear that (im u ′ ) k = AR ( f ) k , for any k ≤ e − 1. Then Theorem 3.1 implies the following consequence, which is essentiallythe same as Lemma 4.12 in [6], which in turn is an extension of the correspondingresult for free curves in Lemma 1.1 in [11]. This result also shed a new light onSaito’s criterion of freeness, see [9], [14], [6], which is used to get the freeness of D in the result below. REENESS VERSUS MAXIMAL DEGREE OF SINGULAR SUBSCHEME FOR SURFACES 9 Corollary 3.4. Let p be the smallest integer such that (im u ′ ) p = AR ( f ) p , i.e. p isthe first degree where some new generator for AR ( f ) should be added besides ρ and ρ . Assume e = d − − e − e ≥ e . Then p ≥ e and the following conditions areequivalent. (1) p = e ; (2) D : f = 0 is a free surface with exponents e , e , e ; (3) ar ( f ) e > (cid:18) e − e + 33 (cid:19) + (cid:18) e − e + 33 (cid:19) ;(4) ar ( f ) e = (cid:18) e − e + 33 (cid:19) + (cid:18) e − e + 33 (cid:19) + 1 . (5) For any k ≥ e one has ar ( f ) k ≥ (cid:18) k − e + 33 (cid:19) + (cid:18) k − e + 33 (cid:19) + (cid:18) k − e + 33 (cid:19) ;(6) For any k ≥ e one has ar ( f ) k = (cid:18) k − e + 33 (cid:19) + (cid:18) k − e + 33 (cid:19) + (cid:18) k − e + 33 (cid:19) . Proof of Theorems 1.2 and 1.3 We have an obvious exact sequence0 → AR ( f ) k − d +1 → S k − d +1 → S k → M ( f ) k → . This implies(4.1) m ( f ) k = (cid:18) k + 33 (cid:19) − (cid:18) k − d + 43 (cid:19) + ar ( f ) k − d +1 , for k ≥ d − Example 4.1. For the surface D : f = 0 in Example 2.7, we have deg Σ = 36, while s − s = 35 . Indeed, in this case e = 1, e = 2 and e = d − − e − e = 4. Hencethe condition of tameness is necessary to have the result in Corollary 1.3.The proof of Theorem 1.3 is more involved, even if in view of Theorem 1.2, theonly point that needs explanation is the fact that the equality s − s = deg Σ impliesthat D : f = 0 is free. Consider the following exact sequence of graded S -modules(4.2) 0 → S ( − e ) ⊕ S ( − e ) u ′ −→ AR ( f ) v ′ −→ B ( f )( − e ) → , coming from Theorem 3.1, where B ( f ) is the Bourbaki ideal of AR ( f ). Using thisand the obvious exact sequence0 → B ( f )( − e ) → S ( − e ) → ( S/B ( f ))( − e ) → , we get ar ( f ) k = (cid:18) k − e + 33 (cid:19) + (cid:18) k − e + 33 (cid:19) + (cid:18) k − e + 33 (cid:19) − dim( S/B ( f )) k − e . Combining this equality with the relation (4.1) written for k replaced by k + d − k large enough,(4.3) dim( S/B ( f )) k − e = Q ( k ) − m ( f ) k + d − , where the polynomial Q ( k ) is given by(4.4) Q ( k ) = X j =1 , (cid:18) k − e j + 33 (cid:19) − (cid:18) k + 33 (cid:19) + (cid:18) k + d + 23 (cid:19) . If we expand binomial coefficients, we get that Q ( k ) is a polynomial in k of degree(at most) one and the coefficient of k is given by s − s . It follows that the equality s − s = deg Σ and the equality (4.3) imply that the subscheme Y of P defined bythe ideal B ( f ) is 0-dimensional or empty. On the other hand, Lemma 3 in [1] showsthat the existence of the exact sequence (4.2) with AR ( f ) a reflexive S -module andcodim Y ≥ Y has pure codimension two, or Y is empty. See alsothe discussion on p. 145 in [2] and note that AR ( f ) is a second syzygy module bydefinition. Hence the only possibility is that Y is empty, i.e. B ( f ) is an m -primaryideal. Then the exact sequence (4.2) translates into the following exact sequence ofsheaves on P → O ( − e ) ⊕ O ( − e ) → F → O ( − e ) → , where F is the sheaf associated to the graded S -module AR ( f ). But this impliesthat F = O ( − e ) ⊕ O ( − e ) ⊕ O ( − e ), i.e. D : f = 0 is free with exponents e , e , e ,since the group of extensions of O ( − e ) by O ( − e ) ⊕ O ( − e ) isExt ( O ( − e ) , O ( − e ) ⊕ O ( − e )) = Ext ( O , O ( e − e ) ⊕ O ( e − e )) == H ( P , O ( e − e ) ⊕ O ( e − e )) = 0 . This completes the proof of Theorem 1.3.5. Two characterizations of nearly free tame surfaces Proposition 5.1. Let D : f = 0 be a reduced surface in P , tame with respect to thepair of syzygies ρ , ρ of degrees e = deg ρ ≤ e = deg ρ . Set e = d − − e − e and assume e ≥ e . Then D is nearly free with exponents e , e , e + 1 if and onlyif dim B ( f ) = 2 .Proof. If D is nearly free, then the equality dim B ( f ) = 2 follows from Proposition2.5, (ii), since the image under v ′ of the syzygies ρ and ρ are, up to a constantfactor, a and − a , in the notation from the proof of Proposition 2.5. Conversely,choose ρ and ρ such that their images a ′ and a ′ under v ′ span B ( f ) . It followsthat the subscheme Y defined by the ideal B ( f ) is contained in the projective line L in P defined by a ′ = a ′ = 0. As in the proof above of Theorem 1.3, using Lemma3 in [1], we see that there are only the following two cases to discuss. REENESS VERSUS MAXIMAL DEGREE OF SINGULAR SUBSCHEME FOR SURFACES 11 Case 1. Y is empty, and this corresponds as we have seen in Theorem 1.3 to thecase D is a free divisor. But for a free divisor one has dim B ( f ) = 4, as the imageof v ′ is the whole ring S . Hence this case is impossible. Case 2. Y has pure codimension 2. It follows that Y = L , and hence B ( f ) isthe ideal spanned by the independent linear forms a ′ and a ′ . This implies usingTheorem 3.1 that the four syzygies ρ , ρ , ρ and ρ span the syzygy module AR ( f ),and hence D : f = 0 is a nearly free surface. (cid:3) Remark 5.2. (i) To check that a given surface is tame with respect to the pair ofsyzygies ρ , ρ without having detailed information on the syzygy module AR ( f ) canbe done using Lemma 2.4 (ii).(ii) In Proposition 5.1, the condition dim B ( f ) = 2 can be replaced by the conditiondim B ( f ) ≥ D is not free. This is the perfect analog of the characterizationof nearly free curves in Theorem 4.1 (ii) in [5], except that for surfaces we need theextra tameness condition. Corollary 5.3. Let D : f = 0 be a reduced surface in P , tame with respect to thepair of syzygies ρ , ρ of degrees e = deg ρ ≤ e = deg ρ . Set e = d − − e − e and assume e ≥ e and that D is not free. Then deg Σ ≤ s − s − , and the equality holds if the surface D is nearly free with exponents e , e , e + 1 .Conversely, if this equality holds and if the syzygy module AR ( f ) is spanned by ρ , ρ and two other syzygies ρ and ρ , then D is a nearly free surface with exponents e , e , e + 1 .Proof. The only point that needs explanation is the last claim. The formula (4.3) im-plies that deg Y , which is the leading coefficient of the Hilbert polynomial H ( S/B ( f )),is one. On the other hand, we know as above that Y has pure codimension 2, sinceit is non empty. Under our assumption, Y is in fact a complete intersection, sincethe ideal B ( f ) is spanned by g j = v ′ ( ρ j ) for j = 3 , 4. The degree of the completeintersection Y is the product deg g · deg g , hence both g and g are linear forms.The claim then follows from Proposition 5.1. (cid:3) The following example shows that the equality s − s − D : f = 0 is nearly free without the tameness assumption. Example 5.4. Consider the surface D : f = x + x y + y z + w ( x w + y z ) . Thissurface is not nearly free since H m ( M ( f )) = 0, see [6]. The generators of AR ( f )have degrees ≥ AR ( f ) = 4. The corresponding Hilbert polynomial is H ( M ( f ))( k ) = 81 k − 507 for k ≥ 21. If we set e = e = 3 and e = d − − e − e = 5,we get s − s − . References [1] C. Bertone, M. Roggero, Positivity of Chern classes for reflexive sheaves on P N , Geom. Dedicata142 (2009), 121–138. 4, 5[2] W. Bruns, E. G. Evans, P. A. Griffith, Syzygies, ideals of height 2, and vector bundles, J.Algebra 67 (1980), 143-162. 4[3] N. Bourbaki, Alg`ebre Commutative, Chapitres I-IX, Hermann, 1961-1983. 3[4] A. D. R. Choudary, A. Dimca, Koszul complexes and hypersurface singularities, Proc. Amer.Math. Soc. 121(1994), 1009–1016. 1[5] A. Dimca, Freeness versus maximal global Tjurina number for plane curves, arXiv:1508.0495.1, 1, 2, 5.2[6] A. Dimca, G. 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