The structure of Koszul algebras defined by four quadrics
aa r X i v : . [ m a t h . A C ] J a n THE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BYFOUR QUADRICS
PAOLO MANTERO AND MATTHEW MASTROENI
Abstract.
Avramov, Conca, and Iyengar ask whether β Si ( R ) ≤ (cid:0) gi (cid:1) for all i when R = S/I is a Koszul algebra minimally defined by g quadrics. In recentwork, we give an affirmative answer to this question when g ≤ g ≤ Introduction
A graded ideal I in a standard graded polynomial ring S over a field k definesa Koszul algebra R = S/I if k ∼ = R/R + has a linear free resolution over R . Thisextraordinary homological condition leads to numerous other restrictions on theHilbert series and Betti numbers of R over S ; see [Fr¨o99] and [Con14]. In particular,a question of Avramov, Conca, and Iyengar asks: Question 1.1 ([ACI10, 6.5]) . If R is Koszul and I is minimally generated by g quadrics, does the following inequality hold for all i ? β Si ( R ) ≤ (cid:18) gi (cid:19) In particular, is pd S R ≤ g ?This question is known to have an affirmative answer when R is LG-quadratic.We say that R or I is G-quadratic if, after a suitable linear change of coordinates ϕ : S → S , the ideal ϕ ( I ) has a Gr¨obner basis consisting of quadrics. We alsosay that R or I is LG-quadratic if R is a quotient of a G-quadratic algebra A by an A -sequence of linear forms. Every G-quadratic algebra is Koszul by uppersemicontinuity of the Betti numbers; see [BC03, 3.13]. Since deforming by a regularsequence of linear forms preserves Koszulness (see Proposition 2.1), every LG-quadratic algebra is also Koszul. Almost all known examples of (commutative) Koszul algebras are LG-quadratic.They include all quotients by quadratic monomial ideals, all quadratic completeintersections [Cav04, 1.2.5], the coordinate rings of Grassmannians in their Pl¨uckerembedding [SW89] and most canonical curves [CRV01], many types of toric rings[Hib87] [BGT97] [OH99], and all suitably high Veronese subrings of any standardgraded algebra [ERT94]. However, there is one notable exception due to Conca.
Example 1.2 ([Con14, 3.8]) . The ring R = k [ x, y, z, w ] / ( xy, xw, ( x − y ) z, z , x + zw )defined by g = 5 quadrics is Koszul by a filtration argument, but it is not LG-quadratic since the numerator of its Hilbert series H R ( t ) = 1 + 2 t − t − t + 2 t (1 − t ) cannot be realized by any 5-generated edge ideal. Nonetheless, the Betti table of R is 0 1 2 3 40 1 – – – –1 – 5 4 – –2 – – 4 6 2so that Question 1.1 still has an affirmative answer in this case.In general, Question 1.1 is known to have an affirmative answer for Koszulalgebras defined by g ≤ I = g −
1) with any number of generators [Mas18]. In[MM20], we give a strong affirmative answer to Question 1.1 for g = 4 quadricsby determining the possible Betti tables of Koszul algebras defined by height twoideals minimally generated by four quadrics. Recording the graded Betti number β Si,i + j ( R ) = dim k Tor Si ( R, k ) i + j in column i and row j of each table and representingzero entries by “ − ” for readability, they are: Theorem 1.3 ([MM20, 4.7]) . Let R = S/I be a Koszul algebra defined by fourquadrics with ht I = 2 . Then the Betti table of R over S is one of the following: (i) 0 1 2 30 1 – – –1 – 4 4 1(ii) 0 1 2 3 40 1 – – – –1 – 4 3 1 –2 – – 3 3 1 (iii) 0 1 2 30 1 – – –1 – 4 3 –2 – – 1 1(iv) 0 1 2 3 40 1 – – – –1 – 4 2 – –2 – – 4 4 1 HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 3
The current paper is the natural continuation of the above work. Our main result(Theorem 3.1) is a structure theorem for the height two ideals generated by fourquadrics defining Koszul algebras over an algebraically closed field. Combining itwith previous structure theorems for Koszul almost complete intersections withany number of generators [Mas18] yields a complete description of the structure ofKoszul algebras defined by g ≤ g = 4 in thetheorem below. Theorem A.
Let S be a polynomial ring over an algebraically closed field k and I ⊆ S be an ideal generated by g = 4 quadrics. Then R = S/I is Koszul if andonly if I has one of the following forms: (1) ( a x, . . . , a x ) for some linear forms x, a , . . . , a with ht( a , . . . , a ) = 4 when ht I = 1 . (2) When ht I = 2 : (i) ( x, y ) ∩ ( z, w ) or ( x, y ) + ( xz + yw ) for independent linear forms x , y , z and w , or ( xy, xz, xw, q ) for independent linear forms y , z , and w and some linear form x and quadric q ∈ ( y, z, w ) \ ( x ) . (ii) ( xy, xz, xw, q ) for independent linear forms y , z , and w and some lin-ear form x and quadric q which is a nonzerodivisor modulo ( xy, xz, xw ) . (iii) ( xz, yz, a x + b y, a x + b y ) for some linear forms x , y , z , a , a , b ,and b such that ht( x, y ) = ht( a x + b y, a x + b y ) = 2 and ht( z, a x + b y, a x + b y, a b − a b ) = 3 . (iv) ( a x, a x, b y, b y ) for some linear forms x , y , a i , b i such that ( a x, a x ) and ( b y, b y ) are transversal ideals with ht( a , a ) = ht( b , b ) = 2 . (3) When ht I = 3 : (i) I = ( xz, xw, q , q ) , where x, z, w are linear forms and q , q are quadricsforming a regular sequence on S/ ( xz, xw ) ; (ii) I = I ( M ) + ( q ) where M is a × matrix of linear forms of S with ht I ( M ) = 2 and q is a quadric regular on S/I ( M ) . (4) I is a complete intersection of quadrics when ht I = 4 . We emphasize that Theorem A applies over an algebraically closed field of ar-bitrary characteristic. It agrees with the computer-aided computations of Roos in[Roo16] for characteristic zero. Out of the 83 different types of quadratic algebrasin 4 variables identified by Roos, there are only four types of Koszul algebras de-fined by height-two ideals of four quadrics with different Koszul homology algebrastructure (Cases 27, 25, 26, 21 in Table 4). These cases correspond precisely toour cases (i)–(iv) respectively.In addition to the intrinsic interest in having a structure theorem, a large partof our interest in the investigation of Koszul algebras defined by four quadrics owesto Conca’s ring R in Example 1.2, which points to unexpected complications in PAOLO MANTERO AND MATTHEW MASTROENI trying to answer Question 1.1 for Koszul algebras defined by g ≥ e ( R ) = 1 and every height-oneideal of quadrics is easily seen to be G-quadratic, Conca’s example is minimal interms of multiplicity and codimension. As a consequence of our results, we showthat Conca’s example is also minimal in terms of the number of quadrics. Theorem B. If R = S/I is a Koszul algebra such that I is minimally generatedby g ≤ quadrics, then, after passing to the algebraic closure of the ground field, R is LG-quadratic. Hence, in hindsight, it is no surprise that Question 1.1 has an affirmative answerfor g ≤
4. We do not know of an example of a Koszul algebra which is notLG-quadratic but which becomes LG-quadratic after extending the ground field.Another question of interest raised in [CI+15] asks whether the absolutely Koszuland Backelin–Roos properties coincide over algebraically closed fields. As a secondapplication of our main theorem, we determine which Koszul algebras defined byat most four equations have the absolutely Koszul property and prove that, in oursetting, the question of Conca et al. has a positive answer (at least if ch( k ) = 0).Finally, we spend a few words on the techniques employed in this paper. To provethe structure theorem, we exploit a number of considerations about syzygies, link-age, annihilators of certain Ext modules, and 1-generic matrices. Deformations to“generic” cases as well as specializations play an important role in every section,including the appendix. In Theorem 3.6, we prove that there are four homolog-ically distinct types of ideals having Betti table (iv) in Theorem 1.3. However,three of them do not define Koszul algebras. Since these cases are numericallyindistinguishable from a Koszul algebra by their Betti table, we show that two ofthese cases are not Koszul using a recently proved condition on the first syzygymap of a Koszul algebra [Mas18, 2.8]. Unfortunately, proving the remaining caseis not Koszul is much more challenging – indeed, the entire appendix is devoted toit. To achieve this goal, we first deform and specialize to a specific algebra whichcan be realized as a symmetric algebra of a certain module over a smaller ring.Exploiting the natural bigrading and invoking a result of [HHO00] about algebraretracts then allows us to significantly reduce the size of the computation neededto show that this ring is not Koszul. Thus, we think that the proofs that thesealgebras are not Koszul may be of independent interest. Notation.
Throughout the remainder of the paper, the following notation will bein force unless specifically stated otherwise. Let k be a fixed algebraically closedground field of arbitrary characteristic, S be a standard graded polynomial ringover k , I ⊆ S be a proper graded ideal, and R = S/I . Recall that the ideal I iscalled nondegenerate if it does not contain any linear forms. We can always reduceto a presentation for R with I nondegenerate by killing a basis for the linear forms HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 5 contained in I , and we will assume that this is the case throughout. We denotethe irrelevant ideal of R by R + = L n ≥ R n .The division of the rest of the paper is as follows. After reviewing some usefulbackground about LG-quadratic algebras in §
2, we complete our classification ofKoszul algebras defined by four quadrics in §
3. Finally, we apply our structuretheorems in § Acknowledgements.
Evidence for the results in this paper was provided by sev-eral computations in Macaulay2 [M2]. We also thank Aldo Conca for helpfulconversations related to this work.2.
Background on LG-Quadratic Algebras
It is useful to know how the Koszul property can be passed to and from quotientrings.
Proposition 2.1 ([CDR13, § . Let S be a standard graded k -algebra and R be a quotient ring of S . (a) If S is Koszul and reg S ( R ) ≤ , then R is Koszul. (b) If R is Koszul and reg S ( R ) is finite, then S is Koszul. Here, the regularity of R over S is defined byreg S ( R ) = max { j | β Si,i + j ( R ) = 0 for some i } (2.1)We will primarily use the above proposition to show that the Koszul property canbe passed back and forth between a ring and its quotient by a regular sequenceconsisting of linear forms or quadrics.The following simple proposition about Hilbert series and its corollary will beuseful for determining that various rings are LG-quadratic. Proposition 2.2 ([Sta78, 3.2]) . Let I ⊆ S be an ideal and L = ( x , . . . x c ) ⊆ S bean ideal generated by linear forms. Then H S/ ( I,L ) ( t ) ≥ (1 − t ) c H S/I ( t ) with equality if and only if the generators of L form a regular sequence mod I . Corollary 2.3.
Let I ⊆ S be an ideal and L = ( x , . . . x c ) ⊆ S be an idealgenerated by linear forms. If the graded Betti numbers of S/I over S are the sameas the graded Betti numbers of S/ ( I, L ) over S/L , then L is generated by a regularsequence mod I .Proof. Since the graded Betti numbers of
S/I determine the numerator P ( t ) ofits Hilbert series, we see that (1 − t ) c H S/I ( t ) = (1 − t ) c P ( t )(1 − t ) dim S = P ( t )(1 − t ) dim S/L = H S/ ( I,L ) ( t ) so that conclusion follows from the preceding proposition. (cid:3) PAOLO MANTERO AND MATTHEW MASTROENI
As a particular application of this fact, we have the following proposition. Re-call that two ideals I , I ⊆ S in a standard graded polynomial ring over k arecalled transversal if I ∩ I = I I . This is easily seen to be equivalent to havingTor S ( S/I , S/I ) = 0 and, hence, Tor Si ( S/I , S/I ) = 0 for all i ≥ Proposition 2.4.
Suppose that R = S/I where I = I + I for transversal heightone ideals I and I generated by quadrics. Then R is LG-quadratic.Proof. We note that I = ( a x, . . . , a s x ) for some linear form x and independentlinear forms a i . Likewise, I = ( b y, . . . , b r y ) for some linear form y and inde-pendent linear forms b j . The minimal free resolutions of S/I and S/I are justthe Koszul complexes of a , . . . , a s and b , . . . , b r respectively, except that the firstdifferential is multiplied by x or y accordingly. Furthermore, since I and I aretransversal, the minimal free resolution of R is just the tensor product of theseresolutions. Thus, the graded Betti numbers of R are uniquely determined by thefact that I and I are transversal height one ideals of quadrics.In the case, where x, y, a , . . . , a s , b , . . . , b r are independent linear forms, R isclearly G-quadratic, since after a suitable linear change of coordinates we mayassume that the given linear forms are variables, and hence, I is a monomialideal. In general, we can consider the G-quadratic algebra A = ˜ S/ ˜ I where ˜ S = S [ z, w, u , . . . , u s , v , . . . , v r ] and ˜ I = ( u z, . . . , u s z, v w, . . . , v r w ). If L = ( u i − a i , v j − b j , z − x, w − y | ≤ i ≤ r, ≤ j ≤ s ) denotes the ideal of linear formsspecializing ˜ I to I , we see that the graded Betti numbers of ˜ S/ ( ˜ I, L ) ∼ = S/I over˜
S/L ∼ = S are the same as those of ˜ S/ ˜ I over ˜ S . Thus, L is generated by a regularsequence mod I by preceding corollary, and R is LG-quadratic. (cid:3) The Structure Theorem
Most of this section is devoted to proving the following structure theorem.
Theorem 3.1.
Let I ⊆ S be an ideal of height 2 generated by g = 4 quadrics.Then R = S/I is Koszul if and only if I has one of the following forms: (i) ( x, y ) ∩ ( z, w ) or ( x, y ) + ( xz + yw ) for independent linear forms x , y , z and w , or ( xy, xz, xw, q ) for independent linear forms y , z , and w andsome linear form x and quadric q ∈ ( y, z, w ) \ ( x ) . (ii) ( xy, xz, xw, q ) for independent linear forms y , z , and w and some linearform x and quadric q which is a nonzerodivisor modulo ( xy, xz, xw ) . (iii) ( xz, yz, a x + b y, a x + b y ) for some linear forms x , y , z , a , a , b , and b such that ht( x, y ) = ht( a x + b y, a x + b y ) = 2 and ht( z, a x + b y, a x + b y, a b − a b ) = 3 . (iv) ( a x, a x, b y, b y ) for some linear forms x , y , a i , b i such that ( a x, a x ) and ( b y, b y ) are transversal ideals with ht( a , a ) = ht( b , b ) = 2 . HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 7
We summarize the proof below.
Proof. (= ⇒ ): If R is Koszul, then R must have one of the Betti tables listed inTheorem 1.3. Tables (i) and (ii) correspond precisely to cases (i) and (ii) respec-tively by [MM20, 3.1, 3.3]. Case (iii) follows by Theorem 3.4. Case (iv) follows byTheorem 3.6 and Lemma 3.9.( ⇐ =): If I has form (i) or (ii), then I is G-quadratic by [MM20, 3.5]. If I hasform (iii) or (iv), then I is LG-quadratic by Corollary 3.5 and Proposition 2.4respectively. (cid:3) As noted above, it remains only to prove structure theorems for the ideals real-izing Betti tables (iii) and (iv) in Theorem 1.3, both of which correspond to idealsof multiplicity one. To aid in the analysis of Betti table (iii), we make use of thefollowing result, which is credited to Eisenbud and Evans in [EHV92] but for whichwe were unable to find an explicit reference. Luckily, it is also essentially a resultof Schenzel on the annihilators of local cohomology.
Lemma 3.2. If R = S/I is a standard graded k -algebra with dim S = d and weset a i = Ann S Ext iS ( R, S ) for each i , then a · · · a d ⊆ I .Proof. By Grothendieck duality, we know that Ext iS ( R, S ) is Matlis dual to H d − iR + ( R )so that a i = Ann S H d − iR + ( R ) for every i . Setting ¯ a i = a i /I for each i , it follows from[Sch79, Prop 3] that ¯ a · · · ¯ a d = 0. Although Schenzel’s result is stated in the localcase, we can reduce to the local case since the ideal ¯ a · · · ¯ a d is graded and, there-fore, it is zero if and only if its localization at the irrelevant ideal is zero. Hence,we must have a · · · a d ⊆ I as wanted. (cid:3) We will also need the following; the interested reader should consult the originalversion for a more general statement. We recall that the unmixed part of an ideal I of height c is the intersection of all the components of height c appearing in anirredundant primary decomposition of I and is denoted by I unm . Proposition 3.3 ([EHV92, 1.1]) . Let R = S/I be a standard graded k -algebrawith ht I = c , and set a i = Ann S Ext iS ( R, S ) for each i . Then ht a i ≥ i for all i ,and a c = I unm . Theorem 3.4.
The ring R = S/I has Betti table if and only if I = ( xz, yz, a x + b y, a x + b y ) for some linear forms x , y , z , a , a , b , b such that ht( x, y ) = ht( a x + b y, a x + b y ) = 2 , and ht( z, a x + b y, a x + b y, a b − a b ) = 3 . PAOLO MANTERO AND MATTHEW MASTROENI
Proof.
By computing the Hilbert series from the Betti table of R , we see that e ( R ) = 1 and ht I = 2. It then follows from the associativity formula for mul-tiplicity [BH93, 4.7.8] that I is contained in a unique minimal prime of heighttwo, which has the form P = ( x, y ) for independent linear forms x and y , andthe length of R P is one so that P must be the P -primary component of I . Set-ting a i = Ann S Ext iS ( R, S ) for each i , it follows from the above proposition that P = I unm = a . By the above lemma, we also know that a a ⊆ I since a i = R for i < iS ( R, S ) = 0. Now, consider the minimal free resolution F • of R over S , which by assumption has the form F • : 0 −→ S ( − ϕ −→ S ( − ⊕ S ( − ϕ −→ S ( − ϕ −→ S (3.1)Since ϕ ∗ : S (3) ⊕ S (4) → S (5) is a presentation for Ext S ( R, S ), we see that a = I ( ϕ ). Hence, there is a linear form z such that P z ⊆ a a ⊆ I . We notethat z = 0; otherwise a would be a complete intersection of 3 quadrics by theabove proposition, and at least one of the rows of the matrix of linear syzygiesof I would be a nontrivial linear syzygy of a , which is impossible. Consequently, xz and yz are independent quadrics so that they are part of a minimal set ofgenerators for I , and we can write I = ( xz, yz, q , q ) where q i = a i x + b i y for somelinear forms a i and b i as I ⊆ P .We claim that ht( q , q ) = 2. Indeed, if this is not the case, then ( q , q ) ⊆ ( ℓ )for a linear form ℓ . If ℓ / ∈ P , then ( ℓ, z ) is a prime ideal of height at most 2 differentfrom P that contains I , which is impossible. If instead ℓ ∈ P , then for any linearform y ′ with P = ( ℓ, y ′ ), we have I = ( ℓz, y ′ z, q , q ). Thus, three of the generatorsof I are divisible by ℓ , which implies e ( R ) = 2 by [MM20, 3.3].Set ∆ = a b − a b . It remains to show that ht( z, q , q , ∆) = 3. We claim thatthe maps ϕ i in (3.1) can be chosen so that ϕ = ∆ − q q z ϕ = y a a − x b b − z q − z − q ϕ = (cid:0) xz yz q q (cid:1) (3.2)Indeed, the columns of ϕ are clearly syzygies of I . We note that ( q , q ) * ( z )since ht( q , q ) = 2. Since the first three columns are easily seen to be independentlinear syzygies and the fourth column is not contained in the S -span of the firstthree columns by the preceding observation, it follows that columns of ϕ are partof a minimal set of generators for Syz S ( I ), and so, our assumption on Betti table of R forces these syzygies to be all of the minimal generators for Syz S ( I ). A similarargument shows that ϕ is the unique syzygy on these generators up to scalarmultiple. With the maps as above, it is clear that ht( z, q , q , ∆) = ht a ≥ z, q , q , ∆) = ( z, I ( M )) where M HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 9 is the submatrix consisting of the first two rows and first three columns of ϕ , andso, we have ht( z, q , q ) ≤ I ( M ) ≤ z, q , q ) = 3.Conversely, when I has the preceding form, we show that R has the desiredBetti table by explicitly constructing the minimal free resolution of R . Considerthe complex F • of free graded S -modules as in (3.1) with differentials as in (3.2).We use the Buchsbaum-Eisenbud acyclicity criterion [BE73] to show that (3.1) isthe minimal free resolution of R . We must check that ht I r i ( ϕ i ) ≥ i for each i ≥ r i = P j ≥ i ( − j − i rank F j . In our case, it is clear that ht I ( ϕ ) = ht I ≥ I ( ϕ ) = ( z, q , q , ∆) has height three by assumption. Finally, one easilychecks by computing minors that ( q , q ) ⊆ I ( ϕ ), and thus, by assumption,ht I ( ϕ ) ≥ (cid:3) Corollary 3.5. If R = S/I is as above, then R is LG-quadratic.Proof. We can consider the generic ideal of the form described by the theoremwhere the linear forms are all variables. By the theorem and Corollary 2.3, anyparticular ideal I of the above form must be obtained by specializing the genericcase by a regular sequence of linear forms. Thus, it suffices to prove that I isG-quadratic in the generic case.In that case, we claim that the generators of I are a Gr¨obner basis for any degreelexicographic order satisfying a > b > b > a > x > y > z . Setting q = xz and q = yz , we apply Buchberger’s Criterion (see [EH12]) by checking that theS-pairs S ( q i , q j ) reduce to zero upon division by q , . . . , q for all i < j . This isclear for ( i, j ) = (1 , , , ,
4) since either the S-pair is zero or theinitial monomials of q i and q j are relatively prime in those cases. In the remainingcases, we see that S ( q , q ) = − b yz and S ( q , q ) = − a xz also clearly reduce tozero so that the generators of I form a Gr¨obner basis as claimed. (cid:3) We now turn our attention to the last remaining Betti table from Theorem 1.3.This is the first instance where the Betti table does not determine whether thering R is Koszul. Theorem 3.6.
The ring R = S/I has Betti table if and only if there are linear forms x , y , a i , and b i with ht( x, y ) = 2 such that I has one of the following forms: (a) ( x , b x, a x + b y, a x + b y ) , where ht( x, b , b ) = 3 and ht( x, y, a , b ) = 4 . (b) ( xy, a x, b y, a x + b y ) , where we have ht( x, b , b ) = ht( y, a , a ) = 3 and ht( x, y, a , b ) = 4 . (c) ( b x, b x, a x + b y, a x + b y ) , where ht( x, b , b ) = 3 , ht( a , a , b , b ) = 4 ,and ht( a x + b y, a x + b y ) = 2 . (d) ( a x, a x, b y, b y ) , where ( a x, a x ) and ( b y, b y ) are transversal idealswith ht( a , a ) = ht( b , b ) = 2 . To simplify the exposition of the proof of the theorem, we first state somepreparatory lemmas.
Remark 3.7. If I , I ⊆ S are transversal ideals and z is a linear form which is anonzerodivisor modulo I = I + I , then the images of I and I in S ′ = S/zS arestill transversal. Indeed, if I ′ j = ( I j , z ) and I ′ = ( I, z ), then applying the SnakeLemma to multiplication by z on each term of the exact sequence0 → S/ ( I ∩ I ) → S/I ⊕ S/I → S/I → → S/ ( I ∩ I , z ) → S/I ′ ⊕ S/I ′ → S/I ′ → I ′ ∩ I ′ = ( I ∩ I , z ). Since I and I are transversal, we see that I ′ ∩ I ′ = ( I ∩ I , z ) =( I I , z ) = I ′ I ′ so that I S ′ ∩ I S ′ = ( I ′ ∩ I ′ ) S ′ = I ′ I ′ S ′ = ( I S ′ )( I S ′ ) as claimed. Lemma 3.8.
Suppose that I = ( a x, a x, b y, b y ) ⊆ S for some linear forms x , y , a i , and b i . Then the Betti table of R = S/I is if and only if ( a x, a x ) and ( b y, b y ) are transversal ideals with ht( a , a ) =ht( b , b ) = ht( x, y ) = 2 .Proof. Let I = ( a x, a x ) and I = ( b y, b y ). If I and I are transversal andht( a , a ) = ht( b , b ) = 2, then the minimal free resolution of S/I j is of the form0 −→ S ( − −→ S ( − −→ S , and the tensor product of the resolutions of S/I and S/J I is a resolution of R so that R has the given Betti table.Conversely, suppose that R has the given Betti table. Then it is easily seenthat ht I = 2, and I is minimally generated by 4 quadrics, from which it followsthat ht( a , a ) = ht( b , b ) = ht( x, y ) = 2. It remains to see that I and I are transversal. This is clear in the generic case where the linear forms are allvariables. By Corollary 2.3, I must be obtained by specializing the generic caseby a regular sequence of linear forms, and so, I and I are transversal by thepreceding remark. (cid:3) HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 11
Lemma 3.9. If I has one of the forms (a)–(c) in Theorem 3.6, then R = S/I hasBetti table and is not Koszul.Proof.
We explicitly construct the minimal free resolution of R for each of theforms (a)–(c) described in Theorem 3.6. In each case, we let q i = a i x + b i y for i = 3 ,
4, and we let J be the ideal generated by the first three generators of I . Case (a): Since ht( x, b , b ) = 3 and ht( x, y, a , b ) = 4, we see that b x, q is aregular sequence so that J = I (cid:18) b − x a y − x (cid:19) , is Cohen–Macaulay of height 2. Hence, S/J has a Hilbert-Burch resolution [BH93,1.4.17], from which it is easily computed that e ( S/J ) = 3. As J is unmixed, itsassociated primes are minimal. Any prime ideal P containing J must contain( x, b ) and ( x, y ), and so, these must be the only associated primes of J . Sinceit is easily seen that J S ( x,y ) = ( x, y ) S ( x,y ) , we know that J = ( x, y ) ∩ L for some( x, b )-primary ideal L , which has multiplicity e ( S/L ) = 2 by the associativityformula. It follows from the height conditions that q / ∈ ( x, b ) so that( J : q ) = (( x, y ) : q ) ∩ ( L : q ) = ( L : q ) = L. By [Eng07, 10], we know that ( x, b ) + ( q ) is also an ( x, b )-primary ideal ofmultiplicity 2, which is contained in ( J : q ) since b q = ∆ b x + ( b b ) q where∆ = a b − b a . Thus, ( J : q ) = L = ( x, b ) + ( q ) by [Eng07, 8].Since the minimal free resolutions of S/ ( J : q ) is well-known (see [Eng07, 10]),we can lift the map S/ ( J : q )( − q → S/J to a chain map between their minimalfree resolutions: S ( − S ( − S ( − S ( − S ( − S ( − S a y − b x ! b a − x b y a − x y − x − b (cid:16) q − a b − a a − b b − q − b a (cid:17) ( x b x b q ) q q − ∆ 00 0 b b q ! q b a − x y − x ! ( x b x q ) Taking the mapping cone of this chain map yields the minimal free resolution of R . In particular, we see that Syz S ( I ) is minimally generated by the columns of the matrix b a q − x y q − ∆ 00 − x b b q − x − b x − b − q If R were Koszul, then Syz S ( I ) would be minimally generated by linear syzygiesand Koszul syzygies on the minimal generators of I by [Mas18, 2.8], but in thiscase, the linear and Koszul syzygies are spanned by all but the fifth column of theabove matrix so that R cannot be Koszul. Case (b): Since ht( x, y, a , b ) = 4, we see that J = ( x, y ) ∩ ( x, b ) ∩ ( y, a ) sothat J is a Cohen–Macaulay ideal of ht J = 2. Clearly q ∈ ( x, y ). On the otherhand, since ht( x, b , b ) = ht( x, y, b ) = ht( x, a , a ) = ht( x, y, a ) = 3, we see that q / ∈ ( x, b ) ∪ ( y, a ). Thus ( J : q ) = ( x, b ) ∩ ( y, a ).Now, the ring S/J has a Hilbert-Burch resolution, and the resolution of S/ ( J : q )is easily constructed using the fact that ht( x, y, a , b ) = 4 (see for example [MM20,3.3]). We lift the map S/ ( J : q )( − q → S/J to a chain map between their minimalfree resolutions as follows S ( − S ( − S ( − S ( − S ( − S ( − S b − a − yx − a − b y b x a − x − y (cid:16) q b b − a b q − a b a a (cid:17) ( xy a x b y a b ) q q a b q a b ! q − a − b y x ! ( xy a x b y ) and taking the mapping cone of this chain map yields the minimal free resolutionof R . In particular, Syz S ( I ) is minimally generated by the columns of the matrix − a − b q y q a b x q a b − xy − a x − b y − a b . As above, the linear and Koszul syzygies are spanned by only the first five columnsof the above matrix so that R cannot be Koszul by [Mas18, 2.8]. Case (c): After a suitable change of generators for I , we may assume thatht( x, a , b , b ) = 4. This is clear if ht( x, a , a , b , b ) = 5. Otherwise, we must haveht( x, a , a , b , b ) = 4 since ht( a , a , b , b ) = 4. In that case, since ht( x, b , b ) =3, we may assume after possibly interchanging q and q that a ∈ ( x, a , b , b )and ht( x, a , b , b ) = 4 as claimed.Since ht( x, y ) = 2 and ht( x, a , b , b ) = ht( a , a , b , b ) = 4, it is clear that q is a nonzerodivisor modulo ( b x, b x ) = ( x ) ∩ ( b , b ). Hence, the minimal free HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 13 resolution of
S/J has the form0 −→ S ( − −→ S ( − ⊕ S ( − −→ S ( − −→ S −→ S/J −→ J : q ) = ( b , b x, q ) = ( b , b x, a x ). Indeed, it is clearthat the colon contains the latter ideal. If q f ∈ ( b x, b x, q ), then b yf ≡ b yg (mod x ) for some g ∈ S so that f ∈ ( b , x ) as ht( x, y ) = 2 and ht( x, b , b ) =3. Similarly, we see that a xf ≡ b , b , a )) so that f ∈ ( b , b , a ) asht( a , a , b , b ) = ht( x, a , b , b ) = 4. And so, it follows that ( J : q ) ⊆ ( b , x ) ∩ ( b , b , a ) = ( b , b x, a x ). As b is a nonzerodivisor modulo ( b x, a x ), we see thatthe minimal free resolution of S/ ( J : q ) has the form0 −→ S ( − −→ S ( − −→ S ( − ⊕ S ( − −→ S −→ S/ ( J : q ) −→ . We obtain a free resolution of R by taking the mapping cone of the chain maplifting multiplication by q in the short exact sequence0 −→ S/ ( J : q )( − q −→ S/J −→ R −→ . The degrees of the free modules in the resolutions of
S/J and S/ ( J : q ) forcethis lift to have entries of positive degree. Hence, the resulting free resolutionis minimal, and R has the desired Betti table. However, unlike cases (a) and (b)above, in this case the first syzygies of I are minimally generated by linear syzygiesand Koszul syzygies on the minimal generators of I , making it more difficult toprove that R is not Koszul. We defer the proof of this fact to Appendix A at theend of this paper. (cid:3) Our approach to the remaining parts of the proof of Theorem 3.6 is based aroundan analysis of the matrix of linear first syzygies on the defining ideal I . Lemma 3.10.
Suppose R = S/I has Betti table:
Then: (a)
There is a unique height two linear prime P containing I , and ( I : P ) isnondegenerate. (b) The matrix of linear syzygies of I has a generalized zero. (c) If the coordinates of any linear syzygy ℓ = ( ℓ , ℓ , ℓ , ℓ ) of I generate an or-der ideal of height 3, then the matrix of linear syzygies of I has a generalizedrow which is zero.Proof. Let q , . . . , q denote the minimal quadratic generators of I . (a) It is easily computed that e ( R ) = 1 and ht I = 2. Hence, there is a uniqueheight two prime P containing I , which is necessarily a linear prime. If ( I : P )is degenerate, then we can write I = ( xz, yz, a x + b y, a x + b y ) for some linearforms x, y, z, a i , b i with P = ( x, y ). However, in that case, the columns of thematrix y a a − x b b − z
00 0 − z are 3 linearly independent linear syzygies on I , contradicting the assumption onthe Betti table of R .(b) Suppose the 4 × M of linear syzygies of I is 1-generic. Then thevector q = ( q , q , q , q ) ∈ S ( − is a syzygy of the module Coker M T . Since M is 1-generic, then I ( M ) is a prime ideal of height 3 generated by quadrics [Eis05,6.4], and the minimal free resolution of Coker M T is given by a Buchsbaum-Rimcomplex [Eis95, A2.13]. It follows from the description of the second differentialof the Buchsbaum-Rim complex that I ⊆ I ( M ). However, I contains a reduciblequadric by [HM+13, 4.7] and [MM20, 4.3]. Since I ( M ) is prime, then I ( M )contains a linear form, which is impossible.(c) After a suitable change of generators for I , we may assume that ℓ = 0 andht( ℓ , ℓ , ℓ ) = 3. We can then write q q q = ℓ ℓ − ℓ ℓ − ℓ − ℓ u u u = ℓ u + ℓ u − ℓ u + ℓ u − ℓ u − ℓ u for some linear forms u i by viewing the q j ’s as a syzygy on the ℓ j ’s. But then I ⊇ ( q , q , q ) = I ( M ), where M = (cid:18) ℓ ℓ ℓ u − u u (cid:19) . If I ( M ) = ( q , q , q ) has height one, then R does not have the assumed Bettitable by [MM20, 3.3]. Hence, I ( M ) has height 2 so that the rows of M representthe two independent linear syzygies of I by [BH93, 1.4.17]. (cid:3) Proof of Theorem 3.6.
By Lemmas 3.8 and 3.9, it remains only to prove that, if R has the given Betti table, then I has one of the forms (a)–(d). As in the precedinglemma, we see that e ( R ) = 1 and ht I = 2, and we let P denote the uniqueheight two linear prime containing I . We consider two cases based on whether thematrix of linear syzygies of I has a generalized row that is zero. This is precisely thedistinction between the possible forms (a)–(b) versus (c)–(d) listed in the theorem. HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 15
Case 1:
Suppose the matrix of linear syzygies of I has a generalized row whichis zero. We can then choose generators q , . . . , q for I such that J = ( q , q , q )has exactly two independent linear syzygies. It then follows from [Mas18, 4.2] that J = I ( M ) for some 3 × M of linear forms. The ideal J is not primebecause it is strictly contained in the height two linear prime P , so the matrix M is not 1-generic by [Eis05, 6.4], and thus M has at least one generalized zero. Wemay then assume M = b d − a c − x for some linear forms a, b, c, d, x . In particular, we see that ht( a, b ) = 2 and x = 0as J = ( ax, bx, ad + bc ) is minimally generated by 3 quadrics.Consider the linear prime ( P, a, b ), which must have height between 2 and 4.We claim that ht(
P, a, b ) = 3. If we have ht(
P, a, b ) = 2, then P = ( a, b ) so that x ∈ ( J : P ) ⊆ ( I : P ), contradicting that the latter ideal is nondegenerate. Inparticular, this would be the case if x / ∈ P , so we must have x ∈ P . Let y be anylinear form such that P = ( x, y ). If ht( P, a, b ) = 4, then ad + bc ∈ J ⊆ P impliesthat (cid:18) dc (cid:19) ≡ λ (cid:18) b − a (cid:19) (mod P )for some some λ ∈ k . Hence, after a suitable change of generators for J , wemay assume J = ( ax, bx, zy ) for some nonzero linear form z ∈ ( a, b ). But then z ∈ ( I : P ), which is a contradiction. Therefore, ht( P, a, b ) = 3, and without lossof generality, we may assume that ht(
P, b ) = 3 and that a ∈ P . As ad + bc ∈ P ,this implies that c ∈ P . Note also that ( a, c ) * ( x ) or else we would have ht J = 1.If a / ∈ ( x ), then P = ( x, a ), and after a suitable change of generators for J we mayassume that c = 0 so that J = ( ax, bx, ad ). Moreover, we note that ht( x, a, b, d ) =4. Indeed, if this were not the case, we would have d ∈ ( x, a, b ) = ( P, b ) so that,after a change of generators for J , we may assume that d ∈ ( a, b ). But then d ∈ ( J : P ) ⊆ ( I : P ), which is a contradiction. Relabeling x, a, b, d as x, y, a , b respectively and writing the last generator of I as a x + b y , we see that I hasthe form (b) above. It remains only to check that ht( x, b , b ) = ht( y, a , a ) = 3.If ht( x, b , b ) <
3, then b ∈ ( x, b ) so that after a suitable change of generatorswe may assume that b = 0. But then R does not have the assumed Betti tableby [MM20, 3.3]. Hence, we must have ht( x, b , b ) = 3, and similarly, we see thatht( y, a , a ) = 3.On the other hand, if a ∈ ( x ), then we must have P = ( x, c ) so that J =( x , bx, dx + bc ). Similarly, we must have ht( x, b, c, d ) = 4. Again, if this were notthe case, we would have d ∈ ( x, b, c ) = ( P, b ) so that, after a change of generatorsfor J , we may assume that d ∈ ( c ) and, hence, we have b + λx ∈ ( J : P ) for some λ ∈ k , which is a contradiction. Relabeling x, b, c, d as x, b , y, a respectively andwriting the last generator of I as a x + b y , we see that I has the form (a), whereht( x, b , b ) = 3 by arguing as above. Case 2:
Suppose I does not have a generalized row that is zero. By thepreceding lemma, the matrix of linear syzygies of I has a generalized zero. Hence, I has a nonzero linear syzygy whose order ideal has height at most 3, and so, itsorder ideal is easily seen to have height two by our assumption and the precedinglemma. Consequently, after a suitable change of generators for I , the matrix oflinear syzygies of I has the form M = d s − c u v − w for some linear forms with ht( c, d ) = 2. If q , . . . , q denote the correspondinggenerators of I , then dq − cq = 0 implies q = cx and q = dx for some nonzerolinear form x . We see that x ∈ P , since otherwise we would have P = ( c, d ) sothat x ∈ ( I : P ), contrary to the preceding lemma. Note also that ht( v, w ) = 2 byour assumption.If ht( s, u, v, w ) = 2, then s, u ∈ ( v, w ). Hence, after a suitable change of gener-ators for I , we may assume that s = u = 0. In that case, arguing as above showsthat q = wy and q = vy for some linear form y ∈ P . We note that P = ( x, y )has height two since ht I = 2. Thus, I has the form (d) by Lemma 3.8, and wemay assume that ht( s, u, v, w ) = 4 for the remainder of the proof.Let y be any linear form such that P = ( x, y ), and write q i = a i x + b i y for somelinear forms a i , b i for i = 3 ,
4. We claim that ht( x, v, w ) = 3. If not, then after asuitable change of generators for I , we may assume that x = w and ht( x, v ) = 2.Then, considering the syzygy represented by the second column of M modulo x ,we have vb y ≡ x ) so that b ≡ x ). Hence, ( q , q , q ) ⊆ ( x ) sothat R cannot have the assumed Betti table by [MM20, 3.3].Since ht( x, v, w ) = 3, considering the same relation modulo ( v, w ) yields scx + udx ≡ v, w )) so that sc + ud ≡ v, w )). Since ht( s, u, v, w ) = 4,we can write (cid:18) cd (cid:19) ≡ λ (cid:18) − us (cid:19) (mod ( v, w ))for some λ ∈ k . If λ = 0, then after subtracting λ − times the first column of M from the second, we may assume that ht( s, u, v, w ) = 2. And so, as above, we seethat I has form (d). Hence, we may assume that λ = 0.In that case, we have c, d ∈ ( v, w ) so that ( c, d ) = ( v, w ) as ht( c, d ) = 2. Then,after a suitable change of generators for I , we may assume that v = d and w = − c . HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 17
Then, considering the syzygy represented by the second column of M modulo x ,we have db y − cb y ≡ x ) so that db − cb ≡ x ) as ht( x, y ) = 2and (cid:18) − b b (cid:19) ≡ α (cid:18) − dc (cid:19) (mod x )as ht( x, c, d ) = 3. If α = 0, then I ⊆ ( x ), contradicting that ht I = 2. Hence, α = 0, and after replacing a , a with different linear forms, we may assume that b = c and b = d . Then the syzygy represented by the second column of M is0 = sb x + ub x + b q − b q = ( s − a ) b x + ( u + a ) b x so that 0 = ( s − a ) b + ( u + a ) b . Since ht( b , b ) = 2, we finally see that (cid:18) a − a (cid:19) = (cid:18) su (cid:19) + β (cid:18) b − b (cid:19) for some β ∈ k so that ( a , a , b , b ) = ( s, u, b , b ) has height four. If ht( q , q ) = 1,then arguing as above shows that I has form (d), and otherwise, ht( q , q ) = 2 sothat I has form (c) of the theorem. (cid:3) Proof of Theorem B.
Suppose R = S/I is a Koszul algebra defined by g ≤ R is LG-quadratic when ht I = g [Cav04, 1.2.5] andwhen ht I = g − I = 1, then I = zJ for some linear form z and linear complete intersection J so that the Betti numbers of R depend only on the number g of minimal gen-erators of I . By Corollary 2.3, any particular height one ideal must be obtainedby specializing the generic case where ht( z, J ) = g + 1 by a regular sequence oflinear forms, and in the generic case, I is a monomial ideal after suitable changeof coordinates and, thus, is G-quadratic.Hence, it only remains to see that R is LG-quadratic when g = 4 and ht I = 2.We consider the possible forms for I described by Theorem A. In cases (i) and(ii), R is LG-quadratic by [MM20, 3.5]. In cases (iii) and (iv), this follows fromCorollary 3.5 and Proposition 2.4 respectively. (cid:3) Applications to the Backelin–Roos Property
As an application of our structure theorem, we prove that all Koszul algebras de-fined by g ≤ I is the sum of two transversal heightone ideals have the Backelin-Roos property. Recall that R has the Backelin–Roosproperty if there exist a surjective Golod homomorphism ϕ : Q → R of standardgraded k -algebras where Q is a complete intersection; the ring R is absolutelyKoszul if every finitely generated graded R -module has finite linearity defect (see[CI+15] for further details). When R is Koszul, it is well-known that Backelin–Roos property Absolutely Koszuland it is an open question whether the two notions are equivalent for Koszulalgebras up to field extensions [CI+15, p. 354]. Using our structure theorem, wegive an affirmative answer to this question for Koszul algebras defined by g ≤ k -algebras ϕ : R → R ′ is an algebra retract if there exists a k -algebra map σ : R ′ → R such that ϕ ◦ σ = Id R ′ . Theorem 4.1.
Let R = S/I be a Koszul algebra defined by g ≤ quadrics overan algebraically closed field k . Then: (a) R has the Backelin-Roos property except possibly when I is the sum of twotransversal height one ideals each generated by two quadrics. (b) If I is the sum of two transversal height one ideals each generated by twoquadrics and ch( k ) = 0 , then R is not even absolutely Koszul.Proof. Clearly, R has the Backelin-Roos property if it is a quadratic completeintersection, in which case ht I = g . By [CI+15, 3.1], R also has the Backelin-Roos property if reg S R ≤
1. In particular, this is the case when ht I = 1. Italso follows that factoring out a regular sequence of linear forms or quadrics pre-serves the Backelin-Roos property by considering one nonzerodivisor at a time. Ifht I = g −
1, then by [Mas18] either I = I ( M ) + ( q , . . . , q g ) for some 3 × M of linear forms with ht I ( M ) = 2 and quadrics q , . . . , q g that form a regularsequence modulo I ( M ), or I = ( a x, a x, q , . . . , q g ) for some linear forms x, a , a and quadrics q , . . . , q g that form a regular sequence modulo ( a x, a x ). Sincereg S S/I ( M ) = reg S S/ ( a x, a x ) = 1, these quotients have Backelin–Roos prop-erty, and thus, so does R . Hence, the proof of the theorem reduces to consideringthe case when ht I = 2 and I is generated by g = 4 quadrics.In that case, we consider the 4 possible forms for I described by Theorem A. Incases (i) and (ii), it follows from [MM20, 3.3] that either reg S R = 1 or R can beobtained by factoring out a quadratic nonzerodivisor from a ring of regularity one.Thus, as above, R has the Backelin-Roos property. Case (iv): In this case, I = ( ax, bx, cy, dy ) is a sum of two transversal heightone ideals each generated by two quadrics. Since the absolutely Koszul prop-erty ascends from quotients of R by a regular sequence of linear forms [CI+15,2.4], it suffices to show that R is not absolutely Koszul in the generic case where a, b, c, d, x, y are all variables of S .Under these circumstances, we claim that R is a bad Koszul algebra in the senseof Roos [Roo05] and, therefore, is not absolutely Koszul [HI05, 1.8]. Being a goodKoszul algebra descends to quotients rings under maps that are large homomor-phisms in the sense of Levin [Roo05, 2.5], of which two important cases are algebra HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 19 retracts and quotients by a regular sequence of linear forms [Lev80, 2.2, 2.3]. Thus,by an algebra retract it first suffices to show that R is a bad Koszul algebra when S = k [ a, b, c, d, x, y ], and then by killing the regular sequence x − a , y − d , wemay further assume that R = k [ a, b, c, d ] / ( a , ab, cd, d ), which Roos shows is abad Koszul algebra in characteristic zero [Roo05, 2.4]. Case (iii): In this case, we have I = ( xz, yz, a x + b y, a x + b y ) for somelinear forms x , y , z , a i , b i such that ht( x, y ) = 2, ht( a x + b y, a x + b y ) = 2,and ht( z, a x + b y, a x + b y, a b − a b ) = 3. Set q i = a i x + b i y for i = 3 , a b − a b , and C = ( q , q ). Then C is a complete intersection, and we claimthe natural surjection ϕ : Q = S/C → R is Golod. Since R is Koszul by Corollary3.5 and since Q is Koszul, then by [CI+15, 3.1] it suffices to show that I/C has a2-linear resolution over Q .We first consider the case where ht( z, q , q ) = 3. In that case, it suffices toshow that I/C = z ( x, y ) Q ∼ = ( x, y ) Q has a linear resolution as a Q -module. Write S = S ′ [ x, y ] where S ′ is a polynomial ring over k . Since C ⊆ ( x, y ), the naturalepimorphism π : Q → Q/ ( x, y ) Q ∼ = S ′ is an algebra retract. Indeed, if Φ : S ′ → Q denotes the composition S ′ ֒ → S → Q , it is easily seen that π ◦ Φ = Id S ′ . Since Q is Koszul, it follows from [CI+15, 2.3] that ld Q Q/ ( x, y ) Q = 0. Thus, ( x, y ) Q hasa linear resolution as a Q -module so that ϕ : Q → R is Golod.In general, consider the ideal ˜ I = ( wx, wy, q , q ) ⊆ ˜ S = S [ w ], which has thesame form as I but where ht( w, q , q ) = 3 since ( w, q , q ) /w ˜ S ∼ = C has height 2.By the preceding paragraph, we know that the surjection ˜ S/C ˜ S → ˜ S/ ˜ I is Golod.By arguing as in [CI+15, 3.9], it suffices to show that w − z is regular mod ˜ I andmod C ˜ S to prove that the induced surjection S/C → R is Golod so that R has theBackelin-Roos property. That w − z is regular on ˜ S/ ˜ I follows from Corollary 2.3since it has the same Betti table over ˜ S as the Betti table of R over S by Theorem3.4. On the other hand, any associated prime of C ˜ S is of the form P ˜ S for someassociated prime P of C in S [Mat89, 23.2], and so, w − z / ∈ P ˜ S since otherwisewe would have w ∈ ( z, P ) ˜ S , which is clearly impossible. Hence, w − z is regularmod C ˜ S . (cid:3) Combining the above theorem with [CI+15, 3.4], we have the following.
Corollary 4.2.
For a Koszul algebra R = S/I defined by g ≤ quadrics over analgebraically closed field of characteristic zero, the following are equivalent: (a) R has the Backelin–Roos property. (b) R is absolutely Koszul. (c) R does not have the Betti table: I is not the sum of two transversal height one ideals each generated by twoquadrics. Remark 4.3.
Roos’ proof in [Roo05] that k [ a, b, c, d ] / ( a , ab, cd, d ) is a bad Koszulalgebra relies on k having characteristic zero. The results of this section would holdin arbitrary characteristic if we knew that this ring is a bad Koszul algebra withoutany restriction on k . Appendix A. The Bad Non-Koszul Algebra
In this section, we show that ideals of the form (c) from Theorem 3.6 neverdefine a Koszul algebra, tying up the remaining loose end in our structure theoremfor Koszul algebras defined by four quadrics.Suppose that R = S/I where I = ( b x, b x, a x + b y, a x + b y ) for some lin-ear forms x, y, a i , b i satisfying ht( x, b , b ) = 3, ht( a , a , b , b ) = 4, and ht( a x + b y, a x + b y ) = 2 as in the theorem. In particular, we can consider the genericideal of this form where the linear forms are all variables. By Lemma 3.9 andCorollary 2.3, any particular ideal I of the above form must be obtained by spe-cializing the generic case by a regular sequence of linear forms, and since theKoszul property ascends and descends along quotients by regular sequences of lin-ear forms by Proposition 2.1, it suffices by first ascending to the generic case andthen specializing again to prove that at least one ideal I of the above form is notKoszul. Notation.
For the remainder of this section, we let R = k [ x, y, a, b ] / ( bx, xy, ax − by, x − y )which is easily seen to have the form (c) from Theorem 3.6. We also let R ′ = k [ x, y ] / ( x − y , xy ) M = Coker R ′ A A = (cid:18) x − y x (cid:19) and note that R ∼ = Sym R ′ ( M ). In particular, R is bigraded by setting deg x =deg y = (1 ,
0) and deg a = deg b = (0 , M has a periodic resolution over R ′ similar to the cokernel of a matrix factorizationover a hypersurface: · · · −→ R ′ ( − A −→ R ′ ( − B −→ R ′ ( − A −→ ( R ′ ) −→ M −→ B = (cid:18) y x y (cid:19) HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 21
We will show that this particular ring R is not Koszul. Thus, not even a sym-metric algebra of a module having a linear resolution over a quadratic completeintersection is necessarily Koszul.Unfortunately, even in this special setup, showing that R is not Koszul seemsto require some technical computation. To simplify the computation as much aspossible, we will exploit the fact that R ′ is an algebra retract of R . Proposition A.1 ([HHO00, 1.4]) . Suppose ϕ : R → R ′ is an algebra retract ofstandard graded k -algebras. Then R is Koszul if and only if R ′ is Koszul and R ′ has a linear resolution as an R -module (via ϕ ). By the above proposition, instead of showing that k does not have a linearresolution over R , we need only show that R ′ does not have a linear resolution over R . This reduces the number of independent syzygies that we need to computefrom at worst 340 (for the residue field) to at worst 80 (for R ′ ). It also has theadvantage that the matrices we need to compute are generally smaller than foralternative approaches such as computing the deviations of R via a minimal modelof R over S ; see [Avr98] for further details.To simplify the computation of the syzygies, we also exploit the bihomogeneousstructure of R . Lemma A.2.
The elements a i b j , xa i , ya i , x for all i, j ≥ form a bihomogeneousbasis for R over k .Proof. It is easily seen that R ′ has 1 , x, y, x as a basis over k so that the polynomialring R ′ [ a, b ] has a bihomogeneous basis consisting of the elements a i b j , xa i b j , ya i b j ,and x a i b j for all i, j ≥
0. We note that R ∼ = R ′ [ a, b ] /J , where J = ( bx, ax − by ),and modulo J , we have xa i b j ≡ j > ya i b j ≡ xa i +1 b j − ≡ j > ya i b ≡ xa i +1 x a i b j ≡ j > x a i ≡ xya i − b ≡ i > R is spanned by the elements a i b j , xa i , ya i , x for all i, j ≥
0. To provethat these elements are k -linearly independent, it suffices to focus on a particularbidegree. For each n ≥
0, the elements of bidegree (0 , n ) are the monomials a i b n − i ,which are all linearly independent because they are even linearly independent in R/ ( x, y ) ∼ = k [ a, b ]. The only monomial in the spanning set of bidegree (2 ,
0) is x ,which is nonzero as it is nonzero even in R/ ( a, b ) ∼ = R ′ . It remains to see that forevery i the two monomials of bidegree (1 , i ) xa i and ya i are linearly independentin R . If not, there are α, β ∈ k not both zero and f , f ∈ R ′ [ a, b ] such that αa i x + βa i y = f bx + f ( ax − by ). We may assume f , f are bihomogeneous of degree(0 , i −
1) so that f , f ∈ k [ a, b ]. Then x ( αa i − f b − f a )+ y ( f b − βa i ) = 0 in R ′ [ a, b ].Since x, y are part of a basis for R ′ [ a, b ] over k [ a, b ], it follows that αa i − f b − f a = 0and f b − βa i = 0. From the latter, we obtain β = 0 = f so that αa i − f b = 0.Consequently, we must have α = 0 = f , which is a contradiction. Thus, xa i and ya i must be linearly independent in R as claimed. (cid:3) Recall that by the above we need to resolve the ideal U = ( a, b ) over R . Inthe following statement, we consider the standard grading of R and determine agraded free resolution of U , while in the proof, to simplify the arguments, we makeuse of the bigrading on R . Lemma A.3.
The first three steps of the minimal free resolution of the ideal U = ( a, b ) over R are R ( − ∂ −→ R ( − ∂ −→ R ( − ∂ −→ R ( − −→ where: ∂ = (cid:18) x b − y x − a (cid:19) ∂ = y b − b − ax y a b x y ∂ = x − b b b a − y x − b − a − b − b x y b x y − a y − x a b x y − b Proof.
Since U is a bihomogeneous ideal, its syzygies will also be generated bybihomogeneous elements. At each step, we consider cases based on the first com-ponent of the bidegree of a syzygy; the number of cases will be small as R isnonzero only in bidegrees (0 , i ), (1 , i ), and (2 , f = ( f , f ) ∈ R (0 , − be a bihomogeneous syzygy of U . If f has bidegree(0 , i + 2), then f i ∈ k [ a, b ] so that f = h ( b, − a ) for some h ∈ k [ a, b ]. If f hasbidegree (1 , i + 1), then we can write f j = c j, xa i + c j, ya i for some c j,r ∈ k so that0 = f a + f b = c , xa i +1 + c , ya i +1 + c , ya i b = ( c , + c , ) xa i +1 + c , ya i +1 Hence, we must have c , = − c , and c , = 0 by the above lemma so that f = c , a i ( x, − y ) + c , a i (0 , x ). If f has bidegree (2 , i + 1), then f = ( c x , c x ) = c ( x ,
0) + c (0 , x ) for some c i ∈ k , and there is no restriction on the c i since x isin the socle of R . However, we note that ( x ,
0) = x ( x, − y ) and (0 , x ) = x (0 , x )belong to the span of the previously found syzygies. Hence, the columns of ∂ HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 23 span the syzygies of U , and since these linear syzygies are easily seen to be linearlyindependent over k , they form a minimal set of generators for the first syzygies.Let f = ( f , f , f ) ∈ R ( − , − ⊕ R (0 , −
2) be a bihomogeneous syzygy on thecolumns of ∂ . If f has bidegree (0 , i + 3), then f = f = 0 and f ∈ k [ a, b ] sothat bf = 0 and, hence, f = 0. If f has bidegree (1 , i + 2), then f , f ∈ k [ a, b ]and f = c , xa i + c , ya i for some c ,r ∈ k . Writing f j = α j, a i +1 + α j, a i b + b f ′ j for some α j,r ∈ k and f ′ j ∈ k [ a, b ], we see that ∂ f = 0 implies0 = f x + f b = ( α , + c , ) xa i +1 − f y + f x − f a = ( − α , + α , − c , ) xa i +1 + ( − α , − c , ) ya i +1 so that α , = − c , and α , = α , − c , . It follows that f is contained in thesubmodule spanned by the columns of the matrix M = b − b − aa b x y and the additional syzygies ( b , ,
0) and (0 , b , b , ,
0) = b ( b, a, − a (0 , b,
0) and (0 , b ,
0) = b (0 , b, , i + 2).If f has bidegree (2 , i + 1), then for j = 1 , f j = c j, xa i + c j, ya i , and f = c x for some c ∈ k so that 0 = f x + f b = c , x a i and 0 = − f y + f x − f a = ( − c , + c , ) x a i . If i >
0, this imposes no restrictions on the f j so that f is in the column span of the matrix xa ya xa ya
00 0 0 0 x = M x y
00 0 0 0 00 x x − x y x where the factorization on the right shows these syzygies are generated by thesyzygies already found in bidegree (1 , i = 0, we must have c , = 0and c , = c , so that f = c , ( y, x,
0) + c , (0 , y, f has bidegree (3 , i + 1), then f = ( c x , c x ,
0) for some c j ∈ k , andthere are no restrictions on the c i . However, the syzygies ( x , ,
0) = y ( y, x,
0) and(0 , x ,
0) = y (0 , y,
0) are generated by those already found in bidegree (2 , ∂ span the syzygies of Im ∂ , and since these linear syzygies areeasily seen to be linearly independent over k , they form a minimal set of generatorsfor the second syzygies of U .Let f ∈ R ( − , − ⊕ R ( − , − be a bihomogeneous syzygy on the columnsof ∂ . If f has bidegree (1 , i + 3), then f = f = 0 and f j ∈ k [ a, b ] for j ≥
3. Then ∂ f = 0 implies that − af + ( f − f ) h = 0 and af + bf = 0 so that ( − f , f − f ) = h ( b, − a ) and ( f , f ) = h ( b, − a ) for some h i ∈ k [ a, b ],and the last component of ∂ f imposes no additional restrictions. Hence, f = h (0 , , , , a, − b ) + h (0 , , b, − a, b, f has bidegree (2 , i + 2), then f , f ∈ k [ a, b ], and for all j ≥
3, we have f j = c j, xa i + c j, ya i for some c j,r ∈ k . Writing f j = α j, a i +1 + α j, a i b + b f ′ j forsome α j,s ∈ k and f ′ j ∈ k [ a, b ] for j = 1 ,
2, we see that ∂ f = 0 implies0 = yf + bf − bf − af = ( α , + c , − c , − c , ) xa i +1 + ( α , − c , ) ya i +1 xf + yf + af + bf = ( α , + α , + c , + c , ) xa i +1 + ( α , + c , ) ya i +1 xf + yf = ( c , + c , ) x a i so that α , = c , , α , = − c , + c , + c , , α , = − c , , α , = − c , − c , − c , ,and if i = 0 we also have c , = − c , . Hence, if i = 0, we have f ′ j = 0 for j = 1 , f is contained in the column span of the matrix M = − b b b a − b − a − b − bx y x y y − x x y When i ≥
1, we see that f is contained in the span of the columns of M plus thefour additional syzygies b a b − ab xa
00 0 0 ya = M a − b b bb − b − b a where the factorization on the right shows these syzygies are generated by thesyzygies already found in bidegree (2 , f has bidegree (3 , i + 1), then f j = c j, xa i + c j, ya i for some c j,r ∈ k for j = 1 , f j = c j x for some c j ∈ k for all j ≥
3. Then ∂ f = 0 implies c , x a i =( c , + c , ) x a i = 0. If i = 0, then f = c , ( x, − y, , , ,
0) + c , (0 , x, , , , i ≥
1, there are no restrictions on f so that f is contained in the column HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 25 span of the matrix xa ya xa ya x x x
00 0 0 0 0 0 0 x = M − y x x − x − y x y y − x − y − y x y where the factorization on the right shows these extra syzygies are generated bythe syzygies already found in bidegree (2 , f has bidegree (4 , i + 1), then i = 0 and f j = 0 for j ≥ f is in the span of ( x , , , , ,
0) = x ( x, − y, , , ,
0) and (0 , x , , , ,
0) = x (0 , x, , , , ∂ span the syzygies of Im ∂ , and sincethese linear syzygies are easily seen to be linearly independent over k , they form aminimal set of generators for the third syzygies of U . (cid:3) Theorem A.4. (a) If ch( k ) = 2 , then the map ∂ of the preceding lemma has a minimal qua-dratic syzygy. (b) If ch( k ) = 2 , then the syzygies of the map ∂ of the preceding lemma arethe image of the map ∂ : R ( − → R ( − given by ∂ = y b − b − b − a − b x y a b y − x − b − a − b a x y b − b − b y − x b − a a
00 0 0 0 0 0 0 0 x y b a x y − b − a − b y − y b − a a x b x y x y and ∂ has a minimal quadratic syzygy.Hence, in either case, R is not Koszul.Proof. Let f ∈ R ( − , − ⊕ R ( − , − ⊕ R ( − , − be a bihomogeneous syzygyon the columns of ∂ . If f has bidegree (1 , i + 4), then f j = 0 for j ≤ f , f ∈ k [ a, b ]. Since ∂ f = 0 implies bf = bf = 0, it follows that f = 0.If f has bidegree (2 , i + 3), then f = f = 0, f j ∈ k [ a, b ] for 3 ≤ j ≤
9, and f j = c j, xa i + c j, ya i for some c j,r ∈ k for j = 10 ,
11. First, we note that ∂ f = 0implies af + b ( − f + f + f ) = 0 and af + b ( f + f + f ) = 0 so that there exist h , h ∈ k [ a, b ] such that: (cid:18) f − f + f + f (cid:19) = h (cid:18) b − a (cid:19) (cid:18) f f + f + f (cid:19) = h (cid:18) b − a (cid:19) . Writing f j = α j, a i +1 + α j, a i b + b f ′ j for some α j,s ∈ k and f ′ j ∈ k [ a, b ], this forces h = α , a i + bf ′ , h = α , a i + bf ′ , and:0 = α , = α , , − f + f + f + ah = ( α , + α , + α , ) a i +1 + ( − α , + α , + α , ) a i b, + ( − f ′ + f ′ + f ′ ) b + abf ′ , f + f + f + ah = ( α , + α , + α , ) a i +1 + ( α , + α , + α , ) a i b, + ( f ′ + f ′ + f ′ ) b + abf ′ . Consequently, the remaining components of ∂ f imply that0 = xf + yf + bf = ( α , + α , + c , ) xa i +1 xf + yf − af = ( α , + α , − c , ) xa i +1 + ( α , − c , ) ya i +1 yf − xf + af + bf = ( α , + c , + c , ) xa i +1 + ( α , + c , ) ya i +1 xf + yf − bf = ( α , + α , − c , ) xa i +1 so that we have the linear equalities: α , = − α , − c , α , = 0 α , = − α , + c , α , = c , α , = − c , α , = − c , − c , α , = − α , + c , α , = 0Moreover, if i = 0, then f ′ j = 0 for all j so that we have the additional equalities α , = − α , − α , α , = α , + α , = α , − c , − c , so that α , = − α , + c , and f is in the column span of the matrix: M = − b − a − b a b − b − bb − a a b a − b − a − b b − a a b x y x y HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 27
On the other hand, if i ≥
1, then we can write f ′ j = α j, a i − + bf ′′ j for some f ′′ j ∈ k [ a, b ] for j = 4 , α , = − α , − α , − α , α , = α , + α , − c , − c , f ′ = − af ′′ − f − α , a i − − bf ′′ f ′ = α , a i − + bf ′′ − f ′ − af ′′ In this case, α , = − α , − α , + c , so that f is contained in the submodulespanned by the columns of M and the matrix − ab − b − a − b − b b ab b b b − b − ab b b b = M b b b a
00 0 0 0 b b − b b b b where the factorization on the right shows these extra syzygies are generated bythe syzygies already found in bidegree (2 , f has bidegree (3 , i + 2), then f , f ∈ k [ a, b ], f j = c j, xa i + c j, ya i for some c j,r ∈ k for 3 ≤ j ≤
9, and f j = c j x for some c j ∈ k for j = 10 ,
11. Writing f j = α j, a i +1 + α j, a i b + b f ′ j for some f ′ j ∈ k [ a, b ] for j = 1 ,
2, we see that ∂ f = 0implies that0 = xf − bf + bf + bf + af = ( α , − c , + c , + c , + c , ) xa i +1 + c , ya i +1 − yf + xf − bf − af − bf − bf = ( − α , + α , − c , − c , − c , − c , ) xa i +1 + ( − α , − c , ) ya i +1 xf + yf + bf = ( c , + c , ) x a i xf + yf − af = ( c , + c , ) x a i yf − xf + af + bf = ( c , − c , ) x a i xf + yf − bf = ( c , + c , ) x a i so that we have the linear equalities: α , = − c , α , = α , − c , − c , − c , c , = 2 c , − c , − c , c , = 0 In addition, if i = 0, then f ′ = f ′ = 0 and c = c = 0 so that we have theadditional equalities c , = − c , c , = − c , c , = c , c , = 0so that c , = 2 c , − c , . Hence, f is in the column span of the matrix: N = b − b − b − a − b a b y − x x y y − x x y x y y − y x On the other hand, if i ≥
1, then f is in the span of the columns of N and thecolumns of the matrix b − a − ab b xa ya xa ya ya − ya − ya xa ya
00 0 0 0 0 0 0 0 xa = (cid:0) N M (cid:1) b b − b − b a b b a a a x − x − x − y x − x − x − y y y where the factorization on the right shows these extra syzygies are generated bythe syzygies already found in bidegrees (3 ,
2) and (2 , f has bidegree (4 , i +1), then we can write f j = c j, xa i + c j, ya i for some c j,r ∈ k for j = 1 , f j = c j x for some c j ∈ k for 3 ≤ j ≤
9, and f = f = 0. Then ∂ f = 0 implies 0 = xf = c , x a i and 0 = − yf + xf = ( c , − c , ) x a i . If i = 0,then c , = 0, c , = c , , and c j = 0 for 3 ≤ j ≤ f = c , ( y, x )+ c , (0 , y ) ∈ R ( − , − . On the other hand, if i ≥
1, then there are no restrictions on the c j,r HE STRUCTURE OF KOSZUL ALGEBRAS DEFINED BY FOUR QUADRICS 29 or c j so that f is contained in the submodule of R ( − , − ⊕ R ( − , − spannedby the columns of the matrix below and the syzygy s = (0 , , , , , , , x , xa ya xa ya x x x x x
00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 x = N y y y x − y x − y y x − x x − y y x x − y x The factorization on the right above shows all of these extra syzygies except pos-sibly s are generated by the syzygies already found in bidegree (3 , k ) = 2,then s = N v where v = (0 , , , , , , , , x, − y ) so that s is also redundant.However, if ch( k ) = 2, it is easily seen that s = (cid:0) N M (cid:1) v has no bihomoge-neous solutions since for every v ∈ R ( − , − ⊕ R ( − , − of bidegree (4 , (cid:0) N M (cid:1) v is zero. Thus, s is a minimal quadraticsyzygy of ∂ , which proves part (a), and so, we may assume that ch( k ) = 2 for theremainder of the proof.If f has bidegree (5 , i + 1), then i = 0 and f j = 0 for j ≥ f ∈ R ( − , − , and f is contained is in the span of ( x ,
0) = y ( y, x ) and (0 , x ) = y (0 , y ) and, therefore, in the span of the syzygies already found in bidegree (4 , ∂ span the syzygies of Im ∂ , and since these syzygies areeasily seen to be linearly independent over k , they form a minimal set of generatorsfor the fourth syzygies of U .To complete the proof of the theorem, we will show that s = (0 , . . . , , x ) ∈ R ( − , − ⊕ R ( − , − is a minimal quadratic syzygy of bidegree (5 ,
2) onthe columns of ∂ . If s were not minimal, then it would be contained in thespan of the syzygies of bidegrees (5 ,
1) and (4 , ,
1) are easily seen tobe ( x, − y ) , (0 , x ) ∈ R ( − , − . If f ∈ R ( − , − ⊕ R ( − , − is a syzygyof bidegree (4 , f j = α j, a + α j, b for some α j,s ∈ k for j = 1 , f j = c j, x + c j, y for some c j,r ∈ k for j ≥
3. It follows from ∂ f = 0 that0 = yf + b ( f − f − f − f ) − af = ( α , + c , − c , − c , − c , − c , ) xa + ( α , − c , ) ya xf + yf + af + bf = ( α , + α , + c , + c , ) xa + ( α , + c , ) ya xf j + yf j +1 = ( c j, + c j +1 , ) x j ∈ { , , } yf − xf = ( c , − c , ) x yf − xf = ( c , − c , ) x yf − yf = (2 c , − c , ) x xf = c , x so that α , = c , = c , α , = − c , + c , + 2 c , + c , α , = − c , α , = − c , − c , − c , c , = c , c , = − c , c , = c , c , = − c , c , = − c , c , = 0and f is contained in the column span of the matrix − b b b b a − b − a − b − bx y x y x y y − x x y x y y − x x y y − x y . If Q denotes the matrix whose columns are the syzygies of bidegree (5 ,
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