Bounds for degrees of syzygies of polynomials defining a grade two ideal
Teresa Cortadellas Benitez, Carlos D'Andrea, Eulalia Montoro
aa r X i v : . [ m a t h . A C ] D ec BOUNDS FOR DEGREES OF SYZYGIES OF POLYNOMIALSDEFINING A GRADE TWO IDEAL
TERESA CORTADELLAS BEN´ITEZ, CARLOS D’ANDREA, AND EUL `ALIA MONTORO
Abstract.
We make explicit the exponential bound on the degrees of thepolynomials appearing in the Effective Quillen-Suslin Theorem, and apply itjointly with the Hilbert-Burch Theorem to show that the syzygy module ofa sequence of m polynomials in n variables defining a complete intersectionideal of grade two is free, and that a basis of it can be computed with boundeddegrees. In the known cases, these bounds improve previous results. Introduction
Let n and m be positive integers, m ≥ , and R := K [ x , . . . , x n ] a poly-nomial ring in n variables with coefficients in an infinite field K . Suppose that a , . . . , a m , p, q ∈ R are such that the following equality of ideals in R holds:(1) I := h a , a , . . . , a m i = h p, q i . In this paper, we will show that Syz( a , . . . , a m ), the R -syzygy module of the se-quence ( a , . . . , a m ) is a free R -module, and we will develop algorithms for a com-putation of a basis of it in terms of the matrices of converting the a i ’s into p, q andvice versa. Our approach will be the use of the Effective Quillen-Suslin Theorempresented in [CCDHKS93]. We will also develop a bound on the degrees of theelements of such a basis as a function of the degrees of the input data.Note that we are not assuming in principle that neither p and q are coprime, northat the ideal they define equals the whole ring R . But our claim can be simplifiedin the first case by removing common factors without changing the syzygy module,and in the second case the problem is solved completely by applying the EffectiveQuillen-Suslin Theorem (cf. [CCDHKS93]) directly to the input data, with betterbounds than those presented below. So we can assume w.l.o.g. for the rest of thetext that gcd( p, q ) = 1 , and that they do not generate the unitary ideal, i.e. thatthe grade of I is equal to two. Let δ a ∈ N be a bound on the total degrees of a , . . . , a m , and δ a bound for the degrees of p and q . The main result of thispaper is the following. Theorem 1.1.
There is an algorithm which computes an R -basis of Syz ( a , . . . , a m ) made by vectors of polynomials of degree bounded by n n ( δ + δ a + 1) n . If in addition I is zero dimensional (for instance when n = 2 ), then anotherbasis of the same module can be computed with the following degree bound: (2) 2 δ + 2 δ a + 2 δ a + δ + 3 mn n (2 δ a + δ a + δ + 1) n . Mathematics Subject Classification.
Primary: 13P20. Secondaries: 13D02,14Q20,68W30.
Key words and phrases.
Effective Quillen-Suslin Theorem, Hilbert-Burch Theorem, syzygyes, µ -bases, degree bounds. The proof of this Theorem is given in Section 5.3 These results can be appliedto the case treated in our ISSAC 2020 paper [CDM20], where n = 2 , m = 4 and p, q a “Shape Lemma” representation of the radical ideal I . Our algorithm andmain result there state that a basis of Syz( a , a , a , a ) can be found with degreebounded by 5 δ (2 δ a + 1) . Note that in this case, δ can be expressed in terms of δ a thanks to B´ezoutTheorem, and we can set δ := δ a . The results in [CDM20] amount then to abound of size δ a . In contrast, the first bound in Theorem 1.1 amounts to a constanttimes δ a , while the second one is of the order of δ a , which is an improvement withrespect to this previous bound. In addition, we will see in Proposition 7.1 that acareful analysis of this situation can actually make the bound of size δ a get smaller,comparable to the results in [CDM20].It should be pointed out, however, that even though the situation presentedhere contains the problem tackled in [CDM20], this paper is not a generalization ofthe results given there, as our methods are slightly different despite the fact thatin both cases we use Hilbert-Burch and Effective Quillen-Suslin theorems. OurISSAC 2020 paper dealt with the case when p and q are “shape” basis of an idealof 4 polynomials in 2 variables, and the results were restricted to that case. Here,we deal with any number of polynomials and variables, and even in the case n = 2we are not assuming that the ideal has a shape basis, just that it is a completeintersection of grade 2. In Section 7 we will compare both approaches.Even though the study of syzygies of ideals of grade 2 does not cover all the casesof interest in the literature -for instance, it is known that if n = 2 the syzygy moduleof a , . . . , a m is always free independently of the fact that I can be generated by 2polynomials, see [Cid19] for general bounds for degrees in this case- the situationpresented in (1) is quite general in the sense that to have Syz( a , . . . , a m ) being afree module, if all the a i ’s are coprime, then Hilbert-Burch Theorem 2.1 impliesthat this ideal must have grade 2 , and hence they should be described -at leastlocally- with two polynomials.The paper is organized as follows: in Section 2 we review both Hilbert-Burchand Effective Quillen-Suslin Theorems (Theorems 2.1 and 2.2 respectively), andgive an explicit bound in Theorem 2.6 for the degree of a unimodular “inverse”matrix to a unimodular one. In Section 3 we show that the matrix of polynomialsconverting ( a . . . a m ) into ( p q ) is unimodular, while the one reversing this processis “almost” unimodular, see Proposition 3.3, but can be replaced by a unimodularone (cf. Proposition 3.4). This result allows a complete characterization of thosedata a , . . . , a m , p, q ∈ R satisfying (1) in terms of a unimodular conversion matrix,see Theorem 3.5.All these results are then used in Section 4 to develop algorithms which com-pute an R -basis of Syz( a , . . . , a m ) based each of them in one of these conversionmatrices.In Section 5 we study degree bounds for the output of the algorithms presented,and also prove Theorem 1.1. In section 6 we show some examples illustrating ourmethods and tools. We conclude the paper by comparing our approach with theone presented in [CDM20] in Section 7. Acknowledgements:
All our computations were done with the aid of the soft-wares Maple [Map20] and Mathematica [Math18]. We also acknowledge useful
OUNDS FOR DEGREES OF SYZYGIES 3 conversations with Mart´ın Sombra while working some of the results of this pa-per. T. Cortadellas was supported by the Spanish MICINN Research projectsMTM2013-40775-P and PID 2019-104844GB-100. C. D’Andrea and E. Montorowere supported by the Spanish MICINN research projects MTM 2015-65361-P andPID2019-104047GB-I00.2.
Hilbert-Burch and Effective Quillen-Suslin Theorems
We start by recalling the well-known Hilbert-Burch Theorem for resolutions oflength 1 . Theorem 2.1. [Eis05, Theorem 3.2]
Suppose that an ideal I in a Noetherian com-mutative ring A admits a free resolution of length as follows: → F G → F → I → . If the rank of the free module F is ℓ, then the rank of F is ℓ + 1 , and there existsa nonzero divisor a ∈ A such that I is equal to a times the ideal of ℓ × ℓ minorsof the matrix G with respect with any given bases of F and F . The generator of I that is the image of the i -th basis vector of F is ± a times the determinant ofthe submatrix of G formed from all except the i -th row. Moreover, the grade of theideal of maximal minors is .Conversely, given an ( ℓ + 1) × ℓ matrix G with entries in A such that the gradeof the ideal of ℓ × ℓ minors of G is at least , and a given nonzero divisor a ∈ A ,the ideal I generated by a times the ℓ × ℓ minors of G admits a free resolution oflength one as above. It has grade if and only if a is a unit. We bring also to the picture the main tool we will use in our paper, namely theEffective Quillen-Suslin Theorem. We recall that R = K [ x , . . . , x n ] is a polynomialring in n variables with coefficients in an infinite field K . Recall that a matrix U ∈ R r × s is called unimodular if the ideal generated by the maximal minors of itequals to the whole ring R. Also, we denote by I r the identity matrix of size r × r .The degree of a matrix equals the maximum of the degrees of its entries. Theorem 2.2. [CCDHKS93, Theorem 3.1]
Assume that F ∈ R r × s is unimodular,with r ≤ s. Then, there exists a square matrix U ∈ R s × s such that (1) U is unimodular, (2) F · U = [ I r , ] ∈ R r × s , (3) deg( U ) = ( rd ) O ( n ) , and (4) U is a product of O ( n s ( rd ) n ) matrices, each of them being elementaryor having the form T ⊕ I s − r − for some T ∈ SL r +1 ( R ) . The proof given in [CCDHKS93] of this result is constructive. In the rest of thissection, we will review some steps of it to make explicit the exponent O ( n ) whichappears in (3).Assume then that a unimodular matrix F ∈ R r × s is given. In a preliminary stepin [CCDHKS93], one has to make a linear change of coordinates, and then multiply F by the unimodular matrix A = ( a ij ) ≤ i,j ≤ s defined by a ij = x n if i = j ≤ r j = i + 11 if i = s, j = 10 everywhere else , TERESA CORTADELLAS BEN´ITEZ, CARLOS D’ANDREA, AND EUL`ALIA MONTORO in such a way that the conditions of Assumption 2.8 in [CCDHKS93] are satisfied,namely that the r × r minor of F made by choosing the first r columns is monic inall the variables x , . . . , x n , and having total degree strictly larger than the degreeof the remaining maximal minors of F . This may increase the value of d in 1 , sowe will have to keep track of this in order to get an explicit bound.Next we will have to deal with a version of Hilbert’s Nullstellensatz presentedin that paper. To do this, consider the s × s matrix Y = ( y ij ) ≤ i,j ≤ s where eachof the y ij is a new indeterminate. Set F Y := F · Y ∈ ( R ⊗ K [ y ij ]) r × s . Denote with D (resp. D ) the determinant of the r × r submatrix of F Y made by choosingits first r columns (resp. the columns 1 , . . . , r − , r + 1), and denote with c ∈ K [ x , . . . , x n − , y ij , ≤ i, j ≤ s ] , the resultant of D and D with respect to x n .We easily verify that(3) deg x ,...,x n − ( c ) ≤ (cid:0) r ( d + 1) (cid:1) , and moreover there exist A , A ∈ K [ x , . . . , x n − , y ij , ≤ i, j ≤ s ] such that(4) c = A D + A D , with deg x ,...,x n − ( A , A ) ≤ (cid:0) r ( d + 1) (cid:1) . Lemma 2.3. [CCDHKS93, Lemma 4.4]
For all ξ ∈ K n − , there exists y ξ ∈ K s × s such that c ( ξ, y ξ ) = 0With this result in hand, we can prove the following effective version of Hilbert’sNullstellensatz. Proposition 2.4.
There exist matrices y , . . . , y n ∈ K s × s such that (5) h c ( x, y ) , . . . , c ( x, y n ) i = R. Proof.
Start by picking any ξ ∈ K n − , and set y y ( ξ ) . Thanks to Lemma2.3 we have that c ( x, y ) = 0 , and hence the variety defined by its zeroes is ahypersurface in K n − . Denote with W , . . . , W ℓ the irreducible components of maximal dimension ofthis hypersurface, and pick χ , . . . , χ ℓ ∈ K n − such that χ j ∈ W j . By Lemma 2.3,none of the polynomials c ( χ j , y ij ) ∈ K [ y ij ] can be identically zero, so the product Q ℓj =1 c ( χ j , y ij ) also is non-zero. As K is infinite, we can choose y ∈ K s × s suchthat Q ℓj =1 c ( χ j , y ) = 0 . With this choice, we have that the variety defined by thezeroes of c ( x, y ) and c ( x, y ) cannot have components of codimension 1 in K n − as the latter polynomial cannot vanish identically in any of the W j .The same argument can be applied recursively as follows: given c ( x, y ) , . . . , c ( x, y i )such that the set of common zeroes of these polynomials has irreducible componentsof dimension at most n − − i, there exist y i +1 ∈ K s × s such that c ( x, y i +1 ) cutsproperly every component of maximal dimension of this algebraic set. For i = n − , the claim follows straightforwardly. (cid:3) We will also need the following refinement of the Effective Nullstellensatz for thedegrees of polynomials involving a B´ezout identity.
Proposition 2.5.
Let c ( x, y ) , . . . , c ( x, y n ) be as in (5) . Then one can have x n = a c ( x, y ) + . . . + a n c ( x, y n ) with a , . . . , a n ∈ R , and deg (cid:0) a i c i ( x, y i ) (cid:1) ≤ (cid:0) r ( d + 1) (cid:1) n − . OUNDS FOR DEGREES OF SYZYGIES 5
Proof.
By the Effective Nullstellensatz (Theorem 1.1 in [Jel05]), there exist b , . . . , b n ∈ K [ x , . . . , x n − ] such that 1 = b c ( x, y ) + . . . + b n c n ( x, y n ) , with deg( b i c ( x, y i )) ≤ r ( d + 1)) n − − . The claim now follows by multiplying by x n both sides of thisequality. (cid:3) Theorem 2.6.
Assume that F ∈ R r × s is unimodular, with r ≤ s. Then, the matrix U ∈ R s × s of Theorem 2.2 can be computed with deg( U ) ≤ n ( r ( d + 1)) n . Proof.
We start by following the steps of the algorithms given in the proof ofProposition 4.1 (Procedure 4.3) in [CCDHKS93] to compute a matrix U n which“eliminates” the variable x n from F by evaluating it to zero, i.e. F · U n = F | x n =0 . • In their step 1, their number N can be replaced by n thanks to Proposition2.4. • In their step 2, we have deg( a i c i ( x, y i )) ≤ (cid:0) r ( d + 1) (cid:1) n − thanks toProposition 2.5. • The degree of what is called E k in their step 3 is bounded bydeg( E k ) ≤ r ( d + 1)(1 + r ( d + 1))2 (cid:0) r ( d + 1) (cid:1) n − ≤ (cid:0) r ( d + 1) (cid:1) n . • To compute the unimodular matrix U n (matrix M in their notation), onehas to multiply N (= n ) of these matrices E k , so we have thatdeg( U n ) ≤ n (cid:0) r ( d + 1) (cid:1) n . By applying this process recursively and eliminating all the variables, we see thatthe unimodular matrix U of Theorem 2.2 can be computed as a product of matrices U n · U n − . . . U · U , where for i > , each U i eliminates the variable x i , and U ∈ K s × s is a matrix of scalars. So, we have then thatdeg( U ) ≤ n (cid:0) r ( d + 1) (cid:1) n , as claimed. (cid:3) Syzygies and unimodularity
In this section, we will relate the matrices converting a , . . . , a m into p, q , andvice versa, with unimodular matrices. This will allow us to use the Effective Quillen-Suslin Theorem 2.2 to produce an R -basis of Syz( a , . . . , a m ) of controlled degree.Let then a , . . . , a m , p, q ∈ R be such that (1) holds. The syzygy module of thesequence ( a , . . . , a m ) is defined asSyz( a , . . . , a m ) := { ( u , . . . , u m ) ∈ R m | u a + . . . + u m a m = 0 } ⊂ R m . Note that we are not claiming that gcd( p, q ) = 1 , but this can be assumed w.l.o.g.as Syz( a , . . . , a m ) does not change after removing a common factor of all thesepolynomials (which would be a common factor of p and q thanks to (1)) ).From (1), we deduce that there exist matrices M ∈ R × m and N ∈ R m × suchthat(6) ( a a . . . a m ) · N = ( p q ) , ( p q ) · M = ( a a . . . a m ) . In principle, there are infinite matrices M and N that satisfy (6). The results thatwe prove in the sequel hold for any of these choices. Denote with K the 2 × M · N . TERESA CORTADELLAS BEN´ITEZ, CARLOS D’ANDREA, AND EUL`ALIA MONTORO
Lemma 3.1.
Assuming that gcd( p, q ) = 1 , there exist e, f ∈ R such that (7) K = (cid:18) − e · q f · qe · p − f · p (cid:19) . Proof.
Write K = (cid:18) α βγ δ (cid:19) . From (6) we deduce straightforwardly that( p q ) · K = ( p q ) · M · N = ( p q ) . So, we have (cid:26) p = α · p + γ · qq = β · p + δ · q . From here we deduce that (cid:26) (1 − α ) p = γ · q (1 − δ ) q = β · p , so there exist λ, ˜ λ in Q ( R ), the field of fractions of R , such that − α = λ · qγ = λ · p − δ = ˜ λ · pβ = ˜ λ · q . As λ · q and λ · p are elements of R and gcd( p, q ) = 1 , then we deduce that λ ∈ R .The same happens with ˜ λ . The claim follows by setting e λ, f ˜ λ. (cid:3) We cannot claim that K is a unimodular matrix. As a matter of fact, with thenotation above we have that det( K ) = 1 − e · q − f · p, which is an element of K ifand only if e · q + f · p ∈ K . If this is the case and det( K ) = 1 , then I is principal,and the Quillen-Suslin Theorem 2.2 shows that Syz( a , . . . , a m ) is a free R -module,and gives bounds for the degrees of a basis of this module, which are better thanthose appearing in Theorem 4.1.In any case, we can modify the matrix M so that we get e = f = 0 . We start bydenoting with(8) ˜ K := (cid:18) − e · q f · q − qe · p − f · p p (cid:19) ∈ R × , the matrix which consists in adding to K the column (cid:18) − qp (cid:19) . Lemma 3.2. ˜ K is a unimodular matrix.Proof. Indeed the 3 maximal minors of ˜ K are p, q and 1 − e · q − f · p. From herethe claim follows straightforwardly. (cid:3)
To connect ˜ K with M and N , we write them down explicitly:(9) N = b c b c ... ... b m c m and M = (cid:18) d d . . . d m e e . . . e m (cid:19) . OUNDS FOR DEGREES OF SYZYGIES 7
Proposition 3.3. N is a unimodular matrix, and so is ˜ M , where (10) ˜ M := (cid:18) d d . . . d m − qe e . . . e m p (cid:19) ∈ R × ( m +1) . Proof.
Set(11) ˜ N = b c b c b m c m
00 0 1 ∈ R ( m +1) × . We clearly have ˜ M · ˜ N = ˜ K. As ˜ K is unimodular, so are ˜ N and ˜ M (this can be seenfor instance by using the Cauchy-Binet formula for computing minors of a productof matrices). The fact that N is unimodular follows just by noting that all thenonzero maximal minors of it are -up to the sign- the nonzero maximal minors of˜ N . (cid:3) Note that Proposition 3.3 states that a ny matrix N as in (6) is unimodular.That does not apply to M, see for instance Example 6.2. However, one can alwaysreplace M with a unimodular one as the following result shows. Proposition 3.4. M can be chosen as in (6) to be unimodular.Proof. Let e, f ∈ R be such that (7) holds. If ( e, f ) = (0 ,
0) we are done. If not,because N is unimodular, its rows generate R as an R -module, so there exist x . . . x m ∈ R such that(12) x · ( b c ) + x · ( b c ) + . . . + x m · ( b m c m ) = ( e − f ) . Set then(13) M ′ = M + (cid:18) x q x q . . . x m q − x p − x p . . . − x m p (cid:19) . We clearly have that M ′ satisfies (6), and an easy computation shows that(14) M ′ · N = (cid:18) M + (cid:18) x q x q . . . x m q − x p − x p . . . − x m p (cid:19)(cid:19) · N = (cid:18) (cid:19) , thanks to (12). So, M ′ is unimodular, as claimed. (cid:3) We conclude by showing a characterization of those ideals of grade 2 having theproperty (1) via the matrix M . Theorem 3.5.
For a . . . , a m , p, q ∈ R , we have that h a , . . . , a m i = h p, q i if andonly if there exists a unimodular matrix M ∈ R × m such that ( p q ) · M = ( a . . . a m ) . Proof.
The “if” part follows from Proposition 3.4. For the converse, apply Theo-rem 2.2 to the unimodular matrix M , and let U ∈ R m × m be such that M · U = (cid:18) . . .
00 1 0 . . . (cid:19) . Set N ∈ R m × to be the matrix consisting of the first twocolumns of U . We then have that (6) holds and so (1), which proves the claim. (cid:3) TERESA CORTADELLAS BEN´ITEZ, CARLOS D’ANDREA, AND EUL`ALIA MONTORO
Remark 3.6.
The role of M and N are different in the characterization of idealssatisfying (1) . Indeed, having N unimodular is not enough to characterize theseideals, as for instance we may have a , . . . , a m ∈ R with m ≥ be such that a / ∈ h a , a i . If we pick set N = ... ... , then it is clear that we will never findan M such that (6) holds. Algorithms
We will now exhibit algorithms to compute R -bases of Syz( a , . . . , a m ) by ap-plying the Effective Quillen-Suslin Theorem 2.2 to ˜ M , M ′ or N .4.1. Working with ˜ M . With notation as above, from Theorem 2.2 we deducethat there exists a square unimodular matrix U ∈ R ( m +1) × ( m +1) such that(15) ˜ M · U = (cid:18) . . .
00 1 0 . . . (cid:19) . Assume w.l.o.g. that det( U ) = 1 . From (6) we deduce that ( p q ) · ˜ M = ( a . . . a m , and hence we must have(16) ( p q ) · ˜ M · U = ( p q . . .
0) = ( a . . . a m · U Let b U ∈ R m × ( m − the submatrix of U consisting in removing its first two columnsand its last row. From (16) we deduce that the columns of b U are syzygies of( a , . . . , a m ) . Our main result is the following.
Theorem 4.1.
The columns of b U are an R -basis of Syz ( a , . . . , a m ) . Proof.
Let U , . . . , U m +1 be the columns of U , and denote with ˜ U ∈ R ( m +1) × m thematrix whose columns are − q · U + p · U , U , . . . , U m +1 , in this order. By applyingCramer’s rule to the last equality of (16), we deduce that the signed maximal minorsof this matrix are a , . . . , a m ,
0. Hence, the ideal generated by these maximal minorsis I, which has grade 2 by our initial assumptions. By applying then the converseof the Hilbert-Burch Theorem 2.1, we deduce then that the columns of ˜ U are abasis of Syz( a , . . . , a m , U − , the inverse matrix of U , is equalto ˜ M .
Indeed, denote then by U − × ( m +1) the submatrix of U − made by these rows.From (15) we deduce that ˜ M · U = U − × ( m +1) · U, and as U is invertible, the claim follows. In particular, the last column of U − is ofthe form − qpr ... r m +1 for suitable r , . . . , r m +1 ∈ R. From the identity U · U − = I m +1 , OUNDS FOR DEGREES OF SYZYGIES 9 we deduce that − q · U + p · U + r · U + . . . + r m +1 · U m +1 = . This implies that if we perform the following column operation in ˜ U : to its firstcolumn (which is equal to − q · U + p · U ) we add r U + . . . + r m +1 U m +1 , thefact that its columns are an R -basis of Syz( a , . . . , a m ,
0) remains unchanged, butnow the first column of the modified matrix is equal to . From here, we can perform new column operations in U , U , . . . , U m +1 in sucha way that we now that the last row of ˜ U equals to (1 0 . . . , and the other coeffi-cients of this matrix have not changed. So, we have shown that Syz( a , . . . , a m , R -basis of the form (cid:18) b U (cid:19) , and from here the claim follows straightforwardly. (cid:3) Working with a unimodular M . If M is already unimodular (which is forinstance the case of the matrix M ′ defined in (13), although we are not requiringthat M · N = I as in (14)), we can apply directly the Effective Quillen-SuslinTheorem 2.2 to M and obtain a unimodular U ∗ ∈ R m × m such that(17) M · U ∗ = (cid:18) . . .
00 1 0 . . . (cid:19) . From (6) we now get that ( p q ) · M = ( a . . . a m ) , and hence(18) ( p q ) · M · U ∗ = ( p q . . .
0) = ( a . . . a m ) · U ∗ . Denote the columns of U ∗ with U ∗ U ∗ . . . , U ∗ m . Let c U ∗ ∈ R m × ( m − be thematrix(19) c U ∗ = (cid:0) qU ∗ − pU ∗ , U ∗ , . . . , U ∗ m (cid:1) . By applying Cramer’s rule to (18), and using the converse of the Hilbert-BurchTheorem 2.1, we deduce straightforwardly that
Theorem 4.2.
The columns of c U ∗ defined in (19) are an R -basis of Syz ( a , . . . , a m ) . Working with N . We can also work directly with the unimodular matrix N from (6) and construct an R basis of Syz( a , . . . , a m ) as follows: denote with N ∗ ∈ R m × m a matrix of determinant equals to 1 such that has N has its first 2columns. This can be done by applying the Quillen-Suslin Theorem 2.2 to N t . Thematrix N ∗ can be taken to be the inverse of matrix U in this claim. Denote itscolumns with N ∗ , . . . , N ∗ m . From (6), we clearly have that ( a . . . a m ) · N ∗ = p, and ( a . . . a m ) · N ∗ = q. As for all i = 1 , . . . , m, we have that ( a . . . a m ) · N ∗ i ∈ h a , . . . , a m i = h p, q i , for i = 3 , . . . , m we write ( a . . . a m ) · N ∗ i = λ i p + δ i q for suitable λ i , δ i ∈ R. Perform then the following elementary column operations in N ∗ : for i = 3 , , . . . , m, replace column N ∗ i with N ∗ i − λ i N ∗ − δ i N ∗ . Call the remaining matrix N ∗∗ . Byconstruction: • det( N ∗∗ ) = 1 , i.e. N ∗∗ is unimodular; • ( a . . . a m ) · N ∗∗ = ( p q . . . N ∗∗ , N ∗∗ , . . . the columns of N ∗∗ , and let b N ∈ R m × ( m − thematrix whose columns are(20) b N = (cid:0) qN ∗∗ − pN ∗∗ N ∗∗ . . . N ∗∗ m (cid:1) Theorem 4.3.
The columns of b N are an R -basis of Syz ( a , . . . , a m ) . Proof.
As before, by taking into account that ( a . . . a m ) · N ∗∗ = ( p q . . . , andapplying Cramer’s rule to this matrix, it is easy to see that the signed maximalminors of b N are a , . . . a m . The converse of the Hilbert-Burch Theorem 2.1 thenproves the claim. (cid:3) Relations among the bases.
In the following, we will show that if one pickconvenient unimodular matrices in the process of computing the several R -basis ofSyz( a , . . . , a m ) described above, the ansatz is essentially the same. We begin byproving the following straightforward relation between U ∗ and N ∗ if M is alreadyunimodular. Proposition 4.4.
Suppose that M · N = I (in particular, we have that M isunimodular). Then, one can find a unimodular m × m matrix which can be used as U ∗ in (17) and also as N ∗∗ from § c U ∗ = b N, i.e. the basesfrom Theorems 4.2 and 4.3 coincide.Proof. Start by picking any U ∗ ∈ R m × m satisfying (17). Denote its columns with U ∗ . . . U ∗ m . As we have M · (cid:0) U ∗ U ∗ (cid:1) = M · N = (cid:18) (cid:19) , we deduce that each of the columns of (cid:0) U ∗ U ∗ (cid:1) − N belongs to the right-kernel of M. Denote also with N and N the columns of N . As M is unimodular, its right-kernel is a free R -module, with basis U ∗ , . . . , U ∗ m . So there exist y i , . . . , y im ∈ R such that U ∗ i − N i = y i U ∗ + . . . + y im U ∗ m , i = 1 , U ∗ to be the matrix obtained from U ∗ by substracting to the column i thefollowing linear combination of columns: y i U ∗ + . . . + y im U ∗ m , i = 1 , . Clearly U ∗ is unimodular, and has N as its first two columns, so the first part of the claimfollows.To see the second part, note that as we have( a . . . a n ) · N ∗ = ( p q ) · M · U ∗ = ( p q . . . , we deduce that λ i , δ i , i = 3 , . . . , m from § N ∗∗ defined in that section equals to N ∗ (which is equal to U ∗ ). Asthe process to convert N ∗∗ into b N and U ∗ into c U ∗ is the same (replace the firsttwo columns with q times the first column minus p times the second column),we deduce straigthforwardly that c U ∗ = b N, which concludes with the proof of theproposition. (cid:3) OUNDS FOR DEGREES OF SYZYGIES 11 If M is not unimodular anymore, we will have to work with matrices ˜ M and ˜ N which were defined in (10) and (11) respectively. The following result then holds. Proposition 4.5.
The matrices U and N ∗ from (15) and § c U ∗ = b N, i.e. the bases from Theorems 4.2 and 4.3 coincide. Note that we are not having any requirement about the result of the productbetween ˜ M and ˜ N as it may be hinted by the situation in Proposition 4.4. Proof.
Denote the columns of N with N , N , and set(21) ˜ N ∗ = (cid:18) N N qN − pN − e f − eq − f p (cid:19) ∈ R ( m +1) × , with e, f ∈ R as in (3.1). As ˜ M · ˜ N = ˜ K, with ˜ K defined in (8), we deduce that˜ M · ˜ N ∗ = (cid:18) (cid:19) . This implies that ˜ N ∗ is unimodular, and hence can be extended to an invertible U ∈ R ( m +1) × ( m +1) such that (15) holds and having ˜ N ∗ as its first columns. Wehave then that U = (cid:18) N N qN − pN N . . . N m +1 − e f − eq − f p r . . . r m +1 (cid:19) for suitable columns N , . . . , N m +1 in R m × and suitable r , . . . , r m +1 ∈ R. As byperforming elementary operations in the first columns of U we can have get , we deduce then that, by picking N ∗ = ( N N N . . . N m +1 ) ∈ R m × m , • N ∗ is unimodular and has as first two columns N and N , i.e. it is a valid N ∗ in the sense of the algorithm described in § • Following the steps of that algorithm, N ∗∗ = N ∗ . • b N from (20) equals to (cid:0) qN − pN N . . . N m +1 (cid:1) = b U from Theorem 4.1.This concludes with the proof of the Proposition. (cid:3) Degree bounds and proof of the main theorem
In this section, we will apply our explicit bound given in Theorem 2.6 to givebounds for deg( b U ) from Theorem 4.1, deg( c U ∗ ) from Theorem 4.2, and deg( b N )from Theorem 4.3. These bounds are given in terms of the degrees of both theinput polynomials, but also of the transition matrices. We will show also somerelations among the input bounds, and conclude by proving Theorem 1.1 from theIntroduction.Recall that we have δ being a bound for the degrees of p and q, and δ a beinga bound on the degrees of a , . . . , a m . In addition, we set δ N (resp. δ M ) being abound for the degrees of the elements in N (resp. M ). Proposition 5.1.
With notation as above, if δ ≥ max { δ M , δ } , we have that deg( b U ) ≤ n n ( δ + 1) n . Proof.
Note that b U is a submatrix of the unimodular matrix U . The result is thenconsequence of Theorem 2.6 with r = 2 . (cid:3) Proposition 5.2.
With notation as above, if M is unimodular, we have that deg( c U ∗ ) ≤ n n ( δ M + 1) n + δ . Proof.
By applying Theorem 2.6 to the unimodular matrix M we obtain U ∗ as in(17) with deg( U ∗ ) ≤ n n ( δ M + 1) n . To get c U ∗ we only have to modify the first column of U ∗ and multiply a couple ofcolumns of the latter by − q and p . From here, the claim follows straigthforwardly. (cid:3) Interestingly, the computation of a basis of syzygies starting from N will give usdegree bounds which can differentiate the degrees of the different actors involved.The drawback is that this bound depends also on m , the number of elements ofthe sequence. Sometimes this leads to lower bounds as in the case of parametriccurves, see Section 7. Proposition 5.3.
With notation as above, we have that deg( b N ) ≤ δ + δ N + δ M + 3 mn n ( δ N + 1) n . Proof.
By applying the Effective Quillen-Suslin procedure to the unimodular matrix N, we deduce that the matrix U of Theorem 2.2 has degree bounded by 3 n n ( δ N +1) n , thanks to Theorem 2.6 with r = 2 . The matrix N ∗ from § U, and can be computed by usingcofactors. Hence, its entries have degree bounded by3 mn n ( δ N + 1) n . As we have ( p q ) · M = ( a a . . . a m ) from (6), we get( a . . . a m ) · N ∗ i = λ i p + δ i q = ( p q ) · M · N ∗ i , and deduce then that deg( λ i ) , deg( δ i ) ≤ δ N + δ M for all i = 3 , . . . , m. By construc-tion, we have that deg( N ∗∗ ) ≤ δ N + δ M + 3 mn n ( δ N + 1) n . To pass from N ∗∗ to b N , the matrix encoding a basis of Syz( a , . . . , a m ) , we onlyneed to modify the first column of N ∗∗ , and we havedeg( b N ) ≤ δ + δ N + δ M + 3 mn n ( δ N + 1) n , as claimed. (cid:3) OUNDS FOR DEGREES OF SYZYGIES 13
From ˜ M to an unimodular M . The bounds on the degrees of the basescomputed via ˜ M or M if the latter is unimodular are not very comparable, forinstance they depend on whether δ > δ M or viceversa. We will compute forcompletion of our study of the degrees appearing in these matrices a bound on theunimodular matrix M ′ defined in (13) in terms of the input degrees of the problem. Proposition 5.4.
With notation as above, a matrix M ′ as in (13) can be computedwith deg( M ′ ) ≤ n n ( δ N + 1) n + δ M + δ N + δ . Proof.
Note that e and f from (7) have degrees bounded by δ M + δ N . Let U N ∈ R m × m the unimodular matrix such that N t · U N = ( I ) ∈ R × m . By Theorem2.6, U N can be computed with deg( U N ) ≤ n n ( δ N + 1) n . As usual, denote thecolumns of U N with U N , . . . U mN . If we set, x x ... x m := eU N + f U N we deduce that (12) is satisfied. By computing explicitly, we getdeg( x i ) ≤ n n ( δ N + 1) n + δ M + δ N . To compute M ′ from M as in (13), we have to add to the latter a matrix withcoefficients then bounded by 3 n n ( δ N + 1) n + δ M + δ N + δ . From here, the claimfollows straightforwardly, as this bound is larger than δ M , which bounds the degreeof M . (cid:3) Relations among bounds.
So far, we have • δ M , a bound on the degree of the elements of M ; • δ N , a bound on the degree of the elements of N ; • δ , a bound on the degrees of p and q ; • δ a , a bound on the degrees of a , . . . , a m . Is there any relation among these bounds? Clearly, from (6) we deduce straight-forwardly that • given δ a and δ N , one can set δ := δ a + δ N ; • given δ and δ M , one can set δ a := δ + δ M . The following relation is less subtle.
Proposition 5.5.
Given δ and δ a , there exists a matrix M such that one can take δ M := δ + δ a Proof.
As we are assuming gcd( p, q ) = 1 , we get these polynomials are an affinecomplete intersection in K n . The result then follows straightforwardly from Corol-lary 5.2 in [DFGS91]. (cid:3) The connection between M and N given via Theorem 3.5 gives a bound for δ N in terms of δ M and δ but not a very optimal one. Proposition 5.6.
Given δ M and δ , there exists a matrix N such that one cantake δ N = 3 n n (max { δ , δ M } + 1) n . Proof.
A possible matrix N can be taken by using the first columns (except the lastrow) of a unimodular U ∈ R ( m +1) × ( m +1) such that (15) holds. Thanks to Theorem2.6, we have that the degree of N is bounded by 3 n n (max { δ , δ M } + 1) n , whichproves the claim. (cid:3) In the zero-dimensional case, one can have a sharper bound for δ N . Proposition 5.7.
If the ideal I is zero-dimensional, given δ and δ a , there existsa matrix N such that one can take δ N = 2 δ a + δ a + δ .Proof. This follows essentially from Theorem 2.5 in [Has09]. (cid:3)
We conclude this section by giving the proof of the main result announced in theIntroduction5.3.
Proof of Theorem 1.1.
The first algorithm essentially consists in computingthe matrix U from (15), and extract the submatrix b U which -thanks to Theorem4.1- encodes an R -basis of Syz( a , . . . , a m ) . A degree bound for b U is given in Proposition 5.1 in terms of the degrees of δ M and δ . By using Proposition 5.5, we can replace this bound by δ + δ a , and thenthe first part of the claim follows straightforwardly.For the second part, we will work with the matrix N as in § b N which -thanks to Theorem 4.3- encodes also an R -basisof Syz( a , . . . , a m ) , having the degree bounds given in Proposition 5.3. Note thatwe are not assuming yet that I is zero-dimensional. This would be used to replace δ N and δ M with the bounds given in Proposition 5.5 and 5.7 to get (2). (cid:3) Examples
We present here the computation of two examples. The first one is the runningexample in [CDM20] and has already M being unimodular, while the second doesnot. To be consistent with the notation of [CDM20], we label the variables as s, t .So in both examples we have that n = 2.6.1. Example 4.1 in [CDM20] . Here we have m = 4 , p = t − s + 2 , q = s + 1 and a ( s, t ) = 11 − s + 3 s + 4 ta ( s, t ) = 5 − s + 2 s + 4 t − st + t a ( s, t ) = 1 + 3 s − s + s ta ( s, t ) = 7 − s + s + 3 t. As it was shown in [CDM20], one can take for this case M = (cid:18) t − s + 2 s
33 1 1 1 (cid:19) . In addition, a simple N is the following:(22) N = −
15 35 − . In this case, as the columns 1 and 4 of M are already an invertible matrix in K [ s, t ] , The same happens with the rows 1 and 4 of N . By pivoting the 2 × OUNDS FOR DEGREES OF SYZYGIES 15 submatrix of M , it is easy to compute a matrix 4 × U ∗ suchthat M · U ∗ = (cid:18) (cid:19) , we get: U ∗ = − − s + t + s + s − t − s − t + − −
10 0 1 0
35 3 s − t − − s − s + t + − s + t + , To pass from U ∗ to c U ∗ we proceed as in (19) and get that c U ∗ = − s + st + s − t − s + s − t − s − t + s − t − − −
10 1 0 s − st − s + t + t + 3 − s − s + t + − s + t + . Thanks to Theorem 4.2, the columns of c U ∗ encode a basis of Syz( a , a , a , a ).Note that what we obtained with this procedure is quite different than the basisobtained in [CDM20]. For instance, the degree of this basis is 2 , which is lowerthan the one obtained in that paper (equal to 5).Now we work with N . From (22) it is easy to extend N to a 4 × N ∗ = −
15 35 − − . To produce the matrix N ∗∗ of the algorithm, we have to modify the columns 3 and4 of N ∗ by using M . We obtain N ∗∗ = −
15 35 s − t + 1 s − − − − s + 3 t + 2 − s . Finally, to obtain b N we must replace the columns 1 and 2 of N ∗∗ with the firstcolumn multiplied by − q plus the second column multiply by p : b N = s − s + t + s − t + 1 s − − − s + s − t − − s + 3 t + 2 − s . By Theorem 4.3, the columns of b N encode another R -basis of Syz( a , a , a , a ) . Note that the bases obtained via b N and c U ∗ are essentially different. For instance,the first column of b N is a relation that only involves a and a , but nothing ofthis nature can be found from the columns of c U ∗ . This is because the relation M · N = (cid:18) (cid:19) does not hold. Another Example.
Set now m = 4 again, and p = t + 2 s + 1 , q = − t − s which are not under the conditions of the Shape Lemma as all the results in[CDM20] are. Consider the following sequence of polynomials: a = s + 3 s + t + 4 st − t a = − s + t + t a = s + 2 s − t a = 1 + s − t. In this case, we can take M = (cid:18) s + t t s − s + t s t (cid:19) , and note that M is not unimodular (setting s = t = 0 makes the rank of M drops).In contrast, we have that˜ M = (cid:18) s + t t s s + 2 t − s + t s t t + 2 s + 1 (cid:19) is unimodular, and by applying Quillen-Suslin to this matrix we obtain U such that˜ M · U = (cid:18) (cid:19) . In our case, we get U = − t − s − s s − t + 11 − − s + t + 1 − s s − t − s − s − t − t − s − t s t − s , and hence b U = t − s − s s − t + 1 − s + t + 1 − s s − t − s − s − t − t is a basis of Syz( a , a , a , a ) . Note that the first columns of U (except the lastrow) encode the matrix N , i.e. we can take this matrix as N = − −
11 0 . From here, we can extend it easily to a unimodular N ∗ = − − , OUNDS FOR DEGREES OF SYZYGIES 17 and then we get N ∗∗ = − − s s − t + 11 − − s s − t − s − t − t . To conclude, we need to replace the first two columns by a combination of them,to get b N = − s + 2 t − − s s − t + 1 − s + t − − s s − ts + 2 t − s − t − t which encodes another basis of Syz( a , a , a , a ) . This basis is quite similar thanthe one we found via U . Indeed, the second and the third columns of both matricescoincide, while the first column of b N equals to the first minus 2 times the lastcolumn of ˜ U .
This “coincidence” is explained by our choice of N from the first twocolumns of the matrix U above.7. Minimal µ -bases of parametric surfaces From a geometric point of view, the sequence ( a , . . . , a m ) of polynomials in K [ x , . . . , x n ] can be regarded as the parametrization of a variety Y ⊂ K m be-ing the image of the map ( a , . . . , a m ) : K n → K m . Understanding the role ofSyz( a , . . . , a m ) in the study of geometric properties of Y is an active area of re-search, and several results and challenges are well identified there, see for instance[Cox03]. In particular, when n = 2 , this map represents typically a surface in K m ,and bounding the degrees of generators of the syzygy module has been posed as anopen problem in [CCL05] for the case m = 4 , i.e. surfaces in 3-dimensional space.A first answer to this problem was posted in [Cid19], where the author exhibitsa general bound of order δ a to solve this problem ( n = 2 , m = 4). In [CDM20] weimproved this bound to δ a in the case the ideal I has a so-called “shape basis”,meaning that one can take p and q in (1) as x − r and s respectively, with r, s ∈ K [ x ] . Indeed, our main result there ([CDM20, Theorem 1.2]) is that a basis ofSyz( a , a , a , a ) in that case can be found with degree bounded by(23) 5 δ (2 δ a + 1) . The approach to solve that problem differ significantly than the one presentedhere. For instance we only used there the “simplified” Efective Quillen-Suslin The-orem presented in [FG90], which is essentially Theorem 2.2 in the case r = 1. Weonly worked with the matrix which is called M in (6), and took advantage of thefact that one of its rows only depends on the variable x in the shape-basis case.From there, after applying the Effective Quillen-Suslin algorithm to that univari-ate row ( r = 1), a tricky manipulation of the remaining row would allow us toperform again the same simplified Effective Quillen-Suslin algorithm to the secondrow. The basis would come then after some simplifications and substitutions ofthese calculations.In the present paper, we only apply once the Effective Quillen-Suslin algorithmdescribed in [CCDHKS93] with r = 2 to the modified matrix ˜ M instead of M .As a result, we obtain the matrix b U which encodes the elements of a basis of Syz( a , . . . , a m ) (Theorem 4.1). In addition, we also get another algorithm over thematrix N which gives another basis for this module (Theorem 4.3). This approachwas not conidered in [CDM20].When applied to the case n = 2 , m = 4 , and using the fact that thanks toB´ezout’s Theorem one can take δ ≤ δ a , the first bound in Theorem 1.1 thenamounts to a constant times δ a , while the second one is of the order of δ a , whichis better than the results obtained in [CDM20] if one substitutes δ with δ a . But we can actually improve the bound of δ a from Theorem 1.1 if we inspectcarefully the structure of a matrix M converting a shape basis into the input se-quence a . . . a , as it was done in [CDM20]. Indeed, we have Proposition 7.1. If a , a , a , a ∈ K [ s, t ] have degrees bounded by δ and theideal generated by them has a shape basis as in [CDM20] of degree δ , a basis ofSyz ( a , a , a , a ) can be found with degree bounded by δ δ a + 1) . Proof.
From the proof of Theorem 1.2 in [CDM20], we get that the matrix M converting the shape basis into the original sequence has δ M ≤ δ a δ . The resultnows follows by applying Theorem 2.6 to ˜ M with n = r = 2 . (cid:3) Remark 7.2.
Note that the bound obtained in Proposition 7.1 is of the same orderthan the one in (23) in terms of δ and δ a . References [CCDHKS93] Caniglia, Leandro; Corti˜nas, Guillermo; Dan´on, Silvia; Heinz, Joos; Krick, Teresa;Solern´o, Pablo.
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OUNDS FOR DEGREES OF SYZYGIES 19
Universitat de Barcelona, Facultat de Educaci´o. Passeig de la Vall d’Hebron 171,08035 Barcelona, Spain
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