Ideals of the form I 1 (XY)
aa r X i v : . [ m a t h . A C ] J a n IDEALS OF THE FORM I ( XY ) JOYDIP SAHA, INDRANATH SENGUPTA, AND GAURAB TRIPATHIA
BSTRACT . In this paper we compute Gr¨obner bases for determinantalideals of the form I ( XY ) , where X and Y are both matrices whoseentries are indeterminates over a field K . We use the Gr¨obner basisstructure to determine Betti numbers for such ideals.
1. I
NTRODUCTION
Let K be a field and { x ij ; 1 ≤ i ≤ m, ≤ j ≤ n } , { y j ; 1 ≤ j ≤ n } be indeterminates over K . Let K [ x ij ] and K [ x ij , y j ] denote the polynomialalgebras over K . Let X denote an m × n matrix such that its entries belongto the ideal h{ x ij ; 1 ≤ i ≤ m, ≤ j ≤ n }i . Let Y = ( y j ) n × be thegeneric n × column matrix. Let I ( XY ) denote the ideal generated bythe × minors or the entries of the m × matrix XY . Ideals of the form I ( XY ) appeared in the work of J. Herzog [9] in 1974. These ideals areclosely related to the notion of Buchsbaum-Eisenbud variety of complexes.A characteristic free study of these varieties can be found in [5], where thedefining equations of these varieties have been described as minors of ma-trices using combinatorial structure of multitableux. It has also been provedthat the varieties are Cohen-Macaulay and Normal. The ideal I ( XY ) is aspecial case of the defining ideal of a variety of complexes, when n = m , n = n , n = 1 , in the notation of [5]. These ideals feature once againin [18], in the study of the structure of a universal ring of a universal pair defined by Hochster. It has been proved in [18] that the set of standardmonomials form a free basis for the universal ring. The initial ideal of thedefining ideal is given by the set of all nonstandard monomials, which forma monomial ideal. A combination of Gr¨obner basis techniques and repre-sentation theory techniques yield the results in [18]. We were not awareof this work when we computed a Gr¨obner basis for the ideal I ( XY ) us-ing very elementary techniques. Our technique uses nothing more than theBuchberger’s criterion and the description of Gr¨obner bases for the idealsof minors of matrices from [4] and [17]. Mathematics Subject Classification.
Primary 13P10; Secondary 13C40, 13D02.
Key words and phrases.
Gr¨obner basis, Betti numbers, determinantal ideals, completelyirreducible systems.The second author is the corresponding author.
Given determinantal ideals I and J , the sum ideal I + J is often difficultto understand and they appear in various contexts. Ideals I ( XY ) + J arespecial in the sense that they occur in several geometric considerations likelinkage and generic residual intersection of polynomial ideals, especiallyin the context of syzygies; see [14], [1], [3], [2], [13]. Some importantclasses of ideals in this category are the Northcott ideals, the Herzog ideals;see Definition 3.4 in [1] and the deviation two Gorenstein ideals defined in[10]. Northcott ideals were resolved by Northcott in [14]. Herzog gave aresolution of a special case of the Herzog ideals in [9]. These results wereextended in [3]. In a similar vein, Bruns-Kustin-Miller [2] resolved theideal I ( XY ) + I min( m,n ) ( X ) , where X is a generic m × n matrix and Y isa generic n × matrix. Johnson-McLoud [13] proved certain properties forthe ideals of the form I ( XY ) + I ( X ) , where X is a generic symmetricmatrix and Y is either generic or generic alternating. One of the recentarticles is [11] which shows connection of ideals of this form with the idealof the dual of the quotient bundle on the Grassmannian G (2 , n ) .Ideals of the form I + J also appear naturally in the study of some natu-ral class of curves; see [8]. While computing Betti numbers for such ideals,a useful technique is often the iterated Mapping Cone. This technique re-quires a good understanding of successive colon ideals between I and J ,which is often difficult to compute. It is helpful if Gr¨obner bases for I and J are known.In this paper our aim is to produce some suitable Gr¨obner bases for idealsof the form I ( XY ) , when Y is a generic column matrix and X is one ofthe following:(1) X is a generic square matrix;(2) X is a generic symmetric matrix;(3) X is a generic ( n + 1) × n matrix.We have also studied I ( XY ) , when(4) X is an ( m × mn ) generic matrix and Y is an ( mn × n ) genericmatrix.Our method is constructive and it would exhibit that the first two casesbehave similarly. Newly constructed Gr¨obner bases will be used to computethe Betti numbers of I ( XY ) . We will see that computing Betti numbers for I ( XY ) in the first two cases is not difficult, while the last two cases are notso straightforward. We will use some results from [15] and [16] which havesome more deep consequences of the Gr¨obner basis computation carriedout in this paper. DEALS OF THE FORM I ( XY )
2. D
EFINING THE PROBLEMS
Let K be a field and { x ij ; 1 ≤ i ≤ n + 1 , ≤ j ≤ n } , { y j ; 1 ≤ j ≤ n } be indeterminates over K . Let R = K [ x ij , y j | ≤ i, j ≤ n ] , b R = K [ x ij , y j | ≤ i ≤ n + 1 , ≤ j ≤ n ] denote polynomial K -algebras.Let X = ( x ij ) n × n , such that X is either generic or generic symmetric.Let b X = ( x ij ) ( n +1) × n and Y = ( y j ) n × be generic matrices. We define I = I ( XY ) and J = I ( b XY ) .Let g i = P nj =1 x ij y j , for ≤ i ≤ n . Then, I = h g , . . . , g n i . Let uschoose the lexicographic monomial order on R given by(1) x > x > · · · > x nn ;(2) x ij , y j < x nn for every ≤ i = j ≤ n .It is an interesting observation that the set { g , . . . , g n } is a Gr¨obner basisfor I with respect to the above monomial order and the elements g , . . . , g n form a regular sequence as well; see Lemma 4.3 and Theorem 6.1. How-ever, this Gr¨obner basis is too small in size to be of much help in applica-tions like computing primary decomposition of I ( XY ) or computing Bettinumbers of ideals of the form I ( XY ) + J , carried out in [15] and [16]respectively. This motivated us to look for a a different Gr¨obner basis for I ; see Theorem 4.1. This construction gives rise to a bigger picture andnaturally generalizes to a Gr¨obner basis for the ideal J = I ( b XY ) . Asan application, we compute the Betti numbers for the ideals I and J ; seesection 6. 3. N OTATION (i) C k := { a = ( a , · · · , a k ) | ≤ a < · · · < a k ≤ n } ; denotesthe collection of all ordered k -tuples from { , · · · , n } . In case of J = I ( b XY ) , the set C k would denote the collection of all ordered k -tuples ( a , · · · , a k ) from { , · · · , n + 1 } .(ii) Given a = ( a , . . . , a k ) ∈ C k ; • X a = [ a , · · · , a k | , , . . . , k ] denotes the k × k minor of thematrix X , with a , . . . , a k as rows and , . . . , k as columns.Similarly, b X a = [ a , · · · , a k | , . . . , k ] denotes the k × k mi-nor of the matrix b X , with a , . . . , a k as rows and , . . . , k ascolumns. • S k := { X a : a ∈ C k } and I k denotes the ideal generated by S k in the polynomial ring R (respectively b R ); • X a ,m := [ a , · · · , a k | , · · · , k − , m ] if m ≥ k ; • f X a = P m ≥ k [ a , · · · , a k | , · · · , k − , m ] y m = P m ≥ k X a ,m y m ; JOYDIP SAHA, INDRANATH SENGUPTA, AND GAURAB TRIPATHI • e S k := { f X a : X a ∈ S k } and e I k denotes the ideal generated by e S k in the polynomial ring R (respectively b R ); • G k = ∪ i ≥ k e S i ; • G = ∪ k ≥ G k ; • X a r := [ a , a , · · · , ˆ a r , a r +1 · · · , a k | , , · · · , k − , if k ≥ .(iii) Suppose that C k = n a < . . . < a ( nk ) o , where < is the lexico-graphic ordering. Given m ≥ k , the map σ m : n X a ,m , . . . , X a ( nk ) ,m o → (cid:26) , · · · , (cid:18) nk (cid:19)(cid:27) is defined by σ m ( X a i ,m ) = i . This is a bijective map. The map σ k will be denoted by σ , which is the bijection from S k to { , · · · , (cid:0) nk (cid:1) } given by σ ( X a i ) = σ k ( X a i ,k ) = i .4. G R ¨ OBNER BASIS FOR I We first construct a Gr¨obner basis for the ideal I . A similar computa-tion works for computing a Gr¨obner basis for the ideal J , which will bediscussed in the next section. Our aim in this section is to prove Theorem 4.1.
The set G k is a reduced Gr¨obner Basis for the ideal e I k , withrespect to the lexicographic monomial order induced by the following orderon the variables: y > y > · · · > y n > x ij for all i, j , such that x ij > x i ′ j ′ if i < i ′ or if i = i ′ and j < j ′ . In particular, G = G is a reduced Gr¨obnerBasis for the ideal e I = I . We first write down the main steps involved in the proof. Let e X a , e X b ∈ G k = ∪ i ≥ k e S i . Then, either X a , X b ∈ S k or X a ∈ S k , X b ∈ S k ′ , for k ′ > k . Our aim is to show that S ( e X a , e X b ) → G k and use Buchberger’scriterion.(A) By Lemma 4.2, we have S ( X a , X b ) −→ S k . We write m a X a + m b X b = S ( X a , X b ) = P ( nk ) t =1 α t X a t −→ S k , such that X a i = X a and X a j = X b , for some i and j . Therefore, by Schreyer’s theoremthe tuples ( α , . . . , α i − m a , . . . , α j − m b , . . . , α r ) generate Syz( I k ) .(B) Syz( I k ) is precisely known by [6].(C) S ( e X a , e X b ) −→ e S k S ( e X a , e X b ) − P ( nk ) t =1 α t e X a t by Lemma 4.8, if X a , X b ∈ S k and by Lemma 4.10, if X a ∈ S k , X b ∈ S k ′ , for k ′ > k .(D) S ( e X a , e X b ) − P ( nk ) t =1 α t e X a t = s ∈ e I k +1 , by Lemma 4.8, if X a , X b ∈ S k . DEALS OF THE FORM I ( XY ) (E) S ( e X a , e X b ) − P ( nk ) t =1 α t e X a t = s ∈ e I k ′ +1 , by Lemma 4.10, if X a ∈ S k , X b ∈ S k ′ , for k ′ > k .(F) s −→ G k , proved in Theorem 4.1 for both the cases.We first prove a number of Lemmas to complete the proof through thesteps mentioned above. Lemma 4.2.
The set S k forms a Gr¨obner basis of I k with respect to thechosen monomial order on R .Proof. We use Buchberger’s criterion for the proof. Let c , d ∈ S k . Supposethat S ( X c , X d ) S k −→ r . Then, S ( X c , X d ) − P a i ∈ C i h i X a i = r .If X is generic (respectively generic symmetric), we know by [17] (re-spectively by [4]) that the set of all k × k minors of the matrix X formsa Gr¨obner basis for the ideal I k ( X ) , with respect to the chosen monomialorder. Therefore, there exists [ a , a , · · · , a k | b , b , · · · , b k ] , such that itsleading term Q ki =1 x a i b i divides Lt ( r ) . We see that if b k = k , the minorbelongs to the set S k and we are done.Let us now consider the case b k ≥ k + 1 . Let X be generic symmetric.Then, a k = k and b k ≥ k + 1 imply that the minor belongs to the set S k .If a k , b k ≥ k + 1 , then x a k b k | Lt ( r ) but x a k b k doesn’t divide any term ofelements in S k . Let X be generic. Then, for any a k and under the condition b k ≥ k + 1 , then x a k b k | Lt ( r ) but x a k b k doesn’t divide any term of elementsin S k . (cid:3) Lemma 4.3.
Let h , h · · · , h n ∈ R be such that with respect to a suit-able monomial order on R , the leading terms of them are pairwise co-prime. Then, h , h · · · , h n is a Gr¨obner basis of the ideal generated by h , h · · · , h n with respect to the same monomial order and they form aregular sequence in R .Proof. . The proof is a routine application of the division algorithm. (cid:3) Lemma 4.4.
Let ≤ k ≤ n . The height of the ideal I k is n − k + 1 , in caseof X .Proof. . Let us consider the case for X . We know that ht ( I k ) ≤ n − k + 1 .It suffices to find a regular sequence of that length in the ideal I k . We claimthat { [1 · · · k | · · · k ] , [2 · · · k +1 | · · · k ] , . . . , [ n − k +1 · · · n | · · · k ] } formsa regular sequence. The leading term of [ a , a , · · · , a k | b , b , · · · , b k ] with respect to the chosen monomial order is Q ki =1 x a i b i . Therefore, leadingterms of the above minors are mutually coprime and we are done by Lemma4.3. (cid:3) JOYDIP SAHA, INDRANATH SENGUPTA, AND GAURAB TRIPATHI
Remark . We now assume that X = ( x ij ) is a generic n × n matrix. Theproof for the symmetric case is exactly the same. Description of generators of
Syz( I k ) . By Lemma 4.4 we conclude that aminimal free resolution of the ideal I k is given by the Eagon-Northcott com-plex. Let us describe the first syzygies of the Eagon-Northcott resolution of I k .Let a = ( a , . . . , a k +1 ) ∈ C k +1 . For ≤ r ≤ k + 1 , we define X a r =[ a , . . . , ˆ a r , . . . , a k +1 | , . . . , k ] . Hence X a r ∈ S k . We define the map φ asfollows. { , , · · · , k } × C k +1 φ −→ R ( nk )( j, a ) α such that α ( i ) = ( ( − r i +1 x ( a ri , j ) if i = σ ( X a r i ) for some r i ;0 otherwise . The map σ is the bijection from S k to { , , · · · , (cid:0) nk (cid:1) } , defined before. Theimage of φ gives a complete list of generators of Syz( I k ) . Example 4.6.
We give an example, by taking k = 3 and n = 5 . Let σ : S −→ { , · · · (cid:0) (cid:1) } be defined by, • [1 , , | , , • [1 , , | , , • [1 , , | , , • [1 , , | , , • [1 , , | , , • [1 , , | , , • [2 , , | , , • [2 , , | , , • [2 , , | , , • [3 , , | , , In our example, φ : { , · · · } × C −→ R ( ) and φ ( j, a ) α . Let j = 2 and a = (1 , , , . Then, X a = [3 , , | , , , X a = [1 , , | , , , X a = [1 , , | , , , X a = [1 , , | , , . Therefore, σ ( X a ) = 10 , σ ( X a ) = 6 , σ ( X a ) = 5 , σ ( X a ) = 4 . Similarly, α (4) = ( − x = − x , α (5) = ( − x = x , α (6) = ( − x = − x , α (10) =( − x = x . Therefore, α = (0 , , , − x , x , − x , , , , x ) . DEALS OF THE FORM I ( XY ) Lemma 4.7.
Let ≤ k ≤ n − and let S k = n X a , . . . , X a ( nk ) o besuch that a < . . . < a ( nk ) with respect to the lexicographic ordering.Suppose that α = ( α , · · · , α ( nk )) ∈ Syz ( I k ) , then P ( nk ) i =1 α i X a i = 0 and P ( nk ) i =1 α i g X a i ∈ e I k +1 .Proof. We have e X a i = P m ≥ k σ − m ( i ) y m . Therefore ( nk ) X i =1 α i e X a i = X i α i ( X m ≥ k σ − m ( i ) y m ) = X m ≥ k ( X i α i σ − m ( i ) y m ) . It is enough to show that P i α i σ − m ( i ) y m ∈ e I k +1 , for every m ≥ k . Wehave α ∈ Syz( I k ) = h Im( φ ) i . Without loss of generality we may as-sume that α ∈ Im( φ ) . There exists ( j, a k +1 ) ∈ { , , · · · k } × C k +1 suchthat φ ( j, a k +1 ) = α . We will show that α i · σ − m ( i ) ∈ I k +1 for every m ≥ k and each i . We have i = σ ( X a k +1 r i ) since α i = 0 . But σ − m ( i ) =[ a , . . . , ˆ a r i , . . . , a k +1 | , . . . , k − , m ] . We have [ a , . . . , a k +1 | j, , . . . , k − , m ] = 0 for j ≤ k − and [ a , . . . , a k +1 | k, , . . . , k − , m ] = ( − k [ a , . . . , a k +1 | , . . . , k, m ] ∈ I k +1 . Therefore, ( nk ) X i =1 α i · σ − m ( i ) = ( nk ) X i =1 ( − r i +1 x ( a ri , j ) [ a , . . . , ˆ a r i , . . . , a k +1 | , . . . , k − , m ]= [ a , . . . , a k +1 | j, , . . . , k − , m ] ∈ I k +1 ; Hence, ( nk ) X i =1 α i g X a i = ( nk ) X i =1 α i · ^ σ − m ( i ) = ( − k ( nk ) X i =1 [ a , . . . , a k +1 | , . . . , k, m ] y m ∈ e I k +1 . (cid:3) Lemma 4.8.
Let X a i , X a j ∈ S k = n X a , . . . , X a ( nk ) o , for i = j . Then,there exist monomials h t in R and a polynomial r ∈ e I k +1 such that(i) S ( X a i , X a j ) = P ( nk ) t =1 h t X a t , upon division by S k ;(ii) S ( e X a i , e X a j ) = P ( nk ) t =1 h t e X a t + r , upon division by e S k .Proof. (i) The expression follows from the observation that S k is a Gr¨obnerbasis for the ideal I k . JOYDIP SAHA, INDRANATH SENGUPTA, AND GAURAB TRIPATHI (ii) We first note that, Lt ( e X a t ) = Lt ( X a t ) y k , for every X a t ∈ S k . Let S ( X a i , X a j ) = cX a i − dX a j , where c = lcm ( Lt ( X a i ) , Lt ( X a j )) X a i and d = lcm ( Lt ( X a i ) , Lt ( X a j )) X a j Hence, S ( e X a i , e X a j ) = c · e X a i − d · e X a i = X m ≥ k [ c · X a i ,m − d · X a j ,m ] y m . It follows immediately that Lt ( S ( e X a i , e X a j )) = y k Lt ( S ( X a i , X a j )) .The set S k is a Gr¨obner basis for the ideal I k . Therefore, we have Lt ( X a t ) | Lt ( S ( X a i , X a j )) , for some t . Then, Lt ( e X a t ) | Lt ( S ( e X a i , e X a j )) and wehave h t = Lt ( S ( X a i , X a j )) Lt ( X a t ) = Lt ( S ( e X a i , e X a j )) Lt ( e X a t ) . We can write r := S ( e X a i , e X a j ) − h t e X a t = X m ≥ k [ c · X a i ,m − d · X a j ,m − h t X a t ,m ] y m = X m>k [ c · X a i ,m − d · X a j ,m − h t X a t ,m ] y m + [ c · X a i − d · X a j − h t X a t ] y k Note that r ∈ e I k and Lt ( r ) = Lt ( S ( e X a i , e X a j ) − h t e X a t ) = y k Lt ( S ( X a i , X a j ) − h t X a t ) . We proceed as before with the polynomial S ( X a i , X a j ) − h t X a t ∈ I k and continue the process to obtain the desired expression involving thepolynomial r .We now show that the polynomial r is in the ideal e I k +1 . Let us write H j = h j + d , H i = h i − c and H t = h t for t = i, j . It follows from S ( X a i , X a j ) = P ( nk ) t =1 h t X a t , that P ( nk ) t =1 H t X a t = 0 . Therefore, H = ( H , . . . , H ( nk )) ∈ Syz( I k ) and by Lemma 4.7 we have P ( nk ) t =1 H t e X a t ∈ e I k +1 . Hence, r = S ( e X a i , e X a j ) − P t = i,j h t e X a t ∈ e I k +1 . (cid:3) Lemma 4.9. (i) Let k ′ > k and a = ( a , . . . , a k ′ ) ∈ C k ′ . Suppose that X a = P b t ∈ C k β b t X b t is the Laplace expansion of X a . Then X b t ∈ C k β b t X b t ,i = [ a , . . . , a k ′ | , . . . , k − , i, k + 1 , . . . , k ′ ] . DEALS OF THE FORM I ( XY ) (ii) Let k ′ > k ; a = ( a , . . . , a k ′ ) ∈ C k ′ , b = ( b , . . . , b k ) ∈ C k .Suppose that X a = P p ∈ C k α p X p and S ( X a , X b ) = cX a − dX b = P p ∈ C k β p X p . Then c X t ≥ k [ a , · · · , a k ′ | , · · · , k − , t, k +1 , · · · , k ′ ] y t − d e X b − X p ∈ C k β p e X p ∈ e I k +1 . Proof. (i) See [12].(ii) We have S ( X a , X b ) = cX a − dX b = P p ∈ C k β p X p . By rearrangingterms we get P p ∈ C k ( cα p − β p ) X p − dX b = 0 and by separating out theterm ( cα b − β b ) X b we get P p = b ( cα p − β p ) X p + ( cα b − β b − d ) X b = 0 .Therefore, P p = b ( cα p − β p ) e X p +( cα b − β b − d ) e X b ∈ e I k +1 , by Lemma 4.7.Hence P t ≥ k P p = b ( cα p − β p ) X p ,t y t + ( cα b − β b − d ) P t ≥ k X b ,t y t ∈ e I k +1 .Now P t ≥ k P p ∈ C k α p X p,t = P t ≥ k [ a , · · · , a k ′ | , · · · , k − , t, k +1 , · · · , k ′ ] by (i). Hence, c X t ≥ k [ a , · · · , a k ′ | , · · · , k − , t, k +1 , · · · , k ′ ] y t − d e X b − X p ∈ C k β p e X p ∈ e I k +1 . (cid:3) Lemma 4.10.
Let k ′ > k ; a = ( a , . . . , a k ′ ) ∈ C k ′ , b = ( b , . . . , b k ) ∈ C k .Suppose that S k = n X a , . . . , X a ( nk ) o , such that a < . . . < a ( nk ) withrespect to the lexicographic ordering. Then, there exist monomials h t ∈ R and a polynomial r ∈ e I k +1 such that(i) S ( X a , X b ) = P ( nk ) t =1 h t X a t , upon division by S k .(ii) S ( e X a , e X b ) = P ( nk ) t =1 ( h t e X a t ) y k ′ + r , upon division by e S k .Proof. (i) The expression follows from the observation that S k is a Gr¨obnerbasis for the ideal I k .(ii) Let S ( X a , X b ) = cX a − dX b , where c = lcm ( Lt ( X a ) , Lt ( X b )) X a and d = lcm ( Lt ( X a ) , Lt ( X b )) X b . Then, S ( e X a , e X b ) = cy k e X a − dy k ′ e X b = cy k X t ≥ k ′ X a ,t y t − dy k ′ X t ≥ k X b ,t y t = y k y k ′ ( cX a − dX b ) + terms devoid of y k . We therefore have Lt ( S ( e X a , e X b )) = y k y k ′ Lt ( S ( X a , X b )) , since y k is thelargest variable appearing in the above expression. The set S k being aGr¨obner basis for the ideal I k , we have Lt ( X a t ) dividing Lt ( S ( X a i , X a j )) for some t . Let h t = Lt ( cX a − dX b ) Lt ( X a t ) , with t = 1 , . . . , (cid:0) nk (cid:1) . Moreover,Lt ( e X a t ) being equal to y k Lt ( X a t ) , it divides Lt ( S ( e X a , e X b )) . Let r := S ( e X a , e X b ) − Lt ( S ( e X a , e X b )) Lt ( e X a t ) e X a t = S ( e X a , e X b ) − y k ′ h t e X a t ∈ e I k . We have r = y k y k ′ ( cX a − dX b ) − y k ′ h t e X a t + terms devoid of y k = y k y k ′ ( cX a − dX b ) − y k ′ h t X i ≥ k X a t ,i y i + terms devoid of y k = y k y k ′ ( cX a − dX b − h t X a t ) + terms devoid of y k = y k y k ′ ( S ( X a , X b ) − h t X a t ) + terms devoid of y k . Hence, Lt ( r ) = Lt ( S ( X a , X b ) − h t X a t ) = y k y k ′ Lt ( S ( X a , X b ) − h t X a t ) .We proceed as before with the polynomial S ( X a , X b ) − h t X a t ∈ I k andcontinue the process to obtain the desired expression involving the polyno-mial r .We now show that the polynomial r is in the ideal e I k +1 . Let us write r = S ( e X a , e X b ) − ( nk ) X t =1 ( h t e X a t ) y k ′ = cy k X l ≥ k ′ X a ,l y l − dy k ′ X l ≥ k X b ,l y l − ( nk ) X t =1 X l ≥ k h t X a t ,l y l y k ′ + T − T ; where T = c P l ≥ k [ a , . . . , a k ′ | , . . . , k − , l, k + 1 , . . . , k ′ ] y l y k ′ . After arearrangement of terms, we may write r = T − ( nk ) X t =1 X l ≥ k h t X a t ,l y l y k ′ − dy k ′ X l ≥ k X b ,l y l + cy k X l ≥ k ′ X a ,l y l − T. Let T ′ = c P l>k [ a , . . . , a k ′ | , . . . , k − , l, k + 1 , . . . , k ′ ] y l y k ′ . Now wenote, cX a − dX b − P ( nk ) t =1 h t X a t = 0 . Hence T − P ( nk ) t =1 P l ≥ k h t X a t ,l y l y k ′ − DEALS OF THE FORM I ( XY ) dy k ′ P l ≥ k X b ,l y l becomes equal to T ′ − ( nk ) X t =1 X l>k h t X a t ,l y l y k ′ − dy k ′ X l>k X b ,l y l . We also have cy k P l ≥ k ′ X a ,l y l − T = cy k P l>k ′ X a ,l y l − T ′ , since theterm for l = k ′ in cy k P l ≥ k ′ X a ,l y l gets cancelled with the term appearingin T for l = k . Hence we write r = T ′ − ( nk ) X t =1 X l>k h t X a t ,l y l y k ′ − dy k ′ X l>k X b ,l y l + cy k X l>k ′ X a ,l y l − T ′ = ( ) + ( ) − T ′ . Clearly, the expression ( ) belongs to e I k +1 , by Lemma 4.9. We note thatno term of ( ) contains y k . So also for T ′ . Hence, the leading term of r isthe leading term of ( ) . By an application of similar argument as above wesee that the expression ( ) , after division by elements of e S k , further reducesto − X l>k ′ X s ≥ k ′ c [ a , . . . , a k ′ | , . . . , k − , s, k + 1 , . . . , k ′ − , l ] y l y s = − X l>k ′ X s>k ′ c [ a , . . . , a k ′ | , . . . , k − , s, k + 1 , . . . , k ′ − , l ] y l y s − X l>k ′ c [ a , . . . , a k ′ | , . . . , k − , k ′ , k + 1 , . . . , k ′ − , l ] y l y k ′ . Moreover, X l>k ′ c [ a , . . . , a k ′ | , . . . , k − , k ′ , k + 1 , . . . , k ′ − , l ] y l y k ′ + T ′ = 0 and X l>k ′ X s>k ′ c [ a , . . . , a k ′ | , . . . , k − , s, k + 1 , . . . , k ′ − , l ] y l y k ′ = 0 . Therefore, after division by elements of e S k , the expression ( ) + ( ) − T ′ reduces to ( ) , which is in e I k +1 . (cid:3) Proof of Theorem 4.1.
We use induction on n − k to prove that G k isa Gr¨obner basis for the ideal e I k . For n − k = 0 ; the set G k = e S n con-tains only one element and hence trivially forms a Gr¨obner basis. We applyBuchberger’s algorithm to prove our claim. Let X a , X b ∈ G k . The follow-ing cases may arise: • X a , X b ∈ S k , for a , b ∈ C k ; • X a ∈ S k ′ and X b ∈ S k where k ′ > k ; a ∈ C k ′ and b ∈ C k .We have proved in Lemmas 4.8 and 4.10 that upon division by e S k , the S -polynomial S ( e X a , e X b ) −→ r for some r ∈ e I k +1 , in both the cases. Byinduction hypothesis, G k +1 is a Gr¨obner basis for e I k +1 . Hence r reduces to modulo G k +1 and hence modulo G k , since G k +1 ⊂ G k .We now show that G k is a reduced Gr¨obner basis for e I k . Let X a ∈ S k ′ and X b ∈ S k where k ′ ≥ k ; a ∈ C k ′ and b ∈ C k . Then, e X a = P i ≥ k ′ X a ,i y i and e X b = P i ≥ k X b ,i y i . If k ′ > k , then y k ′ | Lt ( e X a ) but doesnot divide Lt ( e X b ) . Hence, Lt ( e X a ) does not divide Lt ( e X b ) . If k ′ = k , thenLt ( e X a ) = x ( a , · · · x ( a k ,k ) y k and Lt ( e X b ) = x ( b , · · · x ( b k ,k ) y k . Therefore, e X a | e X b implies that a = b . This proves that the Gr¨obner basis is reduced. (cid:3)
5. G R ¨ OBNER BASIS FOR J Theorem 5.1.
Let us consider the lexicographic monomial order inducedby y > y > · · · > y n > x > x > · · · > x ( n +1) , ( n − > x ( n +1) ,n on b R = K [ x ij , y j | ≤ i ≤ n + 1 , ≤ j ≤ n ] . The set G k is a reducedGr¨obner Basis for the ideal e I k . In particular, G = G is a reduced Gr¨obnerBasis for the ideal e I = J .Proof. The scheme of the proof is the same as that for I , with suitablechanges made for b X in the Lemmas. We only reiterate the last part of theproof where we carry out induction on n − k . For n − k = 0 , the set G k = e S n = { ∆ y n , . . . , ∆ n +1 y n } , where ∆ i = det( b X i ) . We first note thatLt (∆ i ) and Lt (∆ j ) are coprime. Therefore, S (∆ i y n , ∆ j y n ) = Lt (∆ j ) · (∆ i y n ) − Lt (∆ i ) · (∆ j y n )= Lt (∆ j )( Lt (∆ i ) y n + y n p i ) − Lt (∆ i )( Lt (∆ j ) y n − y n p j )= ( Lt (∆ j ) y n ) p i − ( Lt (∆ i ) y n ) p j = (∆ j y n − p j y n ) p i − (∆ i y n − p i y n ) p j = ∆ j y n p i − ∆ i y n p j −→ G n . DEALS OF THE FORM I ( XY ) The rest of the proof is essentially the same as that for Theorem 4.1. (cid:3)
6. B
ETTI N UMBERS OF I AND J Theorem 6.1.
Suppose that X = ( x ij ) n × n is either a generic or a genericsymmetric n × n matrix and Y a generic n × matrix given by Y =( y j ) n × . If X is generic, we write g i = P nj =1 x ij y j and I = I ( XY ) = h g , g , · · · , g n i . If X is generic symmetric, we write g = P nj =1 x j y j , g n = ( P ≤ k ≤ n x kn y k ) and g i = ( P ≤ k x > · · · > x nn ;(2) x ij , y j < x nn for every ≤ i = j ≤ n .Proof. The monomial order chosen is lexicographic order induced by theordering among the variables given by (1) and (2). It is clear from theexpressions of g i that their leading terms are pairwise coprime. Therefore,the proof follows from Lemma 4.3. (cid:3) Corollary 6.2. I is minimally resolved by the Koszul complex G and the i -th Betti number of I is (cid:0) ni (cid:1) . Theorem 6.3.
Suppose that b X = ( x ij ) ( n +1) × n is a generic ( n +1) × n matrixand Y a generic n × matrix given by Y = ( y j ) n × . Let g i = P n +1 j =1 x ij y j and J = I ( b XY ) = h g , · · · , g n +1 i . The total Betti numbers of the ideal J are β = 1 , β = n + 1 , β n +1 = n , β k +1 = (cid:0) nk (cid:1) + (cid:0) nk − (cid:1) + (cid:0) nk +1 (cid:1) for ≤ k < n . We first discuss the scheme of the proof below. We will use the followingobservations to compute the total Betti numbers of J .Step 1. The minimal graded free resolution of I = h g , · · · , g n i is given bythe Koszul Resolution.Step 2. We prove that h g , · · · , g n : g n +1 i = h g , · · · , g n , ∆ i ; where ∆ =det( X ) . This proof requires the fact that h g , · · · , g n , ∆ i is a primeideal, which has been proved in Theorem 5.4 in [15].Step 3. We prove that h g , · · · g n : ∆ i = h y , y , · · · , y n i . Step 4. We construct a graded free resolution of h g , · · · , g n , ∆ i using map-ping cone between resolutions of h g , · · · , g n i and h y , · · · , y n i . Weextract a minimal free resolution from this resolution.Step 5. Finally, we construct a graded free resolution of h g , · · · , g n , g n +1 i using mapping cone between free resolutions of h g , · · · , g n , ∆ i and h g , · · · , g n i . We extract a minimal free resolution from this resolu-tion. Remark . We need detailed information about the ideal h g , · · · , g n , ∆ i ,where ∆ = det( X ) . We need the fact that this ideal is a prime ideal, whichhas been proved in Theorem 5.4 in [15]. We also need a minimal free resolu-tion for this ideal, which has been proved below in Lemma 6.10. We cameto know much later that h g , · · · , g n , ∆ i was defined in [14]. It is knownas the generic Northcott ideal and a minimal free resolution can be foundin [14]. However, we give a different proof here using our Gr¨obner basiscomputation, which also shows the linking of nested complete intersectionideals. Moreover, Northcott’s resolution can perhaps be used to prove that h g , · · · , g n , ∆ i is a prime ideal, although our proof in [15] is absolutelydifferent and uses the result in [7]. Lemma 6.5. ∆ y i = P nj =1 A ji g j , where A ji is the cofactor of x ji in X .Proof. We have ∆ y i = n X j =1 A ji x ji y i = n X j =1 A ji n X k =1 x jk y k ! − n X j =1 A ji X k = i x jk y k ! = n X j =1 A ji g j , since P nj =1 A ji (cid:16)P k = i x jk y k (cid:17) = P k = i (cid:16)P nj =1 A ji x jk (cid:17) y k = 0 . (cid:3) Lemma 6.6. h g , · · · , g n , ∆ i ⊆ h g , · · · , g n : g n +1 i .Proof. We have g i ∈ h g , · · · , g n : g n +1 i , for every ≤ i ≤ n . Moreover, y i ∆ ∈ h g , · · · , g n i , by Lemma 6.5. Hence, g n +1 ∆ ∈ h g , · · · , g n i . (cid:3) Lemma 6.7. h g , · · · , g n : g n +1 i = h g , · · · , g n , ∆ i Proof.
We have proved that h g , · · · , g n , ∆ i ⊆ h g , · · · , g n : g n +1 i in Lemma6.6. We now prove that h g , · · · , g n : g n +1 i ⊆ h g , · · · , g n , ∆ i . Let z ∈h g , · · · , g n : g n +1 i . Then zg n +1 ∈ h g , · · · , g n i ⊂ h g , · · · , g n , ∆ i . It iseasy to see that g n +1 / ∈ h g , · · · , g n , ∆ i . Therefore, z ∈ h g , · · · , g n , ∆ i ,since h g , · · · , g n , ∆ i is a prime ideal by Theorem 5.4 in [15]. (cid:3) Lemma 6.8. h g , · · · , g n : ∆ i = h y , · · · , y n i DEALS OF THE FORM I ( XY ) Proof.
We have y i ∆ ∈ h g , · · · , g n i by Lemma 6.5; which implies that h y , · · · , y n i ⊂ h g , · · · , g n : ∆ i . Let z ∈ h g , · · · , g n : ∆ i . Then z ∆ ∈ h g , · · · , g n i ⊆ h y , · · · , y n i . Therefore, z ∈ h y , · · · , y n i , since ∆ / ∈ h y , · · · , y n i and h y , · · · , y n i is a prime ideal. (cid:3) Mapping Cones.
The resolution for h y , · · · , y n i is given by the Koszulcomplex F (cid:5) . We now give a resolution of h g , · · · , g n , ∆ i by the mappingcone technique. We know that h g , · · · , g n : ∆ i = h y , · · · , y n i , by Lemma6.8. We first construct a connecting homomorphism φ (cid:5) : F (cid:5) −→ G (cid:5) . Let φ denote the multiplication by ∆ . In order to make the map φ a degree zeromap, we set the grading as F ∼ = ( R ( − n )) and G = ( R (0)) . Since F (cid:5) and G (cid:5) are both Koszul resolutions, we set the grading as G i ∼ = ( R ( − i ))( ni ) and F i ∼ = ( R ( − n − i ))( ni ) . Now we see that, i = n implies that − i = − n − i .Hence the image of φ i for i = n is contained in the maximal ideal. Wehave F i = G i , only for i = n . If we can show that the map φ n is not thezero map, then this will be the only free part of the resolution which we cancancel out for obtaining the minimal resolution. Lemma 6.9.
The map φ n is not the zero map.Proof. We refer to [8]. If φ n is the zero map, then φ ( R ) ⊆ δ ( G ) , where δ . denotes the differential of G . The image of δ is the ideal h g , · · · , g n i ,which does not contain φ (1) = ∆ . The map φ n is not the zero map. (cid:3) Therefore, the above discussion proves the following Lemma.
Lemma 6.10.
Hence a minimal graded free resolution of h g , · · · , g n , ∆ i is given by M (cid:5) , such that M i ∼ = ( R ( − n − i + 1))( ni − ) ⊕ ( R ( − i ))( ni ) for < i < n , M ∼ = R (0) and M n ∼ = ( R ( − n )) n . (Proof of Theorem 6.3.) We now find the Betti numbers for the ideal h g , · · · , g n +1 i by constructing the mapping cone between the resolutions M (cid:5) and the resolution G (cid:5) of h g , · · · , g n i . The connecting map ψ is mul-tiplication by g n +1 . Hence to make it degree zero we set, G = ( R (2)) and G i ∼ = ( R (2 − i ))( ni ) for i > . Here we note that − i = − i and − n − i + 1 = 2 − i for ≤ i ≤ n . Hence, for each ≤ i ≤ n , the imageof ψ i is contained in the maximal ideal. This shows that the resolution ob-tained by the mapping cone between M (cid:5) and G (cid:5) is minimal. Hence the totalBetti numbers of J are: β = 1 , β = n + 1 ; β n +1 = n ; β k +1 = (cid:0) nk (cid:1) + (cid:0) nk − (cid:1) + (cid:0) nk +1 (cid:1) for ≤ k < n . (cid:3) Corollary 6.11.
The ring R/ I is Cohen-Macaulay and the ring ˆ R/ J is notCohen-Macaulay.Proof. The polynomial ring R is Cohen-Macaulay and g , . . . , g n is a regu-lar sequence therefore the ring R/ I is Cohen-Macaulay.We have seen that projdim b R b R/ J = n + 1 . Therefore, by the Auslander-Bauchsbaum formula depth b R b R/ J = n ( n + 1) + n − ( n + 1) = n + n − .We have proved in Lemma 5.5 in [15] that h y , . . . , y n i is a minimal primeover J . Therefore, dim b R/ J ≥ dim b R/ h y , . . . , y n i = n + n ; hence thering b R/ J is not Cohen-Macaulay. (cid:3) I ( XY ) , WHERE X IS m × mn GENERIC MATRIX AND Y IS mn × n GENERIC MATRIX
Finally, we consider the case when X = ( x ij ) m × mn is a generic matrixof size m × mn and Y = ( y ij ) mn × n is generic matrix of size mn × n . Wedefine I = I ( XY ) . Let g ij = P mnt =1 x it y tj , with ≤ i ≤ m, ≤ i ≤ n .Then, I = h{ g ij | ≤ i ≤ m, ≤ i ≤ n }i . In this section we construct aGr¨obner basis for the ideal I with respect to a suitable monomial order anduse that to show that the generators g ij , with ≤ i ≤ m , ≤ i ≤ n forma regular sequence. We first set a few notations before we prove the mainresults. • X = (cid:0) A · · · A n (cid:1) , where A s = x m ( s − · · · x ms ) ... ... ... x m ( m ( s − · · · x m ( ms ) is the m × m matrix for every ≤ s ≤ n . • [ X ] s = (cid:16) A s A · · · c A s · · · A n (cid:17) , for every ≤ s ≤ n . • [ Y ] s = y ( m ( s − s ... y ( ms ) s y s ... y ( mn ) s , for every ≤ s ≤ n .We will use Theorem 4.1 for constructing a Gr¨obner basis for the ideal I . A very important reason behind considering this class of ideals is thatwe get some nice examples of transversal intersection of ideals. Two resultsthat would be useful for our purpose are the following: DEALS OF THE FORM I ( XY ) Lemma 7.1.
Let > be a monomial ordering on R . Let I and J be idealsin R , such that m ( I ) and m ( J ) denote unique minimal generating sets fortheir leading ideals Lt ( I ) and Lt ( J ) respectively. Then, I ∩ J = IJ if theset of variables occurring in the set m ( I ) is disjointed from the the set ofvariables occurring in the set m ( J ) .Proof. See Lemma 3.6 in [16]. (cid:3)
Lemma 7.2.
Let I and J be graded ideals in a graded ring R , such that I ∩ J = I · J . Suppose that F (cid:5) and G (cid:5) are minimal free resolutions of I and J respectively. Then F (cid:5) ⊗ G (cid:5) is a minimal free resolution for the gradedideal I + J .Proof. See Lemma 3.7 in [16]. (cid:3)
Theorem 7.3.
Let us choose the lexicographic monomial order on R in-duced by y > y > · · · > y ( mn )1 > y ( m +1)2 > y ( m +2)2 > · · · > y (2 m )2 >y > · · · y ( mn )2 > · · · > y ( m ( n − n > y ( m ( n − n > · · · > y (( mn ) n >y n > · · · y ( m ( n − n > x > x > · · · > x m ( mn ) . Let G s be the re-duced Gr¨obner Basis of the ideal I ([ X ] s [ Y ] s ) for ≤ s ≤ n , obtained byTheorem 4.1. Then G t = ∪ ts =1 G s is a reduced Gr¨obner Basis for the ideal P t = P ts =1 I ([ X ] s [ Y ] s ) for ≤ t ≤ n . In particular, G n is a reducedGr¨obner Basis for the ideal P n = I = I ( XY ) .Proof. We have P t = P ts =1 I ([ X ] s [ Y ] s ) , and we observe that if p ∈ G s and q ∈ G t for ≤ s < t ≤ n , then gcd( Lt ( p ) , Lt ( q )) = 1 . Therefore the S -polynomial of p, q reduces to zero after applying division upon G t . (cid:3) Theorem 7.4.
Let us denote P t = P ts =1 I ([ X ] s [ Y ] s ) , for ≤ t ≤ n − .Then P t ∩ I ([ X ] t +1 [ Y ] t +1 ) = P t · I ([ X ] t +1 [ Y ] t +1 ) . Hence the elements g ij = P mnt =1 x it y tj , ≤ i ≤ m , ≤ i ≤ n form a regular sequence and theKoszul complex resolves R/ I as an R -module minimally.Proof. If p ∈ G s and q ∈ G t , for ≤ s < t ≤ n . Then gcd( Lt ( p ) , Lt ( q )) =1 , therefore by theorem 7.3 and lemma 7.1, we have P t ∩ I ([ X ] t +1 [ Y ] t +1 ) = P t · I ([ X ] t +1 [ Y ] t +1 ) .By Theorem 6.1 the generators of the ideal P form a regular sequenceand also the generators of the ideal I ([ X ] s [ Y ] s ) form a regular sequence foreach ≤ s ≤ n . Hence the Koszul complex resolve R/P and R/I ([ X ] s [ Y ] s ) minimally. Now P t ∩ I ([ X ] t +1 [ Y ] t +1 ) = P t · I ([ X ] t +1 [ Y ] t +1 ) . Hence, byapplication of lemma 7.1 we can conclude that the Koszul complex resolves R/ I minimaly. (cid:3) A CKNOWLEDGEMENTS
The second author is the corresponding author who has been supportedby the research project EMR/2015/000776, sponsored by the SERB, Gov-ernment of India. The third author thanks SERB for the post-doctoral fel-lowship under the said project. The third author thanks CSIR for the SeniorResearch Fellowship for Ph.D. The authors thank the anonymous refereesfor their valuable comments and for drawing their attention to the references[5] and [18], extremely pertinent to this work.R
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Discipline of Mathematics, IIT Gandhinagar, Palaj, Gandhinagar, Gujarat 382355,INDIA.
E-mail address : [email protected] Discipline of Mathematics, IIT Gandhinagar, Palaj, Gandhinagar, Gujarat 382355,INDIA.
E-mail address : [email protected] Department of Mathematics, Jadavpur University, Kolkata, WB 700 032, IN-DIA.
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