Betti Numbers of Certain Sum Ideals

arXiv:1611.04732v1 [math.AC] 15 Nov 2016

BETTI NUMBERS OF CERTAIN SUM IDEALS JOYDIP SAHA, INDRANATH SENGUPTA, AND GAURAB TRIPATHI A BSTRACT. In this paper we compute the Betti numbers for ideals of the form I1 (XY ) + J, where X and Y are matrices and J is the ideal generated by the 2 × 2 minors of the matrix consisting of any two rows of X.

1. I NTRODUCTION Let K be a field and {xij ; 1 ≤ i, j ≤ n}, {yj ; 1 ≤ j ≤ n} be indeterminates over K; n ≥ 2. Let R := K[xij , yj ] denote the polynomial algebra over K. Let X denote an n × n matrix such that its entries belong to the ideal h{xij ; 1 ≤ i ≤ n, 1 ≤ j ≤ n}i. Let Y = (yj )n×1 be the n × 1 column Pnmatrix. Let I1 (XY ) denote the ideal generated by the polynomials gj = i=1 xji yi , j = 1, . . . , n, which are the 1 × 1 minors or entries of the n × n matrix XY . The primality, primary decomposition and Betti numbers of ideals of the form I1 (XY ) have been studied in [9] and [10], with the help of Gröbner bases for I1 (XY ). Ideals of the form I1 (XY ) + J are particularly interesting because they occur in several geometric considerations like linkage and generic residual intersection of polynomial ideals, especially in the context of syzygies. Bruns-Kustin-Miller [1] resolved the ideal I1 (XY )+Imin(m,n) (X), where X is a generic m×n matrix and Y is a generic n×1 matrix. Johnson-McLoud [5] proved certain properties for the ideals of the form I1 (XY ) + I2 (X), where X is a generic symmetric matrix and Y is either generic or generic alternating. We say that I and J intersect transversally if I ∩J = IJ. Suppose that F· resolves R/I and G· resolves R/J minimally. It is interesting to note that if I and J intersect transversally, then the tensor product complex F ⊗R G 2010 Mathematics Subject Classification. Primary 13D02; Secondary 13C40, 13P10, 13D07. Key words and phrases. Gröbner basis, Betti numbers, transversal intersection, mapping cone. The first author thanks UGC for the Senior Research Fellowship. The second author is the corresponding author, who is supported by the the research project EMR/2015/000776 sponsored by the SERB, Government of India. The third author thanks CSIR for the Senior Research Fellowship. 1

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JOYDIP SAHA, INDRANATH SENGUPTA, AND GAURAB TRIPATHI

resolves R/I + J minimally; see Lemma 3.7. Therefore, it is useful to know if two ideals intersect transversally, especially when one is trying to compute minimal free resolutions and Betti numbers for ideals of the form I + J, through iterated techniques; see [4]. 2. N OTATION

M AIN T HEOREM xi1 xi2 · · · xin e • If X is generic and i < j; let Xij = . xj1 xj2 · · · xjn • If X is generic symmetric and i < j; let x1i · · · xii · · · xij · · · xin e Xij = . x1j · · · xij · · · xjj · · · xjn AND THE

eij . • Let Gij denote the set of all 2 × 2 minors of X eij ) denote the ideal generated Gij . • I2 ( X

Our aim in this paper is to prove the following theorem: Theorem 2.1. Let X = (xij ) be either the generic or the generic symmetric matrix of order n. Let 1 ≤ i < j ≤ n. eij ) + hgi , gj i are given by (1) The total Betti numbers for theideal I2 (X n n n b0 = 1, b1 = 2 + 2, b2 = 2 3 + n, bi+1 = i i+1 + (i − 2) ni , for 2 ≤ i ≤ n − 2 and bn = n − 2. (2) Let 1 ≤ k ≤ n−2. Let βk,p denote the p-th total Betti number for the eij )+hgi , gj , gl1 , . . . , gl i, such that 1 ≤ l1 < . . . < lk ≤ n ideal I2 (X k and lt is the smallest in the set {1, 2, . . . , n} \ {i, j, l1 , . . . , lt−1 }, for every 1 ≤ t ≤ k. They are given by βk,0 = 1, βk,p = βk−1,p−1 + βk−1,p for 1 ≤ p ≤ n + k − 1 and βk,n+k = n − 2. eij ) are In particular, the total Betti numbers for the ideal I1 (XY ) + I2 (X βn−2,0 , βn−2,1 , . . . , βn−2,2n−2 . 3. P RELIMINARIES 3.1. Determinantal Ideals. We recall some useful results on determinantal ideals pertaining to our work. We refer to [2], [3], [8] for detailed discussions on these. Theorem 3.1. Let K be a field and let xij : 1 ≤ i ≤ m, 1 ≤ j ≤ n be indeterminates over K. Let A = (xij ) be the m×n matrix of indeterminates and Im (A) denotes the ideal generated by the maximal minors of A. The set of maximal minors of A is a universal Gröbner basis for the ideal Im (A).

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3

Proof. See [2].

The Eagon-Northcott Complex. We present the relevant portion from the book [3] here. Let F = Rf and G = Rg be free modules of finite rank over the polynomial ring R. The Eagon-Northcott complex of a map α : F −→ G (or that of a matrix A representing α) is a complex df −g

df −g+1

EN(α) : 0 → (Symf −g G)∗ ⊗ ∧f F −→ (Symf −g−1 G)∗ ⊗ ∧f −1 F −→ d

d2 g

∧g α

3 · · · −→ (Sym2G)∗ ⊗ ∧g+2 F −→ G∗ ⊗ ∧g+1 F ∧ F −→ ∧g G.

Here Symk G is the k-th symmetric power of G and M ∗ = HomR (M, R). The map dj are defined as follows. First we define a diagonal map (Symk G)∗ → G∗ ⊗ (Symk−1G)∗ X ′ ′′ ui ⊗ ui u 7→ i

as the dual of the multiplication map G ⊗ Symk−1G −→ Symk G in the symmetric algebra of G. Next we define an analogous diagonal map ∧k F −→ F ⊗ ∧k−1 F X ′ ′′ v 7→ vi ⊗ vi i

as the dual of the multiplication in the exterior algebra of F ∗ . Theorem 3.2 (Eagon-Northcott). The Eagon-Northcott complex is a free resolution of R/Ig (α) iff grade(Ig (α)) = f − g + 1 where Ig (α) denotes the g × g minors of the matrix A representing α.

Proof. See [3].

3.2. Mapping Cone. We present the relevant portion from the book [8] ′ ′ here. Let R be the polynomial ring. Let φ· : (U· , d· ) → (U· , d· ) be a map of complexes of finitely generated R-modules. The mapping cone of φ· is the ′ complex W· with differential δ· defined as follows. Let Wi = Ui−1 ⊕ Ui , ′ ′ ′ ′ with δ|Ui−1 = −d + φ : Ui−1 −→ Ui−2 ⊕ Ui−1 and δ|U ′ = d : Ui → Ui−1 i for each i. Theorem 3.3. Let M be an ideal minimally generated by the polynomials f1 , . . . , fr . Set Mi = hf1 , . . . , fi i, for 1 ≤ i ≤ r. Thus, M = Mr . For each i ≥ 1, we have the short exact sequence fi+1

0 −→ S/(Mi : fi+1 ) −→ S/Mi −→ S/Mi+1 −→ 0. If resolutions of S/Mi and S/(Mi : fi+1 ) are known then we can construct a resolution of S/Mi+1 by the mapping cone construction.

4


Proof. See Construction 27.3 in [8].

3.3. Gröbner basis and transversal intersection of Ideals. Lemma 3.4. Let h1 , h2 · · · , hn ∈ R be such that with respect to a suitable monomial order on R, the leading terms of them are mutually coprime. Then, h1 , h2 · · · , hn is a regular sequence in R. Proof. . See Lemma 4.3 in [9].

Lemma 3.5. Suppose that X is either generic or generic symmetric. The eij ), with respect a suitable set Gij is a Gröbner basis for the ideal I2 (X monomial order. Proof. We choose the lexicographic monomial order given by the following ′ ′ ordering among the variables: xst > xs′ t′ if (s , t ) > (s, t) and yn > yn−1 > · · · , y1 > xst for all s, t. We now apply Lemma 4.2 in [9] for the matrix X t and for k = 2. Definition 1. Let T ⊂ R be a set of monomials. We define supp(T ) = {(i, j, 0) | xij divides m for some m ∈ T } ∪ {(0, 0, k) | yk divides m for some m ∈ T }. If T = {m}, then we write supp(m) instead of supp({m}).

Lemma 3.6. Let > be a monomial ordering on R. Let I and J be ideals in R, such that m(I) and m(J) denote unique minimal generating sets for their leading ideals Lt(I) and Lt(J) respectively. Then, I ∩ J = IJ if supp(m(I))∩supp(m(J)) = ∅. In other words, the ideals I and J intersect transversally if the set of variables occurring in the set m(I) is disjointed from the the set of variables occurring in the set m(J). Proof. Let f ∈ I ∩ J; we show that f ∈ IJ. Now f ∈ I ∩ J implies that f ∈ I and therefore Lt(f ) ∈ Lt(I). Hence, mi ∈ m(I) such that mi | Lt(f ). Similarly, there exists monomial mj ∈ m(J) such that mj | Lt(f ). Given that mi and mj are of disjoint support, we have mi mj | Lt(f ) and this (f ) proves that Lt(f ) ∈ Lt(IJ). We replace f by f − Lt . Now the proof mi mj Lt(f ) < Lt(f ). follows by induction since, Lt f − m i mj

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3.4. Homological Lemmas. Lemma 3.7. Let I and J be graded ideals in a graded ring R, such that I ∩ J = I · J. Suppose that F and G are minimal free resolutions of I and J respectively. Then F ⊗ G is a minimal free resolution for the graded ideal I + J. Proof. Consider the short exact sequence 0 −→ I −→ R −→ R/I −→ 0 and tensor it with R/J over R. We get the exact sequence 0 −→ Tor1 (R/I, R/J) −→ I/I · J −→ R/J −→ R/I + J −→ 0. The terms on the left are 0 since R is a flat R module. Moreover, the kernel of the map from I/I · J −→ R/J is I ∩ J/I · J. Therefore Tor1 (R/I, R/J) = 0 if and only if I ∩J = I ·J. By the corollary 1 of theorem 3 proved in [6], Tor1 (R/I, R/J) = 0 implies that Tori (R/I, R/J) = 0 for all i ≥ 1. Therefore, Hi (F ⊗ G ) ≃ Tori (R/I, R/J) ≃ 0 for all i ≥ 1 and H0 (F ⊗ G ) ≃ R/I + J. This proves that F ⊗ G resolves I + J. The resolution is minimal since F and G are minimal. Lemma 3.8. Let

A

A

1 2 Ra1 −→ Ra2 −→ Ra3 be an exact sequence of free modules. Let Q1 , Q2 , Q3 be invertible matrices of sizes a1 , a2 , a3 respectively. Then,

Ra1

Q−1 2 A1 Q 1

Ra2

−→

Q−1 3 A2 Q 2

−→

Ra3

is also an exact sequence of free modules. Proof. The following diagram is a commutative diagram is free modules and the vertical maps are isomorphisms: A1

ROa1

/

RaO 2

Q1 Q−1 2 A1 Q 1

Ra1 Therefore, Ra1 Ra3 is exact.

Q−1 2 A1 Q 1

−→

Ra2

/

R

Q−1 3 A2 Q 2

−→

A2

/

RaO 3

Q2 Q3 Q−1 A2 Q 2 3 a2 / a3

R

A

A

1 2 Ra3 is exact since Ra1 −→ Ra2 −→

Corollary 3.9. Let C

B

A

Ra1 −→ Ra2 −→ Ra3 −→ Ra4 be an exact sequence of free modules. Let P1 , P2 , P3 be invertible matrices of sizes a1 , a2 , a3 respectively. Then, P −1 CP1

BP

AP −1

3 Ra1 2−→ Ra2 −→2 Ra3 −→ Ra4

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is also an exact sequence of free modules. C

B

Proof. Consider the sequence Ra1 −→ Ra2 −→ Ra3 . If we take Q1 = P1 , Q2 = P2 and Q3 = I and apply Lemma ??, we get that the sequence Ra1

P2−1 CP1

−→

BP

Ra2 −→2 Ra3 is exact. We further note that the entire se-

P −1 CP1

BP

A

An+1

A

quence Ra1 2−→ Ra2 −→2 Ra3 −→ Ra4 is exact as well, since Im(B) = BP Im(BP2 ) and P2 is invertible. Let us now consider the sequence Ra2 −→2 A Ra3 −→ Ra4 . We take Q1 = Q3 = I, Q2 = P3−1 and apply Lemma 2.3 to arrive at our conclusion. Lemma 3.10. Let An−1

n · · · −→ Rβn+1 −→ Rβn −→ Rβn−1 −→ Rβn−2 −→ · · ·

be an exact sequence of free R modules. Let aij denote the (i, j)-th entry of An . Suppose that alm = ±1 for some l and m, ali = 0 for i 6= m and ′ ajm = 0 for j 6= l. Let An+1 be the matrix obtained by deleting the m-th ′ row from An+1 , An−1 the matrix obtained by deleting the l-th column from ′ An−1 and An the matrix obtained by deleting the l-th row and m-th column from An . Then, the sequence ′

· · · −→ R

βn+1

An+1

−→ R

′

′

βn −1 An

−→ R

βn−1 −1

An−1

−→ Rβn−2 −→ · · ·

is exact. Proof. The fact that the latter sequence is a complex is self evident. We need to prove its exactness. By the previous lemma we may assume that l = m = 1, for we choose elementary matrices to permute rows and columns and these matrices are always invertible. Now, due to exactness of the first complex we have An−1 An = 0. This implies that the first col′ umn of An−1 = 0, which implies that Im(An−1 ) = Im(An−1 ). Therefore, the right exactness of An+1 is preserved. By a similar argument we can ′ prove that the left exactness of An+1 is preserved. ′

from R. If (x) ∈ ker(A Let (x) denote a tuple with entries n ), then (0, x) ∈ ker(An ). There exists y ∈ Rβn+1 such that An−1 y = (0, x). ′ ′ It follows that An−1 y = (x), proving the left exactness of An . By a ′ similar argument we can prove the right exactness of An . Lemma 3.11. Let A be q × p matrix over R with Aij = ±1, for some i and j. Let C be a p × s matrix and B a r × q matrix over R. There exist an invertible q × q matrix X and an invertible p × p matrix Y , such that

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7

(i) (XAY )kj = δki and (XAY )ik = δjk , that is   ··· 0 ···  ··· 0 ···    ..  .. .  . .  .  .   XAY = 0 · · · 0 1 0 · · · 0 ; 1 at the (i, j)−th spot.  . ..  ..  .. .  .    ··· 0 ···  ··· 0 ··· P (ii) (Y −1 C)kl = Ckl for k 6= j and (Y −1 C)jl = Cjl + t6=i (ait )Ctl P (iii) (BX −1 )kl = Bkl for l 6= i and (BX −1 )ki = Bki + t6=i (atj )Bkt .

Proof. (i) We prove for aij = 1. The other case is similar. We take Y = Πk6=j Ejk (−aik ) and X = Πk6=i Eki (−akj ), where Ekl (α) denotes the matrix E with Ekl = α, Ett = 1 and Eut = 0 for u 6= t and (u, t) 6= (k, l). (ii) and (iii) are easy to verify.

Lemma 3.12. Let A be q × p matrix, C be a p × s matrix and B a r × q matrix over R. The matrices A, B and C satisfy property Pij if they satisfy the following conditions: • Aij = 1 , Aik ∈ m for k 6= j and Akj ∈ m for k 6= i; • Bki ∈ m, for 1 ≤ k ≤ r; • Cjl ∈ m, for 1 ≤ l ≤ s. The matrices XAY , BX −1 and Y −1 C satisfy property Pij , if A, B, C satisfy property Pij . Proof. This follows from the above lemma since aik and akj belong to m. 4. M INIMAL

eij ) + hgi , gj i FREE RESOLUTION OF I2 (X

Lemma 4.1. Let X be generic or generic symmetric. Let i < j. eij )) = n − 1. (i) ht(I2 (X fij ). (ii) The Eagon-Northcott complex minimally resolves the ideal I2 (X

Proof. (i) We show that f1 , . . . , fn−1 , given by fk = xik xj,k+1 − xjk xi,k+1, 1 ≤ k ≤ n − 1 form a regular sequence. Let us first assume that X is generic. We take the lexicographic monomial order induced by the following ordering among the variables: xi1 > e and xi2 > · · · > xin > xj2 > xj3 > · · · > xjn > xj1 not appearing in X the variables yk are smaller than xj1 . Then, Lt(fk ) = xik xj,k+1 and hence

8


gcd(Lt(fk ), Lt(fl )) = 1 for every k 6= l. Therefore, f1 , . . . , fn−1 is a regular e ≥ n − 1. On the other hand, sequence by Lemma 3.4 and hence ht(I2 (X)) e ≤ n − 1, by Theorem [13.10] in [7]. Hence, ht(I2 (X)) e ≤ n − 1. ht(I2 (X)) If X is generic symmetric, then we have to choose the lexicographic monomial order induced by xii > xij > x1i > x2i > · · · > xi−1,i > xi,i+1 > · · · > xin > xjj > · · · > xc ij > · · · > xj−1,j > xj,j+1 > · · · > xjn fij and the variables yp are smaller than and variables xkl not appearing in X xjn . fij ) is n − 1, which is the maximum. Hence, the (ii) The height of I2 (X fij ). Eagon-Northcott complex minimally resolves the ideal I2 (X Lemma 4.2. Let X be generic or generic symmetric. Let i < j. Then, eij ) ∩ hgi i = I2 (X eij ) · hgi i, that is, the ideals I2 (X eij ) and hgi i intersect I2 ( X transversally.

Proof. Let X be generic. We choose the lexicographic monomial order ′ ′ given by the following ordering among the variables: xst > xs′ t′ if (s , t ) > (s, t) and yn > yn−1 > · · · > y1 > xst for all s, t. Then, by Lemma 3.5 eij ). the set of all 2 × 2 minors forms a Gröbner basis for the ideal I2 (X eij ))) doesn’t involve the inClearly, the minimal generating set m(Lt(I2 (X determinates xin and yn , whereas Lt(gi ) = xin yn . Hence, the supports of eij ))) and m(Lt(gi )) are disjoint. Therefore, by Lemma 3.6 we m(Lt(I2 (X are done. Let X be generic symmetric. Once we choose the correct monomial order, the rest of the proof is similar to the generic case. Suppose that (i, j) = (n − 1, n). We choose the lexicographic monomial order given by the following ordering among the variables: y1 > yn > yn−1 > · · · > y2 > > > >

xn−1,n−1 > xn−1,n > x1,n−1 > x2,n−1 > · · · > xn−2,n−1 xnn x1n > · · · > xn−2,n xst for all other s, t.

Suppose that (i, j) 6= (n − 1, n). We choose the lexicographic monomial order given by the following ordering among the variables: yn > yn−1 > · · · > y1 > xii > xij > x1i > x2i > · · · > xi−1,i > xi,i+1 > · · · > xin > xjj > · · · > xj−1,j > xj,j+1 > · · · > xjn > xst for all other s, t.

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9

eij ) + hgi i : gj ) = Lemma 4.3. Let X be generic and i < j. Then, (I2 (X eij ) + hgi i : hxi1 , . . . , xin i. If X is generic symmetric and i < j, then (I2 (X gj ) = hx1i , . . . , xi−1,i , xii , . . . , xin i. P Proof. X be generic. We have xit gj = xjt gi + nk=1 (xit xjk − xik xjt )yk . fij ) + hgi i : gj ). Moreover, I2 (X eij ) + hgi i ⊆ Hence, hxi1 , · · · , xin i ⊆ (I2 (X hxi1 , · · · , xin i and gj ∈ / hxi1 , · · · , xin i. The ideal hxi1 , · · · , xin i being a eij ) + hgi i : gj ). The prime ideal, it follows that hxi1 , · · · , xin i ⊇ (I2 (X proof for the generic symmetric case is similar.

5. R ESOLUTION

OF THE SUM IDEALS

eij ) + Our aim is to construct a minimal free resolution for the ideal I2 (X eij ) and hgi i intersect hg1, . . . , gn i. We have proved that the ideals I2 (X eij ) + hgi i can therefore be resolved transversally; see 4.2. The ideal I2 (X eij ) + minimally by Theorem 3.7. We have also proved that the ideal I2 (X hgi i and the ideal hgj i have linear quotient; see 4.3. Therefore, the ideal eij ) + hgi , gj i can be resolved by the mapping cone construction. A I2 ( X minimal free resolution can then be extracted from this resolution by apeij ) + hgi , gj i plying Lemma 3.12. Next, we will show that the ideal I2 (X intersects transversally with the ideal hgl1 i, if l1 is the minimum in the eij ) + set {1, 2, . . . , n} \ {i, j}; see Lemma 5.4. Therefore, the ideal I2 (X hgi , gj , gl1 i can be resolved minimally by Theorem 3.7. Proceeding in this eij )+hgi , gj , gl1 , . . . , gl i manner, we will be able to show that the ideals I2 (X k and hglk+1 i intersect transversally, if 1 ≤ l1 < . . . < lk < lk+1 ≤ n and lk+1 is the smallest in the set {1, 2, . . . , n} \ {i, j, l1 , . . . , lk }; see Lemma eij ) + 5.4. This finally gives us a minimal free resolution for the ideal I2 (X hgi , gj , gl1 , . . . , gln−2 i, with 1 ≤ l1 < . . . < ln−2 ≤ n and lt ∈ / {i, j} for every t. Let us assume that X is generic and i = 1 and j = 2. The proofs for the general i and j, with i < j would be similar according to the aforesaid scheme. The proofs in the case when X is generic symmetric would be similar as well. Comments for general i < j and the symmetric case have been made whenever necessary.

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e12 ) + hg1 , g2 i. The minimal free 5.1. A minimal free resolution for I2 (X e12 ) is given by the Eagon-Northcott complex, which is resolution of I2 (X the following: δk e −→ 0 Ek−1 −→ · · · E1 −→ E0 −→ R/I2 (X) E : 0 −→ En−1 −→ · · · Ek −→

n where E0 ∼ = R1 , Ek = Rk(k+1) and for each k = 0, . . . , n − 2, the map δk : Ek → Ek−1 is defined as

δk (ei1 ∧ · · · ∧ eik+1 ) ⊗

v2k−1

δk (ei1 ∧ · · · ∧ eik+1 ) ⊗ v1k−1

δk (ei1 ∧ · · · ∧ eik+1 ⊗

v1j v2k−j−1)

=

k+1 X

x2s (ei1 ∧ · · · ∧ eˆs ∧ · · · ∧ eik+1 ) ⊗ v2k−2

=

k+1 X

(−1)s+1 x1s (ei1 ∧ · · · ∧ eˆs ∧ · · · ∧ eik+1 ) ⊗ v1k−2

s=1

s=1

=

k+1 X

(−1)s+1 x1s (ei1 ∧ · · · ∧ eˆs ∧ · · · ∧ eik+1 ) ⊗ v1j−1v2k−j−1

s=1

+

k+1 X

x2s (ei1 ∧ · · · ∧ eˆs ∧ · · · ∧ eik+1 ) ⊗ v1j v2k−j−2

s=1

for every ordered k + 1 tuple (i1 , i2 , · · · , ik+1 ), with 1 ≤ i1 < · · · < ik+1 ≤ n and for every j = 1, 2, · · · , k − 2. A minimal resolution of hg1 i is given by: g1

G : 0 −→ R −→ R −→ R/hg1 i −→ 0. e12 ) and hg1 i intersect transversally, by Lemma 4.2. ThereThe ideals I2 (X e12 ) + hg1 i is given fore, by Lemma 3.7, a minimal free resolution for I2 (X by the tensor product complex ψk+1

e12 )+hg1 i → 0 E ⊗G : 0 → En−1 → · · · → Ek+1 ⊕Ek −→ Ek ⊕Ek−1 → · · · → E0 → R/I2 (X

such that ψk : Ek ⊕ Ek−1 −→ Ek−1 ⊕ Ek−2 is the map defined as j k−j−1 j k−j−1 ψk ei1 ∧ · · · ∧ eik+1 ⊗ v1 v2 = δk (ei1 ∧ · · · ∧ eik+1 ) ⊗ v1 v2 ψk (ei1 ∧ · · · ∧ eik ) ⊗ v1j v2k−j−2 = (−1)k−1g1 (ei1 ∧ · · · ∧ eik ) ⊗ v1j v2k−j−2 j k−j−2 . + δk−1 (ei1 ∧ · · · ∧ eik ) ⊗ v1 v2 e12 ) + hg1, g2 i by mapping Now we find a minimal free resolution for I2 (X e12 ) + cone. Let Ck := (E· ⊗ G· )k . We have proved in Lemma 4.3 that hI2 (X hg1i : g2 i = hx11 , x12 , · · · , x1n i; which is minimally resolved by the Koszul

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11

complex. Let us denote the Koszul Complex by (F· ; σk ), where σk is the kth differential. We first construct the connecting map τ· : F· → E· ⊗ G· . Let n n n us write Fk := R(k ) and Ck := Rk(k+1) ⊕R(k−1)( k ) . The map τk : Fk → Ck is defined as: X τk (ei1 ∧ · · · ∧ eik ) = yj (ei1 ∧· · ·∧eik ∧ej )⊗v1k−1 −(ei1 ∧· · ·∧eik )⊗v1k−2 . j

Let us choose the lexicographic ordering among the k tuples (i1 , . . . , ik ), n such that 1 ≤ i1 < · · · < ik ≤ n in order to write an ordered basis for R(k ) . We define lexicographic ordering among the tuples (i1 , . . . , ik+1 , k − j, j), for j = 0, . . . , k and k = 1, . . . , n − 1 to order the basis elements for n n n Rk(k+1) . Moreover, in the free module Ck = Rk(k+1) ⊕ R(k−1)( k ) , we order n the basis elements in such a way that those for Rk(k+1) appear first. The matrix representation of τk with respect to the chosen ordered bases is the following:   Ak( n )×(n) 0k( n )×(n) k k+1 k k+1          .  −I(n)×(n)   k k    0    (k−1)(nk)×(nk)   0(k−2)(n)×(n) k

k

Theorem 5.1. The following diagram commutes for every k = 1, . . . , n−1: τk

FO k

/

σk+1

CO k ψk+1

Fk+1

τk+1

/

Ck+1

Proof. It suffices to prove the statement for a basis element (ei1 ∧· · ·∧eik+1 ) of Fk+1 . Without loss of generality we consider (e1 ∧ · · · ∧ ek+1 ). We first compute (τk ◦ σk+1 )(e1 ∧ · · · ∧ ek+1 ). σk+1

(e1 ∧ · · · ∧ ek+1 ) 7−→

k+1 X

(−1)j+1 x1j (e1 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 )

j=1

τk

7−→

k+1 X

(−1)

j=1

−

k+1 X j=1

j+1

n X x1j [ ys (e1 ∧ e2 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ es ) ⊗ v1k−1] s=1

(−1)j+1 x1j (e1 ∧ e2 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ) ⊗ v1k−2 .

12


We now compute (ψk+1 ◦ τk+1 )(e1 ∧ · · · ∧ ek+1 ). n τk+1 X ys (e1 ∧ · · · ∧ ek+1 ∧ es ) ⊗ v1k (e1 ∧ · · · ∧ ek+1 ) 7−→ s=1

−(e1 ∧ · · · ∧ ek+1 ) ⊗ v1k−1 n X ψk+1 X 7−→ [ (−1)j+1 ys x1j (e1 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ es ) ⊗ v1k−1 s=1 j=1,2,··· ,k+1,s

−(−1)k g1 (e1 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ ej ) ⊗ v1k−1 −

k+1 X

(−1)j+1 x1j (e1 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ ej ) ⊗ v1k−2

j=1

=

n k+1 X X

[x1j ys (−1)j+1 (e1 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ es ) ⊗ v1k−1]

j=1 s=1 n X

(−1)s+1 ys x1s (e1 ∧ · · · ∧ eˆs ∧ · · · ∧ ek+1 ∧ es ) ⊗ v1k−1

+

s=1

−(−1)k g1 (e1 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ ej ) ⊗ v1k−1 −

k+1 X

(−1)j+1 x1j (e1 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ ej ) ⊗ v1k−2

j=1

=

k+1 X n X

[x1j ys (−1)j+1 (e1 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ es ) ⊗ v1k−1 ]

j=1 s=1 n X

(−1)s+1 (−1)k+1−s ys x1s (e1 ∧ · · · ∧ eˆs ∧ · · · ∧ ek+1 ∧ es ) ⊗ v1k−1

+

s=1

−(−1)k g1 (e1 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ ej ) ⊗ v1k−1 −

k+1 X

(−1)j+1 x1j (e1 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ ej ) ⊗ v1k−2

j=1

=

k+1 X n X


j=1 s=1

+(−1)k g1 (e1 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ ej ) ⊗ v1k−1 −(−1)k g1 (e1 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ ej ) ⊗ v1k−1 −

k+1 X

(−1)j+1 x1j (e1 ∧ e2 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ ej ) ⊗ v1k−2

j=1

=

k+1 X n X


j=1 s=1

−

k+1 X j=1

(−1)j+1 x1j (e1 ∧ · · · ∧ eˆj ∧ · · · ∧ ek+1 ∧ ej ) ⊗ v1k−2.

SUM IDEALS

13

e12 )+ Hence the mapping cone M(E· ⊗G· ; F· ) gives us the resolution for I2 (X hg1, g2 i as described in 3.2. However, this resolution is not minimal. We now construct a minimal free resolution from M(E· ⊗ G· ; F· ). e12 ) + hg1 , g2 i has been constructed in A free resolution for the ideal I2 (X 3.1, which is given by dn+2

dk+1

d

k Dk−1 · · · −→ D1 −→ D0 −→ 0, 0 −→ Dn+2 −→ Dn+1 · · · −→ Dk −→

n n n such that Dk = Fk−1 ⊕ Ck = R(k−1) ⊕ (Rk(k+1) ⊕ R(k−1)( k ) ) and dk = (−σk−1 + τk−1 , ψk ). Let us recall that the map ψ is the differential in the e12 ) + hg1 i, the map σ is the differential in the Koszul free resolution for I2 (X resolution for hx11 , x12 , . . . , x1n i and τ is the connecting homomorphism between the complexes defined in 3.1. Let us order bases for Fk−1 and Ck with respect to the lexicographic ordering. Finally we order basis for Dk in such a way that the basis elements for Fk−1 appear first, followed by the basis elements for Ck . Therefore, the matrix representation for the differential map dk is given by



 −σk−1    τk−1

−σk−1

       0 A     =         ψk τk−1 =     0  

0

0



     −I         0



       .   ψk     

The entries in the matrices representing σk−1 and ψk can only belong to the maximal ideal hxij , yj i, since both are differentials of minimal free resolutions. The block matrix A has also elements in the maximal ideal hxij , yj i. The only block which has elements outside the maximal ideal hxij , yj i is in the identity block appearing in τk−1 . Therefore, it is clear from the matrix representation of the map dk that we can apply Lemma 3.12 repeatedly to get rid of non-minimality. Hence, we get a minimal free resolution and the

14


e12 ) + hg1 , g2 i are total Betti numbers for the ideal I2 (X

b0 = 1, n + 2, b1 = 2 n + n, b2 = 2 · 3 n n n n n − − + + (k − 1) bk+1 = k k k−1 k−1 k k+1 n n , for 2 ≤ k ≤ n − 1, + (k − 2) = k k k+1 bn = n − 2.

eij ) + hg1 , . . . , gn i. 5.2. A minimal free resolution for I2 (X

Lemma 5.2. Let Gk = G12 ∪ {g1 , g2 , . . . , gk }, 1 ≤ k ≤ n, where G12 is the e12 defined in the list of notations in section 2. The set of all 2×2 minors of X e12 ) + hg1 , . . . , gk i with respect set Gk is a Gröbner basis for the ideal I2 (X to a suitable monomial order.

Proof. We take the lexicographic monomial ordering in R induced by the following ordering among the indeterminates: xnn > · · · > xtt > · · · > x33 > y1 > · · · > yn > x11 > · · · > x1n > x21 > · · · > x2n > xst for other s, t. Then, we observe that for every s ≥ 3, Lt(gs ) is coprime with Lt(gt ) for every 1 ≤ t ≤ k; t 6= s and also coprime with Lt(h) for every h ∈ G. e12 ). Therefore, Moreover, by Lemma 3.5, G12 is a Gröbner basis for I2 (X we only have test the S-polynomials S(g1, g2 ), S(g1 , h) and S(g2 , h), for h ∈ G. Pn We can write S(g1 , g2 ) = k=1 [12|1k]yk and note that Lt([12|1k]) ≤ Lt(S(g1 , g2 ) for every 1 ≤ k ≤ n. Hence, S(g1 , g2 ) →G ′ 0. We note that, if i 6= 1 then the leading terms of g1 and [12|st] are mutually coprime and therefore P S(g1 , [12|st]) →Gk 0. Next, the expression S(g1 , [12|1t]) = x1t g2 + s6=t [12|st]ys shows that S(g1 , [12|1t]) →Gk 0. Similarly, if s 6= 1

SUM IDEALS

15

then the leading terms of g2 and [12|st] are mutually coprime and therefore S(g2 , [12|st]) →Gk 0. The proof for S(g2 , [12|1t]) is similar to that of S(g1, [12|st]). Remark. The corresponding result for i < j in general would be the following: Lemma 5.3. Let Gi,j,k = Gij ∪ {gi , gj , gl1 , . . . , glk−2 }, 1 ≤ k ≤ n, 1 ≤ l1 < · · · < lk−2 ≤ n and lt is the smallest in the set {1, 2, . . . , n} \ eij defined {i, j, l1 , . . . , lt−1 }; Gij denotes the set of all 2 × 2 minors of X in the list of notations in section 2. The set Gi,j,k is a Gröbner basis for the eij ) + hg1 , . . . , gk i with respect to a suitable monomial order. ideal I2 (X

Proof. While proving this statement with i < j arbitrary, we have to choose the following monomial orders. The rest of the proof remains similar.

Suppose that X is generic, we choose the lexicographic monomial ordering in R induced by the following ordering among the indeterminates: xnn > · · · > xc cii > · · · > x11 > jj > · · · > x > > >

y1 > · · · > yn xi1 > · · · > xin xj1 > · · · > xjn xst for all other

s, t.

If X is generic symmetric, we choose the lexicographic monomial ordering in R induced by the following ordering among the indeterminates: xnn > · · · > xc cii > · · · > x11 > y1 > · · · > yn jj > · · · > x > xii > xij > x1i > x2i > · · · > xi−1,i > xi,i+1 > · · · > xin > xjj > · · · > xj−1,j > xj,j+1 > · · · > xjn > xst for all other s, t. e12 )+hg1 , . . . , gk i and hgk+1i intersect transverLemma 5.4. The ideals I2 (X sally, for every 2 ≤ k ≤ n − 1.

e12 ) + hg1 , . . . , gk i such Proof. Suppose not, then, there exists hk+1 ∈ / I2 ( X e that hk+1 gk+1 ∈ I2 (X12 ) + hg1, . . . , gk i. Let us choose the same monomial order on R as defined in Lemma 5.2. Upon division by elements of Gk , we may further assume that Lt(h) ∤ Lt(hk+1 ) for every h ∈ Gk , since Gk e12 ) + hg1, . . . , gk i by Lemma 5.2. On is a Gröbner basis for the ideal I2 (X e12 ) + hg1 , . . . , gk i and therefore Lt(h) | the other hand hk+1 gk+1 ∈ I2 (X Lt(hk+1 ), for some h ∈ Gk , since Lt(h) and Lt(gk+1 ) are mutually coprime, - a contradiction.

16


Remark. The corresponding result for i < j in general would be the eij ) + hgi , gj , gl1 , . . . , gl i and hgl i intersect following: The ideals I2 (X k k+1 transversally, if 1 ≤ l1 < . . . < lk < lk+1 ≤ n and lk+1 is the smallest in the set {1, 2, . . . , n} \ {i, j, l1 , . . . , lk }, for every 1 ≤ k ≤ n − 3. The proof is essentially the same as above after we use the Lemma 5.3. Proof of Theorem 2.1. Part (1) of the theorem has been proved in 5.1. We now prove part (2) under the assumption i = 1, j = 2. Let the minimal e12 ) + hg1 , g2 , · · · , gk i be (L , δ ). By Lemma 5.4 free resolution of I2 (X e12 ) + hg1 , . . . , gk+1i is and Lemma 3.7, the minimal free resolution of I2 (X gk+1

given by the tensor product of (L , δ ) and 0 −→ R −→ R −→ 0, and that is precisely (K , ∆ ), with Kp = Lp ⊕Lp−1 and ∆p = (λp , (−1)p gk+1 + λp−1 ). Let βk,p , 0 ≤ p ≤ n + k, denote the p-th total Betti number for the ideal e12 ) + hg1 , . . . , gk i. Then, the total Betti numbers βk+1,p , 0 ≤ p ≤ I2 ( X e12 ) + hg1 , . . . , gk+1i are given by βk+1,0 = 1, n + k + 1 for the ideal I2 (X βk+1,p = βk,p−1 + βk,p for 1 ≤ p ≤ n + k and βk+1,n+k+1 = n − 2. The proof for general i < j follows similarly according to the strategy discussed in the beginning of section 5. eij )+hg1 , . . . , gn i In particular, the total Betti number βn−2,p for the ideal I2 (X are given by βn−2,0 = 1, βn−2,p = βn−3,p−1 + βn−3,p for 1 ≤ p ≤ 2n − 3 and βn−2,2n−2 = n − 2. Example. We show the Betti numbers at each stage for n = 4 and n = 5. 1 6 1 7 n=4: 1 8 1 9 1 10 1 1 1 n=5: 1 1 1

10 11 12 13 14 15

8 3 14 11 3 12 7 2 20 19 9 2 29 39 28 11 2

20 5 4 30 25 9 4 25 25 14 3 37 50 39 17 3 50 87 89 56 20 3 64 137 176 145 76 23 3 R EFERENCES

[1] W. Bruns, A.R. Kustin, M. Miller, The Resolution of the Generic Residual Intersection of a Complete Intersection, Journal of Algebra 128 (1990) 214-239. [2] A. Conca, Emanuela De Negri, Elisa Gorla, Universal Gröbner bases for Maximal Minors, International Mathematics Research Notices 11(2015) 3245-3262.

SUM IDEALS

17

[3] D. Eisenbud, Geometry of Syzygies, Springer-Verlag, NY, 2005. [4] P. Gimenez, I. Sengupta and H. Srinivasan, Minimal graded free resolution for monomial curves defined by arithmetic sequences, Journal of Algebra 388 (2013) 294-310. [5] M.R., Johnson, J. McLoud-Mann, On equations defining Veronese Rings, Arch. Math. (Basel) 86(3)(2006) 205-210. [6] S.Lichtenbaum, On the vanishing of Tor in regular local rings, Illinois J.Math. 10: 220- 226,1966. [7] H. Matsumura, Commutative Ring Theory, Cambridge University Press, NY, 1986. [8] I. Peeva, Graded Syzygies, Springer-Verlag London Limited, 2011. [9] J. Saha, I. Sengupta, G. Tripathi, Ideals of the form I1 (XY ), arXiv:1609.02765 [math.AC] 2016. [10] J. Saha, I. Sengupta, G. Tripathi, Primality of certain Determinanatal ideals, arXiv:1610.00926 [math.AC] 2016.

Department of Mathematics, RKM Vivekananda University, Belur Math, Howrah 711202, 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, India. E-mail address: [email protected]