a minimal free resolution for certain monomial curves

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more important works on free resolution of monomial ideals, subsequent to ..... The proof is divided in three cases, depending upon b = 1,2,3, in the following.
A MINIMAL FREE RESOLUTION FOR CERTAIN MONOMIAL CURVES IN A4 Indranath Sengupta



Department of Mathematics Indian Institute of Science, Bangalore 560 012, India. E-mail : [email protected]

Abstract Let K be a field and m0 < . . . < m3 be coprime positive integers, which form an arithmetic progression. Let C be a monomial curve in the affine 4-space A4K , defined parametrically by X0 = T m0 , . . . , X3 = T m3 . Let A be the coordinate ring of C. In this article, we produce an explicit minimal free resolution for A.

§1 Introduction Let S := K[X1 , . . . , Xn ] be a graded polynomial ring over a field K and M a finite graded S-module. Then, M has a minimal graded free resolution ϑ

ϑ

ϑ

m 1 0 0 −→ S βm −→ S βm−1 −→ · · · −→ S β1 −→ S β0 −→ M −→ 0 ,

such that m ≤ n. The integers m and βi , for 0 ≤ i ≤ m, are well defined for M , in the sense that, these do not depend on the minimal resolution of M we choose. The integer m is called the projective dimension of M as an S-module, βi is called the i-th Betti number of M and the module Mi := ker(ϑi ) is called the i-th syzygy of M . Let I be a homogeneous ideal in S. One of the basic problems in commutative algebra is to construct an explicit minimal free resolution of S/I. Finding an explicit minimal free resolution for an arbitrary monomial ideal still remains an open problem. A nonminimal free resolution was constructed by D. Taylor in her thesis ( see [1; 17.11] ), generalizing Koszul complex in a natural way. Some more important works on free resolution of monomial ideals, subsequent to Taylor’s work, can be found in [2], [3], [4]. We are mainly concerned with certain binomial prime ideals, namely those which arise as the defining ideal for certain affine monomial curves. Bresinsky [5] first gave a minimal free resolution for monomial curves in P3K . Peeva and Sturmfels [6] ∗ This work was done when the author was a CSIR Senior Research Fellow at Indian Institute of Science, Bangalore. Author’s present address is Department of Mathematics, Jadavpur University, Kolkata 700 032, INDIA.

1

2

SENGUPTA

considered lattice ideals of codimension two and generalized Bresinsky’s result. For more work on lattice ideals, which include ideals defining toric varieties, see [7] and [8]. In this article, our aim is to give an explicit minimal free resolution for monomial curves in A4K , defined parametrically by X0 = T m0 , . . . , X3 = T m3 , such that the integers 0 < m0 < . . . < m3 form an arithmetic progression with gcd(m0 , . . . , m3 ) = 1 and they minimally P generate the semigroup Γ := 3i=0 Nmi . Our main result appears in Theorem (3.1).

§2 Preliminaries Before we proceed any further, a few comments are in order. Let e ≥ 3 and m0 , . . . , me−1 be positive integers with gcd(m0 , . . . , me−1 ) = 1. Suppose that, the integers form an almost arithmetic sequence, that is, some e − 1 of these form an arithmetic progression. In [9], certain integers r, r0 , q, q 0 , u, v, w, z, λ, µ, ν have been introduced ( see [9; Lemma(3.1)] ) for giving an explicit description of the standard basis ( see [9] for a definition ) of the semigroup P Γ = e−1 i=0 Nmi . This description has been efficiently used in [10], to construct a minimal set of binomial generators G, for the defining ideal of the affine monomial curve defined by the parametrization X0 = T m0 , . . . , Xe−1 = T me−1 . In [11], with the assumption that P m0 , . . . , me−1 generate Γ = e−1 obner basis with i=0 Nmi minimally, it is shown that G is a Gr¨ respect to a suitable monomial order. When the integers m0 , . . . , me−1 form an arithmetic progression, then the integers r, r0 , q, q 0 , u, v, w, λ, µ, ν can be described in more concrete terms, as shown in [12; Lemma(3.2)] and this has been used to rewrite the set G. In this article, as we have already mentioned, we consider the situation when e = 4 and m0 < . . . < m3 form a minimal arithmetic progression. Therefore, when we refer to the relevant parts of [12], [10], [11], we do it only in the context of the special situation mentioned above. Let m0 < m1 < m2 < m3 be positive integers with gcd(m0 , . . . , m3 ) = 1. We assume P that these integers are minimal, i.e., they generate the semigroup Γ := 3i=0 Nmi minimally. We further assume that m0 , . . . , m3 form an arithmetic progression with common difference d. We write n := m3 and m0 = 3a + b, where a and b are unique integers such that a ≥ 0 and 1 ≤ b ≤ 3. Let K be a field and R := K[X0 , X1 , X2 , Y ] be the polynomial ring over K, graded with respect to the weighted gradation given by wt(Xi ) = mi , wt(Y ) = n. Note that, we have used Y instead of X3 for an indeterminate. The reason for this choice is that, we would like to retain the notations used in [12], [10], [11]. Let T be an indeterminate and η : R → K[T ] be the K-algebra homomorphism, defined by η(X0 ) = T m0 , η(X1 ) = T m1 , η(X2 ) = T m2 , η(Y ) = T n . Let p := ker(η) and A := R/p. Then, with respect to the weighted gradation given on R and the standard gradation on K[T ], η is a graded map, p is a graded ideal in R and A is a graded R-algebra. The ideal p is precisely the defining ideal and A is the coordinate ring for the affine monomial curve defined by the parametrization X0 = T m0 , X1 = T m1 , X2 = T m2 , Y = T n . By an abuse of notation, we would also call this curve as the monomial curve defined by the integers m0 , m1 , m2 , n. (2.1) Definition. A minimal generating set of binomial generators of p is known explicitly. We define ( as in [10] )

MINIMAL FREE RESOLUTION IN A4

3

• ξ11 := X12 − X0 X2 . • ϕi := Xi+1 X2 − Xi Y , for 0 ≤ i ≤ 1. • ψj := Xb+j Y a − X0a+d Xj , if 1 ≤ b ≤ 2 and 0 ≤ j ≤ 2 − b. • θ := Y a+1 − X0a+d X3−b . • G :=

 {ξ11 } ∪ {ϕ0 , ϕ1 } ∪ {ψ0 , ψ1 } ∪ {θ}      

if

b = 1,

{ξ11 } ∪ {ϕ0 , ϕ1 } ∪ {ψ0 } ∪ {θ}

if

b = 2,

{ξ11 } ∪ {ϕ0 , ϕ1 } ∪ {θ}

if

b = 3.

It is known that, G is a minimal generating set for p ( see [10] and [12] ). Let us quickly run through a few basics on Gr¨obner basis, for details we refer to [1]. Let K[X1 , . . . , Xn ] be a polynomial ring over a field K with a monomial order > and let P f ∈ K[X1 , . . . , Xn ], be non-zero with f = cX α + cβ X β , where c, cβ ∈ K and X α is the largest monomial appearing in f with respect to the ordering >. Then, cα X α is called the leading term of f , denoted by Lt(f ) and X α is called the leading monomial of f , denoted by Lm(f ), with respect to >. We now present the notion of the so called S - polynomials, followed by Buchberger’s Criterion. Let f, g ∈ K[X1 , . . . , Xn ] be non-zero with Lt(f ) = cX α , Lt(g) = dX β , for c, d ∈ k and let X γ := lcm(X α , X β ). Then the S - polynomial of f and g, denoted by S(f, g), is the polynomial Xγ Xγ ·f − ·g. S(f, g) := Lt(f ) Lt(g) Given G = {g1 , . . . , gt } ⊆ K[X1 , . . . , Xn ] and f ∈ K[X1 , . . . , Xn ], we say that f reduces to P zero modulo G, denoted by f →G 0, if f can be written as f = ti=1 ai gi , such that whenever ai gi 6= 0, we have Lm(f ) ≥ Lm(ai gi ). Buchberger’s Criterion says that, for an ideal I in K[X1 , . . . , Xn ] and a generating set G = {g1 , . . . , gt } for I, G is a Gr¨obner basis for I iff S(gi , gj ) →G 0, for every i 6= j. (2.2) Definition. For monomials Y α3 2i=0 Xiαi and Y β3 2i=0 Xiβi in R := K[X0 , X1 , X2 , Y ], Q Q with (α3 , . . . , α0 ), (β3 , . . . , β0 ) ∈ N4 , we say that Y α3 2i=0 Xiαi >grevlex Y β3 2i=0 Xiβi , if in P P the difference (α3 − β3 , . . . , α0 − β0 , 3i=0 βi mi − 3i=0 αi mi ) ∈ Z4 , the rightmost non-zero entry is negative. Q

Q

In [11] it is shown that, the set G is a Gr¨obner basis for the ideal p, with respect to the monomial order grevlex on R. This has been done by showing that each S-polynomial reduces to zero modulo G. In the following proposition, we mention some relevant parts of the computation done in [11]. (2.3) Proposition. Let the set G be as defined in (2.1). Then, with respect to the monomial order grevlex on R, (1) S(ξ11 , ϕ0 ) = X2 ξ11 − X1 ϕ0 = −X0 ϕ1

−→G 0.

4

SENGUPTA

(2) S(ξ11 , ϕ1 ) = X22 ξ11 − X12 ϕ1 = X1 Y ξ11 − X0 X2 ϕ1 (3) S(ξ11 , θ) = Y a+1 ξ11 − X12 θ =

−→G 0.

  X a+d X2 ξ11 − X0 X2 θ    0    

if

b = 1,

X0a+d X1 ξ11 − X0 X2 θ

if

b = 2,

X0a+d+1 ξ11 − X0 X2 θ

if

b = 3,

if

b = 1,

X0a+d X1 ϕ0 − X0 Y θ

if

b = 2,

X0a+d+1 ϕ0 − X0 Y θ

if

b = 3,

(4) S(ϕ0 , ϕ1 ) = X2 ϕ0 − X1 ϕ1 = Y ξ11 (5) S(ϕ0 , θ) = Y a+1 ϕ0 − X1 X2 θ =

−→G 0.

  X a+d X2 ϕ0 − X0 Y θ    0    

−→G 0.

 a+d    X0 X2 ϕ1 − X1 Y θ 

(6) S(ϕ1 , θ) = Y a+1 ϕ1 − X22 θ =  X0a+d X1 ϕ1 − X1 Y θ    a+d+1 X0 ϕ1 − X1 Y θ

if

b = 1,

if

b = 2,

if

b = 3,

−→G 0.

−→G 0.

(7) Let b = 1. Then, (i) S(ξ11 , ψ0 ) = Y a ξ11 − X1 ψ0 = −X0 ψ1 a

(ii) S(ξ11 , ψ1 ) = X2 Y ξ11 −

X12 ψ1

=

−→G 0.

X0a+d X1 ξ11

(iii) S(ϕ0 , ψ0 ) = Y a ϕ0 − X2 ψ0 = −X0 θ

− X0 X2 ψ1

−→G 0.

(iv) S(ϕ0 , ψ1 ) = Y a ϕ0 − X1 ψ1 = X0a+d ξ11 − X0 θ

−→G 0.

(v) S(ϕ1 , ψ0 ) = X1 Y a ϕ1 − X22 ψ0 = −X0a+d X2 ξ11 − X12 θ a

(vi) S(ϕ1 , ψ1 ) = Y ϕ1 − X2 ψ1 = −X1 θ (vii) (viii)

−→G 0.

−→G 0.

S(ψ0 , ψ1 ) = X2 ψ0 − X1 ψ1 = X0a+d ξ11 −→G S(ψ0 , θ) = Y ψ0 − X1 θ = X0a+d ϕ0 −→G 0.

(ix) S(ψ1 , θ) = Y ψ1 − X2 θ = X0a+d ϕ1

−→G 0.

0.

−→G 0.

(8) Let b = 2. Then, (i) S(ξ11 , ψ0 ) = X2 Y a ξ11 − X12 ψ0 = X0a+d+1 ξ11 − X0 X2 ψ0 (ii) S(ϕ0 , ψ0 ) = Y a ϕ0 − X1 ψ0 = −X0 θ (iii) (iv)

−→G 0.

−→G 0.

S(ϕ1 , ψ0 ) = Y a ϕ1 − X2 ψ0 = −X0a+d ξ11 − X1 θ S(ψ0 , θ) = Y ψ0 − X2 θ = X0a+d ϕ0 −→G 0.

−→G 0.

The set G is a Gr¨obner basis with respect to grevlex. Proof. The proof of this proposition can be found in [11], where it is proved in the general situation of an almost arithmetic sequence, using the integers p, r, r0 , q, q 0 , u, v, w, z, λ, µ, ν. These integers have been rewritten for a minimal arithmetic progression in [12; Lemma (3.2)]. For making the cross-reference easier, we record these here : p = 2, u = 3, q = 1, r = 1,

MINIMAL FREE RESOLUTION IN A4

5 ( 0

λ = 1, w = 1, v = a + 1, z = 3 − b and q = (

µ= (1) (2) (3) (4) (5) (6)

a+d if 1 ≤ b ≤ 2 , a + d + 1 if b = 3 ,

Use Use Use Use Use Use

0 if 1 ≤ b ≤ 2 , 0 r = 1 if b = 3 ,

(

b if 1 ≤ b ≤ 2 , 1 if b = 3 ,

ν = µ + 1.

k = l = 1, i = 0, r = 1, p = 2, q = 1, λ = 1, w = 1 k = l = 1, i = 1, r = 1, p = 2, q = 1, λ = 1, w = 1 k = l = 1, v = a + 1, r0 ≥ r = 1 in [11; (8.1)]. i = 0, j = 1, r = 1, λ = 1, w = 1 in [11; (9.1)]. i = 0, v = a + 1, r = 1, p = 2, q = 1, r0 ≥ r = 1 in i = 1, v = a + 1, r = 1, p = 2, q = 1, r0 ≥ r = 1 in

in [11; (6.1)]. in [11; (6.3)].

[11; (11.1)]. [11; (11.1)].

(7) For b = 1, we have r = r0 = 1, q 0 = 0, µ = a + d = ν − 1. (i) Use k = l = 1, j = 0, r0 = 1, p = 2, J = {0, 1} in [11; (7.2)]. (ii) Use k = l = 1, j = 1, r0 = 1, q 0 = 0, v = a + 1, w = 1, p = 2 in [11; (7.4)]. (iii) Use i = j = 0, r = r0 = 1, v = a + 1, w = 1, p = 2, q = 1, q 0 = 0, λ = 1 in [11; (10.1)]. (iv) Use i = 0, j = 1, r = r0 = 1, p = 2, q 0 = 0, q = 1, v = a + 1, w = 1, λ = 1 in [11; (10.5)]. (v) Use i = 1, j = 0, r = r0 = 1, p = 2, v = a + 1, w = 1, q 0 = 0, q = 1, λ = 1 in [11; (10.6)]. (vi) Use i = j = 1, r = r0 = 1, v = a + 1, w = 1, p = 2, q 0 = 0, q = 1, λ = 1 in [11; (10.1)]. (vii) Use i = 0, j = 1, r0 = 1 in [11; (12.1)]. (viii) Use j = 0, r = r0 = 1, w = 1, p = 2, q 0 = 0, q = 1 in [11; (13.2)]. (ix) Use j = 1, r = r0 = 1, w = 1, p = 2, q 0 = 0, q = 1 in [11; (13.2)]. (8) For b = 2, we have r0 = 2, q 0 = 0, µ = a + d = ν − 1. (i) Use k = l = 1, j = 0, r0 = 2, q 0 = 0, v = a + 1, w = 1 in [11; (7.4)]. (ii) Use i = j = 0, r = 1, r0 = 2, p = 2, v = a + 1, w = 1, q = 1, q 0 = 0, λ = 1 in [11; (10.5)]. (iii) Use i = 1, j = 0, r = 1, r0 = 2, v − w = a, q = 1, q 0 = 0, p = 2, λ = 1 in [11; (10.1)]. (iv) Use j = 0, 2 = r0 > r = 1, w = 1, q 0 = 0, p = 2, q = 1 in [11; (13.2)]. Hence, G is a Gr¨obner basis with respect to grevlex, by Buchberger’s criterion. A link between Gr¨obner basis and syzygies is established in the following theorem by Schreyer. (2.4) Theorem. Let K be a field, K[X1 , . . . , Xn ] be the polynomial ring and I be an ideal in K[X1 , . . . , Xn ]. Let G := {g1 , . . . , gt } be an ordered set of generators for I, which is a Gr¨obner basis, with respect to some fixed monomial order on K[X1 , . . . , Xn ]. Let Syz(g1 , . . . , gt ) := P {(a1 , . . . , at ) ∈ Rt | ti=1 ai gi = 0}. Suppose that, for i 6= j, S(gi , gj ) = aj gi − ai gj =

t X

hk gk −→G 0 .

k=1

Then, the t-tuples



h1 , · · · , hi − aj , · · · , hj + ai , · · · , ht



generate Syz(g1 , . . . , gt ).

6

SENGUPTA

Proof. See [13; Theorem 10.2.8]. Our aim is to find an explicit minimal free resolution for the R-module A and our method involves, computing minimal generating sets for the successive syzygy modules of A. We appeal to Theorem(2.4), which together with the information on S-polynomials provided in Proposition(2.3), give us the first syzygy module explicitly. For finding a minimal generating set for the second syzygy module, many ideas came from computations using Macaulay [14]. Before we proceed to the main theorem, let us fix some notations which we follow throughout the article. (2.5) Notations. Let f1 , . . . , fm be elements of Rn . • For 1 ≤ i ≤ m and 1 ≤ j ≤ n, let the j-th entry of fi be denoted by fji , i.e., fi = (f1i , . . . , fji , . . . , fni ). • Let hf1 , . . . , fn i denote the R-submodule generated in Rn , by the set {f1 , . . . , fn } and let hf1 , . . . , fˆi , . . . , fn i denote the submodule generated by the set {f1 , . . . , fn } \ {fi }. • For a matrix M , the transpose of M is denoted by M t . • Syz(f1 , . . . , fm ) := {(a1 , . . . , am ) ∈ Rm |

Pm

i=1

ai fi = 0}, for the ordered tuple (f1 , . . . , fm ).

§3 Minimal free resolution (3.1) Theorem. Let m0 = 3a + b, for some integer a ≥ 0 and b ∈ [1, 3]. Let m0 < m1 < m2 < m3 be a minimal, arithmetic sequence of coprime positive integers with common difference d. Then, the sequence of R-modules, ϑ3 (b)

ϑ2 (b)

ϑ1 (b)

ϑ

0 0 −→ Rβ3 (b) −→ Rβ2 (b) −→ Rβ1 (b) −→ R −→ A −→ 0

is a minimal free resolution for the coordinate ring A of the affine monomial curve defined by the integers m0 < . . . < m3 . Here β1 (b), β2 (b), β3 (b) are given by β1 (b) =

 6   

if b = 1 ,

5 if b = 2 ,

  

β2 (b) =

 8   

5 if b = 2 ,

  

4 if b = 3 ,

if b = 1 , and β3 (b) =

 3   

1 if b = 2 ,

  

5 if b = 3 ,

if b = 1 ,

2 if b = 3 .

The map ϑ0 denotes the canonical surjection and ϑ1 (b), ϑ2 (b), ϑ3 (b) are as noted below :        • ϑ1 (1) :=       

X12 − X0 X2



  X12 − X0 X2   X1 X2 − X0 Y   X X −X Y   1 2 0  2  X2 − X1 Y   2  , ϑ1 (2) :=  X − X1 Y   2 X1 Y a − X0a+d+1     X2 Y a − X0a+d+1  a+d a   X2 Y − X0 X1  Y a+1 − X a+d X

Y a+1 − X0a+d X2

0

1

          

MINIMAL FREE RESOLUTION IN A4 

X12 − X0 X2

   X1 X2 − X0 Y and ϑ1 (3) :=    X22 − X1 Y 

Y a+1 − X0a+d+1



−X2

  Y    −Y a    a+d  X0  • ϑ2 (1) :=   0    0    0  

    .   

X1

−X0

0

0

0



−X2

X1

0

0

0

0

0

X1

−X0

0

a

0

0

X1

−X0

−Y a

0

X2

0

−X0

X0a+d

0

         ,         

−Y

−Y

0

X1

0

−Y

a

0

X2

−X1

0

0

X0a+d

0

−Y

X2

−X2

X1

−X0

0

0



−X2

X1

0

0

−Y a

0

X1

−X0

X2

−X1

−Y

X2

         

  Y    0 ϑ2 (2) :=     −X0a+d 

0



0 X0a+d

−Y

−X2

0

Ya



a+d • ϑ3 (1) :=   X0



0 

a

0

  Y    X0a+d+1 − Y a+1 and ϑ2 (3) :=     0 

ϑ3 (2) :=

7

X1

−X0

0



−X2

X1

0

0

0

X12 − X0 X2

X0a+d+1 − Y a+1

0

0

X0a+d+1

     .    

X1 X2 − X0 Y −Y

a+1

X22 − X1 Y

0

−X2

0

X1

0

−X0

0



−Y a

−Y

X2

0

−X1

−X1

X0

X0a+d

 , 

0

−Y

Y

X2

0

−X1

X0a+d X1 − Y a+1 , X0a+d+1 − X2 Y a , X22 − X1 Y , X0 Y − X1 X2 , X0 X2 − X12 

and ϑ3 (3) := 

0

−Y

X0a+d+1 − Y a+1

X0a+d+1 − Y a+1

X2

0

X2

−X1

−X1

X0



 .

Proof. The proof is divided in three cases, depending upon b = 1, 2, 3, in the following subsections.

8

SENGUPTA

§3.A Case : b = 1 In this case, G = {ξ11 } ∪ {ϕ0 , ϕ1 } ∪ {ψ0 , ψ1 } ∪ {θ}, is a minimal generating set and a Gr¨obner basis for p. We fix an ordering as : g1 g2 g3 g4 g5 g6

:= := := := := :=

ξ11 ϕ0 ϕ1 ψ0 ψ1 θ

X12 − X0 X2 , X1 X2 − X0 Y , X22 − X1 Y , X1 Y a − X0a+d+1 , X2 Y a − X0a+d X1 , Y a+1 − X0a+d X2 .

= = = = = =

Let us list down the elements of R6 , which generate Syz(g1 , . . . , g6 ), as indicated in Theorem (2.4). We give a reference of the results we use for obtaining each 6-tuple. G1 G2 G3 G4 G5 G6 G7 G8 G9 G10 G11 G12 G13 G14 G15

:=





:=



:=



−Y a , 0, 0, X1 , −X0 , 0

:=



:=



X0a+d X1 X0a+d X2

:=



:=



:=



X0a+d , −Y a , 0, 0, X1 , −X0

:=



:=



0, X0a+d X2 −X0a+d X2 ,

:=



:=



:=



:=



:=



−X2 , X1 , −X0 , 0, 0, 0

by (2.4), (2.3)(1),

X1 Y − X22 , 0, X12 − X0 X2 , 0, 0, 0



by (2.4), (2.3)(7)(i),

a+1

−Y

X12

, 0, 0, 0, 0,

− X0 X2

a

0, −X1 Y ,

by (2.4), (2.3)(4),

X22 ,

by (2.4), (2.3)(7)(iii), 

by (2.4), (2.3)(7)(iv),

0,

−X12



X0a+d , 0, 0, −X2 , X1 , 0 X0a+d

, 0, −Y , X2



by (2.4), (2.3)(5), by (2.4), (2.3)(7)(v),



by (2.4), (2.3)(7)(vi),

0, 0, X0a+d X2 − Y a+1 , 0, 0, X22 − X1 Y , 0, −Y , 0, X1

by (2.4), (2.3)(7)(ii), by (2.4), (2.3)(3),



− Y a+1 , 0, 0, 0, X1 X2 − X0 Y

0, 0, −Y a , 0, X2 , −X1

X0a+d





0, −Y a , 0, X2 , 0, −X0

0, 0,



− X2 Y a , 0, 0, 0, X12 − X0 X2 , 0

Y , −X2 , X1 , 0, 0, 0

0,

by (2.4), (2.3)(2),







by (2.4), (2.3)(6), by (2.4), (2.3)(7)(vii),



by (2.4), (2.3)(7)(viii),



by (2.4), (2.3)(7)(ix).

We first rename a few elements from the above collection, which are rather distinguished, as will be seen in the following proposition. Let g1 := G1 , g2 := G6 , g3 := G3 , g4 := G8 , g5 := G7 , g6 := G14 , g7 := G11 , g8 := G15 and let g := {gi | 1 ≤ i ≤ 8}. (3.2) Proposition. The R-module Syz(g1 , . . . , g6 ) is minimally generated by the set g. Proof. By Theorem (2.4), the R-module Syz(g1 , . . . , g6 ) is generated by the set {Gi | 1 ≤ i ≤ 15}. Now we note the following identities.

MINIMAL FREE RESOLUTION IN A4

G2 G4 G5 G9 G10 G12 G13

= = = = = = =

9

X2 G1 + X1 G6 , Y a G1 + X1 G8 − X0 G11 , −Y a G6 + X2 G8 − X1 G11 , Y G7 + X2 G14 , X2 G7 − X2 G8 + X1 G11 , Y G11 + X2 G15 , −G7 + G8 .

Hence, Syz(g1 , . . . , g6 ) is generated by the set g. To show the minimality of the set g, we will show that for every 1 ≤ i ≤ 8, there exists a ˆ , . . . , gj(i)8 i. It is easy to check that, j(1) = 1, 1 ≤ j(i) ≤ 6, such that gj(i)i 6∈ hgj(i)1 , . . . , gj(i)i j(2) = 2, j(3) = 5, j(4) = 1, j(5) = 4, j(6) = 2, j(7) = 5 and j(8) = 5 serve our purpose. Hence, g is a minimal generating set for Syz(g1 , . . . , g6 ). Now, to find a minimal generating set for Syz(g1 , . . . , g8 ), let us choose the following elements in R8 , h1 h2 h3

:=



Y a , 0, −X2 , 0 X1 , 0 −X0 , 0

:=



X0a+d , −Y a , −Y , X2 , 0, −X1 , −X1 , X0

:=



0,

X0a+d



,

, 0, −Y , Y , X2 , 0, −X1 ,





,

.

(3.3) Proposition. The elements h1 , h2 , h3 form a R-basis for Syz(g1 , . . . , g8 ). Proof. It is easy to check that h1 , h2 , h3 ∈ Syz(g1 , . . . , g8 ). We first show that h1 , h2 , h3 generate Syz(g1 , . . . , g8 ), which is equivalent to saying that if (f1 , . . . , f8 ) ∈ R8 satisfies the system of equations, (1) (2)

−X2 f1 + Y f2 − Y a f3 + X0a+d f4 X1 f1 − X2 f2 − Y a f4 − Y a f5 + a

X0a+d f6

X0a+d f8

=

0

=

0

=

0

(3)

−X0 f1 + X1 f2 − Y f7 +

(4)

X1 f3 + X2 f5 − Y f6

=

0

(5)

−X0 f3 + X1 f4 + X2 f7 − Y f8

=

0

(6)

−X0 f4 − X0 f5 + X1 f6 − X1 f7 + X2 f8

=

0

then, there exist h1 , h2 , h3 ∈ R, such that fi = h1 hi1 + h2 hi2 + h3 hi3 , for every 1 ≤ i ≤ 8. Our aim is to find such hi ’s. From equation (4), we see that, X1 f3 ∈ hX2 , Y i, which implies that f3 ∈ hX2 , Y i. There exist p1 , p2 ∈ R, such that, f3 = −X2 p1 − Y p2 . We rewrite (4) as, X2 (f5 − X1 p1 ) = Y (f6 + X1 p2 ).

(1)

10

SENGUPTA

This implies that, Y | f5 − X1 p1 , and and X2 | f6 + X1 p2 . Hence, by Eq. (E.1), there exists p3 ∈ R, such that, f5 = X1 p1 + Y p3 and f6 = −X1 p2 + X2 p3 . Now, X0 (5) + X1 (6) yields (X0 X2 − X12 )f7 + (X1 X2 − X0 Y )f8 = X02 f3 + X0 X1 f5 − X12 f6 =

(X0 X2 − X12 )(−X0 p1 − X1 p2 ) + (X1 X2 − X0 Y )(X0 p2 − X1 p3 ) .

Hence, (X0 X2 − X12 )(f7 + X0 p1 + X1 p2 ) = (X1 X2 − X0 Y )(X0 p2 − X1 p3 − f8 ).

(2)

The polynomials X0 X2 − X12 and X1 X2 − X0 Y are irreducible and they do not devide each other. Therefore, it follows from Eq. (E.2) that, there exists q ∈ R, such that, f7 = −X0 p1 − X1 p2 + q(X1 X2 − X0 Y ) and f8 = X0 p2 − X1 p3 − q(X0 X2 − X12 ). Then plugging the values of f3 , f7 and f8 in (5) gives, f4 = −Y p3 + X2 p2 − q(X22 − X1 Y ). Consider Y a (1) + X0a+d (2), we get (X0a+d X1 − X2 Y a )f1 + (Y a+1 − X0a+d X2 )f2 = Y 2a f3 + X0a+d Y a f5 − X02a+2d f6 =

(X0a+d X1 − X2 Y a )(Y a p1 + X0a+d p2 ) + (Y a+1 − X0a+d X2 )(−Y a p2 + X0a+d p3 ) .

Similar steps as above show that there exists r ∈ R, such that f1 = Y a p1 + X0a+d p2 + r(Y a+1 − X0a+d X2 ) and f2 = −Y a p2 + X0a+d p3 − r(X0a+d X1 − X2 Y a ). Substituting the expressions of f1 , f2 , f3 , f4 in (1), we obtain q = r. Now h1 := p1 + qY , h2 := p2 − qX2 and h3 := p3 − qX1 give us the desired result. Now to show the linear independence of h1 , h2 , h3 , let s1 , s2 , s3 ∈ R such that s1 h1 + s2 h2 + s3 h3 = 0. Then it follows that −X2 s1 − Y s2 = 0,

X2 s2 − Y s3 = 0,

X1 s1 + Y s3 = 0.

It is easy to show that s1 = s2 = s3 = 0. Let us now define the matrices ϑ1 (1) :=



g1 , · · · , g6

t

,

ϑ2 (1) :=



gt1 , · · · , gt8

t

,

ϑ3 (1) :=



ht1 , ht2 , ht3

The proof of Theorem (3.1) for b = 1 now follows from Propositions (3.2) and (3.3).

t

.

MINIMAL FREE RESOLUTION IN A4

11

§3.B Case : b = 2 The set G = {ξ11 } ∪ {ϕ0 , ϕ1 } ∪ {ψ0 } ∪ {θ} is a minimal generating set and a Gr¨obner basis for p. We fix an ordering as : g1 g2 g3 g4 g5

:= := := := :=

ξ11 ϕ0 ϕ1 ψ0 θ

= = = = =

X12 − X0 X2 , X1 X2 − X0 Y , X22 − X1 Y , X2 Y a − X0a+d+1 , Y a+1 − X0a+d X1 .

The following elements of R5 , generate Syz(g1 , . . . , g5 ), by Theorem (2.4). G1 G2 G3 G4 G5 G6 G7 G8 G9 G10

:=



:=



:=



:=



:=



:=



:=



0, X0a+d X1 − Y a+1 , 0, 0, X1 X2 − X0 Y

:=



−X0a+d

:=



:=



−X2 , X1 , −X0 , 0, 0



by (2.4), (2.3)(1),

X1 Y − X22 , 0, X12 − X0 X2 , 0, 0



by (2.4), (2.3)(2),

X0a+d+1 − X2 Y a , 0, 0, X12 − X0 X2 , 0



by (2.4), (2.3)(8)(i),

X0a+d X1



by (2.4), (2.3)(3),

− Y a+1 , 0, 0, 0, X12 − X0 X2

Y , −X2 , X1 , 0, 0



0, −Y a , 0, X1 , −X0

0, 0,

by (2.4), (2.3)(4), 

by (2.4), (2.3)(8)(ii),

, 0, −Y a , X2 , −X1

0, X0a+d X1 − Y a+1 , X0a+d , 0, −Y , X2

0, 



X22

− X1 Y



by (2.4), (2.3)(5), by (2.4), (2.3)(8)(iii),



by (2.4), (2.3)(6), by (2.4), (2.3)(8)(iv).

Let g1 := G1 , g2 := G5 , g3 := G6 , g4 := G8 , g5 := G10 and let g := {gi | 1 ≤ i ≤ 5}. (3.4) Proposition. The R-module Syz(g1 , . . . , g5 ) is minimally generated by the set g. Proof. By Theorem (2.4), the R-module Syz(g1 , . . . , g5 ) is generated by the set {Gi | 1 ≤ i ≤ 10}. Now we note the following identities. G2 G3 G4 G7 G9

= = = = =

X2 G1 + X1 G5 , Y a G1 + X1 G6 − X0 G8 , −Y a G5 + X2 G6 − X1 G8 , Y G6 + X1 G10 , X0a+d G5 + Y G8 + X2 G10 .

The above identities show that Syz(g1 , . . . , g5 ) is generated by the set g. Now we show the minimality of the set g. As in the case b = 1, for every 1 ≤ i ≤ 5, ˆ , . . . , gj(i)5 i. It is easy to check we will find a 1 ≤ j(i) ≤ 5, such that gj(i)i 6∈ hgj(i)1 , . . . , gj(i)i that, we can take j(1) = 1, j(2) = 1, j(3) = 2, j(4) = 1, j(5) = 2. Hence, g is minimal.

12

SENGUPTA

(3.5) Proposition. The element h := Syz(g1 , . . . , g5 ) as a R-module.



−θ , −ψ0 , ϕ1 , −ϕ0 , −ξ11



∈ R5 , generates

Proof. It is easy to check that h ∈ Syz(g1 , . . . , g5 ). To show that h1 generates Syz(g1 , . . . , g5 ), let (f1 , . . . , f5 ) ∈ R8 such that it satisfies the system of equations, (1)

−X2 f1 + Y f2 − X0a+d f4 a

X0a+d f5

=

0

=

0

(2)

X1 f1 − X2 f2 − Y f3 +

(3)

−X0 f1 + X1 f2 − Y a f4

=

0

(4)

X1 f3 + X2 f4 − Y f5

=

0

(5)

−X0 f3 − X1 f4 + X2 f5

=

0

Our aim is to find a h ∈ R, such that, (f1 , . . . , f5 ) = h h. Considering X0 (4) + X1 (5), we get (X0 X2 − X12 )f4 = (X0 Y − X1 X2 )f5 . This implies that, X0 X2 − X12 | f5 , since X0 X2 − X12 is an irreducible polynomial and it does not divide X0 Y − X1 X2 . Hence, there exists h ∈ R, such that f5 = (X0 X2 − X12 )h = −ξ11 h. Now the proof follows easily through the following steps : Step 1 : Consider X1 (4) + X2 (5) and substitute the expression of f5 obtained above. This gives, f3 = (X22 − X1 Y )h = ϕ1 h. Step 2 : In X2 (4) + Y (5), we substitute the expression of f3 to get f4 = (X0 Y − X1 X2 )h = −ϕ0 h. Step 3 : In X1 (1) − Y (3), we substitute the expression of f4 to get f1 = (X0a+d X1 − Y a+1 )h = −θ h. Step 4 : Substitute the expression of f4 in X0 (1) − X2 (3), which gives f2 = (X0a+d+1 − X2 Y a )h = −ψ0 h. Hence the proof. We now define the matrices ϑ1 (2) :=



g1 , · · · , g5

t

,

ϑ2 (2) :=



gt1 , · · · , gt5

t

,

ϑ3 (2) := h1 .

The proof of Theorem (3.1), for b = 2, now follows from Propositions (3.4) and (3.5).

MINIMAL FREE RESOLUTION IN A4

13

§3.C Case : b = 3 In this case, a minimal generating set and a Gr¨obner basis for p is G = {ξ11 } ∪ {ϕ0 , ϕ1 } ∪ {θ}. We fix an ordering as : g1 g2 g3 g4

:= := := :=

ξ11 ϕ0 ϕ1 θ

= = = =

X12 − X0 X2 , X1 X2 − X0 Y , X22 − X1 Y , Y a+1 − X0a+d+1 .

The following elements of R4 , generate Syz(g1 , . . . , g4 ), by Theorem (2.4). G1 G2 G3 G4 G5 G6

:=



:=



:=



:=



:=



:=



−X2 , X1 , −X0 , 0 X1 Y −

X22

, 0,

X12



by (2.4), (2.3)(1),

− X0 X2 , 0



by (2.4), (2.3)(2), 

X0a+d+1 − Y a+1 , 0, 0, X12 − X0 X2 Y , −X2 , X1 , 0 0, 0,

by (2.4), (2.3)(3),



X0a+d+1 − Y a+1 , 0, 0, X0a+d+1 − Y a+1 ,

by (2.4), (2.3)(4), 

X1 X2 − X0 Y X22

− X1 Y



by (2.4), (2.3)(5), by (2.4), (2.3)(6).

Let g1 := G1 , g2 := G3 , g3 := G4 , g4 := G5 , g5 := G6 and let g := {gi | 1 ≤ i ≤ 5}. (3.6) Proposition. The R-module Syz(g1 , . . . , g4 ) is minimally generated by the set g. Proof. By Theorem (2.4), the R-module Syz(g1 , . . . , g4 ) is generated by the set {Gi | 1 ≤ i ≤ 6}. Since G2 = X2 G1 + X1 G4 , therefore, Syz(g1 , . . . , g4 ) is in fact generated by g. To show the minimality of the set g, we will find, for every 1 ≤ i ≤ 5, a 1 ≤ j(i) ≤ 4, ˆ , . . . , gj(i)5 i. It is seen easily that, we can take j(1) = 1 and such that, gj(i)i 6∈ hgj(i)1 , . . . , gj(i)i j(2) = 2. Now note that, g43 = ξ11 , g44 = ϕ0 and g45 = ϕ1 . Since ξ11 , ϕ0 and ϕ1 form a part of a minimal generating set, we can therefore choose j(3) = j(4) = j(5) = 4. Hence, g is minimal. Now, to find a minimal generating set for Syz(g1 , . . . , g5 ), we choose the following elements in R5 , h1 h2

:=



0, −Y , X0a+d+1 − Y a+1 , X2 , −X1

:=



X0a+d+1

−Y

a+1

, X2 , 0, −X1 , X0

 

,

.

(3.7) Proposition. The elements h1 and h2 , form a basis for the R-module Syz(g1 , . . . , g5 ). Proof. It is easy to check that h1 , h2 ∈ Syz(g1 , . . . , g5 ). Let (f1 , . . . , f5 ) ∈ R5 satisfy the system of equations,

14

SENGUPTA

(1)

−X2 f1 + (X0a+d+1 − Y a+1 )f2 + Y f3

=

0

(2)

X1 f1 − X2 f3 + (X0a+d+1 − Y a+1 )f4 −X0 f1 + X1 f3 + (X0a+d+1 − Y a+1 )f5 (X12 − X0 X2 )f2 + (X1 X2 − X0 Y )f4 +

=

0

=

0

=

0

(3) (4)

(X22 − X1 Y )f5

To show that h1 , h2 generate Syz(g1 , . . . , g5 ), we have to find h1 , h2 ∈ R, such that, (f1 , . . . , f5 ) = h1 h1 + h2 h2 . Considering X1 (1) + X2 (2), we get (X22 − X1 Y )f3 = (X0a+d+1 − Y a+1 )(X1 f2 + X2 f4 ). This implies that, X0a+d+1 − Y a+1 | f3 since X0a+d+1 − Y a+1 is an irreducible polynomial and it does not divide X1 Y − X22 . Hence, there exists h1 ∈ R, such that f3 = (X0a+d+1 − Y a+1 )h1 . Plugging the expression of f3 in (1) yields X2 f1 = (X0a+d+1 − Y a+1 )(Y h1 + f2 )

(3)

Hence, it follows that, X2 | Y h1 + f2 , and therefore, there exists h2 ∈ R, such that, f2 = −Y h1 + X2 h2 . Now we follow the steps indicated below. Step 1 : Substitute the expression of f2 in Eq. (E.3). This gives f1 = (X0a+d+1 − Y a+1 )h2 . Step 2 : Substituting the expressions of f1 , f3 in (3), we get f5 = −X1 h1 + X0 h2 . Step 3 : Substitute the expression of f1 , f3 in (2), to get f4 = X2 h1 − X1 h2 . This proves that h1 , h2 generate Syz(g1 , . . . , g5 ). To show the linear independence, let p1 , p2 ∈ R be such that p1 h1 + p2 h2 = 0. This implies −Y p1 + X2 p2 = 0 ,

X2 p1 − X1 p2 = 0 ,

−X1 p1 + X0 p2 = 0 .

Now one can easily solve the above system of equations and get p1 = p2 = 0. Hence the proof. We now define the matrices ϑ1 (3) :=



g1 , · · · , g4

t

,

ϑ2 (3) :=



gt1 , · · · , gt5

t

,

ϑ3 (3) :=



ht1 , ht2

t

.

MINIMAL FREE RESOLUTION IN A4

15

The proof of Theorem (3.1), for b = 3, now follows from Propositions (3.6) and (3.7). This completes the proof of the theorem.

Acknowledgements. The author thanks Prof.D. P. Patil for some useful discussions and comments during the preparation of the manuscript.

References [1] Eisenbud, D. Commutative Algebra with a View Toward Algebraic Geometry; Springer Verlag; New York, 1995. [2] Bayer, D. ; Peeva, I ; Sturmfels, B. Monomial resolutions. Mathematical Research Letters. 1998, 5, 31 - 46. [3] Eliahou, S. ; Kervaire, M. Minimal resolution of some monomial ideal. J. Algebra. 1990, 129(1), 163-194. [4] Lyubeznik, G. A new explicit finite free resolution of ideals generated by monomials in an R-sequence. J. Pure Appl. Alg. 1998, 51, 193 - 195. [5] Bresinsky, H. Minimal free resolutions of monomial curves in P3k . Linear Algebra Appl. 1984, 59, 121 - 129. [6] Peeva, I. ; Sturmfels, B. Syzygies of codimension two lattice ideals. Math. Zeitschrift. 1998, 229, 163-194. [7] Bayer, D. ; Sturmfels, B. Cellular resolutions of monomial modules. J. reine angew. Math. 1998, 502, 123 - 140. [8] Peeva, I. ; Sturmfels, B. Generic lattice ideals. Journal of the AMS. 1998, 11(2), 363 373. [9] Patil, D. P. ; Singh, Balwant. Generators for the derivation modules and the relation ideals of certain curves. Manuscripta Math. 1990, 68, 327-335. [10] Patil, D. P. Minimal sets of generators for the relation ideals of certain monomial curves. Manuscripta Math. 1993, 80, 239-248. [11] Sengupta, I. A Gr¨obner basis for certain affine monomial curves. Preprint, to appear in Communications in Algebra. [12] Maloo, A.K. ; Sengupta, I. Criterion for complete intersection for certain monomial curves. Preprint. [13] Vasconcelos, W. Arithmetic of Blowup Algebras; LMS Lecture Note Series 195, Cambridge University Press; UK, 1994. [14] Bayer, D. ; Stillman, M. Macaulay; A system for computation in algebraic geometry and commutative algebra, Source and object code available at zariski. harvard. edu, 1982 1990.