arXiv:1801.03054v2 [] 16 Jan 2018

UNBOUNDEDNESS OF BETTI NUMBERS OF CERTAIN
MONOMIAL CURVES IN A4
RANJANA MEHTA, JOYDIP SAHA, AND INDRANATH SENGUPTA
A BSTRACT. Bresinsky defined a class of monomial curves in A4 with
the property that the minimal number of generators or the first Betti number of the defining ideal is unbounded above. We prove that the same behaviour of unboundedness is true for all the Betti numbers and construct
an explicit minimal free resolution for this class.

1. B RESINSKY ’ S E XAMPLES
Let r ≥ 3 and n1 , . . . , nr be positive integers with gcd(n1 , . . . , nr ) = 1.
Let us assume that the numbers n1 , . . . , nr generate the numerical semigroup
r
X
zj nj | zj nonnegative integers}
Γ(n1 , . . . , nr ) = {
j=1

Pr
minimally, that is if ni =
j=1 zj nj for some non-negative integers zj ,
then zj = 0 for all j 6= i and zi = 1. Let η : k[x1 , . . . , xr ] → k[t] be
the mapping defined by η(xi ) = tni , 1 ≤ i ≤ r, where k is a field. Let
p(n1 , . . . , nr ) = ker(η). Let βi (p(n1 , . . . , nr )) denote the i-th Betti number
of the ideal p(n1 , . . . , nr ). Therefore, β1 (p(n1 , . . . , nr )) denotes the minimal number of generators p(n1 , . . . , nr ). For a given r ≥ 3, let βi (r) =
lub(βi (p(n1 , . . . , nr ))), where is lub is taken over all the sequences of positive integers n1 , . . . , nr . Herzog [8] proved that β1 (3) is 3 and it follows
easily that β2 (3) is a finite integer as well. Bresinsky in [1], [2], [3], [4], [5],
extensively studied relations among the generators n1 , . . . , nr of the numerical semigroup defined by these integers. It was proved in [2] and [3] respectively that, for r = 4 and for certain cases in r = 5, the symmetry condition on the semigroup generated by n1 , . . . , nr imposes an upper bound
on the first Betti number β1 (p(n1 , . . . , nr )). This remains an open question
2010 Mathematics Subject Classification. Primary 13C40, 13P10.
Key words and phrases. Monomial curves, Gröbner bases, Betti numbers.
The second author thanks SERB, Government of India for the Research Associate position at IIT Gandhinagar, through the research project EMR/2015/000776.
The third author is the corresponding author. This research is supported by the research
project IP/IITGN/MATH/IS/201415-13, sponsored by IIT Gandhinagar.
1

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RANJANA MEHTA, JOYDIP SAHA, AND INDRANATH SENGUPTA

in general whether symmetry condition on the numerical semigroup generated by n1 , . . . , nr imposes an upper bound on β1 (p(n1 , . . . , nr )), Bresinsky
[1] constructed a class of monomial curves in A4 to prove that β1 (4) = ∞.
He used this observation to prove that β1 (r) = ∞, for every r ≥ 4. Our
aim in this article is to prove in Theorem 1.1 that for Bresinsky’s examples
βi (4) = ∞, for every 1 ≤ i ≤ 3 and also describe all the syzygies explicitly; 4.1, 4.2. The Betti numbers for this class of curves have been calculated
through the description of the tangent cone in [9] and [10]. While they have
constructed a standard basis, we have constructed a Gr”obner basis and have
used that to describe all the syzygies minimally.
Let us first recall Bresinsky’s example of monomial curves in A4 , as
defined in [1]. Let q2 ≥ 4 be even. q1 = q2 + 1, d1 = q2 − 1. Set
n1 = q1 q2 , n2 = q1 d1 , n3 = q1 q2 + d1 , n4 = q2 d1 . It is clear that
gcd(n1 , n2 , n3 , n4 ) = 1. For the rest of the article let us use the shorthand n to denote Bresinsky’s sequence of integers defined above. Bresinsky [1] proved that the set A = A1 ∪ A2 ∪ {g1 , g2} generates the ideal
p(n1 , . . . , n4 ), where A1 = {fµ |fµ = xµ−1
xq32 −µ − xq22 −µ xµ+1
, 1≤µ≤
1
4
µ2 µ3
ν1 ν4
q2 }, A2 = {f |f = x1 x4 − x2 x3 , ν1 , µ3 < d1 } and g1 = xd11 − x2 q2 ,
g2 = x3 x4 − x2 x1 . We prove the following theorem which not only extracts
a subset of the set A which is a minimal generating set for the ideal p(n)
and also a Gröbner basis with respect to a suitable monomial order. The
main theorem we prove is the following:
(q −m)

′

1
Theorem 1.1. Let S = A1 ∪ A2 ∪ {g1 , g2 }, where A′2 = {hm | xm
−
1 x4
(q2 −m) m
x2
x3 , 1 ≤ m ≤ q2 − 2}.
(i) S is a minimal generating set for the ideal p(n);
(ii) S is a Gröbner basis for p(n) with respect to the lexicographic
monomial order induced by x3 > x2 > x1 > x4 on k[x1 , . . . , x4 ];
(iii) β1 (p(n)) =| S |= 2q2 ;
(iv) β2 (p(n)) = 4(q2 − 1);
(v) β3 (p(n)) = 2q2 − 3.

The proof of the theorem is divided into the following two sections.
2. T HE

FIRST

B ETTI

NUMBER

We first construct the set A′2 ⊂ A2 , in order to extract a minimal generating set out of the generating set constructed by Bresinsly [1]. Let q2 ≥ 4
be an even integer. Considering the binomials occurring in the set A2 , we
arrive at the equation
(2.1)

(q22 + q2 )ν1 + (q22 − q2 )ν4 + (1 − q22 )µ2 = (q22 + 2q2 − 1)µ3 ,

UNBOUNDEDNESS OF BETTI NUMBERS

3

where 0 ≤ ν1 , µ3 ≤ q2 −1 and ν4 , µ2 ≥ 0. Now gcd(q22 −q2 , 1−q22 ) = q2 −1,
hence
q2 − 1 | (q22 + 2q2 − 1)µ3 − (q22 + q2 )ν1 ,
which implies that
q2 − 1 | [(q22 + q2 ) + (q2 − 1)]µ3 − (q22 + q2 )ν1 .
Therefore,
q2 − 1 | (q22 + q2 )(µ3 − ν1 ).
Again gcd(q2 − 1, q22 + q2 ) = 1 (as q2 ≥ 4 is an even integer), so we have
q2 − 1 | µ3 − ν1 .
We have 0 ≤ µ3 , ν1 < q2 − 1, therefore only possibility is
ν1 = µ3 = ℓ, 0 ≤ ℓ ≤ q2 − 2.
Consider the equations
(2.2)

(q22 − q2 )ν4 + (1 − q22 )µ2 = (q22 + 2q2 − 1 − q22 − q2 )ℓ

(2.3)

q2 ν4 − (q2 + 1)µ2 = ℓ

Therefore the solution is given by
ν4 = (1 + q2 )t − ℓ,

where

t∈Z

and

ν4 ≥ 0

µ2 = q2 t − ℓ, where t ∈ Z and µ2 ≥ 0.
Now ν4 ≥ 0 implies that (1 + q2 )t ≥ ℓ. Again µ2 ≥ 0 implies that
q2 t ≥ ℓ. When ℓ = 0 then t ≥ 0 and when 1 ≤ ℓ ≤ q2 − 2 then t ≥ 1.
Therefore the solution is
ν4 = (1 + q2 )t,
ν4 = (1+q2 )t−ℓ,

µ2 = q2 t,

µ2 = q2 t−ℓ,

when
when

ℓ = 0 (and
1 ≤ ℓ ≤ q2 −2

t ≥ 0);
(and

t ≥ 1).

Lemma 2.1. Let us consider the polynomials α1 = x4q2 +1 − xq22 , α2 =
′
xq42 +3 − x2 xq32 −1 . Then α1 , α2 ∈ hSi, where S = A1 ∪ {g1 , g2 } ∪ A2 .
Proof. We have fq2 = xq12 −1 − xq42 +1 ∈ A1 . It is clear that
α1 = (xq12 −1 − xq22 ) − (xq12 −1 − xq42 +1 ) = g1 − fq2 .
Again
α2 = x24 (xq12 −1 − xq22 ) − x24 (xq12 −1 − xq42 +1 ) − x2 (xq32 −1 − xq22 −1 x24 )
= x24 · fq2 − x2 · f1 + x24 · g1 

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RANJANA MEHTA, JOYDIP SAHA, AND INDRANATH SENGUPTA
′

Theorem 2.2. The ideal p(n) is generated by the set S = A1 ∪{g1 , g2 }∪A2 .
Proof. By Bresinsky’s result, the set A1 ∪ A2 ∪ {g1 , g2 } generates the ideal
′
P . It is evident that A2 ⊂ A2 . Therefore it is enough to show that every
element of A2 is in the ideal generated by S.
We divide the proof into two cases.
Case 1: ℓ = 0.
(1+q2 )t

Let Ht = x4

− xq22 t , where t ≥ 0. First note that H0 = 0 and
(q +1)

Hn+1 = x4 2

(q +1)

Hn + xq22 t (x4 2

− xq22 ).

By induction hypothesis and the Lemma 2.1, we get Ht ∈ hSi for all t ≥ 0.
Case 2: 1 ≤ ℓ ≤ q2 − 2.
(1+q2 )t−l

Fix ℓ and let Hlt = xl1 x4

− xq22 t−l xl3 , for t ≥ 1. Then

(1+q2 )−l

Hl1 = xl1 x4

− xq22 −l xl3 ∈ A2
′

and
(q +1)

Hl(n+1) = x4 2

(q +1)

Hln + xq22 t−l xl3 (x4 2

− xq22 ).

By induction hypothesis and the Lemma 2.1 Hlt ∈ hSi for all t ≥ 1.



′

Theorem 2.3. The set S = A1 ∪ {g1 , g2 } ∪ A2 is a minimal generating for
the ideal p(n).
Proof. By lemma 4 of [1] we have A1 as a part of minimal generating set of
ideal p(n). Now it is enough to prove that any element r ∈ T := {g1 , g2 } ∪
′
A2 can not be expressible as the polynomial combination of S \ {r}. Define
maps π1 : k[x1 , x2 , x3 , x4 ] −→ k[x2 , x3 , x4 ], such that π1 (x1 ) = 0 and
π1 (xi ) = xi , for every i = 2, 3, 4. We also define π2 : k[x1 , x2 , x3 , x4 ] −→
k[x1 , x3 , x4 ], such that π2 (x2 ) = 0 and π1 (xi ) = xi , for every i = 1, 3, 4.
P2
P 2 −2
Bm hm , where Aµ , α2 , Bm ∈
Let r = g1 = qµ=1
Aµ fµ + α2 g2 + qm=1
k[x1 , x2 , x3 , x4 ], for 1 ≤ µ ≤ q2 and 1 ≤ m ≤ q2 − 2. Applying π1 on both
side of the above equation we get,
−xq22

=

q2
X

q2 −2

π1 (Aµ )·(−xq22 −µ xµ+1
)+π1 (α2 )·(x3 x4 )+
4

µ=1

which is absurd.

X

m=1

π1 (Bm )·(−xq22 −m xm
3 ),

UNBOUNDEDNESS OF BETTI NUMBERS

5

P2
P 2 −2
Bm hm , where Aµ , α1 , Bm ∈
Let r = g2 = qµ=1
Aµ fµ + α1 g1 + qm=1
k[x1 , x2 , x3 , x4 ] for 1 ≤ µ ≤ q2 and 1 ≤ m ≤ q2 − 2. Applying π2 on the
above equation we get,
x3 x4 =

q2
X

q2 −2

π2 (Aµ )·(xµ−1
xq32 −µ )+π2 (α1 )·(xq12 −1 )+
1

X

q2 +1−m
π2 (Bm )·(xm
),
1 x4

m=1

µ=1

which is absurd.
P2
P 2 −2
′
Let r ∈ A2 , given by r = hm′ = qµ=1
Aµ fµ + α1 g1 + qm=1,m6
=m′ Bm hm .
Applying π2 on the above expression we get,
′ q +1−m′
2
xm
1 x4

=

q2
X

π2 (Aµ ) · (xµ−1
xq32 −µ ) + π2 (α1 ) · (xq12 −1 )
1

µ=1

q2 −2

+ π2 (α2 ) · (x3 x4 ) +

X

q2 +1−m
π2 (Bm ) · (xm
).
1 x4

m=1,m6=m′

Since m′ < (q2 − 1), the above expression is absurd.



Lemma 2.4. β1 (p(n)) =| S |= 2q2 .
′

Proof. It is very easy to observe that the sets A1 , {g1 , g2} and A2 are mu′
tually disjoint. Therefore | S |=| A1 | + | {g1 , g2} | + | A2 |=
q2 + 2 + (q2 − 2) = 2q2 .

3. T HE

SECOND

B ETTI

NUMBER

Theorem 3.1. Consider the lexicographic monomial order induced by x3 >
x2 > x1 > x4 in k[x1 , x2 , x3 , x4 ]. Then,
(i) The set S forms Gröbner basis for the ideal p(n) with respect to the
above monomial order.
(ii) Let D denotes the set of all Schreyer tuples obtained from Gröbner basis S which generate the first syzygy module (see Theorem 1.43 [7]).
Then each entry in the elements of D is either a non-constant polynomial or zero.
′

Proof. At first we order the set S = A1 ∪ {g1 , g2 } ∪ A2 as follows,
(f1 , . . . , fq2 , g1, g2 , h1 , . . . , hq2 −2 ).
Let f, g ∈ S. We consider the S-polynomials S(f, g) and divide the proof
into cases based on the sets f and g belonging to.
Case 1. f, g ∈ A1 .

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RANJANA MEHTA, JOYDIP SAHA, AND INDRANATH SENGUPTA

1(a). f = fµ , g = fµ+1 where, 1 ≤ µ ≤ q2 − 1. We have
S(fµ , fµ+1 ) = x1 · fµ − x3 · fµ+1
= (xq22 −1−µ xµ+1
) · g2 −→S 0.
4
Therefore, the set
T1 = {β µ = (β(µ,1) , · · · , β(µ,2q2 ) ) | 1 ≤ µ ≤ q2 − 1}
give the Schreyer tuples, where
β(µ,µ) = −x1
β(µ,µ+1) = x3
q −(µ+1)

β(µ,q2 +2) = x22
xµ+1
4
β(µ,i) = 0, for i ∈
/ {µ, µ + 1, q2 + 2}.
1(b). f = fµ , g = fµ′ where, µ′ > µ + 1. We have
′

S(fµ , fµ′ ) = x1 µ −µ · fµ − xµ3 −µ · fµ′
′

′

= x2q2 −µ xµ+1
((x3 x4 )µ −µ−1 + (x3 x4 )µ −µ−2 (x1 x2 ) + · · · + (x1 x2 )µ −µ−1 ) · g2
4
−→S 0
′

′

′

Therefore, the set
T2 = {γ µµ′ = (γ(µµ′ ,1) , . . . , γ(µµ′ ,i) , . . . , γ(µµ′ ,2q2 ) )|1 ≤ µ ≤ q2 −2, µ+2 ≤ µ′ ≤ q2 }
give the Schreyer tuples, where
′

γ(µµ′ ,µ) = −x1 (µ −µ)
′

γ(µµ′ ,µ′ ) = x3 (µ −µ)
′

γ(µµ′ ,q2 +2) = x2q2 −µ xµ+1
((x3 x4 )µ −µ−1 + (x3 x4 )µ −µ−2 (x1 x2 ) + · · · + (x1 x2 )µ −µ−1 )
4
γ(µµ′ ,i) = 0 for i ∈
/ {µ, µ′, (q2 + 2)}.
′

′

′

Case 2. f ∈ A1 , g ∈ {g1 , g2 }
2(a). Let f = fµ , g = g1 , where 1 ≤ µ ≤ q2 . We have
xq32 −µ · g1
S(fµ , g1 ) = x2q2 · fµ + xµ−1
1
= xq12 −1 · fµ + xq22 −µ xµ+1
· g1
4

−→S 0

Therefore, the set
T3 = {γ µ : γµ = (γ(µ,1) , . . . , γ(µ,i) , . . . , γ(µ,2q2 ) ) | 1 ≤ µ ≤ q2 − 1}

UNBOUNDEDNESS OF BETTI NUMBERS

give us the Schreyer tuples, where
γ(µ,µ) = xq12 −1 − xq22
γ(µ,q2 +1) = xq22 −µ xµ+1
− xµ−1
xq32 −µ
4
1
γ( µ, i) = 0, for i ∈
/ {µ, (q2 + 1)}, 1 ≤ i ≤ 2q2 .
2(b). Let f = fµ , g = g2 , where 1 ≤ µ ≤ q2 − 1. We have
q −(µ+1)

S(fµ , g2 ) = x4 · fµ − xµ−1
x32
1
= x2 · fµ+1 −→S 0

· g2

Therefore, the set
T4 = {αµ = (α(µ,1) , . . . , α(µ,i) , . . . , α(µ,2q2 ) ) | 1 ≤ µ ≤ q2 − 1}
give us the Schreyer tuples, where
α(µ,µ) = −x4
α(µ,µ+1) = x2
q −(µ+1)

x32
α(µ,q2 +2) = xµ−1
1
α(µ,i) = 0, for i ∈
/ {µ, µ + 1, (q2 + 2)}, 1 ≤ i ≤ 2q2 .
2(c). Let f = fq2 , g = g2 . We have
S(fq2 , g2 ) = x3 x4 · fq2 − xq12 −1 · g2
= x1 x2 · fq2 − xq42 +1 · g2

−→S 0

Therefore, the set
T5 = {α = (α1 , . . . , αi , . . . , α2q2 )}
give us the Schreyer tuples, where
αq2 = x1 x2 − x3 x4
α(q2 +2) = xq12 −1 − xq42 +1
αi = 0, for i ∈
/ {q2 , (q2 + 2)}, 1 ≤ i ≤ 2q2 .
Case 3. Let f = g1 , g = g2 . We have
S(g1, g2 ) = −x3 x4 · g1 − xq22 · g2
= −x1 x2 · g1 − xq12 −1 · g2

−→S 0

Therefore, the set
T6 = {β = (β1 , . . . , βi , . . . , β2q2 )}

7

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RANJANA MEHTA, JOYDIP SAHA, AND INDRANATH SENGUPTA

give us the Schreyer tuples, where
βq2 +1 = x3 x4 − x1 x2
βq2 +2 = xq22 − xq12 −1
βi = 0, for i ∈
/ {(q2 + 1), (q2 + 2)}, 1 ≤ i ≤ 2q2 .
′

Case 4. f ∈ {g1 , g2 }, g ∈ A2 .
4(a). Let f = g1 , g = h1 . We have
S(g1, h1 ) = −x3 · g1 + x2 · h1
= −x1 · fq2 −1 −→S 0
Therefore, the set
T7 = {γ = (γ1 , . . . , γi , . . . , γ2q2 )}
give us the Schreyer tuples, where
γq2 −1
γq2 +1
γq2 +3
γi

=
=
=
=

−x1
x3
−x2
0, for

i∈
/ {(q2 − 1), (q2 + 1), (q2 + 3)}.

4(b). Let f = g1 , g = hm , with 1 < m ≤ (q2 − 2). We have
S(g1 , hm ) = −x3 m · g1 + x2 m · hm
= −x1 m · fq2 −m −→S 0
Therefore, the set
′
′
′
T8 = {α′m = (α(m,1)
, . . . , α(m,i)
, . . . , α(m,2q
|1 < m ≤ q2 − 2}
2)

give us the Schreyer tuples, where
′
α(m,q
= −xm
1
2 −m)
′
α(m,q
= xm
3
2 +2)
′
α(m,q
= −xm
2
2 +2+m)
′
α(m,i)
= 0,

for i ∈
/ {q2 − m, q2 + 1, q2 + 2 + m}.

4(c). Let f = g2 , g = h1 . We have
S(g2 , h1 ) = xq22 −1 · g2 + x4 · h1
= x1 · g1 − x1 · fq2 −→S 0

UNBOUNDEDNESS OF BETTI NUMBERS

Therefore, the set
′
T9 = {α′ = (α1′ , . . . , αi′ , . . . , α2q
)}
2

give us the Schreyer tuples, where
αq′ 2 = −x1
′
α(q
= x1
2 +1)
(q −1)

′
α(q
= −x2 2
2 +2)
′
α(q
= −x4
2 +3)

αi′ = 0,

for i ∈
/ {q2 , q2 + 1, q2 + 2, q2 + 3}.

4(d). Let f = g2 , g = hm , where 1 < m ≤ q2 − 2. We have
· g2 − x4 · hm
S(g2 , hm ) = xq22 −m xm−1
3
= −x1 · hm−1 −→S 0
Therefore, the set
′
′
′
T10 = {β ′m = (β(m,1)
, . . . , β(m,i)
, . . . , β(m,2q
)|1 < m ≤ q2 − 2)}
2)

give us the Schreyer tuples, where
′
= xq22 −m xm−1
β(m,q
3
2 +2)
′
β(m,q
= −x1
2 +m+1)
′
β(m,q
= x4
2 +2+m)
′
β(m,i)
= 0,

for

i∈
/ {(q2 + 2), (q2 + m + 1), (q2 + 2 + m)}.

′

Case 5. f, g ∈ A2
5(a) Let f = hm and g = hm+1 . We have
S(hm , hm+1 ) = −x3 · hm + x2 · hm+1
q2 −m
= xm
· g2
1 x4

−→S 0

Therefore the set
′
′
′
T11 = {γ ′m = (γ(m,1)
, . . . , γ(m,i)
, . . . , γ(m,2q
) | 1 ≤ m ≤ q2 − 3}
2)

give us the Schreyer tuples, where
q2 −m
′
= −xm
γ(m,q
1 x4
2 +2)
′
γ(m,q
= x3
2 +2+m)
′
γ(m,q
= −x2
2 +3+m)
′
γ(m,i)
= 0,

for

i∈
/ {q2 + 2, q2 + m + 2, q2 + m + 3}.

9

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RANJANA MEHTA, JOYDIP SAHA, AND INDRANATH SENGUPTA

5(b). Let f = hm and g = hm′ where m′ > m + 1. We have
′

′

−m
−m
· hm + xm
· hm′
S(hm , hm′ ) = −xm
2
3
′

q2 +1−m
((x3 x4 )m −m−1 + (x3 x4 )m −m−2 (x1 x2 ) + · · · + (x1 x2 )m−m −1 ) · g2
= xm
1 x4
−→S 0
′

′

′

Therefore, the set
′
′
′
T12 = {α′mm′ = (α(mm
′ ,1) , . . . , α(mm′ ,i) , . . . , α(mm′ ,2q ) ) |1 ≤ m ≤ q2 − 4,
2

m + 2 ≤ m′ ≤ q2 − 2}
give us the Schreyer tuples, where
′

′
−m
α(mm
= xm
′ ,m)
3
′

′
−m
α(mm
= −xm
′ ,m′ )
2
′

q2 +1−m
′
α(mm
= xm
((x3 x4 )m −m−1 + (x3 x4 )m −m−2 (x1 x2 ) + · · · + (x1 x2 )m−m −1 )
′ ,q +2)
1 x4
2
′
= 0,
α(mm
′ ,i)

′

′

i∈
/ {m, m′ , q2 + 2}.

for
′

Case 6. f ∈ A and g ∈ A2
6(a). Let f = f1 , g = hq2 −2 . We have
S(f1 , hq2 −2 ) = x22 · f1 + x3 · hq2 −2
= xq12 −2 x24 · g2 + x2 x24 · g1

−→S 0

Therefore, the set
′
T13 = {β ′ = (β1′ , . . . , βi′ , . . . , β2q
)}
2

give us the Schreyer tuples, where
β1′ = −x22
′
β(q
= x2 x24
2 +1)
′
β(q
= xq12 −2 x24
2 +2)
′
β2q
= −x3
2

βi′ = 0,

for i ∈
/ {1, (q2 + 1), (q2 + 2), 2q2 }, 1 ≤ i ≤ 2q2 .

6(b). Let f = f2 , g = hq2 −2 . We have
S(f2 , hq2 −2 ) = x22 · f2 + x1 · hq2 −2
= x34 · g1

−→S 0

′

UNBOUNDEDNESS OF BETTI NUMBERS

11

Therefore, the set
′
T14 = {γ ′ = (γ1′ , . . . , γi′ , . . . , γ2q
)}
2

give us the Schreyer tuples, where
γ2′ = −x22
′
γ(q
= x34
2 +1)
′
γ(2q
= −x1
2)

γi′ = 0,

for

i∈
/ {2, (q2 + 1), (2q2 )}, 1 ≤ i ≤ 2q2 .

6(c). Let f = fµ and g = hm , (µ, m) 6= (1, q2 −2) and (µ, m) 6= (2, q2 −2).
(i) µ + m < q2 . We have
S(fµ , hm ) = xq22 −m · fµ + xµ−1
x3q2 −µ−m · hm
1
= xq42 +1−m · fµ+m − x2q2 −µ−m xµ+1
· (fq2 − g1 )
4

−→S 0

Therefore, the set
′
′
′
T15 = {β ′µm = (β(µm,1)
, . . . , β(µm,i)
, . . . , β(µm,2q
) |1 ≤ µ ≤ q2 − 1,
2)

1 ≤ m < q2 − µ}
give us the Schreyer tuples, where
′
β(µm,µ)
= −xq22 −m
′
β(µm,µ+m)
= xq42 +1−m
′
β(µm,q
= −x2q2 −m−µ xµ+1
4
2)
′
β(µm,q
= x2q2 −m−µ xµ+1
4
2 +1)
′
β(µm,q
= −xµ−1
x3q2 −µ−m
1
2 +2+m)
′
β(m,i)
= 0,

for i ∈
/ {µ, µ + m, q2 , (q2 + 1), (q2 + 2 + m)}.

(ii) Let µ + m = q2 . We have
S(fµ , hm ) = xq22 −m · fµ + xµ−1
· hm
1
· g1
= xµ+1
4

−→S 0

Therefore, the set
′
′
′
T16 = {γ ′µm = (γ(µm,1)
, . . . , γ(µm,i)
, . . . , γ(µm,2q
) | 1 ≤µ ≤ q2 − 1,
2)

m = q2 − µ}

12

RANJANA MEHTA, JOYDIP SAHA, AND INDRANATH SENGUPTA

give us the Schreyer tuples, where
′
γ(µm,µ)
= −xq22 −m
′
γ(µm,q
= xµ+1
4
2 +1)
′
γ(µm,2q
= −xµ−1
1
2 +2−µ)
′
γ(µm,i)
= 0,

for

i∈
/ {µ, (q2 + 1), (2q2 + 2 − µ)}.

(iii) Let µ + m > q2 . We have
2
S(fµ , hm ) = xq22 −m xµ+m−q
· fµ + xµ−1
· hm
3
1
2 q2 +1−m
· fq2
· hµ+m−q2 + xµ+m−q
x4
= xµ+1
4
1

−→S 0

Therefore, the set
′′
′′
′′
T17 = {α′′µm = (α(µm,1)
, . . . , α(µm,i)
, . . . , α(µm,2q
) |1 ≤ µ ≤ q2 − 1,
2)

1 ≤ m < q2 − µ}
give us the Schreyer tuples, where
′′
2
α(µm,µ)
= −xq22 −m xµ+m−q
3
′′
α(µm,q
= x1µ+m−q2 xq42 +1−m
2)
′′
′′
α(µm,µ+m+2)
= xµ+1
, α(µm,q
= −xµ−1
4
1
2 +2+m)
′′
α(m,i)
= 0,

for

i∈
/ {µ, q2 , (q2 + 2 + m), (µ + m + 2)}.

The set D = ∪17
i=1 Ti give us all the Schreyer tuples which form the gen
erating set for the first syzygy module Syz(p(n)).

Theorem 3.2. Let T = T1 ∪T4 ∪T7 ∪T9 ∪T10 ∪T11 ∪T13 ∪T14 . Let M1 denote the first syzygy module Syz(p(n)). Then, the set T ⊂ M1 /mM1 is linearly independent over the field R/m = k, where m = hx1 , x2 , x3 , x4 i. The
set T is a minimal generating set for the first syzygy module and β2 (p(n)) =
4(q2 − 1).
Proof. The module M1 is generated by D, therefore mM1 must be generated
by
L = x1 D ∪ x2 D ∪ x3 D ∪ x4 D.
It follows from the construction of D that each component of elements of L
is either a monomial of total degree greater than one or zero. Now, to show

UNBOUNDEDNESS OF BETTI NUMBERS

13

that T is linearly independent in the vector space M1 /mM1 over the field
R/m = k, we consider the element v ∈ mM1 given by
X

aµ β µ +

X

bm γ ′m + c1 α′ +

m=1

µ=1

+

q2 −2

q2 −3

q2 −1

v=

qX
2 −1

X

cm β ′m

m=2

d µ α µ − l 1 γ + l2 β ′ + l 3 γ ′

µ=1

where aµ , bm , cm , dµ , lm ∈ k. The linear term in the first coordinate v1 is
a1 (−x1 ) + d1 (−x4 ). Each component of elements of L being a monomial
of total degree greater than one or zero, we have a1 (−x1 ) + d1 (−x4 ) = 0,
that is a1 = d1 = 0. Now, the linear term in the i-th coordinate vi , for
1 < i < q2 − 1 is ai−1 (x3 ) + ai (−x1 ) + di−1(x2 ) + di (−x4 ). Once again, by
the same argument we must have ai = di = 0 for 1 ≤ i < q2 − 1. Similarly,
the linear term in vq2 +1 is c1 (x1 ) + l1 (−x3 ). Therefore, for similar reasons
we must have c1 = l1 = 0. We now compute linear term in vq2 −1 and
obtain aq2 −1 (−x1 )+dq2 −1 (−x4 ). Equating this linear term to zero we obtain
P 2 −3
P 2 −2
aq2 −1 = dq2−1 = 0. It turns out that, v = qm=1
bm γ ′m + qm=2
cm β ′m +
l 2 β ′ + l3 γ ′ .
We now compute the linear term in vq2 +3 and obtain b1 x3 + c2 (−x1 ).
Equating this to zero we get b1 = c2 = 0. The linear term in vq2 +2+i , for
1 < i ≤ q2 −3 is bi−1 (−x2 ) + bi x3 + ci x4 + ci+1 (−x1 ). Therefore, bi = ci =
0, for 1 < i < q2 − 2. Finally computing linear term in v2q2 and equating it
to zero we obtain bq2 −3 (−x2 )+cq2 −2 x4 +l2 (−x3 )+l3 (−x1 ) = 0. Therefore,
bq2 −3 = cq2 −2 = l2 = l3 = 0. This proves that the set T ⊂ M1 /mM1 is
linearly independent.
We have T as a part of a minimal generating set for the first syzygies
module, and cardinality of T is
(q2 − 1) + (q2 − 3) + (q2 − 2) + (q2 − 1) + 3 = 4(q2 − 1.)
By Theorem 8.1 [10], the second betti number in the resolution of R/p(n)
is 4(q2 − 1), hence T is a minimal generating set for first syzygy module.
Therefore, β2 (p(n)) =| T |= 4(q2 − 1).

Theorem 3.3. β3 (p(n)) = 2q2 − 3.
Proof. We know that R/p(n) ∼
= k[tn1 , tn2 , tn3 , tn4 ] is a one dimensional integral domain and therefore depth(R/p(n)) = 1. By the Auslander Buchsbaum theorem, projdimR (R/p(n)) = 3 and we have
1 − 2q2 + β2 (p(n)) − β3 (p(n)) = 0.
Therefore, by 3.2, we have β3 (p(n)) = 2q2 − 3.



14

RANJANA MEHTA, JOYDIP SAHA, AND INDRANATH SENGUPTA

4. T HE

SECOND SYZYGY AND A MINIMAL FREE RESOLUTION

Let us order the generating vectors of second syzygy, and consider the
matrix
i
h
′
′
′
′
′
′
′
N = β 1 . . . β q2 −1 | γ 1 . . . γ q2 −3 | α | β 2 . . . β q2 −2 | α1 . . . αq2 −1 | −γ | β | γ

2q2 ×4(q2 −1)

We consider the following sets of vectors.
(i) H1 = {δ µ = (δ(µ,1) , . . . δ(µ,4(q2 −1)) ) | 1 ≤ µ ≤ q2 − 2}, where
δ(µ,µ) =
δ(µ,µ+1) =
δ(µ,(3q2 −6+µ)) =
δ(µ,(3q2 −5+µ)) =
δ(µ,i) =

x4
−x2
−x1
x3
0, for

i∈
/ {µ, µ + 1, (3q2 − 6 + µ), (3q2 − 5 + µ)}.

(ii) H2 = {ξ = (ξ1 , . . . , ξ4(q2 −1) }, where
ξ1 =
ξ2q2 −3 =
ξ4q2 −7 =
ξ4q2 −6 =
ξ4q2 −5 =
ξ4q2 −4 =
ξi =

x22
x2 x24
x1 x24
x34
−x1
x3
0, for

i∈
/ {1, (2q2 − 3), (4q2 − 7), (4q2 − 6), (4q2 − 5), (4q2 − 4)}.

(iii) H3 = {ζ = (ζ1 , . . . , ζ4(q2 −1) )}, where
ζq2 −1
ζq2
ζ2q2 −3
ζ2q2 −2
ζ4q2 −6
ζi

=
=
=
=
=
=

x1
x4
x3
x2
x1
0,

for

i∈
/ {(q2 − 1), q2 , (2q2 − 3), (2q2 − 2), (4q2 − 6)}.

UNBOUNDEDNESS OF BETTI NUMBERS

15

(iv) H3 = {η = (η1 , . . . , η4(q2 −1) )}, where
η2q2 −4
η3q2 −6
η3q2 −5
η4q2 −5
η4q2 −4
ηi

=
=
=
=
=
=

−x1
−x3
x22
−x4
x2
0, for

i∈
/ {(2q2 − 4), (3q2 − 6), (3q2 − 5), (4q2 − 5), (4q2 − 4)}.

(v) H4 = {κµ = (κ(µ,1) , . . . , κ(µ,4(q2 −1)) ) | 1 ≤ µ ≤ q2 − 4}, where
η(µ,(q2 −1+µ))
η(µ,q2 +µ)
η(µ,(2q2 −3+µ))
η(µ,(2q2 −2+µ))
η(µ,i)

=
=
=
=
=

−x1
x4
−x3
x2
0, for i ∈
/ {(q2 − 1 + µ), (q2 + µ), (2q2 − 3 + µ), (2q2 − 2 + µ)}.

Theorem 4.1. Suppose H = H1 ∪ H2 ∪ H3 ∪ H4 . Then H is a minimal
generating set of second syzygy module of p(n).
Proof. We consider the matrix,
P = [δ 1 . . . δ q2 −2 | ξ | ζ | η | κ1 . . . κq2 −4 ]4(q2 −1)×2q3 −3 .
It is easy to check that N · P = 0. Therefore, elements of H are elements
of the second syzygy module. Let M2 denote the second syzygy module
of p(n). We claim that, H ⊂ M2 /mM2 is linearly independent set, where
m = hx1 , x2 , x3 , x4 i. We proceed in the same way as in 3.2, considering the
expression
q2 −4

q2 −2

X
µ=1

pµ δ µ + qξ + wζ + sη +

X

tj κj = u (say),

j=1

where pµ , q, w, s, tj ∈ k for 1 ≤ µ ≤ q2 − 2, 1 ≤ j ≤ q2 − 4 and we
compute linear terms in each coordinate of u. If we compute linear term
in u1 we get, p1 x4 , hence p1 = 0. Next we compute linear term in u2
and we get −p1 x2 + p2 x4 . We have −p1 x2 + p2 x4 = 0, hence p2 = 0.
Proceeding like this, we observe that all coefficients are zero and H is a part
of minimal generating set of M2 . Since the third betti number is 2q2 − 3
and | H |= 2q2 − 3, therefore H is a minimal generating set of the second
syzygy module.


16

RANJANA MEHTA, JOYDIP SAHA, AND INDRANATH SENGUPTA

Corollary 4.2. A minimal free resolution for the ideal p(n) in the polynomial ring R = K[x1 , x2 , x3 , x4 ] is
P

N

0 −→ R2q2 −3 −→ R4(q2 −1) −→ R2q2 −→ R −→ R/p(n) −→ 0,
where
P = [δ 1 . . . δ q2 −2 | ξ | ζ | η | κ1 . . . κq2 −4 ]4(q2 −1)×2q3 −3
and
i
h
′
′
′
′
′
′
′
N = β 1 . . . β q2 −1 | γ 1 . . . γ q2 −3 | α | β 2 . . . β q2 −2 | α1 . . . αq2 −1 | −γ | β | γ

2q2 ×4(q2 −1)

R EFERENCES
[1] H. Bresinsky, On Prime Ideals with Generic Zero xi = tni , Proceedings of the American Mathematical Society, Vol.47, No.2, february 1975.
[2] H. Bresinsky, Symmetric semigroups of integers generated by 4 elements, Manuscripta
Math. 17, (205-219)1975.
[3] H. Bresinsky, Monomial Gorenstein Ideals, Manuscripta Math. 29, (159-181)1979.
[4] H. Bresinsky, Binomial generating sets for monomial curves with applications in A4 ,
Rend. Sem. Mat. Univers. Politecn. Torino, Vol. 46◦ , (3) 1988.
[5] H. Bresinsky and L. T. Hoa, Minimal generating sets for a family of monomial curves
in A4 , Commutative algebra and algebraic geometry (Ferrara), Lecture Notes in Pure
and Appl. Math. 206 (Dekker, New York, 1999) 5–14.
[6] Decker, W.; Greuel, G.-M.; Pfister, G.; Schönemann, H.: S INGULAR 4.0.1 — A
computer algebra system for polynomial computations. http://www.singular.uni-kl.de
(2014).
[7] Decker, W., Lossen, C., Computing in Algebraic Geometry A Quick Start using SINGULAR, Hindustan Book Agency, October 2005.
[8] J. Herzog, Generators and relations of abelian semigroups and semigroup rings,
Manuscripta Mathematica, Vol. 2, No. 3, (1970), 175-193.
[9] J. Herzog, Dumitru I. Stamate, On the defining equations of the tangent cone of a numerical semigroup ring, Journal of Algebra, Volume 418, 15 November 2014, Pages
8-28.
[10] Dumitru I. Stamate, Betti numbers for numerical semigroup rings,
arXiv:1801.00153v1 .

Discipline of Mathematics, IIT Gandhinagar, Palaj, Gandhinagar, Gujarat 382355,
INDIA.
E-mail address: ranjssj16@gmail.com

Discipline of Mathematics, IIT Gandhinagar, Palaj, Gandhinagar, Gujarat 382355,
INDIA.
E-mail address: saha.joydip56@gmail.com

Discipline of Mathematics, IIT Gandhinagar, Palaj, Gandhinagar, Gujarat 382355,
INDIA.
E-mail address: indranathsg@iitgn.ac.in

.

