0. Therefore, in the whole paper we may assume that (♯) as contains at least two non-zero components. This will simplify our calculation. 5
Consider the following system of linear constraints
(5)
yj ≥ a1j x1 + · · · + a(s−1)j xs−1 z ≥ x1 + · · · + xs−1 + 1, y ≥ a x + · · · + a x 1j i1
j
+ asj (z − x1 − · · · − xs−1 − 1), (j = 1, ..., r),
(s−1)j i(s−1)
(i, j z≥
+ asj (z − xi1 − · · · − xi(s−1) ),
= 1, ..., r; j 6= i), xi1 + · · · + xi(s−1) , (i = 1, ..., r), z ≥ 0; y1 ≥ 0, ..., yr ≥ 0; x1 ≥ 0, ..., xs−1 ≥ 0; x11 ≥ 0, ..., xr(s−1) ≥ 0.
For short, we set
u = (u0 , ..., urs+s−1) = (z, y1 , ..., yr , x1 , ..., xs−1 , x11 , ..., xr(s−1) ). By Remark 2.3, a monomial tb ∈ J ∩ I n−1 if and only if the system (5) has an integer solution u∗ such that u∗0 = n, u∗1 = b1 , ..., u∗r = br . The corresponding system of homogeneous linear constraints is
(6)
yj ≥ a1j x1 + · · · + a(s−1)j xs−1 z ≥ x1 + · · · + xs , y ≥ a x + · · · + a x j
(i, j z≥
1j i1
+ asj (z − x1 − · · · − xs−1 ), (j = 1, ..., r),
(s−1)j i(s−1)
+ asj (z − xi1 − · · · − xi(s−1) ),
= 1, ..., r; j 6= i), xi1 + · · · + xi(s−1) , (i = 1, ..., r), z ≥ 0; y1 ≥ 0, ..., yr ≥ 0; x1 ≥ 0, ..., xs−1 ≥ 0; x11 ≥ 0, ..., xr(s−1) ≥ 0.
An integer solution (n, b, x) of this system gives a monomial tb ∈ J ∩I n = I n . Denote the sets of all integer solutions of (5) and (6) by E and S, respectively. Then K[S], K[E] ⊆ K[u] and K[E] is a K[S]-module. Equip K[S] and K[E] with an N-grading by setting deg(uc ) = c0 . Let I be the ideal of K[S] generated by all binomials uα − uβ , such that α0 = β0 , ..., αr = βr . Lemma 2.4. There is an isomorphism of N-graded rings K[S]/I ∼ = R := ⊕n≥0 I n tn . Proof. The above discussion shows that there is an epimorphism of N-graded rings: K[S] → R, uc 7→ tc11 · · · tcrr tc0 .
The kernel of this map is exactly I. The proof is similar to that of Lemma 4.1 in [11], or we can argue directly as follows. By Lemma 1.1, K[S] is generated by a finite number of monomials, say uc1 , ..., ucp . Consider the polynomial ring K[v] of p new variables v = (v1 , ..., vp ). By [11], Lemma 4.1, the kernel of the epimorphism ψ : K[v] → K[S], ψ(vi ) = uci , P
P
is the ideal IA generated by binomials vα − vβ such that pi=1 αi ci = pi=1 βi ci . Such an ideal is called toric ideal associated to the matrix A := {c1 , ..., cp }. Let c′i = (ci0 , ..., cir ) and A′ := {c′1 , ..., c′p }. Again by [11], Lemma 4.1, the kernel of the epimorphism χ : K[v] → R, vi 7→ t1ci1 · · · tcrir tci0 , 6
is IA′ . Clearly IA ⊆ IA′ , ψ(IA′ ) = I. Hence χ induces an isomorphism ϕ : K[S] → R,
such that Ker ϕ = I and ϕ(uci ) = χ(vi ). This implies for all c ∈ S.
ϕ(uc ) = tc11 · · · tcrr tc0
By this isomorphism, we can consider the quotient module K[E]/IK[E] as a module 0 over R. Of course, HmG (G) can be considered as a module over R, too. Lemma 2.5. Let r ≥ 2. Then there is an epimorphism of N-graded modules over R K[E]/IK[E] → ⊕n≥1
J ∩ I n−1 n 0 t = HmG (G). In
Proof. The set M = ⊕n≥1 (J ∩ I n−1 )tn is a module over R and contains the ideal IR. The isomorphism ϕ in the proof of Lemma 2.4 induces a homomorphism K[E]/IK[E] → M, uc 7→ tc11 · · · tcrr tc0 , 0 which is clearly surjective. Since HmG (G) ∼ = M/IR, it is an image of K[E]/IK[E].
Proposition 2.6. Let r ≥ 2 and d be the maximal degree of the generators of I, i.e. 0 d = maxi (ai1 + · · · + air ). Then the R-module HmG (G) is generated by homogeneous elements of degrees less than √ √ B1 := d(rs + s + d)( r)r+1 ( 2d)(r+1)(s−1) . Proof. By Lemma 2.5, it suffices to show that K[E] is generated over K[S] by monomials of degrees less than B1 . The system (5) has rs + s variables. Denote by δ(x) the vector obtained from the coefficient vector of a variable x by deleting already known zero entries. For simplicity we write it in the row form. Then We have
δ(xik ) = (ak1 − as1 , ..., ak(i−1) − as(i−1) , ak(i+1) − as(i+1) , ..., akr − asr , −1). kδ(xik )k2 ≤ 1 + (ak1 − as1 )2 + · · · + (akr − asr )2 ≤ 1 + (a2k1 + · · · + a2kr ) + (a2s1 + · · · + a2sr ) P ≤ 1 + (ak1 + · · · + akr )2 + (as1 + · · · + asr )2 − 2 i0 I n tn .
0 The local cohomology module HR (G) is also a Z-graded R-module. Let + 0 a0 (G) = sup{n| HR (G)n 6= 0}. + 0 (This number is to be taken as −∞ if HR (G) = 0.) It is related to an important invariant + called the Castelnuovo-Mumford regularity of G (see, e.g., [9]). We have
Lemma 2.8. ([6], Proposition 2.4). Ass(I n /I n+1) ⊆ Ass(I n+1 /I n+2 ) for all n > a0 (G). 0 To define HR (G), let us recall the Ratliff-Rush closure of an ideal: +
Ifn = ∪m≥1 I n+m : I m .
This immediately gives
0 Lemma 2.9. For all n > 0 we have HR (G)n−1 ∼ = (Ifn ∩ I n−1 )/I n . +
Recall that I = (ta1 , ..., tas ). Lemma 2.10. For all n > 0 we have Ifn = ∪m≥0 I n+m : (tma1 , ..., tmas ). 8
Proof. Since tmai ∈ I m , the inclusion ⊆ is obvious. To show the inclusion ⊇, let x ∈ I n+m : (tma1 , ..., tmas ). ′
Put m′ = sm and let y be an arbitrary element in I m . Then y = (tmai )y ′ for some i and ′ y ′ ∈ I m −m . We have ′ xy = y ′ (xtmai ) ∈ y ′I n+m ⊆ I n+m . This implies ′ ′ x ∈ I n+m : I m ⊆ Ifn . Proposition 2.11. We have a0 (G) < B2 := s(s + r)4 sr+2 d2 (2d2 )s
2 −s+1
.
Proof. Consider the following system of linear constraints
(7)
yj + aij x ≥ a1j xi1 + · · · + a(s−1)j xi(s−1) z + x ≥ x + · · · + x , i1
(i = 1, ..., s; j z ≥ 0, x ≥ 0;
+ asj (z + x − xi1 − · · · − xi(s−1) ),
i(s−1)
= 1, ..., r), y1 ≥ 0, ..., yr ≥ 0; x11 ≥ 0, ..., xs(s−1) ≥ 0.
By Lemma 2.10 and Remark 2.3 we have tb ∈ Ifn if and only if there is an integer solution u := (u0 , ..., us(s−1)+r+1) := (z, x, y1 , ..., yr , x11 , ..., xs(s−1) )
of (7) such that z = n and b = (u2 , ..., ur+1). This system has s(s − 1) + r + 2 variables. Using the notation in the proof of Proposition 2.6, a straightforward calculation gives kδ(xij )k2 < 2d2 , kδ(yk )k2 = s, kδ(z)k2 < sd2 , kδ(x)k2 < 2sd2.
Let S be the set of all integer solutions of (7). By Lemma 1.1, the ring K[S] is generated by monomials, say uc1 , ..., ucp , with √ √ √ √ r kcj k∗ < (s(s − 1) +√r + 2)( 2d)s(s−1) sd 2sd s √ r+2 2 < [(s + r)2 d( 2d)s −s+1 s ] − 1 =: B3 .
Fix such a generator ucj of K[S]. Let
c′j = (cj2 , ..., cj(r+1) ). Then from (7) we have ′
Since cj1 < B3 , this implies that
(tai )cj1 tcj ∈ I cj0 +cj1 . ′
(tai )B3 tcj ∈ I cj0 +B3 ,
for all i ≤ s. Let
B4 = sB3 .
Since I
sB3
=
s X
(tai )B3 I (s−1)B3 ,
i=1
from the above relationship we get (8)
′
I B4 tcj ⊆ I cj0 +B4 . 9
We will show that Ifn = I n for all n ≥ B4 (B3 + 1). For this aim, let tb ∈ Ifn with n ≥ B4 (B3 + 1).
Then there are α ∈ N and α = (α11 , ..., αs(s−1) ), such that (n, α, b, α) ∈ S. Since c1 , ..., cp generate S, there are nonnegative integers m1 , ..., mp such that (n, α, b, α) = m1 c1 + · · · + mp cp .
This implies
m1 c10 + · · · + mp cp0 = n,
and
′
′
tb = (tc1 )m1 · · · (tcp )mp . Repeated application of (8) gives for all i ≤ s. In other words,
(tai )B4 tb ∈ I n+B4 , tb ∈ I n+B4 : (tB4 a1 , ..., tB4 as ).
Hence there is β = (βij ) ∈ Ns(s−1) such that (n, B4 , b, β) ∈ S. Then one can write (9)
X
(n, B4 , b, β) =
X
m′i ci +
ci1 =0
for some
m′i , m′′i
∈ N. We have
n=
X
m′′i ci ,
ci1 >0
m′i ci0 +
ci1 =0
X
m′′i ci0 .
ci1 >0
Comparing the second components in (9) gives X
X
m′′i ≤
Since ci0 ≤ kci k∗ ≤ B3 , we must have X
m′i ci0 = n −
X
ci1 >0
m′′i ci0 ≥ n − B3
m′′i ci1 = B4 . X
m′′i ≥ B4 (B3 + 1) − B3 B4 = B4 .
In particular, the set of ci with ci1 = 0 in (9) is not empty. From (7) one can see that for ′ such an index i we have tci ∈ I ci0 . Therefore, by (9) one obtains ′
′′
tb = tb tb ,
where ′
and
tb ∈ I ′′
P
tb =
m′i ci0
Y
⊆ I B4 , ′
′′
(tci )mi .
ci1 >0
Repeated application of (8) once more gives us tb ∈ I n . Thus we have shown Since
Ifn = I n .
√ √ r+2 2 B4 (B3 + 1) = sB3 (B3 + 1) < s[(s + r)2 d( 2d)s −s+1 s ]2 ≤ B2 , from Lemma 2.9 we get a0 (G) < B2 . Finally we can prove 10
Theorem 2.12. Let
√ √ 2 B = max{d(rs + s + d)( r)r+1 ( 2d)(r+1)(s−1) , s(s + r)4 sr+2 d2 (2d2 )s −s+1 }.
Then we have
Ass(I n /I n+1 ) = Ass(I B /I B+1 ) for all n ≥ B. Proof. Note that B = max{B1 , B2 }. By Proposition 2.7, Ass(I n /I n+1 ) ⊆ Ass(I B /I B+1 ). By Lemma 2.8 and Proposition 2.11, Ass(I n /I n+1 ) ⊇ Ass(I B /I B+1 ). The number B in the above theorem is very big. However the following examples show that such a number B should depend on d and r. Example 2.13. Let d ≥ 4 and
I = (xd , xd−1 y, xy d−1, y d, x2 y d−2 z) ⊂ K[x, y, z].
A monomial of this type was used in the proof of Theorem 4.1 in [5]. We have I n : (x, y, z)∞ = (xd , xd−1 y, xy d−1, y d, x2 y d−2)n = I n if and only if n ≥ k − 2. Hence Ass(I n−1 /I n ) =
{(x, y, z), (x, y)} {(x, y)}
if n < d − 2, if n ≥ d − 2.
Example 2.14. Let r ≥ 4 and d > r − 3. We put (r−4) (r−3) (r−3) βr−4 d−r+2 tr−2 , u = t1 0 t2 1 · · · tr−30 and v = tβ1 1 · · · tr−3
where
Let
0
if r − 3 − i is even, βi = r−3 2 i if r − 3 − i is odd.
d−r+3 r−3 d−1 r−3 I = (utd1 , utd−1 2 tr , ..., utr−2 tr , utr−1 tr , vtr ), and J be the integral closure I r of I r . Assume that
Ass(J n−1 /J n ) = Ass(J B−1 /I B ) for all n ≥ B. Then
d(d − 1) · · · (d − r + 3) . r(r − 3) Note that in this example J is generated by monomials of degree r(d+2r−3 −1). Thus, if r is fixed, then B is at least O(d(J)r−2), where d(J) is the maximal degree of the generators of J. B≥
Proof. By [13], Corollary 7.60, J is a normal ideal. Using the filtration we get
J n = I rn ⊂ I rn−1 ⊂ · · · ⊂ I r(n−1) = J n−1 ,
Ass(I rn−1 /I rn ) ⊆ Ass(J n−1 /J n ) ⊆ ∪ri=1 Ass(I r(n−1)+i−1 /I r(n−1)+i ). By virtue of [12], Proposition 4, it is shown in the proof of [12], Proposition 16, that m ∈ Ass(I k−1 /I k ) for all k ≫ 0, 11
and
d(d − 1) · · · (d − r + 3) . (r − 3) Hence m ∈ Ass(J n−1 /J n ) for all n ≫ 0. Assume that d(d − 1) · · · (d − r + 3) B< = δ/r. r(r − 3) m 6∈ Ass(I k−1 /I k ) if k < δ :=
Then Br < δ, and hence m 6∈ Ass(I r(B−1)+i−1 /I r(B−1)+i ) for all i ≤ r. This implies m 6∈ Ass(J B−1 /J B ), a contradiction. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
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Institute of Mathematics Hanoi, 18 Hoang Quoc Viet Road, 10307 Hanoi, Vietnam E-mail address: [email protected]
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