• 沒有找到結果。

In this section we shall determine the Hilbert-Kunz function of the hypersurface of the following form : f

N/A
N/A
Protected

Academic year: 2021

Share "In this section we shall determine the Hilbert-Kunz function of the hypersurface of the following form : f"

Copied!
27
0
0

加載中.... (立即查看全文)

全文

(1)

2 The First Form : XaYb + Y cZd

Let K be a field of characteristic p > 0 and

S = K [X1, . . . , Xr, Y1, . . . , Ys, Z1, . . . , Zt] .

In this section we shall determine the Hilbert-Kunz function of the hypersurface of the following form :

f := XaYb+ YcZd

where Xa = X1a1. . . Xrar, Yb = Y1b1. . . Ysbs, Yc = Y1c1. . . Yscs, Zd = Z1d1. . . Ztdt, and r ≥ 1. Let q = pn, J = j | bj > cj , and set R = S/ < f >. Then 0 ≤ |J| := m ≤ s, and w.l.o.g., we assume that b1 > c1, . . . , bm > cm, bm+1 ≤ cm+1, . . . , bs≤ cs. We shall determine the assignment

HKR(q) := dimK

 S < X1q, . . . , Xrq, Y1q, . . . , Ysq, Z1q, . . . , Ztq, f >

 .

Let f + X[q] be the ideal of S generated by all Xiq’s, Yjq’s, Zkq’s, and f . Fix a term order on S, and denote by in f + X[q] the initial ideal of f + X[q]. Then by Lemma 1.1, we get that HKR(q) is equal to dimK



S/in f + X[q]

. By making use of Gr¨obner basis, we understand which monomials one has to add to fill the gap between in f +X[q]

and the ideal in(f )+X[q]. Throughout this section, it is not restrictive to assume that a1 ≥ a2 ≥ · · · ≥ ar > 0 , b1− c1 ≥ b2− c2 ≥ · · · ≥ bm − cm > 0 , cm+1− bm+1 ≥ cm+2 − bm+2 ≥ · · · ≥ cs− bs ≥ 0, and d1 ≥ d2 ≥ · · · ≥ dt> 0. Let u be the maximum of the integers a1 , b1− c1 , cm+1 − bm+1, and d1; that is, u is the greatest integer among all ai’s, (bj − cj)’s, (ch − bh)’s, and dk’s. We also denote by [y] the greatest integer less than or equal to y, and Sin(x) the elementary symmetric polynomial of degree i in n indeterminates x = (x1, . . . , xn). Let Iq be the ideal f + X[q] and define (v)+= max {0, v}.

(2)

In order to make it more easy to determine the Hilbert-Kunz function of R, we shall prove the following lemma.

Lemma 2.1. Let S = K [X1, . . . , Xr, Y1, . . . , Ys], r ≥ 1, s ≥ 1, and the ai, bj, and cj are all positive integers with b1 − c1 ≥ b2 − c2 ≥ · · · ≥ bm − cm > 0, cm+1− bm+1 ≥ cm+2 − bm+2

· · · ≥ cs− bs≥ 0. We denote by G the ideal generated by

X1q, . . . , Xrq, Y1q, . . . , Ysq, X1[q−αa1]+Ye+α(c−b)+, . . . , Xr[q−αar]+Ye+α(c−b)+, Y1[q−α(b1−c1)−c1]+Ye+α(c−b)+, . . . , Ym[q−α(bm−cm)−cm]+Ye+α(c−b)+, and XaYb,

where α is a positive integer, e = (e1, . . . , es), e1 = c1, . . . , em = cm, em+1 = bm+1, . . . , es = bs, (c − b)+= (0, . . . , 0, cm+1− bm+1, . . . , cs− bs). Then the dimension of S/G is equal to

qr+s

r

Y

i=1

(q − ai)

s

Y

j=1

(q − bj) − qr

m

Y

j=1

(q − cj)

s

Y

h=m+1

[q − α(ch− bh) − bh]+

+

r

Y

i=1

(q − ai)

m

Y

j=1

(q − bj)

s

Y

h=m+1

[q − α(ch− bh) − bh]+

+

r

Y

i=1

(q − αai)+

m

Y

j=1

[q − α(bj− cj) − cj ]+

s

Y

h=m+1

[q − α(ch− bh) − bh]+

r

Y

i=1

[q − (α + 1)ai ]+

m

Y

j=1

[q − (α + 1)(bj− cj) − cj ]+

s

Y

h=m+1

[q − α(ch− bh) − bh]+.

Proof : If [q − αai]+ = 0 for some i or [q − α(bj − cj) − cj ]+ = 0 for some j, then G is generated by

X1q, . . . , Xrq, Y1q, . . . , Ysq, Ye+α(c−b)+, and XaYb. Hence,

dimK S/G

= qr+s

r

Y

i=1

(q − ai)

s

Y

j=1

(q − bj) − qr

m

Y

j=1

(q − cj)

s

Y

h=m+1

[q − α(ch− bh) − bh]+

+

r

Y

i=1

(q − ai)

m

Y

j=1

(q − bj)

s

Y

h=m+1

[q − α(ch− bh) − bh]+ .

From now on, we assume q − αai > 0 for each i and q − α(bj− cj) − cj > 0 for each j. Let lα be the minimum of

  q − α(bj− cj) − 1 cj

 ,

 q − 1

bh+ α(ch− bh)



j = 1, . . . , m, h = m + 1, . . . , s

 .

(3)

Then we have q − α(bj0 − cj0) ≤ (lα+ 1)cj0 for some j0 with 1 ≤ j0 ≤ m or

q ≤ (lα+ 1)(bh0 + α(ch0 − bh0)) for some h0 with m + 1 ≤ h0 ≤ s, and q − α(bj− cj) − lαcj ≥ 1 for each j and q − lα(bh+ α(ch− bh)) ≥ 1 for each h.

We consider the ideals Gβ = G : Yβ[e+α(c−b)+], for β = 0, 1, 2, . . . , lα + 1. Since G0 = G, Glα+1 = S, and Gβ+1 = Gβ : Ye+α(c−b)+, we have the exact sequence of K-modules :

0 −→ S/Gβ+1 Ye+α(c−b)+−→ S/Gβ −→ S / < Gβ, Ye+α(c−b)+ > −→ 0.

It follows that

dimK S/G = dimK S/G0 =

lα

X

β=0

dimK



S / < Gβ, Ye+α(c−b)+ >

 .

We determine dimK

S / < Gβ, Ye+α(c−b)+ >

as follows : For β = 0, the ideal < G0, Ye+α(c−b)+ > is generated by

X1q, . . . , Xrq, Y1q, . . . , Ysq, Ye+α(c−b)+, and XaYb. Hence,

dimK

S / < G0, Ye+α(c−b)+ >

= qr+s

r

Y

i=1

(q − ai)

s

Y

j=1

(q − bj) − qr

m

Y

j=1

(q − cj)

s

Y

h=m+1

[q − α(ch− bh) − bh]+

+

r

Y

i=1

(q − ai)

m

Y

j=1

(q − bj)

s

Y

h=m+1

[q − α(ch− bh) − bh]+ .

For 1 ≤ β ≤ lα, the ideal Gβ = G : Yβ[e+α(c−b)+] is generated by

X1q−αa1, . . . , Xrq−αar, Y1q−α(b1−c1)−βc1, . . . , Ymq−α(bm−cm)−βcm, Ym+1q−β[bm+1+α(cm+1−bm+1)], . . . , Ysq−β[bs+α(cs−bs)], and XaY1(b1−βc1)+· · · Ym(bm−βcm)+Ym+1[bm+1−β(bm+1+α(cm+1−bm+1))]+· · · Ys[bs−β(bs+α(cs−bs))]+ .

Hence, the ideal < Gβ, Ye+α(c−b)+ > is generated by

X1q−αa1, . . . , Xrq−αar, Y1q−α(b1−c1)−βc1, . . . , Ymq−α(bm−cm)−βcm, Ym+1q−β[bm+1+α(cm+1−bm+1)], . . . , Ysq−β[bs+α(cs−bs)], XaY1(b1−βc1)+· · · Ym(bm−βcm)+Ym+1[bm+1−β(bm+1+α(cm+1−bm+1))]+· · ·

Ys[bs−β(bs+α(cs−bs))]+, and Ye+α(c−b)+ .

(4)

Thus, dimK

S / < Gβ, Ye+α(c−b)+ >

=

r

Y

i=1

(q − αai)

m

Y

j=1

[q − α(bj − cj) − βcj ]+

s

Y

h=m+1

[q − β(bh+ α(ch− bh))]+

r

Y

i=1

[q − (α + 1)ai]+

m

Y

j=1

q − α(bj − cj) − βcj − (bj − βcj)+

+ s

Y

h=m+1

[q − β(bh+ α(ch− bh))]+

r

Y

i=1

(q − αai)

m

Y

j=1

[q − α(bj− cj) − (β + 1)cj ]+

s

Y

h=m+1

[q − (β + 1)(bh+ α(ch− bh))]+

+

r

Y

i=1

[q − (α + 1)ai]+

m

Y

j=1

[q − α(bj − cj) − βcj − u]+

s

Y

h=m+1

[q − (β + 1)(bh+ α(ch− bh))]+

=

r

Y

i=1

(q − αai) × ( m

Y

j=1

[q − α(bj − cj) − βcj ]+

s

Y

h=m+1

[q − β (bh+ α (ch − bh))]+

m

Y

j=1

[q − α(bj− cj) − (β + 1)cj ]+

s

Y

h=m+1

[q − (β + 1)(bh+ α(ch− bh))]+ )

r

Y

i=1

[q − (α + 1)ai]+× ( m

Y

j=1

q − α(bj− cj) − βcj− (bj − βcj)+

+ s

Y

h=m+1

[q − β(bh + α(ch

bh))]+

m

Y

j=1

[q − α(bj− cj) − βcj− u]+

s

Y

h=m+1

[q − (β + 1)(bh+ α(ch− bh))]+ )

,

where u = max cj , [bj− βcj ]+ . Now, we have

dimK S/G

= qr+s

r

Y

i=1

(q − ai)

s

Y

j=1

(q − bj) − qr

m

Y

j=1

(q − cj)

s

Y

h=m+1

[q − α(ch− bh) − bh]+

+

r

Y

i=1

(q − ai)

m

Y

j=1

(q − bj)

s

Y

h=m+1

[q − α(ch− bh) − bh]+

+

r

Y

i=1

(q − αai) × ( l

α

X

β=1

" m Y

j=1

[q − α(bj − cj) − βcj ]+

s

Y

h=m+1

[q − β(bh+ α(ch− bh))]+

m

Y

j=1

[q − α(bj− cj) − (β + 1)cj ]+

s

Y

h=m+1

[q − (β + 1)(bh+ α(ch− bh))]+

#)

(5)

r

Y

i=1

[q − (α + 1)ai]+ × ( lα

X

β=1

" m Y

j=1

[q − α(bj − cj) − βcj− (bj− βcj)+]+

s

Y

h=m+1

[q − β(bh+

α(ch− bh))]+

m

Y

j=1

[q − α(bj − cj) − βcj − u]+

s

Y

h=m+1

[q − (β + 1)(bh+ α(ch− bh))]+

#) .

Let (∗) be the term

lα

X

β=1

" m Y

j=1

[q − α(bj− cj) − βcj ]+

s

Y

h=m+1

[q − β(bh+ α(ch− bh))]+

m

Y

j=1

[q − α(bj − cj) − (β + 1)cj ]+

s

Y

h=m+1

[q − (β + 1)(bh+ α(ch− bh))]+

# . Since q − α(bj0 − cj0) ≤ (lα+ 1)cj0 for some j0 with 1 ≤ j0 ≤ m or

q ≤ (lα+ 1) bh0+ α(ch0 − bh0)

for some h0 with m + 1 ≤ h0 ≤ s, (∗) is equal to

m

Y

j=1

[q − α(bj− cj) − cj ]+

s

Y

h=m+1

[q − α(ch− bh) − bh]+.

Let (∗∗) be the term

lα

X

β=1

" m Y

j=1

q − α(bj− cj) − βcj− (bj − βcj)+

+ s

Y

h=m+1

[q − β(bh+ α(ch− bh))]+

m

Y

j=1

[q − α(bj − cj) − βcj − u ]+

s

Y

h=m+1

[q − (β + 1)(bh+ α(ch− bh))]+

# . Since

m

Y

j=1

[q − α(bj− cj) − βcj− u ]+

s

Y

h=m+1

[q − (β + 1)(bh+ α(ch− bh))]+

=

m

Y

j=1

q − α(bj − cj) − (β + 1)cj − (bj− (β + 1)cj)+

+ s

Y

h=m+1

[q − (β + 1)(bh+ α(ch− bh))]+,

where β = 1, 2, . . . , lα− 1, the term (∗∗) is equal to

m

Y

j=1

[q − (α + 1)(bj − cj) − cj ]+

s

Y

h=m+1

[q − α(ch − bh) − bh]+

m

Y

j=1

[q − α(bj − cj) − lαcj − ujlα ]+

s

Y

h=m+1

[q − (lα+ 1)(bh+ α(ch− bh))]+.

(6)

Since q − α(bj0 − cj0) ≤ (lα+ 1)cj0 or q ≤ (lα+ 1) bh0+ α(ch0 − bh0), we have

m

Y

j=1

[q − α(bj − cj) − lαcj− ujlα ]+ = 0 or

s

Y

h=m+1

[q − (lα+ 1)(bh+ α(ch− bh))]+= 0 . Thus, (∗∗) is equal to

m

Y

j=1

[q − (α + 1)(bj− cj) − cj ]+

s

Y

h=m+1

[q − α(ch− bh) − bh]+.

So,

dimK S/G

= qr+s

r

Y

i=1

(q − ai)

s

Y

j=1

(q − bj) − qr

m

Y

j=1

(q − cj)

s

Y

h=m+1

[q − α(ch− bh) − bh]+

+

r

Y

i=1

(q − ai)

m

Y

j=1

(q − bj)

s

Y

h=m+1

[q − α(ch − bh) − bh]+

+

r

Y

i=1

(q − αai)+

m

Y

j=1

[q − α(bj− cj) − cj ]+

s

Y

h=m+1

[q − α(ch− bh) − bh]+

r

Y

i=1

[q − (α + 1) ai ]+

m

Y

j=1

[q − (α + 1)(bj− cj) − cj ]+

s

Y

h=m+1

[q − α(ch− bh) − bh]+.



Since u is the maximum of the integers among all ai’s, (bj − cj)’s, (ch− bh)’s, and dk’s, we have

q−v

u  := min (

hq−1 ai

i ,hq−c

j−1 bj−cj

i ,h

q−bh−1 ch−bh

i ,h

q−1 dk

i

1 ≤ i ≤ r, bj − cj > 0, ch− bh > 0 1 ≤ j ≤ m, m + 1 ≤ h ≤ s, 1 ≤ k ≤ t

)

for q  0, where v = 1 or 1 + cj for some j or 1 + bh for some h.

Let lu be the integerq−v

u , and  be the remainder of q − v divided by u. Then lu = q−v−u and one has q − luai > 0, q − lu(bj− cj) − cj > 0, q − lu(ch− bh) − bh > 0, and q − ludk > 0 for all i, j, h, and k. On the other hand, by the definition of lu, at least one of [q − (lu+ 1)ai]+’s, [q − (lu+ 1)(bj − cj) − cj ]+’s, [q − (lu+ 1)(ch− bh) − bh]+’s, and [q − (lu+ 1)dk]+’s must be zero.

(7)

Proposition 2.2. Let f := XaYb+ YcZd. Then HKR(q) = qr+s+t− qt

r

Y

i=1

(q − ai)

s

Y

j=1

(q − bj)

− qr

m

Y

j=1

(q − cj

( l

u

X

α=1

" s Y

h=m+1

[q − α(ch− bh) − bh]

#

×

" t Y

k=1

(q − αdk) −

t

Y

k=1

[q − (α + 1)dk]+

#)

+

r

Y

i=1

(q − ai)

m

Y

j=1

(q − bj

( l

u

X

α=1

" s Y

h=m+1

[q − α(ch− bh) − bh]

#

×

" t Y

k=1

(q − αdk) −

t

Y

k=1

[q − (α + 1)dk]+

#)

+

lu

X

α=1

( r Y

i=1

(q − αai)

m

Y

j=1

[q − α(bj − cj) − cj ]

s

Y

h=m+1

[q − α(ch− bh) − bh]

r

Y

i=1

[q − (α + 1)ai]+

m

Y

j=1

[q − (α + 1)(bj− cj) − cj]+

s

Y

h=m+1

[q − α(ch− bh) − bh] )

× ( t

Y

k=1

(q − αdk) −

t

Y

k=1

[q − (α + 1)dk]+ )

, where lu is the integer  q−v

u , and 0 ≤ m ≤ s .

Proof : Let u be the maximum of the integers a1, b1− c1, cm+1− bm+1, and d1. Let < be the lexicographic order on S and define

ej = cj for j = 1, . . . , m and eh = bh for h = m + 1, . . . , s.

Then XaYb is bigger than YcZd and Ye = Y1e1. . . Yses = Y1c1. . . YmcmYm+1bm+1. . . Ysbs. We determine a Gr¨obner basis of the ideal

Iq =< X1q, . . . , Xrq, Y1q, . . . , Ysq, Z1q, . . . , Ztq, f >,

by means of Buchberger’s algorithm (Algorithm 1.9). By this algorithm, the elements X1q, . . . , Xrq, Y1q, . . . , Ysq, Z1q, . . . , Ztq, Xi(q−δai)+Ye+δ(c−b)+Zδd, i = 1, . . . , r, δ = 1, . . . , l, Yj[q−δ(bj−cj)−cj]+Ye+δ(c−b)+Zδd, j = 1, . . . , m, δ = 1, . . . , l, and XaYb+ YcZd,

form a Gr¨obner basis of the ideal Iq, where l =h

q−1 d1

i

. Thus, the ideal in(Iq) is generated by X1q, . . . , Xrq, Y1q, . . . , Ysq, Z1q, . . . , Ztq, Xi(q−δai)+Ye+δ(c−b)+Zδd, i = 1, . . . , r, δ = 1, . . . , l, Yj[q−δ(bj−cj)−cj]+Ye+δ(c−b)+Zδd, j = 1, . . . , m, δ = 1, . . . , l, and XaYb.

(8)

Now we have to compute the dimension of S/in(Iq). In order to do this, we consider the ideals Kα = in(Iq) : Zαd for α = 0, 1, . . . , l + 1, where Zαd = Z1αd1. . . Ztαdt. Since K0 = in(Iq), Kl+1 = S, and Kα+1 = Kα : Zd, we have the exact sequence of K-modules :

0 −→ S/Kα+1 −→ S/KZd α −→ S / < Kα, Zd> −→ 0 . It follows that

dimK S/in(Iq) = dimK S/K0 =

l

X

α=0

dimK

S / < Kα, Zd> .

We compute dimK

S / < Kα, Zd>

as follows : For α = 0, the ideal < K0, Zd> is generated by

X1q, . . . , Xrq, Y1q, . . . , Ysq, Z1q, . . . , Ztq, XaYb, and Zd. Let S1 = K [X1, . . . , Xr, Y1, . . . , Ys], and S2 = K [Z1, . . . , Zt]. Then

dimK

S / < K0, Zd>

= dimK

S1 / < X1q, . . . , Xrq, Y1q, . . . , Ysq, XaYb >

× dimK

S2 / < Z1q, . . . , Ztq, Zd> 

=

"

qr+s

r

Y

i=1

(q − ai)

s

Y

j=1

(q − bj)

#

×

"

qt

t

Y

k=1

(q − dk)

#

= qr+s+t− qt

r

Y

i=1

(q − ai)

s

Y

j=1

(q − bj) − qr+s

t

Y

k=1

(q − dk) +

r

Y

i=1

(q − ai)

s

Y

j=1

(q − bj)

t

Y

k=1

(q − dk).

For 1 ≤ α ≤ l, the ideal Kα = in(Iq) : Zαd is generated by

X1q, . . . , Xrq, Y1q, . . . , Ysq, Z1q−αd1, . . . , Ztq−αdt, Xi(q−δai)+Ye+δ(c−b)+Z(δ−α)d, i = 1, . . . , r, δ = 1, . . . , l, Yj[q−δ(bj−cj)−cj]+Ye+δ(c−b)+Z(δ−α)d, j = 1, . . . , m, δ = 1, . . . , l, and XaYb.

Hence, the ideal < Kα, Zd> is generated by

X1q, . . . , Xrq, Y1q, . . . , Ysq, Z1q−αd1, . . . , Ztq−αdt, X1(q−αa1)+Ye+α(c−b)+, . . . , Xr(q−αar)+Ye+α(c−b)+, Y1[q−α(b1−c1)−c1]+Ye+α(c−b)+, . . . , Ym[q−α(bm−cm)−cm]+Ye+α(c−b)+, XaYb, and Zd.

參考文獻

相關文件

[This function is named after the electrical engineer Oliver Heaviside (1850–1925) and can be used to describe an electric current that is switched on at time t = 0.] Its graph

In order to establish the uniqueness of a prime factorization, we shall use the alternative form of the Principle of Mathematical Induction.. For the integer 2, we have a unique

6 《中論·觀因緣品》,《佛藏要籍選刊》第 9 冊,上海古籍出版社 1994 年版,第 1

Reading Task 6: Genre Structure and Language Features. • Now let’s look at how language features (e.g. sentence patterns) are connected to the structure

Teachers may consider the school’s aims and conditions or even the language environment to select the most appropriate approach according to students’ need and ability; or develop

利用 determinant 我 們可以判斷一個 square matrix 是否為 invertible, 也可幫助我們找到一個 invertible matrix 的 inverse, 甚至將聯立方成組的解寫下.

Then, we tested the influence of θ for the rate of convergence of Algorithm 4.1, by using this algorithm with α = 15 and four different θ to solve a test ex- ample generated as

Numerical results are reported for some convex second-order cone programs (SOCPs) by solving the unconstrained minimization reformulation of the KKT optimality conditions,