local bézout inequalitieslocal b ezout inequalities azeem khadam abdus salam school of mathematical...

Local Bezout Inequalities

Azeem Khadam

Abdus Salam School of Mathematical SciencesGC University, Lahore

June 04, 2020

Azeem Khadam (ASSMS) June 2020 1 / 50

Outline

1 Bezout’s theorem

2 Local Bezout inequality in the plane

3 Hilbert-Samuel multiplicity

4 Local Bezout inequalities – in general

5 Koszul complexes

6 Euler characteristic

7 Local Bezout inequalities – again


Fundamental theorem of algebra

Throughout: k is an algebraically closed field,A is a Noetherian ring with dimA = t,and for an ideal q ⊆ A, GA(q) = ⊕n≥0q

n/qn+1.

Theorem

Every non-zero polynomial f ∈ k[x ] of degree d has d roots (or zeros),when counted with multiplicities.

In other words, the number of intersection points of the curveC = V (y − f (x)) ⊂ A2

k and the line L = V (y) ⊂ A2k are exactly d , when

counted with multiplicities.




n/qn+1.

Theorem


In other words,

the number of intersection points of the curveC = V (y − f (x)) ⊂ A2






n/qn+1.

Theorem







Example

C : y − x2 = 0 and L : y = 0, then the multiplicity of the intersectionpoint p = (0, 0) is

µ(p; y − x2, y) = 2 ,



Theorem







Theorem



k and the line L = V (y) ⊂ A2k are exactly

d(= deg(f ) · deg(y)), when counted with multiplicities..

Problem: Can we extend FTA to any two curves?



Theorem



k and the line L = V (y) ⊂ A2k are exactly

d(= deg(f ) · deg(y)), when counted with multiplicities..Problem: Can we extend FTA to any two curves?


Bezout’s theorem

Theorem

Let C = V (F ) and D = V (G ) be two curves in the projective plane P2k

such that they have no common component, where F ,G ∈ k[X ,Y ,Z ] arehomogeneous polynomials.

Then the number of intersection points of C and D, counted withmultiplicities, are ∑

P∈F∩G µ(P;F ,G ) = deg F · degG ,

where µ(P;F ,G ) denotes the local intersection multiplicity of P in C ∩D.


Bezout’s theorem

Theorem


such that they have no common component, where F ,G ∈ k[X ,Y ,Z ] arehomogeneous polynomials.Then the number of intersection points of C and D, counted withmultiplicities, are

∑P∈F∩G µ(P;F ,G ) = deg F · degG ,



Bezout’s theorem

Theorem


such that they have no common component, where F ,G ∈ k[X ,Y ,Z ] arehomogeneous polynomials.Then the number of intersection points of C and D, counted withmultiplicities, are ∑




Bezout’s theorem

Theorem






Bezout’s theorem

Theorem





no common component: F and G have no common factor.


Local Bezout inequality in the plane

If P = [0 : 0 : 1] is the origin, it follows that

µ(P;F ,G ) = `A(A/(f , g)) ,

where A = k[x , y ](x ,y) and f = F (x , y , 1), g = G (x , y , 1) denote thepolynomials in A.

Definition

Let (A,m) be a local ring and M an A-module.

A chainM = M0 ⊃ M1 ⊃ · · · ⊃ Mr = 0

of submodules of M is called a composition series of M if everyMi/Mi+1

∼= A/m.

The length r of such a composition series is called the length of Mand denoted by `A(M).



If P = [0 : 0 : 1] is the origin, it follows that

µ(P;F ,G ) = `A(A/(f , g)) (= dimk(A/(f , g))),

where A = k[x , y ](x ,y) and f = F (x , y , 1), g = G (x , y , 1) denote thepolynomials in A.

Proposition

M has a composition series if and only if M satisfies both the ascendingand descending chain conditions.



Remark1 Note that

µ(P;F ,G ) ≥ c · d ,

which is called the local Bezout inequality in the affine plane A2k,

where c and d are initial degrees of f and g respectively.

2 Here the equality occurs if and only if.C and D intersect transversally at the origin (0, 0) if and only if.the initial forms f ? = f + mc+1, g? = g + md+1 is a regular sequencein the form module GA(m) ∼= k[X ,Y ], where m = (x , y)A.



Remark1 Note that

µ(P;F ,G ) ≥ c · d ,



2 Here the equality occurs if and only if.

C and D intersect transversally at the origin (0, 0) if and only if.the initial forms f ? = f + mc+1, g? = g + md+1 is a regular sequencein the form module GA(m) ∼= k[X ,Y ], where m = (x , y)A.



Remark1 Note that

µ(P;F ,G ) ≥ c · d ,



2 Here the equality occurs if and only if.C and D intersect transversally at the origin (0, 0) if and only if.

the initial forms f ? = f + mc+1, g? = g + md+1 is a regular sequencein the form module GA(m) ∼= k[X ,Y ], where m = (x , y)A.



Remark1 Note that

µ(P;F ,G ) ≥ c · d ,






Example

C : y − x2 = 0 and L : y = 0, then the multiplicity of the intersectionpoint p = (0, 0) is

µ(p; y − x2, y) = 2 ,

where initial degree of both C and L is one. Also, C and L do notintersect transversally. Or (y − x2)? = y + m2 = Y , y? = y + m2 = Y isnot a regular sequence.


Local Bezout inequality in the planeProof (1).If B = k[X ,Y ], I = (X ,Y ),

then we have the following diagram with exactrows

B/I c ⊕ B/I dψ−→ B/I c+d φ−→ B/(I c+d , f , g) → 0

↓ ∼=A/(f , g)

π−→ A/(I c+d , f , g) → 0

where ψ(α, β) = f α + gβ, and φ, π are natural surjections.

µ(P;F ,G )= dimk(A/(f , g))

≥ dimk(A/(I c+d , f , g))

= dimk(B/(I c+d , f , g))

= dimk(B/I c+d)− ker(φ)

≥ dimk(B/I c+d)− dimk(B/I c)− dimk(B/I d)

= c · d , since dimk(B/I n) = 1 + 2 + . . .+ n.


Local Bezout inequality in the planeProof (1).If B = k[X ,Y ], I = (X ,Y ),then we have the following diagram with exactrows


↓ ∼=A/(f , g)

π−→ A/(I c+d , f , g) → 0


µ(P;F ,G )= dimk(A/(f , g))







Local Bezout inequality in the planeProof (1).If B = k[X ,Y ], I = (X ,Y ), then we have the following diagram with exactrows


↓ ∼=A/(f , g)

π−→ A/(I c+d , f , g) → 0


µ(P;F ,G ) = dimk(A/(f , g))









↓ ∼=A/(f , g)

π−→ A/(I c+d , f , g) → 0


µ(P;F ,G ) = dimk(A/(f , g))








Remark1 Note that

µ(P;F ,G ) ≥ c · d ,




Question: Provided C and D do not intersect transversally, can weimprove the local Bezout inequality?



Theorem (Bydzovsky 1948, Boda-Schenzel 2017)

With the previous notation, we have

µ(P;F ,G ) ≥ c · d + T

where T is the number of common tangents to f and g when countedwith multiplicities.

Open problem: Can we further improve the above inequality?


Hilbert-Samuel multiplicity

Let (A,m) be a local, and q ⊂ A an m-primary ideal. ThenÀ(A/qn+1) is finite for every n ∈ N0.

In fact, À(A/qn+1) is a polynomial for n� 0. Precisely,

À(A/qn+1) =t∑

i=0

ei (q;A)

(n + t − i

t − i

)for n� 0.

which is called the Hilbert-Samuel polynomial of A w.r.t. the ideal q.

Definition

e0(q;A) is called the Hlbert-Samuel multiplicity of q in A.




In fact, `A(A/qn+1) is a polynomial for n� 0.

Precisely,

`A(A/qn+1) =t∑

i=0

ei (q;A)

(n + t − i

t − i

)for n� 0.


Definition





In fact, `A(A/qn+1) is a polynomial for n� 0. Precisely,

`A(A/qn+1) =t∑

i=0

ei (q;A)

(n + t − i

t − i

)for n� 0.


Definition



System of parameters

Let f = f1, . . . , ft ⊆ q be a system of parameters, or shortly s.o.p.(i.e. `A(A/f A) <∞).

Then e0(f ;A) ≥ e0(q;A) (since I ⊆ J =⇒ e0(I ;A) ≥ e0(J;A)).

Moreover, by Krull’s intersection theorem (∩n≥0qn = 0), there isci ∈ N for i = 1, . . . , t, such that fi ∈ qci \ qci+1.

Then e0(f ;A) ≥ c1 · · · ct · e0(q;A), because (1) (f e11 , . . . , f ett ) ⊂ qc

where c = c1 · · · ct and ei = c/ci =⇒ e0(f e11 , . . . , f ett ;A) ≥ e0(qc ;A),






Then e0(f ;A) ≥ c1 · · · ct · e0(q;A),

because (1) (f e11 , . . . , f ett ) ⊂ qc








where c = c1 · · · ct and ei = c/ci

=⇒ e0(f e11 , . . . , f ett ;A) ≥ e0(qc ;A),





Moreover, by Krull’s intersection theorem, there is ci ∈ N fori = 1, . . . , t, such that fi ∈ qci \ qci+1.


where ei = c/ci =⇒ e0(f e11 , . . . , f ett ;A) ≥ e0(qc ;A) (2)e0(qc ;A) = cte0(q;A)





Moreover, by Krull’s intersection theorem, there is ci ∈ N fori = 1, . . . , t, such that ai ∈ qci \ qci+1.


where ei = c/ci =⇒ e0(f e11 , . . . , f ett ;A) ≥ e0(qc ;A) (2)e0(qc ;A) = cte0(q;A), and (3)e0(f1, . . . , ft−1, xy ;A) = e0(f1, . . . , ft−1, x ;A) + e0(f1, . . . , ft−1, y ;A)[Auslander-Buchsbaum].





Moreover, by Krull’s intersection theorem, there is ci ∈ N fori = 1, . . . , t, such that fi ∈ qci \ qci+1.

Then e0(f ;A) ≥ c1 · · · ct · e0(q;A), because (1) (f e11 , . . . , f etd ) ⊂ qc

where ei = c/ci =⇒ e0(f e11 , . . . , f ett ;A) ≥ e0(qc ;A) (2)e0(qc ;A) = cte0(q;A), and (3)e0(f1, . . . , ft−1, xy ;A) = e0(f1, . . . , ft−1, x ;A) + e0(f1, . . . , ft−1, y ;A)[Auslander-Buchsbaum].



Theorem

The following statements are equivalent:

1 A is Cohen-Macaulay ring;

2 `A(A/f A) = e0(f ;A) for any s.o.p. f = f1, . . . , ft of A;

3 `A(A/f A) = e0(f ;A) for some s.o.p. f = f1, . . . , ft of A.


Local Bezout inequalities

Recall that local Bezout inequality in the plane says that

µ(P;F ,G ) ≥ c · d ,

where µ(P;F ,G ) = `A(A/(f , g)).

Since f , g ∈ A = k[x , y ](x ,y) is a s.o.p. =⇒`A(A/(f , g)) = e0(f , g ;A).




µ(P;F ,G ) ≥ c · d ,

where µ(P;F ,G ) = `A(A/(f , g)).

Since f , g ∈ A = k[x , y ](x ,y) is a s.o.p.

=⇒`A(A/(f , g)) = e0(f , g ;A).




µ(P;F ,G ) ≥ c · d ,

where µ(P;F ,G ) = `A(A/(f , g)).

Since f , g ∈ A = k[x , y ](x ,y) is a s.o.p. =⇒`A(A/(f , g)) = e0(f , g ;A).



So local Bezout inequality in the plane says looks like

e0(f , g ;A) ≥ c · d .

More generally, if A = k[x1, . . . , xt ](x1,...,xt) andq = m = (x1, . . . , xt)A, then we have

e0(f ;A) ≥ c1 · · · ct · e0(m;A)

= c1 · · · ct since e0(m;A) = 1.



We calle0(f ;A) ≥ c1 · · · ct

local Bezout inequality in Atk.

Problem: Can we improve it?.Note that one can prove that equality occurs if and only if f ?1 , . . . , f

?t form

a homogeneous regular sequence in the form ring GA(m) ∼= k[X1, . . . ,Xt ].





Problem: Can we improve it?

.Note that one can prove that equality occurs if and only if f ?1 , . . . , f

?t form






Problem: Can we improve it?.Note that one can prove that equality occurs if and only if f ?1 , . . . , f

?t form



Koszul complexesThe Koszul complex K•(f ;A) of A w.r.t. elements f = f1, . . . , ft is definedas follows:

For i = 0, . . . , t, define Ki (f ;A) = ⊕Aej1···ji be the free A-modulewith basis ej1···ji for 1 ≤ j1 < . . . < ji ≤ t.

The boundary homomorphism Ki (f ;A)→ Ki−1(f ;A) is defined by

dj1...ji : ej1···ji 7→i∑

k=1

(−1)k+1fjk ej1···jk ···ji .

We denote by Hi (f ;A) the i-th homology of the complex K•(f ;A).

Example: t = 2, f1 = f , f2 = g

K•(f , g ;A) : 0→ Aφ→ A⊕ A

ψ→ A→ 0

where φ(z) = (−gz , fz) and ψ(x , y) = fx + gy .


Koszul complexes

Let fi ∈ qci \ qci+1 for i = 1, . . . , t and n ∈ N0.

A subcomplex K•(f , q; n)of K•(f ;A) is defined as follows:

For 0 ≤ i ≤ t, we put Ki (f , q; n) = ⊕1≤j1<...<ji≤tqn−cj1−...−cji and

Ki (f , q; n) = 0 for i > t or i < 0.

The boundary homomorphism Ki (f , q; n)→ Ki−1(f , q; n) is therestriction of the map of dj1...ji : Ki (f ;A)→ Ki−1(f ;A).

We denote by Hi (f , q; n) the i-th homology of the complexK•(f , q; n).

Example: t = 2, f1 = f , f2 = g , c1 = c , c2 = d

K•(f , g , q; n) : 0→ qn−c−dφ→ qn−c ⊕ qn−d

ψ→ qn → 0



Koszul complexes

Let fi ∈ qci \ qci+1 for i = 1, . . . , t and n ∈ N0. A subcomplex K•(f , q; n)of K•(f ;A) is defined as follows:

For 0 ≤ i ≤ t, we put Ki (f , q; n) = ⊕1≤j1<...<ji≤tqn−cj1−...−cji and

Ki (f , q; n) = 0 for i > t or i < 0.

The boundary homomorphism Ki (f , q; n)→ Ki−1(f , q; n) is therestriction of the map of dj1...ji : Ki (f ;A)→ Ki−1(f ;A).

We denote by Hi (f , q; n) the i-th homology of the complexK•(f , q; n).

Example: t = 2, f1 = f , f2 = g , c1 = c , c2 = d

K•(f , g , q; n) : 0→ qn−c−dφ→ qn−c ⊕ qn−d

ψ→ qn → 0



Koszul complexes

K•(f , g , q; n) : 0→ qn−c−d → qn−c ⊕ qn−d → qn → 0↓ ↓ ↓ ↓

K•(f , g ;A) : 0→ A → A⊕ A → A → 0


Koszul complexes

The quotient complex L•(f , q; n) of the embedding K•(f , q; n) ⊂ K•(f ;A)is given by

Li (f , q; n) ∼= ⊕1≤j1<...<ji≤tA/qn−cj1−...−cji , with boundary maps are

those induced by the Koszul complex.

We write Li (f , q : n) for its i-th homology A-module.


Koszul complexes

0 0 0 0↓ ↓ ↓ ↓

K•(f , g , q; n) : 0→ qn−c−d → qn−c ⊕ qn−d → qn → 0↓ ↓ ↓ ↓

K•(f , g ;A) : 0→ A → A⊕ A → A → 0↓ ↓ ↓ ↓

L•(f , g , q; n) : 0→ A/qn−c−d → A/qn−c ⊕ A/qn−d → A/qn → 0↓ ↓ ↓ ↓0 0 0 0


Koszul complexes

The short exact sequence of (Koszul) complexes

0→ K•(f , q; n)→ K•(f ;A)→ L•(f , q; n)→ 0

gives a long exact sequence of (Koszul) homologies

· · · → Hi (f , q; n)→ Hi (f ;A)→ Li (f , q; n)→ · · ·


Euler characteristic

Definition

Let X : 0→ Xn → . . .→ X1 → X0 → 0 denote a bounded complex ofA-modules. Suppose that Hi (X ), i = 0, 1, . . . , n, is an A-module of finitelength.

Then

χA(X ) =n∑

i=0

(−1)i`A(Hi (X ))

is called the Euler characteristic of X .

Lemma

Let 0→ X ′ → X → X ′′ → 0 denote a short exact sequence ofcomplexes such that all the homology modules are of finitelength.Then χA(X ) = χA(X ′) + χA(X ′′).

Let X : 0→ Xn → . . .→ X1 → X0 → 0 be a bounded complex suchthat Xi , i = 0, . . . , n, is of finite length.ThenχA(X ) =

∑ni=0(−1)i`A(Xi ).



Definition

Let X : 0→ Xn → . . .→ X1 → X0 → 0 denote a bounded complex ofA-modules. Suppose that Hi (X ), i = 0, 1, . . . , n, is an A-module of finitelength.Then

χA(X ) =n∑

i=0

(−1)i`A(Hi (X ))


Lemma



∑ni=0(−1)i`A(Xi ).



Definition


χA(X ) =n∑

i=0

(−1)i`A(Hi (X ))


Lemma

Let 0→ X ′ → X → X ′′ → 0 denote a short exact sequence ofcomplexes such that all the homology modules are of finitelength.

Then χA(X ) = χA(X ′) + χA(X ′′).


∑ni=0(−1)i`A(Xi ).



Definition


χA(X ) =n∑

i=0

(−1)i`A(Hi (X ))


Lemma



∑ni=0(−1)i`A(Xi ).



Definition


χA(X ) =n∑

i=0

(−1)i`A(Hi (X ))


Lemma


Let X : 0→ Xn → . . .→ X1 → X0 → 0 be a bounded complex suchthat Xi , i = 0, . . . , n, is of finite length.

ThenχA(X ) =

∑ni=0(−1)i`A(Xi ).



Definition


χA(X ) =n∑

i=0

(−1)i`A(Hi (X ))


Lemma



∑ni=0(−1)i`A(Xi ).



Assume that (A,m) is a local ring and f = f1, . . . , ft ⊂ q is a s.o.p. Then:

Hi (f ;A) are A-module of finite lengths (because (f )Hi (f ;A) = 0, andhence Hi (f ;A) are (finitely generated) module over A/(f )).

Li (f , q; n) are also A-modules of finite lengths by definition.

Hence Hi (f , q; n) are also A-modules of finite lengths by the longexact sequence of homologies.

Theorem (Serre, Auslander-Buchsbaum)

χA(K•(f ;A)) = e0(f ;A) .




Hi (f ;A) are A-module of finite lengths

(because (f )Hi (f ;A) = 0, andhence Hi (f ;A) are (finitely generated) module over A/(f )).




χA(K•(f ;A)) = e0(f ;A) .




Hi (f ;A) are A-module of finite lengths (because (f )Hi (f ;A) = 0, andhence Hi (f ;A) are (finitely generated) module over A/(f )).




χA(K•(f ;A)) = e0(f ;A) .



Proposition

For all n� 0

e0(f ;A) = c1 · · · cd · e0(q;A) + χA(K•(f , q; n)) .

In particular, for all n� 0 the Euler characteristic χA(K•(f , q, ; n)) is apositive constant. We denote it by χA(f , q).

Proof (sketch). From the short exact sequence of complexes

0→ K•(f , q; n)→ K•(f ;A)→ L•(f , q; n)→ 0,

we obtain

χA(K•(f ;A) = χA(L•(f , q, ; n)) + χA(K•(f , q, ; n)).



Proposition

For all n� 0



Proof (sketch).

From the short exact sequence of complexes

0→ K•(f , q; n)→ K•(f ;A)→ L•(f , q; n)→ 0,

we obtain




Proposition

For all n� 0




0→ K•(f , q; n)→ K•(f ;A)→ L•(f , q; n)→ 0,

we obtain




Proposition

For all n� 0




0→ K•(f , q; n)→ K•(f ;A)→ L•(f , q; n)→ 0,

we obtain

e0(f ;A) = χA(L•(f , q, ; n)) + χA(K•(f , q, ; n)).

Also, χA(L•(f , q, ; n)) =∑t

i=0(−1)i∑

1≤j1<...<ji≤t `A(A/qn−cj1−...−cji )= c1 · · · cd · e0(q;A) for n� 0.



Proposition

For all n� 0




0→ K•(f , q; n)→ K•(f ;A)→ L•(f , q; n)→ 0,

we obtain


Also, χA(L•(f , q, ; n)) =∑t

i=0(−1)i∑

1≤j1<...<ji≤t `A(A/qn−cj1−...−cji )

= c1 · · · cd · e0(q;A) for n� 0.



Proposition

For all n� 0




0→ K•(f , q; n)→ K•(f ;A)→ L•(f , q; n)→ 0,

we obtain


Also, χA(L•(f , q, ; n)) =∑t

i=0(−1)i∑

1≤j1<...<ji≤t `A(A/qn−cj1−...−cji )= c1 · · · cd · e0(q;A) for n� 0.


Local Bezout inequalities – again

Proposition

e0(f ;A) = c1 · · · cd · e0(q;A) + χA(f , q)

Problem. In order to study local Bezout inequalities, investigate theconstant χA(f , q).


χA(f , q)

Let f ?1 , . . . , ft−1? be a regular sequence in GA(q). Then:

Li (f , q; n) = 0 for all i > 1 and for all n [A.K. 17].

f1, . . . , ft−1 is a regular sequence in A [Valla-Valabrega], and henceHi (f ;A) = 0 for all i > 1.

Hence by using the long exact sequence Hi (f , q; n) = 0 for all i > 1and for all n, and

0→ H1(f , q; n)→ H1(f ;A)→ · · · → L0(f , q; n)→ 0.

If in addition f = f1, . . . , ft is a regular sequence in A (which is true ifA is CM), then H1(f ;A) = 0, and hence H1(f , q; n) = 0 for all n.

χA(f , q) = `A(H0(f , q; n)) for n� 0.


χA(f , q)



f1, . . . , ft−1 is a regular sequence in A [Valla-Valabrega],

and henceHi (f ;A) = 0 for all i > 1.


0→ H1(f , q; n)→ H1(f ;A)→ · · · → L0(f , q; n)→ 0.


χA(f , q) = `A(H0(f , q; n)) for n� 0.


χA(f , q)





0→ H1(f , q; n)→ H1(f ;A)→ · · · → L0(f , q; n)→ 0.


χA(f , q) = `A(H0(f , q; n)) for n� 0.


χA(f , q)




Hence by using the long exact sequence Hi (f , q; n) = 0 for all i > 1and for all n,

and

0→ H1(f , q; n)→ H1(f ;A)→ · · · → L0(f , q; n)→ 0.


χA(f , q) = `A(H0(f , q; n)) for n� 0.


χA(f , q)





0→ H1(f , q; n)→ H1(f ;A)→ · · · → L0(f , q; n)→ 0.


χA(f , q) = `A(H0(f , q; n)) for n� 0.


χA(f , q)





0→ H1(f , q; n)→ H1(f ;A)→ · · · → L0(f , q; n)→ 0.

If in addition f = f1, . . . , ft is a regular sequence in A (which is true ifA is CM),

then H1(f ;A) = 0, and hence H1(f , q; n) = 0 for all n.

χA(f , q) = `A(H0(f , q; n)) for n� 0.


χA(f , q)





0→ H1(f , q; n)→ H1(f ;A)→ · · · → L0(f , q; n)→ 0.

If in addition f = f1, . . . , ft is a regular sequence in A (which is true ifA is CM), then H1(f ;A) = 0,

and hence H1(f , q; n) = 0 for all n.

χA(f , q) = `A(H0(f , q; n)) for n� 0.


χA(f , q)





0→ H1(f , q; n)→ H1(f ;A)→ · · · → L0(f , q; n)→ 0.


χA(f , q) = `A(H0(f , q; n)) for n� 0.


χA(f , q) – continue

• Note that H0(f , q; n) = qn/∑t

i=1 fiqn−ci ,

and we have a short exactsequence

0→ (t∑

i=1

fiqn−ci + qn+1)/

t∑i=1

fiqn−ci ↪→ H0(f , q; n)

→ qn/(t∑

i=1

fiqn−ci + qn+1)→ 0.

• Further, qn/(∑t

i=1 fiqn−ci + qn+1) ∼= [GA(q)/(f ?)]n, where

f ? = f ?1 , . . . , f?t .



• Note that H0(f , q; n) = qn/∑t

i=1 fiqn−ci , and we have a short exact

sequence

0→ (t∑

i=1

fiqn−ci + qn+1)/

t∑i=1

fiqn−ci ↪→ H0(f , q; n)

→ qn/(t∑

i=1

fiqn−ci + qn+1)→ 0.

• Further, qn/(∑t

i=1 fiqn−ci + qn+1) ∼= [GA(q)/(f ?)]n, where

f ? = f ?1 , . . . , f?t .



Proposition

With previous notations, for n� 0, we have

χA(f , q) = `A([GA(q)/(f ?)]n)

+`A((t∑

i=1

fiqn−ci + qn+1)/

t∑i=1

fiqn−ci ).



Let A = k[x1, . . . , xt ](x1,...,xt), q = m = (x1, . . . , xt)A.

Recall e0(m;A) = 1. Hence:

Corollary

(1) For n� 0, we have

e0(f ;A) ≥ c1 · · · ct + `A([GA(m)/(f ?)]n) .

(2) For t = 2; if f ? and g? do not form a s.o.p., then`A([GA(m)/(f ?, g?)]n) = T for all n� 0. [Bydzovsky ’48, Boda-Schenzel’17].



Let A = k[x1, . . . , xt ](x1,...,xt), q = m = (x1, . . . , xt)A.Recall e0(m;A) = 1.

Hence:

Corollary


e0(f ;A) ≥ c1 · · · ct + `A([GA(m)/(f ?)]n) .




Let A = k[x1, . . . , xt ](x1,...,xt), q = m = (x1, . . . , xt)A.Recall e0(m;A) = 1. Hence:

Corollary


e0(f ;A) ≥ c1 · · · ct + `A([GA(m)/(f ?)]n) .




Proof (2). Assume A = k[x , y ](x ,y),m = (x , y)A,B = GA(m) ∼= k[X ,Y ].

(1) f ?, g? is not a system of parameters for B =⇒ ∃ h, f ′, g ′ ∈ B suchthat f ? = hf ′, g? = hg ′ with gcd(f ′, g ′) = 1 and deg(h) > 0.

(2) =⇒ [B/(f ′, g ′)]n = 0 and [B/(f ′, g ′, h)]n = 0 for all n� 0.

(3) =⇒ The following short exact sequence (where h(f ′, g ′) = (f ?, g?))

0→ (f ′, g ′)/h(f ′, g ′)→ B/h(f ′, g ′)→ B/(f ′, g ′)→ 0,

gives [(f ′, g ′)/h(f ′, g ′)]n = [B/h(f ′, g ′)]n for all n� 0.

(4) =⇒ The following short exact sequence

0→ (f ′, g ′, h)/hB → B/hB → B/(f ′, g ′, h)→ 0,

gives [(f ′, g ′, h)/hB]n = [B/hB]n for all n� 0.




0→ (f ′, g ′) :B h/(f ′, g ′)h→ (f ′, g ′)/h(f ′, g ′)

→ (f ′, g ′)/h((f ′, g ′) :B h)→ 0

gives [(f ′, g ′)/h(f ′, g ′)]n = [(f ′, g ′)/h((f ′, g ′) :B h)]n for all n� 0.

(6) Observe that [(f ′, g ′)/h(f ′, g ′)]n = [(f ′, g ′, h)/hB]n for all n� 0,since(f ′, g ′)/h((f ′, g ′) :B h) = (f ′, g ′)/((f ′, g ′) ∩ hB) ∼= (f ′, g ′, h)/hB.

(7) Combining (3), (4) and (6), we get [B/(f ?, g?)]n = [B/hB]n for alln� 0, whose length is deg(h) which is T .




0→ (f ′, g ′) :B h/(f ′, g ′)h→ (f ′, g ′)/h(f ′, g ′)

→ (f ′, g ′)/h((f ′, g ′) :B h)→ 0


(6) Observe that [(f ′, g ′)/h(f ′, g ′)]n = [(f ′, g ′, h)/hB]n for all n� 0,

since(f ′, g ′)/h((f ′, g ′) :B h) = (f ′, g ′)/((f ′, g ′) ∩ hB) ∼= (f ′, g ′, h)/hB.





0→ (f ′, g ′) :B h/(f ′, g ′)h→ (f ′, g ′)/h(f ′, g ′)

→ (f ′, g ′)/h((f ′, g ′) :B h)→ 0







0→ (f ′, g ′) :B h/(f ′, g ′)h→ (f ′, g ′)/h(f ′, g ′)

→ (f ′, g ′)/h((f ′, g ′) :B h)→ 0



(7) Combining (3), (4) and (6), we get [B/(f ?, g?)]n = [B/hB]n for alln� 0,

whose length is deg(h) which is T .




0→ (f ′, g ′) :B h/(f ′, g ′)h→ (f ′, g ′)/h(f ′, g ′)

→ (f ′, g ′)/h((f ′, g ′) :B h)→ 0






Corollary


e0(f ;A) ≥ c1 · · · ct + `A([GA(m)/(f ?)]n) .


Open problem. For t ≥ 3, what is the geometric interpretation of`A([GA(m)/(f ?)]n) for n� 0?



Remark

For n� 0, the following constants coincide:

(1) À([GA(m)/(f ?)]n);

(2) À([Extt−1GA(m)(GA(m)/(f ?),GA(m))]n);

(3) À([(f ?1 , . . . , ft−1?) : f ?t /(f ?1 , . . . , ft−1

?)]n).


THANK YOU


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