3 The General Case

Before we compute the general case, we first introduce some notations that we will use.

Recall that in Section 2, R = Q[x0, x₁, x₂, w]/hx₀w, w²i and we consider the ˇCech complex with respect to the sequence x₀, x₁, x₂

0 → R→ R^d0 _x0 ⊕ R_x1 ⊕ R_x2 → R^d1 _x0x1 ⊕ R_x0x2 ⊕ R_x1x2 → R^d2 _x0x1x2 → 0.^d3

While introducing the notation f_x^0,2₁ (w), we write R_xˆ0x1ˆx2 for R_x1. Using the notation, we can rewrite the above ˇCech complex as

0 → Rxˆ0xˆ1xˆ2

d0

→ Rx0ˆx1xˆ2⊕ Rxˆ0x1xˆ2⊕ Rxˆ0ˆx1x2

d1

→ Rx0x1xˆ2⊕ Rx0xˆ1x2⊕ Rxˆ0x1x2

d2

→ Rx0x1x2

d3

→ 0.

Next, in the above ˇCech complex, we use e_x0xˆ1ˆx2 to represent the component R_x0xˆ1xˆ2 in Rx0ˆx1xˆ2 ⊕ Rˆx0x1xˆ2 ⊕ Rxˆ0xˆ1x2. Similarly, we use exˆ0x1ˆx2 and eˆx0xˆ1x2 to represent the component R_xˆ0x1ˆx2 and R_xˆ0ˆx1x2 in R_x0ˆx1xˆ2 ⊕ R_ˆx0x1xˆ2 ⊕ R_xˆ0xˆ1x2, respectively. Thus, for example, if (f^1,2, f^0,2, f^0,1) ∈ Rx0xˆ1ˆx2 ⊕ Rxˆ0x1xˆ2 ⊕ Rxˆ0xˆ1x2, we write (f^1,2, f^0,2, f^0,1) = f^1,2e_x0xˆ1xˆ2 + f^0,2e_xˆ0x1ˆx2 + f^0,1e_x0xˆ1xˆ2.

Next, we will introduce how we order the component of Cⁱin the ˇCech complex. Take the above complex as an example. C² = R_x0x1xˆ2 ⊕ R_x0xˆ1x2 ⊕ R_xˆ0x1x2 and there are three components in C². Among these three components,

1. R_x0x1xˆ2 and R_x0xˆ1x2 both have the index x₀ while R_xˆ0x1x2 has the index ˆx₀, so we write R_x0x1xˆ2 and R_x0xˆ1x2 before R_xˆ0x1x2;

2. R_x0x1xˆ2 and R_x0xˆ1x2 both have the index x₀, and R_x0x1xˆ2 has the index x₁ while R_x0ˆx1x2 has the index ˆx₁, so we write R_x0x1xˆ2 before R_x0xˆ1x2.

Hence, we order the components of C² as C² = R_x0x1ˆx2⊕R_x0ˆx1x2⊕R_ˆx0x1x2. Next, in order to make this ordering clearer, we use another example to illustrate it. For the ring R =

Q[x⁰, w, x₁, x₂, x₃, x₄, x₅, x₆]/hx₀w, w²i, we compare the components R_{x0x1x2x3xˆ4x5ˆx6} and R_{x0x1x2xˆ3x4xˆ5x6} of C⁵. Note that they both have the indices x₀, x₁, x₂, and R_{x0x1x2x3xˆ4x5xˆ6} the index x₃while R_{x0x1x2xˆ3x4xˆ5x6} has ˆx₃. Thus we write R_{x0x1x2x3xˆ4x5ˆx6} before R_{x0x1x2xˆ3x4ˆx5x6} when we order the components of C⁵.

Now, we are ready to compute the general case. From now on, R is the ring Q[x⁰, x₁, x₂, . . . , x_n, w]/hx₀w, w²i and p is the maximal ideal hx₀, x₁, x₂, . . . , x_n, wi in R.

Since hx₀, x₁, x₂, . . . , x_ni is p-primary, we consider the ˇCech complex with respect to the sequence x0, x1, x2, . . . , xn.

Similar as in Section 2, we write each component that occurs in the ˇCech complex as an internal direct sum of Q-vector spaces. For C⁰ = R, we write

R = Q[x⁰, x1, . . . , xn] + Q[x¹, x2, . . . , xn]w.

Since x₀w = 0 in R, for C¹ = R_x0xˆ1xˆ2···ˆxn⊕ R_xˆ0x1xˆ2···ˆxn⊕ · · · ⊕ R_ˆx0xˆ1xˆ2···xn, we have

R_x0ˆx1xˆ2···ˆxn = Q[x0, x₁, . . . , x_n] +

∞

X

α0=1

Q[x1, x₂, . . . , x_n]x^−α₀ ⁰, and for i = 1, 2, . . . , n,

Rˆx0xˆ1···ˆxi−1xixˆi+1···ˆxn = Q[x⁰, x1, . . . , xn] + Q[x¹, x2, . . . , xn]w +

∞

X

αi=1

Q[x⁰, x1, . . . , xi−1, xi+1, . . . , xn]x^−α_i ⁱ

+

∞

X

αi=1

Q[x1, x₂, . . . , x_i−1, x_i+1, . . . , x_n]wx^−α_i ⁱ.

In general, for the R-module C^t=L

1≤i1<i2<...<it≤nR_xi1xi2···x_it, there are two categories of components, namely x_i1 = 0 and x_i1 ≥ 1. For example, for the R-module Cⁱ⁺¹, because

x₀w = 0 and so w = 0 in R_{x0x1...xixˆi+1xˆi+2···ˆxn}, we have

R_x0x1···x_ixˆi+1xˆi+2···ˆxn = Q[x0, x₁, x₂, . . . , x_n] +

∞

X

α0=1

Q[x1, x₂, . . . , x_n]x^−α₀ ⁰

+ · · · +

∞

X

αi=1

Q[x0, x₁, . . . , x_i−1, x_i+1, . . . , x_n]x^−α_i ⁱ

+

∞

X

α0=1

∞

X

α1=1

Q[x2, x₃, . . . , x_n]x^−α₀ ⁰x^−α₁ ¹

+ · · · +

∞

X

αi−1=1

∞

X

αi=1

Q[x0, x₁, . . . , x_i−2, x_i+1, . . . , x_n]x^−α_i−1ⁱ⁻¹x^−α_i ⁱ

+ · · · +

∞

X

α0=1

∞

X

α1=1

· · ·

∞

X

αi=1

Q[xⁱ⁺¹, xi+2, . . . , xn]x^−α₀ ⁰x^−α₁ ¹· · · x^−α_i ⁱ;

because w 6= 0 in R_{xˆ0x1x2···xi+1ˆxi+2···ˆxn}, we have

Moreover, we use the same special notation that we use in Section 2. More precisely, we use the upper indices to indicate the ring where the elements come from; we use the lower indices to indicate the variables that have negative powers; we also use the notation (w) to indicate that the elements are multiple of w. For example, if we denote an element as f_x^0,2

1x3(w), we know that

1. this element comes from R_{ˆx0x1xˆ2x3···xn},

2. all terms of this element have negative powers for both x₁ and x₃ and have non-negative powers for x₀, x₂, x₄, . . . , x_n,

3. this element is a multiple of w;

in other words, we know that f_x^0,2

1x2(w) ∈

∞

P

α1=1

∞

P

α3=1

Q[x2, x₄, . . . , x_n]wx^−α₁ ¹x^−α₃ ³∩R_xˆ0x1ˆx2x3···x_n.

Example 3.1 Let R = Q[x0, x₁, x₂, . . . , x_n, w]/hx₀w, w²i and let C be the ˇCech complex with respect to the sequence x₀, x₁, x₂, . . . , x_n. Then we have H⁰(C) = 0.

Proof. Note that H⁰(C) = Ker d₀, where 0 → R_xˆ0xˆ1···ˆxn

d0

→ R_x0xˆ1xˆ2···ˆxn ⊕ R_xˆ0x1xˆ2···ˆxn⊕ · · · ⊕ R_ˆx0xˆ1xˆ2···ˆxn−1xn.

Let f^0,1,...,n + f^0,1,...,n(w) ∈ Rxˆ0ˆx1···ˆxn = Q[x⁰, x1, . . . , xn] + Q[x¹, x2, . . . , xn]w such that d₀(f^0,1,...,n+ f^0,1,...,n(w)) = (0, . . . , 0). Then we have

(f^0,1,...,n, f^0,1,...,n+ f^0,1,...,n(w), . . . , f^0,1,...,n+ f^0,1,...,n(w)) = (0, . . . , 0).

Thus f^0,1,...,n = 0 in Q[x0, x₁, . . . , x_n], and f^0,1,...,n(w) = 0 in Q[x1, . . . , x_n]w, and so f^0,1,...,n+ f^0,1,...,n(w) = 0 in R. Hence we get Ker d₀ = 0, and so H⁰(C) = 0. 2

Before we start the next theorem, we introduce some notations that will be used in the proof. We let I = {x₀, x₁, x₂, . . . , x_n} and for {x_i1, x_i2, . . . , x_it} ⊆ I, we let

I^i1,i2,...,it = I − {x_i1, x_i2, . . . , x_it}.

Also, for x_i ∈ I and J = {x_j1, x_j2, . . . , x_jt} ⊆ I, we let

σ(x_i, J ) =







(−1)^s−1 if j₁ < j₂ < · · · < j_s−1 < i < j_s< · · · < j_t, 0 if x_i ∈ J.

With the above notation, we can rewrite the component R_xi1···x_it → R_xj1xj2···x_jt+1 in Definition 1.1, that gives the differentiation d_t : C^t → C^t+1, as







σ(x_j, J ) · nat : R_xi1···x_it → (R_xi1···x_it)_xj if {x_j1, . . . , x_jt+1} = J ∪ {x_j},

0 otherwise

where J = {x_i1, . . . , x_it}. Note that J = I^A where A = I − {i₁, . . . , i_t}. For exam-ple, the component R_x0x1···x_ixˆi+1xˆi+2...ˆxn → R_x0x1···x_ixˆi+1xi+2xˆi+3...ˆxn is the homomorphism σ(x_i+2, Ii+1,i+2,...,n) · nat : R_x0x1···xi → (R_x0x1···xi)_xi+2.

Example 3.2 Let R = Q[x0, x₁, x₂, . . . , x_n, w]/hx₀w, w²i and let C be the ˇCech complex with respect to the sequence x₀, x₁, x₂, . . . , x_n. Then we have Hⁱ⁺¹(C) = 0 for all 0 ≤ i ≤ n − 2.

Remark 3.3 Note that Hⁱ⁺¹(C) = Ker d_i+1/Im d_i, where

R_{x0x1···xi−1xˆixˆi−1···ˆxn}⊕ · · · ⊕ R_{ˆx0xˆ1···ˆxn−ixn−i+1···xn}

di

→ R_{x0x1···xiˆxi+1xˆi+2···ˆxn}⊕ · · · ⊕ R_{xˆ0xˆ1···ˆxn−i−1xn−i···xn}

di+1

→ R_{x0x1···xi+1ˆxi+2xˆi+3···ˆxn}⊕ · · · ⊕ R_{xˆ0xˆ1···ˆxn−i−2xn−i−1···xn}.

Before we start to compute Hⁱ⁺¹(C), as we did in the proof of Theorem 2.2, we first write R_x0x1···x_i−1xˆixˆi−1···ˆxn, . . ., R_xˆ0ˆx1···ˆxn−ixn−i+1···x_n as internal direct sums and then col-lect the components with the same Q-vector spaces together to get an internal direct sum of Cⁱ = R_x0x1···x_i−1xˆixˆi−1···ˆxn ⊕ · · · ⊕ R_xˆ0ˆx1···ˆxn−ixn−i+1···x_n. Similarly, we also express Cⁱ⁺¹= R_x0x1···x_ixˆi+1xˆi+2···ˆxn⊕ · · · ⊕ R_ˆx0xˆ1···ˆxn−i−1xn−i···xn and Cⁱ⁺²= R_x0x1···x_i+1xˆi+2xˆi+3···ˆxn⊕

· · · ⊕ R_xˆ0ˆx1···ˆxn−i−2xn−i−1···x_n as internal direct sums. Similarly as in (3), d_i+1 sends each component in Cⁱ⁺¹ to the component in Cⁱ⁺²corresponding to the same Q-vector space.

Hence if we express every element a in Cⁱ⁺¹as a sum of terms in the components with the same Q-vector spaces, then we have a ∈ Ker di+1 if and only if every term of a belongs to Ker d_i+1. Similarly, we have a ∈ Im d_i if and only if every term of a belongs to Im d_i. Hence, it is sufficient to show for each component that Ker d_i+1⊆ Im d_i.

We first consider the component which has no negative powers for x₀, x₁, x₂, . . . , x_n and not a multiple of w, i.e., the component with respect to Q[x0, x₁, . . . , x_n]. For f ∈

Cⁱ⁺¹, since

Cⁱ⁺¹ = R_x0x1···x_ixˆi+1ˆxi+2···ˆxn⊕ · · · ⊕ R_x0xˆ1xˆ2···ˆxn−ixn−i+1···xn

⊕ R_xˆ0x1···x_i+1ˆxi+2xˆi+3···ˆxn⊕ · · · ⊕ R_xˆ0ˆx1···ˆxn−i−1xn−i···xn, we write

f = fi+1,i+2,...,n

e_x0x1···xixˆi+1ˆxi+2···ˆxn + · · · + f1,2,...,n−i

e_x0xˆ1xˆ2···ˆxn−ixn−i+1···xn

+ f0,i+2,i+3,...,n

exˆ0x1···xi+1xˆi+2xˆi+3···ˆxn+ · · · + f0,1,...,n−i−1

eˆx0xˆ1···ˆxn−i−1xn−i···xn.

Lemma 3.4 Let f ∈ Cⁱ⁺¹ and let f = fi+1,i+2,...,n

ex0x1···xixˆi+1ˆxi+2···ˆxn+ · · · + f1,2,...,n−i

ex0xˆ1ˆx2···ˆxn−ixn−i+1···xn

+ f0,i+2,i+3,...,n

exˆ0x1···xi+1xˆi+2xˆi+3···ˆxn+ · · · + f0,1,...,n−i−1

exˆ0xˆ1···ˆxn−i−1xn−i···xn, i.e., f is in the component which has no negative powers for x₀, x₁, x₂, . . . , x_n and is not a multiple of w. Then f ∈ Ker di+1 implies f ∈ Im di.

Proof. Since d_i+1(f ) = 0 in Cⁱ⁺²and Cⁱ⁺²= R_x0x1···x_i+1xˆi+2xˆi+3···ˆxn⊕· · ·⊕R_x0xˆ1xˆ2···ˆxn−i−1xn−i···xn⊕ Rxˆ0x1···xi+2xˆi+3xˆi+4···ˆxn⊕· · ·⊕Rxˆ0xˆ1···ˆxn−i−2xn−i−1···xn, for the component ex0x1···xi+1xˆi+2xˆi+3···ˆxn, we have

σ(xi+1, Ii+1,i+2,i+3,...,n

)fi+1,i+2,i+3,...,n

+ · · · + σ(x1, I1,i+2,i+3,...,n

)f1,i+2,i+3,...,n

+ σ(x0, I0,i+2,i+3,...,n

)f0,i+2,i+3,...,n

= 0.

Then we have that the coefficient of the component e_{xˆ0,x1,...,xi+1,ˆxi+2,ˆxi+3,...,ˆxn} of f is f0,i+2,i+3,...,n= −σ(x₀, I0,i+2,i+3,...,n)σ(x_i+1, Ii+1,i+2,i+3,...,n)fi+1,i+2,i+3,...,n

− · · ·

− σ(x₀, I0,i+2,i+3,...,n)σ(x₁, I1,i+2,i+3,...,n)f1,i+2,i+3,...,n.

Similarly for the component e_{x0x1···xixˆi+1xi+2xˆi+3···ˆxn}, we have Repeat the same discussion for all the remaining components with index x₀, i.e., ex0x1···xixˆi+1xˆi+2xi+3ˆxi+4···ˆxn, . . . , ex0xˆ1xˆ2···ˆxn−i−1xn−i···xn, and we will get

From the above relations, we can rewrite f as

f = fi+1,i+2,...,ne_x0x1···x_ixˆi+1ˆxi+2···ˆxn + · · · + f1,2,...,n−i+1e_x0ˆx1xˆ2···ˆxn−i+1xn−i+2···xn

+ − σ(x₀, I0,i+2,i+3,...,n)σ(x_i+1, Ii+1,i+2,i+3,...,n)fi+1,i+2,i+3,...,n− · · ·

− σ(x₀, I0,i+2,i+3,...,n)σ(x₁, I1,i+2,i+3,...,n)f1,i+2,i+3,...,n
e_xˆ0x1···x_i+1xˆi+2ˆxi+3···ˆxn

+ − σ(x₀, I0,i+1,i+3,...,n)σ(x_i+2, Ii+1,i+2,...,n)fi+1,i+2,...,n

− σ(x₀, I0,i+1,i+3,...,n)σ(x_i, Ii,i+1,i+3,...,n)fi,i+1,i+3,...,n

− · · ·

− σ(x₀, I0,i+1,i+3,...,n

)σ(x₁, I1,i+1,i+3,...,n

)f1,i+1,i+3,...,n
e_x0x1···xixˆi+1xi+2ˆxi+3···ˆxn

+ · · ·

+ − σ(x₀, I0,1,2,...,n−i−1

)σ(x_n, I1,2,...,n−i−1,n

)f1,2,...,n−i−1,n− · · ·

− σ(x₀, I0,1,2,...,n−i−1

)σ(x_n−i, I1,2,...,n−i−1,n−i

)f1,2,...,n−i−1,n−i
e_ˆx0xˆ1···ˆxn−i−1xn−i···xn. (7) In order to show that f ∈ Im d_i, we need to find g ∈ Cⁱ such that d_i(g) = f . Note that

for g = gi,i+1,...,ne_{x0x1···xi−1ˆxixˆi+1···ˆxn}+ · · · + g0,1,2,...,n−ie_{xˆ0xˆ1···ˆxn−ixn−i+1···xn}, d_i(gi,i+1,...,ne_{x0x1···xi−1xˆixˆi+1···ˆxn}+ · · · + g0,1,2,...,n−ie_{xˆ0xˆ1···ˆxn−ixn−i+1···xn})

= σ(x_i, Ii,i+1,i+2,...,n)gi,i+1,...,n+ · · · + σ(x₀, I0,i+1,i+2,...,n)g0,i+1,i+2,...,n
ex0x1···xixˆi+1xˆi+2···ˆxn

+ σ(x_i+1, Ii,i+1,i+2,...,n)gi,i+1,i+2,...,n+ σ(x_i−1, Ii−1,i,i+2,...,n)gi−1,i,i+2,...,n

+ · · · + σ(x₀, I0,i,i+2,...,n)g0,i,i+2,...,n
ex0...xi−1xˆixi+1ˆxi+2···ˆxn

+ · · ·

+ σ(x_n, I1,2,...,n−i,n)g1,2,...,n−i,n+ · · · + σ(x_n−i+1, I1,2,...,n−i,n−i+1)g1,2,...,n−i,n−i+1

+ σ(x₀, I0,1,2,...,n−i)g0,1,2,...,n−i
e_{x0xˆ1···ˆxn−i−1xˆn−ixn−i+1···xn} + σ(x_i+1, I0,i+1,i+2,i+3,...,n)g0,i+1,i+2,...,n+ · · ·

+ σ(x₁, I0,1,i+2,i+3,...,n)g0,1,i+2,i+3,...,n
e_ˆx0x1···x_i+1xˆi+2···ˆxn

+ · · ·

+ σ(x_n, I0,1,2,...,n−i−1,n)g0,1,2,...,n−i−1,n+ · · ·

+ σ(x_n−i, I0,1,2,...,n−i−1,n−i)g0,1,2,...,n−i−1,n−i
e_xˆ0xˆ1···ˆxn−i−1xn−i···x_n. (8) We compare (7) with (8). For example, for the component e_ˆx0x1···x_i+1xˆi+2···ˆxn, we want to get

σ(x_i+1, I0,i+1,i+2,i+3,...,n)g0,i+1,i+2,...,n+ · · · + σ(x₁, I0,1,i+2,i+3,...,n)g0,1,i+2,i+3,...,n

= −σ(x₀, I0,i+2,i+3,...,n)σ(x_i+1, Ii+1,i+2,i+3,...,n)fi+1,i+2,i+3,...,n

− · · ·

− σ(x₀, I0,i+2,i+3,...,n)σ(x₁, I1,i+2,i+3,...,n)f1,i+2,i+3,...,n, so we choose

g0,i+1,i+2,...,n = −σ(x_i+1, I0,i+1,i+2,i+3,...,n)σ(x₀, I0,i+2,i+3,...,n) σ(x_i+1, Ii+1,i+2,i+3,...,n)fi+1,i+2,i+3,...,n,

...

g0,1,i+2,i+3,...,n = −σ(x₁, I0,1,i+2,i+3,...,n)σ(x₀, I0,i+2,i+3,...,n) σ(x₁, I1,i+2,i+3,...,n)f1,i+2,i+3,...,n.

We do the same thing for the components e_{xˆ0x1···xixˆi+1xi+2ˆxi+3···ˆxn}, . . . , e_{ˆx0xˆ1···ˆxn−i−1xn−i···xn} and want to choose g^{0,k1,k2,...,kn−i}accordingly for all 0 < k₁ < k₂ < · · · < k_n−i ≤ n. We also choose all the g^{k1,k2,...,kn−i+1} with k₁ > 0 to be zero. Note that the term g0,i+1,i+2,...,n not only contributes to the component e_ˆx0x1···x_i+1ˆxi+2···ˆxn but also contributes to the component e_{xˆ0x1···xixˆi+1xi+2xˆi+3···ˆxn}. More precisely, for the component e_{xˆ0x1···xixˆi+1xi+2ˆxi+3···ˆxn}, we want to get

σ(x_i+2, I0,i+1,i+2,i+3,...,n)g0,i+1,i+2,...,n+ σ(x_i, I0,i,i+1,i+3,...,n)g0,i,i+1,i+3,...,n+ · · · + σ(x₁, I0,1,i+1,i+3,...,n)g0,1,i+1,i+3,...,n

= −σ(x₀, I0,i+1,i+3,...,n)σ(x_i+2, Ii+1,i+2,i+3,...,n)fi+1,i+2,i+3,...,n

− σ(x₀, I0,i+1,i+3,...,n)σ(x_i, Ii,i+1,i+3,...,n)fi,i+1,i+3,...,n

− · · ·

− σ(x₀, I0,i+1,i+3,...,n)σ(x₁, I1,i+2,i+3,...,n)f1,i+1,i+3,...,n, and so we want to choose

g0,i+1,i+2,...,n

= −σ(x_i+2, I0,i+1,i+2,i+3,...,n)σ(x₀, I0,i+1,i+3,...,n)σ(x_i+2, Ii+1,i+2,i+3,...,n)fi+1,i+2,i+3,...,n. However, this is the same as the earlier choice, since

− σ(x_i+1, I0,i+1,i+2,i+3,...,n)σ(x₀, I0,i+2,i+3,...,n)σ(x_i+1, Ii+1,i+2,i+3,...,n)

= −(−1)ⁱ(−1)⁰(−1)ⁱ⁺¹

= (−1)²ⁱ⁺²

= 1 and

− σ(x_i+2, I0,i+1,i+2,i+3,...,n

)σ(x₀, I0,i+1,i+3,...,n

)σ(x_i+2, Ii+1,i+2,i+3,...,n

)

= −(−1)ⁱ(−1)⁰(−1)ⁱ⁺¹

= (−1)²ⁱ⁺²

= 1.

Similarly, g0,i+1,i+2,...,n also contributes to the components e_{xˆ0x1···xiˆxi+1xˆi+2xi+3xˆi+4···ˆxn}, . . . , e_xˆ0x1···x_ixˆi+1···ˆxn−1xn. However, we have for each j = i + 1, . . . , n,

σ(x_j, I0,i+1,...,n) = (−1)ⁱ, σ(x₀, I0,i+1,...,j−1,j+1,...,n) = (−1)⁰, and σ(x_j, I^i+1,...,n) = (−1)ⁱ⁺¹, and so we have

g0,i+1,i+2,...,n

= −σ(xi+1, I0,i+1,i+2,i+3,...,n

)σ(x0, I0,i+2,i+3,...,n

)σ(xi+1, Ii+1,i+2,i+3,...,n

)fi+1,i+2,...,n

= −σ(xi+2, I0,i+1,i+2,i+3,...,n

)σ(x0, I0,i+1,i+3,...,n

)σ(xi+2, Ii+1,i+2,i+3,...,n

)fi+1,i+2,...,n

= · · ·

= −σ(xn, I0,i+1,i+2,i+3,...,n

)σ(x0, I0,i+1,i+3,...,n−1

)σ(xn, Ii+1,i+2,i+3,...,n

)fi+1,i+2,...,n

. Hence, we indeed choose the same g0,i+1,i+2,...,n even when we consider different compo-nents. In general, for all 1 ≤ k₁ < k₂ < · · · < k_n−i≤ n, since for each j = 1, 2, . . . , n − i,

σ(x_kj, I^{0,k1,k2,...,kn−i})σ(x_kj, I^{k1,k2,...,kn−i}) = (−1)^kj−j(−1)^kj−(j−1) = (−1)^2kj−2j+1 = −1, and since σ(x₀, I^{0,k1,...,kj−1,kj+1,...,kn−i}) = (−1)⁰ = 1, we have

g^{0,k1,k2,...,kn−i}

= −σ(xk1, I^{0,k1,k2,k3,...,kn−i})σ(x0, I^{0,k2,k3,...,kn−i})σ(xk1, I^{k1,k2,k3,...,kn−i})f^{k1,k2,...,kn−i}

= −σ(xk2, I^{0,k1,k2,k3,...,kn−i})σ(x0, I^{0,k1,k3,...,kn−i})σ(xk2, I^{k1,k2,k3,...,kn−i})f^{k1,k2,...,kn−i}

= · · ·

= −σ(xkn−i, I^{0,k1,k2,k3,...,kn−i})σ(x0, I^{0,k1,k2,...,kn−i−1})σ(xkn−i, I^{k1,k2,...,kn−i})f^{k1,k2,...,kn−i}. (9) Therefore, the choices made for each g^{0,k1,k2,...,kn−i} while considering different components are in fact all the same. Now we use (9) to show that d_i(g) = f , where

g = g0,i+1,i+2,...,ne_xˆ0x1···x_ixˆi+1xˆi+2···ˆxn + · · · + g0,1,2,...,n−ie_ˆx0xˆ1···ˆxn−ixn−i+1···xn, i.e.,

d_i g0,i+1,i+2,...,n

e_xˆ0x1···xixˆi+1xˆi+2···ˆxn + · · · + g0,1,2,...,n−i

e_ˆx0xˆ1···ˆxn−ixn−i+1···xn

= fi+1,i+2,...,n

e_x0x1···xixˆi+1···ˆxn+ · · · + f1,2,...,n−i

e_x0xˆ1···ˆxn−ixn−i+1···xn

+ f0,i+2,i+3,...,n

e_xˆ0x1···xi+1xˆi+2xˆi+3···ˆxn+ · · · + f0,1,...,n−i−1

e_ˆx0xˆ1···ˆxn−i−1xn−i···xn. (10)

We first take care of the components in the second line of (10), namely the components e_x0x1···x_ixˆi+1···ˆxn, . . . , e_x0xˆ1···ˆxn−ixn−i+1···xn. For the component e_x0x1···x_ixˆi+1···ˆxn, since the components in g that do contribute to it are the components

e_x0x1···x_i−1xˆixˆi+1···ˆxn, e_x0x1···x_i−2xˆi−1xixˆi+1···ˆxn, . . . , e_x0xˆ1x2···x_ixˆi+1···ˆxn, e_xˆ0x1x2···x_iˆxi+1···ˆxn, and since we take

gi,i+1,i+2,...,n

= gi−1,i+1,i+2,...,n

= · · · = g1,i+1,i+2,...,n

= 0, and

g0,i+1,i+2,...,n

= −σ(xi+1, I0,i+1,i+2,i+3,...,n

)σ(x0, I0,i+2,i+3,...,n

)σ(xi+1, Ii+1,i+2,i+3,...,n

)fi+1,i+2,...,n

, we get that the coefficient of e_x0x1···x_iˆxi+1···ˆxn in d_i(g) is

σ(x₀, I0,i+1,i+2,...,n)g0,i+1,i+2,...,n

= σ(x₀, I0,i+1,i+2,...,n) − σ(x_i+1, I0,i+1,i+2,...,n)σ(x₀, I0,i+2,...,n)σ(x_i+1, Ii+1,i+2,i+3,...,n)fi+1,i+2,...,n

= (−1)⁰ − (−1)ⁱ(−1)⁰(−1)ⁱ⁺¹fi+1,i+2,...,n

= fi+1,i+2,...,n.

Similarly, with the same argument for the remaining components e_x0x1···x_i+1ˆxixi+1xˆi+2···ˆxn, . . . , e_{x0ˆx1···ˆxn−ixn−i+1···xn} in the second line of (10), we see that their coefficients are indeed fi,i+2,i+3,...,n, . . . , f1,2,...,n−i, respectively.

Next we check the first component e_ˆx0x1···x_i+1ˆxi+2xˆi+3···ˆxn in the third line of (10). Note that the components in g that do contribute to the component e_{xˆ0x1···xi+1xˆi+2···ˆxn} are the components e_xˆ0x1···x_ixˆi+1xˆi+2···ˆxn, . . . , e_ˆx0xˆ1x2···x_i+1xˆi+2···ˆxn. Moreover, since we take











g0,i+1,i+2,...,n = −σ(x_i+1, I0,i+1,i+2,...,n)σ(x₀, I0,i+2,i+3,...,n)σ(x_i+1, Ii+1,i+2,...,n)fi+1,i+2,...,n, ...

g0,1,i+2,...,n = −σ(x₁, I0,1,i+2,i+3,...,n)σ(x₀, I0,i+2,i+3,...,n)σ(x₁, I1,i+2,...,n)f1,i+2,...,n,

we see that the coefficient of e_{ˆx0x1···xi+1ˆxi+2···ˆxn} in d_i(g) is

σ(x_i+1, I0,i+1,i+2,...,n)g0,i+1,i+2,...,n+ · · · + σ(x₁, I0,1,i+2,i+3,...,n)g0,1,i+2,...,n

= σ(x_i+1, I0,i+1,i+2,...,n) − σ(x_i+1, I0,i+1,i+2,...,n)σ(x₀, I0,i+2,i+3,...,n)σ(x_i+1, Ii+1,i+2,...,n)fi+1,i+2,...,n
+ · · ·

+ σ(x₁, I0,1,i+2,i+3,...,n) − σ(x₁, I0,1,i+2,i+3,...,n)σ(x₀, I0,i+2,i+3,...,n)σ(x₁, I1,i+2,...,n)f1,i+2,...,n

= −σ(x₀, I0,i+2,i+3,...,n)σ(x_i+1, Ii+1,i+2,...,n)fi+1,i+2,...,n− · · ·

− σ(x₀, I0,i+2,i+3,...,n

)σ(x₁, I1,i+2,...,n

)f1,i+2,i+3,...,n

= f0,i+2,i+3,...,n

,

where the last equality follows from (6).

Next we check the second component e_xˆ0x1···x_ixˆi+1xi+2ˆxi+3···ˆxn in the third line of (10).

Since the components in g that do contribute to the component e_xˆ0x1···x_ixˆi+1xi+2xˆi+3···ˆxn are the components e_xˆ0x1···x_ixˆi+1xˆi+2xˆi+3···ˆxn, e_xˆ0x1···x_i−1xˆixˆi+1xi+2xˆi+3···ˆxn, . . . , e_ˆx0xˆ1x2···x_ixˆi+1xi+2xˆi+3···ˆxn. Use (9) and we have



















g0,i+1,i+2,...,n = −σ(x_i+2, I0,i+1,i+2,...,n)σ(x₀, I0,i+1,i+3,...,n)σ(x_i+2, Ii+1,i+2,...,n)fi+1,i+2,...,n

g0,i,i+1,i+3,...,n = −σ(x_i, I0,i,i+1,i+3,...,n)σ(x₀, I0,i+1,i+3,...,n)σ(x_i, Ii,i+1,i+3,...,n)fi,i+1,i+3,...,n

...

g0,1,i+1,i+3,...,n = −σ(x₁, I0,1,i+1,i+3,...,n)σ(x₀, I0,i+1,i+3,...,n)σ(x₁, I1,i+1,i+3,...,n)f1,i+1,i+3,...,n. Note that in the above relations we choose the expression

g0,i+1,i+2,...,n = −σ(x_i+2, I0,i+1,i+2,...,n)σ(x₀, I0,i+1,i+3,...,n)σ(x_i+2, Ii+1,i+2,...,n)fi+1,i+2,...,n, instead of

g0,i+1,i+2,...,n = −σ(x_i+1, I0,i+1,i+2,...,n)σ(x₀, I0,i+2,...,n)σ(x_i+1, Ii+1,i+2,...,n)fi+1,i+2,...,n.

Then we have that the coefficient of e_{ˆx0x1···xixˆi+1xi+2xˆi+3···ˆxn} in d_i(g) is

σ(x_i+2, I0,i+1,i+2,...,n)g0,i+1,i+2,...,n+ σ(x_i, I0,i,i+1,i+3,...,n))g0,i,i+1,i+3,...,n+ · · · + σ(x₁, I0,1,i+1,i+3,...,n)g0,1,i+1,i+3,...,n

= σ(x_i+2, I0,i+1,i+2,...,n) − σ(x_i+2, I0,i+1,i+2,...,n)σ(x₀, I0,i+1,i+3,...,n)σ(x_i+2, Ii+1,i+2,...,n)fi+1,i+2,...,n
+ σ(x_i, I0,i,i+1,i+3,...,n) − σ(x_i, I0,i,i+1,i+3,...,n)σ(x₀, I0,i+1,i+3,...,n)

· σ(x_i, Ii,i+1,i+3,...,n)fi,i+1,i+3,...,n
+ · · ·

+ σ(x₁, I0,1,i+1,i+3,...,n

) − σ(x₁, I0,1,i+1,i+3,...,n

)σ(x₀, I0,i+1,i+3,...,n

)

· σ(x₁, I1,i+1,i+3,...,n

)f1,i+1,i+3,...,n

= −σ(x₀, I0,i+1,i+3,...,n

)σ(x_i+2, Ii+1,i+2,...,n

)fi+1,i+2,...,n

− σ(x₀, I0,i+1,i+3,...,n

)σ(x_i, Ii,i+1,i+3,...,n

)fi,i+1,i+3,...,n

− · · ·

− σ(x₀, I0,i+1,i+3,...,n

)σ(x₁, I1,i+1,i+3,...,n

)f1,i+1,i+3,...,n

= f0,i+1,i+3,...,n

,

where the last equality follows from (6).

Similarly, we do the same argument for the remaining components e_xˆ0x1···x_ixˆi+1xˆi+2xi+3ˆxi+4···ˆxn, . . . , e_xˆ0···ˆxn−i−1xn−i···xn in the third line of (10) and finish the proof. 2

Next, we consider the component which has no negative powers for x₀, x₁, x₂, . . . , x_n and is a multiple of w, i.e., the component with respect to Q[x¹, x2, . . . , xn]w. For f (w) ∈ Cⁱ⁺¹, since

Cⁱ⁺¹= R_{x0x1···xiˆxi+1xˆi+2···ˆxn}⊕ · · · ⊕ R_{x0xˆ1···ˆxn−ixn−i+1···xn}

⊕ R_{xˆ0x1···xi+1xˆi+2xˆi+3···ˆxn}⊕ · · · ⊕ R_{xˆ0x1xˆ2···ˆxn−ixn−i+1···xn}

⊕ R_{xˆ0ˆx1x2···xi+2xˆi+3···ˆxn} ⊕ · · · ⊕ R_{xˆ0xˆ1···ˆxn−i−1xn−i···xn},

and since w = 0 in R_{x0x1···xiˆxi+1xˆi+2···ˆxn}, . . . , R_{x0xˆ1···ˆxn−ixn−i+1···xn}, we write f (w) = f0,i+2,i+3,...,n

(w)e_xˆ0x1···xi+1xˆi+2···ˆxn + · · · + f0,2,...,n−i

(w)e_xˆ0x1ˆx2···ˆxn−ixn−i+1···xn

+ f0,1,i+3,i+4,...,n

(w)e_xˆ0ˆx1x2···xi+2xˆi+3ˆxi+4···ˆxn+ · · · + f0,1,...,n−i−1

(w)e_xˆ0xˆ1···ˆxn−i−1xn−i···xn.

Lemma 3.5 Let f (w) ∈ Cⁱ⁺¹ and let

f (w) = f0,i+2,i+3,...,n(w)e_xˆ0x1···x_i+1xˆi+2···ˆxn + · · · + f0,2,...,n−i(w)e_xˆ0x1ˆx2···ˆxn−ixn−i+1···xn

+ f0,1,i+3,i+4,...,n(w)e_ˆx0xˆ1x2···x_i+2xˆi+3xˆi+4···ˆxn + · · · + f0,1,...,n−i−1(w)e_xˆ0ˆx1···ˆxn−i−1xn−i···xn, i.e., f (w) is contained in the component that has no negative powers for x₀, x₁, x₂, . . . , x_n and is a multiple of w. Then f (w) ∈ Ker di+1 implies f (w) ∈ Im di.

Proof. Since d_i+1(f (w)) = 0 in Cⁱ⁺² and

Cⁱ⁺²= R_x0x1···xi+1ˆxi+2xˆi+3···ˆxn⊕ · · · ⊕ R_x0ˆx1xˆ2···ˆxn−i−1xn−i···xn

⊕ R_xˆ0x1···xi+2xˆi+3xˆi+4···ˆxn⊕ · · · ⊕ R_xˆ0ˆx1···ˆxn−i−2xn−i−1···xn, for the component e_xˆ0x1···x_i+2xˆi+3xˆi+4···ˆxn, we have

σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n(w) + · · · + σ(x₂, I0,2,i+3,...,n)f0,2,i+3,...,n(w) + σ(x₁, I0,1,i+3,...,n)f0,1,i+3,...,n(w) = 0.

Then we have that the coefficient of the component e_ˆx0xˆ1x2···xi+2xˆi+3xˆi+4···ˆxn of f (w) is f0,1,i+3,...,n(w) = −σ(x₁, I0,1,i+3,...,n)σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n(w)

− · · ·

− σ(x₁, I0,1,i+3,...,n)σ(x₂, I0,2,i+3,...,n)f0,2,i+3,...,n(w).

Similarly for the component e_{ˆx0x1···xi+1ˆxi+2xi+3xˆi+4···ˆxn}, we have σ(x_i+3, I0,i+2,i+3,i+4,...,n

)f0,i+2,i+3,i+4,...,n

(w) + σ(x_i+1, I0,i+1,i+2,i+4,...,n

)f0,i+1,i+2,i+4,...,n

(w) + · · ·

+ σ(x₂, I0,2,i+2,i+4,...,n

)f0,2,i+2,i+4,...,n

(w) + σ(x₁, I0,1,i+2,i+4,...,n

)f0,1,i+2,i+4,...,n

(w) = 0.

Then we have that the coefficient of the component e_{ˆx0xˆ1x2···xi+1xˆi+2xi+3ˆxi+4···ˆxn} of f (w) is f0,1,i+2,i+4...,n(w) = −σ(x₁, I0,1,i+2,i+4,...,n)σ(x_i+3, I0,i+2,i+3,i+4,...,n)f0,i+2,i+3,i+4,...,n(w)

− σ(x₁, I0,1,i+2,i+4,...,n)σ(x_i+1, I0,i+1,i+2,i+4,...,n)f0,i+1,i+2,i+4,...,n(w)

− · · ·

− σ(x₁, I0,1,i+2,i+4,...,n)σ(x₂, I0,2,i+2,i+4,...,n)f0,2,i+2,i+4,...,n(w).

Repeat the same discussion for all the remaining components with index ˆx₀ and x₁, i.e., e_xˆ0x1···xi+1xˆi+2xˆi+3xi+4ˆxi+5···ˆxn, . . . , e_{xˆ0x1xˆ2ˆx3}···ˆxn−i−2xˆn−i−1···xn, and we will get



























































f0,1,i+3,...,n(w) = −σ(x1, I0,1,i+3,...,n)σ(xi+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n(w)

− · · ·

−σ(x1, I0,1,i+3,...,n)σ(x2, I0,2,i+3,...,n)f0,2,i+3,...,n(w)

f0,1,i+2,i+4...,n(w) = −σ(x₁, I0,1,i+2,i+4,...,n)σ(x_i+3, I0,i+2,i+3,i+4,...,n)f0,i+2,i+3,i+4,...,n(w)

−σ(x₁, I0,1,i+2,i+4,...,n)σ(x_i+1, I0,i+1,i+2,i+4,...,n)f0,i+1,i+2,i+4,...,n(w)

− · · ·

−σ(x₁, I0,1,i+2,i+4,...,n)σ(x₂, I0,2,i+2,i+4,...,n)f0,2,i+2,i+4,...,n(w) ...

f0,1,2,...,n−i−1(w) = −σ(x₁, I0,1,2,...,n−i−1)σ(x_n, I0,2,3,...,n−i−1,n)f0,2,3,...,n−i−1,n(w)

− · · ·

−σ(x₁, I0,1,2,...,n−i−1)σ(x_n−i, I0,2,3,...,n−i−1,n−i)f0,2,3,...,n−i−1,n−i(w).

(11)

From the above relations, we can rewrite f (w) as f (w)

= f0,i+2,...,n(w)e_ˆx0x1···x_ixi+1ˆxi+2···ˆxn+ · · · + f0,2,3,...,n−i(w)e_{xˆ0x1xˆ2ˆx3}···ˆxn−ixn−i+1···xn

+ − σ(x₁, I0,1,i+3,...,n)σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n(w) − · · ·

− σ(x₁, I0,1,i+3,...,n)σ(x₂, I0,2,i+3,...,n)f0,2,i+3,...,n(w)
e_xˆ0ˆx1x2···x_i+2xˆi+3ˆxi+4···ˆxn

+ − σ(x₁, I0,1,i+2,i+4,...,n)σ(x_i+3, I0,i+2,i+3,i+4,...,n)f0,i+2,i+3,i+4,...,n(w)

− σ(x₁, I0,1,i+2,i+4,...,n

)σ(x_i+1, I0,i+1,i+2,i+4,...,n

)f0,i+1,i+2,i+4,...,n

(w)

− · · ·

− σ(x₁, I0,1,i+2,i+4,...,n

)σ(x₂, I0,2,i+2,i+4,...,n

)f0,2,i+2,i+4,...,n(w)
e_ˆx0xˆ1x2···xi+1xˆi+2xi+3ˆxi+4···ˆxn

+ · · ·

+ − σ(x₁, I0,1,2,3,...,n−i−1

)σ(x_n, I0,2,3,...,n−i−1,n

)f0,2,3,...,n−i−1,n

(w) − · · ·

− σ(x₁, I0,1,2,3,...,n−i−1

)σ(x_n−i, I0,2,3,...,n−i−1,n−i

)f0,2,3,...,n−i−1,n−i(w)
e_xˆ0ˆx1···ˆxn−i−1xn−i···xn. (12) In order to show that f (w) ∈ Im d_i, we need to find g(w) ∈ Cⁱ such that d_i(g(w)) = f (w).

Note that for g(w) = g0,i+1,i+2,...,n(w)e_{xˆ0x1···xiˆxi+1···ˆxn}+· · ·+g0,1,2,...,n−i(w)e_{xˆ0xˆ1···ˆxn−ixn−i+1···xn}, d_i g0,i+1,i+2,...,n(w)e_xˆ0x1···x_ixˆi+1···ˆxn + · · · + g0,1,2,...,n−i(w)e_xˆ0ˆx1···ˆxn−ixn−i+1···xn

= σ(x_i+1, I0,i+1,i+2,...,n)g0,i+1,i+2,...,n(w) + · · ·

+ σ(x₁, I0,1,i+2,...,ng0,1,i+2,...,n(w)
e_ˆx0x1x2···x_i+1xˆi+2···ˆxn

+ · · ·

+ σ(x_n, I0,2,3,...,n−i,n)g0,2,3,...,n−i,n(w) + · · · + σ(x_n−i+1, I0,2,3,...,n−i,n−i+1)g0,2,3,...,n−i,n−i+1(w) + σ(x₁, I0,1,2,3,...,n−i,n

)g0,1,2,...,n−i(w)
e_{xˆ0x1xˆ2xˆ3}···ˆxn−ixn−i+1···xn

+ σ(x_i+2, I0,1,i+2,i+3,...,n

)g0,1,i+2,i+3...,n

(w) + · · · + σ(x₂, I0,1,2,i+3,...,n

)g0,1,2,i+3,...,n(w)
e_ˆx0xˆ1x2···xi+2xˆi+3···ˆxn

+ σ(x_i+3, I0,1,i+2,...,n

)g0,1,i+2,...,n

(w) + σ(x_i+1, I0,1,i+1,i+2,i+4,...,n

)g0,1,i+1,i+2,i+4,...,n

(w) + · · · + σ(x₂, I0,1,2,i+2,i+4,...,n

)g0,1,2,i+2,i+4,...,n(w)
e_xˆ0xˆ1x2···xi+1xˆi+2xi+3xˆi+4···ˆxn

+ · · ·

+ σ(x_n, I0,1,...,n−i−1,n

)g0,1,2,...,n−i−1,n

(w) + · · · + σ(xn−i, I0,1,...,n−i−1,n−i

)g0,1,2,...,n−i−1,n−i(w)
exˆ0xˆ1···ˆxn−i−1xn−i···xn. (13) We compare (12) with (13). For example, for the component e_xˆ0ˆx1x2···x_i+2xˆi+3ˆxi+4···ˆxn, we want to get

σ(x_i+2, I0,1,i+2,i+3,...,n)g0,1,i+2,i+3...,n(w) + · · · + σ(x₂, I0,1,2,i+3,...,n)g0,1,2,i+3,...,n(w)

= −σ(x₁, I0,1,i+3,...,n)σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n(w)

− · · ·

− σ(x₁, I0,1,i+3,...,n)σ(x₂, I0,2,i+3,...,n)f0,2,i+3,...,n(w),

so we choose

g0,1,i+2,i+3,...,n(w)

= −σ(x_i+2, I0,1,i+2,i+3,...,n)σ(x₁, I0,1,i+3,...,n)σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n(w), ...

g0,1,2,i+3,...,n(w)

= −σ(x₂, I0,1,2,i+3,...,n)σ(x₁, I0,1,i+3,...,n)σ(x₂, I0,2,i+3,...,n)f0,2,i+3,...,n(w).

We do the same thing for the components e_xˆ0xˆ1x2···x_i+1xˆi+2xi+3xˆi+4···ˆxn, . . . , e_ˆx0xˆ1···ˆxn−i−1xn−i···x_n

and want to choose g^{0,1,k1,...,kn−i−1}(w) accordingly for all 1 < k₁ < k₂ < · · · < k_n−i−1 ≤ n.

We also choose the other g^{0,k1,k2,...,kn−i}(w) with k₁ > 1 to be zero. Note that the term g0,1,i+2,i+3...,n(w) not only contributes to the component e_ˆx0xˆ1x2···xi+2xˆi+3···ˆxn but also con-tributes to the component e_xˆ0ˆx1x2···x_i+1ˆxi+2xi+3xˆi+4···ˆxn. More precisely, for the component e_xˆ0ˆx1x2···xi+1ˆxi+2xi+3xˆi+4···ˆxn, we want to get

σ(x_i+3, I0,1,i+2,i+3,...,n)g0,1,i+2,i+3,...,n(w) + σ(x_i+1, I0,1,i+1,i+2,i+4,...,n)g0,1,i+1,i+2,i+4,...,n(w) + · · ·

+ σ(x₂, I0,1,2,i+2,i+4,...,n)g0,1,2,i+2,i+4,...,n(w)

= −σ(x₁, I0,1,i+2,i+4,...,n)σ(x_i+3, I0,i+2,i+3,i+4,...,n)f0,i+2,i+3,i+4,...,n(w)

− σ(x₁, I0,1,i+2,i+4,...,n)σ(x_i+1, I0,i+1,i+2,i+4,...,n)f0,i+1,i+2,i+4,...,n(w)

− · · ·

− σ(x₁, I0,1,i+2,i+4,...,n)σ(x₂, I0,2,i+2,i+4,...,n)f0,2,i+2,i+4,...,n(w) and so we want to choose

g0,1,i+2,i+3,...,n(w)

= −σ(x_i+3, I0,1,i+2,i+3,...,n)σ(x₁, I0,1,i+2,i+4,...,n)σ(x_i+3, I0,i+2,i+3,i+4,...,n)f0,i+2,i+3,i+4,...,n(w).

However, this is the same as the earlier choice, since

− σ(x_i+2, I0,1,i+2,i+3,...,n)σ(x₁, I0,1,i+3,i+4,...,n)σ(x_i+2, I0,i+2,i+3,...,n)

= −(−1)ⁱ(−1)⁰(−1)ⁱ⁺¹

= (−1)²ⁱ⁺²

= 1 and

− σ(x_i+3, I0,1,i+2,i+3,...,n)σ(x₁, I0,1,i+2,i+4,...,n)σ(x_i+3, I0,i+2,i+3,i+4,...,n)

= −(−1)ⁱ(−1)⁰(−1)ⁱ⁺¹

= (−1)²ⁱ⁺²

= 1.

Similarly, g0,1,i+2,i+3,...,n(w) also contributes to the components e_xˆ0xˆ1x2···x_i+1xˆi+2xˆi+3xi+4ˆxi+5···ˆxn, . . . , e_{xˆ0ˆx1x2···xi+1xˆi+2···ˆxn−1xn}. However, we have for each j = i + 2, i + 3, . . . , n,

σ(x_j, I0,1,i+2,i+3,...,n) = (−1)ⁱ and σ(x_j, I0,i+2,i+3,...,n) = (−1)ⁱ⁺¹, and so we have

g0,1,i+2,i+3,...,n(w)

= −σ(x_i+2, I0,1,i+2,i+3,...,n)σ(x₁, I0,1,i+3,i+4,...,n)σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n(w)

= −σ(x_i+3, I0,1,i+2,i+3,...,n)σ(x₁, I0,1,i+2,i+4,...,n)σ(x_i+3, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n(w)

= · · ·

= −σ(x_n, I0,1,i+2,i+3,...,n)σ(x₁, I0,1,i+2,i+3,...,n−1)σ(x_n, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n(w).

In general, for all 1 < k₁ < k₂ < · · · < k_n−i−1 ≤ n, since for each j = 1, 2, . . . , n − i − 1, σ(x_kj, I^{0,1,k1,k2,k3,...,kn−i−1})σ(x_kj, I^{0,k1,k2,k3,...,kn−i−1})

= (−1)^kj−(j+1)(−1)^kj−j

= (−1)^2kj−2j−1

= −1,

and σ(x₁, I^{0,1,k1,...,kj−1,kj+1,...,kn−i−1}) = (−1)⁰ = 1, we have g^{0,1,k1,k2,...,kn−i−1}(w)

= −σ(x_k1, I^{0,1,k1,k2,k3,...,kn−i−1})σ(x₁, I^{0,1,k2,k3,...,kn−i−1})σ(x_k1, I^{0,k1,k2,k3,...,kn−i−1})

· f^{0,k1,k2,...,kn−i−1}(w)

= −σ(x_k2, I^{0,1,k1,k2,k3,...,kn−i−1})σ(x₁, I^{0,1,k1,k3,...,kn−i−1})σ(x_k2, I^{0,k1,k2,k3,...,kn−i−1})

· f^{0,k1,k2,...,kn−i−1}(w)

= · · ·

= −σ(x_kn−i−1, I^{0,1,k1,k2,k3,...,kn−i−1})σ(x₁, I^{0,1,k2,k3,...,kn−i−2})σ(x_kn−i−1, I^{0,k1,k2,k3,...,kn−i−1})

· f^{0,k1,k2,...,kn−i−1}(w) (14)

Therefore, the choices made for each g^{0,1,k1,k2,...,kn−i−1}(w) while considering different com-ponents are in fact all the same. Now we use (14) to show that di(g(w)) = f (w), where

g(w) = g0,1,i+2,...,n(w)e_xˆ0xˆ1x2···x_i+1xˆi+2···ˆxn + · · · + g0,1,2,...,n−i(w)e_xˆ0ˆx1···ˆxn−ixn−i+1···x_n, i.e.,

d_i g0,1,i+2,...,n(w)e_{ˆx0xˆ1x2cdotsxi+1xˆi+2}···ˆxn+ · · · + g0,1,2,...,n−i(w)e_ˆx0xˆ1···ˆxn−ixn−i+1···xn

= f0,i+2,i+3,...,n(w)e_ˆx0x1···x_i+1xˆi+2···ˆxn + · · · + f0,2,3,...,n−i(w)e_{ˆx0x1xˆ2xˆ3}···ˆxn−ixn−i+1···xn

+ f0,1,i+3,i+4,...,n(w)e_xˆ0ˆx1x2···x_i+2ˆxi+3xˆi+4···ˆxn+ · · · + f0,1,...,n−i−1(w)e_ˆx0xˆ1···ˆxn−i−1xn−i···xn. (15) We first take care of the components in the second line of (15), namely the components exˆ0x1···xi+1xˆi+2···ˆxn, . . . , exˆ0x1xˆ2xˆ3···ˆxn−ixn−i+1···xn. For the first component exˆ0x1···xi+1xˆi+2···ˆxn, since the components in g(w) that do contribute to it are the components

exˆ0x1···xiˆxi+1xˆi+2···ˆxn, exˆ0x1···xi−1ˆxixi+1xˆi+2···ˆxn, . . . , eˆx0x1xˆ2x3···xi+1ˆxi+2···ˆxn, exˆ0xˆ1x2···xi+1xˆi+2···ˆxn, and since we take

g0,i+1,i+2,...,n(w) = · · · = g0,3,i+2,...,n(w) = g0,2,i+2,...,n(w) = 0

and

g0,1,i+2,i+3,...,n(w)

= −σ(x_i+2, I0,1,i+2,i+3,...,n)σ(x₁, I0,1,i+3,i+4,...,n)σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n(w), we see that the coefficient of e_ˆx0x1···x_i+1ˆxi+2···ˆxn in d_i(g(w)) is

σ(x₁, I0,1,i+2,i+3,...,n

)g0,1,i+2,i+3,...,n

(w)

= σ(x1, I0,1,i+2,i+3,...,n

) − σ(xi+2, I0,1,i+2,i+3,...,n

)σ(x1, I0,1,i+3,i+4,...,n

)

· σ(xi+2, I0,i+2,i+3,i+4,...,n

)f0,i+2,i+3,i+4,...,n

(w)

= (−1)⁰ − (−1)ⁱ(−1)⁰(−1)ⁱ⁺¹f0,i+2,i+3,i+4,...,n

(w)

= (−1)²ⁱ⁺²f0,i+2,i+3,i+4,...,n

(w)

= f0,i+2,i+3,i+4,...,n

(w).

Similarly, with the same argument for the remaining components e_xˆ0x1···x_iˆxi+1xi+2xˆi+3···ˆxn, . . . , exˆ0x1xˆ2xˆ3···ˆxn−i···xn, in the second line of (15), we see that their coefficients are indeed f0,i+1,i+3,...,n, . . . , f0,2,...,n−i, respectively.

Next we check the first component in the third line of (15). Note that the components in g(w) that do contribute to the component exˆ0xˆ1x2···xi+2xˆi+3xˆi+4...ˆxn are the components e_xˆ0ˆx1x2···x_i+1ˆxi+2xˆi+3xˆi+4···ˆxn, . . . , e_{xˆ0ˆx1x2xˆ3x4}···x_i+2xˆi+3xˆi+4···ˆxn, e_{ˆx0xˆ1xˆ2x3}···x_i+2xˆi+3xˆi+4···ˆxn. More-over, since we take























g0,1,i+2,i+3...,n(w) = −σ(x_i+2, I0,1,i+2,i+3,...,n)σ(x₁, I0,1,i+3,...,n)

· σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n(w), ...

g0,1,2,i+3,...,n(w) = −σ(x₂, I0,1,2,i+3,...,n)σ(x₁, I0,1,i+3,...,n)

· σ(x₂, I0,2,i+3,...,n)f0,2,i+3,...,n(w),

we see that the coefficient of e_{ˆx0xˆ1x2···xi+2xˆi+3xˆi+4...ˆxn} in d_i(g(w)) is

σ(x_i+2, I0,1,i+2,i+3,...,n)g0,1,i+2,i+3,...,n(w) + · · · + σ(x₂, I0,1,2,i+3...,n)g0,1,2,i+3,...,n(w)

= σ(x_i+2, I0,1,i+2,i+3,...,n) − σ(x_i+2, I0,1,i+2,i+3,...,n)σ(x₁, I0,1,i+3,...,n)σ(x_i+2, I0,i+2,i+3,...,n)

· f0,i+2,i+3,...,n(w)
+ · · ·

+ σ(x₂, I0,1,2,i+3,...,n) − σ(x₂, I0,1,2,i+3,...,n)σ(x₁, I0,1,i+3,...,n)σ(x₂, I0,2,i+3,...,n)f0,2,i+3,...,n(w)

= −σ(x₁, I0,1,i+3,...,n)σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n(w)

− · · ·

− σ(x₁, I0,1,i+3,...,n)σ(x₂, I0,2,i+3,...,n)f0,2,i+3,...,n(w)

= f0,1,i+3,...,n(w),

where the last equality follows from (11).

Next we check the second component e_xˆ0ˆx1x2···xi+1xˆi+2xi+3xˆi+4···ˆxnin the third line of (15).

Since the components in g(w) that do contribute to the components e_ˆx0xˆ1x2···x_i+1xˆi+2xi+3ˆxi+4···ˆxn

are the components

e_{xˆ0xˆ1x2···xi+1xˆi+2xˆi+3xˆi+4···ˆxn}, e_{xˆ0xˆ1x2···xixˆi+1xˆi+2xi+3ˆxi+4···ˆxn}, . . . , e_{xˆ0xˆ1xˆ2x3···xi+1xˆi+2xi+3xˆi+4···ˆxn}. Use (14) and we have



































g0,1,i+2,i+3,i+4,...,n(w) = −σ(x_i+3, I0,1,i+2,i+3,i+4,...,n)σ(x₁, I0,1,i+2,i+4,...,n)

· σ(xi+3, I0,i+2,i+3,i+4,...,n)f0,i+2,i+3,i+4,...,n(w) g0,1,i+1,i+2,i+4,...,n(w) = −σ(x_i+1, I0,1,i+1,i+2,i+4,...,n)σ(x₁, I0,1,i+2,i+4,...,n)

· σ(xi+1, I0,i+1,i+2,i+4,...,n)f0,i+1,i+2,i+4,...,n(w) ...

g0,1,2,i+2,i+4,...,n(w) = −σ(x2, I0,1,2,i+2,i+4,...,n)σ(x1, I0,1,i+2,i+4,...,n)

· σ(x₂, I0,2,i+2,i+4,...,n)f0,2,i+2,i+4,...,n(w).

Note that in the above relation we choose the expression g0,1,i+2,i+3,i+4,...,n

(w)

= −σ(x_i+3, I0,1,i+2,i+3,i+4,...,n

)σ(x₁, I0,1,i+2,i+4,...,n

)σ(x_i+3, I0,i+2,i+3,i+4,...,n

)f0,i+2,i+3,i+4,...,n

(w),

instead of

g0,1,i+2,i+3,i+4,...,n(w)

= −σ(x_i+2, I0,1,i+2,i+3,i+4,...,n)σ(x₁, I0,1,i+3,i+4,...,n)σ(x_i+2, I0,i+2,i+3,i+4,...,n)f0,i+2,i+3,i+4,...,n(w).

Then we have that the coefficient of e_ˆx0xˆ1x2···x_i+1xˆi+2xi+3ˆxi+4···ˆxn in d_i(g(w)) is σ(x_i+3, I0,1,i+2,i+3,i+4,...,n

)g0,1,i+2,i+3,i+4,...,n

(w) + σ(x_i+1, I0,1,i+1,i+2,i+4,...,n

)g0,1,i+1,i+2,i+4,...,n

(w) + · · ·

+ σ(x2, I0,1,2,i+2,i+4,...,n

)g0,1,2,i+2,i+4,...,n

(w)

= σ(xi+3, I0,1,i+2,i+3,i+4,...,n

) − σ(xi+3, I0,1,i+2,i+3,i+4,...,n

)σ(x1, I0,1,i+2,i+4,...,n

)

· σ(xi+3, I0,i+2,i+3,i+4,...,n

)f0,i+2,i+3,i+4,...,n

(w)
+ σ(xi+1, I0,1,i+1,i+2,i+4,...,n

) − σ(xi+1, I0,1,i+1,i+2,i+4,...,n

)σ(x1, I0,1,i+2,i+4,...,n

)

· σ(xi+1, I0,i+1,i+2,i+4,...,n

)f0,i+1,i+2,i+4,...,n

(w)
+ · · ·

+ σ(x2, I0,1,2,i+2,i+4,...,n

) − σ(x2, I0,1,2,i+2,i+4,...,n

)σ(x1, I0,1,i+2,i+4,...,n

)

· σ(x2, I0,2,i+2,i+4,...,n

)f0,2,i+2,i+4,...,n

(w)

= −σ(x1, I0,1,i+2,i+4,...,n

)σ(xi+3, I0,i+2,i+3,i+4,...,n

)f0,i+2,i+3,i+4,...,n

(w)

− σ(x1, I0,1,i+2,i+4,...,n

)σ(xi+1, I0,i+1,i+2,i+4,...,n

)f0,i+1,i+2,i+4,...,n

(w)

− · · ·

− σ(x₁, I0,1,i+2,i+4,...,n)σ(x₂, I0,2,i+2,i+4,...,n)f0,2,i+2,i+4,...,n(w)

= f0,1,i+2,i+4,...,n(w),

where the last equality follows from (11).

Similarly, we do the same argument for the remaining components e_xˆ0xˆ1x2···x_i+1ˆxi+2xˆi+3xi+4xˆi+5···ˆxn, . . . , exˆ0ˆx1···ˆxn−i−1xn−i···xn in the third line of (15) and finish the proof. 2

Next, for 0 ≤ j < i, we consider the component which has negative powers for

x₀, x₁, . . . , x_j and is not a multiple of w, i.e., the component with respect to

∞

X

α0=1

∞

X

α1=1

· · ·

∞

X

αj=1

Q[xj+1, . . . , x_n]x^−α₀ ⁰x^−α₁ ¹· · · x^−α_j ^j.

For f_x0x1···x_j ∈ Cⁱ⁺¹, since

Cⁱ⁺¹= R_x0x1···x_ixˆi+1xˆi+2···ˆxn ⊕ · · · ⊕ R_x0x1···x_j+1xˆj+2···ˆxn+j−i+1xn+j−i+2···xn

⊕ R_x0x1···x_jxˆj+1xj+2···x_i+1xˆi+2···ˆxn⊕ · · · ⊕ R_x0x1x2···x_jxˆj+1···ˆxn+j−ixn+j−i+1···xn

⊕ R_x0x1···x_j−1xˆjxj+1···x_i+1xˆi+2···ˆxn⊕ · · · ⊕ R_xˆ0xˆ1···ˆxn−i−1xn−i···xn, and since none of the components in the last line of the above equation has

∞

X

α0=1

∞

X

α1=1

· · ·

∞

X

αj=1

Q[xj+1, . . . , x_n]x^−α₀ ⁰x^−α₁ ¹· · · x^−α_j ^j as a subspace, we write

f_x0x1···x_j = fi+1,i+2,...,n

x0x1...xj e_x0x1···x_ixˆi+1···ˆxn+ · · · + fj+2,...,n+j−i+1

x0x1...xj e_x0x1···x_j+1xˆj+2···ˆxn+j−i+1xn+j−i+2···x_n

+ fj+1,i+2,...,n

x0x1...xj e_x0x1···x_jxˆj+1xj+2···x_i+1xˆi+2···ˆxn+ · · · + fj+1,...,n+j−i

x0x1...xj e_x0x1x2···x_jxˆj+1···ˆxn+j−ixn+j−i+1···x_n.

Lemma 3.6 Let f_x0x1···x_j ∈ Cⁱ⁺¹, and let f_x0x1···xj = fi+1,i+2,...,n

x0x1...xj e_{x0x1···xixˆi+1···ˆxn}+ · · · + fj+2,...,n+j−i+1

x0x1...xj e_{x0x1···xj+1xˆj+2···ˆxn+j−i+1xn+j−i+2···xn} + fj+1,i+2,...,n

x0x1...xj e_{x0x1···xjˆxj+1xj+2···xi+1ˆxi+2···ˆxn} + · · · + fj+1,...,n+j−i

x0x1...xj e_{x0x1x2···xjxˆj+1···ˆxn+j−ixn+j−i+1···xn},

i.e., f_x0x1···x_j is in the component which has negative powers for x₀, x₁, . . . , x_j and is not a multiple of w. Then f_x0x1···xj ∈ Ker d_i+1 implies f_x0x1···xj ∈ Im d_i.

Proof. Since d_i+1(f_x0x1x2···x_j) = 0 in Cⁱ⁺² and

Cⁱ⁺²= R_x0x1···xixi+1xˆi+2···ˆxn⊕ · · · ⊕ R_x0x1···xj+1xˆj+2···ˆxn+j−ixn+j−i+1···xn

⊕ R_x0x1···xjxˆj+1xj+2···xi+2xˆi+3···ˆxn⊕ · · · ⊕ R_xˆ0xˆ1···ˆxn−i−2xn−i−1···xn

for the component e_{x0x1···xj···xixi+1ˆxi+2···ˆxn}, we have σ(x_i+1, Ii+1,i+2,...,n)fi+1,i+2,...,n

x0x1...xj + · · · + σ(x_j+2, Ij+2,i+2,...,n)fj+2,i+2,...,n x0x1...xj

+ σ(x_j+1, Ij+1,i+2,...,n)fj+1,i+2,...,n x0x1...xj = 0.

Then we have that the coefficient of the component e_x0x1···x_jxˆj+1xj+2···x_i+1xˆi+2···ˆxn of f_x0x1...xj is

fj+1,i+2,...,n

x0x1...xj = −σ(x_j+1, Ij+1,i+2,...,n)σ(x_i+1, Ii+1,i+2,...,n)fi+1,i+2,...,n x0x1...xj

− · · ·

− σ(x_j+1, Ij+1,i+2,...,n)σ(x_j+2, Ij+2,i+2,...,n)fj+2,i+2,...,n x0x1...xj . Similarly for the component ex0x1···xj···xixˆi+1xi+2xˆi+3···ˆxn, we have

σ(x_i+2, I^i+1,...,n)f_x^i+1,...,n_0x1...xj + σ(x_i, Ii,i+1,i+3,...,n)fi,i+1,i+3,...,n x0x1...xj + · · · + σ(x_j+2, Ij+2,i+1,i+3,...,n)fj+2,i+1,i+3,...,n

x0x1...xj + σ(x_j+1, Ij+1,i+1,i+3,...,n)fj+1,i+1,i+3,...,n x0x1...xj = 0.

Then we have that the coefficient of the component e_{x0···xjˆxj+1xj+2···xixˆi+1xi+2xˆi+3···ˆxn} of f_x0x1...xj is

fj+1,i+1,i+3,...,n

x0x1...xj = −σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_i+2, I^i+1,...,n)f_x^i+1,...,n

0x1...xj

− σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_i, Ii,i+1,i+3,...,n)fi,i+1,i+3,...,n x0x1...xj

− · · ·

− σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_j+2, Ij+2,i+1,i+3,...,n)fj+2,i+1,i+3,...,n x0x1...xj . Repeat the same discussion for all the remaining components with index x₀, x₁, . . . , x_j,

i.e., e_{x0x1···xj···xixˆi+1xˆi+2xi+3xˆi+4···ˆxn}, . . . , e_{x0x1x2···xj+1xˆj+2···ˆxn+j−ixn+j−i+1···xn} and we will get



























































fj+1,i+2,...,n

x0x1...xj = −σ(x_j+1, Ij+1,i+2,...,n)σ(x_i+1, Ii+1,i+2,...,n)fi+1,i+2,...,n x0x1...xj

− · · ·

−σ(x_j+1, Ij+1,i+2,...,n)σ(x_j+2, Ij+2,i+2,...,n)fj+2,i+2,...,n x0x1...xj

fj+1,i+1,i+3,...,n

x0x1...xj = −σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_i+2, I^i+1,...,n)f_x^i+1,...,n

0x1...xj

−σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_i, Ii,i+1,i+3,...,n)fi,i+1,i+3,...,n x0x1...xj

− · · ·

−σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_j+2, Ij+2,i+1,i+3,...,n)fj+2,i+1,i+3,...,n x0x1...xj

... fj+1,j+2,...,n+j−i

x0x1...xj = −σ(x_j+1, Ij+1,j+2,...,n+j−i)σ(x_n, Ij+2,...,n+j−i,n)fj+2,...,n+j−i,n x0x1...xj

− · · ·

−σ(x_j+1, Ij+1,j+2,...,n+j−i)σ(x_n+j−i+1, Ij+2,...,n+j−i+1)fj+2,...,n+j−i+1 x0x1...xj .

(16)

From the above relations, we can rewrite f_x0x1...xj as and since none of the components in the second line above has

∞ as a subspace, we only need to consider

g_x0x1...xj = gi,i+1,...,n

x0x1...xj e_{x0x1···xi−1xˆi···ˆxn}+ · · · + gj+1,j+2,...,n+j−i+1

x0x1...xj e_{x0···xjˆxj+1···ˆxn+j−i+1xn+j−i+2···xn}.

Note that d_i(gi,i+1,...,n

x0x1...xje_x0x1···x_i−1xˆi···ˆxn+ · · · + gj+1,j+2,...,n+j−i+1

x0x1...xj e_x0···x_jxˆj+1···ˆxn+j−i+1xn+j−i+2···xn)

= σ(x_i, Ii,i+1,...,n)gi,i+1,...,n

x0x1···xj + · · · + σ(x_j+1, Ij+1,i+1,...,n)gj+1,i+1,...,n

x0x1···xj
e_x0···x_ixˆi+1xˆi+2···ˆxn

+ · · ·

+ σ(x_n, Ij+2,...,n+j−i+1,n)gj+2,...,n+j−i+1,n x0x1···xj + · · ·

+ σ(x_n, Ij+2,...,n+j−i+1,n+j−i+2)gj+2,...,n+j−i+1,n+j−i+2 x0x1···xj

+ σ(x_j+1, Ij+1,j+2,...,n+j−i+1

)gj+1,j+2,...,n+j−i+1

x0x1···xj
e_x0···xj+1ˆxj+2···ˆxn+j−i+1xn+j−i+2···xn

+ σ(x_i+1, Ij+1,i+1,i+2,...,n

)gj+1,i+1,i+2,...,n

x0x1···xj + · · · + σ(x_j+3, Ij+1,j+3,i+2,...,n

)gj+1,j+3,i+2,...,n x0x1···xj

+ σ(x_j+2, Ij+1,j+2,i+2,...,n

)gj+1,j+2,i+2,...,n

x0x1···xj
e_x0···xjˆxj+1xj+2···xi+1xˆi+2···ˆxn

+ σ(x_i+2, Ij+1,i+1,i+2,i+3,...,n

)gj+1,i+1,i+2,i+3,...,n

+ σ(x_i, Ij+1,i,i+1,i+3,...,n

)gj+1,i,i+1,i+3,...,n

+ · · ·

+ σ(x_j+2, Ij+1,j+2,i+1,i+3,...,n

)gj+1,j+2,i+1,i+3,...,n
e_x0x1···xjxˆj+1xj+2···xixˆi+1xi+2xˆi+3···ˆxn

+ · · ·

+ σ(xn, Ij+1,...,n+j−i,n

)gj+1,...,n+j−i,n

x0x1···xj + · · · + σ(xn+j−i+2, Ij+1,...,n+j−i,n+j−i+2

)gj+1,...,n+j−i,n+j−i+2 x0x1···xj

+ σ(xn+j−i+1, Ij+1,...,n+j−i,n+j−i+1

)gj+1,...,n+j−i,n+j−i+1

x0x1···xj
ex0···xjˆxj+1···ˆxn+j−ixn+j−i+1···xn. (18) We compare (17) with (18). For example, for the component e_x0···x_jxˆj+1xj+2···x_i+1xˆi+2···ˆxn, we want to get

σ(x_i+1, Ij+1,i+1,i+2,...,n

)gj+1,i+1,i+2,...,n

x0x1···xj + · · · + σ(x_j+2, Ij+1,j+2,i+2,...,n

)gj+1,j+2,i+2,...,n x0x1···xj

= −σ(x_j+1, Ij+1,i+2,...,n

)σ(x_i+1, Ii+1,i+2,...,n

)fi+1,i+2,...,n x0x1...xj

− · · ·

− σ(x_j+1, Ij+1,i+2,...,n

)σ(x_j+2, Ij+2,i+2,...,n

)fj+2,i+2,...,n x0x1...xj ,

so we choose gj+1,i+1,i+2,...,n

x0x1···xj

= −σ(x_i+1, Ij+1,i+1,i+2,...,n)σ(x_j+1, Ij+1,i+2,...,n)σ(x_i+1, Ii+1,i+2,...,n)fi+1,i+2,...,n x0x1...xj , ...

gj+1,j+2,i+2,...,n x0x1···xj

= −σ(x_j+2, Ij+1,j+2,i+2,...,n

)σ(x_j+1, Ij+1,i+2,...,n

)σ(x_j+2, Ij+2,i+2,...,n

)fj+2,i+2,...,n x0x1...xj . We do the same thing for the components e_x0···x_jˆxj+1xj+2···x_ixˆi+1xi+2xˆi+3···ˆxn, . . . ,

e_x0···xjxˆj+1···ˆxn+j−ixn+j−i+1···xn, and want to choose g^j+1,kx0x1···x^1,k_j^2,...,kn−i accordingly for all j + 1 <

k₁ < k₂ < · · · < k_n−i ≤ n. We also choose all the g^{k1,k2,...,kn+i+1} with k₁ > j + 1 to be zero. Note that the term gj+1,i+1,i+2,...,n

x0x1···xj not only contributes to the component e_x0···x_jxˆj+1xj+2···x_i+1xˆi+2···ˆxnbut also contributes to the component e_x0···x_jˆxj+1xj+2···x_ixˆi+1xi+2xˆi+3···ˆxn. More precisely, for the component e_x0···xjxˆj+1xj+2···xiˆxi+1xi+2xˆi+3···ˆxn, we want to get

σ(x_i+2, Ij+1,i+1,i+2,i+3,...,n)gj+1,i+1,i+2,i+3,...,n

x0x1...xj + σ(x_i, Ij+1,i,i+1,i+3,...,n)gj+1,i,i+1,i+3,...,n x0x1...xj

+ · · ·

+ σ(x_j+2, Ij+1,j+2,i+1,i+3,...,n)gj+1,j+2,i+1,i+3,...,n x0x1...xj

= −σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_i+2, I^i+1,...,n)f_x^i+1,...,n_0x1...x

j

− σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_i, Ii,i+1,i+3,...,n)fi,i+1,i+3,...,n x0x1...xj

− · · ·

− σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_j+2, Ij+2,i+1,i+3,...,n)fj+2,i+1,i+3,...,n x0x1...xj

and so we want to choose gj+1,i+1,i+2,...,n

x0x1···xj = −σ(x_i+2, Ij+1,i+1,i+2,...,n)σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_i+2, Ii+1,i+2,...,n)fi+1,i+2,...,n x0x1...xj . However, this is the same as the earlier choice, since

− σ(x_i+1, Ij+1,i+1,i+2,...,n

)σ(x_j+1, Ij+1,i+2,...,n

)σ(x_i+1, Ii+1,i+2,...,n

)

= −(−1)ⁱ(−1)^j+1(−1)ⁱ⁺¹

= (−1)^2i+j+3

and

− σ(x_i+2, Ij+1,i+1,i+2,...,n)σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_i+2, Ii+1,i+2,...,n)

= −(−1)ⁱ(−1)^j+1(−1)ⁱ⁺¹

= (−1)^2i+j+3. Similarly, gj+1,i+1,i+2,...,n

x0x1···x_j also contributes to the components e_x0···x_jxˆj+1xj+2···x_ixˆi+1xˆi+2xi+3xˆi+4···ˆxn, . . . , e_x0···xjxˆj+1xj+2···xixˆi+1···ˆxn−1xn. However, we have for each l = i + 1, . . . , n,

σ(x_l, Ij+1,i+1,i+2,...,n) = (−1)ⁱ, σ(x_j+1, Ij+1,...,l−1,l+1,...,n) = (−1)^j+1, and σ(x_l, Ii+1,i+2,...,n) = (−1)ⁱ⁺¹,

so we have

gj+1,i+1,i+2,...,n x0x1···xj

= −σ(x_i+1, Ij+1,i+1,i+2,...,n)σ(x_j+1, Ij+1,i+2,...,n)σ(x_i+1, Ii+1,i+2,...,n)fi+1,i+2,...,n x0x1...xj

= −σ(x_i+2, Ij+1,i+1,i+2,...,n)σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_i+2, Ii+1,i+2,...,n)fi+1,i+2,...,n x0x1...xj

= · · ·

= −σ(x_n, Ij+1,i+1,i+2,...,n)σ(x_j+1, Ij+1,i+1,i+3,...,n−1)σ(x_n, Ii+1,i+2,...,n)fi+1,i+2,...,n x0x1...xj

Hence, we indeed choose the same gj+1,i+1,i+2,...,n

x0x1···x_j even when we consider different compo-nents. In general, for all j+2 ≤ k₁ < k₂ < · · · < k_n−i ≤ n, since for each l = 1, 2, . . . , n−i,

σ(x_kl, I^{j+1,k1,k2,k3,...,kn−i})σ(x_kl, I^{k1,k2,k3,...,kn−i})

= (−1)^kl−l(−1)^kl−(l−1)

= (−1)^2kl−2l+1

= −1,

and σ(x_j+1, I^{j+1,k1,k2,...,kl−1,kl+1,...,kn−i}) = (−1)^j+1, we have g^j+1,k_x ^{1,k2,...,kn−i}

0x1···x_j = −σ(x_k1, I^{j+1,k1,k2,k3,...,kn−i})σ(x_j+1, I^{j+1,k2,k3,...,kn−i})

· σ(x_k1, I^{k1,k2,k3,...,kn−i})f_x^{k1,k2,...,kn−i}

0x1···x_j

= −σ(x_k2, I^{j+1,k1,k2,k3,...,kn−i})σ(x_j+1, I^{j+1,k1,k3,...,kn−i})

· σ(x_k2, I^{k1,k2,k3,...,kn−i})f_x^k₀¹_x^,k₁²_···x^,...,kn−i

j

= · · ·

= −σ(x_kn−i, I^{j+1,k1,k2,k3,...,kn−i})σ(x_j+1, I^{j+1,k1,k2,...,kn−i−1})

· σ(x_kn−i, I^{k1,k2,...,kn−i})f_x^k₀¹_x^,k₁²_···x^,...,kn−i

j . (19)

Therefore, the choices made for each gx^j+1,k0x1···x^1,kj^2,...,kn−i while considering different compo-nents are in fact all the same. Now we use (19) to claim that d_i(g_x0x1···xj) = f_x0x1···xj, where

g_x0x1···x_j

= gj+1,i+1,i+2,...,n

x0x1···xj e_x0···x_jxˆj+1xj+2···x_ixˆi+1···ˆxn+ · · · + gj+1,j+2,...,n+j−i+1

x0x1...xj e_x0···x_jxˆj+1···ˆxn+j−i+1xn+j−i+2···xn, i.e.,

d_i(gj+1,i+1,i+2,...,n

x0x1···x_j e_{x0···xjˆxj+1xj+2···xiˆxi+1···ˆxn} + · · · + gj+1,j+2,...,n+j−i+1

x0x1...xj e_{x0···xjxˆj+1···ˆxn+j−i+1xn+j−i+2···xn})

= fi+1,i+2,...,n

x0x1...xj e_{x0x1···xixˆi+1···ˆxn}+ · · · + fj+2,j+3,...,n+j−i+1

x0x1...xj e_{x0x1···xj+1xˆj+2···ˆxn+j−i+1xn+j−i+2···xn} + fj+1,i+2,...,n

x0x1...xj e_{x0x1···xjxˆj+1xj+2···xi+1ˆxi+2···ˆxn}+ · · · + fj+1,j+2,...,n+j−i

x0x1...xj e_{x0···xjxˆj+1···ˆxn+j−ixn+j−i+1···xn}. (20) We first take care of the components in the second line of (20), namely the components e_{x0x1···xixˆi+1···ˆxn}, . . . , e_{x0x1···xj+1ˆxj+2···ˆxn+j−i+1xn+j−i+2···xn}. For the component e_{x0x1···xixˆi+1···ˆxn}, since the components in g_x0x1···x_j that do contribute to it are the components

e_x0x1···x_i−1xˆixˆi+1···ˆxn, e_x0x1···x_i−2ˆxi−1xiˆxi+1···ˆxn, . . . , e_x0x1···x_j+1xˆj+2xj+3···x_ixˆi+1···ˆxn, e_x0x1···x_jˆxj+1xj+2···x_ixˆi+1···ˆxn, and since we take

gi,i+1,...,n

x0x1···x_j = gi−1,i+1,...,n

x0x1···x_j = · · · = gj+2,i+1,...,n x0x1···x_j = 0,

and (20). Note that the components in g_x0x1...xj that do contribute to the component e_x0x1···x_jxˆj+1xj+2···x_i+1ˆxi+2···ˆxn are the components

we see that the coefficient of e_{x0x1···xjxˆj+1xj+2···xi+1xˆi+2···ˆxn} is σ(x_i+1, Ij+1,i+1,i+2,...,n)gj+1,i+1,i+2,...,n

x0x1...xj + σ(x_i, Ij+1,i,i+2,...,n)gj+1,i,i+2,...,n x0x1...xj

+ · · · + σ(x_j+2, Ij+1,j+2,i+2,...,n)gj+1,j+2,i+2,...,n x0x1...xj

= σ(x_i+1, Ij+1,i+1,i+2,...,n) − σ(x_i+1, Ij+1,i+1,i+2,...,n)σ(x_j+1, Ij+1,i+2,...,n)

· σ(x_i+1, Ii+1,i+2,...,n)fi+1,i+2,...,n x0x1...xj

+ σ(x_i, Ij+1,i,i+2,...,n) − σ(x_i, Ij+1,i,i+2,...,n)σ(x_j+1, Ij+1,i+2,...,n)

· σ(x_i, Ii,i+2,...,n

)fi,i+2,...,n x0x1...xj

+ · · ·

σ(x_j+2, Ij+1,j+2,i+2,...,n

) − σ(x_j+2, Ij+1,j+2,i+2,...,n

)σ(x_j+1, Ij+1,i+2,...,n

)

· σ(x_j+2, Ij+2,i+2,...,n

)fj+2,i+2,...,n x0x1...xj

= −σ(x_j+1, Ij+1,i+2,...,n

)σ(x_i+1, Ii+1,i+2,...,n

)fi+1,i+2,...,n x0x1...xj

− σ(x_j+1, Ij+1,i+2,...,n

)σ(x_i, Ii,i+2,...,n

)fi,i+2,...,n x0x1...xj

− · · ·

− σ(xj+1, Ij+1,i+2,...,n

)σ(xj+2, Ij+2,i+2,...,n

)fj+2,i+2,...,n x0x1...xj

= fj+1,i+2,...,n x0x1...xj ,

where the last equality follows from (16).

Next we check the second component e_x0x1···x_jxˆj+1xj+2···x_ixˆi+1xi+2xˆi+3···ˆxn in the third line of (20). Since the components in g_x0x1...xj that do contribute to the component e_x0x1···x_jxˆj+1xj+2···x_iˆxi+1xi+2xˆi+3···ˆxn are the components e_x0x1···x_jˆxj+1xj+2···x_iˆxi+1xˆi+2xˆi+3···ˆxn, e_x0x1···xjxˆj+1xj+2···xi−1xˆiˆxi+1xi+2xˆi+3···ˆxn, . . . , e_x0x1···xjxˆj+1xˆj+2xj+3···xiˆxi+1xi+2xˆi+3···ˆxn. Use (19) we

have



































gj+1,i+1,i+2,i+3,...,n

x0x1...xj = −σ(xi+2, Ij+1,i+1,i+2,i+3,...,n)σ(xj+1, Ij+1,i+1,i+3,...,n)

· σ(x_i+2, Ii+1,i+2,i+3,...,n)fi+1,i+2,i+3,...,n x0x1...xj

gj+1,i,i+1,i+3,...,n

x0x1...xj = −σ(xi, Ij+1,i,i+1,i+3,...,n)σ(xj+1, Ij+1,i+1,i+3,...,n)

· σ(x_i, Ii,i+1,i+3,...,n)fi,i+1,i+3,...,n x0x1...xj

... gj+1,j+2,i+1,i+3,...,n

x0x1...xj = −σ(x_j+2, Ij+1,j+2,i+1,i+3,...,n)σ(x_j+1, Ij+1,i+1,i+3,...,n)

· σ(xj+2, Ij+2,i+1,i+3,...,n)fj+2,i+1,i+3,...,n x0x1...xj . Note that in the above relation we choose the expression

gj+1,i+1,i+2,i+3,...,n x0x1...xj

= −σ(xi+2, Ij+1,i+1,i+2,i+3,...,n

)σ(xj+1, Ij+1,i+1,i+3,...,n

)σ(xi+2, Ii+1,i+2,i+3,...,n

)fi+1,i+2,i+3,...,n x0x1...xj , instead of

gj+1,i+1,i+2,i+3,...,n x0x1...xj

= −σ(x_i+1, Ij+1,i+1,i+2,i+3,...,n)σ(x_j+1, Ij+1,i+2,i+3,...,n)σ(x_i+1, Ii+1,i+2,i+3,...,n)fi+1,i+2,i+3,...,n x0x1...xj .

Then the coefficient of e_{x0x1···xjˆxj+1xj+2···xiˆxi+1xi+2xˆi+3···ˆxn} in d_i(g_x0x1...xj) is σ(x_i+2, Ij+1,i+1,i+2,i+3,...,n)gj+1,i+1,i+2,i+3,...,n

x0x1...xj + σ(x_i, Ij+1,i,i+1,i+3,...,n)gj+1,i,i+1,i+3,...,n x0x1...xj

+ · · ·

+ σ(x_j+2, Ij+1,j+2,i+1,i+3,...,n)gj+1,j+2,i+1,i+3,...,n x0x1...xj

= σ(x_i+2, Ij+1,i+1,i+2,i+3,...,n) − σ(x_i+2, Ij+1,i+1,i+2,i+3,...,n)σ(x_j+1, Ij+1,i+1,i+3,...,n)

· σ(x_i+2, Ii+1,i+2,i+3,...,n)fi+1,i+2,i+3,...,n x0x1...xj

+ σ(x_i, Ij+1,i,i+1,i+3,...,n) − σ(x_i, Ij+1,i,i+1,i+3,...,n)σ(x_j+1, Ij+1,i+1,i+3,...,n)

· σ(x_i, Ii,i+1,i+3,...,n)fi,i+1,i+3,...,n x0x1...xj

+ · · ·

+ σ(x_j+2, Ij+1,j+2,i+1,i+3,...,n) − σ(x_j+2, Ij+1,j+2,i+1,i+3,...,n)σ(x_j+1, Ij+1,i+1,i+3,...,n)

· σ(x_j+2, Ij+2,i+1,i+3,...,n)fj+2,i+1,i+3,...,n x0x1...xj

= −σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_i+2, Ii+1,i+2,i+3,...,n)fi+1,i+2,i+3,...,n x0x1...xj

− σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_i, Ii,i+1,i+3,...,n)fi,i+1,i+3,...,n x0x1...xj

− · · ·

− σ(x_j+1, Ij+1,i+1,i+3,...,n)σ(x_j+2, Ij+2,i+1,i+3,...,n)fj+2,i+1,i+3,...,n x0x1...xj

= fj+1,i+1,i+3,...,n x0x1...xj ,

where the last equality follows from (16).

Similarly, we do the same argument for the remaining components

e_x0x1···x_jxˆj+1xj+2···x_ixˆi+1xˆi+2xi+3ˆxi+4···ˆxn, . . . , e_x0···x_jxˆj+1···ˆxn+j−ixn+j−i+1···xn

in the third line of (20) and finish the proof. 2

In general, for all 0 ≤ j < i and 0 ≤ k₀ < · · · < k_j ≤ n, as in the proof of Lemma 3.6, we can repeat the same argument for the component which has negative powers for x_k0, x_k1, . . . , x_kj and is not a multiple of w, except with a much more complicated notation and indexing. Hence, we see that for every component which has negative powers for

some of the variables and is not a multiple of w, if f_xk0···xkj is contained in this component, f_xk0···x_kj ∈ Ker d_i+1 implies f_xk0···x_kj ∈ Im d_i.

Next, we consider the component which has negative powers for x₁, x₂, . . . , x_j with 1 ≤ j < i and is a multiple of w, i.e., the component with respect to

∞

X

α1=1

∞

X

α2=1

· · ·

∞

X

αj=1

Q[xj+1, x_j+2, . . . , x_n]wx^−α₁ ¹x^−α₂ ²· · · x^−α_j ^j. For f_x1x2···xj(w) ∈ Cⁱ⁺¹, since

Cⁱ⁺¹= R_{x0x1···xixˆi+1···ˆxn}⊕ · · · ⊕ R_{x0ˆx1···ˆxn−ixn−i+1···xn}

⊕ R_{xˆ0x1···xi+1xˆi+2···ˆxn} ⊕ · · · ⊕ R_{xˆ0x1···xj+1xˆj+2···ˆxn+j−ixn+j−i+1···xn}

⊕ R_{xˆ0x1···xjxˆj+1xj+2···xi+2xˆi+3···ˆxn}⊕ · · · ⊕ R_{xˆ0x1···xjˆxj+1ˆxj+2···ˆxn+j−i−1xn+j−i···xn}

⊕ R_{xˆ0x1···xj−1xˆjxj+1···xi+2ˆxi+3···ˆxn}⊕ · · · ⊕ R_{ˆx0xˆ1···ˆxn−i−1xn−i···xn},

and since w = 0 in R_x0x1···x_iˆxi+1···ˆxn, . . . , R_x0xˆ1···ˆxn−i···x_n, and since none of the components in the last line of the above equation has

∞

P

α1=1

∞

P

α2=1

· · ·

∞

P

αj=1Q[xj+1, x_j+2, . . . , x_n]wx^−α₁ ¹x^−α₂ ²· · · x^−α_j ^j as a subspace, we write

f_x1x2···x_j(w)

= f0,i+2,...,n

x1x2···x_j (w)e_xˆ0x1···x_i+1xˆi+2···ˆxn+ · · · + f0,j+2,...,n+j−i

x1x2···x_j (w)e_xˆ0x1···x_j+1xˆj+2···ˆxn+j−ixn+j−i+1···x_n

+ f0,j+1,i+3,...,n

x1x2···x_j (w)e_xˆ0x1···x_jxˆj+1xj+2···x_i+2xˆi+3···ˆxn

+ · · ·

+ f0,j+1,...,n+j−i−1

x1x2···x_j (w)e_xˆ0x1···x_jxˆj+1ˆxj+2···ˆxn+j−i−1xn+j−i···x_n.

Lemma 3.7 Let f_x1x2···x_j(w) ∈ Cⁱ⁺¹, and let f_x1x2···xj(w)

= f0,i+2,...,n

x1x2···xj (w)e_xˆ0x1···xi+1xˆi+2···ˆxn+ · · · + f0,j+2,...,n+j−i

x1x2···xj (w)e_xˆ0x1···xj+1xˆj+2···ˆxn+j−ixn+j−i+1···xn

+ f0,j+1,i+3,...,n

x1x2···xj (w)e_xˆ0x1···xjxˆj+1xj+2···xi+2xˆi+3···ˆxn

+ · · ·

+ f0,j+1,...,n+j−i−1

x1x2···xj (w)e_xˆ0x1···xjxˆj+1xˆj+2···ˆxn+j−i−1xn+j−i···xn,

i.e., f_x1x2···xj(w) is contained in the component which has negative powers for x₁, x₂, . . . , x_j and is a multiple of w. Then f_x1x2···x_j(w) ∈ Ker d_i+1 implies f_x1x2···x_j(w) ∈ Im d_i.

Proof. Since d_i+1(f_x1x2...xj(w)) = 0 in Cⁱ⁺² and

Cⁱ⁺² = R_x0x1···x_i+1xˆi+2···ˆxn ⊕ · · · ⊕ R_x0xˆ1···ˆxn−i−1xn−i···x_n

⊕ R_ˆx0x1···x_i+2ˆxi+3···ˆxn ⊕ · · · ⊕ R_ˆx0x1···x_j+1xˆj+2···ˆxn+j−i−1xn+j−i···x_n

⊕ R_ˆx0x1···x_jxˆj+1xj+2···x_i+3xˆi+4···ˆxn ⊕ · · · ⊕ R_xˆ0xˆ1···ˆxn−i−2xn−i−1···x_n, for the component e_xˆ0x1···x_i+2xˆi+3···ˆxn, we have

σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n

x1x2...xj (w) + · · · + σ(x_j+2, I0,j+2,i+3,...,n)f0,j+2,i+3,...,n x1x2...xj (w) + σ(x_j+1, I0,j+1,i+3,...,n)f0,j+1,i+3,...,n

x1x2...xj (w) = 0.

Then we have that the coefficient of the component e_xˆ0x1···x_jxˆj+1xj+2···x_i+2xˆi+3···ˆxnof f_x1x2...xj(w) is

f0,j+1,i+3,...,n

x1x2...xj (w) = −σ(xj+1, I0,j+1,i+3,...,n

)σ(xi+2, I0,i+2,i+3,...,n

)f0,i+2,i+3,...,n x1x2...xj (w)

− · · ·

− σ(x_j+1, I0,j+1,i+3,...,n)σ(x_j+2, I0,j+2,i+3,...,n)f0,j+2,i+3,...,n x1x2...xj (w).

Similarly for the component e_{ˆx0x1···xi+1ˆxi+2xi+3xˆi+4···ˆxn}, we have σ(x_i+3, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n

x1x2...xj (w) + σ(x_i+1, I0,i+1,i+2,i+4,...,n)f0,i+1,i+2,i+4,...,n x1x2...xj (w) + · · · + σ(x_j+2, I0,j+2,i+2,i+4,...,n)f0,j+2,i+2,i+4,...,n

x1x2...xj (w) + σ(x_j+1, I0,j+1,i+2,i+4,...,n)f0,j+1,i+2,i+4,...,n

x1x2...xj (w) = 0,

and we have that the coefficient of the component e_xˆ0x1x2···x_jˆxj+1xj+2···x_i+1ˆxi+2xi+3xˆi+4···ˆxn of f_x1x2...xj(w) is

f0,j+1,i+2,i+4,...,n

x1x2...xj (w) = −σ(x_j+1, I0,j+1,i+2,i+4,...,n

)σ(x_i+3, I0,i+2,i+3,...,n

)f0,i+2,i+3,...,n x1x2...xj (w)

− σ(x_j+1, I0,j+1,i+2,i+4,...,n

)σ(x_i+1, I0,i+1,i+2,i+4,...,n

)f0,i+1,i+2,i+4,...,n x1x2...xj (w)

− · · ·

− σ(x_j+1, I0,j+1,i+2,i+4,...,n

)σ(x_j+2, I0,j+2,i+2,i+4,...,n

)f0,j+2,i+2,i+4,...,n x1x2...xj (w).

Repeat the same discussion for all the remaining components with index ˆx₀, x₁, x₂, . . . , x_j+1,

− σ(x_j+1, I0,j+1,j+2,...,n+j−i−1)σ(x_n+j−i, I0,j+2,...,n+j−i−1,n+j−i)f0,j+2,...,n+j−i−1,n+j−i

x1x2...xj (w).

(21)

From the above relations, we can rewrite f_x1x2...xj(w) as f_x1x2...xj(w)

= f0,i+2,...,n

x1x2···xj (w)e_ˆx0x1···x_i+1ˆxi+2···ˆxn + · · · + f0,j+2,...,n+j−i

x1x2···xj (w)e_xˆ0x1···x_j+1xˆj+2···ˆxn+j−ixn+j−i+1···xn

+ − σ(x_j+1, I0,j+1,i+3,...,n)σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n x1x2...xj (w)

− · · ·

− σ(x_j+1, I0,j+1,i+3,...,n)σ(x_j+2, I0,j+2,i+3,...,n)f0,j+2,i+3,...,n

x1x2...xj (w)
e_xˆ0x1···xjxˆj+1xj+2...xi+2xˆi+3···ˆxn

+ − σ(x_j+1, I0,j+1,i+2,i+4,...,n

)σ(x_i+3, I0,i+2,...,n

)f0,i+2,i+3,...,n x1x2...xj (w)

− σ(x_j+1, I0,j+1,i+2,i+4,...,n

)σ(x_i+1, I0,i+1,i+2,i+4,...,n

)f0,i+1,i+2,i+4,...,n x1x2...xj (w)

− · · ·

− σ(x_j+1, I0,j+1,i+2,i+4,...,n

)σ(x_j+2, I0,j+2,i+2,i+4,...,n

)f0,j+2,i+2,i+4,...,n x1x2...xj (w)

· e_xˆ0x1x2···xjxˆj+1xj+2···xi+1xˆi+2xi+3xˆi+4···ˆxn

+ · · ·

+ − σ(x_j+1, I0,j+1,j+2,...,n+j−i−1

)σ(x_n, I0,j+2,...,n+j−i−1,n

)f0,j+2,...,n+j−i−1,n x1x2...xj (w)

− · · ·

− σ(xj+1, I0,j+1,j+2,...,n+j−i−1

)σ(xn+j−i, I0,j+2,...,n+j−i−1,n+j−i

)f0,j+2,...,n+j−i−1,n+j−i

x1x2...xj (w)

· ex0...xjxˆj+1...ˆxn+j−i−1xn+j−i···xn. (22)

In order to show that f_x1x2...xj(w) ∈ Im d_i, we need to find g_x1x2...xj(w) ∈ Cⁱ such that di(gx1x2...xj(w)) = fx1x2...xj(w). Note that this desired gx1x2...xj(w) ∈ Cⁱ must be contained in the component with respect to

∞

P

α1=1

∞

P

α2=1

· · ·

∞

P

αj=1Q[xj+1, x_j+2, . . . , x_n]wx^−α₁ ¹x^−α₂ ²· · · x^−α_j ^j. Since

Cⁱ = R_x0···x_i−1xˆi···ˆxn⊕ · · · ⊕ R_x0xˆ1···ˆxn−i+1xn−i+2···xn

⊕ R_xˆ0x1···x_ixˆi+1···ˆxn⊕ · · · ⊕ R_xˆ0x1···x_jxˆj+1···ˆxn+j−ixn+j−i+1···xn

⊕ R_xˆ0x1···x_j−1ˆxjxj+1···x_i+1xˆi+2···ˆxn ⊕ · · · ⊕ R_xˆ0···ˆxn−ixn−i+1···xn,

and since w = 0 in the components in the first line of the above equation and since none

of the components in the last line of the above equation has

as a subspace, we only need to consider g_x1x2...xj(w) = g0,i+1,...,n

we want to get

σ(x_i+2, I0,j+1,i+2,i+3,...,n)g0,j+1,i+2,i+3,...,n

x1x2···xj (w) + · · · + σ(x_j+2, I0,j+1,j+2,i+3,...,n)g0,j+1,j+2,i+3,...,n x1x2···xj (w)

= −σ(x_j+1, I0,j+1,i+3,...,n)σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n x1x2...xj (w)

− · · ·

− σ(x_j+1, I0,j+1,i+3,...,n)σ(x_j+2, I0,j+2,i+3,...,n)f0,j+2,i+3,...,n x1x2...xj (w), so we choose

g0,j+1,i+2,i+3,...,n

x1x2···x_j (w) = −σ(x_i+2, I0,j+1,i+2,i+3,...,n)σ(x_j+1, I0,j+1,i+3,...,n)

· σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n x1x2...xj (w), ...

g0,j+1,j+2,i+3,...,n

x1x2···xj (w) = −σ(x_j+2, I0,j+1,j+2,i+3,...,n)σ(x_j+1, I0,j+1,i+3,...,n)

· σ(x_j+2, I0,j+2,i+3,...,n)f0,j+2,i+3,...,n x1x2...xj (w).

We do the same thing for the components e_ˆx0x1···x_jxˆj+1xj+2···x_i+1xˆi+2xi+3ˆxi+4···ˆxn, . . . ,

exˆ0x1···xjxˆj+1···ˆxn+j−i−1xn+j−i···xn and want to choose g^{0,j+1,k1,...,kn−i−1}(w) accordingly for all j + 1 < k₁ < · · · < k_n−i−1 ≤ n. We also choose all the g^{0,k1,...,kn−i} with k₁ > j + 1 to be zero. Note that the term g0,j+1,i+2,i+3,...,n

x1x2···x_j (w) not only contributes to the component

e_xˆ0x1···x_jxˆj+1xj+2···x_i+2ˆxi+3···ˆxnbut also contributes to the component e_xˆ0x1···x_jxˆj+1xj+2···x_i+1xˆi+2xi+3xˆi+4···ˆxn. More precisely, for the component eˆx0x1···xjxˆj+1xj+2···xi+1xˆi+2xi+3xˆi+4···ˆxn, we want to get

σ(x_i+3, I0,j+1,i+2,...,n)g0,j+1,i+2,...,n

x1x2···xj (w) + σ(x_i+1, I0,j+1,i+1,i+2,i+4,...,n)g0,j+1,i+1,i+2,i+4,...,n x1x2···xj (w) + · · ·

+ σ(x_j+2, I0,j+1,j+2,i+2,i+4,...,n)g0,j+1,j+2,i+2,i+4,...,n x1x2···xj (w)

= −σ(x_j+1, I0,j+1,i+2,i+4,...,n)σ(x_i+3, I0,i+2,...,n)f0,i+2,i+3,...,n x1x2...xj (w)

− σ(x_j+1, I0,j+1,i+2,i+4,...,n)σ(x_i+1, I0,i+1,i+2,i+4,...,n)f0,i+1,i+2,i+4,...,n x1x2...xj (w)

− · · ·

− σ(x_j+1, I0,j+1,i+2,i+4,...,n)σ(x_j+2, I0,j+2,i+2,i+4,...,n)f0,j+2,i+2,i+4,...,n x1x2...xj (w)

and so we choose

g0,j+1,i+2,...,n

x1x2···x_j (w) = −σ(x_i+3, I0,j+1,i+2,...,n)σ(x_j+1, I0,j+1,i+2,i+4,...,n)

· σ(x_i+3, I0,i+2,...,n)f0,i+2,i+3,...,n x1x2...xj (w).

However, this is the same as the earlier choice, since

− σ(x_i+2, I0,j+1,i+2,i+3,...,n)σ(x_j+1, I0,j+1,i+3,...,n)σ(x_i+2, I0,i+2,i+3,...,n)

= −(−1)ⁱ(−1)^j(−1)ⁱ⁺¹

= (−1)^2i+j+2

= (−1)^j and

− σ(xi+3, I0,j+1,i+2,i+3,...,n

)σ(xj+1, I0,j+1,i+2,i+4,...,n

)σ(xi+3, I0,i+2,...,n

)

= −(−1)ⁱ(−1)^j(−1)ⁱ⁺¹

= (−1)^2i+j+2

= (−1)^j. Similarly, g0,j+1,i+2,...,n

x1x2···x_j (w) also contributes to the components

exˆ0x1···xjxˆj+1xj+2···xi+1xˆi+2xˆi+3xi+4ˆxi+5···ˆxn, . . . , exˆ0x1···xjxˆj+1xj+2···xi+1xˆi+2···ˆxn−1xn. However, we have for each l = i + 2, . . . , n,

σ(x_l, I0,j+1,i+2,i+3,...,n

) = (−1)^l, σ(x_j+1, I0,j+1,...,n

) = (−1)^j+1 and σ(x_l, I0,i+2,i+3,...,n

) = (−1)^l+1, so we have

g0,j+1,i+2,i+3,...,n x1x2···xj (w)

= −σ(x_i+2, I0,j+1,i+2,i+3,...,n

)σ(x_j+1, I0,j+1,i+3,...,n

)σ(x_i+2, I0,i+2,i+3,...,n

)f0,i+2,i+3,...,n x1x2...xj (w)

= −σ(x_i+3, I0,j+1,i+2,i+3,...,n

)σ(x_j+1, I0,j+1,i+2,i+4,...,n

)σ(x_i+3, I0,i+2,i+3,...,n

)f0,i+2,i+3,...,n x1x2...xj (w)

= · · ·

= −σ(x_n, I0,j+1,i+2,i+3,...,n

)σ(x_j+1, I0,j+1,i+2,...,n−1

)σ(x_n, I0,i+2,i+3,...,n

)f0,i+2,i+3,...,n x1x2...xj (w).

Hence, we indeed choose the same g0,j+1,i+2,i+3,...,n

x1x2···x_j (w) even when we consider different components. In general, for all j + 1 ≤ k₁ < k₂ < · · · < k_n−i−1 ≤ n, since for each l = 1, 2, . . . , n − i − 1,

σ(x_kl, I^{0,j+1,k1,k2,k3,...,kn−i−1})σ(x_kl, I^{0,k1,k2,k3,...,kn−i−1})

= (−1)^kl−(l+1)(−1)^kl−l

= (−1)^2kl−2l−1

= −1,

and σ(x_j+1, I^{j+1,k1,...,kl−1kl+1,...,kn−i−1}) = (−1)^j+1, we have g^0,j+1,k_x1···xj ^{1,k2,...,kn−i−1}(w)

= −σ(xk1, I^{0,j+1,k1,k2,k3,...,kn−i−1})σ(xj+1, I^{j+1,k2,k3,...,kn−i−1})σ(xk1, I^{0,k1,k2,k3,...,kn−i−1})

· f_x^0,k_1···x^1,k_j^{2,...,kn−i−1}(w)

= −σ(xk2, I^{0,j+1,k1,k2,k3,...,kn−i−1})σ(xj+1, I^{j+1,k1,k3,...,kn−i−1})σ(xk2, I^{0,k1,k2,k3,...,kn−i−1})

· f_x^0,k_1···x^1,k_j^{2,...,kn−i−1}(w)

= · · ·

= −σ(xkn−i−1, I^{0,j+1,k1,k2,k3,...,kn−i−1})σ(xj+1, I^{0,j+1,k1,k2,...,kn−i−2})σ(xkn−i−1, I^{0,k1,k2,...,kn−i−1})

· f_x^k₁¹_···x^,k2,...,k_j ⁿ⁻ⁱ⁻¹(w). (24)

Therefore, the choices made for each g^0,j+1,kx0x1···x¹j^{,k2,...,kn−i−1}(w) while considering different components are in fact all the same. Now we use (24) to show that d_i(g_x1x2...xj(w)) = f_x1x2...xj(w), where

g_x1x2...xj(w) = g0,j+1,i+2,...,n

x1x2···x_j (w)e_{ˆx0x1···xjxˆj+1xj+2···xi+1xˆi+2···ˆxn}+ · · · + g0,j+1,j+2,...,n+j−i

x1x2···x_j (w)e_{ˆx0x1···xjxˆj+1···ˆxn+j−ixn+j−i+1···xn},

i.e.,

d_i(g0,j+1,i+2,...,n

x1x2···xj (w)e_ˆx0x1···x_jxˆj+1xj+2···x_i+1xˆi+2···ˆxn+ · · · + g0,j+1,j+2,...,n+j−i

x1x2···xj (w)e_xˆ0x1···x_jˆxj+1···ˆxn+j−ixn+j−i+1···xn)

= f0,i+2,...,n

x1x2···xj (w)e_xˆ0x1···x_i+1xˆi+2···ˆxn+ · · · + f0,j+2,...,n+j−i

x1x2···xj (w)e_xˆ0x1···x_j+1xˆj+2···ˆxn+j−ixn+j−i+1···xn

+ f0,j+1,i+3,...,n

x1x2···xj (w)e_xˆ0x1···x_jxˆj+1xj+2...xi+2xˆi+3···ˆxn + · · · + f0,j+1,j+2,...,n+j−i−1

x1x2···xj (w)e_{xˆ0x1...xjxˆj+1...ˆxn+j−i−1xn+j−i}···xn. (25) We first take care of the components in the third line of (25), namely the components e_xˆ0x1···xi+1xˆi+2···ˆxn, . . . , e_xˆ0x1···xj+1xˆj+2···ˆxn+j−ixn+j−i+1···xn. For the component e_xˆ0x1···xi+1xˆi+2···ˆxn, since the components in g_x1x2···x_j(w) that do contribute are the components e_xˆ0x1···x_ixˆi+1xˆi+2···ˆxn, e_xˆ0x1···xi−1xˆixi+1xˆi+2···ˆxn, . . . , e_xˆ0x1···xj+1ˆxj+2xj+3···xi+1xˆi+2···ˆxn, e_ˆx0x1···xjxˆj+1xj+2···xi+1xˆi+2···ˆxn, and since we take

g0,i+1,...,n

x1x2···xj (w) = g0,i,i+2,...,n

x1x2···xj (w) = · · · = g0,j+2,i+2,...,n

x1x2···xj (w) = 0, and

g0,j+1,i+2,i+3,...,n

x1x2···x_j (w) = −σ(x_i+2, I0,j+1,i+2,i+3,...,n)σ(x_j+1, I0,j+1,i+3,...,n)

· σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n x1x2...xj (w), we get that the coefficient of e_xˆ0x1···x_i+1ˆxi+2···ˆxn in d_i(g_x1x2...xj(w)) is

σ(x_j+1, I0,j+1,i+2,i+3,...,n)g0,j+1,i+2,i+3,...,n x1x2···x_j (w)

= σ(x_j+1, I0,j+1,i+2,i+3,...,n) − σ(x_i+2, I0,j+1,i+2,i+3,...,n)σ(x_j+1, I0,j+1,i+3,...,n)

· σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n x1x2...xj (w)

= (−1)^j − (−1)ⁱ(−1)^j(−1)ⁱ⁺¹f0,i+2,i+3,...,n x1x2...xj (w)

= f0,i+2,i+3,...,n x1x2...xj (w).

Similarly, with the same argument for the remaining components exˆ0x1···xiˆxi+1xi+2xˆi+3···ˆxn, . . . , e_xˆ0x1···x_j+1xˆj+2···ˆxn+j−ixn+j−i+1···x_n in the third line of (25), we see that their coefficients are indeed f0,i+1,i+3,...,n

x1x2···xj (w), . . . , f0,j+2,...,n+j−i

x1x2···xj (w), respectively.

Next we check the first component e_{xˆ0x1···xjxˆj+1xj+2...xi+2xˆi+3···ˆxn} in the fourth line of (25). Note that the components in g_x1x2···x_j(w) that do contribute to the component e_{xˆ0x1···xjxˆj+1xj+2...xi+2ˆxi+3···ˆxn} are the components e_{xˆ0x1···xjxˆj+1xj+2...xi+1xˆi+2xˆi+3···ˆxn}, . . . , e_xˆ0x1···x_jxˆj+1xˆj+2xj+3...xi+2xˆi+3···ˆxn. Moreover, since we take























g0,j+1,i+2,i+3,...,n

x1x2···x_j (w) = −σ(x_i+2, I0,j+1,i+2,i+3,...,n)σ(x_j+1, I0,j+1,i+3,...,n)

· σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n x1x2···xj (w), ...

g0,j+1,j+2,i+3,...,n

x1x2···xj (w) = −σ(x_j+2, I0,j+1,j+2,i+3,...,n)σ(x_j+1, I0,j+1,i+3,...,n)

· σ(x_j+2, I0,j+2,i+3,...,n)f0,j+2,i+3,...,n x1x2···x_j (w), we see that the coefficient of e_ˆx0x1···x_jxˆj+1xj+2...xi+2xˆi+3···ˆxn in d_i(g_x1x2···x_j(w)) is

σ(x_i+2, I0,j+1,i+2,i+3,...,n)g0,j+1,i+2,i+3,...,n

x1x2···x_j (w) + · · · + σ(x_j+2, I0,j+1,j+2,i+3,...,n)g0,j+1,j+2,i+3,...,n x1x2···x_j (w)

= σ(x_i+2, I0,j+1,i+2,i+3,...,n) − σ(x_i+2, I0,j+1,i+2,i+3,...,n)σ(x_j+1, I0,j+1,i+3,...,n)

· σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n x1x2···x_j (w)
+ · · ·

+ σ(x_j+2, I0,j+1,j+2,i+3,...,n) − σ(x_j+2, I0,j+1,j+2,i+3,...,n)σ(x_j+1, I0,j+1,i+3,...,n)

· σ(x_j+2, I0,j+2,i+3,...,n)f0,j+2,i+3,...,n x1x2···x_j (w)

= −σ(x_j+1, I0,j+1,i+3,...,n)σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n x1x2...xj (w)

− · · ·

− σ(x_j+1, I0,j+1,i+3,...,n)σ(x_j+2, I0,j+2,i+3,...,n)f0,j+2,i+3,...,n x1x2...xj (w)

= f0,j+1,i+3,...,n x1x2...xj (w),

where the last equality follows from (21).

Next we check the second component e_xˆ0x1···x_jˆxj+1xj+2···x_i+1ˆxi+2xi+3xˆi+4···ˆxn in the fourth line of (25). Since the components in g_x1x2...xj(w) that do contribute to the compo-nent e_xˆ0x1···x_jxˆj+1xj+2···x_i+1xˆi+2xi+3xˆi+4···ˆxn are the components e_xˆ0x1···x_jxˆj+1xj+2···x_i+1xˆi+2···ˆxn, e_xˆ0x1···x_jxˆj+1xj+2···x_iˆxi+1xˆi+2xi+3xˆi+4···ˆxn, . . . , e_ˆx0x1···x_jxˆj+1xˆj+2xj+3···x_i+1xˆi+2xi+3xˆi+4···ˆxn. Use (24)

and we have



































g0,j+1,i+2,i+3,...,n

x1x2...xj (w) = −σ(xi+3, I0,j+1,i+2,i+3,...,n)σ(xj+1, I0,j+1,i+2,i+4,...,n)

· σ(x_i+3, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n x1x2...xj (w) g0,j+1,i+1,i+2,i+4,...,n

x1x2...xj (w) = −σ(xi+1, I0,j+1,i+1,i+2,i+4,...,n)σ(xj+1, I0,j+1,i+2,i+4,...,n)

· σ(x_i+1, I0,i+1,i+2,i+4,...,n)f0,i+1,i+2,i+4,...,n x1x2...xj (w) ...

g0,j+1,j+2,i+2,i+4,...,n

x1x2...xj (w) = −σ(x_j+2, I0,j+1,j+2,i+2,i+4,...,n)σ(x_j+1, I0,j+1,i+2,i+4,...,n)

· σ(xj+2, I0,j+2,i+2,i+4,...,n)f0,j+2,i+2,i+4,...,n x1x2...xj (w).

Note that in the above relation we choose the expression g0,j+1,i+2,i+3,...,n

x1x2...xj (w) = −σ(xi+3, I0,j+1,i+2,i+3,...,n

)σ(xj+1, I0,j+1,i+2,i+4,...,n

)

· σ(xi+3, I0,i+2,i+3,...,n

)f0,i+2,i+3,...,n x1x2...xj (w), instead of

g0,j+1,i+2,i+3,...,n

x1x2···x_j (w) = −σ(xi+2, I0,j+1,i+2,i+3,...,n

)σ(xj+1, I0,j+1,i+3,...,n

)

· σ(x_i+2, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n x1x2···x_j (w).

Then the coefficient of e_{xˆ0x1···xjˆxj+1xj+2···xi+1ˆxi+2xi+3xˆi+4···ˆxn} in d_i(g_x1x2···xj(w)) is σ(x_i+3, I0,j+1,i+2,i+3,...,n)g0,j+1,i+2,i+3,...,n

x1x2...xj (w) + σ(x_i+1, I0,j+1,i+1,i+2,i+4,...,n)g0,j+1,i+1,i+2,i+4,...,n x1x2...xj (w) + · · ·

+ σ(x_j+2, I0,j+1,j+2,i+2,i+4,...,n)g0,j+1,j+2,i+2,i+4,...,n x1x2...xj (w)

= σ(x_i+3, I0,j+1,i+2,i+3,...,n) − σ(x_i+3, I0,j+1,i+2,i+3,...,n)σ(x_j+1, I0,j+1,i+2,i+4,...,n)

· σ(x_i+3, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n x1x2...xj (w)

+ σ(x_i+1, I0,j+1,i+1,i+2,i+4,...,n) − σ(x_i+1, I0,j+1,i+1,i+2,i+4,...,n)σ(x_j+1, I0,j+1,i+2,i+4,...,n)

· σ(x_i+1, I0,i+1,i+2,i+4,...,n)f0,i+1,i+2,i+4,...,n x1x2...xj (w)
+ · · ·

+ σ(x_j+2, I0,j+1,j+2,i+2,i+4,...,n) − σ(x_j+2, I0,j+1,j+2,i+2,i+4,...,n)σ(x_j+1, I0,j+1,i+2,i+4,...,n)

· σ(x_j+2, I0,j+2,i+2,i+4,...,n)f0,j+2,i+2,i+4,...,n x1x2...xj (w)

= −σ(x_j+1, I0,j+1,i+2,i+4,...,n)σ(x_i+3, I0,i+2,i+3,...,n)f0,i+2,i+3,...,n x1x2...xj (w)

− σ(x_j+1, I0,j+1,i+2,i+4,...,n)σ(x_i+1, I0,i+1,i+2,i+4,...,n)f0,i+1,i+2,i+4,...,n x1x2...xj (w)

− · · ·

− σ(x_j+1, I0,j+1,i+2,i+4,...,n)σ(x_j+2, I0,j+2,i+2,i+4,...,n)f0,j+2,i+2,i+4,...,n x1x2...xj (w)

= f0,j+1,i+2,i+4,...,n x1x2...xj (w),

where the last equality follows from (21).

Similarly, we do the same argument for the remaining components

e_xˆ0x1···x_jxˆj+1xj+2···x_i+1ˆxi+2xˆi+3xi+4xˆi+5···ˆxn, . . . , e_{xˆ0x1...xjxˆj+1...ˆxn+j−i−1xn+j−i}···xn

in the fourth and fifth lines of (25) and finish the proof. 2

In general, for all 1 ≤ j < i and 1 ≤ k₁ < · · · < k_j ≤ n, as in the proof of Lemma 3.7, we can repeat the same argument for the component which has negative powers for x_k1, x_k2, . . . , x_kj and is a multiple of w, except with a much more complicated notation and indexing. Hence, we see that for every component which has negative powers for some

of the variables and is a multiple of w, if f_xk1xk2...xkj(w) is contained in this component, f_xk1xk2...xkj(w) ∈ Ker d_i+1 implies f_xk1xk2...xkj(w) ∈ Im d_i.

Finally, by Remark 3.3, Theorem 3.2 follows from Lemma 3.4, 3.5, 3.6, and 3.7.

Hence, we see that Hⁱ⁺¹(C) = 0 for all 0 ≤ i ≤ n − 2.

Next, we compute Hⁿ(C).

Theorem 3.8 Let R = Q[x⁰, x1, x2, . . . , xn, w]/hx0w, w²i and let C be the ˇCech complex with respect to the sequence x₀, x₁, x₂, . . . , x_n. Then we have Hⁿ(C) 6= 0. More precisely,

Hⁿ(C) =

∞

X

α1=1

∞

X

α2=1

· · ·

∞

X

αn=n

Qwx^−α1 ¹x^−α₂ ²· · · x^−α_n ⁿexˆ0x1x2···xn+ Im dn−1
/Im dn−1.

By Remark 3.3, we can separate our computation into four parts, as we do for the proof of Theorem 3.2.

We first consider the component which has no negative powers for x₀, x₁, x₂, . . . , x_n and is not a multiple of w, i.e., the component with respect to Q[x0, x₁, . . . , x_n]. For f ∈ Cⁿ, since Cⁿ = R_x0x1···xn−1xˆn ⊕ · · · ⊕ R_x0xˆ1x2···xn⊕ R_xˆ0x1x2···xn, we write

f = fⁿe_{x0x1···xn−1xˆn} + · · · + f¹e_{x0xˆ1x2···xn−1xn}+ f⁰e_{xˆ0x1x2···xn−1xn}.

Lemma 3.9 Let f ∈ Cⁿand let f = fⁿe_x0x1···x_n−1ˆxn+· · ·+f¹e_x0ˆx1x2···x_n−1xn+f⁰e_xˆ0x1x2···x_n−1xn, i.e., f is contained in the component which has no negative powers for x₀, x₁, . . . , x_n and is not a multiple of w. Then f ∈ Ker d_n implies f ∈ Im d_n−1.

Proof. Since d_n(f ) = 0 in Cⁿ⁺¹ and Cⁿ⁺¹ = R_x0x1···xn, we have σ(xn, Iⁿ)fⁿ+ · · · + σ(x1, I¹)f¹+ σ(x0, I⁰)f⁰ = 0.

Then we have that the coefficient of the component e_ˆx0x1x2···x_n−1xn of f is

f⁰ = −σ(x₀, I⁰)σ(x_n, Iⁿ)fⁿ− · · · − σ(x₀, I⁰)σ(x₁, I¹)f¹. (26)

From the above relation, we can rewrite f as f = fⁿe_x0x1···x_n−1xˆn + · · · + f¹e_x0xˆ1x2···x_n−1xn

+ − σ(x₀, I⁰)σ(x_n, Iⁿ)fⁿ− · · · − σ(x₀, I⁰)σ(x₁, I¹)f¹
e_xˆ0x1x2···x_n−1xn. (27) In order to show that f ∈ Im dn−1, we need to find g ∈ Cⁿ⁻¹ such that dn−1(g) = f . Note that for g = g^n−1,ne_x0···x_n−2xˆn−1xˆn+ · · · + g^0,1e_xˆ0xˆ1x2···xn,

d_n−1(g^n−1,ne_x0···xn−2xˆn−1xˆn + · · · + g^0,1e_xˆ0xˆ1x2···xn)

= σ(xn−1, I^n−1,n)g^n−1,n+ · · · + σ(x0, I^0,n)g^0,n
ex0···xn−1ˆxn

+ · · ·

+ σ(xn, I^1,n)g^1,n+ · · · + σ(x2, I^1,2)g^1,2+ σ(x0, I^0,1)g^0,1
ex0ˆx1x2···xn

+ σ(xn, I^0,n)g^0,n+ · · · + σ(x2, I^0,2)g^0,2+ σ(x1, I^0,1)g^0,1
exˆ0x1···xn. (28) We compare (27) with (28). For the component e_xˆ0x1···x_n, we want to get

σ(x_n, I^0,n)g^0,n+ · · · + σ(x₂, I^0,2)g^0,2+ σ(x₁, I^0,1)g^0,1

= −σ(x₀, I⁰)σ(x_n, Iⁿ)fⁿ− · · · − σ(x₀, I⁰)σ(x₂, I²)f²− σ(x₀, I⁰)σ(x₁, I¹)f¹, so we choose

g^0,n = −σ(x_n, I^0,n)σ(x₀, I⁰)σ(x_n, Iⁿ)fⁿ, ...

g^0,2 = −σ(x₂, I^0,2)σ(x₀, I⁰)σ(x₂, I²)f², g^0,1 = −σ(x₁, I^0,1)σ(x₀, I⁰)σ(x₁, I¹)f¹.

We also choose all the g^k1,k2 with k₁ > 0 to be zero. Now we check that d_n−1(g) = f , where

g = g^0,ne_{ˆx0x1x2···xn−1xˆn} + · · · + g^0,2e_{xˆ0x1xˆ2x3···xn} + g^0,1e_{ˆx0xˆ1x2···xn}, i.e.,

d_n−1(g^0,ne_xˆ0x1x2···xn−1ˆxn+ · · · + g^0,2e_ˆx0x1xˆ2x3···xn+ g^0,1e_xˆ0xˆ1x2···xn)

= fⁿe_x0x1···xn−1xˆn + · · · + f¹e_x0xˆ1x2···xn−1xn

+ − σ(x₀, I⁰)σ(x_n, Iⁿ)fⁿ− · · · − σ(x₀, I⁰)σ(x₁, I¹)f¹
e_ˆx0x1x2···xn−1xn. (29)

We first take care of the components in the second line of (29), namely the components e_x0x1···xn−1xˆn, . . . , e_x0xˆ1x2···xn−1xn. For the component e_x0x1···x_j−1xˆjxj+1···xn with 1 ≤ j ≤ n, since the components in g that do contribute are the components

e_x0···x_j−1xˆjxj+1···xn−1ˆxn, . . . , e_x0···x_j−1ˆxjˆxj+1xj+2···xn, e_x0···x_j−2ˆxj−1xˆjxj+1···xn, . . . , e_xˆ0x1···x_j−1ˆxjxj+1···xn, and since we take

g^j,n = · · · = g^j,j+1= g^j−1,j = · · · = g^1,j = 0 and

g^0,j = −σ(x_j, I^0,j)σ(x₀, I⁰)σ(x_j, I^j)f^j, we get that the coefficient of e_x0x1···xj−1xˆjxj+1···xn in d_n−1(g) is

σ(x₀, I^0,j)g^0,j = σ(x₀, I^0,j) − σ(x_j, I^0,j)σ(x₀, I⁰)σ(x_j, I^j)f^j

= (−1)⁰ − (−1)^j−1(−1)⁰(−1)^jf^j

= (−1)^2jf^j

= f^j.

Hence, the coefficients of ex0x1···xn−1xˆn, . . . , ex0ˆx1x2···xn−1xn are indeed fⁿ, . . . , f¹, respec-tively.

Next we check the component e_ˆx0x1x2···x_n−1xn in the third line of (29). Note that the components in g that do contribute to the component eˆx0x1x2···xn−1xn in dn−1(g) are the component e_xˆ0x1x2···x_n−1ˆxn, . . . , e_xˆ0x1xˆ2x3···x_n−1xn, e_xˆ0xˆ1x2···x_n−1xn. Moreover, since we take











g^0,n = −σ(x_n, I^0,n)σ(x₀, I⁰)σ(x_n, Iⁿ)fⁿ, ...

g^0,1 = −σ(x₁, I^0,1)σ(x₀, I⁰)σ(x₁, I¹)f¹,

we see that the coefficient of e_{ˆx0x1x2···xn−1xn} in d_n−1(g) is σ(x_n, I^0,n)g^0,n+ · · · + σ(x₀, I^0,1)g^0,1

= σ(x_n, I^0,n) − σ(x_n, I^0,n)σ(x₀, I⁰)σ(x_n, Iⁿ)fⁿ
+ · · ·

+ σ(x₁, I^0,1) − σ(x₁, I^0,1)σ(x₀, I⁰)σ(x₁, I¹)f¹

= −σ(x₀, I⁰)σ(x_n, Iⁿ)fⁿ− · · · − σ(x₀, I⁰)σ(x₁, I¹)f¹

= f⁰,

where the last equality follows from (26). This completes the proof. 2

Next, we consider the component which has no negative powers for x0, x1, x2, . . . , xn

and is a multiple of w, i.e., the component with respect to Q[x1, x₂, . . . , x_n]w. For f (w) ∈ Cⁿ, since Cⁿ= Rx0x1···xn−1ˆxn⊕ · · · ⊕ Rx0xˆ1x2···xn⊕ Rxˆ0x1x2···xn, and since w = 0 in R_x0x1···x_n−1ˆxn, . . . , R_x0xˆ1x2···xn, we write f (w) = f⁰(w)e_xˆ0x1x2···xn.

Lemma 3.10 Let f (w) ∈ Cⁿ, and let f (w) = f⁰(w)exˆ0x1x2···xn, i.e., f (w) is contained in the component which has no negative powers for x₀, x₁, . . . , x_n and is a multiple of w.

Then f (w) ∈ Ker dn and f (w) ∈ Im dn−1.

Proof. Since w = 0 in R_x0x1···x_n, d_n(f (w)) = d_n(f⁰(w)e_ˆx0x1x2···x_n) = σ(x₀, I⁰)f⁰(w)e_x0x1···x_n

= 0, and so f (w) ∈ Ker dn. Moreover, take g(w) = σ(x1, I^0,1)f⁰(w)exˆ0xˆ1x2···xn ∈ Cⁿ⁻¹. Then

dn−1(g(w)) = σ(x1, I^0,1) σ(x0, I^0,1)f⁰(w)ex0ˆx1x2···xn+ σ(x1, I^0,1)f⁰(w)exˆ0x1···xn

= f⁰(w)exˆ0x1···xn = f (w),

where the second equality is because w = 0 in R_x0xˆ1x2···xn and thus f⁰(w) = 0 in R_{x0ˆx1x2···xn}. Therefore, f (w) ∈ Im d_n−1. 2

Next, for 0 ≤ i < n, we consider the component which has negative powers for x₀, x₁, . . . , x_i and is not a multiple of w, i.e., the component with respect to

∞

X

α0=1

∞

X

α1=1

· · ·

∞

X

αi=1

Q[xi+1, . . . , x_n]x^−α₀ ⁰x^−α₁ ¹· · · x^−α_i ⁱ. For f_x0x1···x_i ∈ Cⁿ, since

Cⁿ= R_x0x1···xn−1xˆn⊕ · · · ⊕ R_x0···xi+1xˆi+2xi+3···xn⊕ R_x0···xiˆxi+1xi+2···xn

⊕ R_x0···xi−1xˆixi+1···xn ⊕ · · · ⊕ R_xˆ0x1x2···xn,

and since none of the components in the second line of the above equation has

∞

X

α0=1

∞

X

α1=1

· · ·

∞

X

αi=1

Q[xⁱ⁺¹, . . . , x_n]x^−α₀ ⁰x^−α₁ ¹· · · x^−α_i ⁱ as a subspace, we write

f_x0x1···x_i = f_xⁿ_0x1···xie_x0x1···x_n−1xˆn+· · ·+f_xⁱ⁺²_0x1···xie_x0···x_i+1ˆxi+2xi+3···xn+f_xⁱ⁺¹_0x1···xie_x0···x_ixˆi+1xi+2···xn.

Lemma 3.11 Let f_x0x1···x_i ∈ Cⁿ and let f_x0x1···xi = f_xⁿ

0x1···x_ie_{x0x1···xn−1xˆn}+· · ·+f_xⁱ⁺²

0x1···x_ie_{x0···xi+1ˆxi+2xi+3···xn}+f_xⁱ⁺¹

0x1···x_ie_{x0···xixˆi+1xi+2···xn}, i.e., f_x0x1···x_i is contained in the component which has negative powers for x₀, x₁, x₂, . . . , x_i and is not a multiple of w. Then f_x0x1···xi ∈ Ker d_n implies f_x0x1···xi ∈ Im d_n−1.

Proof. Since d_n(f_x0x1···xi) = 0 in Cⁿ⁺¹ and Cⁿ⁺¹ = R_x0x1···xn, we have

σ(x_n, Iⁿ)f_xⁿ_0x1···xi + · · · + σ(x_i+2, Iⁱ⁺²)f_xⁱ⁺²_0x1···xi + σ(x_i+1, Iⁱ⁺¹)f_xⁱ⁺¹_0x1···xi = 0.

Then we have that the coefficient of the component e_x0···x_ixˆi+1xi+2···x_n of f_x0x1···x_i is f_xⁱ⁺¹_0x1···xi = −σ(x_i+1, Iⁱ⁺¹)σ(x_n, Iⁿ)f_xⁿ_0x1···xi − · · · − σ(x_i+1, Iⁱ⁺¹)σ(x_i+2, Iⁱ⁺²)f_xⁱ⁺²_0x1···xi.

(30)

From the above relation, we can rewrite f_x0x1···xi as

f_x0x1···x_i = f_xⁿ_0x1···xie_x0x1···xn−1ˆxn+ · · · + f_xⁱ⁺²_0x1···xie_x0···x_i+1ˆxi+2xi+3···xn

+ − σ(x_i+1, Iⁱ⁺¹)σ(x_n, Iⁿ)f_xⁿ_0x1···xi

− · · ·

− σ(x_i+1, Iⁱ⁺¹)σ(x_i+2, Iⁱ⁺²)f_xⁱ⁺²_0x1···xi
e_x0···x_ixˆi+1xi+2···xn. (31) In order to show that f_x0x1···x_i ∈ Im d_n−1, we need to find g_x0x1···x_i ∈ Cⁿ⁻¹ such that

− σ(x_i+1, Iⁱ⁺¹)σ(x_i+2, Iⁱ⁺²)f_xⁱ⁺²_0x1···xi
e_x0···x_ixˆi+1xi+2···xn. (31) In order to show that f_x0x1···x_i ∈ Im d_n−1, we need to find g_x0x1···x_i ∈ Cⁿ⁻¹ such that