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(2) . 致謝詞 我能順利完成這篇論文，首先要感謝的就是劉容真老師，老師對 於學術上嚴謹認真的態度，都將做為我往後努力學習的目標。過去的 兩年，不管是在課業上還是生活上，劉容真老師都給予我相當大的支 持與鼓勵，當老師的學生真的很幸福。感謝洪有情老師以及林惠雯老 師在百忙之中抽空擔任我的口試委員，老師們口試時給的建議讓我獲 益良多。 感謝同門的好友小島、韋達，陪我一起努力、奮鬥；感謝文傑、 宜容、建勳、婷婷、承緯這兩年的陪伴；特別是晉宇、雅芳，在碩二 下學期陪我走過最黑暗的時期，在我難過時陪我聊天散心，真的真 的…謝謝；也謝謝小金出現在這最後兩個月，給了我許多正面的想法 並且相信自己；最後我要感謝我的家人，有你們在背後支持我，才能 讓我能無後顧之憂的完成課業。 .
(3) Contents 1 Introduction. 1. 2 A Special Case. 6. 3 The General Case. 20. References. 81.
(4) 1. Introduction. Let (A, m, k) be a Noetherian local ring with maximal ideal m and let M be an A-module. We define Γm (M ) = {y ∈ M | mk y = 0 for some k ≥ 0}. It is not difficult to check that Γm (−) is a left exact additive functor on the category n of A-modules. The right derived functor associate to Γm (−), denoted by Hm (−), is. called the local cohomology functor. In other words, take an injective resolution I of M and delete M , then we get a cochain complex Γm (I) by applying the functor Γm n to every term in I. Then the nth local cohomology module Hm (M ) of M is the nth n cohomology H n (Γm (I)) associate to Γm (I), i.e., Hm (M ) = H n (Γm (I)). In this thesis, i we compute the local cohomology modules Hm (A) for the Noetherian local ring A = Rp. where R = Q[x0 , x1 , . . . , xn , w]/hx0 w, w2 i and p = hx0 , x1 , . . . , xn , wiR. Our strategy is ˇ making use of Cech complexes.. ˇ Definition 1.1 Let x1 , x2 , . . . , xn be a sequence of elements of a ring R. The Cech complex with respect to the sequence x1 , x2 , . . . , xn is the cochain complex d. d. dn−1. 0 1 C : 0 −→ C 0 −→ C 1 −→ · · · −→ C n −→ 0. where C t =. L. 1≤i1 <i2 <...<it ≤n. Rxi1 xi2 ···xit and the differentiation dt : C t → C t+1 is given. on the component Rxi1 ···xit → Rxj1 xj2 ···xjt+1 to be (−1)s−1 · nat : R ˆ xi1 ···xit → (Rxi1 ···xit )xjs if {i1 , . . . , it } = {j1 , . . . , js , . . . , jt+1 }, 0 otherwise. Note that nat : Rxi1 ···xit → (Rxi1 ···xit )xjs is the natural homomorphism defined by r xki r 7→ . (xj1 · · · xjt )k (xj1 · · · xjt xi )k We use Example 1.2 to illustrate this definition. 1.
(5) ˇ Example 1.2 Let R = Q[x0 , x1 , w]/hx0 w, w2 i, and we consider the Cech complex with respect to the sequence x0 , x1 d. d. d. 0 → R →0 Rx0 ⊕ Rx1 →1 Rx0 x1 →2 0. We think of R as Q[x0 , x1 ] + Q[x1 ]w. Note that x0 w = 0 in R, so w = 0 in Rx0 and Rx0 x1 . Thus we think of Rx0 , Rx1 , and Rx0 x1 as ∞ X 0 , Rx0 = Q[x0 , x1 ] + Q[x1 ]x−α 0 α0 =1. Rx1 = Q[x0 , x1 ] + Q[x1 ]w +. ∞ X. 1 Q[x0 ]x−α 1. Rx0 x1 = Q[x0 , x1 ] +. α0 =1. 0 Q[x1 ]x−α 0. 1 , Qwx−α 1. α1 =1. α1 =1 ∞ X. +. ∞ X. +. ∞ X. 1 Q[x0 ]x−α 1. α1 =1. +. ∞ X ∞ X. 0 −α1 x1 . Qx−α 0. α0 =1 α1 =1. 1. Note that the map nat : R → Rx0 sends every element in Q[x0 , x1 ] to itself and sends every element in Q[x1 ]w to zero. Indeed, for a ∈ Q[x1 ]w in R, since a is a multiple of w and since x0 w = 0 in R, nat(a) = a/1 = ax0 /x0 = 0. Moreover, the map nat : R → Rx1 sends every element in Q[x0 , x1 ] or Q[x1 ]w to itself. Therefore, for α + β ∈ R with α ∈ Q[x0 , x1 ] and β ∈ Q[x1 ]w, d0 (α + β) = d0 (α) + d0 (β) = (α, α) + (0, β) = (α, α + β) ∈ Rx0 ⊕ Rx1 . ∞ P. 2. Note that the map nat : Rx0 → Rx0 x1 sends every element in Q[x0 , x1 ] or. 0 Q[x1 ]x−α 0. α0 =1. to itself. Moreover, the map nat : Rx1 → Rx0 x1 sends every element in Q[x0 , x1 ] ∞ ∞ P P 1 1 or Q[x0 ]x−α to itself, and it sends every element in Q[x ]w or Qwx−α to 1 1 1 α1 =1. zero. Therefore, for α1 +α2 ∈ Rx0 with α1 ∈ Q[x0 , x1 ] and α2 ∈. ∞ P α0 =1. for β1 + β2 + β3 + β4 ∈ Rx1 with β1 ∈ Q[x0 , x1 ], β2 ∈ Q[x1 ]w, β3 ∈ and β4 ∈. ∞ P. α1 =1 0 Q[x1 ]x−α , and 0. ∞ P. Q[x0 ]x1−α1 ,. α1 =1 1 Qwx−α , 1. α1 =1. d1 (α1 + α2 , β1 + β2 + β3 + β4 ) = d1 (α1 , 0) + d1 (α2 , 0) + d1 (0, β1 ) + d1 (0, β2 ) + d1 (0, β3 ) + d1 (0, β4 ) = −α1 − α2 + β1 + β3 ∈ Rx0 x1 . 2.
(6) It is well-know that if (A, m, k) is a Noetherian local ring with maximal ideal m n (A) ' H n (CA ) for all and if x1 , x2 , . . . , xn ∈ A generate an m-primary ideal, then Hm. n ≥ 0, where CA is the cochain complex with respect to the sequence x1 , x2 , . . . , xn . A proof with complete detail for this fact can be found in [C]. Moreover, for a cochain complex C over a ring R, taking cohomology commutes with tensoring a flat R-module, i.e., H i (C) ⊗ M ' H i (C ⊗ M ) if M is a flat R-module.. Theorem 1.3 Let C be a cochain complex of R-modules and let M be a flat R-module. Then H n (C) ⊗R M ' H n (C ⊗R M ), for all n.. Proof. First note that if N1 is a submodule of an R-module N , N1 ⊗R M can be thought as a submodule of N ⊗ M . More precisely, since the inclusion map ι : N1 → N is one-to-one and since M is a flat R-module, the map ι ⊗ 1M : N1 ⊗R M −→ N ⊗R M is one-to-one and so we can identify N1 ⊗R M with the image of ι ⊗ 1M in N ⊗R M . dn−2. dn−1. dn+1. d. n Since C : · · · −→ C n−1 −→ C n −→ C n+1 −→ · · · is a cochain complex of R-modules,. we have the following short exact sequence ι. π. 0 −→ Im dn−1 −→ Ker dn −→ Ker dn /Im dn−1 −→ 0, where ι is the inclusion map and π is the canonical projection. Since M is a flat R-module, we have that ι⊗1. π⊗1. M 0 −→ Im dn−1 ⊗ M −→ Ker dn ⊗ M −→M (Ker dn /Im dn−1 ) ⊗ M −→ 0. is also exact. Hence, we have Ker (π ⊗ 1M ) = Im (ι ⊗ 1M ) = Im dn−1 ⊗ M = Im (dn−1 ⊗ 1M ). Furthermore, by the First Isomorphism Theorem, we have (Ker dn /Im dn−1 ) ⊗ M ' (Ker dn ⊗ M )/Ker (π ⊗ 1M ) = (Ker dn ⊗ M )/Im (dn−1 ⊗ 1M ). 3.
(7) Next, we claim that Ker dn ⊗ M = Ker (dn ⊗ 1M ). We consider the short exact sequence ι. d. n Im dn+1 −→ 0 0 −→ Ker dn −→ C n −→. where ι : Ker dn → C n is the inclusion map. Since M is a flat R-module, we have that dn ⊗1. ι⊗1. M 0 −→ Ker dn ⊗ M −→ C n ⊗ M −→M Im dn ⊗ M −→ 0. is also exact. Thus, Ker (dn ⊗ 1M ) = Im (ι ⊗ 1M ) = Ker dn ⊗ M . Hence, we have (Ker dn /Im dn−1 ) ⊗ M ' Ker (dn ⊗ 1M )/Im (dn−1 ⊗ 1M ), i.e., H n (C) ⊗R M ' H n (C ⊗R M ). 2. In particular, if S is a multiplicatively closed subset of a ring R, then S −1 R is a flat dn−2. dn−1. dn+1. d. n R-module. Hence, if C : · · · −→ C n−1 −→ C n −→ C n+1 −→ · · · is a cochain complex. and if we let CS −1 R denote the cochain complex S −1 dn−2. S −1 dn−1. S −1 dn+1. S −1 d. CS −1 R : · · · −→ S −1 C n−1 −→ S −1 C n −→n S −1 C n+1 −→ · · · , then H n (CS −1 R ) ' H n (C ⊗R S −1 R) ' H n (C) ⊗R S −1 R ' S −1 H n (C). Hence, taking cohomology commutes with taking localization. Therefore, in order to compute i the local cohomology modules Hm (A) for the Noetherian local ring A = Rp where. R = Q[x0 , x1 , . . . , xn , w]/hx0 w, w2 i and p = hx0 , x1 , . . . , xn , wiA, we first compute the ˇ cohomology modules H t (C), where C is the Cech complex with respect to the sequence x0 , x1 , . . . , xn in R, and then compute H t (C)p for each t, which is isomorphic to H t (Cp ). Since Cp0 = Rp = A and since for each t = 1, 2, . . . , n + 1 Cpt =. M. Rxi1 xi2 ···xit. 0≤i1 <···<it ≤n. =. M. p. =. M. Rxi1 xi2 ···xit. 0≤i1 <···<it ≤n. p. =. M. (Rp )xi1 xi2 ···xit. 0≤i1 <···<it ≤n. Axi1 xi2 ···xit ,. 0≤i1 <···<it ≤n. ˇ Cp is indeed the Cech complex with respect to the sequence x0 , x1 , . . . , xn in Rp = A. Moreover, since x0 , x1 , . . . , xn generate an m = pp -primary ideal in A, for each t, we have t Hm (A) ' H t (Cp ) ' H t (C)p .. 4.
(8) In Section 3, we show that H t (C) = 0 for all t = 0, 1, . . . , n − 1 and n. H (C) = H n+1 (C) =. ∞ X ∞ X. ···. α1 =1 α2 =1 ∞ X ∞ X α0 =1 α1 =1. ∞ X αn =n ∞ X. ···. 2 n exˆ0 x1 x2 ···xn + Im dn−1 /Im dn−1 6= 0, · · · x−α Qwx1−α1 x−α 2 n 0 −α1 n + Im dn /Im dn 6= 0. x1 · · · x−α Qx−α 0 n. αn =1. n+1 n t (A) 6= 0. Also, because (A) 6= 0, Hm (A) = 0 for all t = 0, 1, . . . , n − 1 and Hm Hence, Hm. the notation and computation is very complicated, we do the special case, n = 2, in Section 2 first.. 5.
(9) 2. A Special Case. Because the notation and the computation for the general case is very complicated, in this section, we do a special case first. Let R = Q[x0 , w, x1 , x2 ]/hx0 w, w2 i, and let p be the maximal ideal hx0 , x1 , x2 , wi in R. ˇ Since hx0 , x1 , x2 i is p-primary, we consider the Cech complex with respect to the sequence x0 , x1 , x2 d. d. d. d. 0 → R →0 Rx0 ⊕ Rx1 ⊕ Rx2 →1 Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 →2 Rx0 x1 x2 →3 0. We think of R as Q[x0 , x1 , x2 ] + Q[x1 , x2 ]w, which is an internal direct sum of Q-vector. 6.
(10) spaces. Similarly, we have ∞ X. Rx0 = Q[x0 , x1 , x2 ] +. 0 , Q[x1 , x2 ]x−α 0. α0 =1. Rx1 = Q[x0 , x1 , x2 ] + Q[x1 , x2 ]w + Rx2 = Q[x0 , x1 , x2 ] + Q[x1 , x2 ]w +. ∞ X α1 =1 ∞ X. 1 Q[x0 , x2 ]x−α 1. +. 1 + Q[x0 , x1 ]x−α 2. Rx0 x1 = Q[x0 , x1 , x2 ] + Rx0 x2 = Q[x0 , x1 , x2 ] +. α0 =1 ∞ X. 0 Q[x1 , x2 ]x−α 0. +. 0 + Q[x1 , x2 ]x−α 0. Rx1 x2 = Q[x0 , x1 , x2 ] + Q[x1 , x2 ]w + +. 2 Q[x1 ]wx−α + 2. ∞ X. Rx0 x1 x2 = Q[x0 , x1 , x2 ] +. ∞ X. ∞ X. + +. α1 =1 ∞ X. α1 =1 ∞ X. +. 2 + Q[x0 , x1 ]x−α 2. ∞ X ∞ X α0 =1 α1 =1 ∞ X ∞ X. 0 −α1 x1 , Q[x2 ]x−α 0. 0 −α2 x2 , Q[x1 ]x−α 0. α0 =1 α2 =1. +. 1 −α2 Q[x0 ]x−α x2 + 1. ∞ X. 1 Q[x2 ]wx−α 1. α1 =1 ∞ ∞ X X. +. ∞ X. 2 Q[x0 , x1 ]x−α 2. α2 =1 1 −α2 Qwx−α x2 , 1. α1 =1 α2 =1. 0 Q[x1 , x2 ]x−α + 0. 1 −α2 Q[x0 ]x−α x2 + 1. α1 =1 α2 =1 ∞ X ∞ X ∞ X. 1 Q[x0 , x2 ]x−α 1. 1 Q[x0 , x2 ]x−α 1. ∞ X. 1 Q[x0 , x2 ]x−α + 1. α1 =1. α0 =1 ∞ X ∞ X. ∞ X. α1 =1 α2 =1. α2 =1. 2 , Q[x1 ]wx−α 2. α2 =1. α0 =1. ∞ X. α1 =1 ∞ X. 1 , Q[x2 ]wx−α 1. α2 =1. α2 =1 ∞ X. ∞ X. ∞ X ∞ X α0 =1 α2 =1. 0 −α2 Q[x1 ]x−α x2 + 0. ∞ X. 2 Q[x0 , x1 ]x−α 2. α2 =1 ∞ X ∞ X. 0 −α1 Q[x2 ]x−α x1 0. α0 =1 α1 =1. 0 −α1 −α2 Qx−α x1 x2 . 0. α0 =1 α1 =1 α2 =1. Note that x0 w = 0 in R, so w = 0 in Rx0 , Rx0 x1 , Rx0 x2 , and Rx0 x1 x2 . For convenience, we will use some special notations. We will 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 F ∈ Rx1 , we write F = f 0,2 + f 0,2 (w) +. 7.
(11) fx0,2 + fx0,2 (w), where 1 1 f 0,2 ∈ Rx1 ∩ Q[x0 , x1 , x2 ], f 0,2 (w) ∈ Rx1 ∩ Q[x1 , x2 ]w, fx0,2 1. ∞ X. ∈ Rx1 ∩. 1 , Q[x0 , x2 ]x−α 1. α1 =1 ∞ X. fx0,2 (w) ∈ Rx1 ∩ 1. 1 . Q[x2 ]wx−α 1. α1 =1. The upper index 0, 2 in fx0,2 (w) means that fx0,2 (w) comes from Rxb0 x1 xb2 = Rx1 ; the lower 1 1 index x1 in fx0,2 (w) means that all terms of fx0,2 (w) have negative powers for x1 and have 1 1 nonnegative powers for x0 and x2 ; (w) in fx0,2 (w) means that fx0,2 (w) is a multiple of w. 1 1 Also, if F ∈ Rx0 x1 , we write F = f 2 + fx20 + fx21 + fx20 x1 , where f 2 ∈ Rx0 x1 ∩ Q[x0 , x1 , x2 ], fx20. ∈ Rx0 x1 ∩. fx21 ∈ Rx0 x1 ∩ fx20 x1. ∞ X α0 =1 ∞ X. 0 Q[x1 , x2 ]x−α , 0. 1 Q[x0 , x2 ]x−α , 1. α1 =1 ∞ X. ∈ Rx0 x1 ∩. ∞ X. 0 −α1 Q[x2 ]x−α x1 . 0. α0 =1 α1 =1. The upper index 2 in fx20 x1 means that fx20 x1 comes from Rx0 x1 xb2 = Rx0 x1 ; the lower index x0 x1 in fx20 x1 means that all terms of fx20 x1 have negative powers for both x0 and x1 and have nonnegative powers for x2 . Now we start to compute H i (C). The next theorem shows that H 0 (C) = 0. ˇ Theorem 2.1 Let R = Q[x0 , w, x1 , x2 ]/hx0 w, w2 i and consider the Cech complex with respect to the sequence x0 , x1 , x2 d. d. d. d. C : 0 → R →0 Rx0 ⊕ Rx1 ⊕ Rx2 →1 Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 →2 Rx0 x1 x2 →3 0. 8.
(12) Then we have H 0 (C) = 0.. Proof. Let f 0,1,2 + f 0,1,2 (w) ∈ R such that d0 (f 0,1,2 + f 0,1,2 (w)) = (0, 0, 0). Then we have (f 0,1,2 , f 0,1,2 + f 0,1,2 (w), f 0,1,2 + f 0,1,2 (w)) = (0, 0, 0). Thus f 0,1,2 = 0 in Q[x0 , x1 , x2 ], and f 0,1,2 (w) = 0 in Q[x1 , x2 ]w, and so f 0,1,2 + f 0,1,2 (w) = 0 in R. Hence we get Ker d0 = 0, 2. and so H 0 (C) = 0.. In next theorem, we show that H 1 (C) = 0.. ˇ Theorem 2.2 Let R = Q[x0 , w, x1 , x2 ]/hx0 w, w2 i and consider the Cech complex with respect to the sequence x0 , x1 , x2 d. d. d. d. C : 0 → R →0 Rx0 ⊕ Rx1 ⊕ Rx2 →1 Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 →2 Rx0 x1 x2 →3 0. Then we have H 1 (C) = 0.. Proof. In order to compute H 1 (C), we expand Rx0 , Rx1 , Rx2 as internal direct sums and. 9.
(13) then collect the components with the same Q-vector spaces together as the following: Rx0 ⊕ Rx1 ⊕ Rx2 = Q[x0 , x1 , x2 ] +. ∞ X. 0 Q[x1 , x2 ]x−α 0. . α0 =1. ⊕ Q[x0 , x1 , x2 ] + Q[x1 , x2 ]w + ⊕ Q[x0 , x1 , x2 ] + Q[x1 , x2 ]w +. ∞ X α1 =1 ∞ X. 1 Q[x0 , x2 ]x−α 1. +. 2 + Q[x0 , x1 ]x−α 2. α2 =1. ∞ X α1 =1 ∞ X. 1 Q[x2 ]wx−α 1. . 2 Q[x1 ]wx−α 2. . α2 =1. = Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] + 0 ⊕ Q[x1 , x2 ]w) ⊕ Q[x1 , x2 ]w +. ∞ X. 0 ⊕0⊕0 Q[x1 , x2 ]x−α 0. α0 =1. + 0⊕ + 0⊕. ∞ X α1 =1 ∞ X. 1 Q[x0 , x2 ]x−α ⊕0 1 1 Q[x2 ]wx−α ⊕0 1. α1 =1. + 0⊕0⊕ + 0⊕0⊕. ∞ X α2 =1 ∞ X. 2 Q[x0 , x1 ]x−α 2. . 2 Q[x1 ]wx−α . 2. (1). α2 =1. Note that in the last expression of the above equation, the + are indeed internal direct. 10.
(14) sums. We also express Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 similarly and get Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 = Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] + 0 ⊕ 0 ⊕ Q[x1 , x2 ]w + +. ∞ X. 0 ⊕ Q[x1 , x2 ]x−α 0. α0 =1 ∞ X. ∞ X. . 0 ⊕0 Q[x1 , x2 ]x−α 0. α0 =1 ∞ X. 1 ⊕0⊕ Q[x0 , x2 ]x−α 1. 1 Q[x0 , x2 ]x−α 1. . α1 =1. α1 =1. + 0⊕0⊕. ∞ X. 1 Q[x2 ]wx−α 1. . α1 =1 ∞ X. + 0⊕. 0 Q[x0 , x1 ]x−α 2. ⊕. +. ∞ X. 2 Q[x0 , x1 ]x−α 2. . α2 =1. α2 =1. + 0⊕0⊕. ∞ X. ∞ X. 2 Q[x1 ]wx−α 2. . α2 =1 ∞ X. 0 −α1 Q[x2 ]x−α x ⊕ 0 ⊕ 0 0 1. α0 =1 α1 =1 ∞ X ∞ X. + 0⊕. 0 −α2 Q[x1 ]x−α x ⊕ 0 0 2. α0 =1 α2 =1 ∞ X ∞ X. + 0⊕0⊕ + 0⊕0⊕. α1 =1 α2 =1 ∞ X ∞ X α1 =1 α2 =1. 11. 1 −α2 Q[x0 ]x−α x2 1. 1 −α2 Qwx−α x2 , 1. . (2).
(15) which is also an internal direct sum. Moreover, note that d1 Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] ⊆ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ], d1 0 ⊕ Q[x1 , x2 ]w) ⊕ Q[x1 , x2 ]w ⊆ 0 ⊕ 0 ⊕ Q[x1 , x2 ]w, d1. ∞ X. 0 Q[x1 , x2 ]x−α 0. ∞ ∞ X X −α0 0 ⊕ 0, Q[x1 , x2 ]x−α ⊕0⊕0 ⊆ Q[x1 , x2 ]x0 ⊕ 0. α0 =1. d1 0 ⊕ d1 0 ⊕. ∞ X α1 =1 ∞ X. 1 Q[x0 , x2 ]x−α 1. ⊕0 ⊆. α0 =1 ∞ X. d1 0 ⊕ 0 ⊕. 1 Q[x0 , x2 ]x−α 1. ⊕0⊕. 1 ⊕0 ⊆0⊕0⊕ Q[x2 ]wx−α 1 ∞ X α2 =1 ∞ X. 2 Q[x0 , x1 ]x−α 2. . ⊆0⊕. ∞ X. 1 , Q[x0 , x2 ]x−α 1. 1 , Q[x2 ]wx−α 1. α1 =1 ∞ X. 0 Q[x0 , x1 ]x−α 2. α2 =1 ∞ X. 2 Q[x1 ]wx−α ⊆0⊕0⊕ 2. α2 =1. ∞ X α1 =1. α1 =1. α1 =1. d1 0 ⊕ 0 ⊕. α0 =1. ⊕. ∞ X. 2 , Q[x0 , x1 ]x−α 2. α2 =1 2 Q[x1 ]wx−α . 2. (3). α2 =1. Hence, if we express every element a in Rx0 ⊕Rx1 ⊕Rx2 as a = b1 +b2 +b3 +b4 +b5 +b6 +b7 , where b1 ∈ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ], b2 ∈ 0 ⊕ Q[x1 , x2 ]w) ⊕ Q[x1 , x2 ]w, ∞ ∞ P P 0 1 b3 ∈ Q[x1 , x2 ]x−α ⊕ 0 ⊕ 0, b4 ∈ 0 ⊕ Q[x0 , x2 ]x−α ⊕ 0, 0 1 α0 =1. b5 ∈ 0 ⊕. ∞ P α1 =1. b7 ∈ 0 ⊕ 0 ⊕. α1 =1 1 Q[x2 ]wx−α 1. ∞ P. ⊕ 0,. b6 ∈ 0 ⊕ 0 ⊕. ∞ P. 2 Q[x0 , x1 ]x−α , 2. α2 =1 2 Q[x1 ]wx−α , 2. α2 =1. then we have a ∈ Ker d1 ⇔ b1 , b2 , b3 , b4 , b5 , b6 , b7 ∈ Ker d1 . Similarly, because R = Q[x0 , x1 , x2 ] + Q[x1 , x2 ]w and because d0 Q[x0 , x1 , x2 ] ⊆ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ], d0 Q[x1 , x2 ]w ⊆ 0 ⊕ Q[x1 , x2 ]w ⊕ Q[x1 , x2 ]w, with the notation as above, we have a ∈ Im d0 ⇔ b1 , b2 ∈ Im d0 and b3 = b4 = b5 = b6 = b7 = 0. Hence, in order to show Ker d1 ⊆ Im d0 , it suffices to show that for each 1 ≤ i ≤ 7, if bi ∈ Ker d1 , then bi ∈ Im d0 . 1. For the component Q[x0 , x1 , x2 ]⊕Q[x0 , x1 , x2 ]⊕Q[x0 , x1 , x2 ] in Rx0 ⊕Rx1 ⊕Rx2 , we let (f 1,2 , f 0,2 , f 0,1 ) ∈ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] in Rx0 ⊕ Rx1 ⊕ Rx2 such 12.
(16) that d1 (f 1,2 , f 0,2 , f 0,1 ) = (0, 0, 0) in Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 , i.e., (f 1,2 , f 0,2 , f 0,1 ) ∈ Ker d1 . Then we have (−f 1,2 + f 0,2 , −f 1,2 + f 0,1 , −f 0,2 + f 0,1 ) = (0, 0, 0), so we get f 1,2 = f 0,2 = f 0,1 in Q[x0 , x1 , x2 ]. Hence (f 1,2 , f 0,2 , f 0,1 ) = (f 1,2 , f 1,2 , f 1,2 ) = d0 (f 1,2 ) ∈ Im d0 . 2. For the component 0 ⊕ Q[x1 , x2 ]w ⊕ Q[x1 , x2 ]w in Rx0 ⊕ Rx1 ⊕ Rx2 , we let (0, f 0,2 (w), f 0,1 (w)) ∈ 0 ⊕ Q[x1 , x2 ]w ⊕ Q[x1 , x2 ]w in Rx0 ⊕ Rx1 ⊕ Rx2 such that d1 (0, f 0,2 (w), f 0,1 (w)) = (0, 0, 0) in Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 , i.e., (0, f 0,2 (w), f 0,1 (w)) ∈ Ker d1 . Then we have (0, 0, −f 0,2 (w) + f 0,1 (w)) = (0, 0, 0), so we get f 0,2 (w) = f 0,1 (w) in Q[x1 , x2 ]w. Hence (0, f 0,2 (w), f 0,1 (w)) = (0, f 0,2 (w), f 0,2 (w)) = d0 (f 0,2 (w)) ∈ Im d0 . 3. For the component. ∞ P α0 =1. ∞ P α0 =1. 0 ⊕ 0 ⊕ 0 in Rx0 ⊕ Rx1 ⊕ Rx2 , we let (fx1,2 , 0, 0) ∈ Q[x1 , x2 ]x−α 0 0. 0 Q[x1 , x2 ]x−α ⊕ 0 ⊕ 0 in Rx0 ⊕ Rx1 ⊕ Rx2 such that d1 (fx1,2 , 0, 0) = (0, 0, 0) in 0 0. Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 , i.e., (fx1,2 , 0, 0) ∈ Ker d1 . Then we have (−fx1,2 , −fx1,2 , 0) = 0 0 0 ∞ P 0 (0, 0, 0), so we get fx1,2 = 0 in Q[x1 , x2 ]x−α . Hence (fx1,2 , 0, 0) = (0, 0, 0) ∈ 0 0 0 α0 =1. Im d0 . 4. For the component 0 ⊕ 0⊕. ∞ P α1 =1. ∞ P α1 =1. 1 Q[x0 , x2 ]x−α 1. 1 Q[x0 , x2 ]x−α ⊕ 0 in Rx0 ⊕ Rx1 ⊕ Rx2 , we let (0, fx0,2 , 0) ∈ 1 1. ⊕ 0 in Rx0 ⊕ Rx1 ⊕ Rx2 such that d1 (0, fx0,2 , 0) = (0, 0, 0) in 1. Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 , i.e., (0, fx0,2 , 0) ∈ Ker d1 . Then we have (fx0,2 , 0, −fx0,2 ) = 1 1 1 ∞ P 1 (0, 0, 0), so we get fx0,2 = 0 in Q[x0 , x2 ]x−α . Hence (0, fx0,2 , 0) = (0, 0, 0) ∈ 1 1 1 α1 =1. Im d0 . 5. For the component 0⊕ 0⊕. ∞ P α1 =1. ∞ P α1 =1. 1 Q[x2 ]wx−α ⊕0 in Rx0 ⊕Rx1 ⊕Rx2 , we let (0, fx0,2 (w), 0) ∈ 1 1. 1 Q[x2 ]wx−α ⊕ 0 in Rx0 ⊕ Rx1 ⊕ Rx2 such that d1 (0, fx0,2 (w), 0) = (0, 0, 0) in 1 1. Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 , i.e., (0, fx0,2 (w), 0) ∈ Ker d1 . Then we have (0, 0, −fx0,2 (w)) = 1 1 ∞ P 1 Q[x2 ]wx−α . Hence (0, fx0,2 (w), 0) = (0, 0, 0) ∈ (0, 0, 0), so we get fx0,2 (w) = 0 in 1 1 1 α1 =1. Im d0 . 6. For the component 0 ⊕ 0 ⊕. ∞ P α2 =1. 2 Q[x0 , x1 ]x−α in Rx0 ⊕ Rx1 ⊕ Rx2 , we let (0, 0, fx0,1 )∈ 2 2. 13.
(17) 0⊕0⊕. ∞ P α2 =1. 2 in Rx0 ⊕ Rx1 ⊕ Rx2 such that d1 (0, 0, fx0,1 ) = (0, 0, 0) in Q[x0 , x1 ]x−α 2 2. ) = (0, 0, 0), , fx0,1 ) ∈ Ker d1 . Then we have (0, fx0,1 Rx0 x1 ⊕Rx0 x2 ⊕Rx1 x2 , i.e., (0, 0, fx0,1 2 2 2 ∞ P 2 . Hence (0, 0, fx0,1 ) = (0, 0, 0) ∈ Im d0 . so we get fx0,1 = 0 in Q[x0 , x1 ]x−α 2 2 2 α2 =1. ∞ P. 7. For the component 0⊕0⊕ 0⊕0⊕. α2 =1. ∞ P. 2 Q[x1 ]wx−α 2. α2 =1. 2 in Rx0 ⊕Rx1 ⊕Rx2 , we let (0, 0, fx0,1 (w)) ∈ Q[x1 ]wx−α 2 2. (w)) = (0, 0, 0) in in Rx0 ⊕ Rx1 ⊕ Rx2 such that d1 (0, 0, fx0,1 2. Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 , i.e., (0, 0, fx0,1 (w)) ∈ Ker d1 . Then we have (0, 0, fx0,1 (w)) = 2 2 ∞ P 2 (w)) = (0, 0, 0) ∈ . Hence (0, 0, fx0,1 (w) = 0 in Q[x1 ]wx−α (0, 0, 0), so we get fx0,1 2 2 2 α2 =1. Im d0 .. From the above discussion, we have Ker d1 ⊆ Im d0 , and so H 1 (C) = 0.. In next theorem, we show that H 2 (C) = 0⊕0⊕. ∞ P ∞ P. 2. 1 −α2 Qwx−α x2 +Im d1 /Im d1 . 1. α1 =1 α2 =1. ˇ Theorem 2.3 Let R = Q[x0 , w, x1 , x2 ]/hx0 w, w2 i and consider the Cech complex with respect to the sequence x0 , x1 , x2 d. d. d. d. C : 0 → R →0 Rx0 ⊕ Rx1 ⊕ Rx2 →1 Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 →2 Rx0 x1 x2 →3 0. Then we have H 2 (C) = (0 ⊕ 0 ⊕. ∞ P ∞ P. 1 −α2 Qwx−α x2 + Im d1 )/Im d1 . 1. α1 =1 α2 =1. Proof. Recall that we express Rx0 x1 x2 as ∞ X. Rx0 x1 x2 = Q[x0 , x1 , x2 ] +. 0 Q[x1 , x2 ]x−α 1. α0 =1. + +. ∞ X. ∞ X. 1 −α2 Q[x0 ]x−α x2 + 1. α1 =1 α2 =1 ∞ X ∞ X ∞ X. ∞ X. +. ∞ X. α1 =1 ∞ X. 1 Q[x0 , x2 ]x−α 1. +. 0 −α2 Q[x1 ]x−α x2 + 0. α0 =1 α2 =1 0 −α1 −α2 Qx−α x1 x2 . 0. ∞ X. 2 Q[x0 , x1 ]x−α 2. α2 =1 ∞ ∞ X X. 0 −α2 Q[x2 ]x−α x1 0. α0 =1 α1 =1. (4). α0 =1 α1 =1 α2 =1. 14.
(18) We will use expressions (1) and (2) to compute H 2 (C). Note that d2 (Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ]) ⊆ Q[x0 , x1 , x2 ], d2 (0 ⊕ 0 ⊕ Q[x1 , x2 ]w) = 0, d2 (. ∞ X. 0 ⊕ Q[x1 , x2 ]x−α 0. ∞ X α0 =1. α0 =1 ∞ X. 1 ⊕0⊕ Q[x0 , x2 ]x−α 1. d2 (. ∞ X. 1 )⊆ Q[x0 , x2 ]x−α 1. ∞ X. d2 (0 ⊕ 0 ⊕. α0 =1 ∞ X. 0 , Q[x1 , x2 ]x−α 0. 1 , Q[x0 , x2 ]x−α 1. α1 =1. α1 =1. α1 =1. ∞ X. 0 ⊕ 0) ⊆ Q[x1 , x2 ]x−α 0. 1 ) = 0, Q[x2 ]wx−α 1. α1 =1 ∞ X. d2 (0 ⊕. 2 Q[x0 , x1 ]x−α 2. ⊕. ∞ X. 2 ) Q[x0 , x1 ]x−α 2. ⊆. ∞ X. d2 (0 ⊕ 0 ⊕. 0 , Q[x0 , x1 ]x−α 2. α2 =1. α2 =1. α2 =1. ∞ X. 2 Q[x1 ]wx−α ) = 0, 2. α2 =1. d2 (. ∞ X ∞ X. 0 −α1 Q[x2 ]x−α x1 0. α0 =1 α1 =1 ∞ X ∞ X. d2 (0 ⊕. 0 −α2 Q[x1 ]x−α x2 ⊕ 0) ⊆ 0. α0 =1 α2 =1 ∞ X ∞ X. d2 (0 ⊕ 0 ⊕. d2 (0 ⊕ 0 ⊕. ⊕ 0 ⊕ 0) ⊆. α1 =1 α2 =1 ∞ X ∞ X. 1 −α2 Q[x0 ]x−α x2 ) ⊆ 1. ∞ X ∞ X α0 =1 α1 =1 ∞ X ∞ X α0 =1 α2 =1 ∞ X ∞ X. 0 −α1 Q[x2 ]x−α x1 , 0. 0 −α2 Q[x1 ]x−α x2 , 0. 1 −α2 Q[x0 ]x−α x2 , 1. α1 =1 α2 =1 1 −α2 Qwx−α x2 ) = 0. 1. (5). α1 =1 α2 =1. Hence, if we express every element a in Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 as a = b1 + b2 + b3 + b4 + b5 + b6 + b7 + b8 + b9 + b10 + b11 , where b1 ∈ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ], ∞ ∞ P P 0 0 b3 ∈ Q[x1 , x2 ]x−α ⊕ Q[x1 , x2 ]x−α ⊕ 0, 0 0 α0 =1. ∞ P. b5 ∈ 0 ⊕ 0 ⊕ b7 ∈ 0 ⊕ 0 ⊕. α1 =1 ∞ P. α0 =1. α1 =1. 1 Q[x2 ]wx−α , 1. b6 ∈ 0 ⊕. 2 Q[x1 ]wx−α , 2. b8 ∈. α2 =1. b9 ∈ 0 ⊕. ∞ P ∞ P. α0 =1 α2 =1 ∞ P. b11 ∈ 0 ⊕ 0 ⊕. b2 ∈ 0 ⊕ 0 ⊕ Q[x1 , x2 ]w, ∞ ∞ P P 1 1 b4 ∈ Q[x0 , x2 ]x−α ⊕0⊕ Q[x0 , x2 ]x−α , 1 1 ∞ P. α2 =1 ∞ P ∞ P α0 =1 α1 =1. 0 −α2 Q[x1 ]x−α x2 0. ∞ P. ⊕ 0,. b10 ∈ 0 ⊕ 0 ⊕. 0 Q[x0 , x1 ]x−α ⊕ 2. α1 =1 α2 =1. 15. 2 Q[x0 , x1 ]x−α , 2. α2 =1 0 −α1 Q[x2 ]x−α x1 ⊕ 0 ⊕ 0, 0. ∞ P ∞ P α1 =1 α2 =1. 1 −α2 Qwx−α x2 , 1. α1 =1 ∞ P. 1 −α2 Q[x0 ]x−α x2 , 1.
(19) then we have a ∈ Ker d2 ⇔ b1 , b2 , b3 , b4 , b5 , b6 , b7 , b8 , b9 , b10 , b11 ∈ Ker d2 . Similarly, from (3), with the notation as above, we have a ∈ Im d1 ⇔ b1 , b2 , b3 , b4 , b5 , b6 , b7 ∈ Im d1 and b8 = b9 = b10 = b11 = 0. In the following discussion, we will see that for 1 ≤ i ≤ 10, if ∞ P ∞ P 1 −α2 x2 , bi ∈ Ker d2 , then bi ∈ Im d1 , and in the last component b11 ∈ 0⊕0⊕ Qwx−α 1 α1 =1 α2 =1. b11. ˘ ∈ Ker d2 . Hence, instead of considering the Cech complex d. d. Rx0 ⊕ Rx1 ⊕ Rx2 →1 Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 →2 Rx0 x1 x2 , we consider the restricted complex for each component.. d. 1. For the component Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] →1 Q[x0 , x1 , x2 ] ⊕ d. Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] →2 Q[x0 , x1 , x2 ], we let (f 2 , f 1 , f 0 ) ∈ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] ⊕ Q[x0 , x1 , x2 ] in Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 such that d2 (f 2 , f 1 , f 0 ) = 0 in Rx0 x1 x2 , i.e., (f 2 , f 1 , f 0 ) ∈ Ker d2 . Then we have f 2 − f 1 + f 0 = 0, so we get f 0 = −f 2 + f 1 in Q[x0 , x1 , x2 ]. Hence (f 2 , f 1 , f 0 ) = (f 2 , f 1 , −f 2 + f 1 ) = d1 (0, f 2 , f 1 ) ∈ Im d1 . d. d. 2. For the component 0 ⊕ Q[x1 , x2 ]w ⊕ Q[x1 , x2 ]w →1 0 ⊕ 0 ⊕ Q[x1 , x2 ]w →2 0, we let (0, 0, f 0 (w)) ∈ 0⊕0⊕Q[x1 , x2 ]w in Rx0 x1 ⊕Rx0 x2 ⊕Rx1 x2 such that d2 (0, 0, f 0 (w)) = 0 in Rx0 x1 x2 , i.e., (0, 0, f 0 (w)) ∈ Ker d2 . Then we have (0, 0, f 0 (w)) = d1 (0, 0, f 0 (w)) ∈ Im d1 . 3. For the component d2. 0→. ∞ P. ∞ P α0 =1. 0 Q[x1 , x2 ]x−α , 0. α0 =1. ∞ P. d. 0 Q[x1 , x2 ]x−α ⊕0⊕0 →1 0. 0 Q[x1 , x2 ]x−α ⊕ 0. α0 =1. we let (fx20 , fx10 , 0) ∈. ∞ P. ∞ P. Q[x1 , x2 ]x0−α0 ⊕. α0 =1. 0 Q[x1 , x2 ]x−α ⊕ 0. α0 =1. ∞ P. 0 Q[x1 , x2 ]x−α ⊕ 0. α0 =1. 0 in Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 such that d2 (fx20 , fx10 , 0) = 0 in Rx0 x1 x2 , i.e., (fx20 , fx10 , 0) ∈ ∞ P 0 Ker d2 . Then we have fx20 − fx10 = 0, so we get fx10 = fx20 in Q[x1 , x2 ]x−α . 0 α0 =1. Hence (fx20 , fx10 , 0) = (fx20 , fx20 , 0) = d1 (−fx20 , 0, 0) ∈ Im d1 . 4. For the component 0⊕ ∞ P α0 =1. ∞ P. d. 1 Q[x0 , x2 ]x−α ⊕0 →1 1. α1 =1 ∞ P. d 0 Q[x1 , x2 ]x−α ⊕0 →2 0. ∞ P. 1 Q[x0 , x2 ]x−α ⊕0⊕ 1. α1 =1 1 Q[x0 , x2 ]x−α , 1. α1 =1. 16. we let (fx21 , 0, fx01 ) ∈. ∞ P α1 =1. ∞ P. 1 Q[x0 , x2 ]x−α ⊕ 1. α1 =1 1 Q[x0 , x2 ]x−α ⊕ 1.
(20) ∞ P. 0⊕. α1 =1. 1 in Rx0 x1 ⊕Rx0 x2 ⊕Rx1 x2 such that d2 (fx21 , 0, fx01 ) = 0 in Rx0 x1 x2 , Q[x0 , x2 ]x−α 1. i.e., (fx21 , 0, fx01 ) ∈ Ker d2 . Then we have fx21 + fx01 = 0, so we get fx00 = −fx20 in ∞ P 0 Q[x1 , x2 ]x−α . Hence (fx21 , 0, fx01 ) = (fx21 , 0, −fx20 ) = d1 (0, fx20 , 0) ∈ Im d1 . 0 α0 =1. ∞ P. 5. For the component 0 ⊕. d. 1 ⊕ 0 →1 0 ⊕ 0 ⊕ Q[x2 ]wx−α 1. α1 =1. we let. (0, 0, fx01 (w)). d. 1 Q[x2 ]wx−α →2 0, 1. α1 =1. ∞ P. ∈ 0⊕0⊕. ∞ P. 1 Q[x2 ]wx−α 1. in Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 such. α1 =1. that d2 (0, 0, fx01 (w)) = 0 in Rx0 x1 x2 , i.e., (0, 0, fx01 (w)) ∈ Ker d2 . Then we have (0, 0, fx01 (w)) = d1 (0, −fx01 (w), 0) ∈ Im d1 . 6. For the component 0⊕0⊕ ∞ P. ∞ P α2 =1. 2 , Q[x0 , x1 ]x−α 2. we let (0, fx12 , fx02 ) ∈ 0 ⊕. α2 =1. ∞ P. d. 2 Q[x0 , x1 ]x−α →1 0⊕ 2. ∞ P. 0 ⊕ Q[x0 , x1 ]x−α 2. α2 =1 0 Q[x0 , x1 ]x−α 2. ⊕. ∞ P. ∞ P. d. 2 Q[x0 , x1 ]x−α →2 2. α2 =1. Q[x0 , x1 ]x2−α2. α2 =1. α2 =1. in Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 such that d2 (0, fx12 , fx02 ) = 0 in Rx0 x1 x2 , i.e., (0, fx12 , fx02 ) ∈ ∞ P 0 Ker d2 . Then we have −fx12 + fx02 = 0, so we get fx02 = fx12 in Q[x0 , x1 ]x−α . 2 α2 =1. Hence. (0, fx12 , fx02 ). =. (0, fx12 , fx12 ). =. ∞ P. 7. For the component 0 ⊕ 0 ⊕. ∈ Im d1 . d. 2 Q[x1 ]wx−α →1 0 ⊕ 0 ⊕ 2. α2 =1. we let (0, 0, fx02 (w)) ∈ 0 ⊕ 0 ⊕. d1 (0, 0, fx12 ). ∞ P. d. 2 Q[x1 ]wx−α →2 0, 2. α2 =1. ∞ P. 2 Q[x1 ]wx−α 2. in Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 such. α2 =1. that d2 (0, 0, fx02 (w)) = 0 in Rx0 x1 x2 , i.e., (0, 0, fx02 (w)) ∈ Ker d2 . Then we have (0, 0, fx02 (w)) = d1 (0, 0, fx02 (w)) ∈ Im d1 . d. 8. For the component 0⊕0⊕0 →1 2 , 0, 0) ∈ we let (f0,1. ∞ P ∞ P. ∞ P ∞ P. d. 0 −α1 Q[x2 ]x−α x1 ⊕0⊕0 →2 0. α0 =1 α1 =1. ∞ P ∞ P. 0 −α1 Q[x2 ]x−α x1 , 0. α0 =1 α1 =1. 0 −α1 Q[x2 ]x−α x1 ⊕ 0 ⊕ 0 in Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 such 0. α0 =1 α1 =1 2 2 that d2 (f0,1 , 0, 0) = 0 in Rx0 x1 x2 , i.e., (f0,1 , 0, 0) ∈ Ker d2 . Then we have fx20 x1 = 0 ∞ ∞ P P 2 0 −α1 in Q[x2 ]x−α x1 . Hence (f0,1 , 0, 0) = (0, 0, 0) ∈ Im d1 . 0 α0 =1 α1 =1 d. 9. For the component 0⊕0⊕0 →1 0⊕ 1 we let (0, f0,2 , 0) ∈ 0 ⊕. ∞ P ∞ P. ∞ P ∞ P. α0 =1 α2 =1. d. 0 −α2 Q[x1 ]x−α x2 ⊕0 →2 0. ∞ P ∞ P. 0 −α1 Q[x2 ]x−α x1 , 0. α0 =1 α1 =1. 0 −α2 Q[x1 ]x−α x2 ⊕ 0 in Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 such 0. α0 =1 α2 =1 1 1 that d2 (0, f0,2 , 0) = 0 in Rx0 x1 x2 , i.e., (0, f0,2 , 0) ∈ Ker d2 . Then we have −fx10 x2 = 0, ∞ ∞ P P 1 0 −α2 so we get fx10 x2 = 0 in Q[x1 ]x−α x2 . Hence (0, f0,2 , 0) = (0, 0, 0) ∈ Im d1 . 0 α0 =1 α2 =1. 17.
(21) d. 10. For the component 0⊕0⊕0 →1 0⊕0⊕. ∞ P ∞ P. d. 1 −α2 x2 →2 Q[x0 ]x−α 1. 0 we let (0, 0, f1,2 ) ∈ 0⊕0⊕. 1 −α2 x2 , Q[x0 ]x−α 1. α1 =1 α2 =1. α1 =1 α2 =1. ∞ P ∞ P. ∞ P ∞ P. 1 −α2 x2 Q[x0 ]x−α 1. in Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 such. α1 =1 α2 =1 0 0 ) ∈ Ker d2 . Then we have fx10 x2 = 0 ) = 0 in Rx0 x1 x2 , i.e., (0, 0, f1,2 that d2 (0, 0, f1,2 ∞ P ∞ P 0 1 −α2 ) = (0, 0, 0) ∈ Im d1 . x2 . Hence (0, 0, f1,2 in Q[x0 ]x−α 1 α1 =1 α2 =1 d. 11. For the component 0 ⊕ 0 ⊕ 0 →1 0 ⊕ 0 ⊕ to see that Ker d2 = 0 ⊕ 0 ⊕. ∞ P ∞ P. d. 1 −α2 x2 →2 0, it is not hard Qwx−α 1. α1 =1 α2 =1. ∞ P ∞ P. 1 −α2 x2 and Im d1 = 0. Thus, for this Qwx−α 1. α1 =1 α2 =1 ∞ P. component, Ker d2 /Im d1 = 0 ⊕ 0 ⊕. ∞ P. 1 −α2 x2 . Qwx−α 1. α1 =1 α2 =1. From the above discussion, we get H 2 (C) = (0 ⊕ 0 ⊕. ∞ P ∞ P. 1 −α2 x2 + Im d1 )/Im d1 . Qwx−α 1. α1 =1 α2 =1. 2. In next theorem, we show that H 3 (C) = (. ∞ P ∞ P ∞ P. 0 −α1 −α2 Qx−α x1 x2 +Im d2 )/Im d2 . 0. α0 =1 α1 =1 α2 =1. ˇ Theorem 2.4 Let R = Q[x0 , w, x1 , x2 ]/hx0 w, w2 i and consider the Cech complex with respect to the sequence x0 , x1 , x2 d. d. d. d. C : 0 → R →0 Rx0 ⊕ Rx1 ⊕ Rx2 →1 Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 →2 Rx0 x1 x2 →3 0. Then we have H 3 (C) =. ∞ P ∞ P ∞ P. 0 −α1 −α2 Qx−α x1 x2 + Im d2 /Im d2 . 0. α0 =1 α1 =1 α2 =1. 18.
(22) Proof. It is clear that Ker d3 = Rx0 x1 x2 . Next we compute Im d2 . Im d2 = d2 (Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 ) = d2 (Rx0 x1 ⊕ 0 ⊕ 0) + d2 (0 ⊕ Rx0 x2 ⊕0 ) + d2 (0 ⊕ 0 ⊕ Rx1 x2 ) = Q[x0 , x1 , x2 ] +. ∞ X. α0 =1 ∞ X. + Q[x0 , x1 , x2 ] +. + Q[x0 , x1 , x2 ] +. ∞ X. 0 + Q[x1 , x2 ]x−α 0. α0 =1 ∞ X. α1 =1 ∞ X. 0 + Q[x1 , x2 ]x−α 0. 1 + Q[x0 , x2 ]x−α 1. = Q[x0 , x1 , x2 ] +. 0 + Q[x1 , x2 ]x−α 0. +. 0 −α1 Q[x2 ]x−α x1 + 0. α0 =1 α1 =1. ∞ X. ∞ X ∞ X. 0 −α2 x2 Q[x1 ]x−α 0. 2 Q[x0 , x1 ]x−α 2. ∞ X ∞ X. 1 −α2 x2 Q[x0 ]x−α 1. α1 =1 α2 =1. 1 + Q[x0 , x2 ]x−α 1. ∞ X. 2 Q[x0 , x1 ]x−α 2. α2 =1 ∞ X ∞ X. 1 −α2 Q[x0 ]x−α x2 . 1. α1 =1 α2 =1. Hence, compare the expression (4) with the above equation and then we have H 3 (C) = Ker d3 /Im d2 =. ∞ X ∞ X ∞ X α0 =1 α1 =1 α2 =1. 2. 19. . α0 =1 α2 =1. 0 −α2 Q[x1 ]x−α x2 + 0. α0 =1 α2 =1. 0 −α1 x1 Q[x2 ]x−α 0. 2 + Q[x0 , x1 ]x−α 2. α1 =1. α0 =1 ∞ X ∞ X. α2 =1 ∞ X. ∞ X ∞ X. α0 =1 α1 =1 ∞ X ∞ X. α2 =1. α1 =1 ∞ X. 1 + Q[x0 , x2 ]x−α 1. 0 −α1 −α2 Qx−α x1 x2 + Im d2 /Im d2 . 0. . .
(23) 3. The General Case. Before we compute the general case, we first introduce some notations that we will use. ˇ Recall that in Section 2, R = Q[x0 , x1 , x2 , w]/hx0 w, w2 i and we consider the Cech complex with respect to the sequence x0 , x1 , x2 d. d. d. d. 0 → R →0 Rx0 ⊕ Rx1 ⊕ Rx2 →1 Rx0 x1 ⊕ Rx0 x2 ⊕ Rx1 x2 →2 Rx0 x1 x2 →3 0. (w), we write Rxˆ0 x1 xˆ2 for Rx1 . Using the notation, we While introducing the notation fx0,2 1 ˇ can rewrite the above Cech complex as d. d. d. d. 0 → Rxˆ0 xˆ1 xˆ2 →0 Rx0 xˆ1 xˆ2 ⊕ Rxˆ0 x1 xˆ2 ⊕ Rxˆ0 xˆ1 x2 →1 Rx0 x1 xˆ2 ⊕ Rx0 xˆ1 x2 ⊕ Rxˆ0 x1 x2 →2 Rx0 x1 x2 →3 0. ˇ Next, in the above Cech complex, we use ex0 xˆ1 xˆ2 to represent the component Rx0 xˆ1 xˆ2 in Rx0 xˆ1 xˆ2 ⊕ Rxˆ0 x1 xˆ2 ⊕ Rxˆ0 xˆ1 x2 . Similarly, we use exˆ0 x1 xˆ2 and exˆ0 xˆ1 x2 to represent the component Rxˆ0 x1 xˆ2 and Rxˆ0 xˆ1 x2 in Rx0 xˆ1 xˆ2 ⊕ Rxˆ0 x1 xˆ2 ⊕ Rxˆ0 xˆ1 x2 , respectively. Thus, for example, if (f 1,2 , f 0,2 , f 0,1 ) ∈ Rx0 xˆ1 xˆ2 ⊕ Rxˆ0 x1 xˆ2 ⊕ Rxˆ0 xˆ1 x2 , we write (f 1,2 , f 0,2 , f 0,1 ) = f 1,2 ex0 xˆ1 xˆ2 + f 0,2 exˆ0 x1 xˆ2 + f 0,1 ex0 xˆ1 xˆ2 . ˇ Next, we will introduce how we order the component of C i in the Cech complex. Take the above complex as an example. C 2 = Rx0 x1 xˆ2 ⊕ Rx0 xˆ1 x2 ⊕ Rxˆ0 x1 x2 and there are three components in C 2 . Among these three components,. 1. Rx0 x1 xˆ2 and Rx0 xˆ1 x2 both have the index x0 while Rxˆ0 x1 x2 has the index xˆ0 , so we write Rx0 x1 xˆ2 and Rx0 xˆ1 x2 before Rxˆ0 x1 x2 ; 2. Rx0 x1 xˆ2 and Rx0 xˆ1 x2 both have the index x0 , and Rx0 x1 xˆ2 has the index x1 while Rx0 xˆ1 x2 has the index xˆ1 , so we write Rx0 x1 xˆ2 before Rx0 xˆ1 x2 .. Hence, we order the components of C 2 as C 2 = Rx0 x1 xˆ2 ⊕Rx0 xˆ1 x2 ⊕Rxˆ0 x1 x2 . Next, in order to make this ordering clearer, we use another example to illustrate it. For the ring R = 20.
(24) Q[x0 , w, x1 , x2 , x3 , x4 , x5 , x6 ]/hx0 w, w2 i, we compare the components Rx0 x1 x2 x3 xˆ4 x5 xˆ6 and Rx0 x1 x2 xˆ3 x4 xˆ5 x6 of C 5 . Note that they both have the indices x0 , x1 , x2 , and Rx0 x1 x2 x3 xˆ4 x5 xˆ6 the index x3 while Rx0 x1 x2 xˆ3 x4 xˆ5 x6 has xˆ3 . Thus we write Rx0 x1 x2 x3 xˆ4 x5 xˆ6 before Rx0 x1 x2 xˆ3 x4 xˆ5 x6 when we order the components of C 5 . Now, we are ready to compute the general case.. From now on, R is the ring. Q[x0 , x1 , x2 , . . . , xn , w]/hx0 w, w2 i and p is the maximal ideal hx0 , x1 , x2 , . . . , xn , wi in R. ˇ Since hx0 , x1 , x2 , . . . , xn i is p-primary, we consider the Cech complex with respect to the sequence x0 , x1 , x2 , . . . , xn . ˇ Similar as in Section 2, we write each component that occurs in the Cech complex as an internal direct sum of Q-vector spaces. For C 0 = R, we write R = Q[x0 , x1 , . . . , xn ] + Q[x1 , x2 , . . . , xn ]w. Since x0 w = 0 in R, for C 1 = Rx0 xˆ1 xˆ2 ···ˆxn ⊕ Rxˆ0 x1 xˆ2 ···ˆxn ⊕ · · · ⊕ Rxˆ0 xˆ1 xˆ2 ···xn , we have Rx0 xˆ1 xˆ2 ···ˆxn = Q[x0 , x1 , . . . , xn ] +. ∞ X. 0 Q[x1 , x2 , . . . , xn ]x−α , 0. α0 =1. and for i = 1, 2, . . . , n, Rxˆ0 xˆ1 ···ˆxi−1 xi xˆi+1 ···ˆxn = Q[x0 , x1 , . . . , xn ] + Q[x1 , x2 , . . . , xn ]w + +. ∞ X αi =1 ∞ X. i Q[x0 , x1 , . . . , xi−1 , xi+1 , . . . , xn ]x−α i. i Q[x1 , x2 , . . . , xi−1 , xi+1 , . . . , xn ]wx−α . i. αi =1. In general, for the R-module C t =. L. 1≤i1 <i2 <...<it ≤n. Rxi1 xi2 ···xit , there are two categories of. components, namely xi1 = 0 and xi1 ≥ 1. For example, for the R-module C i+1 , because. 21.
(25) x0 w = 0 and so w = 0 in Rx0 x1 ...xi xˆi+1 xˆi+2 ···ˆxn , we have Rx0 x1 ···xi xˆi+1 xˆi+2 ···ˆxn = Q[x0 , x1 , x2 , . . . , xn ] +. ∞ X. 0 Q[x1 , x2 , . . . , xn ]x−α 0. α0 =1. + ··· + +. ∞ X αi =1 ∞ X. i Q[x0 , x1 , . . . , xi−1 , xi+1 , . . . , xn ]x−α i. ∞ X. 0 −α1 x1 Q[x2 , x3 , . . . , xn ]x−α 0. α0 =1 α1 =1. + ··· +. ∞ ∞ X X. −α. i Q[x0 , x1 , . . . , xi−2 , xi+1 , . . . , xn ]xi−1i−1 x−α i. αi−1 =1 αi =1. + ··· +. ∞ X ∞ X α0 =1 α1 =1. ···. ∞ X. i 0 −α1 ; Q[xi+1 , xi+2 , . . . , xn ]x−α x1 · · · x−α 0 i. αi =1. 22.
(26) because w 6= 0 in Rxˆ0 x1 x2 ···xi+1 xˆi+2 ···ˆxn , we have Rxˆ0 x1 x2 ···xi+1 xˆi+2 ···ˆxn = Q[x0 , x1 , x2 , . . . , xn ] + Q[x1 , x2 , . . . , xn ]w +. ∞ X. 1 Q[x0 , x2 , x3 , . . . , xn ]x−α 1. +. ∞ X. 1 Q[x2 , x3 , . . . , xn ]wx−α 1. α1 =1. α1 =1. + ··· +. +. + +. ∞ X αi+1 =1 ∞ X. −α. Q[x0 , x1 , x2 , . . . , xi , xi+2 , . . . , xn ]xi+1i+1 −α. Q[x1 , x2 , . . . , xi , xi+2 , . . . , xn ]wxi+1i+1. αi+1 =1 ∞ X ∞ X α1 =1 α2 =1 ∞ X ∞ X. 1 −α2 x2 Q[x0 , x3 , x4 , . . . , xn ]x−α 1. 1 −α2 Q[x3 , x4 , . . . , xn ]wx−α x2 1. α1 =1 α2 =1. + ··· +. +. ∞ ∞ X X. −α. i xi+1i+1 Q[x0 , x1 , . . . , xi−1 , xi+2 , . . . , xn ]x−α i. αi =1 αi+1 =1 ∞ ∞ X X. −α. i xi+1i+1 Q[x1 , x2 , ..., xi−1 , xi+2 , . . . , xn ]wx−α i. αi =1 αi+1 =1. + ··· +. +. ∞ X ∞ X α1 =1 α2 =1 ∞ X ∞ X α1 =1 α2 =1. ··· ···. ∞ X αi+1 =1 ∞ X. −α. 1 −α2 Q[x0 , xi+2 , . . . , xn ]x−α x2 · · · xi+1i+1 1. −α. 1 −α2 Q[xi+2 , . . . , xn ]wx−α x2 · · · xi+1i+1 . 1. αi+1 =1. 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 fx0,2 (w), we know that 1 x3 1. this element comes from Rxˆ0 x1 xˆ2 x3 ···xn , 23.
(27) 2. all terms of this element have negative powers for both x1 and x3 and have nonnegative powers for x0 , x2 , x4 , . . . , xn , 3. this element is a multiple of w; in other words, we know that fx0,2 (w) ∈ 1 x2. ∞ P ∞ P. 1 −α3 x3 ∩Rxˆ0 x1 xˆ2 x3 ···xn . Q[x2 , x4 , . . . , xn ]wx−α 1. α1 =1 α3 =1. ˇ Example 3.1 Let R = Q[x0 , x1 , x2 , . . . , xn , w]/hx0 w, w2 i and let C be the Cech complex with respect to the sequence x0 , x1 , x2 , . . . , xn . Then we have H 0 (C) = 0. Proof. Note that H 0 (C) = Ker d0 , where d. 0 → Rxˆ0 xˆ1 ···ˆxn →0 Rx0 xˆ1 xˆ2 ···ˆxn ⊕ Rxˆ0 x1 xˆ2 ···ˆxn ⊕ · · · ⊕ Rxˆ0 xˆ1 xˆ2 ···ˆxn−1 xn . Let f 0,1,...,n + f 0,1,...,n (w) ∈ Rxˆ0 xˆ1 ···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 , . . . , xn ], and f 0,1,...,n (w) = 0 in Q[x1 , . . . , xn ]w, and so f 0,1,...,n + f 0,1,...,n (w) = 0 in R. Hence we get Ker d0 = 0, and so H 0 (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 , . . . , xn } and for {xi1 , xi2 , . . . , xit } ⊆ I, we let I i1 ,i2 ,...,it = I − {xi1 , xi2 , . . . , xit }. Also, for xi ∈ I and J = {xj1 , xj2 , . . . , xjt } ⊆ I, we let (−1)s−1 if j < j < · · · < j 1 2 s−1 < i < js < · · · < jt , σ(xi , J) = 0 if x ∈ J. i. With the above notation, we can rewrite the component Rxi1 ···xit → Rxj1 xj2 ···xjt+1 in Definition 1.1, that gives the differentiation dt : C t → C t+1 , as σ(x , J) · nat : R j xi1 ···xit → (Rxi1 ···xit )xj if {xj1 , . . . , xjt+1 } = J ∪ {xj }, 0 otherwise 24.
(28) where J = {xi1 , . . . , xit }. Note that J = I A where A = I − {i1 , . . . , it }. For example, the component Rx0 x1 ···xi xˆi+1 xˆi+2 ...ˆxn → Rx0 x1 ···xi xˆi+1 xi+2 xˆi+3 ...ˆxn is the homomorphism σ(xi+2 , I i+1,i+2,...,n ) · nat : Rx0 x1 ···xi → (Rx0 x1 ···xi )xi+2 .. ˇ Example 3.2 Let R = Q[x0 , x1 , x2 , . . . , xn , w]/hx0 w, w2 i and let C be the Cech complex with respect to the sequence x0 , x1 , x2 , . . . , xn . Then we have H i+1 (C) = 0 for all 0 ≤ i ≤ n − 2.. Remark 3.3 Note that H i+1 (C) = Ker di+1 /Im di , where Rx0 x1 ···xi−1 xˆi xˆi−1 ···ˆxn ⊕ · · · ⊕ Rxˆ0 xˆ1 ···ˆxn−i xn−i+1 ···xn d. →i Rx0 x1 ···xi xˆi+1 xˆi+2 ···ˆxn ⊕ · · · ⊕ Rxˆ0 xˆ1 ···ˆxn−i−1 xn−i ···xn di+1. → Rx0 x1 ···xi+1 xˆi+2 xˆi+3 ···ˆxn ⊕ · · · ⊕ Rxˆ0 xˆ1 ···ˆxn−i−2 xn−i−1 ···xn .. Before we start to compute H i+1 (C), as we did in the proof of Theorem 2.2, we first write Rx0 x1 ···xi−1 xˆi xˆi−1 ···ˆxn , . . ., Rxˆ0 xˆ1 ···ˆxn−i xn−i+1 ···xn as internal direct sums and then collect the components with the same Q-vector spaces together to get an internal direct sum of C i = Rx0 x1 ···xi−1 xˆi xˆi−1 ···ˆxn ⊕ · · · ⊕ Rxˆ0 xˆ1 ···ˆxn−i xn−i+1 ···xn . Similarly, we also express C i+1 = Rx0 x1 ···xi xˆi+1 xˆi+2 ···ˆxn ⊕ · · · ⊕ Rxˆ0 xˆ1 ···ˆxn−i−1 xn−i ···xn and C i+2 = Rx0 x1 ···xi+1 xˆi+2 xˆi+3 ···ˆxn ⊕ · · · ⊕ Rxˆ0 xˆ1 ···ˆxn−i−2 xn−i−1 ···xn as internal direct sums. Similarly as in (3), di+1 sends each component in C i+1 to the component in C i+2 corresponding to the same Q-vector space. Hence if we express every element a in C i+1 as 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 di+1 . Similarly, we have a ∈ Im di if and only if every term of a belongs to Im di . Hence, it is sufficient to show for each component that Ker di+1 ⊆ Im di .. We first consider the component which has no negative powers for x0 , x1 , x2 , . . . , xn and not a multiple of w, i.e., the component with respect to Q[x0 , x1 , . . . , xn ]. For f ∈. 25.
(29) C i+1 , since C i+1 = Rx0 x1 ···xi xˆi+1 xˆi+2 ···ˆxn ⊕ · · · ⊕ Rx0 xˆ1 xˆ2 ···ˆxn−i xn−i+1 ···xn ⊕ Rxˆ0 x1 ···xi+1 xˆi+2 xˆi+3 ···ˆxn ⊕ · · · ⊕ Rxˆ0 xˆ1 ···ˆxn−i−1 xn−i ···xn , we write f = f i+1,i+2,...,n ex0 x1 ···xi xˆi+1 xˆi+2 ···ˆxn + · · · + f 1,2,...,n−i ex0 xˆ1 xˆ2 ···ˆxn−i xn−i+1 ···xn + f 0,i+2,i+3,...,n exˆ0 x1 ···xi+1 xˆi+2 xˆi+3 ···ˆxn + · · · + f 0,1,...,n−i−1 exˆ0 xˆ1 ···ˆxn−i−1 xn−i ···xn .. Lemma 3.4 Let f ∈ C i+1 and let f = f i+1,i+2,...,n ex0 x1 ···xi xˆi+1 xˆi+2 ···ˆxn + · · · + f 1,2,...,n−i ex0 xˆ1 xˆ2 ···ˆxn−i xn−i+1 ···xn + f 0,i+2,i+3,...,n exˆ0 x1 ···xi+1 xˆi+2 xˆi+3 ···ˆxn + · · · + f 0,1,...,n−i−1 exˆ0 xˆ1 ···ˆxn−i−1 xn−i ···xn , i.e., f is in the component which has no negative powers for x0 , x1 , x2 , . . . , xn and is not a multiple of w. Then f ∈ Ker di+1 implies f ∈ Im di .. Proof. Since di+1 (f ) = 0 in C i+2 and C i+2 = Rx0 x1 ···xi+1 xˆi+2 xˆi+3 ···ˆxn ⊕· · ·⊕Rx0 xˆ1 xˆ2 ···ˆxn−i−1 xn−i ···xn ⊕ Rxˆ0 x1 ···xi+2 xˆi+3 xˆi+4 ···ˆxn ⊕· · ·⊕Rxˆ0 xˆ1 ···ˆxn−i−2 xn−i−1 ···xn , for the component ex0 x1 ···xi+1 xˆi+2 xˆi+3 ···ˆxn , we have σ(xi+1 , I i+1,i+2,i+3,...,n )f i+1,i+2,i+3,...,n + · · · + σ(x1 , I 1,i+2,i+3,...,n )f 1,i+2,i+3,...,n + σ(x0 , I 0,i+2,i+3,...,n )f 0,i+2,i+3,...,n = 0. Then we have that the coefficient of the component exˆ0 ,x1 ,...,xi+1, xˆi+2 ,ˆxi+3 ,...,ˆxn of f is f 0,i+2,i+3,...,n = −σ(x0 , I 0,i+2,i+3,...,n )σ(xi+1 , I i+1,i+2,i+3,...,n )f i+1,i+2,i+3,...,n − ··· − σ(x0 , I 0,i+2,i+3,...,n )σ(x1 , I 1,i+2,i+3,...,n )f 1,i+2,i+3,...,n .. 26.
(30) Similarly for the component ex0 x1 ···xi xˆi+1 xi+2 xˆi+3 ···ˆxn , we have σ(xi+2 , I i+1,i+2,...,n )f i+1,i+2,...,n + σ(xi , I i,i+1,i+3,...,n )f i,i+1,i+3,...,n + σ(xi−1 , I i−1,i+1,i+3,...,n )f i−1,i+1,i+3,...,n + · · · + σ(x1 , I 1,i+1,i+3,...,n )f 1,i+1,i+3,...,n + σ(x0 , I 0,i+1,i+3,...,n )f 0,i+1,i+3,...,n = 0, and we have the coefficient of the component exˆ0 ,x1 ,...,xi xˆi+1 ,xi+2 xˆi+3 ,...,ˆxn of f is f 0,i+1,i+3,...,n = −σ(x0 , I 0,i+1,i+3,...,n )σ(xi+2 , I i+1,i+2,...,n )f i+1,i+2,...,n − σ(x0 , I 0,i+1,i+3,...,n )σ(xi , I i,i+1,i+3,...,n )f i,i+1,i+3,...,n − ··· − σ(x0 , I 0,i+1,i+3,...,n )σ(x1 , I 1,i+1,i+3,...,n )f 1,i+1,i+3,...,n . Repeat the same discussion for all the remaining components with index x0 , i.e., ex0 x1 ···xi xˆi+1 xˆi+2 xi+3 xˆi+4 ···ˆxn , . . . , ex0 xˆ1 xˆ2 ···ˆxn−i−1 xn−i ···xn , and we will get f 0,i+2,i+3,...,n = −σ(x0 , I 0,i+2,i+3,...,n )σ(xi+1 , I i+1,i+2,i+3,...,n )f i+1,i+2,i+3,...,n −··· −σ(x0 , I 0,i+2,i+3,...,n )σ(x1 , I 1,i+2,i+3,...,n )f 1,i+2,i+3,...,n f 0,i+1,i+3,...,n = −σ(x0 , I 0,i+1,i+3,...,n )σ(xi+2 , I i+1,i+2,...,n )f i+1,i+2,...,n −σ(x0 , I 0,i+1,i+3,...,n )σ(xi , I i,i+1,i+3,...,n )f i,i+1,i+3,...,n −···. f 0,1,2,...,n−i−1 . −σ(x0 , I 0,i+1,i+3,...,n )σ(x1 , I 1,i+1,i+3,...,n )f 1,i+1,i+3,...,n .. . = −σ(x0 , I 0,1,2,...,n−i−1 )σ(xn , I 1,2,...,n−i−1,n )f 1,2,...,n−i−1,n −··· −σ(x0 , I 0,1,2,...,n−i−1 )σ(xn−i , I 1,2,...,n−i−1,n−i )f 1,2,...,n−i−1,n−i .. 27. (6).
(31) From the above relations, we can rewrite f as f = f i+1,i+2,...,n ex0 x1 ···xi xˆi+1 xˆi+2 ···ˆxn + · · · + f 1,2,...,n−i+1 ex0 xˆ1 xˆ2 ···ˆxn−i+1 xn−i+2 ···xn + − σ(x0 , I 0,i+2,i+3,...,n )σ(xi+1 , I i+1,i+2,i+3,...,n )f i+1,i+2,i+3,...,n − · · · − σ(x0 , I 0,i+2,i+3,...,n )σ(x1 , I 1,i+2,i+3,...,n )f 1,i+2,i+3,...,n exˆ0 x1 ···xi+1 xˆi+2 xˆi+3 ···ˆxn + − σ(x0 , I 0,i+1,i+3,...,n )σ(xi+2 , I i+1,i+2,...,n )f i+1,i+2,...,n − σ(x0 , I 0,i+1,i+3,...,n )σ(xi , I i,i+1,i+3,...,n )f i,i+1,i+3,...,n − ··· − σ(x0 , I 0,i+1,i+3,...,n )σ(x1 , I 1,i+1,i+3,...,n )f 1,i+1,i+3,...,n ex0 x1 ···xi xˆi+1 xi+2 xˆi+3 ···ˆxn + ··· + − σ(x0 , I 0,1,2,...,n−i−1 )σ(xn , I 1,2,...,n−i−1,n )f 1,2,...,n−i−1,n − · · · − σ(x0 , I 0,1,2,...,n−i−1 )σ(xn−i , I 1,2,...,n−i−1,n−i )f 1,2,...,n−i−1,n−i exˆ0 xˆ1 ···ˆxn−i−1 xn−i ···xn . (7) In order to show that f ∈ Im di , we need to find g ∈ C i such that di (g) = f . Note that. 28.
(32) for g = g i,i+1,...,n ex0 x1 ···xi−1 xˆi xˆi+1 ···ˆxn + · · · + g 0,1,2,...,n−i exˆ0 xˆ1 ···ˆxn−i xn−i+1 ···xn , di (g i,i+1,...,n ex0 x1 ···xi−1 xˆi xˆi+1 ···ˆxn + · · · + g 0,1,2,...,n−i exˆ0 xˆ1 ···ˆxn−i xn−i+1 ···xn ) = σ(xi , I i,i+1,i+2,...,n )g i,i+1,...,n + · · · + σ(x0 , I 0,i+1,i+2,...,n )g 0,i+1,i+2,...,n ex0 x1 ···xi xˆi+1 xˆi+2 ···ˆxn + σ(xi+1 , I i,i+1,i+2,...,n )g i,i+1,i+2,...,n + σ(xi−1 , I i−1,i,i+2,...,n )g i−1,i,i+2,...,n + · · · + σ(x0 , I 0,i,i+2,...,n )g 0,i,i+2,...,n ex0 ...xi−1 xˆi xi+1 xˆi+2 ···ˆxn + ··· + σ(xn , I 1,2,...,n−i,n )g 1,2,...,n−i,n + · · · + σ(xn−i+1 , I 1,2,...,n−i,n−i+1 )g 1,2,...,n−i,n−i+1 + σ(x0 , I 0,1,2,...,n−i )g 0,1,2,...,n−i ex0 xˆ1 ···ˆxn−i−1 xˆn−i xn−i+1 ···xn + σ(xi+1 , I 0,i+1,i+2,i+3,...,n )g 0,i+1,i+2,...,n + · · · + σ(x1 , I 0,1,i+2,i+3,...,n )g 0,1,i+2,i+3,...,n exˆ0 x1 ···xi+1 xˆi+2 ···ˆxn + ··· + σ(xn , I 0,1,2,...,n−i−1,n )g 0,1,2,...,n−i−1,n + · · · + σ(xn−i , I 0,1,2,...,n−i−1,n−i )g 0,1,2,...,n−i−1,n−i exˆ0 xˆ1 ···ˆxn−i−1 xn−i ···xn .. (8). We compare (7) with (8). For example, for the component exˆ0 x1 ···xi+1 xˆi+2 ···ˆxn , we want to get σ(xi+1 , I 0,i+1,i+2,i+3,...,n )g 0,i+1,i+2,...,n + · · · + σ(x1 , I 0,1,i+2,i+3,...,n )g 0,1,i+2,i+3,...,n = −σ(x0 , I 0,i+2,i+3,...,n )σ(xi+1 , I i+1,i+2,i+3,...,n )f i+1,i+2,i+3,...,n − ··· − σ(x0 , I 0,i+2,i+3,...,n )σ(x1 , I 1,i+2,i+3,...,n )f 1,i+2,i+3,...,n , so we choose g 0,i+1,i+2,...,n. = −σ(xi+1 , I 0,i+1,i+2,i+3,...,n )σ(x0 , I 0,i+2,i+3,...,n ) σ(xi+1 , I i+1,i+2,i+3,...,n )f i+1,i+2,i+3,...,n , .. .. g 0,1,i+2,i+3,...,n = −σ(x1 , I 0,1,i+2,i+3,...,n )σ(x0 , I 0,i+2,i+3,...,n ) σ(x1 , I 1,i+2,i+3,...,n )f 1,i+2,i+3,...,n . 29.
(33) We do the same thing for the components exˆ0 x1 ···xi xˆi+1 xi+2 xˆi+3 ···ˆxn , . . . , exˆ0 xˆ1 ···ˆxn−i−1 xn−i ···xn and want to choose g 0,k1 ,k2 ,...,kn−i accordingly for all 0 < k1 < k2 < · · · < kn−i ≤ n. We also choose all the g k1 ,k2 ,...,kn−i+1 with k1 > 0 to be zero. Note that the term g 0,i+1,i+2,...,n not only contributes to the component exˆ0 x1 ···xi+1 xˆi+2 ···ˆxn but also contributes to the component exˆ0 x1 ···xi xˆi+1 xi+2 xˆi+3 ···ˆxn . More precisely, for the component exˆ0 x1 ···xi xˆi+1 xi+2 xˆi+3 ···ˆxn , we want to get σ(xi+2 , I 0,i+1,i+2,i+3,...,n )g 0,i+1,i+2,...,n + σ(xi , I 0,i,i+1,i+3,...,n )g 0,i,i+1,i+3,...,n + · · · + σ(x1 , I 0,1,i+1,i+3,...,n )g 0,1,i+1,i+3,...,n = −σ(x0 , I 0,i+1,i+3,...,n )σ(xi+2 , I i+1,i+2,i+3,...,n )f i+1,i+2,i+3,...,n − σ(x0 , I 0,i+1,i+3,...,n )σ(xi , I i,i+1,i+3,...,n )f i,i+1,i+3,...,n − ··· − σ(x0 , I 0,i+1,i+3,...,n )σ(x1 , I 1,i+2,i+3,...,n )f 1,i+1,i+3,...,n , and so we want to choose g 0,i+1,i+2,...,n = −σ(xi+2 , I 0,i+1,i+2,i+3,...,n )σ(x0 , I 0,i+1,i+3,...,n )σ(xi+2 , I i+1,i+2,i+3,...,n )f i+1,i+2,i+3,...,n . However, this is the same as the earlier choice, since − σ(xi+1 , I 0,i+1,i+2,i+3,...,n )σ(x0 , I 0,i+2,i+3,...,n )σ(xi+1 , I i+1,i+2,i+3,...,n ) = −(−1)i (−1)0 (−1)i+1 = (−1)2i+2 =1 and − σ(xi+2 , I 0,i+1,i+2,i+3,...,n )σ(x0 , I 0,i+1,i+3,...,n )σ(xi+2 , I i+1,i+2,i+3,...,n ) = −(−1)i (−1)0 (−1)i+1 = (−1)2i+2 = 1. 30.
(34) Similarly, g 0,i+1,i+2,...,n also contributes to the components exˆ0 x1 ···xi xˆi+1 xˆi+2 xi+3 xˆi+4 ···ˆxn , . . . , exˆ0 x1 ···xi xˆi+1 ···ˆxn−1 xn . However, we have for each j = i + 1, . . . , n, σ(xj , I 0,i+1,...,n ) = (−1)i , σ(x0 , I 0,i+1,...,j−1,j+1,...,n ) = (−1)0 , and σ(xj , I i+1,...,n ) = (−1)i+1 , and so we have g 0,i+1,i+2,...,n = −σ(xi+1 , I 0,i+1,i+2,i+3,...,n )σ(x0 , I 0,i+2,i+3,...,n )σ(xi+1 , I i+1,i+2,i+3,...,n )f i+1,i+2,...,n = −σ(xi+2 , I 0,i+1,i+2,i+3,...,n )σ(x0 , I 0,i+1,i+3,...,n )σ(xi+2 , I i+1,i+2,i+3,...,n )f i+1,i+2,...,n = ··· = −σ(xn , I 0,i+1,i+2,i+3,...,n )σ(x0 , I 0,i+1,i+3,...,n−1 )σ(xn , I i+1,i+2,i+3,...,n )f i+1,i+2,...,n . Hence, we indeed choose the same g 0,i+1,i+2,...,n even when we consider different components. In general, for all 1 ≤ k1 < k2 < · · · < kn−i ≤ n, since for each j = 1, 2, . . . , n − i, σ(xkj , I 0,k1 ,k2 ,...,kn−i )σ(xkj , 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 di (g) = f , where g = g 0,i+1,i+2,...,n exˆ0 x1 ···xi xˆi+1 xˆi+2 ···ˆxn + · · · + g 0,1,2,...,n−i exˆ0 xˆ1 ···ˆxn−i xn−i+1 ···xn , i.e., di g 0,i+1,i+2,...,n exˆ0 x1 ···xi xˆi+1 xˆi+2 ···ˆxn + · · · + g 0,1,2,...,n−i exˆ0 xˆ1 ···ˆxn−i xn−i+1 ···xn. . = f i+1,i+2,...,n ex0 x1 ···xi xˆi+1 ···ˆxn + · · · + f 1,2,...,n−i ex0 xˆ1 ···ˆxn−i xn−i+1 ···xn + f 0,i+2,i+3,...,n exˆ0 x1 ···xi+1 xˆi+2 xˆi+3 ···ˆxn + · · · + f 0,1,...,n−i−1 exˆ0 xˆ1 ···ˆxn−i−1 xn−i ···xn . 31. (10).
(35) We first take care of the components in the second line of (10), namely the components ex0 x1 ···xi xˆi+1 ···ˆxn , . . . , ex0 xˆ1 ···ˆxn−i xn−i+1 ···xn . For the component ex0 x1 ···xi xˆi+1 ···ˆxn , since the components in g that do contribute to it are the components ex0 x1 ···xi−1 xˆi xˆi+1 ···ˆxn , ex0 x1 ···xi−2 xˆi−1 xi xˆi+1 ···ˆxn , . . . , ex0 xˆ1 x2 ···xi xˆi+1 ···ˆxn , exˆ0 x1 x2 ···xi xˆi+1 ···ˆxn , and since we take g i,i+1,i+2,...,n = g i−1,i+1,i+2,...,n = · · · = g 1,i+1,i+2,...,n = 0, and g 0,i+1,i+2,...,n = −σ(xi+1 , I 0,i+1,i+2,i+3,...,n )σ(x0 , I 0,i+2,i+3,...,n )σ(xi+1 , I i+1,i+2,i+3,...,n )f i+1,i+2,...,n , we get that the coefficient of ex0 x1 ···xi xˆi+1 ···ˆxn in di (g) is σ(x0 , I 0,i+1,i+2,...,n )g 0,i+1,i+2,...,n = σ(x0 , I 0,i+1,i+2,...,n ) − σ(xi+1 , I 0,i+1,i+2,...,n )σ(x0 , I 0,i+2,...,n )σ(xi+1 , I i+1,i+2,i+3,...,n )f i+1,i+2,...,n = (−1)0 − (−1)i (−1)0 (−1)i+1 f i+1,i+2,...,n = f i+1,i+2,...,n . Similarly, with the same argument for the remaining components ex0 x1 ···xi+1 xˆi xi+1 xˆi+2 ···ˆxn , . . . , ex0 xˆ1 ···ˆxn−i xn−i+1 ···xn in the second line of (10), we see that their coefficients are indeed f i,i+2,i+3,...,n , . . . , f 1,2,...,n−i , respectively. Next we check the first component exˆ0 x1 ···xi+1 xˆi+2 xˆi+3 ···ˆxn in the third line of (10). Note that the components in g that do contribute to the component exˆ0 x1 ···xi+1 xˆi+2 ···ˆxn are the components exˆ0 x1 ···xi xˆi+1 xˆi+2 ···ˆxn , . . . , exˆ0 xˆ1 x2 ···xi+1 xˆi+2 ···ˆxn . Moreover, since we take g 0,i+1,i+2,...,n = −σ(xi+1 , I 0,i+1,i+2,...,n )σ(x0 , I 0,i+2,i+3,...,n )σ(xi+1 , I i+1,i+2,...,n )f i+1,i+2,...,n , .. . g 0,1,i+2,...,n = −σ(x , I 0,1,i+2,i+3,...,n )σ(x , I 0,i+2,i+3,...,n )σ(x , I 1,i+2,...,n )f 1,i+2,...,n , 1. 0. 32. 1. .
(36) we see that the coefficient of exˆ0 x1 ···xi+1 xˆi+2 ···ˆxn in di (g) is σ(xi+1 , I 0,i+1,i+2,...,n )g 0,i+1,i+2,...,n + · · · + σ(x1 , I 0,1,i+2,i+3,...,n )g 0,1,i+2,...,n = σ(xi+1 , I 0,i+1,i+2,...,n ) − σ(xi+1 , I 0,i+1,i+2,...,n )σ(x0 , I 0,i+2,i+3,...,n )σ(xi+1 , I i+1,i+2,...,n )f i+1,i+2,...,n + ··· + σ(x1 , I 0,1,i+2,i+3,...,n ) − σ(x1 , I 0,1,i+2,i+3,...,n )σ(x0 , I 0,i+2,i+3,...,n )σ(x1 , I 1,i+2,...,n )f 1,i+2,...,n. . = −σ(x0 , I 0,i+2,i+3,...,n )σ(xi+1 , I i+1,i+2,...,n )f i+1,i+2,...,n − · · · − σ(x0 , I 0,i+2,i+3,...,n )σ(x1 , I 1,i+2,...,n )f 1,i+2,i+3,...,n = f 0,i+2,i+3,...,n , where the last equality follows from (6). Next we check the second component exˆ0 x1 ···xi xˆi+1 xi+2 xˆi+3 ···ˆxn in the third line of (10). Since the components in g that do contribute to the component exˆ0 x1 ···xi xˆi+1 xi+2 xˆi+3 ···ˆxn are the components exˆ0 x1 ···xi xˆi+1 xˆi+2 xˆi+3 ···ˆxn , exˆ0 x1 ···xi−1 xˆi xˆi+1 xi+2 xˆi+3 ···ˆxn , . . . , exˆ0 xˆ1 x2 ···xi xˆi+1 xi+2 xˆi+3 ···ˆxn . Use (9) and we have g 0,i+1,i+2,...,n = −σ(xi+2 , I 0,i+1,i+2,...,n )σ(x0 , I 0,i+1,i+3,...,n )σ(xi+2 , I i+1,i+2,...,n )f i+1,i+2,...,n g 0,i,i+1,i+3,...,n = −σ(x , I 0,i,i+1,i+3,...,n )σ(x , I 0,i+1,i+3,...,n )σ(x , I i,i+1,i+3,...,n )f i,i+1,i+3,...,n i 0 i . .. g 0,1,i+1,i+3,...,n = −σ(x , I 0,1,i+1,i+3,...,n )σ(x , I 0,i+1,i+3,...,n )σ(x , I 1,i+1,i+3,...,n )f 1,i+1,i+3,...,n . 1 0 1 Note that in the above relations we choose the expression g 0,i+1,i+2,...,n = −σ(xi+2 , I 0,i+1,i+2,...,n )σ(x0 , I 0,i+1,i+3,...,n )σ(xi+2 , I i+1,i+2,...,n )f i+1,i+2,...,n , instead of g 0,i+1,i+2,...,n = −σ(xi+1 , I 0,i+1,i+2,...,n )σ(x0 , I 0,i+2,...,n )σ(xi+1 , I i+1,i+2,...,n )f i+1,i+2,...,n .. 33. .
(37) Then we have that the coefficient of exˆ0 x1 ···xi xˆi+1 xi+2 xˆi+3 ···ˆxn in di (g) is σ(xi+2 , I 0,i+1,i+2,...,n )g 0,i+1,i+2,...,n + σ(xi , I 0,i,i+1,i+3,...,n ))g 0,i,i+1,i+3,...,n + · · · + σ(x1 , I 0,1,i+1,i+3,...,n )g 0,1,i+1,i+3,...,n = σ(xi+2 , I 0,i+1,i+2,...,n ) − σ(xi+2 , I 0,i+1,i+2,...,n )σ(x0 , I 0,i+1,i+3,...,n )σ(xi+2 , I i+1,i+2,...,n )f i+1,i+2,...,n + σ(xi , I 0,i,i+1,i+3,...,n ) − σ(xi , I 0,i,i+1,i+3,...,n )σ(x0 , I 0,i+1,i+3,...,n ) · σ(xi , I i,i+1,i+3,...,n )f i,i+1,i+3,...,n + ··· + σ(x1 , I 0,1,i+1,i+3,...,n ) − σ(x1 , I 0,1,i+1,i+3,...,n )σ(x0 , I 0,i+1,i+3,...,n ) · σ(x1 , I 1,i+1,i+3,...,n )f 1,i+1,i+3,...,n = −σ(x0 , I 0,i+1,i+3,...,n )σ(xi+2 , I i+1,i+2,...,n )f i+1,i+2,...,n − σ(x0 , I 0,i+1,i+3,...,n )σ(xi , I i,i+1,i+3,...,n )f i,i+1,i+3,...,n − ··· − σ(x0 , I 0,i+1,i+3,...,n )σ(x1 , I 1,i+1,i+3,...,n )f 1,i+1,i+3,...,n = f 0,i+1,i+3,...,n , where the last equality follows from (6). Similarly, we do the same argument for the remaining components exˆ0 x1 ···xi xˆi+1 xˆi+2 xi+3 xˆi+4 ···ˆxn , . . . , exˆ0 ···ˆxn−i−1 xn−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 , . . . , xn and is a multiple of w, i.e., the component with respect to Q[x1 , x2 , . . . , xn ]w. For f (w) ∈ C i+1 , since C i+1 = Rx0 x1 ···xi xˆi+1 xˆi+2 ···ˆxn ⊕ · · · ⊕ Rx0 xˆ1 ···ˆxn−i xn−i+1 ···xn ⊕ Rxˆ0 x1 ···xi+1 xˆi+2 xˆi+3 ···ˆxn ⊕ · · · ⊕ Rxˆ0 x1 xˆ2 ···ˆxn−i xn−i+1 ···xn ⊕ Rxˆ0 xˆ1 x2 ···xi+2 xˆi+3 ···ˆxn ⊕ · · · ⊕ Rxˆ0 xˆ1 ···ˆxn−i−1 xn−i ···xn ,. 34. .
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