Cech complexes vs universal connected sequences ˇ

In Theorem 3.3.5, we will show that {Hⁿ(− ⊗ C), Eⁿ}n≥0 is a universal connected se-quence. Before we prove Theorem 3.3.5, we present some lemmas that we need in the proof.

Lemma 3.3.1. Let M be an R-module and let S be a multiplicative closed set in R.

Then as RS-modules, E(M)S is an essential extension of MS.

Because R is a Noetherian ring, Σ has a maximal element. Let Ann(ty) with t ∈ S be a maximal element in Σ. Then we have that

RSx = RSy = RSty .

Since E(M) is an essential extension of M and since ty 6= 0, R(ty) ∩ M = I(ty) 6= 0, where I = (M :R ty) is an ideal of R. Again by the fact that R is a Noetherian ring, I = (a1, a2, · · · , an) for some elements a1, a2, · · · , an ∈ R. Now we show that there exists ai such that s[ai(ty)] 6= 0 for all s ∈ S. Suppose that for each i = 1, 2, · · · , n, there exists si ∈ S such that s_i[ai(ty)] = 0. Take s = Qn

i=1si, then s[ai(ty)] = 0 for all i = 1, 2, · · · , n. Thus we have that ai ∈ Ann(sty) for all i = 1, 2, · · · , n. On the other hand, since Ann(ty) ⊆ Ann(sty) and since Ann(ty) is a maximal element in Σ, Ann(ty) = Ann(sty). So we get ai ∈ Ann(ty) for all i = 1, 2, · · · , n. Therefore, I ⊆ Ann(ty), i.e., I(ty) = 0. It contradicts to I(ty) 6= 0, so there exists ai such that s[ai(ty)] 6= 0 for all s ∈ S. Hence ^ai(ty)₁ ∈ (R(ty) ∩ M)S = RS(ty) ∩ MS = RSx ∩ MS and

ai(ty)

1 6= 0 in E(M)_S. Therefore, RSx ∩ MS 6= 0 for all nonzero x ∈ E(M)_S. 2 Lemma 3.3.2. Let p∈ SpecR and let y ∈ R.

(1) If y ∈ p, then E(R/p)y = 0.

(2) If y /∈ p, then yE(R/p) = E(R/p).

Proof. For (1), because y ∈ p, (R/p)y = 0. By Lemma 3.3.1, E(R/p)y is an essential extension of (R/p)_y. Therefore E(R/p)_y = 0.

For (2), consider the R-module homomorphism f : E(R/p) → E(R/p) defined by f (a) = ya for all a ∈ E(R/p). Now we claim that f is one-to-one, i.e., Ker f = 0.

Suppose Ker f 6= 0. Because Ker f is a nonzero submodule of E(R/p) and E(R/p) is an essential extension of R/p, Ker f ∩ R/p 6= 0. However for a ∈ Ker f ∩ R/p, a = r + p for some r ∈ R. Thus 0 = f (a) = f (r + p) = yr + p, and so we have yr ∈ p. Since p is a prime ideal and since y /∈ p, r ∈ p, i.e., a = r + p = 0 in R/p.

Hence Ker f ∩ R/p = 0 and we get a contradiction. Thus Ker f = 0, i.e., f is one-to-one. By the First Isomorphism Theorem, E(R/p) ∼= Im f = yE(R/p). Then yE(R/p)

is injective since E(R/p) is injective. Moreover, because yE(R/p) is a submodule of E(R/p), yE(R/p) is a direct summand of E(R/p), i.e., there exists a submodule M1 of E(R/p) such that E(R/p) = yE(R/p) ⊕ M₁. However, E(R/p) is indecomposable, by Proposition 3.2.9, so M1 = 0. Therefore, yE(R/p) = E(R/p). 2

Remark 3.3.3. In the proof of Lemma 3.3.2(2), we have that the R-module homomor-phism f : E(R/p) → E(R/p), defined by f (a) = ya for all a ∈ E(R/p), is one-to-one, and Im f = yE(R/p). Therefore, for every a ∈ E(R/p), there exists a unique element b ∈ E(R/p) such that a = yb. Similarly, because y /∈ p and p is prime, y^s ∈ p for all/ s ∈ N. Thus for every a ∈ E(R/p) and for every s ∈ N, there exists a unique element b ∈ E(R/p) such that a = y^sb. In particular, if a ∈ E(R/p) such that y^sa = 0, then a = 0.

Notation 3.3.4. In the proof of Theorem3.3.5, we need to use some special notations.

(1) For t > 1 and 1 ≤ i1 < i2 < . . . < it≤ n, we let e_i1i2...it to represent the component Rx_i1x_i2···x_it in C^t =L

1≤i1<i2<...<it≤nRx_i1x_i2···x_it. Hence, we can write C^t=L

1≤i1<i2<...<it≤nRx_i1x_i2···x_it =P

1≤i1<i2<...<it≤nRx_i1x_i2···x_itei1i2...it. Similarly, we also write

L

1≤i1<i2<...<it≤nE(R/p)x_i1x_i2···x_it =P

1≤i1<i2<...<it≤nE(R/p)x_i1x_i2···x_itei1i2...it, where p ∈ Spec(R). We also use the convention that e_j1j2...jt = ei1i2...it as long as {j1, j2, . . . , jt} = {i1, i2, . . . , it}.

(2) For two disjoint subsets X and Y of {1, 2, . . . , n}, we let δ(X, Y ) = (−1)^|Z|, where Z = {(a, b) ∈ X × Y | a < b}. Note that if X = X1 ∪ X2 is a disjoint union, then Z = {(a, b) ∈ X × Y | a < b} = {(a, b) ∈ X1 × Y | a < b} ∪ {(a, b) ∈ X2×Y | a < b} is a disjoint union and so δ(X, Y ) = δ(X1, Y )·δ(X2, Y ). Moreover, if i, j ∈ {1, 2, . . . , n} are distinct, then δ({i}, {j}) · δ({j}, {i}) = −1. With this

new notation, the componentR_xi1···xit → R_{xj1xj2···xjt+1}, that gives the differentiation because the tth term C^t in C is a flat R-module, the diagram

0 ^//A ⊗ C^{t //} is commutative with both rows exact. In other words, the diagram

0 ^//A ⊗C ^// is commutative. By Lemma 3.1.6, we have that the diagram

H^t(C ⊗ C) ^{Et //}

Next, we show that {H^t(− ⊗ C), E^t}_t≥0 is universal. Let E : 0 → A → B → C → 0 be a short exact sequence of R-modules. For all t = 0, 1, · · · , n, because the tth term C^t in C is a flat R-module, the sequence

0 ^//A ⊗ C^{t //}B ⊗ C^{t //}C ⊗ C^{t //}0 is exact. Hence the sequence of complexes

0 ^//A ⊗C ^//B ⊗C ^//C ⊗C ^//0

is exact. By Lemma 3.1.5, there is a long exact sequence of cohomology

· · · → H^t(A ⊗ C) → H^t(B ⊗ C) → H^t(C ⊗ C)→ H^{Et t+1}(A ⊗ C) → · · · .

By Corollary 2.3.4, it remains to show that if I is an injective R-module, then H^t(I ⊗ C) = 0 for all t > 0. Let I be an injective R-module. From Proposition 3.2.9 and Proposition 3.2.11, we know that I =L

j∈JE(R/pj), where {pj | j ∈ J} is a family of prime ideals of R. Note that L

j∈JE(R/pj)
⊗ A ∼= L

j∈J E(R/pj) ⊗ A
for all R-module A. Hence, we only need to take care of the case where I = E(R/p) for some p ∈ Spec(R). In other words, it suffices to show that if p ∈ Spec(R), then H^t(E(R/p)⊗C) = 0 for all t > 0. We separate the discussion into two situations.

Suppose that p = m. Because the ideal x = (x1, x2, . . . , xn) is m-primary, we have xi1xi2 · · · x_it ∈ m for t > 1 and 1 ≤ i₁ < i2 < . . . < it ≤ n. By Lemma 3.3.2 (1), E(R/p)x_i1x_i2···x_it = 0 for t > 1 and 1 ≤ i1 < i2 < . . . < it ≤ n. Hence for all t > 1, we have that

E(R/p) ⊗ C^t = E(R/p) ⊗ L

1≤i1<i2<...<it≤nRx_i1x_i2···x_it

=L

1≤i1<i2<...<it≤nE(R/p)_{xi1xi2···xit}

= 0.

Therefore the complex E(R/p) ⊗ C is

0 → E(R/p) → 0 → 0 → · · · and so H^t(E(R/p) ⊗ C) = 0 for all t > 0.

Suppose that p 6= m. Because m is the unique maximal ideal of R, p ( m. Since x = (x1, x2, . . . , xn) is m-primary, there exists xj such that xj ∈ p. Now we fix the/ index j and consider the natural R-module homomorphism ι : E(R/p) → E(R/p)_xj, i.e., ι(a) = ^a₁ for all a ∈ E(R/p). Note that

a ∈ Ker ι ⇒ a

1 = 0 in E(R/p)xj ⇒ x_j^ka = 0 in E(R/p) ⇒ a = 0 in E(R/p), where the last implication follows from Remark 3.3.3. Hence the natural R-module homomorphism ι : E(R/p) → E(R/p)xj is one-to-one. Moreover, since E(R/p) is an injective R-module, there exists an R-module homomorphism g : E(R/p)xj → E(R/p) such that gι = 1E(R/p). More precisely, by Remark 3.3.3, for all s ∈ N and for all

the diagram (1) is just the diagram We let σ¹be the composition of the canonical R-module homomorphismLn

i=1E(R/p)xi → i.e., the case of t = 1. Note that with the notation we mention in Notation 3.3.4, we have

n

where the third equality follows form the fact that j /∈ {i, w} for all w ∈ L \ {j}. Hence we mention in Notation 3.3.4(1), we have that

L

1≤i1<i2<...<it≤nE(R/p)x_i1x_i2···x_it =P

1≤i1<i2<...<it≤nE(R/p)x_i1x_i2···x_itei1i2...it. Hence, in order to show that d^t−1σ^t+ σ^t+1d^t = 1, we only need to show that for all 1 ≤ i1 < i2 < . . . < it≤ n, (d^t−1σ^t+ σ^t+1d^t)(αei1i2...it) = αei1i2...it for all α ∈ E(R/p)x_i1x_i2···x_it. We separate our discussion into two cases.

Case 1: j /∈ {i1, i2, . . . , it}. Let_(x ^a

i1x_i2···x_it)^k ∈ E(R/p)x_i1x_i2···x_it. Since j /∈ {i1, i2, . . . , it}, we have

d^t−1σ^t a

(xi1xi2· · · xit)^kei1i2...it
= d^t−1(0) = 0.

Let J = {1, 2, . . . , n} \ {i₁, i₂, . . . , i_t}. Then j ∈ J and we have

Note that for all w ∈ V , we have

δ(U, {w}) · δ(Um∪ {w}, {j})

= δ(U_m, {w})δ({j}, {w})
· δ(Um, {j})δ({w}, {j})

= δ(Um, {w})δ(Um, {j})δ({j}, {w})δ({w}, {j})

= (−1) · δ(U_m, {w})δ(U_m, {j}).

Hence (d^t−1σ^t+ σ^t+1d^t)(_(x ^a

i1x_i2···x_it)^kei1i2...it) = _(x ^a

i1x_i2···x_it)^kei1i2...it.

Thus d^t−1σ^t+ σ^t+1d^t= 1 for all t > 0. Hence by Definition 3.1.1 (1), the cochain map L E(R/p)xi //

1



L E(R/p)x_i1x_i2 //

1



· · · ^//L E(R/p)x_i1···x_in−1 //

1



E(R/p)x1···xn //

1



0

L E(R/p)xi //L E(R/p)x_i1x_i2 //· · · ^//L E(R/p)x_i1···x_in−1 //E(R/p)x1···xn //0 is null homotopic. Hence by Remark 3.1.2, the induced map 1^∗ : H^t(E(R/p) ⊗ C) → H^t(E(R/p) ⊗ C) is the zero map for all t > 0. Therefore H^t(E(R/p) ⊗ C) = 0 for all t > 0, and this completes the proof. 2

Theorem 3.3.6. Let x1, x2, · · · , xn ∈ R such that the ideal x = (x₁, x2, . . . , xn) is m-primary and let C be the ˇCech complex with respect to the sequence x1, x2, . . . , xn. Then H_m^t (A) ∼= H^t(A ⊗ C), for all R-modules A and t ≥ 0.

Proof. By Theorem 3.1.10, {H_m^t (−), E^t}t≥0is a universal connected sequence with initial Hm⁰ (−) = Γm(−). By Theorem 3.3.5, {H^t(− ⊗ C), E^t}_t≥0 is also a universal connected sequence. Moreover, from Lemma 3.2.12, we have that H⁰(− ⊗ C) = Γm(−). Therefore, by Lemma 2.3.5, H_m^t (A) ∼= H^t(A ⊗ C) for all R-modules A and t ≥ 0. 2

References

[1] M. F. Atiyah and I. G. Macdonald, Introduction to Commutative Algebra, Addison Wesley, 1969.

[2] W. Bruns and J. Herzog, Cohen-Macaulay Rings, Cambridge University Press, 1998.

[3] D. Katz, Lecture Notes on Homological Algebra, University of Kansas.

[4] S. MacLane, Homology, Springer, Berlin, 1963.