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**Example text**

Then there exists an injective resolution I • such that 0→ 0→ M′ ↓ I •′ → M ↓ → I• → → M ′′ ↓ I • ′′ →0 →0 is a commutative diagram, in which the bottom row is a short exact sequence of complexes. Proof. Let the maps in the short exact sequence be i : M ′ → M and p : M → M ′′ , and let the maps on I • ′ be δ • ′ , and the maps on I • ′′ be δ • ′′ . Consider the diagram in which the rows are exact: 0→ 0→ M′ ↓= M′ → i M δ ′−1 I ′0 → δ ′′−1 ◦p −→ I ′′0 δ ′′0 → I ′′1 δ ′′1 → I ′′2 → ··· δ ′0 I ′1 → δ ′1 I ′2 → δ ′2 I ′3 → ··· → By the Comparison Theorem for injectives, there exist the maps as below that make all squares commute: 0→ 0→ M′ ↓= M′ i → δ ′−1 → M ↓f0 I ′0 δ ′′−1 ◦p −→ δ ′0 → I ′′0 ↓f1 I ′1 δ ′′0 → δ ′1 → I ′′1 ↓f2 I ′2 δ ′′1 → δ ′2 → I ′′2 ↓f3 I ′3 → ··· → ··· Now define I n = I ′n ⊕ I ′′n , δ −1 : M → I 0 by δ −1 (m) = (−f 0 (m), δ ′′−1 ◦ p(m)), and δ n : I n → I n+1 by δ n (a, b) = (δ ′n (b) + (−1)n f n+1 (a), δ ′′n (a)).

Proof. Let q be a prime ideal minimal over J contained in P . By the generalized Krull principal ideal theorem, q = P . Suppose that q = J . Let a ∈ q \ J . Since q ⊆ P , we can write a = j1 + a1 xd for some j1 ∈ J and some a1 in R. Then a1 xd ∈ q, and since xd ∈ q, it follows that a1 ∈ q. Then we can write a1 in a similar way as a, and an iteration of this gives that for all n ≥ 1, a = jn + an xnd for some jn ∈ J and some an in R. But then since R is local, a ∈ J , which is a contradiction. This proves that J is a prime ideal, strictly contained in P .

Xn; M ) = 0 for i = n, n − 1, . . , n − l + 1, then there exists y1 , . . , yn ∈ R such that (x1 , . . , xn ) = (y1 , . . , yn ) and y1 , . . , yl is a regular sequence on M . Proof. If l = 0, there is nothing to prove. So we may assume that l > 0. 3 that says that Hn (K• (x1 , . . , xn; M )) = annM (x1 , . . , xn) for all x1 , . . , xn and M . By assumption, annM (x1 , . . , xn ) = 0, so that (x1 , . . , xn ) is not contained in any prime ideal that is associated to M . Thus by the strengthened form of Prime Avoidance, there exists y1 ∈ (x1 , .