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Proof. Let P be the kernel of f . Since f maps into a field, P is a prime ideal of R. 2), there is a prime ideal Q of S such that Q ∩ R = P . By the factor theorem, f induces an injective ring homomorphism f : R/P → C, which extends in the natural way to the fraction field K of R/P . Let L be the fraction field of S/Q. 11), S/Q is integral over R/P , hence L is an algebraic extension of K. Since C is algebraically closed, f extends to a monomorphism g : L → C. If p : S → S/Q is the canonical epimorphism and g = g ◦ p, then g is the desired extension of f , because g extends f and f ◦ p|R = f .

With R-module homomorphisms θn : Mn → Mn−1 , n ≥ 1. (We are restricting to the countable case to simplify the notation, but the ideas carry over to an arbitrary family of modules, indexed by a directed set. If i ≤ j, we have a homomorphism fij from Mj to Mi . )The collection of modules and maps is called an inverse system. Mi is said to be coherent if it respects the A sequence (xi ) in the direct product maps θn in the sense that for every i we have θi+1 (xi+1 ) = xi . The collection M of all coherent sequences is called the inverse limit of the inverse system.

Suppose that a ∈ I, x ∈ M , and (1 + a)x = 0. Then x = −ax = −a(−ax) = a2 x = a2 (−ax) = −a3 x = a4 x = · · · , hence x ∈ I n M for all n ≥ 0. Conversely, we must show that for some a ∈ I, 1 + a annihilates everything in the kernel N . 8), for some n we have, for all k ≥ 0, I k ((I n M ) ∩ N ) = (I n+k M ) ∩ N. Set k = 1 to get I((I n M ) ∩ N ) = (I n+1 M ) ∩ N. But N ⊆ I n+1 M ⊆ I n M , so the above equation says that IN = N . 1), there exists a ∈ I such that (1 + a)N = 0. 3 Corollary n If I is a proper ideal of the Noetherian integral domain R, then ∩∞ n=0 I = 0.

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