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Suppose g1 , . . , gr is a basis for F and let mi = β(gi ) ∈ M . Suppose mi = j mij , where the mij ∈ M are homogeneous, and let F be the graded free module with generators gij satisfying deg gij = deg mij . Then, if β : F → M is the morphism of graded modules taking gij to mij , we have a map α : F → F taking gi to j gij such that β = β ◦ α. It is clear that α is an injection. Now, given a finitely generated submodule K ⊂ ker β, we get a finitely generated submodule α(K) ⊂ ker β , and a graded free module G with a map γ : F → G such that α(K) ⊂ ker γ such that a diagram such as in (6) commutes.

Proof. Both statements are easy. For the second, note that if s/u ∈ U −1 S, and s satisfies the monic equation sn + an−1 sn−1 + . . + a0 = 0 over R, then s/u satisfies the monic equation (s/u)n + (an−1 /u)(s/u)n−1 + . . + a0 /un = 0 over U −1 R. 3. 1. Reduced Rings. 1. We will say that a ring R is of dimension 0 if every maximal ideal of R is also minimal. Reduced rings of dimension 0 are easy to describe. 2. The only reduced local rings of dimension 0 are fields. 58 4. INTEGRALITY: THE COHEN-SEIDENBERG THEOREMS Proof.

THE HILBERT-SAMUEL POLYNOMIAL 33 Proof. For every r ∈ N, we have F r M ⊃ qr M . Hence, it suffices to show that M/qr M is Artinian, for every r ∈ N. But this follows from the Lemma above, and the fact that V (q) = V (qr ), for any r ∈ N. Consider the graded associated ring grq (R): this is generated in degree 1 by the s generators of q. Moreover, if n0 ∈ N is such that qF n M = F n+1 M , for all n ≥ n + 0 (this exists, since M is stable), then grF (M ) is generated over grq (R) by the finitely generated R/q-submodule M/F 1 M ⊕ .

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Commutative Algebra by Keerthi Madapusi

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Categories: Algebraic Geometry