Commutative Algebra, Edition: version 18 Jun 2007 by Keerthi Madapusi

By Keerthi Madapusi

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Additional resources for Commutative Algebra, Edition: version 18 Jun 2007

Example text

Let R be a graded ring and let M be a graded R-module. Then the following are equivalent: (1) M is flat. (2) TorR 1 (N, M ) = 0, for every graded R-module N . (3) TorR 1 (R/I, M ) = 0, for every homogeneous ideal I ⊂ R. lat-star-local-ring-free 49 (4) For every homogeneous ideal I ⊂ M , the map I ⊗ M → M is a monomorphism. (5) Given any relation i ni mi = 0 ∈ M , with ni ∈ R and mi ∈ M homogeneous, we can find homogeneous elements aij ∈ R and mj ∈ M such that aij mj = mi , for all i; j aij ni = 0, for all j.

A ring R is normal if it is reduced and is integrally closed in its total quotient ring K(R). The following Proposition gives us a ready bank of normal domains. 11. Every UFD is normal. Proof. Let R be a UFD, and let r/s ∈ K(R) be integral over R satisfying a monic equation n n−1 (r/s) + an−1 (r/s) + . . + a0 = 0. We can assume that r/s is reduced so that r and s are relatively prime. Now, multiply the equation above by sn to get rn + an−1 srn−1 + . . + a0 sn = 0, which implies that r ∈ (s), contradicting the fact that r and s are relatively prime.

Tensor this with M to obtain the following diagram with exact rows: I n /I n+1 ⊗R M > αn I/I n+1 ⊗R M > (I/I n ) ⊗R M σn ∨ 0 > > 0 > 0 σn−1 ∨ I n M/I n+1 M > Mn ∨ > Mn−1 By hypothesis, αn is an isomorphism. Observe, moreover that, for all n ≥ 0, we have I/I n+1 ⊗R M ∼ = IRn ⊗Rn Mn . Therefore, by induction, σn−1 is a monomorphism. Now, it’s a simple application of the Snake Lemma to see that the map in the middle is also a monomorphism. Now, assume that M is I-adically ideal separated. (7) ⇒ (1): We will show that, for every ideal a ⊂ R, the map ϕ : a ⊗R M → M is an injection.

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