A Course in Commutative Algebra (Graduate Texts in by Gregor Kemper

By Gregor Kemper

This textbook deals an intensive, glossy advent into commutative algebra. it truly is intented quite often to function a consultant for a process one or semesters, or for self-study. The conscientiously chosen subject material concentrates at the techniques and effects on the middle of the sphere. The ebook keeps a relentless view at the common geometric context, permitting the reader to achieve a deeper figuring out of the cloth. even though it emphasizes idea, 3 chapters are dedicated to computational elements. Many illustrative examples and routines improve the textual content.

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The following lemma gives a connection between ideal powers and radical ideals. 6 (Ideal powers and radical ideals). Let R be a ring and I, J ⊆ R ideals. If I is finitely generated, then I⊆ √ J ⇐⇒ there exists k ∈ N0 such that I k ⊆ J. √ Proof. We have I = (a1 , . . , an ). Suppose that I ⊆ J. Then there exists m > 0 with am i ∈ J for i = 1, . . , n. Set k := n · (m − 1) + 1. We need to show that the product of k arbitrary elements from I lies in J. So let x1 , . . , xk ∈ I and write n xi = ri,j aj with ri,j ∈ R.

Conversely, assume that I is a prime ideal. ,xn ] (∅) = K[x1 , . . , xn ]. To show that X is irreducible, let X = X1 ∪X2 with Xi closed in X, so Xi = X ∩VK n (Ii ) with Ii ⊆ K[x1 , . . , xn ] ideals. 1(a) for the equality. ,xn ] (X) = I. Since I is a prime ideal, there exists i with Ii ⊆ I, so X ⊆ VK n (I) ⊆ VK n (Ii ). This implies Xi = X. Therefore X is irreducible. (b) The proof of this part is obtained from the proof of part (a) by changing K[x1 , . . ” The following theorem allows us to view irreducible spaces as the “atoms” of a Noetherian space.

2 Spectra 37 (b) Let M be a nonempty set of subsets of R. Then VSpec(R) (S) = VSpec(R) S∈M S . S∈M (c) Let X ⊆ Spec(R) be a subset. Then IR (X) is a radical ideal of R. (d) Let I ⊆ R be an ideal. Then √ IR VSpec(R) (I) = I. (e) We have a pair of inverse bijections between the set of radical ideals of R and the set of closed subsets of Spec(R), given by VSpec(R) and IR . Both bijections are inclusion-reversing. Proof. (a) If P ∈ VSpec(R) (S), then S ⊆ P , so also (S)R ⊆ P and (S)R ∩ (T )R ⊆ P . The same follows if P ∈ VSpec(R) (T ), so in both cases P ∈ (b) (c) (d) (e) VSpec(R) (S)R ∩ (T )R .

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