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.

**Read Online or Download A Course in Commutative Algebra (Graduate Texts in Mathematics, Volume 256) PDF**

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**Additional resources for A Course in Commutative Algebra (Graduate Texts in Mathematics, Volume 256)**

**Sample text**

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 .