# Get A Course in Commutative Algebra PDF

By Gregor Kemper

ISBN-10: 3642035442

ISBN-13: 9783642035449

ISBN-10: 3642035450

ISBN-13: 9783642035456

This textbook deals a radical, glossy advent into commutative algebra. it really is intented in most cases to function a advisor for a process one or semesters, or for self-study. The rigorously chosen subject material concentrates at the recommendations and effects on the middle of the sector. The e-book continues a continuing view at the average geometric context, permitting the reader to achieve a deeper realizing of the cloth. even though it emphasizes idea, 3 chapters are dedicated to computational features. Many illustrative examples and workouts increase the text.

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**Additional info for A Course in Commutative Algebra**

**Example text**

By assumption, Y itself is not irreducible, so Y = Y1 ∪ Y2 with Y1 , Y2 Y closed subsets. By the minimality of Y , the Yi are ﬁnite unions of closed irreducible subsets, so the same is true for Y . This is a contradiction. Hence in particular, X = Y1 ∪ · · · ∪ Ym with Yi ⊆ X closed and irreducible (where m = 0 if X = ∅). We may assume the Yi to be pairwise distinct. 1). 1) is satisﬁed, and let Z ⊆ X be an irreducible subset. Then Z = (Z ∩ Z1 ) ∪ · · · ∪ (Z ∩ Zn ), so Z = Z ∩ Zi for some i, which implies Z ⊆ Zi .

Specmax (R). Setting Ii := ij=1 mj yields a descending chain of ideals, so by hypothesis n there exists n such that In+1 = In . This implies j=1 mj ⊆ mn+1 , so there exists j ≤ n with mj = mn+1 , a contradiction. We conclude that there exist ﬁnitely many maximal ideals m1 , . . , mk . Setting I := m1 · · · mk , we obtain a descending chain of ideals I i , i ∈ N0 , so there exists n ∈ N0 with I i = I n =: J for i ≥ n. 1) By way of contradiction, assume J = {0}. Then the set M := {J ⊆ R | J is an ideal and J J = {0}} is nonempty.

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 . Conversely, let P ∈ VSpec(R) (S)R ∩ (T )R and assume S ⊆ P . So there exists f ∈ S \ P . Let g ∈ T .

### A Course in Commutative Algebra by Gregor Kemper

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