Download e-book for kindle: A Course in Commutative Algebra (Graduate Texts in by Gregor Kemper

By Gregor Kemper

ISBN-10: 3642035442

ISBN-13: 9783642035449

This textbook deals an intensive, sleek creation into commutative algebra. it's intented commonly to function a consultant for a process one or semesters, or for self-study. The conscientiously chosen subject material concentrates at the options and effects on the heart of the sphere. The publication keeps a relentless view at the traditional geometric context, allowing the reader to achieve a deeper realizing of the fabric. even though it emphasizes idea, 3 chapters are dedicated to computational features. Many illustrative examples and workouts enhance the textual content.

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2). (2) A ring homomorphism ϕ: R → S induces a morphism ϕ∗ : Spec(S) → Spec(R), Q → ϕ−1 (Q) of spectra. In the special case that R ⊆ S and ϕ is the inclusion, we have ϕ∗ (Q) = R ∩ Q. Notice that the correspondence between ring homomorphisms and morphisms of spectra is not bijective. If ψ: S → T is a further ring homomorphism, then (ψ ◦ ϕ)∗ = ϕ∗ ◦ ψ ∗ . In general, ϕ∗ does not restrict to a map Specmax (S) → Specmax (R); but if ϕ is a homomorphism of affine K-algebras, it does. 1 corresponding to ϕ.

1) R and C with the usual Euclidean topology are neither Noetherian nor irreducible. (2) Every finite space is Noetherian. (3) Every singleton is irreducible. (4) If K is an infinite field, then X = K 1 with the Zariski topology is irreducible, since the closed subsets are X and its finite subsets. More generally, we will see that K n is irreducible. 10). The topological spaces that we normally deal with in analysis are almost never Noetherian or irreducible. 7). However, the following two theorems show that the situation is much better when we consider spaces with the Zariski topology.

An,mn )R for i > n. 3) By the definition of Ji , there exist polynomials fi,j ∈ I of degree at most i whose ith coefficient is ai,j . Set I := fi,j | i = 0, . . , n, j = 1, . . , mi R[x] ⊆ I. We claim that I = I . To prove the claim, consider a polynomial f = d i i=0 bi x ∈ I with deg(f ) = d. We use induction on d. We first consider 30 2 Noetherian and Artinian Rings the case d ≤ n. 2) and write bd = with rj ∈ R. Then md j=1 rj ad,j md f := f − rj fd,j j=1 lies in I and has degree less than d, so by induction f ∈ I .

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A Course in Commutative Algebra (Graduate Texts in Mathematics, Volume 256) by Gregor Kemper

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