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 an intensive, sleek advent into commutative algebra. it truly is intented often 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 heart of the sector. The ebook continues a relentless view at the usual geometric context, allowing the reader to realize a deeper realizing of the fabric. even though it emphasizes thought, 3 chapters are dedicated to computational features. Many illustrative examples and workouts increase the text.

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Example text

11 has statements on existence and uniqueness, which justifies the following definition. 12. Let X be a Noetherian topological space. 11 are called the irreducible components of X. 13. Let K be an algebraically closed field and g ∈ K[x1 , . . , xn ] a nonzero polynomial. Let p1 , . . , pr be the distinct prime factors of g. Then r VK n (g) = VK n (pi ). i=1 The (pi ) ⊆ K[x1 , . . ,xn ] (VK n (pi )) = (pi ). 10(a), the VK n (pi ) are the irreducible components of the hypersurface VK n (g). We make the obvious convention of calling a prime ideal P ∈ Spec(R) minimal if for all Q ∈ Spec(R) the inclusion Q ⊆ P implies Q = P .

An } ⊆ A of size n of an algebra A over a field K is called algebraically independent if for all nonzero polynomials f ∈ K[x1 , . . , xn ] we have f (a1 , . . , an ) = 0. 3. Let A be an algebra over a field K. Then the transcendence degree of A is defined as trdeg(A) := sup {|T | | T ⊆ A is finite and algebraically independent} . So trdeg(A) ∈ N0 ∪ {−1, ∞}, where −1 occurs if A = {0} is the zero ring. ) Our next goal is to show that the dimension and the transcendence degree of an affine algebra coincide.

Ik }. Then all mj lie in Mn , so for i ≥ n we have Mi ⊆ N = (m1 , . . , mk )R ⊆ Mn ⊆ Mi , which implies equality. Therefore (a) holds. 9 implies that every Noetherian module over a ring is finitely generated. This raises the question whether the converse is true, too. But this is clearly false in general: If R is a non-Noetherian ring, then R is not Noetherian as a module over itself, but it is finitely generated (with 1 the only generator). The following theorem shows that if the converse does not go wrong in this very simple way, then in fact it holds.

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A Course in Commutative Algebra by Gregor Kemper

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