By Gregor Kemper

This textbook bargains an intensive, sleek advent into commutative algebra. it really is intented frequently to function a advisor for a process one or semesters, or for self-study. The rigorously chosen subject material concentrates at the options and effects on the heart of the sector. The booklet continues a relentless view at the common geometric context, allowing the reader to achieve a deeper knowing of the cloth. even though it emphasizes thought, 3 chapters are dedicated to computational points. Many illustrative examples and routines enhance the textual content.

**Read 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**

Assume dim(X) = 0. Since X is a subset of a Noetherian space, X is Noetherian, too. 11(a), X is a ﬁnite union of closed, irreducible subsets Zi . Choose xi ∈ Zi . Then {xi } ⊆ Zi is a chain of closed, irreducible subsets, so Zi = {xi }. It follows that X is ﬁnite. Conversely, if X is ﬁnite, then the irreducible subsets are precisely the subsets of size 1, so dim(X) = 0. 11: equidimensional algebras whose dimension is only 1 less than the number of generators. These correspond to equidimensional aﬃne varieties in K n of dimension n − 1.

Then dim(A) = trdeg(A). We will prove the theorem together with the following proposition, which often facilitates the computation of the transcendence degree since the set S can be taken to be ﬁnite. 10 (Calculating the transcendence degree). Let A be an afﬁne algebra, and let S ⊆ A be a generating set. Then trdeg(A) = sup {|T | | T ⊆ S is ﬁnite and algebraically independent} . 10. 6 we have dim(A) ≤ sup {|T | | T ⊆ S is ﬁnite and algebraically independent} , and this supremum is clearly less than or equal to trdeg(A).

An ∈ A are algebraically independent over L. By induction, dim(A ) ≥ n − 1, so there exists a chain P0 P1 ··· Pn−1 in Spec(A ). Set Pi := A ∩ Pi ∈ Spec(A). Then Pi−1 ⊆ Pi for i = 1, . . , n−1. These inclusions are strict since clearly L · Pi = Pi for all i. Moreover, L ∩ Pn−1 = {0}, since otherwise Pn−1 would contain an invertible element from L, leading to Pn−1 = A . It follows that a1 + Pn−1 ∈ A/Pn−1 is not algebraic over K. 1(b), A/Pn−1 is not a ﬁeld, so Pn−1 is not a maximal ideal. Let Pn ⊂ A be a maximal ideal containing Pn−1 .