By Gregor Kemper

ISBN-10: 3642035442

ISBN-13: 9783642035449

This textbook deals a radical, smooth creation into commutative algebra. it's intented ordinarily to function a advisor for a process one or semesters, or for self-study. The conscientiously chosen subject material concentrates at the thoughts and effects on the middle of the sphere. The booklet continues a relentless view at the common geometric context, allowing the reader to achieve a deeper realizing of the fabric. even though it emphasizes thought, 3 chapters are dedicated to computational features. 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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**Extra resources for A Course in Commutative Algebra (Graduate Texts in Mathematics, Volume 256)**

**Example text**

Xn ] a nonconstant polynomial. Proof. 22. Let M ⊆ Spec(K[x1 , . . , xn ]) be the set of all prime ideals that are minimal over I. 14(d), and the minimal prime ideals of A are the P/I, P ∈ M. 2) dim (K[x1 , . . , xn ]/P ) = dim (A/(P/I)) = n − 1. 7 that P = {0}.

In fact, the Zariski topology is the coarsest topology such that all polynomials are continuous (assuming that {0} ⊂ K 1 is closed). , the function C → C, x → x (complex conjugation). (h) The Zariski-open subsets of K n are unions of solution sets of polynomial inequalities. 1 Aﬃne Varieties 35 open sets U1 and U2 with Pi ∈ Ui . If K is an inﬁnite ﬁeld, then K n with the Zariski topology is never Hausdorﬀ. In fact, it is not hard to see that two nonempty open subsets U1 , U2 ⊆ K n always intersect.

By the deﬁnition of n, all a ∈ S are algebraic over L := Quot (K[a1 , . . , an ]), so Quot(A) is algebraic over L, too. There exists a nonzero element a ∈ P1 . We have a nonzero k i polynomial G = i=0 gi x ∈ L[x] with G(a) = 0. Since a = 0, we may assume g0 = 0. Furthermore, we may assume gi ∈ K[a1 , . . , an ]. Then k g0 = − gi ai ∈ P1 , i=1 so viewing g0 as a polynomial in n indeterminates over K, we obtain g0 (a1 + P1 , . . , an +P1 ) = 0, contradicting the algebraic independence of the ai +P1 ∈ A/P1 .

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

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