Solutions to Atiyah-Macdonald, Chapter 1. Dave Karpuk. May 19, Exercise 1. Let x be a nilpotent element of a ring A. Show that 1+x is a unit of A. Deduce. Trial solutions to. Introduction to Commutative Algebra. ( & I.G. MacDonald) by M. Y.. This document was transferred to. Atiyah and Macdonald “provided exercises at the end of each chapter.” They and complete solutions are given at the end of the book.

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## Solutions to Atiyah and MacDonald’s Introduction to Commutative Algebra

A E must vanish. A M are vector spaces of the same dimension over k. Conversely, it follows immediately from the de?

macdonwld But Y is Noetherian, hence quasi-compact in the sense of the bookso Y is covered by a? Therefore may restrict our attention to open subsets V contained in U.

This implies that, f1 A [x1 ], f2 A [x2 ]. Otherwise, the process continues to yield new ideals and, as n? Yi contains an open set of Xfor some i. The inclusion functor D? Then, B is an integral domain and its zero ideal is not the intersection of maximal ideals; in particular, its Jacobson atiya is nonzero.

But then x bx? This completes the proof of the statement. The integrality of B over f A implies, by theorem 5. E is an injection, so N is? Spec B if and only if? Since any A-module is a direct limit of its? Y is open in X. The above implies, in particular, that the completions of A with respect to the two different topologies fail to coincide.

Let f be integral over A[x]; it will satisfy a relation of the form: But condition iii implies that q is equal to the intersection of all maximal ideals that contain it strictly, a contradiction, as desired. So we may assume that A is an integral domain and f is injective.

### Solutions to Atiyah and MacDonald’s Introduction to Commutative Algebra – PDF Free Download

Therefore, G is necessarily empty, or E is open in Xas desired. Therefore, we may equivalently describe F as the minimal subspace of X that contains all open and closed subsets of X and is closed under finite intersections and unions.

macdoonald Construction of an algebraic closure of a field E. These results show that the space Spec A of all prime ideals of A can be endowed with a topology – the Zariski topology – if we define the V E to be its closed sets.

For that, we will perform induction on the degree of f. Therefore S is minimal.

In this case, the tower K? Kand since K is? Spec B ; then, by the equivalent condition c of chapter 5, exercise 10, we have that V q? Therefore, M is Noetherian. However, a chain of prime soluutions p0? This implies that G acts faithfully, as desired.

Thus A is Noetherian, as desired. A M will also be injective. Conversely, let m be closed in the a-adic topology.

### Papaioannou A.-Solutions to Atiyah and MacDonald’s Introduction to Commutative Algebra ()_百度文库

Therefore, we may equivalently describe F as the minimal subspace of X that contains all open and closed subsets of X and is closed under? With the above notation, we have the following: Thus Bf is not a? By assumption, f macdnald a dense image and thus by chapter 1, exercise 21, f is injective. From the proof it now follows that we may choose the yi to be linear combinations of sloutions xi. Therefore, p is minimal.

A, let n be the minimal degree of all polynomials in A[x] that y satis? Therefore X is homeomorphic to Max C X.