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.

Author: | Jujas Yoramar |

Country: | Venezuela |

Language: | English (Spanish) |

Genre: | Art |

Published (Last): | 3 September 2009 |

Pages: | 175 |

PDF File Size: | 14.96 Mb |

ePub File Size: | 13.17 Mb |

ISBN: | 293-3-43616-164-8 |

Downloads: | 83335 |

Price: | Free* [*Free Regsitration Required] |

Uploader: | Akilabar |

We claim that A is a valuation ring of K and conversely. Yi contains an open set of Xfor some i. A A[x] will also be Macdonalx by Proposition 7.

Thus, p-adic completion does not commute with taking kernels, and therefore it is not a right-exact functor on the category of all Z-modules, as desired.

X be both open and closed. If Spec B is a Noetherian space, then the converse is also true. Since m was arbitrary, m? In particular and this is obvious as wellthe image of Spec Af is the basic open set Xf.

A c, we merely need to show that b? Maxdonald the converse were not true, then there would be a non-empty set F? Hence all fibres have one element; in particular, they are finite. But, if r Sp 0? Hence we must have a? Therefore, X is quasi-compact.

But then x bx? Thus a is prime.

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

By this process, we show that all the coefficients will have to be nilpotent. This is, in fact, an isomorphism and yield the desired result. Hence the original map is an isomorphism, as desired. It is straightforward that if there exists such an x, then p?

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

Let a be an ideal of A and let p1 be a minimal prime ideal that contains a. Sn implies S1 a?

In this case, the generator xn is algebraic over k [x1. Solutikns shown sokutions the book, G is an Artinian Z-module, and its annihilator obviously equals 0.

We thus have the following we use freely the results of chapter 1, proposition 1. Therefore, any subspace of X is Noetherian. Macdonale x0 be a non-zero element of a and let m1m2. For example, consider the situation of chapter 1, exercise If S were any saturated and multiplicatively closed subset of S that contained S, then its complement in A would contain at least one prime ideal that has 0 non-trivial intersection with S.

Letting pi be the minimal prime ideal that contains bi gives rise to an increasing sequence Mpi of submodules of M, which stabilizes because M is Noetherian.

We claim of R is of course R that Solytions is an integral domain, but its m-adic completion is not. It is always true that c f g?

Now, with the notation of the problem, let A? Therefore, the characteristic soluitons K equals some prime p and K is a? Therefore, x is a unit and since x was arbitrary, we conclude that A is a field. The above lemma implies that we may remove the tensor by the?

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

If it were a field, then it would be finite over B, hence integral over B, hence B would be a field by the Nullstellensatz, a contradiction. Were Bn integral over Amthe 1 would satisfy an equation of the form: Given the universal property of K ATheorem The strictly ascending chain of ideals 0?

We will just sketch the construction, which is identical to the one for fields. Chapter 10 Completions Sa contradiction. Since A is Artinian, it will be of finite length as an A-module, and we conclude that it will be a finite k-algebra, as desired.

Now, since since f is continuoushence we macdonals replace Z by Z above.