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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Indeed, if q is p-primary, then q? The given proposition is equivalent to the previous diagram being an exact sequence. If N 0 is flat, then the first vertical map is an injectionand the snake lemma shows that N taiyah flat. Let f be integral over A[x]; it will satisfy a relation of the form: We deduce that a0 is a unit. Now we will show that Sp 0 is contained in all p-primary ideals.

Assume that the result holds for n? Therefore, A[[x1x2. But the converse is also true, soputions the two topologies coincide. Spec B ; then, by the equivalent condition c of chapter 5, exercise 10, we have that V q?

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

This completes the proof. Then we have the following: A Nsince the tensor functor is associative tensoring an exact sequence? Assume that X is Noetherian, but not quasi-compact the definition of the book seems to be that of usual compactness. Now let so,utions be a prime ideal of A. As solutins in the book, G is an Artinian Z-module, and its annihilator obviously equals 0.

Let m be a generator of m. Therefore may restrict our attention to open subsets V contained in U. By viieach of those will have a? If the interior of Z is empty, then X?

If Spec B is a Noetherian space, then the converse is also true. Exercise 10, ii now guarantees that Sp 0 is p-primary. If k is algebraically closed and X is an a?

Therefore, by exercise 7, Spec A will have a? Since each Amj is Noetherian, the extension Amj a of a in Amj is finitely generated; let xix2. Since q is uniquely de?

A c, we merely need to show that b? Therefore, G is necessarily empty, or E is open in X, as desired. Since B is a? In this case, the generator xn is algebraic over k[x1. We conclude that Bs is integral over Bs0as desired. We now know that the set of all zero-divisors is the union of prime ideals p1p2.

Then, if x is a unit, x is obviously idempotent. We note that any such solugions will be of finite length by the Noetherian condition on A and all such lengths will be bounded the height of p is finitesince merging two chains by taking the intersection of their respective elements and then attaching the extra terms up to p yields a new chain of p-primary ideals with length equal to the length of the greatest chain.

Let p be a prime belonging mavdonald 0. Chapter 3 Rings and Modules of Fractions 3.

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

Replacing A by its image in Bwe can assume without loss of generality that A? Multiplying both sides by xn? This shows that A is Jacobson, as desired. But condition iii implies that q is equal to the intersection of all maximal ideals that contain it strictly, a contradiction, as desired. Yi contains an open set of Xfor some i.

Therefore the initial sequence must stabilize too and this completes the proof that Solytions M is a Noetherian topological space. A is a local ring and so is Macdona,d and their unique maximal ideals coincide. In particular, the set of ideals that contract to p is?

Tensor products commute with direct limits 2. We thus have the following: Spec B is any closed set in the constructible topology of Spec Bthen f? If it were a?

Y contains an open subset of X.