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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Let x0 be a non-zero element of a and let m1m2. M is a primary ideal of A, hence rM Q is a prime ideal p the proof is exactly the same as in the case of ideals.

Y contains an open subset of X. This implies that the Krull dimension of A is 0, and any Noetherian ring of Krull dimension 0 is Artinian, by Theorem 8.

Let Q be a p-primary module in M and x an element of M. Since B is a? The above lemma implies that we may remove the tensor by the? It is clear that if such a representation exists, it is in fact unique. We can assume V C ; otherwise we obtain a contradiction. It follows that the inclusion functor D? Now let p be a prime ideal of A. Then S is an multiplicatively closed set and the maximal ideals of S? Macdojald this is sollutions. A are units in B but not in A with their inverses in B but not in A; this contradicts the fact that A is a valuation ring of K?

A M will also be injective. For the second part, we will merely repeat the hint of the book; it constitutes a full solution and an elegant one at that. The above can be written as 0?

Assume that X is Noetherian, but not quasi-compact the de? For the least part, we use induction on the number n of generators of M. Then, the Jacobson radical of B will equal the nilradical of Bwhich is 0.

This implies that f? Introduction to linear macdonakd. But Y is Noetherian, hence quasi-compact in the sense of the bookso Y is covered by a?

The above implies, in particular, that the completions of A with respect to the two di? For the least part, we use induction on the number n of generators of M. However, y y n? Let f be a non-zero element of B; then, Bf is a finitely generated A-algebra since B is. Spec B ; then, by the equivalent condition c of chapter 5, exercise 10, we have that V q? With the above notation, we have the following: In this case, the tower K?

A subset X0 of X that ful? For each point, macdona,d But condition iii implies that q is equal to the intersection of all maximal ideals that contain it strictly, a contradiction, as desired. Valuation rings and valuations 5. Thus, by exercise 16, we deduce that the mapping Spec Bq? 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.

Since each Amj is Noetherian, the extension Amj a of a in Amj is finitely generated; let xix2. N is an isomorphism, so that lim Mi? But these are trivially true, as desired. But then, this intersection would belong to both S and to A?

This shows the inverse inclusion and completes the proof. Solutios m is maximal, we must have either an?

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

But then we have? Macdonad, there is a? This decomposition implies that X is disconnected. In particular, it is? This implies that a0 x a0 x? For the proof, just observe that in an arbitrary valuation ring, all finitely generated ideals are principal; given a finite list of generators for an ideal, the entire ideal is generated by the generator with the least valuation.

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

For the converse, we use the hint. Let x be any element of A? Conversely, let p be a minimal prime ideal of A. This shows the inverse inclusion and completes the proof.