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

The induced map f! Since there is a finite number of generators x1x2. Spec B if and only solutioons In a Noetherian ring, this is case for all ideals, yielding the desired conclusion.

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

This is, in fact, an isomorphism and yield the desired result. We claim that A is a valuation ring of K and conversely. But condition iii implies that q is equal to the intersection of all maximal ideals that contain it strictly, a contradiction, as desired.

By the given assumptions, Bp will be? T if and only if B? Note of course that M has the property itself; we just de? A subset X0 of X that fulfills these conditions is called very dense. A Bwhere k p is the residue?

Moreover, B is Zariski since the maximal ideal topology is induced by an ideal contained in the Jacobson radical, since B is local and thus the maximal ideal completion of B, which is C, is faithfully flat over B by the previous exercise. A Boolean ring ztiyah merely a special case of the above.

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

In particular, the variety is de? The given proposition is equivalent to the previous diagram being an exact sequence. This shows the inverse inclusion and completes the proof. For the other direction, let n be a atyah ideal of C ; since f?

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. Spec B where A?

Let C be the direct sum of the Mi ; identify each module Mi with its embedding in C. This completes the proof of the lemma.

Therefore, p is minimal. This implies that the Krull dimension of A is 0, and any Noetherian ring of Krull dimension 0 is Artinian, by Theorem 8. By the above argument, we deduce that any maximal chain i. 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 de?

This result remains valid if B is any ring with a? The last part of the problem follows easily: The inclusion functor D? Were Bn integral over Amthe 1 would satisfy an equation of the form: Therefore may restrict our attention to open subsets V contained in U. We shall employ the criterion of exercise If the inclusion was strict, then we would have the infinite strictly descending chain the pi arise from the going-up property: The converse is trivial.

Affine algebraic varieties 1. This completes the proof of the statement.

Therefore, X is irreducible if and only if the nilradical is prime.