Atiyah-Macdonald Part II

by ayushkhaitan3437

Let me start by first talking about the proof of the fact that the intersection of all prime ideals in a ring is the nilradical. The proof is constructive, and hence perhaps non-trivial. I will attempt to generalize the methodology of the proof.

This proof deals with creating a sort of border around the number objects that satisfy a certain property. Then it takes two external objects (that clearly don’t satisfy the property), and gets a third object. Now this third object may or may not lie inside the order. It is easily proven that it does not. Hence, _theorem_

Also, the mechanism of the proof could be used to prove that the maximal ideal containing all odd numbers is prime. The maximal ideal containing all even numbers is prime. The maximal ideal containing all powers of 2 is prime, etc.

12. Let a\in R be an idempotent element. As (a) is a proper ideal of R, it is contained within M, which is the unique maximal ideal. Hence, as a is part of the Jacobson ring, 1-a is a unit. Now (1-a)=(1-a^2). This implies that a=0, which is obviously a contradiction.

15. i) Easy

ii) Easy

iii) It is easy to prove. Note that this is not true only for the set of **prime** ideals containing \{\bigcup E_i\}. Let W(A) be the set of ideals (any ideal) containing the set A. Then W(\bigcup E_i)=\bigcap W(E_i). This is an important generalization.

iv) I have proved this in an earlier post. For what kinds of ideals do we have V(a\cap b)=V(ab)? Thinking about these things is what will give your thinking depth. The set of ideals which satisfy the following property: if m^2\in I\implies m\in I. Prime ideals and even radical ideals are vastly specialized examples of this very general condition.

For what kinds of ideals do we have V(a)\cup V(b)=V(ab)? Ideals which exhibit the following property: if a\cap b\subseteq I, then a\subseteq I or b\subseteq I. Prime ideals are just one of the many types of ideals which may potentially fulfil this property. In fact, one may name a new class of ideals which are defined in such a way so as to fulfil this property.

Also, what kinds of sets mimic the behaviour of closed sets? Infinite intersections should be closed, finite unions should be closed, \emptyset and the whole space X (arbitrary name) should be closed, etc.

We see that V(E) exhibits the properties of a closed set. How do we ensure that the infinite union of such closed sets is not closed? Refer to this question I asked on stackexchange- http://math.stackexchange.com/questions/846020/a-question-about-the-zariski-topology/846034#846034. The argument does not work for an infinite number of ideals because….well let me first give the argument for a finite number of ideals. Assume we have the ideals a_1,a_2,\dots,a_n, and V(a_1a_2\dots a_n) does not contain a_1,a_2,\dots,a_{n-1}. Then it must contain a_n. Now if we have an infinite number of ideals, can we claim that V does not contain all but one ideal? Yes. But can we prove that the ideal left aside is indeed contained within V? For that we need to deal with infinite products of elements. Such products are not defined. Hence, this argument does not translate to the infinite case.

Also note that the infinite union of closed sets may indeed be closed! Take the union of [a_k,a_{k+1}], where a_i is the i^{th} natural number. However, the infinite union need not be closed. Take closed sets of the form [\frac{1}{2^k},\frac{1}{2^{k-1}}] for example.

17. i) X_f\cup X_g=(V(f)\cup V(g))'=(V(fg))'=X_{fg}

A little word about **why** V(f)\cup V(g)=V(fg) is true. If a prime ideal contains fg, and does not contain f, it HAS to contain g! This is also true if f and g are ideals, as mentioned in an earlier problem.

ii) If every prime ideal contains f, then f has to be nilpotent. If f is not nilpotent, there is a prime ideal which does not contain it, as the intersection of all prime ideals is the nilradical.

iii) If no prime ideal contains f, then f is a unit. This is because if f is not a unit, then (f) is a proper ideal, and hence contained inside some maximal ideal. Every such maximal ideal is prime.

A little detour here. Why is every maximal ideal prime? Let ab\in M, where M is maximal, and a\notin M. Then as+m=1, where m\in M and s\in R (R is the commutative ring under consideration). Now multiply by b to get abs+mb=b. Note that abs and mb, both are in M, and M is closed under addition (like every other ideal). Hence, b\in M.

iv) If X_f=X_g, then V(f)=V(g). Now every ideal (not just prime ideals) containing f will also contain (f). Also, every prime ideal containing (f) will also contain r((f)). Is that all? Will the intersection not contain any other element? Yes, we’re sure that the intersection will not contain any other element. Which brings me to my proof of this fact (not mentioned in Atiyah-Macdonald).

Let us consider an ideal containing (f), which does not contain any power of a. Is this set nonempty? No, (f) is one such ideal. Let us assume that the intersection of all prime ideals containing f contains an element a\notin r(f). It is easy to check that Zorn’s lemma can be used to create a maximum ideal not containing any power of a. It is also easy to prove that this maximum ideal is prime. Hence there is a prime ideal which does not contain a.

Now if r((f))=r((g)), then every prime ideal containing f (and hence containing r((f))) will contain r((g)), and hence g. Hence V(f)\subseteq V(g). We may similarly prove V(g)\subseteq V(f).

v) The first thing to note here is that V(f,g)=V(f)\cap V(g). Also: let \{X_f,X_g,X_h,\dots\} be an open covering of X. Then V(f)\cap V(g)\cap V(h)\cap\dots is empty. Note that the infinite intersection of closed sets is again a closed set. Also, that closed set is V(f,g,h,\dots). If there is no prime ideal which contains ALL these elements, then there must be a finite number of them which add up to give 1. Let those elements be a_1,a_2,a_3,\dots a_n. Then X_{a_1},X_{a_2},\dots X_{a_n} also covers X.

vi) Let X_f have a covering \{X_m,X_n,\dots\}. Then V(m)\cap V(n)\cap\dots contains no ideal from X_f. Hence every ideal it contains contains f. In other words, V(m,n,\dots) contains f. If that is true, then a finite number of them must add up to form f. Hence, the intersection of a finite number of V(a) will lie completely inside V(f). Taking their complements will give a finite cover of X_f.

What really is the gist of this argument? When we’re talking about an open set X_f, we’re really just talking about its complement, or the prime ideals which contain f. When we’re talking about an open cover, we’re really just talking about the intersection of the prime ideals in the complement. Hence, when one is a subset of another, we have a useful result. All prime ideals which contain these many elements also contain f…something like that. At the end of the day, all we’re doing is talking about various prime ideals containing certain elements, and what we can extract from that.

What if we had an open set of the form X_f\cup X_g? Then the complement is V(f,g). If X_f\cup X_g has an open cover, then there are finite expressions in V(f,g) that are equal to f and g. Let a_1+a_2+\dots a_m=f and b_1+b_2+\dots b_m=g. Then taking taking complement of V(a_1,a_2,\dots a_m,b_1,b_2,\dots b_n) gives an open cover of X_f\cup X_g.

vii)

18. i) The set is closed if it contains all ideals containing a certain element, say f. If x was not maximal, then we could have made it bigger, and that bigger ideal would still contain f.

ii) This is by definition. It can be rather misguiding that there is no topological “proof” for this. For example, \overline\{a,b,c\}=V(a,b,c) for any a,b,c in commutative ring R.

iii) \overline{\{x\}} is the set of all prime ideals containing p_x. Hence if y\in\overline{\{x\}}, then p_y must contain p_x.

iv) Let us take two prime ideals m and n. Also, suppose that X_m contains X_n. This implies that n

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