This page covers section 3.5 (“More Ideals and Quotient Rings”). Throughout, is a ring.
Lemma 3.5.1: Let be commutative and unital and only have the ideals and . Then is a field.
Definition: An ideal is a maximal ideal if any other ideal such that is either or .
The prime ideals of are exactly the ideals for a prime integer. See also: problem 3.4.8.
Theorem 3.5.1: Let be commutative and unital and let be an ideal of . Then is maximal if and only if is a field.
The problems below are paraphrased from/inspired by those given in Topics in Algebra by Herstein. The solutions are my own unless otherwise noted. I will generally try, in my solutions, to stick to the development in the text. This means that problems will not be solved using ideas and theorems presented further on in the book.
Let be an element of and consider which was shown to be a right ideal in problem 3.4.14. Because , we have that is a non-zero element of . By the conditions of the problem, we conclude that . In particular, there exists such that . We would like to show also that . If that is the case, then we have exhibited a multiplicative inverse in for an arbitrary non-zero ring element.
Considering and arguing as above, we find an element such that . However, we see that while so that . Therefore, and we have proved that every non-zero element of has a multiplicative inverse, so that is a division ring.
This is related to problem 3.5.1 in that we only relax the condition on being unital. My solution was heavily influenced by the nice solution by AllTheCheese in this forum thread.
Case 1: there exists some such that . In that case, there exists an element such that . We would like to show that this is the multiplicative unit for the ring, which will put us in the realm of problem 3.5.1. First recall the annihilator shown in problem 3.4.17 to be a right ideal of . As , we must have that , so annihilates nothing but the zero element. However, right-multiplying the equation by , we find
In light of the comment about , we have that .
Now take any other element ; we would like to show that so that is truly the unit element. Recalling that , there exists such that , so . It is more difficult to show that .
Let and multiply on the right by to see that (, see below for the proof*). Then . Recall problem 3.4.18 which tells us that for all is a right ideal. Putting and , we see that , so there exists at least one element of which doesn’t belong to . With our strict condition on ideals, this means that . In other words, there does not exist a non-zero element for which for all . Letting , and assuming , we see that there is at least some such that , which forces us to conclude that . However, we have already seen that , a contradiction. The only way out is to have that , which shows that , after all. Now we have shown that case 1 leads to the ring having a unit, and therefore being a division ring by problem 3.5.1.
Case 2: there is no such that ; this is the complement of case 1. Then always so that for any . Now, let be a non-zero element. If we consider the set , we see that it is a right ideal because multiplication from the outside always yields which belongs to , while closure under addition is clear. Also, contains the non-zero element , so . Now, if is finite, the sequence will yield distinct elements and then loop around to elements it has already visited. In particular, if is composite, say with integers, then is non-zero, and is a right ideal with a non-zero element, but it is a proper subset of . By the condition of the problem, such a may not exist. Therefore if is in case 2 and is finite, then must be prime.
Finally, we must also consider case 2 with infinite. In fact, this is not possible. Simply consider the ideal which is a proper subset of with a non-zero element. As before, this is a contradiction of the terms of the problem.
Here is a brief digression from the main course of the proof to show that for the alleged identity element of . Recall that the non-zero element satisfied and implied . Now, implies that , so . In particular, there exists with . Multiplying on the left by gives . As only annihilates the zero element, we must have that , so as claimed.
(a) There is a natural homomorphism to consider, namely
given by , where on the right hand side is reduced modulo so as to live in . To show that this is well-defined, let be such that . Then so they differ by a multiple of and the map is well-defined. It is also easily seen to be a homomorphism. Because the kernel is trivial, is an isomorphism.
(b) By problem 3.4.8, we know that is a maximal ideal of , and by theorem 3.5.1, we therefore have that is a field, because it is a commutative unital ring () modded out by a maximal ideal.
First we note that an invertible element in this ring is one which has no zeroes. The multiplicative identity element is the constant function , and if has no roots then it has a multiplicative inverse given by . If an ideal contains an invertible element, then it is all of . Therefore every element of a proper ideal must have a root.
The sketch of the proof is as follows:
(1) Let . We have that , and therefore are also in because it is an ideal. Suppose that and share no root. Then at any root of , we have . Similarly at any root of . Of course, elsewhere. Therefore if and share no root, then contains the invertible element , and hence . We are interested in proper ideals of , in which, as we have just seen, any pair of elements has a root in common.
We can say something more general than the pairwise statement, though. Given any finite subset of a proper ideal , say , the same argument shows that all members of that subset share at least one zero. Otherwise would be an invertible element which forces .
With and writing for the (non-empty) set of zeroes of , we have just shown that the collection has the finite intersection property. Any finite subcollection of that collection has non-empty intersection.
(2) We should now like to extend the statement above to cover the existence of a root shared by everything in the ideal. That is, we would like to prove that
It is not true in general that the finite intersection property extends to a non-empty intersection on the whole collection. However, we have additional conditions at our disposal.
Let . Because is continuous, we know that the inverse image under of an open set is open. In particular, is open. Its complement, , is therefore closed in . That is, every zero set that we consider is topologically closed.
Additionally, is a compact set. In a compact metric space such as , any collection of closed sets with the finite intersection property has a non-empty intersection (proof). This follows simply by taking the definition that “every open cover of a compact set has a finite subcover” and turning it into a statement about the closed sets given by the complements of the sets in the cover. Therefore we have that, for any proper ideal of , there is a nonempty set on which every member of takes the value zero.
(3) Now we want to determine what a maximal ideal looks like. If , then may not be maximal. If , then we could take a proper, non-empty subset and construct the proper ideal . If , then , so we also have and therefore . In other words, reducing the size of the zero set makes the ideal bigger — the bigger ideal contains all the functions that were zero on the expanded zero set, but it also contains new functions that take other values on the roots that were excluded. On a side note, it may be true that zero sets are in one-to-one correspondences with ideals, but I do not know at this time. For the purposes of this exercise, it suffices to construct one ideal from a given zero set, as we constructed earlier in this paragraph.
Thus it must be the case that a maximal ideal has exactly one shared root among its elements, . Calling that root , we must have as a subset of . However, we easily verify that is itself an ideal, so any proper subset of it would clearly not be maximal. Therefore, our maximal ideal is and the claim is proven.