Zorn's Lemma - An Example Application

An Example Application

We will go over a typical application of Zorn's lemma: the proof that every nontrivial ring R with unity contains a maximal ideal. The set P here consists of all (two-sided) ideals in R except R itself, which is not empty since it contains at least the trivial ideal {0}. This set is partially ordered by set inclusion. We are done if we can find a maximal element in P. The ideal R was excluded because maximal ideals by definition are not equal to R.

We want to apply Zorn's lemma, and so we take a non-empty totally ordered subset T of P and have to show that T has an upper bound, i.e. that there exists an ideal I ⊆ R which is bigger than all members of T but still smaller than R (otherwise it would not be in P). We take I to be the union of all the ideals in T. Because T contains at least one element, and that element contains at least 0, the union I contains at least 0 and is not empty. Now to prove that I is an ideal: if a and b are elements of I, then there exist two ideals J, K ∈ T such that a is an element of J and b is an element of K. Since T is totally ordered, we know that J ⊆ K or K ⊆ J. In the first case, both a and b are members of the ideal K, therefore their sum a + b is a member of K, which shows that a + b is a member of I. In the second case, both a and b are members of the ideal J, and we conclude similarly that a + b ∈ I. Furthermore, if r ∈ R, then ar and ra are elements of J and hence elements of I. We have shown that I is an ideal in R.

Now comes the heart of the proof: why is I smaller than R? The crucial observation is that an ideal is equal to R if and only if it contains 1. (It is clear that if it is equal to R, then it must contain 1; on the other hand, if it contains 1 and r is an arbitrary element of R, then r1 = r is an element of the ideal, and so the ideal is equal to R.) So, if I were equal to R, then it would contain 1, and that means one of the members of T would contain 1 and would thus be equal to R – but we explicitly excluded R from P.

The condition of Zorn's lemma has been checked, and we thus get a maximal element in P, in other words a maximal ideal in R.

Note that the proof depends on the fact that our ring R has a multiplicative unit 1. Without this, the proof wouldn't work and indeed the statement would be false. For example, the ring with as additive group and trivial multiplication (i. e. for all ) has no maximal ideal (and of course no 1): Its ideals are precisely the additive subgroups. The factor group by a proper subgroup is a divisible group, hence certainly not finitely generated, hence has a proper non-trivial subgroup, which gives rise to a subgroup and ideal containing .