Let be a commutative ring with 1. Then there exists a *maximal* ideal in i.e. a proper ideal such that the only ideal that contains is itself.

We did not discuss this in the class, and the main reason is that it requires a form of the Axiom of Choice. We will use its form, known as Zorn’s Lemma

Every partially ordered set in which every chain (i.e. totally ordered subset) has an upper bound contains at least one maximal element.

We take the set of all proper ideals of as the set ordered by inclusion, i.e. iff for

Any chain is then a set of ideals such that for any either or We need to check that has an upper bound, i.e. such that for any Set

We need to check that i.e. that it is a proper ideal. As we only need to check that it is an ideal. It suffices, and is easy, to see that

- for any there exists such that and thus (Indeed, there exists such that and w.l.o.g as is a chain)
- for any and here exists such that and thus (Indeed, there exists such that and as is an ideal.)

Thus Zorn’s Lemma is applicable to , so it has maximal element, that is obviously a maximal proper ideal in QED.

As a corollary, we see that every proper ideal is contained in a maximal proper ideal Indeed, contains a maximal proper ideal and for a maximal ideal

In particular, any non-invertible element is contained in such an ideal, by applying the latter construction to

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Tags: algebra, MAS313, maths

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