## Maximal Ideals (Pt. II)

**Point of Post: **This is a continuation of this post.

Just as was the case for prime ideals there is a theorem for maximal ideals that says there are ‘enough of them’. More directly, this will say something to the effect that every ideal is contained in a maximal ideal (for unital rings). This theorem is famous enough that it warrants a name-attachment, in particular to Wolfgang Krull who first stated the theorem in 1929. As we shall see (and it really should be apparent to you since we are trying to prove the existence of a maximal element of a poset) that we are going to use Zorn’s lemma. More surprising perhaps is the fact that Krull’ls lemma is equivalent to the Axiom of Choice (for more information one can see the great book *Axiom of Choice*by the renowned set theorist Jech).

**Theorem (Krull):** *Let be a unital ring. Then, given any ideal of , there exists a maximal ideal containing .*

**Proof: **We are evidently trying to show that the poset of all proper ideals containing has a maximal element. To do this, let be a chain in and let be it’s union. As was checked in the similar argument for prime ideals, we have that is an ideal, and it evidently contains . Thus, to see that it suffices to check that it’s proper. But, from first principles we know that if and only if , but since is not contained in any element of it evidently isn’t contained in their union. Thus, , so . Since was arbitrary we may conclude from Zorn’s lemma that has a maximal element , which is evidently the desired maximal ideal.

Applying this lemma to gives:

**Corollary: ***Every non-zero unital ring contains a maximal ideal.*

*Remark: *There do exist rings that don’t have maximal ideals, but they are fairly ugly, and go too far off into a tangent to pursue.

Another thing this theorem does is illuminate the connection between non-units of and . Namely, the maximal ideals of form the entirety of the nonunits of . Said more directly:

**Theorem: ***Let be a commutative unital ring, then*

**Proof: **Evidently all and so it suffices to prove that the non-units are the union over . To see this we merely note that every nonunit is contained in a maximal ideal since, by the previous theorem, we know that is contained in some maximal ideal and since the conclusion follows. Conversely, we know that if is an element of the union over then it cannot be unit, otherwise the ideal it belongs to would be equal to contradictory to assumption.

An obvious question one may ask is, we already have a set of ideals of which are, in some way, special–prime ideals. So, how do these maximal ideals relate to them? Well, the long and short of it is that:

**Theorem: ***Let be a commutative unital ring, then .*

**Proof: **This follows immediately from the fact that if then is a field, and so is, in particular, a prime ideal. It follows that .

So, now we’d like to see for some classes of, or specific rings. For example, it should not be hard to see, similar to the case for prime ideals, that:

**Theorem:** *Let be finitely many rings, then*

Perhaps more useful is the fact that the fourth isomorphism theorem clearly implies that:

**Theorem: ***Let be a ring and an ideal of . Then, .*

Perhaps more fun is the following gem:

**Theorem: ***Let be, as usual, the set of continuous maps and let, for , . Then, .*

**Proof: **Evidently is maximal (as discussed before) since . Conversely, let be a maximal ideal, and suppose that for each there existed a such that . Since is continuous there exists some open neighborhood of such that is non-vanishing on . Since is evidently the union of the we may, by compactness, produce a finite subcover . Considering then gives a non-vanishing element of , but a non-vanishing element of is invertible, which creates an obvious contradiction.

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