# Some Notions of Dimension

It is time to pull together some ideas we’ve built up, and show that they actually correlate how we want them to.

Recall that we have a notion of dimension for a ring called the Krull dimension. Review this if necessary, but essentially you just take the sup of the heights of all prime ideals in your ring.

For a topological space, we first define “irreducible.” Irreducible simply means that you can’t express the space as the union of two nonempty closed sets. To familiarize yourself with this definition some more you can try to prove that it is equivalent to the definition that any two non-empty open sets intersect non-trivially, or any non-empty open set is dense. So note right away that irreducible is a pretty rough condition. Almost none of the spaces I usually talk about are irreducible, since there are tons of non-dense open sets. And by the other criterion, any Hausdorff space is reducible.

Moving on, we now can define the dimension of a topological space to be $sup\{n : Z_0\supsetneq Z_1\supsetneq \cdots \supsetneq Z_n\}$ where the $Z_i$‘s are irreducible subspaces.

Naturally, we are now interested to see if the topological notion of dimension on the topological space $Spec(R)$ is the same as the Krull dimension of the ring R.

First, we’ll need a quick lemma:

A subspace $E\subset Spec(R)$ is irreducible if and only if $I(E)$ is a prime ideal. Note that this is really exactly what we wanted to happen, since prime ideals are points in Spec(R). When we say something is “irreducible,” what we are talking about are the smallest things that cannot be broken apart, i.e. points. I confess I am far oversimplifying this idea of “points” as we will see before the week is over if all goes as planned. (At this point you might want to review spec).

Proof of Lemma: I promise to fill this in later in the week. I just realized that it is an assignment to be turned in on Friday, and not all the readers of this blog that are in my commutative algebra class have this done yet.

Now let’s actually check dimensions. Theorem: $Krulldim(R)=dim(Spec(R))$. Just write it out now:

$dim(Spec(R))=sup\{n: Z_0\supsetneq Z_1\supsetneq \cdots \supsetneq Z_n , \ Z_i \ irreducible\}$
$=sup\{n: p_0\subsetneq p_1\subsetneq \cdots \subsetneq p_n , \ p_i=I(Z_i)\}$
$= Krulldim(R)$ where that switching comes from the Lemma. The correspondence is 1-1.

So for some future posts, I want to clarify some more on dimension and what “points” really are. I also want to do the Nullstellensatz and talk about why it is so important, but I may not. I’ll do some hunting to see what other math bloggers have done on it. I’m pretty sure it has been posted on extensively already, in which I’ll just point people in those directions and add some things that I personally find interesting.