# Finishing up Dimensions

We are now on the last inequality: $\dim R\geq \delta(R)$. Recall we’re supposing $(R, \frak{m})$ is Noetherian and local. Let $\dim R=d$, then the inequality is saying we can find an ideal, $\frak{q}$ that is an $\frak{m}$-primary ideal and is generated by $d$ elements: $x_1, \ldots, x_d$.

Let’s construct these elements inductively. The way we want to do it is so that at each step any prime ideal containing $(x_1, \ldots, x_i)$ has height bigger than or equal to $i$ to force the dimension of $R$ to be big.

Suppose the $x_1, \ldots, x_{i-1}$ have been constructed in the given way. Let $\{p_j\}$ be the minimal prime ideals of $(x_1, \ldots, x_{i-1})$ with height exactly $i-1$. But we have that $\frak{m}\neq p_j$ for any $j$ since $ht(\frak{m})=\dim R=d>i-1=ht(p_j)$. Thus $\frak{m}\neq \cup p_j$ (there is a well-known fact that if any ideal $\frak{a}\subset \cup p_j$, then in fact $\frak{a}\subset p_j$ for some $j$).

Now pick some element $x_i\in \frak{m}\setminus (\cup p_j)$, and let $\frak{q}$ be a prime ideal containing $(x_1, \ldots, x_i)$. We definitely have that $ht(\frak{q})\geq i$, since $\frak{q}$ contains a minimal prime ideal of $(x_1, \ldots , x_{i-1})$, say $\frak{p}$. If for some $j$, $\frak{p}=p_j$, then since $x_i\in\frak{q}\setminus \frak{p}$, we have strictly $\frak{q}\supset \frak{p}$ increasing the height. If $\frak{p}\neq p_j$ for any $j$, then since $\{p_j\}$ are all minimal primes of height $i-1$, we have $ht(\frak{q})>i-1$.

All that is left is to show that $\frak{q}=(x_1, \ldots, x_d)$ is $\frak{m}$-primary. Let $\frak{p}$ be any prime ideal of $\frak{q}$. Then if $\frak{p}\subsetneq \frak{m}$ we have that $ht(\frak{p})< ht(\frak{m})=d$. So this is impossible and we have $\frak{p}=\frak{m}$. Thus $\frak{q}$ is $\frak{m}$-primary.

Over the last couple of posts we have finally completed the first goal. We have $\delta(R)=d(R)=\dim R$. In other words, for Noetherian local rings we have an equivalence between the maximum length of chains of prime ideals, the degree of the Hilbert polynomial, and the least number of generators of an $\frak{m}$-primary ideal.

Next time I'll derive some results directly from this including the Principal Ideal Theorem. Then we'll move on to something different (only for awhile, then we'll return).

### Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

### 8 thoughts on “Finishing up Dimensions”

1. Is there a connection between this “Hilbert polynomial” way of proving things that you use and the other proof of the principal ideal theorem given, e.g., by Eisenbud?

2. I can’t really tell if there is an underlying connection. I guess in some sense it is a “theorem” when done as in chapter 10 of Eisenbud, whereas it is more like a “corollary” to the equivalences given in this post. Eisenbud does eventually end up proving essentially everything in these posts in almost an identical manner.

Also, I can’t quite figure out what happened, but at some point I adopted a sort of non-standard convention. The equivalence is usually stated, krull dimension = degree of Hilbert poly + 1 = least number of generators of an m-primary ideal. As soarerz pointed out (which confused me at first) I think my convention what to call the degree of the polynomial (which is the same as Atiyah/MacDonald) is just different and it isn’t actually wrong.

3. I’m sorry that the comment I left in the post is probably wrong – I did the computations again and d(R) should be the same as the order of the pole at t = 1. I don’t know why but every time I do this kind of thing, the ans seems to be off by 1 very often.

Anyway, degree of Hilbert polynomial for $G_m(R)$ = order of pole – 1. So the degree of Hilbert polynomial for your $l$ function = order of pole, no problem.

4. 2 more things:

1. excerpt from 4th paragraph: “If for some j, $\frak{p}=p_j$, then since $x_i\in\frak{q}\setminus \frak{p}$, we have strictly $\frak{q}\supset \frak{p}$ increasing the height. If $\frak{p}\neq p_j$ for any j, then since $\{p_j\}$ are all minimal primes of height i-1, we have $ht(\frak{q})>i-1$.”

The case division looks unnecessary, because by construction $\frak{q}$ strictly contains $\frak{p}$, and we know that $ht(\frak{p}) \ge i - 1$, so $ht(\frak{q})$ must be at least i.

2. Is there any motivation for this proof at all? (And on this issue, the motivation for the two equivalent definitions other than Krull dimension) I’ve also read it on Atiyah before, but it just seems to be a bunch of logic deduction to me.

5. Em I think I know how to figure out this proof now.
The key is actually to find an element $x \in \frak{m}$ such that $\mathrm{dim} \frac{R}{(x)}$ is of lower dimension. For this, one has to avoid elements lying in the minimal prime ideals.

6. It looks like you are a true pro. Did you study about the topic? lawl