We are now on the last inequality: . Recall we’re supposing is Noetherian and local. Let , then the inequality is saying we can find an ideal, that is an -primary ideal and is generated by elements: .

Let’s construct these elements inductively. The way we want to do it is so that at each step any prime ideal containing has height bigger than or equal to to force the dimension of to be big.

Suppose the have been constructed in the given way. Let be the minimal prime ideals of with height exactly . But we have that for any since . Thus (there is a well-known fact that if any ideal , then in fact for some ).

Now pick some element , and let be a prime ideal containing . We definitely have that , since contains a minimal prime ideal of , say . If for some , , then since , we have strictly increasing the height. If for any , then since are all minimal primes of height , we have .

All that is left is to show that is -primary. Let be any prime ideal of . Then if we have that . So this is impossible and we have . Thus is -primary.

Over the last couple of posts we have finally completed the first goal. We have . 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 -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).

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November 8, 2009 at 2:20 pm

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?

November 9, 2009 at 8:58 pm

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.

November 12, 2009 at 8:43 am

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.

November 12, 2009 at 8:56 am

2 more things:

1. excerpt from 4th paragraph: “If for some j, , then since , we have strictly increasing the height. If for any j, then since are all minimal primes of height i-1, we have .”

The case division looks unnecessary, because by construction strictly contains , and we know that , so 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.

November 13, 2009 at 11:36 pm

Em I think I know how to figure out this proof now.

The key is actually to find an element such that is of lower dimension. For this, one has to avoid elements lying in the minimal prime ideals.

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December 31, 2009 at 1:09 pm

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