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).