# Intuition on Krull Principal Ideal Theorem Proof's Major Ingredients

The krull principal ideal theorem I am interested in is the one involving that over Noetherian ring, minimal primes over the principal ideal must be height at most 1.

In the proof, it involves

1. symbolic power of prime ideals

2. localization at the same prime ideal such that symbolic power of prime ideals is identified as extension of powers of the prime ideal

3. Quotienting out the principal ideal results in artinian ring

4. Nakayama Lemma shows any $Q'$ such that $Q'\subset Q\subset P$ where $P$ minimal over $(x)$, $Q$ and $Q'$ are primes, $Q'=0$.

I could reproduce the proof of PIT. I think 4 is necessary for noetherian ring. $2$ is consequence of $1$. $2$ combining $3$ deduces $4$.

Q1. Was there alternative versions of proof which does not invoke symbolic power of prime ideals? I have seen Eisenbud, Atiyah's proof on PIT and they all use symbolic power of prime ideals. The usage of symbolic power of prime ideals is too much ad hoc.

Q2. Symbolic power of prime ideal is more or less generalization of primary ideal. What is the motivation for adopting it and adapting it to the proof?

Q3. What are usage of symbolic power of prime ideals? I have not encountered a problem that needs to use it unless the problem requires PIT application which will quote symbolic power.

• One can prove Krull by using Poincare series; see the last chapter of Atiyah & Macdonald. – Angina Seng Aug 29 '17 at 17:45
• Late to the party: In Kemper‘s book, it is proven using properties of Artinianity. – Qi Zhu Jul 16 '19 at 5:10
• I'm not an expert and am happy to be corrected. The way that I think about it: primary decomposition corresponds in the geometric setting to breaking down a variety into its irreducible components. I guess sort of implicitly in primary decomposition there's always a consideration of an ideal is $\mathfrak{p}$-primary, and the decomposition and uniqueness theorems tell us what the irreducible decomposition of the variety becomes. PIT says that taking out (irreducible) hypersurfaces corresponds to moving 1 dimension down. – mi.f.zh Jun 5 '20 at 1:24
• The n-th symbolic power corresponds to "functions with zeroes of order $n$ along that variety". In the proofs I know of the PIT that use the symbolic powers, at one point in time there's an equality which looks like $\mathfrak{p}^{(n)} = \mathfrak{p}^{(n+1)} + \mathfrak{m}\mathfrak{p}^{(n)}$, where $\mathfrak{m}$ is the maximal ideal in the local ring. Nakayama's lemma says that eventually $\mathfrak{p}^{(n)} = \mathfrak{p}^{(n+1)}$, so that functions with zeroes of order $n$ have order $n+1$, and vice versa. – mi.f.zh Jun 5 '20 at 1:24
• This equality is brought back to the setting of localisation, and symbolic powers (by their algebraic definition) become 'normal' powers. So to me, symbolic powers are necessary in this sense since we want to see that arbitrary powers of functions vanish, allowing us to pass freely between the various local rings/localisations. But I might be wrong in all of this... – mi.f.zh Jun 5 '20 at 1:24