I asked this question last year in MSE, but I didn't get an answer.

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I took a commutative algebra course last semester (using Kaplansky's book), and I learned about Krull's intersection theorem. In the course, we proved it without using the Artin-Rees Lemma. I have heard that the standard proof uses the lemma.

Are there any other simple or intuitive proofs for the theorem? I have heard that there is an intuitive reason for coordinate rings in 
[this post](https://mathoverflow.net/questions/109161/a-geometric-insight-into-a-proof-of-krull-intersection-theorem), that *the only function which vanishes to arbitrarily high order at a point is the zero function*. In case of polynomials, this is easy to accept if we consider the polynomial as a Taylor series expansion (with finitely many terms). Using this, I tried to prove it for a local ring, as follows.

Let $R$ be a Noetherian ring. Suppose there exists $D:R\to R$ satisfying $D(ab)=(Da)b+a(Db)$ and $D(a+b)=Da+Db$, that is, it is a derivation on $R$. Define $\mathrm {Poly}(R):=\{r\in R\,| \,D^{k}r=0$ for some $k>0\}$, which means that the set of elements in $R$ on which $D$ behaves as a polynomial. 

We check that $\mathrm {Poly}(R)$ is a subring of $R$, and I wonder whether $R=\mathrm {Poly}(R)$. Then I want to *approximate* elements in $R$ via their Taylor expansions, but it was not easy to formalize this. I think that the evaluation map corresponds to quotients by maximal ideal; and then it would be possible to find a function $r^{*}:k\to k$ where $k=R/\mathfrak{m}$ is a field and $r^{*}$ is induced by an element $r\in R$. 

All these things are possible in the case of polynomial ring, but it is very hard to do it for general Noetherian rings (or Noetherian local rings). I also noticed that there aren't any nontrivial derivations on the  ring $\mathbf{Z}$. Is there a proof that uses these ideas or versions of them?